Решение:

a)

$$\begin{aligned} \left(\frac{x}{x+1} + 1\right) \cdot \frac{1+x}{2x-1} &= \left(\frac{x + (x+1)}{x+1}\right) \cdot \frac{1+x}{2x-1} \\ &= \frac{2x+1}{x+1} \cdot \frac{x+1}{2x-1} \\ &= \frac{2x+1}{2x-1} \end{aligned}$$

Ответ: $$\frac{2x+1}{2x-1}$$

б)

$$\begin{aligned} \left(\frac{5y^2}{1-y^2}\right) : \left(1 - \frac{1}{1-y}\right) &= \frac{5y^2}{1-y^2} : \left(\frac{1-y-1}{1-y}\right) \\ &= \frac{5y^2}{1-y^2} : \frac{-y}{1-y} \\ &= \frac{5y^2}{(1-y)(1+y)} \cdot \frac{1-y}{-y} \\ &= \frac{5y^2 (1-y)}{-y (1-y)(1+y)} \\ &= -\frac{5y}{1+y} \end{aligned}$$

Ответ: $$\frac{-5y}{1+y}$$

в)

$$\begin{aligned} \left(\frac{4a}{2-a} - a\right) : \frac{a+2}{a-2} &= \left(\frac{4a - a(2-a)}{2-a}\right) : \frac{a+2}{a-2} \\ &= \frac{4a - 2a + a^2}{2-a} : \frac{a+2}{a-2} \\ &= \frac{a^2 + 2a}{2-a} : \frac{a+2}{a-2} \\ &= \frac{a(a+2)}{2-a} \cdot \frac{a-2}{a+2} \\ &= \frac{a(a+2)(-(2-a))}{(2-a)(a+2)} \\ &= -a \end{aligned}$$

Ответ: $$-a$$

г)

$$\begin{aligned} \frac{x-2}{x-3} \cdot \left(x + \frac{x}{2-x}\right) &= \frac{x-2}{x-3} \cdot \left(\frac{x(2-x) + x}{2-x}\right) \\ &= \frac{x-2}{x-3} \cdot \frac{2x - x^2 + x}{2-x} \\ &= \frac{x-2}{x-3} \cdot \frac{3x - x^2}{2-x} \\ &= \frac{(x-2)(x(3-x))}{-(x-3)(x-2)} \\ &= \frac{x(3-x)}{-(x-3)} \\ &= \frac{-x(x-3)}{-(x-3)} \\ &= x \end{aligned}$$

Ответ: $$x$$