Giải thích các bước giải:

1.ĐKXĐ: $x\ne 1$

Ta có:

$A=(\dfrac{x-1}{3x+(x-1)^2}-\dfrac{1-3x+x^2}{x^3-1}-\dfrac1{1-x}):\dfrac{x^2+2x+1}{x-1}$

$\to A=(\dfrac{x-1}{x^2+x+1}-\dfrac{1-3x+x^2}{(x-1)(x^2+x+1)}+\dfrac1{x-1})\cdot \dfrac{x-1}{x^2+2x+1}$

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

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

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

$\to A=\dfrac{1}{x^2+x+1}$

2.Để $A\in Z$

$\to \dfrac1{x^2+x+1}\in Z$

Mà $x^2+x+1=(x+\dfrac12)^2+\dfrac34>0$

$\to 0<\dfrac1{x^2+x+1}\le \dfrac1{0+\dfrac34}$

$\to 0<A<\dfrac43$

Do $A\in Z$

$\to A=1$

$\to \dfrac2A=2\in Z$

$\to A=1$ Chọn

$\to \dfrac1{x^2+x+1}=1$

$\to x^2+x+1=1$

$\to x^2+x=0$

$\to x(x+1)=0$

$\to x\in\{0; -1\}$