Đáp án:

$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ne 1\\
x \ne  - 1\\
x \ne 2
\end{array} \right.\\
A = \dfrac{{{x^2} - x}}{{1 - {x^2}}} = \dfrac{{x\left( {x - 1} \right)}}{{\left( {1 - x} \right)\left( {x + 1} \right)}} = \dfrac{{ - x}}{{x + 1}}\\
B = \dfrac{{1 - 2x}}{{{x^2} - 3x + 2}} + \dfrac{{x + 1}}{{x - 2}}\\
 = \dfrac{{1 - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{x - 2}}\\
 = \dfrac{{1 - 2x + \left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{1 - 2x + {x^2} - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{{x^2} - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{x\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \dfrac{x}{{x - 1}}\\
b)\left| {x - 2} \right| = 3\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 3 \Leftrightarrow x = 5\left( {tm} \right)\\
x - 2 =  - 3 \Leftrightarrow x =  - 1\left( {ktm} \right)
\end{array} \right.\\
 + Khi:x = 5\\
 \Leftrightarrow A = \dfrac{{ - x}}{{x + 1}} = \dfrac{{ - 5}}{{5 + 1}} = \dfrac{{ - 5}}{6}\\
c)C = A - B\\
 = \dfrac{{ - x}}{{x + 1}} - \dfrac{x}{{x - 1}}\\
 = \dfrac{{ - x\left( {x - 1} \right) - x\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{ - {x^2} + x - {x^2} - x}}{{{x^2} - 1}}\\
 = \dfrac{{ - 2{x^2}}}{{{x^2} - 1}}\\
d)C = \dfrac{{ - 2{x^2}}}{{{x^2} - 1}} = \dfrac{{ - 2{x^2} + 2 - 2}}{{{x^2} - 1}}\\
 = \dfrac{{ - 2\left( {{x^2} - 1} \right) - 2}}{{{x^2} - 1}}\\
 =  - 2 - \dfrac{2}{{{x^2} - 1}}\\
C \in Z\\
 \Leftrightarrow \dfrac{2}{{{x^2} - 1}} \in Z\\
 \Leftrightarrow \left( {{x^2} - 1} \right) \in \left\{ { - 1;1;2} \right\}\\
 \Leftrightarrow {x^2} \in \left\{ {0;2;3} \right\}\\
 \Leftrightarrow x \in \left\{ {0;\sqrt 2 ; - \sqrt 2 ;\sqrt 3 ; - \sqrt 3 } \right\}\\
Do:x \in Z\\
 \Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0
\end{array}$