Question

Solve:$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Answer 1

We have,

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

$=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}$

$=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{3-2}$

$={\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\left(\sqrt{3}\right)\left(\sqrt{2}\right)$

$=3+2+2\sqrt{6}$

$=5+2\sqrt{6}$

Hence,

We get $(5+2\sqrt{6})$.