$a)$ Ta có: $A=\dfrac{3x-2}{x}-\dfrac{x-7}{x-5}-\dfrac{10}{x^2-5x}$ ($x\ne0$; $x\ne5$)

$=\dfrac{3x-2}{x}-\dfrac{x-7}{x-5}-\dfrac{10}{x(x-5)}$

$=\dfrac{(3x-2)(x-5)-x(x-7)-10}{x(x-5)}$

$=\dfrac{3x^2-15x-2x+10-x^2+7x-10}{x(x-5)}$

$=\dfrac{2x^2-10x}{x(x-5)}$

$=\dfrac{2x(x-5)}{x(x-5)}$

$=2$

Vậy $A=2$ với $x\ne0$; $x\ne5$

$b)$ Ta có: $B=A.\dfrac{x+1}{x-1}$ ($x\ne0$; $x\ne1$; $x\ne5$)

$\Rightarrow$ $B=2.\dfrac{x+1}{x-1}=\dfrac{2x+2}{x-1}=\dfrac{2(x-1)+4}{x-1}=2+\dfrac{4}{x-1}$

Vì $x$ $\in$ $\mathbb{Z}$ $\Rightarrow$ $x+1$ $\in$ $\mathbb{Z}$

$\Rightarrow$ $B$ $\in$ $\mathbb{Z}$ khi $\dfrac{4}{x-1}$ $\in$ $\mathbb{Z}$

Tức là $4$ $\vdots$ $x-1$ $\Rightarrow$ $x-1$ $\in$ `Ư_4={-4; -2; -1; 1; 2; 4}`

$\Rightarrow$ $x$ $\in$ `{-3; -1; 0; 2; 3; 5}`

Mà $x\ne0$; $x\ne1$; $x\ne5$; $x$ $\in$ $\mathbb{Z}$

$\Rightarrow$ $x$ $\in$ `{-3; -1; 2; 3}`

Vậy $x$ $\in$ `{-3; -1; 2; 3}` thì $B$ có giá trị nguyên