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