Question

Solve:$\frac{3+\sqrt{6}}{17\sqrt{3}-2\sqrt{32}+3\sqrt{18}-4\sqrt{48}}$

Answer 1

$\frac{3+\sqrt{6}}{17\sqrt{3}-2\sqrt{32}+3\sqrt{18}-4\sqrt{48}}$

$\Rightarrow \frac{3+\sqrt{6}}{17\sqrt{3}-2\sqrt{2\times 2\times 2\times 2\times 2}+3\sqrt{3\times 3\times 2}-4\sqrt{2\times 2\times 2\times 2\times 3}}$

$\Rightarrow \frac{3+\sqrt{6}}{17\sqrt{3}-2\times 2\times 2\sqrt{2}+3\times 3\sqrt{2}-4\times 2\times 2\sqrt{3}}$

$\Rightarrow \frac{3+\sqrt{6}}{17\sqrt{3}-8\sqrt{2}+9\sqrt{2}-16\sqrt{3}}$

$\Rightarrow \frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}$

$\Rightarrow \frac{\sqrt{3}\times \sqrt{3}+\sqrt{3}\times \sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$\Rightarrow \frac{\sqrt{3}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}$

$\Rightarrow \sqrt{3}$ Answer