СРОЧНО ДАЮ 80 БАЛЛОВ ЗАВТРА КОНТРОЛЬНАЯ ПО АЛГЕБРЕ

Решите задачу:

$1)\; \; a=\frac{1}{2000}\; ,\; \; b=\frac{1}{1999}\\\\(a^{-1}+b^{-1})^2-4a^{-1}b^{-1}=(\frac{1}{a}+\frac{1}{b})^2-\frac{4}{ab}=\\\\=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}-\frac{4}{ab}=\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}=(\frac{1}{a}-\frac{1}{b})^2=\\\\=(\frac{1}{1/2000}-\frac{1}{1/1999})^2=(2000-1999)^2=1^2=1$

$2)\; \; ( \frac{x+y}{x-y} - \frac{x-y}{x+y} )^{-1}:( \frac{x^2-y^2}{2y} \cdot (2x)^{-1})=\\\\= (\frac{(x+y)^2-(x-y)^2}{(x-y)(x+y)} )^{-1}: \frac{(x-y)(x+y)}{2y\cdot 2x} =\\\\= (\frac{(x+y-x+y)(x+y+x-y)}{(x-y)(x+y)} )^{-1}\cdot \frac{4xy}{(x-y)(x+y)} = (\frac{2y\cdot 2x}{(x-y)(x+y)} )^{-1}\cdot \frac{4xy}{(x-y)(x+y)} =\\\\= \frac{(x-y)(x+y)}{4xy}\cdot \frac{4xy}{(x-y)(x+y)}=1$