Question

Освободитесь от иррациональности в знаменателе дроби:

Answer 1

$\frac{6}{\sqrt{10}+\sqrt{6}+5-\sqrt{15}} = \frac{6}{(\sqrt{10}+\sqrt{6}) + (5-\sqrt{15})} = \\ \\ \\ = \frac{6}{(\sqrt{10}+\sqrt{6}) + (5-\sqrt{15})} * \frac{(\sqrt{10}+\sqrt{6}) - (5-\sqrt{15})}{(\sqrt{10}+\sqrt{6}) - (5-\sqrt{15})} = \\ \\ \\ = \frac{6((\sqrt{10}+\sqrt{6}) - (5-\sqrt{15}))}{(\sqrt{10}+\sqrt{6})^2 - (5-\sqrt{15})^2} = \\ \\ \\ = \frac{6((\sqrt{10}+\sqrt{6}) - (5-\sqrt{15}))}{(16+2\sqrt{60}) - (40-10\sqrt{15})} =$

$= \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})}{14\sqrt{15} - 24} = \\ \\ \\ = \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})}{14\sqrt{15} - 24} * \frac{{14 \sqrt{15} + 24}}{14 \sqrt{15} + 24}} = \\ \\ \\ = \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})*({14 \sqrt{15} + 24})}{(14\sqrt{15})^2 - 24^2} =$

$= \frac{6(\sqrt{10}+\sqrt{6} -5+\sqrt{15})*({14 \sqrt{15} + 24})}{2364}$