Избавьте от иррациональности в знаменателе 3/(2-√2+√3-√6)

Решите задачу:

$\frac{3}{2-\sqrt{2}+\sqrt{3}-\sqrt{6}}=\frac{3}{\sqrt{2}(\sqrt{2}-1)+\sqrt{3}(1-\sqrt{2})}=\frac{3}{\sqrt{2}(\sqrt{2}-1)-\sqrt{3}(\sqrt{2}-1)}=$

$=\frac{3}{(\sqrt{2}-\sqrt{3})*(\sqrt{2} -1)}=$

$=\frac{3*(\sqrt{2}+\sqrt{3})*(\sqrt{2}+1)}{(\sqrt{2}-\sqrt{3})(\sqrt{2}-1)*(\sqrt{2}+\sqrt{3})*(\sqrt{2}+1)}=$

$=\frac{3*(\sqrt{2}+\sqrt{3})*(\sqrt{2}+1)}{(2-3)(2-1)}=$

$=\frac{3*(2+\sqrt{6}+\sqrt{2}+\sqrt{3})}{-1*1}=$

$=\frac{6+3\sqrt{6}+3\sqrt{2}+3\sqrt{3}}{-1}= -({6+3\sqrt{6}+3\sqrt{2}+3\sqrt{3})$