#### (i) Rationalise the denominator of the following : $\frac{2}{3\sqrt{3}}$ (ii)Rationalise the denominator of the following : $\frac{\sqrt{40}}{\sqrt{3}}$ (iii) Rationalise the denominator of the following : $\frac{3+\sqrt{2}}{4\sqrt{2}}$ (iv)Rationalise the denominator of the following :$\frac{16}{\sqrt{41}-5}$ (v) Rationalise the denominator of the following : $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ (vi) Rationalise the denominator of the following  : $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ (vii) Rationalise the denominator of the following : $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$(viii) Rationalise the denominator of the following : $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$(ix) Rationalise the denominator of the following : $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

(i) Answer.  $\frac{2\sqrt{3}}{9}$
Solution.         We have, $\frac{2}{3\sqrt{3}}$
Rationalising the denominator, we get:
$\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2\sqrt{3}}{3\sqrt{3}\sqrt{3}}$
$= \frac{2\sqrt{3}}{9}$
Hence the answer is $\frac{2\sqrt{3}}{9}$

(ii) Answer.  $\frac{2\sqrt{30}}{3}$
Solution. We have ,$\frac{\sqrt{40}}{\sqrt{3}}$
We know that, 40 = (2) (2) (10)
$\frac{\sqrt{40}}{\sqrt{3}}= \frac{\sqrt{2\cdot 2\cdot 10}}{\sqrt{3}}= \frac{2\sqrt{10}}{\sqrt{3}}$

Rationalising the denominator, we get:
$= \frac{2\sqrt{10}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2\sqrt{10}\sqrt{3}}{\sqrt{3}\sqrt{3}}$
$= \frac{2\sqrt{30}}{3}$

Hence the answer is: $\frac{2\sqrt{30}}{3}$

(iii) Answer.  $\frac{3\sqrt{2}+2}{8}$
Solution.  We have $\frac{3+\sqrt{2}}{4\sqrt{2}}$
Rationalising the denominator, we get:
$\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\left ( 3+\sqrt{2} \right )\sqrt{2}}{4\sqrt{2}\sqrt{2}}$
$= \frac{3\sqrt{2}+2}{8}$
Hence the answer is $\frac{3\sqrt{2}+2}{8}$

(iv) Answer. $\sqrt{41}+5$
Solution. We have $\frac{16}{\sqrt{41}-5}$
Rationalising the denominator, we get:
$\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41}-5 \right )\sqrt{41}+5}$
Using  the identity (a – b) (a + b) = a2 – b2
We get:
$= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41} \right )^{2}-\left ( 5 \right )^{2}}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{41-25}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{16}$
$= \sqrt{41}+5$
Hence the answer is $\sqrt{41}+5$

(v) Answer.    $7+4\sqrt{3}$
Solution. We have, $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Rationalising the denominator, we get:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2-\sqrt{3}}$
$= \frac{\left ( 2+\sqrt{3} \right )^{2}}{\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}$

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
$= \frac{2^{2}+\left ( \sqrt{3} \right )^{2}+2\cdot 2\cdot \sqrt{3}}{2^{2}-\left ( \sqrt{3} \right )^{2}}$
$= \frac{4+3+4\sqrt{3}}{4-3}$
$= 7+4\sqrt{3}$
Hence the answer is $7+4\sqrt{3}$

(vi)Answer.  $3\sqrt{2}-2\sqrt{3}$
Solution.   We have, $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
Rationalising the denominator, we get:
$= \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$= \frac{\sqrt{6}\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right )}$
Using  the identity (a – b) (a + b) = a2 – b2
We get:
$= \frac{\sqrt{18}-\sqrt{12}}{\left ( \sqrt{3} \right )^{2}\left ( \sqrt{2} \right )^{2}}$
$= \frac{\sqrt{3\cdot 3\cdot 2}-\sqrt{2\cdot 2\cdot 3}}{3-2}$
$= \frac{3\sqrt{2}-2\sqrt{3}}{1}$
$= 3\sqrt{2}-2\sqrt{3}$
Hence the answer $3\sqrt{2}-2\sqrt{3}$

(vii) Answer. $5+2\sqrt{6}$
Solution. We have, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Rationalising the denominator, we get:
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$= \frac{\left ( \sqrt{3} +\sqrt{2}\right )^{2}}{\left ( \sqrt{3} -\sqrt{2} \right )\left ( \sqrt{3} +\sqrt{2} \right )}$

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
$= \frac{\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{2} \right )^{2}+2\sqrt{3}\sqrt{2}}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{2} \right )^{2}}$
$= \frac{3+2+2\sqrt{6}}{3-2}$
$= 5+2\sqrt{6}$
Hence the answer is $5+2\sqrt{6}$

(viii)

Answer: $9+2 \sqrt{15}$

Solution:

We have $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Rationalize

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

$=\frac{15+3 \sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4 \sqrt{15}}{2}$

$=9+2 \sqrt{15}$

(ix) Answer: $\frac{9+4 \sqrt{6}}{15}$

Solution:

We have $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

$=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$

Rationalize

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

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

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

$=\frac{18+8 \sqrt{6}}{30}$

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