Question

2 (root2+root6)÷3 {root(2+root3)} is equal to??

Answer 1

$$\begin{gathered} Wearegiven \\ \frac{2\left(\sqrt{2}+\sqrt{6}\right)}{3\left(\sqrt{2+\sqrt{3}}\right)}=\frac{2\left(\sqrt{2}+\sqrt{2\times 3}\right)}{3\left(\sqrt{2+\sqrt{3}}\right)}=\frac{2\sqrt{2}\left(1+\sqrt{3}\right)}{3\left(\sqrt{2+\sqrt{3}}\right)} \\ =\frac{2\sqrt{2}\left(1+\sqrt{3}\right)}{3\left(\sqrt{2+\sqrt{3}}\right)}\times \frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}} \\ =\frac{2\sqrt{2}\times \sqrt{2-\sqrt{3}}\left(1+\sqrt{3}\right)}{3\left(\sqrt{4-3}\right)}=\frac{2\sqrt{2}\times \sqrt{2-\sqrt{3}}\left(1+\sqrt{3}\right)}{3\left(1\right)} \\ =\frac{2\sqrt{2}\times \sqrt{2-\sqrt{3}}\left(1+\sqrt{3}\right)}{3} \\ \mathrm{Therefore}, \\ \frac{2\left(\sqrt{2}+\sqrt{6}\right)}{3\left(\sqrt{2+\sqrt{3}}\right)}=\frac{2\sqrt{2}\times \sqrt{2-\sqrt{3}}\left(1+\sqrt{3}\right)}{3} \end{gathered}$$