Question

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

Answer 1

$\bigg( \dfrac{2x+y}{2 x^{2} y- xy^{2} } - \dfrac{2}{ y^{2}+2xy} \bigg) : \dfrac{(6x+ y) ^{2} }{ 4x^{3}- y^{2} x} =\\\\\\=\bigg( \dfrac{2x+y}{xy(2x-y)} - \dfrac{2}{y(2x+y)} \bigg)\cdot \dfrac{x(4x^2-y^2)}{(6x+y)^2} =\\\\\\= \dfrac{((2x+y)\cdot(2x+y)-2x\cdot (2x-y))}{xy(2x-y)(2x+y)} \cdot \dfrac{x(2x-y)(2x+y)}{(6x+y)^2}=\\\\\\= \dfrac{4x^2+4xy+y^2-4x^2+2xy}{y(6x+y)^2}= \dfrac{y(6x+y)}{y(6x+y)^2} = \dfrac{1}{6x+y}$