Question

Please help me in this question

Answer 1

$\bold{Given: \frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}} }$

We have to rationalize the denominator.

Consider,

$\frac{1}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}$

$=\frac{1}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}\times\frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5})-\sqrt{11}}$

$=\frac{(\sqrt{6}+\sqrt{5})+\sqrt{11}}{(\sqrt{6}+\sqrt{5}-\sqrt{11})(\sqrt{6}+\sqrt{5}+\sqrt{11})}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}+\sqrt{5})^2-(\sqrt{11})^2}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6})^2+(\sqrt{5})^2+2\sqrt{6\times5}-11}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}$

$=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}\times\frac{\sqrt{30}}{\sqrt{30}}$

$=\frac{(\sqrt{6}+\sqrt{5}+\sqrt{11})\times\sqrt{30}}{2(\sqrt{30})^2}$

$=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{2\times30}$

$=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}$