a) $M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}$

$\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}$

$\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}$

$\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}$

$\Leftrightarrow M=\dfrac{2}{1-2x}$

b) $M=\dfrac{2}{1-2x}=\dfrac{-2}{3}$

$\Rightarrow2.3=\left(1-2x\right).\left(-2\right)$

$\Rightarrow6=-2+4x$

$\Rightarrow4x=6-\left(-2\right)$

$\Rightarrow4x=6+2$

$\Rightarrow4x=8$

$\Rightarrow x=8:4$

$\Rightarrow x=2$

Vậy $M=\dfrac{-2}{3}$ thì $x=2$

c) Để $M=\dfrac{2}{1-2x}\in Z$ $\Leftrightarrow2⋮1-2x$

$\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}$

$\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.$

Mà $x\in Z$

$\Rightarrow x\in\left\{1;0\right\}$

Vậy $x=1$ hoặc $x=0$ thì $M\in Z$