Giải thích các bước giải:

a.ĐKXĐ: $x\ne \pm1$

Ta có:

$A=\dfrac{x+1}{2x-2}+\dfrac3{x^2-1}-\dfrac{x+3}{2x+2}$

$\to A=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\dfrac{6}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}$

$\to A=\dfrac{\left(x+1\right)^2+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}$

$\to A=\dfrac{10}{2\left(x-1\right)\left(x+1\right)}$

$\to A=\dfrac{5}{\left(x-1\right)\left(x+1\right)}$

b.(Thêm đk: $x\in Z$)

Ta có:

$M=AB$

$\to M=\dfrac{5}{\left(x-1\right)\left(x+1\right)}\cdot \dfrac{2x^2+3x+1}5$

$\to M=\dfrac{2x^2+3x+1}{(x+1)(x-1)}$

$\to M=\dfrac{(2x+1)(x+1)}{(x+1)(x-1)}$

$\to M=\dfrac{2x+1}{x-1}$

Để $M\in Z$

$\to 2x+1\quad\vdots\quad x-1$

$\to 2(x-1)+3\quad\vdots\quad x-1$

$\to 3\quad\vdots\quad x-1$

$\to x-1\in U(3)$

$\to x-1\in\{1, 3, -1, -3\}$

$\to x\in\{2, 4, 0, -2\}$