#### (i) Simplify the following : $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$ (ii) Simplify the following : $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$ (iii) Simplify the following : $4\sqrt{12}\times 7\sqrt{6}$ (iv) Simplify the following : $4\sqrt{28}\div 3\sqrt{7}\div 3\sqrt{7}$ (v) Simplify the following : $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$ (vi) Simplify the following : $\left ( \sqrt{3}-\sqrt{2} \right )^{2}$ (vii) Simplify the following : $\sqrt[4]{81}- 8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{255}$ (viii) Simplify the following : $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$ (ix) Simplify the following : $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$

(i) Answer. $\sqrt{5}$
Solution.    $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$
We know that,
45 = $3\times 3\times 5$
20 = $2\times 2\times 5$
So we get
$\sqrt{3\times3\times5 }-3\sqrt{2\times2\times5}+4\sqrt{5}$
$= 3\sqrt{5}-3\left ( 2\sqrt{5} \right )+4\sqrt{5}$
$= 3\sqrt{5}-6\sqrt{5}+4\sqrt{5}$
$= 7\sqrt{5}-6\sqrt{5}$
$= \sqrt{5}$
Hence the answer is $\sqrt{5}$

(ii)  Answer. $\frac{7\sqrt{6}}{12}$
Solution. We have, $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$
We know that,

$24= 6\times 4= 3\times 2\times 2\times 2$
$54= 9\times 6= 3\times 3\times 3\times 2$
So we get
$\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}= \frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}$
$= \frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}$
Taking LCM (3,4) = 12

$= \frac{3\sqrt{6}+4\sqrt{6}}{12}$
$= \frac{7\sqrt{6}}{12}$

(iii) Answer.  $\sqrt[28]{2^{18} \times 3^{11}}$
Solution.   We have
$\sqrt[4]{12}\times \sqrt[7]{6}$
We know that
12 = $2\times 2\times 3$
6 = $2\times 3$
So we get,
=$\sqrt[4]{2\times 2\times 3}\times \sqrt[7]{2\times 3}$
$=2^{1 / 4} \cdot 2^{1 / 4} \cdot 3^{1 / 4} \cdot 2^{1 / 7} \cdot 3^{1 / 7}$
$=2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}} \times 3^{\frac{1}{4}+\frac{1}{7}}$

$=2^{9 / 14} \times 3^{11 / 28}$

=$\sqrt[28]{2^{18} \times 3^{11}}$
Hence the number is $\sqrt[28]{2^{18} \times 3^{11}}$.

(iv)Answer.    $\frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$
Solution.   We have, $4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}$
We know that
28 = $4\times 7$
So we can write,
$4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}= \left [ \frac{4\sqrt{28}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$
= $\left [ \frac{4\sqrt{4\times 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$

= $\left [ \frac{4\times 2\sqrt{ 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$

$= \frac{8}{3}\div 7^{\frac{1}{3}}$
$= \frac{8}{\left ( 3\times 7^{\frac{1}{3}} \right )}$
$= \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$
Hence the answer is $\frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$

(v) Answer.  $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$
We know that
27 = $3\times 3\times 3$
So, $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$ = $3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}$

$= 3\sqrt{3}+2\left ( 3\sqrt{3} \right )+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
(Rationalising the denominator)

$= 3\sqrt{3}+6\left ( \sqrt{3} \right )+\frac{7\sqrt{3}}{3}$

$= \left ( 3+6+\frac{7}{3} \right )\sqrt{3}$         (Taking $\sqrt{3}$ common)
Now LCM (1,1,3) = 3
$= \left ( \frac{9+18+7}{3} \right )\sqrt{3}$
$= \frac{34}{3}\sqrt{3}$
$= 19\cdot 63$

(vi) Answer. $5-2\sqrt{6}$
Solution. Given, $\left ( \sqrt{3}-\sqrt{2} \right )^{2}$
We know that (a + b)2 = a2 – 2ab + b2
Comparing the given equation with the identity, we get:

$\left ( \sqrt{3}-\sqrt{2} \right )^{2}= \left ( \sqrt{3} \right )^{2}-2\left ( \sqrt{3} \right )\left ( \sqrt{2} \right )+\left ( \sqrt{2} \right )^2$
= 3 + 2 – $2\sqrt{3\times 2}$
$= 5-2\sqrt{6}$
Hence the answer is $5-2\sqrt{6}$

Solution. We have, $\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
We know that
81 = $3\times 3\times3\times3$
216 = $6\times 6\times6$
32 = $2\times 2\times2\times2\times2$
225 = $15\times 15$
So,$\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
$= \sqrt[4]{3\times3\times3\times3 }-8\sqrt[3]{6\times6\times6}+15\sqrt[5]{2\times2\times2\times2\times2}+\sqrt{15\times15}$
= 3 – 8 × 6 + 15 × 2 + 15
= 3 – 48 + 30 + 15
= – 45 + 45
= 0

(viii) Answer.   $\frac{5}{2\sqrt{2}}$

Solution.   We have, $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$
We know that, 8 =$2\times 2\times 2$
So,

$\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$ $= \frac{3}{\sqrt{2\times 2\times 2}}+\frac{1}{\sqrt{2}}$

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

$= \frac{3}{2\sqrt{2}}+\frac{2}{2\sqrt{2}}$
$= \frac{5}{2\sqrt{2}}$
Hence the answer is $\frac{5}{2\sqrt{2}}$

(ix) Answer.     $\frac{\sqrt{3}}{2}$
Solution.     We have, $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$
LCM (3,6) = 6

$\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}= \frac{4\sqrt{3}}{6}-\frac{\sqrt{3}}{6}$

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

$= \frac{3\sqrt{3}}{6}$
$= \frac{\sqrt{3}}{2}$
Hence the answer is $\frac{\sqrt{3}}{2}$.