Question

Question 2
Simplify:
$\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$

Answer 1

We have, $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}.....\left(i\right)$
Now, $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}\times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}=\frac{7\sqrt{30}-21}{(\sqrt{10}{)}^2-(\sqrt{3}{)}^2}$ [by rationalisation]
[Using identity, $(a-b)(a+b)=a^2-b^2]$
$\frac{7\sqrt{30}-21}{10-3}=\frac{7(\sqrt{30}-3)}{7}=\sqrt{30}-3\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}\times \frac{(\sqrt{6}-\sqrt{5})}{\sqrt{6}-\sqrt{5}}$ [by rationalisation]
$=\frac{2\sqrt{30}-10}{(\sqrt{6}{)}^2-(\sqrt{5}{)}^2}$ [Using identity ,$(a-b)(a+b)=a^2-b^2]$
$=\frac{2\sqrt{30}-10}{6-5}=2\sqrt{30}-10$
and $\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}\times \frac{\sqrt{15}-3\sqrt{2}}{\sqrt{15}-3\sqrt{2}}$ [by rationlisation]
$=\frac{3(\sqrt{30}-6)}{15-18}=\frac{3(\sqrt{30}-6)}{-3}=(-\sqrt{30}+6)=6-\sqrt{30}$
From eq.(i). $\frac{7\sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3\sqrt{2}}{\sqrt{15}+3\sqrt{2}}$
$=(\sqrt{30}-3)-(2\sqrt{30}-10)-(6-\sqrt{30})=\sqrt{30}-3-2\sqrt{30}+10-6+\sqrt{30}=2\sqrt{30}-2\sqrt{30}+10-9=1$