Question

Rationalise the denominator of $\frac{3}{\sqrt{3}-\sqrt{2}+\sqrt{5}}$.

Answer 1

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

$=\frac{3(\sqrt{3}-\sqrt{2}-\sqrt{5})}{(\sqrt{3}-\sqrt{2}{)}^2-(\sqrt{5}{)}^2}$

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

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

$=\frac{3(\sqrt{3}-\sqrt{2}-\sqrt{5})}{-2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}$

$=\frac{3(\sqrt{18}-\sqrt{12}-\sqrt{30})}{-12}$

$=\frac{3\sqrt{2}-2\sqrt{3}-\sqrt{30}}{-4}$

$=\frac{2\sqrt{3}+\sqrt{30}-3\sqrt{2}}{4}$