a/ ĐKXĐ: $x\ne 3$
$\dfrac{x-3}{x+1}=-2$
$↔x-3=-2x-2$
$↔x-3+2x+2=0$
$↔3x=1$
$↔x=\dfrac{1}{3}$
b/ $\dfrac{x+1}{x-3}=0$
$↔x+1=0$
$↔x=-1$
c/ $\dfrac{x-2}{x-1}=-2$
$↔x-2=-2x+2$
$↔x-2+2x-2=0$
$↔3x=4$
$↔x=\dfrac{4}{3}$
d/ $A-B$
$=\dfrac{x+1}{x-3}-\dfrac{x-2}{x-1}$
$=\dfrac{(x+1)(x-1)}{(x-3)(x-1)}-\dfrac{(x-2)(x-3)}{(x-1)(x-3)}$
$=\dfrac{x²-1-x²+5x-6}{(x-1)(x-3)}$
$=\đfrac{5x-7}{x²-4x+3}$
e/ $B\in\mathbb{Z}$
$↔\dfrac{x-2}{x-1}\in\mathbb{Z}$
$↔\dfrac{x-1-1}{x-1}\in\mathbb Z$
$↔1-\dfrac{1}{x-1}\in\mathbb Z$
$↔1\vdots x-1$
$↔x-1in Ư(1)=\{±1\}$
$↔x\in \{2;0\}$