№ 3. Упростить выражение: а) \(\left(\frac{2x}{2x+y}-\frac{4x^2}{4x^2+4xy+y^2}\right)\div\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\); б) \(\left(\frac{m}{m-6}-\frac{2m}{m^2-12m+36}\right)\cdot\frac{36-m^2}{m-8}+\frac{12m}{m-6}\).

а)

Так как \(4x^2+4xy+y^2=(2x+y)^2\), то

\[\frac{2x}{2x+y}-\frac{4x^2}{(2x+y)^2}=\frac{2x(2x+y)-4x^2}{(2x+y)^2}=\frac{2xy}{(2x+y)^2}.\]

Далее:

\[\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}=\frac{2x}{(2x-y)(2x+y)}-\frac{1}{2x-y}\]

\[=\frac{2x-(2x+y)}{(2x-y)(2x+y)}=-\frac{y}{(2x-y)(2x+y)}.\]

Поэтому

\[\frac{2xy}{(2x+y)^2}\div\left(-\frac{y}{(2x-y)(2x+y)}\right)=\frac{2x(y-2x)}{2x+y}.\]

б)

\[\left(\frac{m}{m-6}-\frac{2m}{(m-6)^2}\right)\cdot\frac{36-m^2}{m-8}+\frac{12m}{m-6}\]

\[=\frac{m(m-6)-2m}{(m-6)^2}\cdot\frac{(6-m)(6+m)}{m-8}+\frac{12m}{m-6}\]

\[=\frac{m(m-8)}{(m-6)^2}\cdot\frac{36-m^2}{m-8}+\frac{12m}{m-6}\]

\[=\frac{m(36-m^2)}{(m-6)^2}+\frac{12m(m-6)}{(m-6)^2}\]

\[=\frac{m(36-m^2+12m-72)}{(m-6)^2}=\frac{-m(m-6)^2}{(m-6)^2}=-m.\]

Ответ: а) \(\frac{2x(y-2x)}{2x+y}\); б) \(-m\).

Ограничения: а) \(2x+y\ne0\), \(2x-y\ne0\), \(y-2x\ne0\), \(y\ne0\); б) \(m\ne6\), \(m\ne8\).