Question

Simplify by rationalising the denominator $\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}$

Answer 1

Given that: $\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}$

Here R.F of $\sqrt{6}-\sqrt{3}$ is $\sqrt{6}+\sqrt{3}$

So, $\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}-\sqrt{3}}\times \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}$
$=\frac{(\sqrt{6}+\sqrt{3}{)}^2}{(\sqrt{6}{)}^2-(\sqrt{3}{)}^2}[\because (a-b\left)\right(a+b)=a^2-b^2]$
$=\frac{(\sqrt{6}{)}^2+(\sqrt{3}{)}^2+2\sqrt{6}\times \sqrt{3}}{6-3}[\because (a+b{)}^2=a^2+2ab+b^2]$
$=\frac{6+3+2\sqrt{18}}{3}=\frac{9+6\sqrt{2}}{3}=\frac{3(3+2\sqrt{2})}{3}=3+2\sqrt{2}$
$\therefore \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}=3+2\sqrt{2}$