Помогите освободить от иррациональности

Решите задачу:

$1)\; \; \frac{1}{\sqrt2+\sqrt5-\sqrt3}= \frac{\sqrt2+\sqrt5+\sqrt3}{(\sqrt2+\sqrt5)^2-3}= \frac{\sqrt2+\sqrt5+\sqrt3}{7+2\sqrt{10}-3} =\frac{\sqrt2+\sqrt5+\sqrt3}{2(\sqrt{10}+2)}=\\\\= \frac{(\sqrt2+\sqrt5+\sqrt3)(\sqrt{10}-2)}{2\cdot (10-4)}= \frac{(\sqrt2+\sqrt5+\sqrt3)(\sqrt{10}-2)}{12}=\\\\= \frac{(\sqrt2+\sqrt5+\sqrt3)(\sqrt5-\sqrt2)\cdot \sqrt2}{12}= \frac{\sqrt2\cdot ((5-2)+\sqrt3(\sqrt5-\sqrt2))}{12}=\\\\= \frac{\sqrt2(3+\sqrt{15}-\sqrt6)}{12}$

$2)\; \; \frac{6}{\sqrt{10}+\sqrt6+5-\sqrt{15}}=\frac{6\cdot ((\sqrt{10}+\sqrt6)-(5-\sqrt{15}))}{(\sqrt{10}+\sqrt6)^2-(5-\sqrt{15})^2}=\frac{6\cdot (\sqrt{10}+\sqrt6-5+\sqrt{15})}{16+2\sqrt{60}-40+10\sqrt{15}}=\\\\= \frac{6(\sqrt{10}+\sqrt6-5+ \sqrt{15})}{14\sqrt{15}-24}=\frac{6(\sqrt{10}+\sqrt6-5+\sqrt{15})}{2(7\sqrt{15}-12)}=\\\\=\frac{3(\sqrt{10}+\sqrt6-5+\sqrt{15})(7\sqrt{15}+12)}{735-144}=- \frac{3(\sqrt{10}+\sqrt6-5+\sqrt{15})(7\sqrt5+12)}{591}$