Question

\(\left(\frac{2\sqrt{32}}{\sqrt{3}}-1\right):\left(7-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}}\right)\)

Answer 1

\(\left(\frac{2\sqrt{32}}{\sqrt{3}}-1\right):\left(7-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}}\right)\)

\(=\left[\frac{2\sqrt{96}}{3}-\frac{3}{3}\right]:\left[\frac{14}{2}-\frac{\left(\sqrt{2}-\sqrt{3}\right)\sqrt{2}}{2}\right]\)

\(=\left[\frac{8\sqrt{6}-3}{3}\right]:\left[\frac{14-2+\sqrt{6}}{2}\right]\)

\(=\left[\frac{\sqrt{3}\left(8\sqrt{2}-\sqrt{3}\right)}{3}\right]:\left[\frac{12+\sqrt{6}}{2}\right]\)

\(=\left[\frac{\sqrt{3}\left(8\sqrt{2}-\sqrt{3}\right)}{3}\right]:\left[\frac{\sqrt{3}\left(2\sqrt{6}+1\right)}{\sqrt{2}}\right]\)

\(=\frac{8\sqrt{2}-\sqrt{3}}{\sqrt{3}}.\frac{\sqrt{2}}{\sqrt{3}\left(2\sqrt{6}+1\right)}\)

\(=\frac{16-\sqrt{6}}{6\sqrt{6}+3}\)