Đáp án:
$\begin{array}{l}
Dkxd:x \ne 0;x \ne 2;x \ne - 2;x \ne 1\\
a)\\
C = \left( {\dfrac{{x + 2}}{{2 - x}} + \dfrac{{4{x^2}}}{{4 - {x^2}}} - \dfrac{{2 - x}}{{x + 2}}} \right):\dfrac{{{x^2} - x}}{{2x - {x^2}}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x + 2} \right) + 4{x^2} - \left( {2 - x} \right)\left( {2 - x} \right)}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}\\
.\dfrac{{x\left( {2 - x} \right)}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{{x^2} + 4x + 4 + 4{x^2} - 4 + 4x - {x^2}}}{{2 + x}}.\dfrac{1}{{x - 1}}\\
= \dfrac{{4{x^2} + 8x}}{{2 + x}}.\dfrac{1}{{x - 1}}\\
= \dfrac{{4x\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}\\
= \dfrac{{4x}}{{x - 1}}\\
b)C < 4\\
\Leftrightarrow \dfrac{{4x}}{{x - 1}} - 4 < 0\\
\Leftrightarrow \dfrac{{4x - 4\left( {x - 1} \right)}}{{x - 1}} < 0\\
\Leftrightarrow \dfrac{{4x - 4x + 4}}{{x - 1}} < 0\\
\Leftrightarrow \dfrac{4}{{x - 1}} < 0\\
\Leftrightarrow x - 1 < 0\\
\Leftrightarrow x < 1\\
Vậy\,x < 1;x \ne 0;x \ne - 2\\
c)C = \dfrac{{4x}}{{x - 1}} = \dfrac{{4x - 4 + 4}}{{x - 1}}\\
= 4 + \dfrac{4}{{x - 1}} \in Z\\
\Leftrightarrow \left( {x - 1} \right) \in \left\{ { - 4; - 2; - 1;1;2;4} \right\}\\
\Leftrightarrow x \in \left\{ { - 3; - 1;0;2;3;5} \right\}\\
Do:x \ne 0;x \ne 1;x \ne 2;x \ne - 2\\
\Leftrightarrow x \in \left\{ { - 3; - 1;3;5} \right\}\\
Vậy\,x \in \left\{ { - 3; - 1;3;5} \right\}
\end{array}$
