Question

решите пожалуйста по буквой б))

Answer 1

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

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

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

$= \frac{y-2x-2x}{2x^2y^2} = \frac{y-4x}{2x^2y^2}$