Question

Данные выражение : 4ху/у^2-х^2 : (1/у^2-х^2 + 1 / х^2+2ху + у ^ 2 ) 2 ) ( х-2у/х^2-2ху - 1/х^2-4у^2: х+2у/(2у-х)^2)*(х+2у)^2/4у^2

Answer 1

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

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

$=\frac{(x+2y)^2-x(x-2y)}{x(x+2y)^2} \frac{(x+2y)^2}{4y^2}=\frac{x^2+4xy+4y^2-x^2+2xy}{4xy^2}=\frac{6xy+4y^2}{4xy^2}=\frac{3x+2y}{2xy}$