Ответ:
а)
\[\left(\frac{x}{x+1}+1\right)\cdot\frac{x+1}{2x-1}=\frac{x+x+1}{x+1}\cdot\frac{x+1}{2x-1}=\frac{2x+1}{2x-1}.\]
ОДЗ: \(x\ne-1\), \(x\ne\frac12\).
б)
\[\frac{5y^2}{1-y^2}:\left(1-\frac1{1-y}\right)=\frac{5y^2}{(1-y)(1+y)}:\frac{-y}{1-y}.\]
\[\frac{5y^2}{(1-y)(1+y)}\cdot\frac{1-y}{-y}=-\frac{5y}{1+y}.\]
ОДЗ: \(y\ne-1\), \(y\ne0\), \(y\ne1\).
в)
\[\left(\frac{4a}{2-a}-a\right):\frac{a+2}{a-2}=\frac{4a-a(2-a)}{2-a}\cdot\frac{a-2}{a+2}.\]
\[\frac{a^2+2a}{2-a}\cdot\frac{a-2}{a+2}=\frac{a(a+2)}{-(a-2)}\cdot\frac{a-2}{a+2}=-a.\]
ОДЗ: \(a\ne2\), \(a\ne-2\).
г)
\[\frac{x-y}{x}-\frac{5y}{x^2}\cdot\frac{x^2-xy}{5y}=\frac{x-y}{x}-\frac{5y}{x^2}\cdot\frac{x(x-y)}{5y}.\]
\[\frac{x-y}{x}-\frac{x-y}{x}=0.\]
ОДЗ: \(x\ne0\), \(y\ne0\).
Ответ: а) \(\frac{2x+1}{2x-1}\); б) \(-\frac{5y}{1+y}\); в) \(-a\); г) \(0\).
