Question

Simplify of the following:
$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}}$.

Answer 1

We take the $LCM$ of the denominators of $(\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}})$ and then simplify as follows:

$\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{2})+\sqrt{5}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{\left(\sqrt{2\times 3}\right)+\left(\sqrt{2\times 2}\right)+\left(\sqrt{5\times 3}\right)-\left(\sqrt{5\times 2}\right)}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=\frac{\sqrt{6}+\sqrt{4}+\sqrt{15}-\sqrt{10}}{3-2}=\frac{\sqrt{6}+2+\sqrt{15}-\sqrt{10}}{1}$

$=2+\sqrt{6}+\sqrt{15}-\sqrt{10}$ .

Hence, $\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}}=2+\sqrt{6}+\sqrt{15}-\sqrt{10}$ .