Question

СРОЧНО ДАЮ 80 БАЛЛОВ
ЗАВТРА КОНТРОЛЬНАЯ ПО АЛГЕБРЕ

Answer 1

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

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