Question

Evaluate $\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}}-\frac{3\sqrt{3}}{\sqrt{5}+\sqrt{2}}+\frac{2\sqrt{2}}{\sqrt{5}+\sqrt{3}}$

Answer 1

$\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}}-\frac{3\sqrt{3}}{\sqrt{5}+\sqrt{2}}+\frac{2\sqrt{2}}{\sqrt{5}+\sqrt{3}}$

$=\frac{\sqrt{5}\times (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})\times (\sqrt{3}-\sqrt{2})}-\frac{3\sqrt{3}\times (\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})\times (\sqrt{5}-\sqrt{2})}$ $+\frac{2\sqrt{2}\times (\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})\times (\sqrt{5}-\sqrt{3})}\left[\text{On rationalising}\right]$

$=\frac{\sqrt{15}-\sqrt{10}}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}-\frac{(3\sqrt{15}-3\sqrt{6})}{(\sqrt{5}{)}^2-(\sqrt{2}{)}^2}+\frac{2\sqrt{10}-2\sqrt{6}}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}$

$\text{Using this identity above,}$ $(\sqrt{a}\times \sqrt{b}=\sqrt{a\times b})$ $\text{and}$ $(a+b)(a-b)=(a^2-b^2)$

$=\frac{\sqrt{15}-\sqrt{10}}{3-2}-\frac{3(\sqrt{15}-\sqrt{6})}{5-2}+\frac{2(\sqrt{10}-\sqrt{6})}{5-3}$

$=\sqrt{15}-\sqrt{10}-\frac{3(\sqrt{15}-\sqrt{6})}{3}+\frac{2(\sqrt{10}-\sqrt{6})}{2}$

$=\sqrt{15}-\sqrt{10}-\sqrt{15}+\sqrt{6}+\sqrt{10}-\sqrt{6}=0$