Question

Rationalise the denominator of $\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}$

Answer 1

Given: $\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}$
We have to rationalize the denominator.
$\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}\times \frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5})+\sqrt{11}}=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}+\sqrt{5}{)}^2-(\sqrt{11}{)}^2}$ [Since, $(a+b)(a-b)=a^2-b^2$ ]
$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}{)}^2+(\sqrt{5}{)}^2+2\left(\sqrt{6}\right)\left(\sqrt{5}\right)-11}$
$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{6+5+2\left(\sqrt{6}\right)\left(\sqrt{5}\right)-11}$
$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{11+2\left(\sqrt{30}\right)-11}$
$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\left(\sqrt{30}\right)}\times \frac{\sqrt{30}}{\sqrt{30}}$
$=\frac{\sqrt{6\times 30}+\sqrt{5\times 30}+\sqrt{11\times 30}}{2(\sqrt{30}{)}^2}$
$=\frac{\sqrt{6\times 6\times 5}+\sqrt{5\times 6\times 5}+\sqrt{11\times 6\times 5}}{2\left(30\right)}$
$=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}$ is the required answer.