Question

√$\frac{3 \sqrt{2} }{ \sqrt{6}+ \sqrt{3} } + \frac{2 \sqrt{3} }{ \sqrt{6} +2} - \frac{4 \sqrt{3} }{ \sqrt{6}- \sqrt{2} }$

Answer 1

$\frac{3\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})} + \frac{2\sqrt{3}(\sqrt{6}-2)}{(\sqrt{6}+2)(\sqrt{6}-2)} - \frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} \\ \\ \\ \frac{3\sqrt{2}\sqrt{2}\sqrt{3}-3\sqrt{2}\sqrt{3}}{6-3} + \frac{2\sqrt{3}\sqrt{2}\sqrt{3} - 2* \sqrt{3}*2}{6 - 2^2} - \frac{4\sqrt{3}\sqrt{2}\sqrt{3}+4\sqrt{3}\sqrt{2}}{6 - 2} \\ \\ 4\sqrt{3} - \sqrt{2}\sqrt{3} + 3\sqrt{2} - 2 \sqrt{3} - 3\sqrt{2} - \sqrt{3}\sqrt{2}$

ADD and subtract corresponding terms :

$2\sqrt{3} - 2\sqrt{6}$