Simplify the following

Solution:

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}}$$\ldots($ i)

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, $\left.(a-b)(a+b)=a^{2}-b^{2}\right]$

$=\frac{2 \sqrt{30}-10}{(\sqrt{6})^{2}-(\sqrt{5})^{2}}$      [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 rationalisation]

$=\frac{3 \sqrt{30}-18}{(\sqrt{15})^{2}-(3 \sqrt{2})^{2}}$         [using identity, $(a-b)(a+b)=a^{2}-b^{2}$ ]

$=\frac{3(\sqrt{30}-6)}{15-18}=\frac{3(\sqrt{30}-6)}{-3}=(-\sqrt{30}+6)$

$=6-\sqrt{30}$

From Eq. (i),          $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$

$=(\sqrt{30}-3)-(2 \sqrt{30}-10)-(6-\sqrt{30})$

$=\sqrt{30}-3-2 \sqrt{30}+10-6+\sqrt{30}$

$=2 \sqrt{30}-2 \sqrt{30}+10-9=1$