Đáp án:
a) $A = -\dfrac{2}{x+1}$
b) $x \in\{-3;-2;0\}$
Giải thích các bước giải:
$\begin{array}{l}\quad A =\left(\dfrac{1}{x-1} - \dfrac{1}{x+1}\right):\left(1 +\dfrac{x}{1-x}\right)\qquad (x \ne \pm1)\\ a)\quad A =\dfrac{x+1 - (x-1)}{(x-1)(x+1)}:\dfrac{1-x+x}{1-x}\\ \to A = \dfrac{2}{(x-1)(x+1)}:\dfrac{1}{1-x}\\ \to A = \dfrac{2}{(x-1)(x+1)}\cdot(1-x)\\ \to A = \dfrac{-2}{x+1}\\ b)\quad A \in \Bbb Z\\ \to -\dfrac{2}{x+1}\in \Bbb Z\\ \to x+1 \in Ư(2)=\{-2;-1;1;2\}\\ \to x \in \{-3;-2;0;1\}\\ mà\,\,x \ne \pm 1\\ nên \,\,x \in\{-3;-2;0\} \end{array}$
