Chapter 1 Number Systems (Class 9 - Maths NCERT Exemplar Solutions)

Solution:

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The given expression is $\left[\left(\frac {5}{6}\right)^\frac{1}{5}\right]^{-\frac{1}{6}}$.

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We use the property of exponents $(a^m)^n = a^{mn}$. Here, $a = \frac{5}{6}$, $m = \frac{1}{5}$, and $n = -\frac{1}{6}$.

Simplifying the given expression:

$\left[\left(\frac {5}{6}\right)^\frac{1}{5}\right]^{-\frac{1}{6}} = \left(\frac {5}{6}\right)^{\frac{1}{5} \times \left(-\frac{1}{6}\right)}$

$\left(\frac {5}{6}\right)^{\frac{1}{5} \times \left(-\frac{1}{6}\right)} = \left(\frac {5}{6}\right)^{-\frac{1 \times 1}{5 \times 6}} = \left(\frac {5}{6}\right)^{-\frac{1}{30}}$

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Now, let's evaluate each option:

Option (A): $\left( \frac{5}{6} \right)^{\frac{1}{5}-\frac{1}{6}}$

Simplify the exponent: $\frac{1}{5} - \frac{1}{6} = \frac{6 - 5}{30} = \frac{1}{30}$.

So, option (A) is $\left( \frac{5}{6} \right)^{\frac{1}{30}}$.

This is not equal to $\left( \frac{5}{6} \right)^{-\frac{1}{30}}$.

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Option (B): $ \frac{1}{\left[ \left( \frac{5}{6} \right)^{\frac{1}{5}} \right]^\frac{1}{6}} $

Simplify the denominator using $(a^m)^n = a^{mn}$:

$\left[ \left( \frac{5}{6} \right)^{\frac{1}{5}} \right]^\frac{1}{6} = \left(\frac{5}{6}\right)^{\frac{1}{5} \times \frac{1}{6}} = \left(\frac{5}{6}\right)^{\frac{1}{30}}$

So, option (B) is $ \frac{1}{\left(\frac{5}{6}\right)^{\frac{1}{30}}} $. Using the property $\frac{1}{a^n} = a^{-n}$, this becomes $\left(\frac{5}{6}\right)^{-\frac{1}{30}}$.

This is equal to the given expression.

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Option (C): $\left(\frac{6}{5} \right)^{\frac{1}{30}}$

We know that $\frac{6}{5} = \left(\frac{5}{6}\right)^{-1}$.

Using the property $(a^m)^n = a^{mn}$:

$\left(\frac{6}{5} \right)^{\frac{1}{30}} = \left(\left(\frac{5}{6}\right)^{-1}\right)^{\frac{1}{30}} = \left(\frac{5}{6}\right)^{-1 \times \frac{1}{30}} = \left(\frac{5}{6}\right)^{-\frac{1}{30}}$

This is equal to the given expression.

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Option (D): $\left(\frac{5}{6} \right)^{-\frac{1}{30}}$

This is exactly the simplified form of the given expression.

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Comparing the simplified given expression $\left(\frac {5}{6}\right)^{-\frac{1}{30}}$ with the simplified options, we see that option (A) is $\left( \frac{5}{6} \right)^{\frac{1}{30}}$, while options (B), (C), and (D) are all equal to $\left( \frac{5}{6} \right)^{-\frac{1}{30}}$.

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Therefore, the expression which is not equal to $\left[\left(\frac {5}{6}\right)^\frac{1}{5}\right]^{-\frac{1}{6}}$ is $\left( \frac{5}{6} \right)^{\frac{1}{5}-\frac{1}{6}}$.

Solution:

Rational numbers are defined as numbers that can be expressed in the form $p/q$, where $p$ and $q$ are integers and $q \neq 0$.

The set of Real numbers consists of both rational and irrational numbers. Therefore, every rational number belongs to the set of real numbers.

On the other hand, a rational number like $1/2$ is neither a natural number, nor an integer, nor a whole number.

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The correct option is (C).

Solution:

According to the denseness property of rational numbers, between any two distinct rational numbers, there exist infinitely many other rational numbers.

For example, between $0.1$ and $0.2$, we can find $0.11, 0.111, 0.12, 0.125$, and so on.

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The correct option is (C).

Solution:

A rational number has a decimal expansion that is either terminating (e.g., $1/2 = 0.5$) or non-terminating repeating (e.g., $1/3 = 0.333...$).

Decimal representations that are non-terminating and non-repeating are characteristic of irrational numbers (e.g., $\pi$ or $\sqrt{2}$).

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The correct option is (D).

Solution:

Let's consider two cases for the product of irrational numbers:

Case 1: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ (Irrational)

Case 2: $\sqrt{2} \times \sqrt{2} = 2$ (Rational)

This demonstrates that the product can result in either a rational or an irrational number depending on the values chosen.

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The correct option is (D).

Solution:

The number $\sqrt{2}$ is an irrational number.

By definition, the decimal expansion of any irrational number is non-terminating and non-recurring (non-repeating).

$\sqrt{2} = 1.41421356...$

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The correct option is (D).

Solution:

Let's evaluate each option:

(A) $\displaystyle \sqrt{\frac{4}{9}} = \frac{2}{3}$ (Rational)

(B) $\displaystyle \frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2$ (Rational)

(C) $\sqrt{7}$ is the square root of a non-perfect square, so it is Irrational.

(D) $\sqrt{81} = 9$ (Rational)

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The correct option is (C).

Solution:

(A) $0.14$ is a terminating decimal, hence rational.

(B) $0.14\overline{16}$ is non-terminating but repeating, hence rational.

(C) $0.\overline{1416}$ is non-terminating but repeating, hence rational.

(D) $0.4014001400014...$ is non-terminating and non-repeating (the pattern changes by adding an extra zero), hence it is irrational.

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The correct option is (D).

Solution:

We know the approximate values of these irrational numbers:

$\sqrt{2} \approx 1.414$

$\sqrt{3} \approx 1.732$

We need a rational number between $1.414$ and $1.732$.

Checking the options:

(A) and (B) involve $\sqrt{2}$ and $\sqrt{3}$, making them irrational.

(C) $1.5$ is rational and $1.414 < 1.5 < 1.732$.

(D) $1.8$ is rational but $1.8 > 1.732$.

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The correct option is (C).

Solution:

Let the given number be $x$.

Multiplying both sides by $10$:

Subtracting equation (i) from equation (ii):

$10x - x = (19.999...) - (1.999...)$

$9x = 18$

$\displaystyle x = \frac{\cancel{18}^2}{\cancel{9}_1}$

$x = 2$

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The correct option is (C).

Solution:

We can add these terms as they are like irrational terms (both contain $\sqrt{3}$).

Treating $\sqrt{3}$ as a variable:

$2\sqrt{3} + 1\sqrt{3}$

Taking $\sqrt{3}$ common:

$(2 + 1)\sqrt{3}$

$3\sqrt{3}$

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The correct option is (C).

Solution:

To find the product of two square roots, we can use the identity $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$:

$\sqrt{10} \times \sqrt{15} = \sqrt{10 \times 15}$

$\sqrt{10} \times \sqrt{15} = \sqrt{150}$

Now, let us find the prime factors of 150 to simplify the radical:

$\begin{array}{c|cc} 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$\sqrt{150} = \sqrt{2 \times 3 \times 5 \times 5}$

The correct option is (B).

Solution:

To rationalise the denominator, we multiply both the numerator and the denominator by the conjugate of $(\sqrt{7} - 2)$, which is $(\sqrt{7} + 2)$.

$\displaystyle \frac{1}{\sqrt{7} - 2} = \frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2}$

Using the identity $(a - b)(a + b) = a^2 - b^2$ in the denominator:

$\displaystyle \frac{\sqrt{7} + 2}{(\sqrt{7})^2 - (2)^2}$

$\displaystyle \frac{\sqrt{7} + 2}{7 - 4}$

The correct option is (A).

Solution:

First, simplify the square roots in the denominator:

$\sqrt{9} = 3$

$\sqrt{8} = \sqrt{2 \times 2 \times 2} = 2\sqrt{2}$

The expression becomes: $\displaystyle \frac{1}{3 - 2\sqrt{2}}$

Now, rationalise by multiplying with the conjugate $(3 + 2\sqrt{2})$:

$\displaystyle \frac{1}{3 - 2\sqrt{2}} \times \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}}$

$\displaystyle \frac{3 + 2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2}$

$\displaystyle \frac{3 + 2\sqrt{2}}{9 - 8}$

The correct option is (D).

Solution:

To rationalise, we multiply the denominator by its conjugate $(3\sqrt{3} + 2\sqrt{2})$.

The denominator will be in the form $(a - b)(a + b) = a^2 - b^2$:

$\text{Denominator} = (9 \times 3) - (4 \times 2)$

$\text{Denominator} = 27 - 8$

The correct option is (B).

Solution:

Simplify each radical term:

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

Substitute these into the fraction:

$\displaystyle \frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$

Factor out common terms from numerator and denominator:

$\displaystyle \frac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})}$

The correct option is (B).

Solution:

Rationalise the expression inside the square root first:

$\displaystyle \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}}$

$\displaystyle \sqrt{\frac{(\sqrt{2} - 1)^2}{(\sqrt{2})^2 - (1)^2}}$

$\displaystyle \sqrt{\frac{(\sqrt{2} - 1)^2}{2 - 1}} = \sqrt{(\sqrt{2} - 1)^2}$

$\sqrt{2} - 1$

Substitute the value $\sqrt{2} = 1.4142$:

$\begin{array}{cc} & 1 & . & 4 & 1 & 4 & 2 \\ - & 1 & . & 0 & 0 & 0 & 0 \\ \hline & 0 & . & 4 & 1 & 4 & 2 \\ \hline \end{array}$

The correct option is (C).

Solution:

Using the law of exponents $\sqrt[n]{a} = a^{1/n}$ and $(a^m)^n = a^{mn}$:

The expression is $\left( (2^2)^{1/3} \right)^{1/4}$

Multiply the powers:

$\displaystyle 2^{2 \times \frac{1}{3} \times \frac{1}{4}}$

$\displaystyle 2^{\frac{2}{12}}$

The correct option is (C).

Solution:

Convert the radical terms into exponential form:

$\sqrt[3]{2} = 2^{1/3}$

$\sqrt[4]{2} = 2^{1/4}$

$\sqrt[12]{32} = (2^5)^{1/12} = 2^{5/12}$

Now, multiply using the rule $a^m \cdot a^n = a^{m+n}$:

$\displaystyle 2^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}}$

Take the LCM of 3, 4, and 12, which is 12:

$\displaystyle 2^{\frac{4 + 3 + 5}{12}}$

$\displaystyle 2^{\frac{12}{12}}$

The correct option is (B).

Solution:

Simplify 81 as $3^4$ or $9^2$:

$\displaystyle \left( (9^2)^{-2} \right)^{1/4}$

Using the property $(a^m)^n = a^{mn}$:

$\displaystyle 9^{2 \times (-2) \times \frac{1}{4}}$

$\displaystyle 9^{-4 \times \frac{1}{4}}$

$9^{-1}$

The correct option is (A).

Solution:

Add the exponents since the bases are identical:

$(256)^{0.16 + 0.09} = (256)^{0.25}$

$\text{Since } 0.25 = \frac{1}{4}$:

$(256)^{1/4}$

Now, let us find the prime factors of 256:

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

We see that $256 = 2^8$, or $4^4$:

$(4^4)^{1/4} = 4^1$

The correct option is (A).

Solution:

Let's evaluate option (C):

$\text{The expression is } ( (x^3)^{1/2} )^{2/3}$

Multiply the powers:

$\displaystyle x^{3 \times \frac{1}{2} \times \frac{2}{3}}$

$\displaystyle x^{\frac{6}{6}}$

Checking others:

(A) Cannot be simplified to $x$.

(B) $x^{4 \times \frac{1}{3} \times \frac{1}{12}} = x^{1/9}$.

(D) $x^{12/7 + 7/12} \neq x$.

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The correct option is (C).

Solution:

Yes, there exist pairs of irrational numbers whose sum and product are both rational numbers.

To justify this, let us consider two irrational numbers that are conjugates of each other.

Let the first irrational number be:

Let the second irrational number be:

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Case 1: Sum of the numbers

Sum $= a + b$

Sum $= (2 + \sqrt{3}) + (2 - \sqrt{3})$

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Case 2: Product of the numbers

Product $= a \times b$

Product $= (2 + \sqrt{3})(2 - \sqrt{3})$

Using the algebraic identity $(x + y)(x - y) = x^2 - y^2$:

Product $= (2)^2 - (\sqrt{3})^2$

Product $= 4 - 3$

Thus, we have found two irrational numbers whose sum and product are both rational.

Solution:

The given statement is True.

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Justification with Example:

Let us assume a number $x$ such that it is the fourth root of a non-perfect square rational number.

Let:

Now, let us calculate the square of this number ($x^2$):

$x^2 = (\sqrt[4]{2})^2$

$x^2 = (2^{1/4})^2$

$x^2 = 2^{2/4}$

Now, let us calculate the fourth power of this number ($x^4$):

$x^4 = (\sqrt[4]{2})^4$

$x^4 = (2^{1/4})^4$

Therefore, for $x = \sqrt[4]{2}$, the value of $x^2$ is irrational but the value of $x^4$ is rational.

Solution:

Yes, if $x$ is a rational number and $y$ is an irrational number, then $x + y$ is necessarily an irrational number.

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Justification:

Suppose $x + y$ is a rational number $r$. Then we can write:

Since the difference of two rational numbers ($r$ and $x$) is always rational, this implies that $y$ must be a rational number. However, this contradicts our given condition that $y$ is irrational.

Therefore, $x + y$ must be irrational.

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Example:

Let $x = 2$ (a rational number) and $y = \sqrt{3}$ (an irrational number).

The sum $x + y = 2 + \sqrt{3}$ is an irrational number because its decimal expansion is non-terminating and non-recurring.

Solution:

No, $xy$ is not necessarily irrational.

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Justification:

The product of a rational number and an irrational number is irrational except when the rational number is zero.

If we take the rational number $x = 0$, then the product $xy$ will always be $0$, which is a rational number.

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Example:

Let the rational number be $x = 0$.

Let the irrational number be $y = \sqrt{2}$.

Now, finding the product $xy$:

Since $0$ is a rational number (can be written as $0/1$), the product is not irrational in this case.

Solution:

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Let's evaluate each statement and provide justification.

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(i) $\frac{\sqrt{2}}{3}$ is a rational number.

This statement is False.

Justification: $\sqrt{2}$ is an irrational number, and 3 is a non-zero rational number. The quotient of an irrational number and a non-zero rational number is always an irrational number. Therefore, $\frac{\sqrt{2}}{3}$ is an irrational number.

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(ii) There are infinitely many integers between any two integers.

This statement is False.

Justification: The integers are ... , -2, -1, 0, 1, 2, ... . Between any two distinct integers, there is a finite number of integers. For example, between 2 and 5, the integers are 3 and 4 (a finite number). Between any two consecutive integers, such as 2 and 3, there are no integers at all.

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(iii) Number of rational numbers between 15 and 18 is finite.

This statement is False.

Justification: 15 and 18 are both rational numbers. A property of rational numbers is that they are dense in the real numbers, meaning that between any two distinct rational numbers, there are infinitely many other rational numbers. For instance, if $r_1$ and $r_2$ are two rational numbers, then $\frac{r_1+r_2}{2}$ is a rational number between them. We can repeat this process indefinitely, generating infinitely many rational numbers between $r_1$ and $r_2$.

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(iv) There are numbers which cannot be written in the form $\frac{p}{q}$, q ≠ 0 , p, q both are integers.

This statement is True.

Justification: Numbers that cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, are called irrational numbers. Examples of such numbers include $\sqrt{2}$, $\sqrt{3}$, $\pi$, and $e$. These numbers have non-terminating, non-repeating decimal expansions and thus cannot be expressed as a fraction of two integers.

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(v) The square of an irrational number is always rational.

This statement is False.

Justification: Consider the irrational number $\sqrt{2} + 1$. The sum of an irrational number ($\sqrt{2}$) and a rational number (1) is irrational, so $\sqrt{2} + 1$ is irrational.

Now, let's find the square of $\sqrt{2} + 1$:

$(\sqrt{2} + 1)^2 = (\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2$

$(\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1$

$(\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}$

The number $3 + 2\sqrt{2}$ is the sum of a rational number (3) and an irrational number ($2\sqrt{2}$). The sum of a rational and a non-zero irrational number is irrational. Therefore, $(\sqrt{2} + 1)^2$ is irrational. This is a counterexample to the statement.

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(vi) $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.

This statement is False.

Justification: Let's simplify the expression $\frac{\sqrt{12}}{\sqrt{3}}$:

$\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4}$

$\sqrt{4} = 2$

The value of the expression is 2. The number 2 is a rational number because it can be written as $\frac{2}{1}$, where 2 and 1 are integers and $1 \neq 0$. The fact that $\sqrt{12}$ and $\sqrt{3}$ are not integers does not mean their quotient is not rational; the rationality of the quotient depends on the simplified value of the quotient itself.

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(vii) $\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form p/q , q ≠ 0 and so it is a rational number.

This statement is False.

Justification: While the expression is presented as a fraction, for a number to be rational when written in the form $\frac{p}{q}$, it is required that both $p$ and $q$ must be integers and $q \neq 0$.

Let's simplify the given expression:

$\frac{\sqrt{15}}{\sqrt{3}} = \sqrt{\frac{15}{3}} = \sqrt{5}$

The number $\sqrt{5}$ is an irrational number because 5 is not a perfect square, and its decimal expansion is non-terminating and non-repeating. Although the original expression looks like a fraction, the numerator ($\sqrt{15}$) and the denominator ($\sqrt{3}$) are not integers. After simplification, the number is $\sqrt{5}$, which is irrational. Therefore, the original expression represents an irrational number.

Solution:

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We need to classify each given number as rational or irrational and provide justification.

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(i) $\sqrt{196}$

$\sqrt{196} = 14$.

Since 14 is an integer, it can be written as $\frac{14}{1}$, which is in the form $\frac{p}{q}$ where $p=14$ and $q=1$ are integers and $q \neq 0$.

Therefore, $\sqrt{196}$ is a rational number.

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(ii) $3\sqrt{18}$

Simplify the expression: $3\sqrt{18} = 3 \times \sqrt{9 \times 2} = 3 \times \sqrt{9} \times \sqrt{2} = 3 \times 3 \times \sqrt{2} = 9\sqrt{2}$.

Since $\sqrt{2}$ is an irrational number and 9 is a non-zero rational number, their product $9\sqrt{2}$ is irrational.

Therefore, $3\sqrt{18}$ is an irrational number.

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(iii) $\sqrt{\frac{9}{27}}$

Simplify the expression: $\sqrt{\frac{9}{27}} = \sqrt{\frac{1}{3}} = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

To rationalize the denominator, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$: $\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

Since $\sqrt{3}$ is an irrational number and 3 is a non-zero rational number, their quotient $\frac{\sqrt{3}}{3}$ is irrational.

Therefore, $\sqrt{\frac{9}{27}}$ is an irrational number.

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(iv) $\frac{\sqrt{28}}{\sqrt{343}}$

Simplify the expression: $\frac{\sqrt{28}}{\sqrt{343}} = \sqrt{\frac{28}{343}}$.

Simplify the fraction inside the square root: $\frac{28}{343} = \frac{4 \times 7}{49 \times 7} = \frac{4}{49}$.

So, $\sqrt{\frac{28}{343}} = \sqrt{\frac{4}{49}} = \frac{\sqrt{4}}{\sqrt{49}} = \frac{2}{7}$.

Since $\frac{2}{7}$ is in the form $\frac{p}{q}$ where $p=2$ and $q=7$ are integers and $q \neq 0$, it is rational.

Therefore, $\frac{\sqrt{28}}{\sqrt{343}}$ is a rational number.

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(v) $-\sqrt{0.4}$

Rewrite the decimal as a fraction: $0.4 = \frac{4}{10} = \frac{2}{5}$.

The expression is $-\sqrt{\frac{2}{5}} = -\frac{\sqrt{2}}{\sqrt{5}}$. This is the negative of the quotient of two irrational numbers $\sqrt{2}$ and $\sqrt{5}$. The quotient $\frac{\sqrt{2}}{\sqrt{5}} = \sqrt{\frac{2}{5}} = \sqrt{0.4}$. Since 0.4 is not a perfect square, $\sqrt{0.4}$ is irrational. The negative of an irrational number is irrational.

Alternatively, $-\frac{\sqrt{2}}{\sqrt{5}} = -\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = -\frac{\sqrt{10}}{5}$. Since $\sqrt{10}$ is irrational and 5 is a non-zero rational, $-\frac{\sqrt{10}}{5}$ is irrational.

Therefore, $-\sqrt{0.4}$ is an irrational number.

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(vi) $\frac{\sqrt{12}}{\sqrt{75}}$

Simplify the expression: $\frac{\sqrt{12}}{\sqrt{75}} = \sqrt{\frac{12}{75}}$.

Simplify the fraction inside the square root: $\frac{12}{75} = \frac{3 \times 4}{3 \times 25} = \frac{4}{25}$.

So, $\sqrt{\frac{12}{75}} = \sqrt{\frac{4}{25}} = \frac{\sqrt{4}}{\sqrt{25}} = \frac{2}{5}$.

Since $\frac{2}{5}$ is in the form $\frac{p}{q}$ where $p=2$ and $q=5$ are integers and $q \neq 0$, it is rational.

Therefore, $\frac{\sqrt{12}}{\sqrt{75}}$ is a rational number.

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(vii) 0.5918

This is a terminating decimal. Terminating decimals can always be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

$0.5918 = \frac{5918}{10000}$. Since 5918 and 10000 are integers and $10000 \neq 0$, it is rational.

Therefore, 0.5918 is a rational number.

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(viii) $(1 + \sqrt{5}) - (4 + \sqrt{5})$

Simplify the expression: $(1 + \sqrt{5}) - (4 + \sqrt{5}) = 1 + \sqrt{5} - 4 - \sqrt{5}$.

Combine like terms: $(1 - 4) + (\sqrt{5} - \sqrt{5}) = -3 + 0 = -3$.

Since -3 is an integer, it can be written as $\frac{-3}{1}$, which is in the form $\frac{p}{q}$ where $p=-3$ and $q=1$ are integers and $q \neq 0$.

Therefore, $(1 + \sqrt{5}) - (4 + \sqrt{5})$ is a rational number.

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(ix) 10.124124 ...

This decimal can be written as $10.\overline{124}$. It is a non-terminating decimal with a repeating block of digits '124'.

Any non-terminating repeating decimal represents a rational number because it can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

Therefore, 10.124124 ... is a rational number.

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(x) 1.010010001…

This is a non-terminating decimal. The pattern of digits is 01, then 001, then 0001, and so on, with an increasing number of zeros between the ones. There is no fixed block of digits that repeats periodically.

This is a non-terminating, non-repeating decimal. Numbers with non-terminating, non-repeating decimal expansions are irrational by definition.

Therefore, 1.010010001… is an irrational number.

To Locate:

The position of $\sqrt{13}$ on the number line.

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Given:

The irrational number $\sqrt{13}$.

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Construction Required:

1. Draw a horizontal line and mark a point $O$ on it representing the origin ($0$).

2. Mark a point $A$ on the line such that $OA = 3$ units.

3. At point $A$, construct a perpendicular $AB$ of length $2$ units.

4. Join the points $O$ and $B$.

5. Using a compass with $O$ as the centre and $OB$ as the radius, draw an arc that intersects the number line at a point $P$.

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Solution:

We can express $13$ as the sum of the squares of two integers:

$13 = 9 + 4$

In the right-angled triangle $\triangle OAB$, where $\angle OAB = 90^\circ$:

By using the Pythagoras Theorem:

$OB^2 = OA^2 + AB^2$

Substituting the constructed values $OA = 3$ and $AB = 2$:

$OB^2 = 3^2 + 2^2$

$OB^2 = 9 + 4$

By construction, the arc is drawn with $O$ as the centre and $OB$ as the radius to meet the number line at $P$.

Therefore, point $P$ on the number line represents the value of $\sqrt{13}$.

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Alternate Method:

One can also take the base $OA = 2$ units and the perpendicular height $AB = 3$ units. The resulting hypotenuse will remain $OB = \sqrt{2^2 + 3^2} = \sqrt{13}$ units, leading to the same location on the number line.

To Express: $0.12\overline{3}$ in the form $\frac{p}{q}$.

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Solution:

Let $x$ be the given decimal number:

Since there are two digits (1 and 2) after the decimal point that do not repeat, we first multiply both sides by $100$:

Now, since only one digit (3) is repeating, we multiply equation (ii) by $10$:

Subtracting equation (ii) from equation (iii):

$1000x - 100x = (123.333...) - (12.333...)$

$900x = 111$

$\displaystyle x = \frac{111}{900}$

On simplifying the fraction by dividing both numerator and denominator by 3:

$\displaystyle x = \frac{\cancel{111}^{37}}{\cancel{900}_{300}}$

Therefore, the form $\frac{p}{q}$ of $0.12\overline{3}$ is $\frac{37}{300}$.

To Simplify: $(3\sqrt{5} - 5\sqrt{2})(4\sqrt{5} + 3\sqrt{2})$

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Solution:

We use the distributive property of multiplication (FOIL method) to expand the expression:

$(3\sqrt{5} - 5\sqrt{2})(4\sqrt{5} + 3\sqrt{2}) = 3\sqrt{5}(4\sqrt{5} + 3\sqrt{2}) - 5\sqrt{2}(4\sqrt{5} + 3\sqrt{2})$

Multiply each term:

$= (3\sqrt{5} \times 4\sqrt{5}) + (3\sqrt{5} \times 3\sqrt{2}) - (5\sqrt{2} \times 4\sqrt{5}) $$ - (5\sqrt{2} \times 3\sqrt{2})$

$= (12 \times 5) + 9\sqrt{10} - 20\sqrt{10} - (15 \times 2)$

$= 60 + 9\sqrt{10} - 20\sqrt{10} - 30$

Group the rational numbers and the like irrational terms together:

$= (60 - 30) + (9\sqrt{10} - 20\sqrt{10})$

$= 30 - 11\sqrt{10}$

The simplified value is $30 - 11\sqrt{10}$.

To Find: The value of $a$ in $\displaystyle \frac{6}{3\sqrt{2} - 2\sqrt{3}} = 3\sqrt{2} - a\sqrt{3}$

---

Solution:

First, we rationalise the denominator of the Left Hand Side (LHS) by multiplying with the conjugate $(3\sqrt{2} + 2\sqrt{3})$:

$\displaystyle \text{LHS} = \frac{6}{3\sqrt{2} - 2\sqrt{3}} \times \frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$

Using the identity $(x-y)(x+y) = x^2 - y^2$ in the denominator:

$\displaystyle \text{LHS} = \frac{6(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2})^2 - (2\sqrt{3})^2}$

$\displaystyle \text{LHS} = \frac{6(3\sqrt{2} + 2\sqrt{3})}{(9 \times 2) - (4 \times 3)}$

$\displaystyle \text{LHS} = \frac{6(3\sqrt{2} + 2\sqrt{3})}{18 - 12}$

$\displaystyle \text{LHS} = \frac{\cancel{6}(3\sqrt{2} + 2\sqrt{3})}{\cancel{6}}$

Now, equate the simplified LHS to the Right Hand Side (RHS):

$3\sqrt{2} + 2\sqrt{3} = 3\sqrt{2} - a\sqrt{3}$

Comparing the coefficients of $\sqrt{3}$ on both sides:

$a = -2$

The value of $a$ is $-2$.

To Simplify: $\left[ 5\left( 8^{\frac{1}{3}} + 27^{\frac{1}{3}} \right)^3 \right]^{\frac{1}{4}}$

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Solution:

Let's simplify the terms inside the parentheses first using prime factorisation:

$8 = 2^3$

$27 = 3^3$

The expression becomes:

$\displaystyle \left[ 5\left( (2^3)^{\frac{1}{3}} + (3^3)^{\frac{1}{3}} \right)^3 \right]^{\frac{1}{4}}$

Using the law of exponents $(a^m)^n = a^{mn}$:

$\displaystyle \left[ 5\left( 2^{3 \times \frac{1}{3}} + 3^{3 \times \frac{1}{3}} \right)^3 \right]^{\frac{1}{4}}$

$\displaystyle \left[ 5\left( 2^1 + 3^1 \right)^3 \right]^{\frac{1}{4}}$

$\displaystyle \left[ 5\left( 5 \right)^3 \right]^{\frac{1}{4}}$

$\displaystyle \left[ 5^1 \times 5^3 \right]^{\frac{1}{4}}$

Using $a^m \times a^n = a^{m+n}$:

$\displaystyle \left[ 5^4 \right]^{\frac{1}{4}}$

$5^{4 \times \frac{1}{4}} = 5^1$

The simplified value is 5.

Solution:

---

We are asked to classify the variables x, y, z, and u as rational or irrational based on the given equations.

---

(i) $x^2 = 5$

Solving for x, we get $x = \pm\sqrt{5}$.

The number 5 is not a perfect square (i.e., there is no integer $k$ such that $k^2 = 5$). The square root of a positive integer that is not a perfect square is an irrational number.

Therefore, $\sqrt{5}$ is an irrational number, and $-\sqrt{5}$ is also an irrational number.

The variable x represents an irrational number.

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(ii) $y^2 = 9$

Solving for y, we get $y = \pm\sqrt{9}$.

Since $3^2 = 9$, $\sqrt{9} = 3$.

So, $y = \pm 3$.

The numbers 3 and -3 are integers. Integers can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g., $3 = \frac{3}{1}$, $-3 = \frac{-3}{1}$).

Therefore, 3 and -3 are rational numbers.

The variable y represents a rational number.

---

(iii) $z^2 = 0.04$

Solving for z, we get $z = \pm\sqrt{0.04}$.

We can write 0.04 as a fraction: $0.04 = \frac{4}{100}$.

So, $z = \pm\sqrt{\frac{4}{100}}$.

Using the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$z = \pm \frac{\sqrt{4}}{\sqrt{100}} = \pm \frac{2}{10}$

Simplifying the fraction: $\frac{2}{10} = \frac{1}{5}$.

So, $z = \pm \frac{1}{5}$ (or $\pm 0.2$).

The numbers $\frac{1}{5}$ and $-\frac{1}{5}$ are in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g., $p=1, q=5$ or $p=-1, q=5$).

Therefore, $\frac{1}{5}$ and $-\frac{1}{5}$ are rational numbers.

The variable z represents a rational number.

---

(iv) $u^2 = \frac{17}{4}$

Solving for u, we get $u = \pm\sqrt{\frac{17}{4}}$.

Using the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$u = \pm \frac{\sqrt{17}}{\sqrt{4}} = \pm \frac{\sqrt{17}}{2}$.

The number 17 is not a perfect square, so $\sqrt{17}$ is an irrational number.

The number $\frac{\sqrt{17}}{2}$ is the quotient of an irrational number ($\sqrt{17}$) and a non-zero rational number (2). The quotient of an irrational number and a non-zero rational number is always irrational.

Therefore, $\frac{\sqrt{17}}{2}$ is an irrational number, and $-\frac{\sqrt{17}}{2}$ is also an irrational number.

The variable u represents an irrational number.

---

In summary:

x is irrational.

y is rational.

z is rational.

u is irrational.

Solution:

---

To find rational numbers between two given rational numbers, we can use various methods, such as finding the average of the two numbers repeatedly or by expressing the numbers with a common denominator and then finding fractions between them.

---

(i) Between –1 and –2

The given numbers are -1 and -2. Both are rational numbers.

Let's find three rational numbers between them.

Method 1: Using Decimal Form

-1 can be written as -1.0.

-2 can be written as -2.0.

We need numbers between -2.0 and -1.0. We can easily list terminating decimals between them, which are rational numbers. For example:

-1.1, -1.2, -1.3, -1.4, -1.5, -1.6, -1.7, -1.8, -1.9.

Let's choose three of these:

$-1.1 = -\frac{11}{10}$

$-1.5 = -\frac{15}{10} = -\frac{3}{2}$

$-1.9 = -\frac{19}{10}$

Three rational numbers between –1 and –2 are $-\frac{11}{10}$, $-\frac{3}{2}$, $-\frac{19}{10}$.

---

(ii) Between 0.1 and 0.11

The given numbers are 0.1 and 0.11. Both are terminating decimals, so they are rational numbers.

0.1 can be written as 0.100.

0.11 can be written as 0.110.

We need numbers between 0.100 and 0.110. We can list terminating decimals between them:

0.101, 0.102, 0.103, 0.104, ..., 0.109.

Let's choose three of these:

$0.101 = \frac{101}{1000}$

$0.105 = \frac{105}{1000} = \frac{21}{200}$

$0.109 = \frac{109}{1000}$

Three rational numbers between 0.1 and 0.11 are $\frac{101}{1000}$, $\frac{21}{200}$, $\frac{109}{1000}$.

---

(iii) Between $\frac{5}{7}$ and $\frac{6}{7}$

The given numbers are $\frac{5}{7}$ and $\frac{6}{7}$. Both are rational numbers.

Method: Increase the denominator.

To find rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$, we can multiply the numerator and denominator of both fractions by a number greater than 1, say 10.

$\frac{5}{7} = \frac{5 \times 10}{7 \times 10} = \frac{50}{70}$

$\frac{6}{7} = \frac{6 \times 10}{7 \times 10} = \frac{60}{70}$

Now we can easily find fractions with denominator 70 and numerators between 50 and 60.

For example, $\frac{51}{70}, \frac{52}{70}, \frac{53}{70}, \dots, \frac{59}{70}$.

Let's choose three of these:

$\frac{51}{70}$

$\frac{55}{70} = \frac{11}{14}$ (simplified)

$\frac{59}{70}$

Three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{51}{70}$, $\frac{11}{14}$, $\frac{59}{70}$.

---

(iv) Between $\frac{1}{4}$ and $\frac{1}{5}$

The given numbers are $\frac{1}{4}$ and $\frac{1}{5}$. Both are rational numbers.

Method: Find a common denominator and then increase it.

The least common multiple of the denominators 4 and 5 is 20. Express both fractions with denominator 20:

$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$

$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$

We need numbers between $\frac{4}{20}$ and $\frac{5}{20}$. Multiply numerator and denominator of both fractions by a number greater than 1, say 10, to create more space between the numerators.

$\frac{4}{20} = \frac{4 \times 10}{20 \times 10} = \frac{40}{200}$

$\frac{5}{20} = \frac{5 \times 10}{20 \times 10} = \frac{50}{200}$

Now we can find fractions with denominator 200 and numerators between 40 and 50.

For example, $\frac{41}{200}, \frac{42}{200}, \frac{43}{200}, \dots, \frac{49}{200}$.

Let's choose three of these:

$\frac{41}{200}$

$\frac{45}{200} = \frac{9}{40}$ (simplified)

$\frac{49}{200}$

Three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{41}{200}$, $\frac{9}{40}$, $\frac{49}{200}$.

Solution:

---

We need to insert one rational number and one irrational number between each given pair of numbers.

Recall that a rational number can be expressed as a terminating or non-terminating repeating decimal. An irrational number has a non-terminating and non-repeating decimal expansion.

---

(i) 2 and 3

Rational number: We can take the average of 2 and 3: $\frac{2+3}{2} = \frac{5}{2} = 2.5$. $2 < 2.5 < 3$. Since 2.5 is a terminating decimal, it is a rational number.

Irrational number: We can construct a non-terminating, non-repeating decimal between 2 and 3. For example, $2.1010010001...$ (where the number of zeros between consecutive '1's increases by one each time). This number is greater than 2 and less than 3. Since its decimal expansion is non-terminating and non-repeating, it is an irrational number.

---

(ii) 0 and 0.1

Rational number: We can choose a terminating decimal between 0 and 0.1, for example, 0.05. $0 < 0.05 < 0.1$. Since 0.05 is a terminating decimal, it is a rational number ($0.05 = \frac{5}{100} = \frac{1}{20}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0 and 0.1. For example, $0.01010010001...$ (where the number of zeros between consecutive '1's increases by one each time). This number is greater than 0 and less than 0.1. Since its decimal expansion is non-terminating and non-repeating, it is an irrational number.

---

(iii) $\frac{1}{3}$ and $\frac{1}{2}$

We can approximate the decimal values: $\frac{1}{3} \approx 0.333...$ and $\frac{1}{2} = 0.5$. We need a number between 0.333... and 0.5.

Rational number: We can choose a terminating decimal like 0.4. $0.333... < 0.4 < 0.5$. Since 0.4 is a terminating decimal, it is rational ($0.4 = \frac{4}{10} = \frac{2}{5}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.333... and 0.5. For example, $0.414114111...$ (where the number of '1's between consecutive '4's increases by one each time). This number is greater than 0.333... and less than 0.5. Since its decimal expansion is non-terminating and non-repeating, it is an irrational number.

---

(iv) $\frac{-2}{5}$ and $\frac{1}{2}$

We can convert these to decimals: $\frac{-2}{5} = -0.4$ and $\frac{1}{2} = 0.5$. We need a number between -0.4 and 0.5.

Rational number: We can choose an integer like 0. $-0.4 < 0 < 0.5$. Since 0 is an integer, it is rational ($0 = \frac{0}{1}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between -0.4 and 0.5. For example, $0.1010010001...$ (pattern of increasing zeros between 1s). $-0.4 < 0.101001... < 0.5$. This decimal is non-terminating and non-repeating, so it is irrational.

---

(v) 0.15 and 0.16

We need a number between 0.15 and 0.16.

Rational number: We can choose a terminating decimal like 0.153. $0.15 < 0.153 < 0.16$. Since 0.153 is a terminating decimal, it is rational ($0.153 = \frac{153}{1000}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.15 and 0.16. For example, $0.15110111001111000...$ (pattern of increasing 1s followed by increasing 0s). $0.15 < 0.1511011100... < 0.16$. This decimal is non-terminating and non-repeating, so it is irrational.

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(vi) $\sqrt{2}$ and $\sqrt{3}$

We can approximate the decimal values: $\sqrt{2} \approx 1.414...$ and $\sqrt{3} \approx 1.732...$. We need a number between 1.414... and 1.732....

Rational number: We can choose a terminating decimal like 1.5. $\sqrt{2} \approx 1.414 < 1.5 < 1.732 \approx \sqrt{3}$. Since 1.5 is a terminating decimal, it is rational ($1.5 = \frac{15}{10} = \frac{3}{2}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 1.414... and 1.732.... For example, $1.5050050005...$ (pattern of increasing zeros between 5s). $\sqrt{2} < 1.505005... < \sqrt{3}$. This decimal is non-terminating and non-repeating, so it is irrational.

---

(vii) 2.357 and 3.121

These are terminating decimals (and thus rational numbers). We need a number between 2.357 and 3.121.

Rational number: We can choose a terminating decimal like 2.5. $2.357 < 2.5 < 3.121$. Since 2.5 is a terminating decimal, it is rational ($2.5 = \frac{25}{10} = \frac{5}{2}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 2.357 and 3.121. For example, $2.414114111...$ (pattern of increasing 1s between 4s). $2.357 < 2.414114... < 3.121$. This decimal is non-terminating and non-repeating, so it is irrational.

---

(viii) 0.0001 and 0.001

These are terminating decimals (and thus rational numbers). We need a number between 0.0001 and 0.001.

Rational number: We can choose a terminating decimal like 0.0005. $0.0001 < 0.0005 < 0.001$. Since 0.0005 is a terminating decimal, it is rational ($0.0005 = \frac{5}{10000} = \frac{1}{2000}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.0001 and 0.001. For example, $0.000120120012000...$ (pattern of '12' followed by increasing zeros). $0.0001 < 0.0001201200... < 0.001$. This decimal is non-terminating and non-repeating, so it is irrational.

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(ix) 3.623623 and 0.484848

The numbers are 3.623623 and 0.484848. Both are terminating decimals (and thus rational numbers). Note that $0.484848 < 3.623623$. We need a number between them.

Rational number: We can choose a terminating decimal like 1. $0.484848 < 1 < 3.623623$. Since 1 is an integer, it is rational ($1 = \frac{1}{1}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.484848 and 3.623623. For example, $1.121121112...$ (pattern of increasing 1s between 2s). $0.484848 < 1.121121112... < 3.623623$. This decimal is non-terminating and non-repeating, so it is irrational.

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(x) 6.375289 and 6.375738

These are terminating decimals (and thus rational numbers). We need a number between 6.375289 and 6.375738.

Rational number: We can choose a terminating decimal like 6.3753. $6.375289 < 6.3753 < 6.375738$. Since 6.3753 is a terminating decimal, it is rational ($6.3753 = \frac{63753}{10000}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 6.375289 and 6.375738. For example, $6.3754040040004...$ (pattern of increasing zeros between 4s). $6.375289 < 6.375404004... < 6.375738$. This decimal is non-terminating and non-repeating, so it is irrational.

Given:

A set of numbers to be represented on the number line:

$7$, $7.2$, $\displaystyle \frac{-3}{2}$, and $\displaystyle \frac{-12}{5}$

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To Represent:

Locate and plot the given integers, decimals, and rational numbers on a single number line.

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Solution:

First, we convert the given rational numbers (fractions) into decimal form to identify their exact positions relative to integers.

1. For the first number: $7$

This is a positive integer. We move $7$ units to the right from the origin ($0$).

2. For the second number: $7.2$

This is a positive decimal. It lies between the integers $7$ and $8$. To plot this, we divide the unit length between $7$ and $8$ into $10$ equal parts and mark the 2nd point.

3. For the third number: $\displaystyle \frac{-3}{2}$

Let's convert this into decimal form:

This number is negative and lies exactly midway between $-1$ and $-2$ on the left side of the origin.

4. For the fourth number: $\displaystyle \frac{-12}{5}$

Let's convert this into decimal form:

This number lies between the negative integers $-2$ and $-3$. To plot this, we divide the unit length between $-2$ and $-3$ into $10$ equal parts and mark the 4th point from $-2$ towards $-3$.

---

Graphical Representation:

The following diagram shows the positions of all four points on the number line:

In the number line above:

Point A represents $7$.

Point B represents $7.2$.

Point C represents $\displaystyle \frac{-3}{2}$ (i.e., $-1.5$).

Point D represents $\displaystyle \frac{-12}{5}$ (i.e., $-2.4$).

To Locate:

Represent the irrational numbers $\sqrt{5}$, $\sqrt{10}$, and $\sqrt{17}$ on the number line.

---

General Principle:

To locate these irrational numbers, we use the Pythagoras Theorem. For any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides ($c^2 = a^2 + b^2$).

We represent the number $\sqrt{n}$ by finding two numbers $a$ and $b$ such that $n = a^2 + b^2$.

---

1. Locating $\sqrt{5}$ on the Number Line:

We can write $5$ as the sum of the squares of $2$ and $1$:

Construction Required:

1. Take a number line and mark point $O$ as $0$.

2. Mark point $A$ such that $OA = 2$ units.

3. Draw a perpendicular $AB = 1$ unit at point $A$.

4. Join $OB$. By Pythagoras Theorem in $\triangle OAB$:

5. With $O$ as the centre and $OB$ as the radius, draw an arc intersecting the number line at point $P$.

Point $P$ represents $\sqrt{5}$.

---

2. Locating $\sqrt{10}$ on the Number Line:

We can write $10$ as the sum of the squares of $3$ and $1$:

Construction Required:

1. Mark point $A$ on the number line such that $OA = 3$ units.

2. Draw a perpendicular $AB = 1$ unit at point $A$.

3. Join $OB$. In $\triangle OAB$:

4. With $O$ as the centre and $OB$ as the radius, draw an arc intersecting the number line at point $Q$.

Point $Q$ represents $\sqrt{10}$.

---

3. Locating $\sqrt{17}$ on the Number Line:

We can write $17$ as the sum of the squares of $4$ and $1$:

Construction Required:

1. Mark point $A$ on the number line such that $OA = 4$ units.

2. Draw a perpendicular $AB = 1$ unit at point $A$.

3. Join $OB$. In $\triangle OAB$:

4. With $O$ as the centre and $OB$ as the radius, draw an arc intersecting the number line at point $R$.

Point $R$ represents $\sqrt{17}$.

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Proof:

In all the above cases, by the definition of a circle/arc:

Therefore, the distances on the number line from the origin to points $P$, $Q$, and $R$ are exactly $\sqrt{5}$, $\sqrt{10}$, and $\sqrt{17}$ units respectively.

To Find:

Geometrically represent the square roots of the decimal numbers $\sqrt{4.5}$, $\sqrt{5.6}$, $\sqrt{8.1}$, and $\sqrt{2.3}$ on the number line.

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General Method of Construction:

To represent $\sqrt{x}$ geometrically for any positive real number $x$, we follow these steps:

1. Draw a line and mark a point $A$. Mark a point $B$ such that $AB = x$ units.

2. From $B$, mark a point $C$ such that $BC = 1$ unit.

3. Find the midpoint of $AC$ and mark it as $O$.

4. With $O$ as the centre and $OA$ (or $OC$) as the radius, draw a semicircle.

5. Draw a line perpendicular to $AC$ passing through point $B$ and let it intersect the semicircle at point $D$.

6. The length $BD$ represents $\sqrt{x}$.

7. To represent it on the number line, treat $B$ as the origin ($0$). With $B$ as the centre and $BD$ as the radius, draw an arc that intersects the number line at point $P$.

8. Point $P$ represents the value $\sqrt{x}$ on the number line.

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(i) Representation of $\sqrt{4.5}$

Construction Required:

Following the general method, we take $x = 4.5$ units.

1. Draw $AB = 4.5$ units and $BC = 1$ unit.

2. Draw a semicircle on $AC$ as diameter.

3. Draw perpendicular $BD$ at $B$.

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(ii) Representation of $\sqrt{5.6}$

Construction Required:

Following the general method, we take $x = 5.6$ units.

1. Draw $AB = 5.6$ units and $BC = 1$ unit.

2. Draw a semicircle on $AC$ as diameter.

3. Draw perpendicular $BD$ at $B$.

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(iii) Representation of $\sqrt{8.1}$

Construction Required:

Following the general method, we take $x = 8.1$ units.

1. Draw $AB = 8.1$ units and $BC = 1$ unit.

2. Draw a semicircle on $AC$ as diameter.

3. Draw perpendicular $BD$ at $B$.

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(iv) Representation of $\sqrt{2.3}$

Construction Required:

Following the general method, we take $x = 2.3$ units.

1. Draw $AB = 2.3$ units and $BC = 1$ unit.

2. Draw a semicircle on $AC$ as diameter.

3. Draw perpendicular $BD$ at $B$.

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Proof of the Method:

In $\triangle OBD$, let $AB = x$ and $BC = 1$. Then $AC = x + 1$.

The radius of the semicircle $OA = OC = OD = \frac{x+1}{2}$.

Now, we find the length of $OB$:

$OB = OC - BC$

$\displaystyle OB = \frac{x+1}{2} - 1 = \frac{x-1}{2}$

In right-angled triangle $\triangle OBD$, using Pythagoras Theorem:

$BD^2 = OD^2 - OB^2$

$\displaystyle BD^2 = \left(\frac{x+1}{2}\right)^2 - \left(\frac{x-1}{2}\right)^2$

$\displaystyle BD^2 = \frac{x^2 + 2x + 1}{4} - \frac{x^2 - 2x + 1}{4}$

$\displaystyle BD^2 = \frac{4x}{4}$

This confirms that the length $BD$ constructed always represents the square root of the chosen value $x$.

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(i) Express 0.2 in the form $\frac{p}{q}$:

$0.2 = \frac{2}{10}$

Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2.

$0.2 = \frac{\cancel{2}^1}{\cancel{10}_5}$

Thus, $0.2 = \frac{1}{5}$.

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(ii) Express 0.888... in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since one digit is repeating after the decimal point, multiply equation (i) by 10.

Subtract equation (i) from equation (ii).

Divide both sides by 9.

Thus, $0.888... = \frac{8}{9}$.

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(iii) Express 5.$\overline{2}$ in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since one digit is repeating after the decimal point, multiply equation (i) by 10.

Subtract equation (i) from equation (ii).

Divide both sides by 9.

Thus, 5.$\overline{2}$ = $\frac{47}{9}$.

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(iv) Express 0.$\overline{001}$ in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since three digits are repeating after the decimal point, multiply equation (i) by $10^3 = 1000$.

Subtract equation (i) from equation (ii).

Divide both sides by 999.

Thus, 0.$\overline{001}$ = $\frac{1}{999}$.

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(v) Express 0.2555... in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since there is one non-repeating digit (2) after the decimal point, multiply equation (i) by 10.

Since there is one repeating digit (5) after the decimal point in equation (ii), multiply equation (ii) by 10.

Subtract equation (ii) from equation (iii).

Divide both sides by 90.

Thus, 0.2555... = $\frac{23}{90}$.

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(vi) Express 0.1$\overline{34}$ in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since there is one non-repeating digit (1) after the decimal point, multiply equation (i) by 10.

Since there are two repeating digits (34) after the decimal point in equation (ii), multiply equation (ii) by $10^2 = 100$.

Subtract equation (ii) from equation (iii).

Divide both sides by 990.

Thus, 0.1$\overline{34}$ = $\frac{133}{990}$.

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(vii) Express 0.00323232… in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since there are two non-repeating digits (00) after the decimal point, multiply equation (i) by $10^2 = 100$.

Since there are two repeating digits (32) after the decimal point in equation (ii), multiply equation (ii) by $10^2 = 100$.

Subtract equation (ii) from equation (iii).

Divide both sides by 9900.

Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4.

Thus, 0.00323232… = $\frac{8}{2475}$.

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(viii) Express 0.404040... in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since two digits are repeating after the decimal point, multiply equation (i) by $10^2 = 100$.

Subtract equation (i) from equation (ii).

Divide both sides by 99.

Thus, 0.404040... = $\frac{40}{99}$.

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Let $x$ be the given number.

We can write this as:

The repeating block of digits is $142857$, which consists of 6 digits.

Multiply equation (i) by $10^6 = 1000000$.

We can write this as:

Subtract equation (i) from equation (ii).

Now, solve for $x$ by dividing both sides by 999999.

To simplify the fraction, we can observe that $142857 \times 7 = 999999$.

So, we can divide both the numerator and the denominator by 142857.

Thus, we have shown that $0.142857142857... = \frac{1}{7}$.

To Simplify:

The following mathematical expressions involving surds and radicals.

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(i) $\sqrt{45}$ – 3$\sqrt{20}$ + 4$\sqrt{5}$

First, we simplify the square roots by identifying perfect square factors:

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$

$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

Now, substitute these into the expression:

$3\sqrt{5} - 3(2\sqrt{5}) + 4\sqrt{5}$

$3\sqrt{5} - 6\sqrt{5} + 4\sqrt{5}$

Taking $\sqrt{5}$ as a common factor:

$(3 - 6 + 4)\sqrt{5}$

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(ii) $\frac{\sqrt{24}}{8}$ + $\frac{\sqrt{54}}{9}$

Simplify the numerators:

$\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$

$\sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}$

Substitute into the fraction:

$\displaystyle \frac{2\sqrt{6}}{8} + \frac{3\sqrt{6}}{9}$

Cancel common factors:

$\displaystyle \frac{\cancel{2}^1\sqrt{6}}{\cancel{8}_4} + \frac{\cancel{3}^1\sqrt{6}}{\cancel{9}_3}$

$\displaystyle \frac{\sqrt{6}}{4} + \frac{\sqrt{6}}{3}$

Take the Least Common Multiple (LCM) of 4 and 3, which is 12:

$\displaystyle \frac{3\sqrt{6} + 4\sqrt{6}}{12}$

$\displaystyle \frac{7\sqrt{6}}{12}$

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(iii) $4\sqrt {12} \times 7\sqrt {6}$

Note: The original expression $\sqrt[4]{12} \times \sqrt[7]{6}$ is a misprint. We are providing the solution for the corrected expression $4\sqrt {12} \times 7\sqrt {6}$.

Multiply the coefficients and the radicals separately:

$(4 \times 7) \times (\sqrt{12} \times \sqrt{6})$

$28 \times \sqrt{72}$

Simplify $\sqrt{72}$:

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

$28 \times 6\sqrt{2}$

$168\sqrt{2}$

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(iv) $4\sqrt{28}$ ÷ $3\sqrt{7}$

Note: The original expression $4\sqrt{28} ÷ 3\sqrt{7} ÷ \sqrt[3]{7}$ is a misprint. We are providing the solution for the corrected expression $4\sqrt{28} ÷ 3\sqrt{7}$.

$\displaystyle \frac{4\sqrt{28}}{3\sqrt{7}}$

Using the property $\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$:

$\displaystyle \frac{4}{3} \times \sqrt{\frac{28}{7}}$

$\displaystyle \frac{4}{3} \times \sqrt{4}$

$\displaystyle \frac{4}{3} \times 2$

$\displaystyle \frac{8}{3}$

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(v) $3\sqrt{3}$ + $2\sqrt{27}$ + $\frac{7}{\sqrt{3}}$

Simplify $\sqrt{27}$ and rationalise the third term:

$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

$\displaystyle \frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{7\sqrt{3}}{3}$

Substitute back:

$3\sqrt{3} + 2(3\sqrt{3}) + \frac{7\sqrt{3}}{3}$

$3\sqrt{3} + 6\sqrt{3} + \frac{7\sqrt{3}}{3}$

$9\sqrt{3} + \frac{7\sqrt{3}}{3}$

Take LCM as 3:

$\displaystyle \frac{27\sqrt{3} + 7\sqrt{3}}{3}$

$\displaystyle \frac{34\sqrt{3}}{3}$

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(vi) $\left( \sqrt{3} - \sqrt{2} \right)^{2}$

Using the identity $(a - b)^2 = a^2 + b^2 - 2ab$:

$3 + 2 - 2\sqrt{6}$

$5 - 2\sqrt{6}$

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(vii) $\sqrt [4] {81}$ - $8\sqrt [3] {216}$ + $15\sqrt [5] {32}$ + $\sqrt{225}$

Simplify each term using prime factorisation:

$\begin{array}{c|cc} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \implies 81 = 3^4 \implies \sqrt[4]{81} = 3$

$\begin{array}{c|cc} 6 & 216 \\ \hline 6 & 36 \\ \hline 6 & 6 \\ \hline & 1 \end{array} \implies 216 = 6^3 \implies \sqrt[3]{216} = 6$

$\begin{array}{c|cc} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} \implies 32 = 2^5 \implies \sqrt[5]{32} = 2$

$\sqrt{225} = 15$

Now, substitute the values:

$3 - 8(6) + 15(2) + 15$

$3 - 48 + 30 + 15$

$48 - 48$

$0$

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(viii) $\frac{3}{\sqrt{8}}$ + $\frac{1}{\sqrt{2}}$

Simplify $\sqrt{8}$:

$\sqrt{8} = 2\sqrt{2}$

$\displaystyle \frac{3}{2\sqrt{2}} + \frac{1}{\sqrt{2}}$

Take LCM as $2\sqrt{2}$:

$\displaystyle \frac{3 + 2}{2\sqrt{2}} = \frac{5}{2\sqrt{2}}$

Rationalise by multiplying with $\sqrt{2}$:

$\displaystyle \frac{5}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{2}}{2 \times 2}$

$\displaystyle \frac{5\sqrt{2}}{4}$

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(ix) $\frac{2\sqrt{3}}{3} - \frac{\sqrt{3}}{6}$

Take the LCM of 3 and 6, which is 6:

$\displaystyle \frac{2(2\sqrt{3}) - \sqrt{3}}{6}$

$\displaystyle \frac{4\sqrt{3} - \sqrt{3}}{6}$

$\displaystyle \frac{3\sqrt{3}}{6}$

$\displaystyle \frac{\cancel{3}^1\sqrt{3}}{\cancel{6}_2}$

$\displaystyle \frac{\sqrt{3}}{2}$

To Rationalise: The denominators of the given irrational expressions.

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(i) $\displaystyle \frac{2}{3\sqrt{3}}$

To rationalise the denominator, we multiply both the numerator and the denominator by $\sqrt{3}$.

$\displaystyle \frac{2}{3\sqrt{3}} = \frac{2}{3\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\displaystyle \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{3 \times (\sqrt{3} \times \sqrt{3})}$

$\displaystyle \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{3 \times 3}$

$\displaystyle \frac{2\sqrt{3}}{9}$

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(ii) $\displaystyle \frac{\sqrt{40}}{\sqrt{3}}$

First, we simplify $\sqrt{40}$:

$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$

Now, multiply the numerator and denominator by $\sqrt{3}$:

$\displaystyle \frac{2\sqrt{10}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\displaystyle \frac{2\sqrt{10 \times 3}}{3}$

$\displaystyle \frac{2\sqrt{30}}{3}$

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(iii) $\displaystyle \frac{3 \;+\; \sqrt{2}}{4\sqrt{2}}$

Multiply both the numerator and the denominator by $\sqrt{2}$:

$\displaystyle \frac{3 + \sqrt{2}}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\displaystyle \frac{\sqrt{2}(3 + \sqrt{2})}{4 \times 2}$

$\displaystyle \frac{3\sqrt{2} + (\sqrt{2})^2}{8}$

$\displaystyle \frac{3\sqrt{2} + 2}{8}$

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(iv) $\displaystyle \frac{16}{\sqrt{41} \;-\; 5}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{41} + 5)$:

$\displaystyle \frac{16}{\sqrt{41} - 5} \times \frac{\sqrt{41} + 5}{\sqrt{41} + 5}$

Using the identity $(a - b)(a + b) = a^2 - b^2$:

$\displaystyle \frac{16(\sqrt{41} + 5)}{(\sqrt{41})^2 - (5)^2}$

$\displaystyle \frac{16(\sqrt{41} + 5)}{41 - 25}$

$\displaystyle \frac{\cancel{16}^1(\sqrt{41} + 5)}{\cancel{16}_1}$

$\sqrt{41} + 5$

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(v) $\displaystyle \frac{2 \;+\; \sqrt{3}}{2 \;-\; \sqrt{3}}$

Multiply the numerator and denominator by the conjugate $(2 + \sqrt{3})$:

$\displaystyle \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$

Using $(a + b)^2 = a^2 + b^2 + 2ab$ in the numerator and $(a - b)(a + b) = a^2 - b^2$ in the denominator:

$\displaystyle \frac{(2 + \sqrt{3})^2}{(2)^2 - (\sqrt{3})^2}$

$\displaystyle \frac{(2)^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})}{4 - 3}$

$\displaystyle \frac{4 + 3 + 4\sqrt{3}}{1}$

$7 + 4\sqrt{3}$

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(vi) $\displaystyle \frac{\sqrt{6}}{\sqrt{2} \;+\; \sqrt{3}}$

To avoid negative denominators, we can write the denominator as $(\sqrt{3} + \sqrt{2})$ and multiply by $(\sqrt{3} - \sqrt{2})$:

$\displaystyle \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$\displaystyle \frac{\sqrt{6 \times 3} - \sqrt{6 \times 2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$\displaystyle \frac{\sqrt{18} - \sqrt{12}}{3 - 2}$

$\displaystyle \frac{3\sqrt{2} - 2\sqrt{3}}{1}$

$3\sqrt{2} - 2\sqrt{3}$

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(vii) $\displaystyle \frac{\sqrt{3} \;+\; \sqrt{2}}{\sqrt{3} \;-\; \sqrt{2}}$

Multiply both parts by $(\sqrt{3} + \sqrt{2})$:

$\displaystyle \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

$\displaystyle \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$\displaystyle \frac{3 + 2 + 2\sqrt{6}}{3 - 2}$

$5 + 2\sqrt{6}$

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(viii) $\displaystyle \frac{3\sqrt{5} \;+\; \sqrt{3}}{\sqrt{5} \;-\; \sqrt{3}}$

Multiply by $(\sqrt{5} + \sqrt{3})$:

$\displaystyle \frac{3\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

$\displaystyle \frac{3\sqrt{5}(\sqrt{5} + \sqrt{3}) + \sqrt{3}(\sqrt{5} + \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2}$

$\displaystyle \frac{15 + 3\sqrt{15} + \sqrt{15} + 3}{5 - 3}$

$\displaystyle \frac{18 + 4\sqrt{15}}{2}$

Divide each term by 2:

$9 + 2\sqrt{15}$

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(ix) $\displaystyle \frac{4\sqrt{3} \;+\; 5\sqrt{2}}{\sqrt{48} \;+\; \sqrt{18}}$

First, simplify the radicals in the denominator:

$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

The expression becomes: $\displaystyle \frac{4\sqrt{3} + 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}}$

Now multiply by the conjugate $(4\sqrt{3} - 3\sqrt{2})$:

$\displaystyle \frac{4\sqrt{3} + 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}} \times \frac{4\sqrt{3} - 3\sqrt{2}}{4\sqrt{3} - 3\sqrt{2}}$

$\displaystyle \text{Denominator} = (4\sqrt{3})^2 - (3\sqrt{2})^2 = 48 - 18 = 30$

$\text{Numerator} = 4\sqrt{3}(4\sqrt{3} - 3\sqrt{2}) + 5\sqrt{2}(4\sqrt{3} - 3\sqrt{2})$

$\text{Numerator} = 48 - 12\sqrt{6} + 20\sqrt{6} - 30$

$\text{Numerator} = 18 + 8\sqrt{6}$

Final fraction: $\displaystyle \frac{18 + 8\sqrt{6}}{30}$

Simplify by dividing by 2:

$\displaystyle \frac{9 + 4\sqrt{6}}{15}$

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(i) Find the values of a and b in $\frac{5 \;+\; 2\sqrt{3}}{7 \;+\; 4\sqrt{3}} = a - 6\sqrt{3}$:

Consider the Left Hand Side (LHS): $\frac{5 \;+\; 2\sqrt{3}}{7 \;+\; 4\sqrt{3}}$.

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $7 - 4\sqrt{3}$.

LHS $= \frac{5 \;+\; 2\sqrt{3}}{7 \;+\; 4\sqrt{3}} \times \frac{7 \;-\; 4\sqrt{3}}{7 \;-\; 4\sqrt{3}}$

LHS $= \frac{(5 \;+\; 2\sqrt{3})(7 \;-\; 4\sqrt{3})}{(7 \;+\; 4\sqrt{3})(7 \;-\; 4\sqrt{3})}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

Denominator $= (7)^2 - (4\sqrt{3})^2 = 49 - (4^2 \times (\sqrt{3})^2) $$ = 49 - (16 \times 3) = 49 - 48 = 1$.

Using distribution (FOIL) in the numerator:

Numerator $= 5(7) + 5(-4\sqrt{3}) + (2\sqrt{3})(7) + (2\sqrt{3})(-4\sqrt{3})$

Numerator $= 35 - 20\sqrt{3} + 14\sqrt{3} - 8(\sqrt{3})^2$

Numerator $= 35 - 6\sqrt{3} - 8(3)$

Numerator $= 35 - 6\sqrt{3} - 24$

Numerator $= (35 - 24) - 6\sqrt{3} = 11 - 6\sqrt{3}$.

So, LHS $= \frac{11 - 6\sqrt{3}}{1} = 11 - 6\sqrt{3}$.

Now equate LHS to RHS:

Comparing the rational parts on both sides:

Comparing the irrational parts on both sides:

This equation is satisfied and does not involve a parameter $b$ to be found. The structure of the RHS $a - 6\sqrt{3}$ implies that the coefficient of $\sqrt{3}$ is fixed at -6.

Thus, the value of $a$ is 11.

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(ii) Find the values of a and b in $\frac{3 \;-\; \sqrt{5}}{3 \;+\; 2\sqrt{5}} = a\sqrt{5} - \frac{19}{11}$:

Consider the Left Hand Side (LHS): $\frac{3 \;-\; \sqrt{5}}{3 \;+\; 2\sqrt{5}}$.

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $3 - 2\sqrt{5}$.

LHS $= \frac{3 \;-\; \sqrt{5}}{3 \;+\; 2\sqrt{5}} \times \frac{3 \;-\; 2\sqrt{5}}{3 \;-\; 2\sqrt{5}}$

LHS $= \frac{(3 \;-\; \sqrt{5})(3 \;-\; 2\sqrt{5})}{(3 \;+\; 2\sqrt{5})(3 \;-\; 2\sqrt{5})}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

Denominator $= (3)^2 - (2\sqrt{5})^2 = 9 - (4 \times 5) = 9 - 20 = -11$.

Using distribution (FOIL) in the numerator:

Numerator $= 3(3) + 3(-2\sqrt{5}) + (-\sqrt{5})(3) + (-\sqrt{5})(-2\sqrt{5})$

Numerator $= 9 - 6\sqrt{5} - 3\sqrt{5} + 2(\sqrt{5})^2$

Numerator $= 9 - 9\sqrt{5} + 2(5)$

Numerator $= 9 - 9\sqrt{5} + 10$

Numerator $= (9 + 10) - 9\sqrt{5} = 19 - 9\sqrt{5}$.

So, LHS $= \frac{19 - 9\sqrt{5}}{-11} = -\frac{19}{11} + \frac{9\sqrt{5}}{11}$.

Now equate LHS to RHS:

Rearrange the RHS to group rational and irrational parts:

Comparing the rational parts on both sides:

This is an identity and does not help find $a$ or $b$.

Comparing the irrational parts on both sides:

Divide both sides by $\sqrt{5}$ (since $\sqrt{5} \neq 0$):

Thus, the value of $a$ is $\frac{9}{11}$. There is no parameter $b$ in the given equation to find.

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(iii) Find the values of a and b in $\frac{\sqrt{2} \;+\; \sqrt{3}}{3\sqrt{2} \;-\; 2\sqrt{3}} = 2 - b\sqrt{6}$:

Consider the Left Hand Side (LHS): $\frac{\sqrt{2} \;+\; \sqrt{3}}{3\sqrt{2} \;-\; 2\sqrt{3}}$.

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $3\sqrt{2} + 2\sqrt{3}$.

LHS $= \frac{\sqrt{2} \;+\; \sqrt{3}}{3\sqrt{2} \;-\; 2\sqrt{3}} \times \frac{3\sqrt{2} \;+\; 2\sqrt{3}}{3\sqrt{2} \;+\; 2\sqrt{3}}$

LHS $= \frac{(\sqrt{2} \;+\; \sqrt{3})(3\sqrt{2} \;+\; 2\sqrt{3})}{(3\sqrt{2} \;-\; 2\sqrt{3})(3\sqrt{2} \;+\; 2\sqrt{3})}$

Using the identity $(x-y)(x+y) = x^2 - y^2$ in the denominator:

Denominator $= (3\sqrt{2})^2 - (2\sqrt{3})^2 = (3^2 \times (\sqrt{2})^2) - (2^2 \times (\sqrt{3})^2) $$ = (9 \times 2) - (4 \times 3) = 18 - 12 = 6$.

Using distribution (FOIL) in the numerator:

Numerator $= (\sqrt{2})(3\sqrt{2}) + (\sqrt{2})(2\sqrt{3}) + (\sqrt{3})(3\sqrt{2}) + (\sqrt{3})(2\sqrt{3})$

Numerator $= 3(\sqrt{2})^2 + 2\sqrt{2 \times 3} + 3\sqrt{3 \times 2} + 2(\sqrt{3})^2$

Numerator $= 3(2) + 2\sqrt{6} + 3\sqrt{6} + 2(3)

Numerator $= 6 + 5\sqrt{6} + 6$

Numerator $= (6 + 6) + 5\sqrt{6} = 12 + 5\sqrt{6}$.

So, LHS $= \frac{12 + 5\sqrt{6}}{6} = \frac{12}{6} + \frac{5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6}$.

Now equate LHS to RHS:

Comparing the rational parts on both sides:

This is an identity and does not help find $a$ or $b$.

Comparing the irrational parts on both sides:

Divide both sides by $\sqrt{6}$ (since $\sqrt{6} \neq 0$):

Multiply both sides by -1:

Thus, the value of $b$ is $-\frac{5}{6}$. There is no parameter $a$ in the given equation to find its specific value.

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(iv) Find the values of a and b in $\frac{(7 \;+\; \sqrt{5})}{(7 \;-\; \sqrt{5})} - \frac{(7 \;-\; \sqrt{5})}{(7 \;+\; \sqrt{5})} = a + \frac{7}{11}\sqrt{5}b$:

Consider the Left Hand Side (LHS): $\frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}} - \frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}}$.

Rationalize the first term $\frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}}$:

$\frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}} = \frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}} \times \frac{7 \;+\; \sqrt{5}}{7 \;+\; \sqrt{5}} = \frac{(7+\sqrt{5})^2}{7^2 - (\sqrt{5})^2} = \frac{49 + 14\sqrt{5} + 5}{49 - 5} = \frac{54 + 14\sqrt{5}}{44}$

Simplify the fraction by dividing numerator and denominator by 2:

$= \frac{\cancel{54}^{27} + \cancel{14}^{7}\sqrt{5}}{\cancel{44}^{22}} = \frac{27 + 7\sqrt{5}}{22}$.

Rationalize the second term $\frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}}$:

$\frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}} = \frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}} \times \frac{7 \;-\; \sqrt{5}}{7 \;-\; \sqrt{5}} = \frac{(7-\sqrt{5})^2}{7^2 - (\sqrt{5})^2} = \frac{49 - 14\sqrt{5} + 5}{49 - 5} = \frac{54 - 14\sqrt{5}}{44}$

Simplify the fraction by dividing numerator and denominator by 2:

$= \frac{\cancel{54}^{27} - \cancel{14}^{7}\sqrt{5}}{\cancel{44}^{22}} = \frac{27 - 7\sqrt{5}}{22}$.

Now subtract the second simplified term from the first simplified term:

LHS $= \frac{27 + 7\sqrt{5}}{22} - \frac{27 - 7\sqrt{5}}{22}$

Since the denominators are the same, subtract the numerators:

LHS $= \frac{(27 + 7\sqrt{5}) - (27 - 7\sqrt{5})}{22}$

LHS $= \frac{27 + 7\sqrt{5} - 27 + 7\sqrt{5}}{22}$

LHS $= \frac{(27 - 27) + (7\sqrt{5} + 7\sqrt{5})}{22}$

LHS $= \frac{0 + 14\sqrt{5}}{22} = \frac{14\sqrt{5}}{22}$

Simplify the fraction by dividing numerator and denominator by 2:

LHS $= \frac{\cancel{14}^{7}\sqrt{5}}{\cancel{22}^{11}} = \frac{7\sqrt{5}}{11}$.

Now equate LHS to RHS:

Rewrite the LHS to clearly show the rational and irrational parts:

Comparing the rational parts on both sides:

Comparing the irrational parts on both sides:

Divide both sides by $\frac{7}{11}\sqrt{5}$ (since $\frac{7}{11}\sqrt{5} \neq 0$):

Thus, the value of $a$ is 0 and the value of $b$ is 1.

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Given that $a = 2 + \sqrt{3}$.

First, let's find the value of $\frac{1}{a}$.

$\frac{1}{a} = \frac{1}{2 + \sqrt{3}}$

To simplify $\frac{1}{2 + \sqrt{3}}$, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, which is $2 - \sqrt{3}$.

$\frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

Denominator $= (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1$.

Numerator $= 1 \times (2 - \sqrt{3}) = 2 - \sqrt{3}$.

So, $\frac{1}{a} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$.

Now, we need to find the value of $a - \frac{1}{a}$.

Substitute the values of $a$ and $\frac{1}{a}$ into the expression:

$a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3})$

Remove the parentheses, remembering to change the signs of the terms inside the second parenthesis:

$a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3}$

Combine like terms (rational terms and irrational terms):

$a - \frac{1}{a} = (2 - 2) + (\sqrt{3} + \sqrt{3})$

$a - \frac{1}{a} = 0 + 2\sqrt{3}$

$a - \frac{1}{a} = 2\sqrt{3}$

Thus, the value of $a - \frac{1}{a}$ is $2\sqrt{3}$.

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(i) Simplify and evaluate $\frac{4}{\sqrt{3}}$:

Rationalise the denominator by multiplying the numerator and denominator by $\sqrt{3}$.

$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{(\sqrt{3})^2} = \frac{4\sqrt{3}}{3}$

Now, substitute the approximate value $\sqrt{3} \approx 1.732$.

$\frac{4\sqrt{3}}{3} \approx \frac{4 \times 1.732}{3}$

Calculate the numerator:

$4 \times 1.732 = 6.928$

Now divide by 3:

$\frac{6.928}{3} \approx 2.30933...$

Rounding to three places of decimal, the value is $2.309$.

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(ii) Simplify and evaluate $\frac{6}{\sqrt{6}}$:

Rationalise the denominator by multiplying the numerator and denominator by $\sqrt{6}$.

$\frac{6}{\sqrt{6}} = \frac{6}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{6\sqrt{6}}{(\sqrt{6})^2} = \frac{6\sqrt{6}}{6}$

Simplify the fraction:

$\frac{\cancel{6}\sqrt{6}}{\cancel{6}} = \sqrt{6}$

Now, substitute the approximate values $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$.

$\sqrt{6} \approx 1.414 \times 1.732$

Calculate the product:

$1.414 \times 1.732 \approx 2.449048$

Rounding to three places of decimal, the value is $2.449$.

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(iii) Simplify and evaluate $\frac{\sqrt{10} \;-\; \sqrt{5}}{2}$:

The denominator is already a rational number (2), so no rationalization is needed.

We can rewrite $\sqrt{10}$ as $\sqrt{2 \times 5} = \sqrt{2} \times \sqrt{5}$.

So, the expression is $\frac{\sqrt{2}\sqrt{5} \;-\; \sqrt{5}}{2}$.

Now, substitute the approximate values $\sqrt{2} \approx 1.414$ and $\sqrt{5} \approx 2.236$.

$\frac{\sqrt{10} \;-\; \sqrt{5}}{2} \approx \frac{(1.414 \times 2.236) - 2.236}{2}$

Calculate the term inside the parenthesis:

$1.414 \times 2.236 \approx 3.162224$

Now perform the subtraction in the numerator:

$3.162224 - 2.236 = 0.926224$

Now divide by 2:

$\frac{0.926224}{2} = 0.463112$

Rounding to three places of decimal, the value is $0.463$.

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(iv) Simplify and evaluate $\frac{\sqrt{2}}{2 \;+\; \sqrt{2}}$:

Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2 - \sqrt{2}$.

$\frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$

$= \frac{\sqrt{2}(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2}$

$= \frac{2\sqrt{2} - (\sqrt{2})^2}{4 - 2}$

$= \frac{2\sqrt{2} - 2}{2}$

Factor out 2 from the numerator and simplify:

$= \frac{\cancel{2}(\sqrt{2} - 1)}{\cancel{2}}$

$= \sqrt{2} - 1$

Now, substitute the approximate value $\sqrt{2} \approx 1.414$.

$\sqrt{2} - 1 \approx 1.414 - 1$

$= 0.414$

The value is already in three decimal places, so the value is $0.414$.

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(v) Simplify and evaluate $\frac{1}{\sqrt{3} \;+\; \sqrt{2}}$:

Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - \sqrt{2}$.

$\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$= \frac{1 \times (\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$= \frac{\sqrt{3} - \sqrt{2}}{3 - 2}$

$= \frac{\sqrt{3} - \sqrt{2}}{1}$

$= \sqrt{3} - \sqrt{2}$

Now, substitute the approximate values $\sqrt{3} \approx 1.732$ and $\sqrt{2} \approx 1.414$.

$\sqrt{3} - \sqrt{2} \approx 1.732 - 1.414$

$= 0.318$

The value is already in three decimal places, so the value is $0.318$.

To Simplify:

The following expressions involving powers and indices using the laws of exponents.

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(i) $(1^{3} + 2^{3} + 3^{3})^{\frac{1}{2}}$

First, we calculate the values of the cubes inside the parentheses:

$1^3 = 1$

$2^3 = 8$

$3^3 = 27$

Substitute these values back into the expression:

$\text{Sum} = 1 + 8 + 27 = 36$

The expression becomes:

$(36)^{\frac{1}{2}}$

Since $36 = 6^2$:

$(6^2)^{\frac{1}{2}} = 6^{2 \times \frac{1}{2}} = 6^1$

The simplified value is 6.

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(ii) $\left( \frac{3}{5} \right)^{4} . \left( \frac{8}{5} \right)^{-12} . \left( \frac{32}{5} \right)^{6}$

First, we use the law $a^{-n} = (\frac{1}{a})^n$ to make the exponents positive and express all numbers in terms of prime factors ($8 = 2^3$, $32 = 2^5$):

$\displaystyle \left( \frac{3}{5} \right)^4 \cdot \left( \frac{5}{2^3} \right)^{12} \cdot \left( \frac{2^5}{5} \right)^6$

Expand using $( \frac{a}{b} )^n = \frac{a^n}{b^n}$:

$\displaystyle \frac{3^4}{5^4} \cdot \frac{5^{12}}{(2^3)^{12}} \cdot \frac{(2^5)^6}{5^6}$

$\displaystyle \frac{3^4 \cdot 5^{12} \cdot 2^{30}}{5^4 \cdot 2^{36} \cdot 5^6}$

Combine the bases using $a^m \cdot a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$:

Base 3: $3^4$

Base 5: $5^{12 - (4 + 6)} = 5^{12 - 10} = 5^2$

Base 2: $2^{30 - 36} = 2^{-6} = \frac{1}{2^6}$

$\text{Result} = \frac{3^4 \cdot 5^2}{2^6} = \frac{81 \times 25}{64}$

The simplified value is $\frac{2025}{64}$.

---

(iii) $\left( \frac{1}{27} \right)^{-\frac{2}{3}}$

First, remove the negative exponent by taking the reciprocal of the base:

$(27)^{\frac{2}{3}}$

Express 27 as $3^3$:

$(3^3)^{\frac{2}{3}}$

Using $(a^m)^n = a^{mn}$:

$3^{3 \times \frac{2}{3}} = 3^2$

The simplified value is 9.

---

(iv) $\left( \left( \left( 625 \right)^{-\frac{1}{2}} \right)^{-\frac{1}{4}} \right)^{2}$

Using the law $(a^m)^n = a^{mn}$, we multiply all the exponents together:

$\text{Total Exponent} = -\frac{1}{2} \times -\frac{1}{4} \times 2$

$\text{Total Exponent} = \frac{1}{8} \times 2 = \frac{1}{4}$

Now, simplify the base $625$ using prime factorisation:

$\begin{array}{c|cc} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \implies 625 = 5^4$

The expression becomes:

$(5^4)^{\frac{1}{4}} = 5^{4 \times \frac{1}{4}} = 5^1$

The simplified value is 5.

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(v) $\displaystyle \frac{9^{\frac{1}{3}} \;\times\; 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \;\times\; 3^{-\frac{2}{3}}}$

Express all bases in terms of prime number 3 ($9 = 3^2, 27 = 3^3$):

$\displaystyle \frac{(3^2)^{\frac{1}{3}} \cdot (3^3)^{-\frac{1}{2}}}{3^{\frac{1}{6}} \cdot 3^{-\frac{2}{3}}}$

$\displaystyle \frac{3^{\frac{2}{3}} \cdot 3^{-\frac{3}{2}}}{3^{\frac{1}{6}} \cdot 3^{-\frac{2}{3}}}$

Combine all powers of base 3:

$\displaystyle 3^{\frac{2}{3} - \frac{3}{2} - \frac{1}{6} + \frac{2}{3}}$

Find the LCM of 3, 2, and 6, which is 6:

$\displaystyle 3^{\frac{4 - 9 - 1 + 4}{6}}$

$\displaystyle 3^{-\frac{2}{6}} = 3^{-\frac{1}{3}}$

The simplified value is $\frac{1}{3^{\frac{1}{3}}}$ or $\frac{1}{\sqrt [3] {3}}$.

---

(vi) $64^{-\frac{1}{3}} \left ( 64^{\frac{1}{3}} - 64^{\frac{2}{3}} \right )$

Distribute the term outside the parentheses:

$(64^{-\frac{1}{3}} \cdot 64^{\frac{1}{3}}) - (64^{-\frac{1}{3}} \cdot 64^{\frac{2}{3}})$

Apply $a^m \cdot a^n = a^{m+n}$:

$64^{-\frac{1}{3} + \frac{1}{3}} - 64^{-\frac{1}{3} + \frac{2}{3}}$

$64^0 - 64^{\frac{1}{3}}$

We know $a^0 = 1$ and $64 = 4^3$:

$1 - (4^3)^{\frac{1}{3}}$

$1 - 4$

The simplified value is -3.

---

(vii) $\displaystyle \frac{8^{\frac{1}{3}} \;\times\; 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$

Move the term with negative exponent to the numerator:

$8^{\frac{1}{3}} \cdot 16^{\frac{1}{3}} \cdot 32^{\frac{1}{3}}$

Since all terms have the same power ($1/3$), we can multiply the bases together:

$(8 \times 16 \times 32)^{\frac{1}{3}}$

Express in terms of base 2 ($8 = 2^3, 16 = 2^4, 32 = 2^5$):

$(2^3 \cdot 2^4 \cdot 2^5)^{\frac{1}{3}}$

$(2^{3+4+5})^{\frac{1}{3}} = (2^{12})^{\frac{1}{3}}$

$2^{12 \times \frac{1}{3}} = 2^4$

The simplified value is 16.

Given:

$a = 5 + 2\sqrt{6}$

$b = \frac{1}{a}$

To Find:

The value of $a^2 + b^2$.

---

Solution:

We are given $a = 5 + 2\sqrt{6}$.

We are also given that $b = \frac{1}{a}$. Let's find the value of $b$ by substituting the value of $a$:

$b = \frac{1}{5 + 2\sqrt{6}}$

To simplify this expression and rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $5 - 2\sqrt{6}$.

$b = \frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}}$

In the denominator, we use the difference of squares formula: $(x+y)(x-y) = x^2 - y^2$. Here, $x=5$ and $y=2\sqrt{6}$.

$b = \frac{5 - 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2}$

$b = \frac{5 - 2\sqrt{6}}{25 - (2^2 \times (\sqrt{6})^2)}$

$b = \frac{5 - 2\sqrt{6}}{25 - (4 \times 6)}$

$b = \frac{5 - 2\sqrt{6}}{25 - 24}$

$b = \frac{5 - 2\sqrt{6}}{1}$

$b = 5 - 2\sqrt{6}$

So, we have $a = 5 + 2\sqrt{6}$ and $b = 5 - 2\sqrt{6}$.

We need to find the value of $a^2 + b^2$. We can use the algebraic identity $a^2 + b^2 = (a+b)^2 - 2ab$.

First, let's find the sum $a+b$:

$a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6})$

$a + b = 5 + 2\sqrt{6} + 5 - 2\sqrt{6}$

$a + b = (5 + 5) + (2\sqrt{6} - 2\sqrt{6})$

$a + b = 10 + 0$

$a + b = 10$

Next, let's find the product $ab$:

$ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6})$

Using the identity $(x+y)(x-y) = x^2 - y^2$ again:

$ab = (5)^2 - (2\sqrt{6})^2$

$ab = 25 - (4 \times 6)$

$ab = 25 - 24$

$ab = 1$

Now, substitute the values of $(a+b)$ and $ab$ into the identity $a^2 + b^2 = (a+b)^2 - 2ab$:

$a^2 + b^2 = (10)^2 - 2(1)$

$a^2 + b^2 = 100 - 2$

$a^2 + b^2 = 98$

---

Alternate Solution:

We can also calculate $a^2$ and $b^2$ separately and then add them.

Given $a = 5 + 2\sqrt{6}$. Calculate $a^2$ using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$a^2 = (5 + 2\sqrt{6})^2$

$a^2 = (5)^2 + 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$a^2 = 25 + 20\sqrt{6} + (4 \times 6)$

$a^2 = 25 + 20\sqrt{6} + 24$

$a^2 = 49 + 20\sqrt{6}$

From the primary solution, we found $b = 5 - 2\sqrt{6}$. Calculate $b^2$ using the identity $(x-y)^2 = x^2 - 2xy + y^2$:

$b^2 = (5 - 2\sqrt{6})^2$

$b^2 = (5)^2 - 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$b^2 = 25 - 20\sqrt{6} + (4 \times 6)$

$b^2 = 25 - 20\sqrt{6} + 24$

$b^2 = 49 - 20\sqrt{6}$

Now, add $a^2$ and $b^2$:

$a^2 + b^2 = (49 + 20\sqrt{6}) + (49 - 20\sqrt{6})$

$a^2 + b^2 = 49 + 20\sqrt{6} + 49 - 20\sqrt{6}$

$a^2 + b^2 = (49 + 49) + (20\sqrt{6} - 20\sqrt{6})$

$a^2 + b^2 = 98 + 0$

$a^2 + b^2 = 98$

---

The value of $a^2 + b^2$ is 98.

---

We need to express each term in the sum in the form $\frac{p}{q}$.

Term 1: 0.6

$0.6 = \frac{6}{10}$

Simplify the fraction:

$0.6 = \frac{\cancel{6}^3}{\cancel{10}_5} = \frac{3}{5}$

---

Term 2: 0.$\overline{7}$

Let $x = 0.\overline{7}$

Multiply equation (i) by 10 (since one digit is repeating):

Subtract equation (i) from equation (ii):

Solve for $x$:

So, $0.\overline{7} = \frac{7}{9}$.

---

Term 3: 0.4$\overline{7}$

Let $y = 0.4\overline{7}$

Multiply equation (iii) by 10 (to move the non-repeating digit past the decimal):

Multiply equation (iv) by 10 (since one digit is repeating after the decimal in (iv)):

Subtract equation (iv) from equation (v):

Solve for $y$:

So, $0.4\overline{7} = \frac{43}{90}$.

---

Adding the fractions:

Now we need to calculate the sum: $0.6 + 0.\overline{7} + 0.4\overline{7} = \frac{3}{5} + \frac{7}{9} + \frac{43}{90}$.

Find the Least Common Multiple (LCM) of the denominators 5, 9, and 90.

Prime factorization of the denominators:

$5 = 5^1$

$9 = 3^2$

$90 = 2 \times 45 = 2 \times 9 \times 5 = 2^1 \times 3^2 \times 5^1$

LCM$(5, 9, 90) = 2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90$.

Convert each fraction to have a denominator of 90:

$\frac{3}{5} = \frac{3 \times 18}{5 \times 18} = \frac{54}{90}$

$\frac{7}{9} = \frac{7 \times 10}{9 \times 10} = \frac{70}{90}$

$\frac{43}{90}$ (already has the denominator 90)

Add the fractions:

$\frac{54}{90} + \frac{70}{90} + \frac{43}{90} = \frac{54 + 70 + 43}{90}$

Calculate the sum in the numerator:

$54 + 70 = 124$

$124 + 43 = 167$

The sum is $\frac{167}{90}$.

The fraction $\frac{167}{90}$ is in the form $\frac{p}{q}$, where $p=167$ and $q=90$. Both are integers and $q \neq 0$. The fraction is already in its simplest form as 167 is a prime number and 90 is not divisible by 167.

Thus, $0.6 + 0.\overline{7} + 0.4\overline{7} = \frac{167}{90}$.

---

We need to simplify the given expression by rationalizing the denominator of each term.

Term 1: $\frac{7\sqrt{3}}{\sqrt{10} \;+\; \sqrt{3}}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{10} - \sqrt{3}$.

$\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} \times \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} = \frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{(\sqrt{10})^2 - (\sqrt{3})^2}$

$= \frac{7\sqrt{30} - 7(\sqrt{3})^2}{10 - 3} = \frac{7\sqrt{30} - 7(3)}{7} = \frac{7\sqrt{30} - 21}{7}$

$= \frac{7(\sqrt{30} - 3)}{7} = \sqrt{30} - 3$

---

Term 2: $\frac{2\sqrt{5}}{\sqrt{6} \;+\; \sqrt{5}}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{6} - \sqrt{5}$.

$\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{(\sqrt{6})^2 - (\sqrt{5})^2}$

$= \frac{2\sqrt{30} - 2(\sqrt{5})^2}{6 - 5} = \frac{2\sqrt{30} - 2(5)}{1} = 2\sqrt{30} - 10$

---

Term 3: $\frac{3\sqrt{2}}{\sqrt{15} \;+\; 3\sqrt{2}}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{15} - 3\sqrt{2}$.

$\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \times \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} = \frac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{(\sqrt{15})^2 - (3\sqrt{2})^2}$

$= \frac{3\sqrt{30} - 3(3\sqrt{2})(\sqrt{2})}{15 - (3^2 \times (\sqrt{2})^2)} = \frac{3\sqrt{30} - 9(2)}{15 - (9 \times 2)} = \frac{3\sqrt{30} - 18}{15 - 18}$

$= \frac{3\sqrt{30} - 18}{-3} = \frac{3(\sqrt{30} - 6)}{-3} = -(\sqrt{30} - 6) = -\sqrt{30} + 6$

---

Now, substitute the simplified terms back into the original expression:

$(\sqrt{30} - 3) - (2\sqrt{30} - 10) - (-\sqrt{30} + 6)$

Remove the parentheses, being careful with the signs:

$= \sqrt{30} - 3 - 2\sqrt{30} + 10 + \sqrt{30} - 6$

Group the terms with $\sqrt{30}$ and the constant terms:

$= (\sqrt{30} - 2\sqrt{30} + \sqrt{30}) + (-3 + 10 - 6)$

Combine the coefficients of $\sqrt{30}$:

$= (1 - 2 + 1)\sqrt{30} + (-3 + 10 - 6)$

$= 0\sqrt{30} + (7 - 6)$

$= 0 + 1$

$= 1$

The simplified value of the expression is 1.

Given:

Expression: $\frac{4}{(3\sqrt{3} \;-\; 2\sqrt{2})}+\frac{3}{(3\sqrt{3} \;+\; 2\sqrt{2})}$

Approximate values: $\sqrt{2} = 1.414$, $\sqrt{3} = 1.732$

To Find:

The value of the expression using the given approximations, rounded to three decimal places.

---

Solution:

First, we simplify the given expression by combining the two fractions. The denominators are conjugates of each other, so we can find a common denominator by multiplying them.

Common denominator = $(3\sqrt{3} - 2\sqrt{2})(3\sqrt{3} + 2\sqrt{2})$

Using the identity $(a-b)(a+b) = a^2 - b^2$:

$(3\sqrt{3})^2 - (2\sqrt{2})^2 = (3^2 \times (\sqrt{3})^2) - (2^2 \times (\sqrt{2})^2)$

$= (9 \times 3) - (4 \times 2) = 27 - 8 = 19$

Now, combine the fractions over the common denominator:

$\frac{4(3\sqrt{3} \;+\; 2\sqrt{2}) \;+\; 3(3\sqrt{3} \;-\; 2\sqrt{2})}{(3\sqrt{3} \;-\; 2\sqrt{2})(3\sqrt{3} \;+\; 2\sqrt{2})}$

$= \frac{4(3\sqrt{3}) \;+\; 4(2\sqrt{2}) \;+\; 3(3\sqrt{3}) \;-\; 3(2\sqrt{2})}{19}$

$= \frac{12\sqrt{3} \;+\; 8\sqrt{2} \;+\; 9\sqrt{3} \;-\; 6\sqrt{2}}{19}$

Group like terms in the numerator (terms with $\sqrt{3}$ and terms with $\sqrt{2}$):

$= \frac{(12\sqrt{3} \;+\; 9\sqrt{3}) \;+\; (8\sqrt{2} \;-\; 6\sqrt{2})}{19}$

$= \frac{(12 + 9)\sqrt{3} \;+\; (8 - 6)\sqrt{2}}{19}$

$= \frac{21\sqrt{3} \;+\; 2\sqrt{2}}{19}$

---

Now, substitute the given approximate values $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$ into the simplified expression.

Value $\approx \frac{21(1.732) \;+\; 2(1.414)}{19}$

Calculate the products in the numerator:

$21 \times 1.732 = 36.372$

$2 \times 1.414 = 2.828$

Add these values:

$36.372 + 2.828 = 39.200$

Now, divide the sum by 19:

Value $\approx \frac{39.200}{19}$

Perform the division:

$39.200 \div 19 \approx 2.06315...$

Rounding the result to three places of decimal:

Value $\approx 2.063$

The value of the expression, rounded to three decimal places, is $2.063$.

Given:

$a = \frac{3 \;+\; \sqrt{5}}{2}$

To Find:

The value of $a^2 + \frac{1}{a^2}$.

---

Solution:

We are given the value of $a$. To find $a^2 + \frac{1}{a^2}$, we can first find the value of $\frac{1}{a}$.

$\frac{1}{a} = \frac{1}{\frac{3 \;+\; \sqrt{5}}{2}}$

$\frac{1}{a} = \frac{2}{3 \;+\; \sqrt{5}}$

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $3 - \sqrt{5}$.

$\frac{1}{a} = \frac{2}{3 \;+\; \sqrt{5}} \times \frac{3 \;-\; \sqrt{5}}{3 \;-\; \sqrt{5}}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

$\frac{1}{a} = \frac{2(3 \;-\; \sqrt{5})}{3^2 \;-\; (\sqrt{5})^2}$

$\frac{1}{a} = \frac{2(3 \;-\; \sqrt{5})}{9 \;-\; 5}$

$\frac{1}{a} = \frac{2(3 \;-\; \sqrt{5})}{4}$

Simplify the fraction:

$\frac{1}{a} = \frac{\cancel{2}^1(3 \;-\; \sqrt{5})}{\cancel{4}_2}$

$\frac{1}{a} = \frac{3 \;-\; \sqrt{5}}{2}$

Now we have $a = \frac{3 + \sqrt{5}}{2}$ and $\frac{1}{a} = \frac{3 - \sqrt{5}}{2}$.

We can find $a^2 + \frac{1}{a^2}$ using the identity $a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$.

First, calculate the sum $a + \frac{1}{a}$:

$a + \frac{1}{a} = \frac{3 \;+\; \sqrt{5}}{2} + \frac{3 \;-\; \sqrt{5}}{2}$

Since the denominators are the same, add the numerators:

$a + \frac{1}{a} = \frac{(3 \;+\; \sqrt{5}) \;+\; (3 \;-\; \sqrt{5})}{2}$

$a + \frac{1}{a} = \frac{3 \;+\; \sqrt{5} \;+\; 3 \;-\; \sqrt{5}}{2}$

Combine the terms in the numerator:

$a + \frac{1}{a} = \frac{(3 \;+\; 3) \;+\; (\sqrt{5} \;-\; \sqrt{5})}{2}$

$a + \frac{1}{a} = \frac{6 \;+\; 0}{2}$

$a + \frac{1}{a} = \frac{6}{2}$

$a + \frac{1}{a} = 3$

Now, substitute the value of $a + \frac{1}{a}$ into the identity $a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$:

$a^2 + \frac{1}{a^2} = (3)^2 - 2$

$a^2 + \frac{1}{a^2} = 9 - 2$

$a^2 + \frac{1}{a^2} = 7$

---

Alternate Solution:

We can directly calculate $a^2$ and $\frac{1}{a^2}$ and then add them.

$a = \frac{3 \;+\; \sqrt{5}}{2}$

$a^2 = \left(\frac{3 \;+\; \sqrt{5}}{2}\right)^2 = \frac{(3 \;+\; \sqrt{5})^2}{2^2}$

$a^2 = \frac{3^2 \;+\; 2(3)(\sqrt{5}) \;+\; (\sqrt{5})^2}{4}$

$a^2 = \frac{9 \;+\; 6\sqrt{5} \;+\; 5}{4}$

$a^2 = \frac{14 \;+\; 6\sqrt{5}}{4}$

Simplify:

$a^2 = \frac{\cancel{2}(7 \;+\; 3\sqrt{5})}{\cancel{4}_2}$

$a^2 = \frac{7 \;+\; 3\sqrt{5}}{2}$

From the first method, we found $\frac{1}{a} = \frac{3 \;-\; \sqrt{5}}{2}$.

So, $\frac{1}{a^2} = \left(\frac{1}{a}\right)^2 = \left(\frac{3 \;-\; \sqrt{5}}{2}\right)^2 = \frac{(3 \;-\; \sqrt{5})^2}{2^2}$

$\frac{1}{a^2} = \frac{3^2 \;-\; 2(3)(\sqrt{5}) \;+\; (\sqrt{5})^2}{4}$

$\frac{1}{a^2} = \frac{9 \;-\; 6\sqrt{5} \;+\; 5}{4}$

$\frac{1}{a^2} = \frac{14 \;-\; 6\sqrt{5}}{4}$

Simplify:

$\frac{1}{a^2} = \frac{\cancel{2}(7 \;-\; 3\sqrt{5})}{\cancel{4}_2}$

$\frac{1}{a^2} = \frac{7 \;-\; 3\sqrt{5}}{2}$

Now, add $a^2$ and $\frac{1}{a^2}$:

$a^2 + \frac{1}{a^2} = \frac{7 \;+\; 3\sqrt{5}}{2} + \frac{7 \;-\; 3\sqrt{5}}{2}$

$a^2 + \frac{1}{a^2} = \frac{(7 \;+\; 3\sqrt{5}) \;+\; (7 \;-\; 3\sqrt{5})}{2}$

$a^2 + \frac{1}{a^2} = \frac{7 \;+\; 3\sqrt{5} \;+\; 7 \;-\; 3\sqrt{5}}{2}$

$a^2 + \frac{1}{a^2} = \frac{(7 \;+\; 7) \;+\; (3\sqrt{5} \;-\; 3\sqrt{5})}{2}$

$a^2 + \frac{1}{a^2} = \frac{14 \;+\; 0}{2}$

$a^2 + \frac{1}{a^2} = \frac{14}{2}$

$a^2 + \frac{1}{a^2} = 7$

The value of $a^2 + \frac{1}{a^2}$ is 7.

Given:

$x = \frac{\sqrt{3} \;+\; \sqrt{2}}{\sqrt{3} \;-\; \sqrt{2}}$

$y = \frac{\sqrt{3} \;-\; \sqrt{2}}{\sqrt{3} \;+\; \sqrt{2}}$

To Find:

The value of $x^2 + y^2$.

---

Solution:

First, let's simplify the expressions for $x$ and $y$ by rationalizing their denominators.

For $x = \frac{\sqrt{3} \;+\; \sqrt{2}}{\sqrt{3} \;-\; \sqrt{2}}$, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + \sqrt{2}$.

$x = \frac{(\sqrt{3} \;+\; \sqrt{2})}{(\sqrt{3} \;-\; \sqrt{2})} \times \frac{(\sqrt{3} \;+\; \sqrt{2})}{(\sqrt{3} \;+\; \sqrt{2})}$

$x = \frac{(\sqrt{3} \;+\; \sqrt{2})^2}{(\sqrt{3})^2 \;-\; (\sqrt{2})^2}$

Using $(a+b)^2 = a^2 + 2ab + b^2$ in the numerator and $(a-b)(a+b) = a^2 - b^2$ in the denominator:

$x = \frac{(\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2}{3 - 2}$

$x = \frac{3 + 2\sqrt{6} + 2}{1}$

$x = 5 + 2\sqrt{6}$

For $y = \frac{\sqrt{3} \;-\; \sqrt{2}}{\sqrt{3} \;+\; \sqrt{2}}$, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - \sqrt{2}$.

$y = \frac{(\sqrt{3} \;-\; \sqrt{2})}{(\sqrt{3} \;+\; \sqrt{2})} \times \frac{(\sqrt{3} \;-\; \sqrt{2})}{(\sqrt{3} \;-\; \sqrt{2})}$

$y = \frac{(\sqrt{3} \;-\; \sqrt{2})^2}{(\sqrt{3})^2 \;-\; (\sqrt{2})^2}$

Using $(a-b)^2 = a^2 - 2ab + b^2$ in the numerator and $(a+b)(a-b) = a^2 - b^2$ in the denominator:

$y = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2}{3 - 2}$

$y = \frac{3 - 2\sqrt{6} + 2}{1}$

$y = 5 - 2\sqrt{6}$

So, we have $x = 5 + 2\sqrt{6}$ and $y = 5 - 2\sqrt{6}$.

We can find $x^2 + y^2$ using the algebraic identity $x^2 + y^2 = (x+y)^2 - 2xy$.

First, calculate the sum $x+y$:

$x + y = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6})$

$x + y = 5 + 2\sqrt{6} + 5 - 2\sqrt{6}$

$x + y = (5+5) + (2\sqrt{6} - 2\sqrt{6})$

$x + y = 10 + 0 = 10$

Next, calculate the product $xy$:

$xy = (5 + 2\sqrt{6})(5 - 2\sqrt{6})$

Using the identity $(a+b)(a-b) = a^2 - b^2$:

$xy = (5)^2 - (2\sqrt{6})^2$

$xy = 25 - (4 \times 6)$

$xy = 25 - 24$

$xy = 1$

Now, substitute the values of $(x+y)$ and $xy$ into the identity $x^2 + y^2 = (x+y)^2 - 2xy$:

$x^2 + y^2 = (10)^2 - 2(1)$

$x^2 + y^2 = 100 - 2$

$x^2 + y^2 = 98$

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Alternate Solution:

We can also calculate $x^2$ and $y^2$ directly and then add them.

We have $x = 5 + 2\sqrt{6}$.

$x^2 = (5 + 2\sqrt{6})^2$

Using $(a+b)^2 = a^2 + 2ab + b^2$:

$x^2 = 5^2 + 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$x^2 = 25 + 20\sqrt{6} + (4 \times 6)$

$x^2 = 25 + 20\sqrt{6} + 24$

$x^2 = 49 + 20\sqrt{6}$

We have $y = 5 - 2\sqrt{6}$.

$y^2 = (5 - 2\sqrt{6})^2$

Using $(a-b)^2 = a^2 - 2ab + b^2$:

$y^2 = 5^2 - 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$y^2 = 25 - 20\sqrt{6} + (4 \times 6)$

$y^2 = 25 - 20\sqrt{6} + 24$

$y^2 = 49 - 20\sqrt{6}$

Now, add $x^2$ and $y^2$:

$x^2 + y^2 = (49 + 20\sqrt{6}) + (49 - 20\sqrt{6})$

$x^2 + y^2 = 49 + 20\sqrt{6} + 49 - 20\sqrt{6}$

$x^2 + y^2 = (49 + 49) + (20\sqrt{6} - 20\sqrt{6})$

$x^2 + y^2 = 98 + 0$

$x^2 + y^2 = 98$

The value of $x^2 + y^2$ is 98.

To Simplify:

$(256)^{-\left( 4^{-\frac{3}{2}} \right)}$

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Solution:

To simplify the expression, we first evaluate the exponent of the base 256. The exponent is $-4^{-\frac{3}{2}}$.

Step 1: Simplify the inner exponent $4^{-\frac{3}{2}}$.

We use the law of exponents $a^{-n} = \frac{1}{a^n}$:

$4^{-\frac{3}{2}} = \frac{1}{4^{\frac{3}{2}}}$

$\text{Since } 4 = 2^2 \text{, we have:}$

$\displaystyle \frac{1}{(2^2)^{\frac{3}{2}}} = \frac{1}{2^{2 \times \frac{3}{2}}}$

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Step 2: Substitute the value from (i) back into the main expression.

The expression now becomes:

$(256)^{-\frac{1}{8}}$

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Step 3: Simplify the base 256 using prime factorisation.

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

From the above calculation, we get:

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Step 4: Evaluate the final power.

$(2^8)^{-\frac{1}{8}}$

Using the identity $(a^m)^n = a^{mn}$:

$\displaystyle 2^{8 \times \left( -\frac{1}{8} \right)}$

$\displaystyle 2^{\cancel{8} \times \left( -\frac{1}{\cancel{8}} \right)}$

$2^{-1}$

$\displaystyle \frac{1}{2}$

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Final Answer:

The simplified value of the given expression is $\frac{1}{2}$ or 0.5.

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Alternate Solution:

We can also write 256 as $4^4$ or $16^2$. Using $256 = (16)^2$:

$(16^2)^{-\frac{1}{8}} = 16^{2 \times -\frac{1}{8}} = 16^{-\frac{1}{4}}$

$\text{Since } 16 = 2^4 \text{, we have:}$

$(2^4)^{-\frac{1}{4}} = 2^{4 \times -\frac{1}{4}} = 2^{-1} = \frac{1}{2}$

To Find:

The value of the expression: $\displaystyle \frac{4}{(216)^{-\frac{2}{3}}} + \frac{1}{(256)^{-\frac{3}{4}}} + \frac{2}{(243)^{-\frac{1}{5}}}$

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Solution:

To find the value of the given expression, we will first simplify each term individually using the laws of exponents.

The primary laws used are:

1. $\displaystyle \frac{1}{a^{-n}} = a^n$

2. $(a^m)^n = a^{mn}$

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Step 1: Simplify the first term

First, we perform the prime factorisation of 216:

$\begin{array}{c|cc} 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $216 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$.

Now, simplify the term:

$\displaystyle \frac{4}{(216)^{-\frac{2}{3}}} = 4 \times (216)^{\frac{2}{3}}$

$\displaystyle = 4 \times (6^3)^{\frac{2}{3}}$

$\displaystyle = 4 \times 6^{3 \times \frac{2}{3}}$

$\displaystyle = 4 \times 6^2 = 4 \times 36$

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Step 2: Simplify the second term

First, we perform the prime factorisation of 256:

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $256 = 2^8$, which can also be written as $(2^2)^4 = 4^4$.

Now, simplify the term:

$\displaystyle \frac{1}{(256)^{-\frac{3}{4}}} = (256)^{\frac{3}{4}}$

$\displaystyle = (4^4)^{\frac{3}{4}}$

$\displaystyle = 4^{4 \times \frac{3}{4}}$

$\displaystyle = 4^3$

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Step 3: Simplify the third term

First, we perform the prime factorisation of 243 (Note: In the standard context of this problem, the base is 243):

$\begin{array}{c|cc} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So, $243 = 3^5$.

Now, simplify the term:

$\displaystyle \frac{2}{(243)^{-\frac{1}{5}}} = 2 \times (243)^{\frac{1}{5}}$

$\displaystyle = 2 \times (3^5)^{\frac{1}{5}}$

$\displaystyle = 2 \times 3^{5 \times \frac{1}{5}}$

$\displaystyle = 2 \times 3^1$

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Final Addition:

Now, substituting the values from (i), (ii), and (iii) back into the expression:

The final value of the expression is 214.