Exercise 1F RS Aggarwal Class 9 Number Systems Best Solutions

Exercise 1F RS Aggarwal Class 9 contains a total of twenty five questions. The questions are based on the following topic.

Rationalisation: Rationalization is the process of converting an irrational number into an equivalent expression with a rational denominator by multiplying its numerator and denominator by a suitable number.

Exercise 1F RS Aggarwal Class 9 Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16 Q.17 Q.18 Q.19 Q.20 Q.21 Q.22 Q.23 Q.24 Q.25

The first question in Exercise 1F RS Aggarwal Class 9 revolves around rationalizing the denominator of an expression.

1. Write the rationalizing factor of the denominator in $\frac{1}{\sqrt{2} + \sqrt{3}}$ .

Answer

The rationalizing factor of an irrational expression is the factor that, when multiplied to the irrational expression, transforms it into a rational number.

The denominator of the given expression is $\sqrt{2} + \sqrt{3}$ .
$\sqrt{2} - \sqrt{3}$ is its rationalising factor because $(\sqrt{2} + \sqrt{3})$ × $(\sqrt{2} - \sqrt{3})$ = ${(\sqrt{2})}^{2} - {(\sqrt{3})}^{2}$ = 2 − 3 = −1 = a rational number.

The second question of Exercise 1F RS Aggarwal Class 9 revolves around rationalizing the denominator of an expression.

2. Rationalise the denominator of each of the following.

(i) $\frac{1}{\sqrt{7}}$

Answer

We have
$\frac{1}{\sqrt{7}}$
Multiplying by $\sqrt{7}$ in both numerator and denominator, we get
$\frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}$ = $\frac{\sqrt{7}}{7}$ .
Hence, the denominator is rationalized.

(ii) $\frac{\sqrt{5}}{2 \sqrt{3}}$

Answer

The given expression is $\frac{\sqrt{5}}{2 \sqrt{3}}$ .
Multiplying by $\sqrt{3}$ in the denominator and numerator both, we get
$\frac{\sqrt{5} \times \sqrt{3}}{2 \sqrt{3} \times \sqrt{3}}$ = $\frac{\sqrt{5 \times 3}}{2 \times 3}$ = $\frac{\sqrt{15}}{6}$ .
Hence, the denominator is rationalized.

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

Answer

The given expression is $\frac{1}{2 + \sqrt{3}}$ .

Multiplying by $(2 - \sqrt{3})$ in the numerator and denominator both, we get.

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

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

= $\frac{2 - \sqrt{3}}{4 - 3}$

= $2 - \sqrt{3}$ .

Hence, the denominator is rationalized.

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

Answer

The given expression is $\frac{1}{\sqrt{5} - 2}$ .

Multiplying by $(\sqrt{5} + 2)$ in the numerator and denominator both, we get.

$\frac{1 \times (\sqrt{5} + 2)}{(\sqrt{5} - 2) (\sqrt{5} + 2)}$

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

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

= $(\sqrt{5} + 2)$

Hence, the denominator is rationalized.

(v) $\frac{1}{5 + 3 \sqrt{2}}$

Answer

The given expression is $\frac{1}{5 + 3 \sqrt{2}}$ .

Multiplying by $(5 - 3 \sqrt{2})$ in both numerator and denominator, we get.

$\frac{1 \times (5 - 3 \sqrt{2})}{(5 + 3 \sqrt{2}) (5 - 3 \sqrt{2})}$

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

= $\frac{(5 - 3 \sqrt{2})}{25 - 18}$

= $\frac{(5 - 3 \sqrt{2})}{7}$ .

Hence, the denominator is rationalized.

(vi) $\frac{1}{\sqrt{7} - \sqrt{6}}$

Answer

The given expression is $\frac{1}{\sqrt{7} - \sqrt{6}}$ .

Multiplying by $(\sqrt{7} + \sqrt{6})$ in numerator and denominator both, we get.

$\frac{1 \times (\sqrt{7} + \sqrt{6})}{(\sqrt{7} - \sqrt{6}) (\sqrt{7} + \sqrt{6})}$

= $\frac{(\sqrt{7} + \sqrt{6})}{{(\sqrt{7})}^{2} - {(\sqrt{6})}^{2}}$

= $\frac{(\sqrt{7} + \sqrt{6})}{7 - 6}$

= $(\sqrt{7} + \sqrt{6})$ .

Hence, the denominator is rationalized.

(vii) $\frac{4}{\sqrt{11} - \sqrt{7}}$

Answer

The given expression is $\frac{4}{\sqrt{11} - \sqrt{7}}$ .

Multiplying by $(\sqrt{11} + \sqrt{7})$ in numerator and denominator both.

$\frac{4 \times (\sqrt{11} + \sqrt{7})}{(\sqrt{11} - \sqrt{7}) (\sqrt{11} + \sqrt{7})}$

= $\frac{4 (\sqrt{11} + \sqrt{7})}{{(\sqrt{11})}^{2} - {(\sqrt{7})}^{2}}$

= $\frac{4 (\sqrt{11} + \sqrt{7})}{11 - 7}$

= $\frac{4 (\sqrt{11} + \sqrt{7})}{4}$

= $(\sqrt{11} + \sqrt{7})$ .

Hence, the denominator is rationalized.

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

Answer

The given expression is $\frac{1 + \sqrt{2}}{2 - \sqrt{2}}$ .

Multiplying by $(2 + \sqrt{2})$ in numerator and denominator both, we get.

$\frac{(1 + \sqrt{2}) \times (2 + \sqrt{2})}{(2 - \sqrt{2}) \times (2 + \sqrt{2})}$

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

= $\frac{4 + 3 \sqrt{2}}{4 - 2}$

= $\frac{4 + 3 \sqrt{2}}{2}$ .

Hence, the denominator is rationalized.

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

Answer

The given expression is $\frac{3 - 2 \sqrt{2}}{3 + 2 \sqrt{2}}$ .

Multiplying by $3 - 2 \sqrt{2}$ in numerator and denominator both, we get.

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

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

= $\frac{3^{2} - 2 \times 3 \times 2 \sqrt{2} + {(2 \sqrt{2})}^{2}}{9 - 8}$

= $\frac{9 - 12 \sqrt{2} + 8}{1}$

= $17 - 12 \sqrt{2}$ .

Hence, the denominator is rationalized.

In Exercise 1F RS Aggarwal Class 9, the third question asks us to figure out the value of an expression when some of the numbers are given.

3. It being given that $\sqrt{2}$ = 1.414, $\sqrt{3}$ = 1.732, $\sqrt{5}$ = 2.236 and $\sqrt{10}$ = 3.162, find the value to three places of decimals, of each of the following.

(i) $\frac{2}{\sqrt{5}}$

Answer

The given expression is $\frac{2}{\sqrt{5}}$ .

Rationalising the denominator by multiplying with $\sqrt{5}$ in the numerator and denominator both, we get.

$\frac{2 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}$ .

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

= $\frac{2 \times 2.236}{5}$ ... ( $∵ \sqrt{5} = 2.236$ ).

= $\frac{4.472}{5}$

= 0.8944 = 0.894 (approx. three decimal places).

(ii) $\frac{2 - \sqrt{3}}{\sqrt{3}}$

Answer

The given expression is $\frac{2 - \sqrt{3}}{\sqrt{3}}$ .

Rationalising the denominator by multiplying by $\sqrt{3}$ in the numerator and denominator both, we get.

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

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

= $\frac{2 \times 1.732 - 3}{3}$ ... ( $∵ \sqrt{3} = 1.732$ )

= $\frac{3.464 - 3}{3}$

= $\frac{0.464}{3}$

= 0.154666...

= 0.155 (approx. to three decimal places).

(iii) $\frac{\sqrt{10} - \sqrt{5}}{\sqrt{2}}$

Answer

The given expression is $\frac{\sqrt{10} - \sqrt{5}}{\sqrt{2}}$ .

Rationalising the denominator by multiplying by $\sqrt{2}$ in both numerator and denominator both, we get.

$\frac{(\sqrt{10} - \sqrt{5}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$

= $\frac{\sqrt{10} \times \sqrt{2} - \sqrt{5} \times \sqrt{2}}{2}$

= $\frac{3.162 \times 1.414 - 2.236 \times 1.414}{2}$

= $\frac{4.471068 - 3.161704}{2}$

= $\frac{1.309364}{2}$

= 0.654682

= 0.655 (approx. to three decimal places).

In Exercise 1F RS Aggarwal Class 9, the fourth question asks us to figure out the values of $a$ and $b$ from a given condition.

4. Find the rational numbers $a$ and $b$ such that

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

Answer

The given equation is $\frac{\sqrt{2} - 1}{\sqrt{2} + 1} = a + b \sqrt{2}$ .
Multiplying numerator and denominator of LHS of the equation by $(\sqrt{2} - 1)$ to rationalize the denominator.
$\Rightarrow \frac{(\sqrt{2} - 1) \times (\sqrt{2} - 1)}{(\sqrt{2} + 1) \times (\sqrt{2} - 1)} = a + b \sqrt{2}$ .
$\Rightarrow \frac{{(\sqrt{2} - 1)}^{2}}{{(\sqrt{2})}^{2} - 1^{2}} = a + b \sqrt{2}$ .
$\Rightarrow \frac{{(\sqrt{2})}^{2} - 2 \times \sqrt{2} \times 1 + 1^{2}}{2 - 1} = a + b \sqrt{2}$ .
$\Rightarrow \frac{2 - 2 \sqrt{2} + 1}{1} = a + b \sqrt{2}$ .
$\Rightarrow 3 - 2 \sqrt{2}$ $= a + b \sqrt{2}$ .
$\Rightarrow 3 + (- 2) \sqrt{2} = a + b \sqrt{2}$ .
On comparing like terms on both sides, we get
$a = 3, b = - 2$ .

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

Answer

The given equation is
$\frac{2 - \sqrt{5}}{2 + \sqrt{5}} = a \sqrt{5} + b$ .
Multiplying numerator and denominator of LHS of the equation by $(2 - \sqrt{5})$ to rationalize the denominator.
$\Rightarrow \frac{(2 - \sqrt{5}) \times (2 - \sqrt{5})}{(2 + \sqrt{5}) \times (2 - \sqrt{5})} = a \sqrt{5} + b$ .
$\Rightarrow \frac{{(2 - \sqrt{5})}^{2}}{2^{2} - {(\sqrt{5})}^{2}} = a \sqrt{5} + b$ .
$\Rightarrow \frac{2^{2} - 2 \times 2 \times \sqrt{5} + {(\sqrt{5})}^{2}}{4 - 5} = a \sqrt{5} + b$ .
$\Rightarrow \frac{4 - 4 \sqrt{5} + 5}{- 1} = a \sqrt{5} + b$ .
$\Rightarrow \frac{9 - 4 \sqrt{5}}{- 1} = a \sqrt{5} + b$ .
$\Rightarrow 4 \sqrt{5} - 9 = a \sqrt{5} + b$ .
$\Rightarrow 4 \sqrt{5} + (- 9) = a \sqrt{5} + b$ .
On comparing like terms on both sides, we get.
$a = 4, b = - 9$ .

(iii) $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = a + b \sqrt{6}$

Answer

The given equation is
$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = a + b \sqrt{6}$ .
Multiplying numerator and denominator of LHS of the equation by $(\sqrt{3} + \sqrt{2})$ to rationalize the denominator.
$\Rightarrow \frac{(\sqrt{3} + \sqrt{2}) \times (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2}) \times (\sqrt{3} + \sqrt{2})} = a + b \sqrt{6}$ .
$\Rightarrow \frac{{(\sqrt{3} + \sqrt{2})}^{2}}{{(\sqrt{3})}^{2} - {(\sqrt{2})}^{2}} = a + b \sqrt{6}$ .
$\Rightarrow \frac{{(\sqrt{3})}^{2} + 2 \times \sqrt{3} \times \sqrt{2} + {(\sqrt{2})}^{2}}{3 - 2} = a + b \sqrt{6}$ .
$\Rightarrow \frac{3 + 2 \sqrt{6} + 2}{1} = a + b \sqrt{6}$ .
$\Rightarrow \frac{5 + 2 \sqrt{6}}{1} = a + b \sqrt{6}$ .
$\Rightarrow 5 + 2 \sqrt{6} = a + b \sqrt{6}$ .
On comparing like terms on both sides, we get.
$a = 5, b = 2$ .

(iv) $\frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a + b \sqrt{3}$

Answer

The given equation is
$\frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a + b \sqrt{3}$ .
Multiplying numerator and denominator of LHS of the equation by $(7 - 4 \sqrt{3})$ to rationalize the denominator.
$\Rightarrow \frac{(5 + 2 \sqrt{3}) \times (7 - 4 \sqrt{3})}{(7 + 4 \sqrt{3}) \times (7 - 4 \sqrt{3})} = a + b \sqrt{3}$ .
$\Rightarrow \frac{5 (7 - 4 \sqrt{3}) + 2 \sqrt{3} (7 - 4 \sqrt{3})}{7^{2} - {(4 \sqrt{3})}^{2}} = a + b \sqrt{3}$ .
$\Rightarrow \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48} = a + b \sqrt{3}$ .
$\Rightarrow \frac{(35 - 24) - \sqrt{3} (20 - 14)}{1} = a + b \sqrt{3}$ .
$\Rightarrow \frac{11 - 6 \sqrt{3}}{1} = a + b \sqrt{3}$ .
$\Rightarrow 11 - 6 \sqrt{3} = a + b \sqrt{3}$ .
$\Rightarrow 11 + (- 6) \sqrt{3} = a + b \sqrt{3}$ .
On comparing the like terms on both sides, we get.
$a = 11, b = - 6$ .

In Exercise 1F RS Aggarwal Class 9, the fifth question asks us to figure out the value of an expression when some of the numbers are given.

5. It being given that $\sqrt{3}$ = 1.732, $\sqrt{5}$ = 2.236, $\sqrt{6}$ = 2.449 and $\sqrt{10}$ = 3.162, find to three places of decimal, the value of each of the following.

(i) $\frac{1}{\sqrt{6} + \sqrt{5}}$

Answer

The given expression is $\frac{1}{\sqrt{6} + \sqrt{5}}$ .
Multiplying by $(\sqrt{6} - \sqrt{5})$ in both numerator and denominator to rationalise it, we get.
$\frac{1 \times (\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5}) \times (\sqrt{6} - \sqrt{5})}$
= $\frac{1 \times (\sqrt{6} - \sqrt{5})}{{(\sqrt{6})}^{2} - {(\sqrt{5})}^{2}}$
= $\frac{1 \times (\sqrt{6} - \sqrt{5})}{6 - 5}$
= $(\sqrt{6} - \sqrt{5})$
= $(2.449 - 2.236)$
= 0.213

(ii) $\frac{6}{\sqrt{5} + \sqrt{3}}$

Answer

The given expression is $\frac{6}{\sqrt{5} + \sqrt{3}}$ .
Multiplying by $(\sqrt{5} - \sqrt{3})$ in both numerator and denominator to rationalise it, we get.
$= \frac{6 \times (\sqrt{5} - \sqrt{3})}{(\sqrt{5} + \sqrt{3}) \times (\sqrt{5} - \sqrt{3})}$
$= \frac{6 \times (\sqrt{5} - \sqrt{3})}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}}$
$= \frac{6 \times (\sqrt{5} - \sqrt{3})}{5 - 3}$
$= \frac{6 \times (\sqrt{5} - \sqrt{3})}{2}$
$= 3 \times (\sqrt{5} - \sqrt{3})$
$= 3 \times (2.236 - 1.732)$
$= 3 \times 0.504$
= $1.512$

(iii) $\frac{1}{4 \sqrt{3} - 3 \sqrt{5}}$

Answer

The given expression is $\frac{1}{4 \sqrt{3} - 3 \sqrt{5}}$ .
Multiplying by $(4 \sqrt{3} + 3 \sqrt{5})$ in both numerator and denominator to rationalise it, we get.
$= \frac{1 \times (4 \sqrt{3} + 3 \sqrt{5})}{(4 \sqrt{3} - 3 \sqrt{5}) \times (4 \sqrt{3} + 3 \sqrt{5})}$
$= \frac{1 \times (4 \sqrt{3} + 3 \sqrt{5})}{{(4 \sqrt{3})}^{2} - {(3 \sqrt{5})}^{2}}$
$= \frac{(4 \sqrt{3} + 3 \sqrt{5})}{48 - 45}$
$= \frac{(4 \sqrt{3} + 3 \sqrt{5})}{3}$
$= \frac{4 \times 1.732 + 3 \times 2.236}{3}$
$= \frac{6.928 + 6.708}{3}$
$= \frac{13.636}{3}$
$= 4.545333 . . .$
= 4.545

(iv) $\frac{3 + \sqrt{5}}{3 - \sqrt{5}}$

Answer

The given expression is $\frac{3 + \sqrt{5}}{3 - \sqrt{5}}$ .
Multiplying by $(3 + \sqrt{5})$ in both numerator and denominator to rationalise it, we get.
$\frac{(3 + \sqrt{5}) \times (3 + \sqrt{5})}{(3 - \sqrt{5}) \times (3 + \sqrt{5})}$
$= \frac{{(3 + \sqrt{5})}^{2}}{{(3)}^{2} - {(\sqrt{5})}^{2}}$
$= \frac{3^{2} + 2 \times 3 \times \sqrt{5} + {(\sqrt{5})}^{2}}{9 - 5}$
$= \frac{9 + 6 \sqrt{5} + 5}{4}$
$= \frac{14 + 6 \sqrt{5}}{4}$
$= \frac{14 + 6 \times 2.236}{4}$
$= \frac{14 + 13.416}{4}$
$= \frac{27.416}{4}$
= 6.854

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

Answer

The given expression is $\frac{1 + 2 \sqrt{3}}{2 - \sqrt{3}}$ .
Multiplying by $(2 + \sqrt{3})$ in both numerator and denominator to rationalise it, we get.
$\frac{(1 + 2 \sqrt{3}) \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})}$
$= \frac{1 (2 + \sqrt{3}) + 2 \sqrt{3} (2 + \sqrt{3})}{{(2)}^{2} - {(\sqrt{3})}^{2}}$
$= \frac{2 + \sqrt{3} + 4 \sqrt{3} + 6}{4 - 3}$
$= \frac{8 + 5 \sqrt{3}}{1}$
$= 8 + 5 \sqrt{3}$
$= 8 + 5 \times 1.732$
$= 8 + 8.660$
= 16.660

(vi) $\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

Answer

The given expression is $\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ .
Multiplying by $(\sqrt{5} + \sqrt{2})$ in both numerator and denominator to rationalise it, we get.
$\frac{(\sqrt{5} + \sqrt{2}) \times (\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2}) \times (\sqrt{5} + \sqrt{2})}$
$= \frac{{(\sqrt{5} + \sqrt{2})}^{2}}{{(\sqrt{5})}^{2} - {(\sqrt{2})}^{2}}$
$= \frac{{(\sqrt{5})}^{2} + 2 \times \sqrt{5} \times \sqrt{2} + {(\sqrt{2})}^{2}}{5 - 2}$
$= \frac{5 + 2 \sqrt{10} + 2}{3}$
$= \frac{7 + 2 \sqrt{10}}{3}$
$= \frac{7 + 2 \times 3.162}{3}$
$= \frac{7 + 6.324}{3}$
$= \frac{13.324}{3}$
= 4.441333...
= 4.441 (approx. to three decimal places).

In Exercise 1F RS Aggarwal Class 9, the sixth question asks us to rationalize the denominator of the given expression.

6. Simplify by rationalizing the denominator.

(i) $\frac{7 \sqrt{3} - 5 \sqrt{2}}{\sqrt{48} + \sqrt{18}}$

Answer

The given expression is $\frac{7 \sqrt{3} - 5 \sqrt{2}}{\sqrt{48} + \sqrt{18}}$ .
$\frac{7 \sqrt{3} - 5 \sqrt{2}}{\sqrt{48} + \sqrt{18}}$
$= \frac{7 \sqrt{3} - 5 \sqrt{2}}{\sqrt{4 \times 4 \times 3} + \sqrt{3 \times 3 \times 2}}$
$= \frac{7 \sqrt{3} - 5 \sqrt{2}}{4 \sqrt{3} + 3 \sqrt{2}}$
Multiplying by $(4 \sqrt{3} - 3 \sqrt{2})$ in both numerator and denominator, we get.
$= \frac{(7 \sqrt{3} - 5 \sqrt{2}) \times (4 \sqrt{3} - 3 \sqrt{2})}{(4 \sqrt{3} + 3 \sqrt{2}) \times (4 \sqrt{3} - 3 \sqrt{2})}$
$= \frac{7 \sqrt{3} (4 \sqrt{3} - 3 \sqrt{2}) - 5 \sqrt{2} (4 \sqrt{3} - 3 \sqrt{2})}{{(4 \sqrt{3})}^{2} - {(3 \sqrt{2})}^{2}}$
$= \frac{84 - 21 \sqrt{6} - 20 \sqrt{6} + 30}{48 - 18}$
= $\frac{114 - 41 \sqrt{6}}{30}$ .

(ii) $\frac{2 \sqrt{6} - \sqrt{5}}{3 \sqrt{5} - 2 \sqrt{6}}$

Answer

The given expression is $\frac{2 \sqrt{6} - \sqrt{5}}{3 \sqrt{5} - 2 \sqrt{6}}$ .
$\frac{2 \sqrt{6} - \sqrt{5}}{3 \sqrt{5} - 2 \sqrt{6}}$
Multiplying by $(3 \sqrt{5} + 2 \sqrt{6})$ in numerator and denominator, we get.
$= \frac{(2 \sqrt{6} - \sqrt{5}) \times (3 \sqrt{5} + 2 \sqrt{6})}{(3 \sqrt{5} - 2 \sqrt{6}) \times (3 \sqrt{5} + 2 \sqrt{6})}$
$= \frac{2 \sqrt{6} (3 \sqrt{5} + 2 \sqrt{6}) - \sqrt{5} (3 \sqrt{5} + 2 \sqrt{6})}{{(3 \sqrt{5})}^{2} - {(2 \sqrt{6})}^{2}}$
$= \frac{6 \sqrt{30} + 24 - 15 - 2 \sqrt{30}}{45 - 24}$
= $\frac{4 \sqrt{30} + 9}{21}$

In Exercise 1F RS Aggarwal Class 9, the seventh question asks us to simplify the given expression.

7. Simplify.

(i) $\frac{4 + \sqrt{5}}{4 - \sqrt{5}} + \frac{4 - \sqrt{5}}{4 + \sqrt{5}}$

Answer

We have
$\frac{4 + \sqrt{5}}{4 - \sqrt{5}} + \frac{4 - \sqrt{5}}{4 + \sqrt{5}}$
Solving by taking LCM of denominators, we get.
$= \frac{(4 + \sqrt{5}) (4 + \sqrt{5}) + (4 - \sqrt{5}) (4 - \sqrt{5})}{(4 - \sqrt{5}) (4 + \sqrt{5})}$
$= \frac{{(4 + \sqrt{5})}^{2} + {(4 - \sqrt{5})}^{2}}{4^{2} - {(\sqrt{5})}^{2}}$
$= \frac{4^{2} + 2 \times 4 \times \sqrt{5} + {(\sqrt{5})}^{2} + 4^{2} - 2 \times 4 \times \sqrt{5} + {(\sqrt{5})}^{2}}{14 - 5}$
$= \frac{16 + 8 \sqrt{5} + 5 + 16 - 8 \sqrt{5} + 5}{11}$
$= \frac{16 + 8 \sqrt{5} + 5 + 16 - 8 \sqrt{5} + 5}{11}$
= $\frac{42}{11}$ .

(ii) $\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{2}{\sqrt{5} - \sqrt{3}} + \frac{3}{\sqrt{5} - \sqrt{2}}$

Answer

The given expression is
$\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{2}{\sqrt{5} - \sqrt{3}} + \frac{3}{\sqrt{5} - \sqrt{2}}$
Multiplying by $(\sqrt{3} + \sqrt{2})$ , $(\sqrt{5} + \sqrt{3})$ and $(\sqrt{5} + \sqrt{2})$ in both numerators and denominators of first term, second term and third term respectively, we get.
$= \frac{1 \times (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2}) (\sqrt{3} + \sqrt{2})} - \frac{2 \times (\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3}) (\sqrt{5} + \sqrt{3})} + \frac{3 \times (\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2})}$
$= \frac{(\sqrt{3} + \sqrt{2})}{{(\sqrt{3})}^{2} - {(\sqrt{2})}^{2}} - \frac{2 (\sqrt{5} + \sqrt{3})}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}} + \frac{3 (\sqrt{5} + \sqrt{2})}{{(\sqrt{5})}^{2} - {(\sqrt{2})}^{2}}$
$= \frac{(\sqrt{3} + \sqrt{2})}{3 - 2} - \frac{2 (\sqrt{5} + \sqrt{3})}{5 - 3} + \frac{3 (\sqrt{5} + \sqrt{2})}{5 - 2}$
$= \frac{(\sqrt{3} + \sqrt{2})}{1} - \frac{2 (\sqrt{5} + \sqrt{3})}{2} + \frac{3 (\sqrt{5} + \sqrt{2})}{3}$
$= (\sqrt{3} + \sqrt{2}) - (\sqrt{5} + \sqrt{3}) + (\sqrt{5} + \sqrt{2})$
$= \sqrt{3} + \sqrt{2} - \sqrt{5} - \sqrt{3} + \sqrt{5} + \sqrt{2}$
$= \sqrt{3} + \sqrt{2} - \sqrt{5} - \sqrt{3} + \sqrt{5} + \sqrt{2}$
= $2 \sqrt{2}$ .

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

Answer

The given expression is $\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .
Solving first two terms by taking LCM of
their denominators and the third term by
multiplying it with conjugate of its denominator.
$= \frac{(2 + \sqrt{3}) (2 + \sqrt{3}) + (2 - \sqrt{3}) (2 - \sqrt{3})}{(2 + \sqrt{3}) (2 - \sqrt{3})} + \frac{(\sqrt{3} - 1) (\sqrt{3} - 1)}{(\sqrt{3} + 1) (\sqrt{3} - 1)}$
$= \frac{{(2 + \sqrt{3})}^{2} + {(2 - \sqrt{3})}^{2}}{2^{2} - {(\sqrt{3})}^{2}} + \frac{{(\sqrt{3} - 1)}^{2}}{{(\sqrt{3})}^{2} - 1^{2}}$
$= \frac{2^{2} + {(\sqrt{3})}^{2} + 2 \times 2 \times \sqrt{3} + 2^{2} - 2 \times 2 \times \sqrt{3} + {(\sqrt{3})}^{2}}{4 - 3} + \frac{{(\sqrt{3})}^{2} - 2 \times \sqrt{3} \times 1 + 1^{2}}{3 - 1}$
$= \frac{4 + 3 + 4 \sqrt{3} + 4 - 4 \sqrt{3} + 3}{1} + \frac{3 - 2 \sqrt{3} + 1}{2}$
$= \frac{4 + 3 + 4 \sqrt{3} + 4 - 4 \sqrt{3} + 3}{1} + \frac{4 - 2 \sqrt{3}}{2}$
$= 14 + \frac{2 (2 - \sqrt{3})}{2}$
$= 14 + \frac{2 (2 - \sqrt{3})}{2}$
$= 14 + 2 - \sqrt{3}$
= $16 - \sqrt{3}$ .

(iv) $\frac{2 \sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{6 \sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{8 \sqrt{3}}{\sqrt{6} + \sqrt{2}}$

Answer

The given expression is $\frac{2 \sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{6 \sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{8 \sqrt{3}}{\sqrt{6} + \sqrt{2}}$ .
Multiplying by $(\sqrt{2} - \sqrt{3})$ , $(\sqrt{6} - \sqrt{3})$ and $(\sqrt{6} - \sqrt{2})$
in both numerators and denominators of first term,
second term and third term respectively, we get.
$\frac{2 \sqrt{6} \times (\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3}) (\sqrt{2} - \sqrt{3})} + \frac{6 \sqrt{2} \times (\sqrt{6} - \sqrt{3})}{(\sqrt{6} + \sqrt{3}) (\sqrt{6} - \sqrt{3})} - \frac{8 \sqrt{3} \times (\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2}) (\sqrt{6} - \sqrt{2})}$
$= \frac{2 \sqrt{6} \times (\sqrt{2} - \sqrt{3})}{{(\sqrt{2})}^{2} - {(\sqrt{3})}^{2}} + \frac{6 \sqrt{2} \times (\sqrt{6} - \sqrt{3})}{{(\sqrt{6})}^{2} - {(\sqrt{3})}^{2}} - \frac{8 \sqrt{3} \times (\sqrt{6} - \sqrt{2})}{{(\sqrt{6})}^{2} - {(\sqrt{2})}^{2}}$
$= \frac{2 \sqrt{6} \times (\sqrt{2} - \sqrt{3})}{2 - 3} + \frac{6 \sqrt{2} \times (\sqrt{6} - \sqrt{3})}{6 - 3} - \frac{8 \sqrt{3} \times (\sqrt{6} - \sqrt{2})}{6 - 2}$
$= \frac{2 \sqrt{6} \times (\sqrt{2} - \sqrt{3})}{- 1} + \frac{6 \sqrt{2} \times (\sqrt{6} - \sqrt{3})}{3} - \frac{8 \sqrt{3} \times (\sqrt{6} - \sqrt{2})}{4}$
$= - 2 \sqrt{6} \times (\sqrt{2} - \sqrt{3})$ + $2 \sqrt{2} \times (\sqrt{6} - \sqrt{3})$ − $2 \sqrt{3} \times (\sqrt{6} - \sqrt{2})$
$= \frac{2 \sqrt{6} \times (\sqrt{2} - \sqrt{3})}{- 1} + \frac{6 \sqrt{2} \times (\sqrt{6} - \sqrt{3})}{3} - \frac{8 \sqrt{3} \times (\sqrt{6} - \sqrt{2})}{4}$
$= - 2 \sqrt{12} + 2 \sqrt{18} + 2 \sqrt{12} - 2 \sqrt{6} - 2 \sqrt{18} + 2 \sqrt{6}$
$= - 2 \sqrt{12} + 2 \sqrt{18} + 2 \sqrt{12} - 2 \sqrt{6} - 2 \sqrt{18} + 2 \sqrt{6}$
= $0$

In Exercise 1F RS Aggarwal Class 9, the eighth question asks us to prove the given equation.

8. Prove that.

(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$

Answer

The given equality is
$\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$ .
We will take the expression in the left hand side (LHS) and simplify to get the value equal to the value in the right hand side.
Taking LHS, we get.
$\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1}$
$= \frac{1 \times (3 - \sqrt{7})}{(3 + \sqrt{7}) \times (3 - \sqrt{7})} + \frac{1 \times (\sqrt{7} - \sqrt{5})}{(\sqrt{7} + \sqrt{5}) \times (\sqrt{7} - \sqrt{5})} + \frac{1 \times (\sqrt{5} - \sqrt{3})}{(\sqrt{5} + \sqrt{3}) \times (\sqrt{5} - \sqrt{3})} + \frac{1 \times (\sqrt{3} - 1)}{(\sqrt{3} + 1) (\sqrt{3} - 1)}$
$= \frac{(3 - \sqrt{7})}{3^{2} - {(\sqrt{7})}^{2}} + \frac{(\sqrt{7} - \sqrt{5})}{{(\sqrt{7})}^{2} - {(\sqrt{5})}^{2}} + \frac{(\sqrt{5} - \sqrt{3})}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}} + \frac{(\sqrt{3} - 1)}{{(\sqrt{3})}^{2} - 1^{2}}$
$= \frac{(3 - \sqrt{7})}{9 - 7} + \frac{(\sqrt{7} - \sqrt{5})}{7 - 5} + \frac{(\sqrt{5} - \sqrt{3})}{5 - 3} + \frac{(\sqrt{3} - 1)}{3 - 1}$
$= \frac{(3 - \sqrt{7})}{2} + \frac{(\sqrt{7} - \sqrt{5})}{2} + \frac{(\sqrt{5} - \sqrt{3})}{2} + \frac{(\sqrt{3} - 1)}{2}$
$= \frac{3 - \sqrt{7} + \sqrt{7} - \sqrt{5} + \sqrt{5} - \sqrt{3} + \sqrt{3} - 1}{2}$
$= \frac{3 - \sqrt{7} + \sqrt{7} - \sqrt{5} + \sqrt{5} - \sqrt{3} + \sqrt{3} - 1}{2}$
$= \frac{3 - 1}{2}$
$= \frac{2}{2} = 1 = R H S$
Hence, proved!

(ii) $\frac{1}{1 + \sqrt{2}}$ + $\frac{1}{\sqrt{2} + \sqrt{3}}$ + $\frac{1}{\sqrt{3} + \sqrt{4}}$ + $\frac{1}{\sqrt{4} + \sqrt{5}}$ + $\frac{1}{\sqrt{5} + \sqrt{6}}$ + $\frac{1}{\sqrt{6} + \sqrt{7}}$ + $\frac{1}{\sqrt{7} + \sqrt{8}}$ + $\frac{1}{\sqrt{8} + \sqrt{9}}$ = 2

Answer

The given equality is
$\frac{1}{1 + \sqrt{2}}$ + $\frac{1}{\sqrt{2} + \sqrt{3}}$ + $\frac{1}{\sqrt{3} + \sqrt{4}}$ + $\frac{1}{\sqrt{4} + \sqrt{5}}$ + $\frac{1}{\sqrt{5} + \sqrt{6}}$ + $\frac{1}{\sqrt{6} + \sqrt{7}}$ + $\frac{1}{\sqrt{7} + \sqrt{8}}$ + $\frac{1}{\sqrt{8} + \sqrt{9}}$ = 2
$= \frac{1 \times (1 - \sqrt{2})}{(1 + \sqrt{2}) \times (1 - \sqrt{2})} + \frac{1 \times (\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3}) \times (\sqrt{2} - \sqrt{3})} + \frac{1 \times (\sqrt{3} - \sqrt{4})}{(\sqrt{3} + \sqrt{4}) \times (\sqrt{3} - \sqrt{4})} + \frac{1 \times (\sqrt{4} - \sqrt{5})}{(\sqrt{4} + \sqrt{5}) \times (\sqrt{4} - \sqrt{5})} + \frac{1 \times (\sqrt{5} - \sqrt{6})}{(\sqrt{5} + \sqrt{6}) \times (\sqrt{5} - \sqrt{6})} + \frac{1 \times (\sqrt{6} - \sqrt{7})}{(\sqrt{6} + \sqrt{7}) \times (\sqrt{6} - \sqrt{7})} + \frac{1 \times (\sqrt{7} - \sqrt{8})}{(\sqrt{7} + \sqrt{8}) \times (\sqrt{7} - \sqrt{8})} + \frac{1 \times (\sqrt{8} - \sqrt{9})}{(\sqrt{8} + \sqrt{9}) \times (\sqrt{8} - \sqrt{9})}$
$= \frac{(1 - \sqrt{2})}{1^{2} - {(\sqrt{2})}^{2}} + \frac{(\sqrt{2} - \sqrt{3})}{{(\sqrt{2})}^{2} - {(\sqrt{3})}^{2}} + \frac{(\sqrt{3} - \sqrt{4})}{{(\sqrt{3})}^{2} - {(\sqrt{4})}^{2}} + \frac{(\sqrt{4} - \sqrt{5})}{{(\sqrt{4})}^{2} - {(\sqrt{5})}^{2}} + \frac{(\sqrt{5} - \sqrt{6})}{{(\sqrt{5})}^{2} - {(\sqrt{6})}^{2}} + \frac{(\sqrt{6} - \sqrt{7})}{{(\sqrt{6})}^{2} - {(\sqrt{7})}^{2}} + \frac{(\sqrt{7} - \sqrt{8})}{{(\sqrt{7})}^{2} - {(\sqrt{8})}^{2}} + \frac{(\sqrt{8} - \sqrt{9})}{{(\sqrt{8})}^{2} - {(\sqrt{9})}^{2}}$
$= \frac{(1 - \sqrt{2})}{1 - 2} + \frac{(\sqrt{2} - \sqrt{3})}{2 - 3} + \frac{(\sqrt{3} - \sqrt{4})}{3 - 4} + \frac{(\sqrt{4} - \sqrt{5})}{4 - 5} + \frac{(\sqrt{5} - \sqrt{6})}{5 - 6} + \frac{(\sqrt{6} - \sqrt{7})}{6 - 7} + \frac{(\sqrt{7} - \sqrt{8})}{7 - 8} + \frac{(\sqrt{8} - \sqrt{9})}{8 - 9}$
$= \frac{(1 - \sqrt{2})}{- 1} + \frac{(\sqrt{2} - \sqrt{3})}{- 1} + \frac{(\sqrt{3} - \sqrt{4})}{- 1} + \frac{(\sqrt{4} - \sqrt{5})}{- 1} + \frac{(\sqrt{5} - \sqrt{6})}{- 1} + \frac{(\sqrt{6} - \sqrt{7})}{- 1} + \frac{(\sqrt{7} - \sqrt{8})}{- 1} + \frac{(\sqrt{8} - \sqrt{9})}{- 1}$
$= - (1 - \sqrt{2}) - (\sqrt{2} - \sqrt{3}) - (\sqrt{3} - \sqrt{4}) - (\sqrt{4} - \sqrt{5}) - (\sqrt{5} - \sqrt{6}) - (\sqrt{6} - \sqrt{7}) - (\sqrt{7} - \sqrt{8}) - (\sqrt{8} - \sqrt{9})$
$= - 1 + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{4} - \sqrt{4} + \sqrt{5} - \sqrt{5} + \sqrt{6} - \sqrt{6} + \sqrt{7} - \sqrt{7} + \sqrt{8} - \sqrt{8} + \sqrt{9}$
$= - 1 + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{4} - \sqrt{4} + \sqrt{5} - \sqrt{5} + \sqrt{6} - \sqrt{6} + \sqrt{7} - \sqrt{7} + \sqrt{8} - \sqrt{8} + \sqrt{9}$
$= - 1 + 3$
= 2 = RHS.
Hence, proved!

In Exercise 1F RS Aggarwal Class 9, the ninth question asks us to find the value of $a$ and $b$ from the given equation.

9. Find the values of $a$ and $b$ if
$\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}} = a + b \sqrt{5}$ .

Answer

We have
$\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}} = a + b \sqrt{5}$ .
Simplifying left hand side by taking
LCM of the denominators
$\frac{(7 + 3 \sqrt{5}) (3 - \sqrt{5}) - (7 - 3 \sqrt{5}) (3 + \sqrt{5})}{(3 + \sqrt{5}) (3 - \sqrt{5})} = a + b \sqrt{5}$
$\Rightarrow \frac{[21 - 7 \sqrt{5} + 9 \sqrt{5} - 15] - [21 + 7 \sqrt{5} - 9 \sqrt{5} - 15]}{9 - 5} = a + b \sqrt{5}$
$\Rightarrow \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5} + 9 \sqrt{5} + 15}{9 - 5} = a + b \sqrt{5}$
$\Rightarrow \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5} + 9 \sqrt{5} + 15}{4} = a + b \sqrt{5}$
$\Rightarrow \frac{- 7 \sqrt{5} + 9 \sqrt{5} - 7 \sqrt{5} + 9 \sqrt{5}}{4} = a + b \sqrt{5}$
$\Rightarrow \frac{4 \sqrt{5}}{4} = a + b \sqrt{5}$
$\Rightarrow \frac{4 \sqrt{5}}{4} = a + b \sqrt{5}$
$\Rightarrow \sqrt{5} = a + b \sqrt{5}$
$\Rightarrow 0 + 1 \sqrt{5} = a + b \sqrt{5}$
Comparing like terms on both sides, we get.
$a = 0, b = 1$

In Exercise 1F RS Aggarwal Class 9, the tenth question asks us to simplify an expression.

10. Simplify $\frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}} + \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ .

Answer

We have
$\frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}} + \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ .
Simplifying by taking the LCM of denominators
$\frac{(\sqrt{13} - \sqrt{11}) \times (\sqrt{13} - \sqrt{11}) + (\sqrt{13} + \sqrt{11}) \times (\sqrt{13} + \sqrt{11})}{(\sqrt{13} + \sqrt{11}) (\sqrt{13} - \sqrt{11})}$
$= \frac{{(\sqrt{13} - \sqrt{11})}^{2} + {(\sqrt{13} + \sqrt{11})}^{2}}{{(\sqrt{13})}^{2} - {(\sqrt{11})}^{2}}$
$= \frac{{(\sqrt{13})}^{2} - 2 \times \sqrt{13} \times \sqrt{11} + {(\sqrt{11})}^{2} + {(\sqrt{13})}^{2} + 2 \times \sqrt{13} \times \sqrt{11} + {(\sqrt{11})}^{2}}{13 - 11}$
$= \frac{13 - 2 \sqrt{143} + 11 + 13 + 2 \sqrt{143} + 11}{13 - 11}$
$= \frac{13 - 2 \sqrt{143} + 11 + 13 + 2 \sqrt{143} + 11}{13 - 11}$
$= \frac{48}{2}$
= 24

In Exercise 1F RS Aggarwal Class 9, the tenth question is about rational number verification.

11. If $x = 3 + 2 \sqrt{2}$ , check whether $x + \frac{1}{x}$ is rational or irrational.

Answer

It is given that
$x = 3 + 2 \sqrt{2}$ .
$∴ x + \frac{1}{x}$
$= (3 + 2 \sqrt{2}) + \frac{1}{(3 + 2 \sqrt{2})}$
Rationalising the denominator of the second term, we get
$= (3 + 2 \sqrt{2}) + \frac{1 \times (3 - 2 \sqrt{2})}{(3 + 2 \sqrt{2}) (3 - 2 \sqrt{2})}$
$= (3 + 2 \sqrt{2}) + \frac{(3 - 2 \sqrt{2})}{3^{2} - {(2 \sqrt{2})}^{2}}$
$= (3 + 2 \sqrt{2}) + \frac{(3 - 2 \sqrt{2})}{9 - 8}$
$= (3 + 2 \sqrt{2}) + \frac{(3 - 2 \sqrt{2})}{1}$
$= (3 + 2 \sqrt{2}) + (3 - 2 \sqrt{2})$
$= 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2}$
$= 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2}$
= 9.
Hence, $x + \frac{1}{x}$ is a rational number.

In Exercise 1F RS Aggarwal Class 9, the 11th question asks us to find the value of an expression.

12. If $x = 2 - \sqrt{3}$ , find the value of ${(x - \frac{1}{x})}^{3}$

Answer

It is given that
$x = 2 - \sqrt{3}$
$∴ (x - \frac{1}{x})^{3}$
Putting the value of $x$ , we get.
$= {[(2 - \sqrt{3}) - \frac{1}{(2 - \sqrt{3})}]}^{3}$
Rationalising the denominator of the second expression, we get.
$= {[(2 - \sqrt{3}) - \frac{1 \times (2 + \sqrt{3})}{(2 - \sqrt{3}) (2 + \sqrt{3})}]}^{3}$
$= {[(2 - \sqrt{3}) - \frac{(2 + \sqrt{3})}{2^{2} - {(\sqrt{3})}^{2}}]}^{3}$
$= {[(2 - \sqrt{3}) - \frac{(2 + \sqrt{3})}{4 - 3}]}^{3}$
$= {[(2 - \sqrt{3}) - \frac{(2 + \sqrt{3})}{1}]}^{3}$
$= {[(2 - \sqrt{3}) - (2 + \sqrt{3})]}^{3}$
$= {[2 - \sqrt{3} - 2 - \sqrt{3}]}^{3}$
$= {[2 - \sqrt{3} - 2 - \sqrt{3}]}^{3}$
$= {[- 2 \sqrt{3}]}^{3}$
= $- 24 \sqrt{3}$

In Exercise 1F RS Aggarwal Class 9, the 13th question asks us to find the value of an expression.

13. If $x = 9 - 4 \sqrt{5}$ , find the value of $x^{2} + \frac{1}{x^{2}}$ .

Answer

It is given that
$x = 9 - 4 \sqrt{5}$ .
$∴ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5}}$
= $\frac{1 \times (9 + 4 \sqrt{5})}{(9 - 4 \sqrt{5}) (9 + 4 \sqrt{5})}$
= $\frac{(9 + 4 \sqrt{5})}{9^{2} - {(4 \sqrt{5})}^{2}}$
= $\frac{(9 + 4 \sqrt{5})}{81 - 80}$
= $9 + 4 \sqrt{5}$
We will use the following identity to calculate the value of $x^{2} + \frac{1}{x^{2}}$ .
$∵ {(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = {(x + \frac{1}{x})}^{2} - 2$ ... (i)
Putting the value of $x$ and $\frac{1}{x}$ in equation (i), we get.
$x^{2} + \frac{1}{x^{2}} = {(9 + 4 \sqrt{5} + 9 - 4 \sqrt{5})}^{2} - 2$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = {(9 + 4 \sqrt{5} + 9 - 4 \sqrt{5})}^{2} - 2$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 18^{2} - 2$
$\Rightarrow x^{2} + \frac{1}{x^{2}} = 324 - 2$
$\Rightarrow$ $x^{2} + \frac{1}{x^{2}} = 322$ .

In Exercise 1F RS Aggarwal Class 9, the 14th question asks us to find the value of an expression.

14. If $x = \frac{5 - \sqrt{21}}{2}$ , find the value of $x + \frac{1}{x}$ .

Answer

Multiplying numerator and denominator of LHS of the equation by $(\sqrt{2} - 1)$ to rationalize the denominator.
It is given that
$x = \frac{5 - \sqrt{21}}{2}$ .
$∴ \frac{1}{x} = \frac{2}{5 - \sqrt{21}}$
$= \frac{2 \times (5 + \sqrt{21})}{(5 - \sqrt{21}) \times (5 + \sqrt{21})}$
$= \frac{2 \times (5 + \sqrt{21})}{5^{2} - {(\sqrt{21})}^{2}}$
$= \frac{2 \times (5 + \sqrt{21})}{25 - 21}$
$= \frac{2 \times (5 + \sqrt{21})}{4}$
$= \frac{5 + \sqrt{21}}{2}$
Now, we have
$x + \frac{1}{x}$
$= \frac{5 - \sqrt{21}}{2} + \frac{5 + \sqrt{21}}{2}$
$= \frac{5 - \sqrt{21} + 5 + \sqrt{21}}{2}$
$= \frac{10}{2}$
= $5$

In Exercise 1F RS Aggarwal Class 9, the 15th question asks us to find the value of an expression.

15. If $a = 3 - 2 \sqrt{2}$ , find the value of $a^{2} - \frac{1}{a^{2}}$ .

Answer

We have
$a = 3 - 2 \sqrt{2}$ .
$∴ \frac{1}{a} = \frac{1}{3 - 2 \sqrt{2}}$
$= \frac{1 \times (3 + 2 \sqrt{2})}{(3 - 2 \sqrt{2}) \times (3 + 2 \sqrt{2})}$
$= \frac{(3 + 2 \sqrt{2})}{3^{2} - {(2 \sqrt{2})}^{2}}$
$= \frac{(3 + 2 \sqrt{2})}{9 - 8}$
$= \frac{(3 + 2 \sqrt{2})}{1}$
$= (3 + 2 \sqrt{2})$
Now, we use the following identity.
$a^{2} - \frac{1}{a^{2}} = (a + \frac{1}{a}) (a - \frac{1}{a})$
$\Rightarrow a^{2} - \frac{1}{a^{2}} = (3 - 2 \sqrt{2} + 3 + 2 \sqrt{2}) (3 - 2 \sqrt{2} - 3 - 2 \sqrt{2})$
$\Rightarrow a^{2} - \frac{1}{a^{2}} = (6) (- 4 \sqrt{2})$
$\Rightarrow a^{2} - \frac{1}{a^{2}} = - 24 \sqrt{2}$

In Exercise 1F RS Aggarwal Class 9, the 16th question asks us to find the value of an expression.

16. If $x = \sqrt{13} + 2 \sqrt{3}$ , find the value of $x - \frac{1}{x}$ .

Answer

We have
$x = \sqrt{13} + 2 \sqrt{3}$
$∴ \frac{1}{x} = \frac{1}{\sqrt{13} + 2 \sqrt{3}}$
$= \frac{6 \times (\sqrt{5} - \sqrt{3})}{(\sqrt{5} + \sqrt{3}) \times (\sqrt{5} - \sqrt{3})}$
$= \frac{1 \times (\sqrt{13} - 2 \sqrt{3})}{(\sqrt{13} + 2 \sqrt{3}) (\sqrt{13} - 2 \sqrt{3})}$
$= \frac{(\sqrt{13} - 2 \sqrt{3})}{{(\sqrt{13})}^{2} - {(2 \sqrt{3})}^{2}}$
$= \frac{(\sqrt{13} - 2 \sqrt{3})}{13 - 12}$
$= (\sqrt{13} - 2 \sqrt{3})$ .
Now, we have
$x - \frac{1}{x} = (\sqrt{13} + 2 \sqrt{3}) - (\sqrt{13} - 2 \sqrt{3})$
$\Rightarrow x - \frac{1}{x} = \sqrt{13} + 2 \sqrt{3} - \sqrt{13} + 2 \sqrt{3}$
$\Rightarrow x - \frac{1}{x} = 4 \sqrt{3}$ .

In Exercise 1F RS Aggarwal Class 9, the 17th question asks us to find the value of an expression.

17. If $x = 2 + \sqrt{3}$ , find the value of $x^{3} + \frac{1}{x^{3}}$

Answer

Multiplying numerator and denominator of LHS of the equation by $(\sqrt{2} - 1)$ to rationalize the denominator.
We have
$x = 2 + \sqrt{3}$
$∴ \frac{1}{x} = \frac{1}{2 + \sqrt{3}}$
$= \frac{1 \times (2 - \sqrt{3})}{(2 + \sqrt{3}) (2 - \sqrt{3})}$
$= \frac{2 - \sqrt{3}}{2^{2} - {(\sqrt{3})}^{2}}$
$= \frac{2 - \sqrt{3}}{4 - 3}$
$= \frac{2 - \sqrt{3}}{1}$
= $2 - \sqrt{3}$
$∴ x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4$ ... (1)
We use the following identity to find the value of $x^{3} + \frac{1}{x^{3}}$ .
${(x + \frac{1}{x})}^{3} = x^{3} + \frac{1}{x^{3}} + 3 \times (x + \frac{1}{x})$
$\Rightarrow x^{3} + \frac{1}{x^{3}} = {(x + \frac{1}{x})}^{3} - 3 \times (x + \frac{1}{x})$
$\Rightarrow x^{3} + \frac{1}{x^{3}} = {(4)}^{3} - 3 \times (4)$ ... (from 1)
$\Rightarrow x^{3} + \frac{1}{x^{3}} = 64 - 12$
$\Rightarrow x^{3} + \frac{1}{x^{3}} = 52$ .

In Exercise 1F RS Aggarwal Class 9, the 18th question asks us to show an equality.

18. If $x = \frac{5 - \sqrt{3}}{5 + \sqrt{3}}$ and $y = \frac{5 + \sqrt{3}}{5 - \sqrt{3}}$ , show that $x - y = - \frac{10 \sqrt{3}}{11}$ .

Answer

We have
$x = \frac{5 - \sqrt{3}}{5 + \sqrt{3}}$ , $y = \frac{5 + \sqrt{3}}{5 - \sqrt{3}}$ .
$x - y = \frac{5 - \sqrt{3}}{5 + \sqrt{3}} - \frac{5 + \sqrt{3}}{5 - \sqrt{3}}$
$\Rightarrow x - y = \frac{{(5 - \sqrt{3})}^{2} - {(5 + \sqrt{3})}^{2}}{(5 + \sqrt{3}) (5 - \sqrt{3})}$
$\Rightarrow x - y = \frac{5^{2} - 2 \times 5 \times \sqrt{3} + {(\sqrt{3})}^{2} - 5^{2} - 2 \times 5 \times \sqrt{3} - {(\sqrt{3})}^{2}}{5^{2} - {(\sqrt{3})}^{2}}$
$\Rightarrow x - y = \frac{25 - 10 \sqrt{3} + 3 - 25 - 10 \sqrt{3} - 3}{25 - 3}$
$\Rightarrow x - y = \frac{- 20 \sqrt{3}}{22}$
$\Rightarrow x - y = \frac{- 10 \sqrt{3}}{11}$ .

In Exercise 1F RS Aggarwal Class 9, the 19th question asks us to show an equality.

19. If $a = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $b = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$ , show that $3 a^{2} + 4 a b - 3 b^{2} = 4 + \frac{56}{3} \sqrt{10}$

Answer

We have
$a = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ , $b = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$ .
$∴ 3 a^{2} + 4 a b - 3 b^{2}$
$= 3 a^{2} - 3 b^{2} + 4 a b$
$= 3 (a^{2} - b^{2}) + 4 a b$
$= 3 (a + b) (a - b) + 4 a b$
$= 3 (\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} + \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}) (\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} - \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}) + 4 (\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}) (\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}})$
$= 3 [\frac{{(\sqrt{5} + \sqrt{2})}^{2} + {(\sqrt{5} - \sqrt{2})}^{2}}{(\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2})}] [\frac{{(\sqrt{5} + \sqrt{2})}^{2} - {(\sqrt{5} - \sqrt{2})}^{2}}{(\sqrt{5} - \sqrt{2}) (\sqrt{5} + \sqrt{2})}] + 4 \times 1$
$= 3 [\frac{(5 + 2 + 2 \sqrt{10}) + (5 + 2 - 2 \sqrt{10})}{5 - 2}] [\frac{(5 + 2 + 2 \sqrt{10}) - (5 + 2 - 2 \sqrt{10})}{5 - 2}] + 4$
$= 3 [\frac{5 + 2 + 5 + 2}{3}] [\frac{2 \sqrt{10} + 2 \sqrt{10}}{3}] + 4$
$= 3 [\frac{14}{3}] [\frac{4 \sqrt{10}}{3}] + 4$
$= 14 \times [\frac{4 \sqrt{10}}{3}] + 4$
$= \frac{56 \sqrt{10}}{3} + 4$ .
Hence, proved.

In Exercise 1F RS Aggarwal Class 9, the 20th question asks us to show an equality.

20. If $a = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ and $b = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ , find the value of $a^{2} + b^{2} - 5 a b$ .

Answer

We have
$a = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ , $b = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$
$∴ a^{2} + b^{2} - 5 a b$
$= {(a + b)}^{2} - 2 a b - 5 a b$
$= {(a + b)}^{2} - 7 a b$
$= {(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} + \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}})}^{2} - 7 \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$
$= {[\frac{{(\sqrt{3} - \sqrt{2})}^{2} + {(\sqrt{3} + \sqrt{2})}^{2}}{(\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2})}]}^{2} - 7 \times 1$
$= {[\frac{{(\sqrt{3})}^{2} + {(\sqrt{2})}^{2} - 2 \times \sqrt{3} \times \sqrt{2} + {(\sqrt{3})}^{2} + {(\sqrt{2})}^{2} + 2 \times \sqrt{3} \times \sqrt{2}}{{(\sqrt{3})}^{2} - {(\sqrt{2})}^{2}}]}^{2} - 7 \times 1$
$= {[\frac{3 + 2 - 2 \sqrt{6} + 3 + 2 + 2 \sqrt{6}}{3 - 2}]}^{2} - 7$
$= {[\frac{3 + 2 + 3 + 2}{1}]}^{2} - 7$
$= 10^{2} - 7$
$= 100 - 7$
= $93$ .

In Exercise 1F RS Aggarwal Class 9, the 21st question asks us to find the value of an expression.

21. If $p = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}$ and $q = \frac{3 + \sqrt{5}}{3 - \sqrt{5}}$ , find the value of $p^{2} + q^{2}$ .

Answer

We have
$p = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}$ , $q = \frac{3 + \sqrt{5}}{3 - \sqrt{5}}$ .
$∵ p^{2} + q^{2} = {(p + q)}^{2} - 2 p q$
$= {(\frac{3 - \sqrt{5}}{3 + \sqrt{5}} + \frac{3 + \sqrt{5}}{3 - \sqrt{5}})}^{2} - 2 \times \frac{3 - \sqrt{5}}{3 + \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 - \sqrt{5}}$
$= {[\frac{{(3 - \sqrt{5})}^{2} + {(3 + \sqrt{5})}^{2}}{(3 + \sqrt{5}) (3 - \sqrt{5})}]}^{2} - 2 \times 1$
$= {[\frac{3^{2} + {(\sqrt{5})}^{2} - 2 \times 3 \times \sqrt{5} + 3^{2} + {(\sqrt{5})}^{2} + 2 \times 3 \times \sqrt{5}}{3^{2} - {(\sqrt{5})}^{2}}]}^{2} - 2$
$= {[\frac{9 + 5 - 6 \sqrt{5} + 9 + 5 + 6 \sqrt{5}}{9 - 5}]}^{2} - 2$
$= {[\frac{9 + 5 + 9 + 5}{4}]}^{2} - 2$
$= {[\frac{28}{4}]}^{2} - 2$
$= 7^{2} - 2$
$= 49 - 2$
= $47$

In Exercise 1F RS Aggarwal Class 9, the 22nd question asks us to rationalize the given expression.

22. Rationalise the denominator of each of the following.

(i) $\frac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13}}$

Answer

We have
$\frac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13}}$
$= \frac{1 \times [(\sqrt{7} + \sqrt{6}) + \sqrt{13}]}{[(\sqrt{7} + \sqrt{6}) - \sqrt{13}] \times [(\sqrt{7} + \sqrt{6}) + \sqrt{13}]}$
$= \frac{[(\sqrt{7} + \sqrt{6}) + \sqrt{13}]}{[{(\sqrt{7} + \sqrt{6})}^{2} - {(\sqrt{13})}^{2}]}$
$= \frac{[(\sqrt{7} + \sqrt{6}) + \sqrt{13}]}{[{(\sqrt{7})}^{2} + {(\sqrt{6})}^{2} + 2 \times \sqrt{7} \times \sqrt{6} - 13]}$
$= \frac{[(\sqrt{7} + \sqrt{6}) + \sqrt{13}]}{[7 + 6 + 2 \sqrt{42} - 13]}$
$= \frac{\sqrt{7} + \sqrt{6} + \sqrt{13}}{[13 + 2 \sqrt{42} - 13]}$
$= \frac{\sqrt{7} + \sqrt{6} + \sqrt{13}}{2 \sqrt{42}}$
$= \frac{\sqrt{7} + \sqrt{6} + \sqrt{13}}{2 \sqrt{42}} \times \frac{\sqrt{42}}{\sqrt{42}}$
$= \frac{(\sqrt{7} + \sqrt{6} + \sqrt{13}) \times \sqrt{42}}{2 \times \sqrt{42} \times \sqrt{42}}$
$= \frac{\sqrt{7} \times \sqrt{42} + \sqrt{6} \times \sqrt{42} + \sqrt{13} \times \sqrt{42}}{84}$
$= \frac{\sqrt{7 \times 42} + \sqrt{6 \times 42} + \sqrt{13 \times 42}}{84}$
$= \frac{\sqrt{7 \times 7 \times 6} + \sqrt{6 \times 6 \times 7} + \sqrt{546}}{84}$
= $\frac{7 \sqrt{6} + 6 \sqrt{7} + \sqrt{546}}{84}$ .

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

Answer

We have
$\frac{3}{\sqrt{3} + \sqrt{5} - \sqrt{2}}$
$= \frac{3 \times [(\sqrt{3} + \sqrt{5}) + \sqrt{2}]}{[(\sqrt{3} + \sqrt{5}) - \sqrt{2}] \times [(\sqrt{3} + \sqrt{5}) + \sqrt{2}]}$
$= \frac{3 \times [(\sqrt{3} + \sqrt{5}) + \sqrt{2}]}{[{(\sqrt{3} + \sqrt{5})}^{2} - {(\sqrt{2})}^{2}]}$
$= \frac{3 \times [(\sqrt{3} + \sqrt{5}) + \sqrt{2}]}{[{(\sqrt{3})}^{2} + {(\sqrt{5})}^{2} + 2 \times \sqrt{3} \times \sqrt{5} - 2]}$
$= \frac{3 \times [(\sqrt{3} + \sqrt{5}) + \sqrt{2}]}{[3 + 5 + 2 \sqrt{15} - 2]}$
$= \frac{(3 \sqrt{3} + 3 \sqrt{5} + 3 \sqrt{2})}{6 + 2 \sqrt{15}}$
$= \frac{(3 \sqrt{3} + 3 \sqrt{5} + 3 \sqrt{2})}{(6 + 2 \sqrt{15})} \times \frac{(6 - 2 \sqrt{15})}{(6 - 2 \sqrt{15})}$
$= \frac{3 \sqrt{3} (6 - 2 \sqrt{15}) + 3 \sqrt{5} (6 - 2 \sqrt{15}) + 3 \sqrt{2} (6 - 2 \sqrt{15})}{6^{2} - (2 \sqrt{15})^{2}}$
$= \frac{18 \sqrt{3} - 18 \sqrt{5} + 18 \sqrt{5} - 30 \sqrt{3} + 18 \sqrt{2} - 6 \sqrt{30}}{36 - 60}$
$= \frac{- 12 \sqrt{3} + 18 \sqrt{2} - 6 \sqrt{30}}{- 24}$
$= \frac{- 6 (2 \sqrt{3} + 3 \sqrt{2} - \sqrt{30})}{- 24}$
= $\frac{(2 \sqrt{3} + 3 \sqrt{2} - \sqrt{30})}{4}$

(iii) $\frac{4}{2 + \sqrt{3} + \sqrt{7}}$

Answer

We have
$\frac{4}{2 + \sqrt{3} + \sqrt{7}}$
$= \frac{4 \times [(2 + \sqrt{3}) - \sqrt{7}]}{[(2 + \sqrt{3}) + \sqrt{7}] \times [(2 + \sqrt{3}) - \sqrt{7}]}$
$= \frac{4 \times [(2 + \sqrt{3}) - \sqrt{7}]}{[{(2 + \sqrt{3})}^{2} - {(\sqrt{7})}^{2}]}$
$= \frac{4 \times [(2 + \sqrt{3}) - \sqrt{7}]}{[2^{2} + {(\sqrt{3})}^{2} + 2 \times 2 \times \sqrt{3} - 7]}$
$= \frac{4 \times [(2 + \sqrt{3}) - \sqrt{7}]}{[4 + 3 + 4 \sqrt{3} - 7]}$
$= \frac{4 \times (2 + \sqrt{3} - \sqrt{7})}{4 \sqrt{3}}$
$= \frac{4 \times (2 + \sqrt{3} - \sqrt{7})}{4 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{(2 + \sqrt{3} - \sqrt{7})}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2 \sqrt{3} + 3 - \sqrt{21}}{3}$
= $\frac{2 \sqrt{3} - \sqrt{21} + 3}{3}$

In Exercise 1F RS Aggarwal Class 9, the 23rd question asks us to find the value of an expression correct to three places of decimal.

23. Given $\sqrt{2}$ = 1.414 and $\sqrt{6}$ = 2.449, find the value of $\frac{1}{\sqrt{3} - \sqrt{2} - 1}$ correct to 3 places of decimal.

Answer

The given expression is
$\frac{1}{\sqrt{3} - \sqrt{2} - 1}$
First, we rationalise the denominator and then put the given values.
We note that value of $\sqrt{3}$ is not given in the question. So, we will try to eliminate it from the expression. For this, we will regroup the terms of the denominator in the following way.
$\frac{1}{\sqrt{3} - (\sqrt{2} + 1)}$
$= \frac{1}{[\sqrt{3} - (\sqrt{2} + 1)]} \times \frac{[\sqrt{3} + (\sqrt{2} + 1)]}{[\sqrt{3} + (\sqrt{2} + 1)]}$
$= \frac{[\sqrt{3} + (\sqrt{2} + 1)]}{[\sqrt{3} - (\sqrt{2} + 1)] \times [\sqrt{3} + (\sqrt{2} + 1)]}$
$= \frac{[\sqrt{3} + \sqrt{2} + 1]}{[{(\sqrt{3})}^{2} - {(\sqrt{2} + 1)}^{2}]}$
$= \frac{[\sqrt{3} + \sqrt{2} + 1]}{[3 - (2 + 1 + 2 \sqrt{2})]}$
$= \frac{[\sqrt{3} + \sqrt{2} + 1]}{[3 - (3 + 2 \sqrt{2})]}$
$= \frac{[\sqrt{3} + \sqrt{2} + 1]}{[3 - 3 - 2 \sqrt{2}]}$
$= \frac{\sqrt{3} + \sqrt{2} + 1}{- 2 \sqrt{2}}$
$= \frac{\sqrt{3} + \sqrt{2} + 1}{- 2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\sqrt{6} + 2 + \sqrt{2}}{- 4}$
Now, putting the respective values, we get.
$= \frac{2.449 + 2 + 1.414}{- 4}$
$= \frac{5.863}{- 4}$
$= - 1.46575$ $≃$ $- 1.466$

In Exercise 1F RS Aggarwal Class 9, the 24th question asks us to find the value of an expression.

24. If $x = \frac{1}{2 - \sqrt{3}}$ , find the value of $x^{3} - 2 x^{2} - 7 x + 5$ .

Answer

We are given that
$x = \frac{1}{2 - \sqrt{3}}$ .
$\Rightarrow x = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$
$\Rightarrow x = \frac{2 + \sqrt{3}}{2^{2} - {(\sqrt{3})}^{2}}$
$\Rightarrow x = \frac{2 + \sqrt{3}}{4 - 3}$
$\Rightarrow x = 2 + \sqrt{3}$
Now, putting the value of $x$ in the given expression
$x^{3} - 2 x^{2} - 7 x + 5$
$= {(2 + \sqrt{3})}^{3} - 2 {(2 + \sqrt{3})}^{2} - 7 (2 + \sqrt{3}) + 5$
$= 2^{3} + {(\sqrt{3})}^{3} + 3 \times 2^{2} \times \sqrt{3} + 3 \times 2 \times {(\sqrt{3})}^{2} - 2 \times [2^{2} + {(\sqrt{3})}^{2} + 2 \times 2 \times \sqrt{3}] - 14 - 7 \sqrt{3} + 5$
$= 8 + 3 \sqrt{3} + 12 \sqrt{3} + 18 - 8 - 6 - 8 \sqrt{3} - 14 - 7 \sqrt{3} + 5$
$= 18 - 6 - 14 + 5$
= 3

In Exercise 1F RS Aggarwal Class 9, the 25th question asks us to evaluate an expression.

25. Evaluate $\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$ , it being given that $\sqrt{5}$ = 2.236 and $\sqrt{10}$ = 3.162.

Answer

We have
$\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$
$= \frac{15}{\sqrt{10} + 2 \sqrt{5} + 2 \sqrt{10} - \sqrt{5} - 4 \sqrt{5}}$
$= \frac{15}{3 \sqrt{10} - 3 \sqrt{5}}$
$= \frac{5}{(\sqrt{10} - \sqrt{5})}$
$= \frac{5}{(\sqrt{10} - \sqrt{5})} \times \frac{(\sqrt{10} + \sqrt{5})}{(\sqrt{10} + \sqrt{5})}$
$= \frac{5 (\sqrt{10} + \sqrt{5})}{{(\sqrt{10})}^{2} - {(\sqrt{5})}^{2}}$
$= \frac{5 (\sqrt{10} + \sqrt{5})}{10 - 5}$
$= \sqrt{10} + \sqrt{5}$
$= 3.162 + 2.236$
= $5.398$

Exercise 1F RS Aggarwal Class 9 Solutions

1. Write the rationalizing factor of the denominator in 12+3.

Answer

2. Rationalise the denominator of each of the following.

(i) 17

Answer

(ii) 523

Answer

(iii) 12+3

Answer

(iv) 15−2

Answer

(v) 15+32

Answer

(vi) 17−6

Answer

(vii) 411−7

Answer

(viii) 1+22−2

Answer

(ix) 3−223+22

Answer

3. It being given that 2 = 1.414, 3 = 1.732, 5 = 2.236 and 10 = 3.162, find the value to three places of decimals, of each of the following.

(i) 25

Answer

(ii) 2−33

Answer

(iii) 10−52

Answer

4. Find the rational numbers a and b such that

(i) 2-12+1=a+b2

Answer

(ii) 2-52+5=a5+b

Answer

(iii) 3+23−2=a+b6

Answer

(iv) 5+237+43=a+b3

Answer

5. It being given that 3 = 1.732, 5 = 2.236, 6 = 2.449 and 10 = 3.162, find to three places of decimal, the value of each of the following.

(i) 16+5

Answer

(ii) 65+3

Answer

(iii) 143−35

Answer

(iv) 3+53−5

Answer

(v) 1+232−3

Answer

(vi) 5+25−2

Answer

6. Simplify by rationalizing the denominator.

(i) 73-5248+18

Answer

(ii) 26-535-26

Answer

7. Simplify.

(i) 4+54−5+4−54+5

Answer

(ii) 13−2-25-3+35−2

Answer

(iii) 2+32−3+2−32+3+3−13+1

Answer

(iv) 262+3+626+3−836+2

Answer

8. Prove that.

(i) 13+7+17+5+15+3+13+1=1

Answer

(ii) 11+2 +12+3 +13+4 +14+5 +15+6 +16+7 +17+8 +18+9 = 2

Answer

9. Find the values of a and b if 7+353+5−7−353−5=a+b5.

Answer

10. Simplify 13−1113+11+13+1113−11.

Answer

11. If x=3+22, check whether x+1x is rational or irrational.

Answer

12. If x=2−3, find the value of (x−1x)3

Answer

13. If x=9−45, find the value of x2+1x2.

Answer

1. Write the rationalizing factor of the denominator in $\frac{1}{\sqrt{2} + \sqrt{3}}$ .

(i) $\frac{1}{\sqrt{7}}$

(ii) $\frac{\sqrt{5}}{2 \sqrt{3}}$

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

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

(v) $\frac{1}{5 + 3 \sqrt{2}}$

(vi) $\frac{1}{\sqrt{7} - \sqrt{6}}$

(vii) $\frac{4}{\sqrt{11} - \sqrt{7}}$

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

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

3. It being given that $\sqrt{2}$ = 1.414, $\sqrt{3}$ = 1.732, $\sqrt{5}$ = 2.236 and $\sqrt{10}$ = 3.162, find the value to three places of decimals, of each of the following.

(i) $\frac{2}{\sqrt{5}}$

(ii) $\frac{2 - \sqrt{3}}{\sqrt{3}}$

(iii) $\frac{\sqrt{10} - \sqrt{5}}{\sqrt{2}}$

4. Find the rational numbers $a$ and $b$ such that

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

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

(iii) $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = a + b \sqrt{6}$

(iv) $\frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a + b \sqrt{3}$

5. It being given that $\sqrt{3}$ = 1.732, $\sqrt{5}$ = 2.236, $\sqrt{6}$ = 2.449 and $\sqrt{10}$ = 3.162, find to three places of decimal, the value of each of the following.

(i) $\frac{1}{\sqrt{6} + \sqrt{5}}$

(ii) $\frac{6}{\sqrt{5} + \sqrt{3}}$

(iii) $\frac{1}{4 \sqrt{3} - 3 \sqrt{5}}$

(iv) $\frac{3 + \sqrt{5}}{3 - \sqrt{5}}$

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

(vi) $\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

(i) $\frac{7 \sqrt{3} - 5 \sqrt{2}}{\sqrt{48} + \sqrt{18}}$

(ii) $\frac{2 \sqrt{6} - \sqrt{5}}{3 \sqrt{5} - 2 \sqrt{6}}$

(i) $\frac{4 + \sqrt{5}}{4 - \sqrt{5}} + \frac{4 - \sqrt{5}}{4 + \sqrt{5}}$

(ii) $\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{2}{\sqrt{5} - \sqrt{3}} + \frac{3}{\sqrt{5} - \sqrt{2}}$

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

(iv) $\frac{2 \sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{6 \sqrt{2}}{\sqrt{6} + \sqrt{3}} - \frac{8 \sqrt{3}}{\sqrt{6} + \sqrt{2}}$

(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$

(ii) $\frac{1}{1 + \sqrt{2}}$ + $\frac{1}{\sqrt{2} + \sqrt{3}}$ + $\frac{1}{\sqrt{3} + \sqrt{4}}$ + $\frac{1}{\sqrt{4} + \sqrt{5}}$ + $\frac{1}{\sqrt{5} + \sqrt{6}}$ + $\frac{1}{\sqrt{6} + \sqrt{7}}$ + $\frac{1}{\sqrt{7} + \sqrt{8}}$ + $\frac{1}{\sqrt{8} + \sqrt{9}}$ = 2

9. Find the values of $a$ and $b$ if
$\frac{7 + 3 \sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3 \sqrt{5}}{3 - \sqrt{5}} = a + b \sqrt{5}$ .

10. Simplify $\frac{\sqrt{13} - \sqrt{11}}{\sqrt{13} + \sqrt{11}} + \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ .

11. If $x = 3 + 2 \sqrt{2}$ , check whether $x + \frac{1}{x}$ is rational or irrational.

12. If $x = 2 - \sqrt{3}$ , find the value of ${(x - \frac{1}{x})}^{3}$

13. If $x = 9 - 4 \sqrt{5}$ , find the value of $x^{2} + \frac{1}{x^{2}}$ .

14. If $x = \frac{5 - \sqrt{21}}{2}$ , find the value of $x + \frac{1}{x}$ .

15. If $a = 3 - 2 \sqrt{2}$ , find the value of $a^{2} - \frac{1}{a^{2}}$ .

16. If $x = \sqrt{13} + 2 \sqrt{3}$ , find the value of $x - \frac{1}{x}$ .

17. If $x = 2 + \sqrt{3}$ , find the value of $x^{3} + \frac{1}{x^{3}}$

18. If $x = \frac{5 - \sqrt{3}}{5 + \sqrt{3}}$ and $y = \frac{5 + \sqrt{3}}{5 - \sqrt{3}}$ , show that $x - y = - \frac{10 \sqrt{3}}{11}$ .

19. If $a = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $b = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$ , show that $3 a^{2} + 4 a b - 3 b^{2} = 4 + \frac{56}{3} \sqrt{10}$

20. If $a = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$ and $b = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$ , find the value of $a^{2} + b^{2} - 5 a b$ .

21. If $p = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}$ and $q = \frac{3 + \sqrt{5}}{3 - \sqrt{5}}$ , find the value of $p^{2} + q^{2}$ .

(i) $\frac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13}}$

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

(iii) $\frac{4}{2 + \sqrt{3} + \sqrt{7}}$

23. Given $\sqrt{2}$ = 1.414 and $\sqrt{6}$ = 2.449, find the value of $\frac{1}{\sqrt{3} - \sqrt{2} - 1}$ correct to 3 places of decimal.

24. If $x = \frac{1}{2 - \sqrt{3}}$ , find the value of $x^{3} - 2 x^{2} - 7 x + 5$ .

25. Evaluate $\frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$ , it being given that $\sqrt{5}$ = 2.236 and $\sqrt{10}$ = 3.162.