Помогите срочно!! ответ лучишим

Решите задачу:

$\frac{1}{\sqrt3+\sqrt2+1} = \frac{\sqrt3+\sqrt2-1}{(\sqrt3+\sqrt2+1)(\sqrt3+\sqrt2-1)} =\frac{\sqrt3+\sqrt2-1}{(\sqrt3+\sqrt2)^2-1} = \frac{\sqrt3+\sqrt2-1}{5+2\sqrt6-1}=\\\\=\frac{\sqrt3+\sqrt2-1}{2\sqrt2(\sqrt3+\sqrt2)} = \frac{(\sqrt3+\sqrt2-1)(\sqrt3-\sqrt2)}{2\sqrt2(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}= \frac{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)-(\sqrt3-\sqrt2)}{2\sqrt2(3-2)} =$

$= \frac{3-2-\sqrt3+\sqrt2}{2\sqrt2}=\frac{1+\sqrt2-\sqrt3}{2\sqrt2} =\frac{(1+\sqrt2-\sqrt3)\sqrt2}{2\sqrt2\cdot \sqrt2}=\frac{\sqrt2+2-\sqrt6}{4 }$