Помогите пожалуйста с вычислением:1) 〖(√(3-2√2) -√(3+2√2) )〗^2 ; 2) 3/(5-2√6)+3/(5+2√6)...

Решите задачу:

$((\sqrt{3-2\sqrt{2}} -\sqrt{3+2\sqrt{2}})^2=\\\ =(\sqrt{3-2\sqrt{2}})^2-2\sqrt{3-2\sqrt{2}}*\sqrt{3+2\sqrt{2}}+(\sqrt{3+2\sqrt{2}})^2=\\\ =3-2\sqrt{2}}-2\sqrt{(3-2\sqrt{2})(3+2\sqrt{2})}+3+2\sqrt{2}}=\\\ =6-2\sqrt{9-8}=6-2=4$

$\frac{3}{5-2\sqrt{6}}+\frac{3}{5+2\sqrt{6}}=\frac{3(5+2\sqrt{6})+3(5-2\sqrt{6})}{(5-2\sqrt{6})(5+2\sqrt{6})}=\frac{15+6\sqrt{6}+15-6\sqrt{6}}{25-24}=30$