У меня немного не сходится. Помогите 536 г и е

[latex]\frac{1}{\sqrt5+\sqrt3-\sqrt2}*\frac{\sqrt5+\sqrt3+\sqrt2}{\sqrt5+\sqrt3+\sqrt2}=\frac{\sqrt5+\sqrt3+\sqrt2}{(\sqrt5+\sqrt3)^2-2}=\frac{\sqrt5+\sqrt3+\sqrt2}{6+2\sqrt15}*\frac{6-2\sqrt15}{6-2\sqrt15}=\\=-\frac{6\sqrt5+6\sqrt3+2\sqrt6-2\sqrt{75}-2\sqrt{45}+2\sqrt{30}}{24}=\frac{6\sqrt5+6\sqrt3+2\sqrt6-2\sqrt{75}-2\sqrt{45}-2\sqrt{30}}{-24}=\\=\frac{6\sqrt5+6\sqrt3+2\sqrt6-10\sqrt3-6\sqrt5-2\sqrt{30}}{-24}=\frac{-4\sqrt3+2\sqrt6-2\sqrt{30}}{-24}=\frac{2\sqrt3-\sqrt6+\sqrt{30}}{12}[/latex] [latex]\frac{1}{\sqrt{15}-\sqrt{10}+\sqrt3-\sqrt2}=\frac{1}{\sqrt5(\sqrt3-\sqrt2)+\sqrt3-\sqrt2}=\frac{1}{(\sqrt5+1)(\sqrt3-\sqrt2)}*\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\\=\frac{\sqrt3+\sqrt2}{\sqrt5+1}*\frac{\sqrt5-1}{\sqrt5-1}=\frac{(\sqrt3+\sqrt2)(\sqrt5-1)}{4}=\frac{\sqrt15-\sqrt3+\sqrt10-\sqrt2}{4}[/latex]