Đáp án:

$\begin{array}{l}
a)Dkxd:x \ne 1;x \ne  - 1;x \ne 2\\
A = \frac{{2{x^2} + 2x}}{{1 - {x^2}}} = \frac{{2x\left( {x + 1} \right)}}{{\left( {1 - x} \right)\left( {x + 1} \right)}} = \frac{{2x}}{{1 - x}}\\
B = \frac{{1 - 2x}}{{{x^2} - 3x + 2}} + \frac{{x + 1}}{{x - 2}}\\
 = \frac{{1 - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \frac{{x + 1}}{{x - 2}}\\
 = \frac{{1 - 2x + \left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{{1 - 2x + {x^2} - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{{{x^2} - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{{x\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{x}{{x - 1}}\\
b)\left| {x - 2} \right| = 3\\
 \Rightarrow \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 =  - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 5\left( {tmdk} \right)\\
x =  - 1\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow A = \frac{{2x}}{{1 - x}} = \frac{{2.5}}{{1 - 5}} = \frac{{ - 10}}{4} = \frac{{ - 5}}{2}\\
c)C = A - B\\
 = \frac{{2x}}{{1 - x}} - \frac{x}{{x - 1}}\\
 = \frac{{2x + x}}{{1 - x}}\\
 = \frac{{3x}}{{1 - x}}\\
d)C = \frac{{3x}}{{1 - x}} = \frac{{ - 3 + 3x + 3}}{{1 - x}} =  - 3 + \frac{3}{{1 - x}} =  - 3 - \frac{3}{{x - 1}}\\
C \in Z\\
 \Rightarrow \frac{3}{{x - 1}} \in Z\\
 \Rightarrow \left( {x - 1} \right) \in U\left( 3 \right)\\
 \Rightarrow \left( {x - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow x \in \left\{ { - 2;0;2;4} \right\}\\
Do:x \ne 2\\
 \Rightarrow a)Dkxd:x \ne 1;x \ne  - 1;x \ne 2\\
A = \frac{{2{x^2} + 2x}}{{1 - {x^2}}} = \frac{{2x\left( {x + 1} \right)}}{{\left( {1 - x} \right)\left( {x + 1} \right)}} = \frac{{2x}}{{1 - x}}\\
B = \frac{{1 - 2x}}{{{x^2} - 3x + 2}} + \frac{{x + 1}}{{x - 2}}\\
 = \frac{{1 - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \frac{{x + 1}}{{x - 2}}\\
 = \frac{{1 - 2x + \left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{{1 - 2x + {x^2} - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{{{x^2} - 2x}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{{x\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \frac{x}{{x - 1}}\\
b)\left| {x - 2} \right| = 3\\
 \Rightarrow \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 =  - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 5\left( {tmdk} \right)\\
x =  - 1\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow A = \frac{{2x}}{{1 - x}} = \frac{{2.5}}{{1 - 5}} = \frac{{ - 10}}{4} = \frac{{ - 5}}{2}\\
c)C = A - B\\
 = \frac{{2x}}{{1 - x}} - \frac{x}{{x - 1}}\\
 = \frac{{2x + x}}{{1 - x}}\\
 = \frac{{3x}}{{1 - x}}\\
d)C = \frac{{3x}}{{1 - x}} = \frac{{ - 3 + 3x + 3}}{{1 - x}} =  - 3 + \frac{3}{{1 - x}} =  - 3 - \frac{3}{{x - 1}}\\
C \in Z\\
 \Rightarrow \frac{3}{{x - 1}} \in Z\\
 \Rightarrow \left( {x - 1} \right) \in U\left( 3 \right)\\
 \Rightarrow \left( {x - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow x \in \left\{ { - 2;0;2;4} \right\}
\end{array}$

Mà x khác 2

Vậy $x \in \left\{ { - 2;0;4} \right\}$