Question

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

Answer 1

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

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

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

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

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

$=\sqrt{3}$

Hence, this is the answer.