Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x\# 2\\
x\# - 2\\
x\# 0\\
x\# 3
\end{array} \right.\\
A = \dfrac{{2 + x}}{{2 - x}} - \dfrac{{4{x^2}}}{{{x^2} - 4}} - \dfrac{{2 - x}}{{2 + x}}\\
= \dfrac{{\left( {2 + x} \right)\left( {2 + x} \right) + 4{x^2} - \left( {2 - x} \right)\left( {2 - x} \right)}}{{\left( {2 + x} \right)\left( {2 - x} \right)}}\\
= \dfrac{{{x^2} + 4x + 4 + 4{x^2} - 4 + 4x - {x^2}}}{{\left( {2 + x} \right)\left( {2 - x} \right)}}\\
= \dfrac{{4{x^2} + 8x}}{{\left( {2 + x} \right)\left( {2 - x} \right)}}\\
= \dfrac{{4x\left( {x + 2} \right)}}{{\left( {2 + x} \right)\left( {2 - x} \right)}}\\
= \dfrac{{4x}}{{2 - x}}\\
B = \dfrac{{{x^2} - 3x}}{{2{x^2} - {x^3}}} = \dfrac{{x\left( {x - 3} \right)}}{{{x^2}\left( {2 - x} \right)}}\\
= \dfrac{{x - 3}}{{x\left( {2 - x} \right)}}\\
P = A.\dfrac{1}{B}\\
= \dfrac{{4x}}{{2 - x}}.\dfrac{{x\left( {2 - x} \right)}}{{x - 3}} = \dfrac{{4{x^2}}}{{x - 3}}\\
b)P \vdots 4\\
\Leftrightarrow \dfrac{{4{x^2}}}{{x - 3}} \vdots 4\\
\Leftrightarrow 4.\dfrac{{{x^2}}}{{x - 3}} \vdots 4\\
\Leftrightarrow \dfrac{{{x^2}}}{{x - 3}} \in Z\\
\Leftrightarrow \dfrac{{{x^2} - 9 + 9}}{{x - 3}} \in Z\\
\Leftrightarrow x + 3 + \dfrac{9}{{x - 3}} \in Z\\
\Leftrightarrow \dfrac{9}{{x - 3}} \in Z\\
\Leftrightarrow \left( {x - 3} \right) \in \left\{ { - 9; - 3; - 1;1;3;9} \right\}\\
\Leftrightarrow x \in \left\{ { - 6;0;2;4;6;12} \right\}\\
Do:x\# 0;x\# 2\\
\Leftrightarrow x \in \left\{ { - 6;4;6;12} \right\}\\
Vay\,x \in \left\{ { - 6;4;6;12} \right\}
\end{array}$