Question

2 ROOT 6 BY ROOT 2 + ROOT 3

Answer 1

$$\begin{gathered} \frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}} \\ =\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \\ =\frac{2\sqrt{12}-2\sqrt{18}}{2-3} \\ =2\sqrt{18}-2\sqrt{12} \\ =6\sqrt{2}-4\sqrt{3} \end{gathered}$$