Question

2 root 6 by root 2 + root 3 + 6 root 2 by root 6 + root 3 what is the answer.

Answer 1

$$\begin{gathered} \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}} \\ \mathrm{On}\mathrm{rationalizing},\mathrm{we}\mathrm{get} \\ =\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}\times \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}} \\ =\frac{2\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{3-2}+\frac{6\sqrt{2}\left(\sqrt{6}-\sqrt{3}\right)}{6-3} \\ =2\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)+\frac{6\sqrt{2}\left(\sqrt{6}-\sqrt{3}\right)}{3} \\ =2\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)+2\sqrt{2}\left(\sqrt{6}-\sqrt{3}\right) \\ =2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6} \\ =2\sqrt{18}-2\sqrt{6} \\ =6\sqrt{2}-2\sqrt{6}\mathrm{ANS}... \end{gathered}$$