√(1+√2/1-√2)^2+2+(1-√2/1+√2)^2 Срочно пжпжпжпжпжпжп

Решите задачу:

$(\frac{1+\sqrt{2}}{1-\sqrt{2}} )^2+2+(\frac{1-\sqrt{2}}{1+\sqrt{2}})^2= (\frac{(1+\sqrt{2})^2}{(1-\sqrt{2})(1+\sqrt{2})})^2+2+(\frac{(1-\sqrt{2})(1-\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})})^2= (\frac{3+2\sqrt{2}}{1-2})^2+2+(\frac{3-2\sqrt{2}}{1-2})^2= (-(3+2\sqrt{2} ))^2+2+(-(3-2\sqrt{2} ))^2= (3+2\sqrt{2} )^2+2+(3-2\sqrt{2} )^2= 9+12\sqrt{2}+ 8+2 + 9-12\sqrt{2}+ 8 =36$