Đáp án:
$\begin{array}{l}
Dkxd:x \ne 0;x \ne \pm 2;x \ne 3\\
A = \left( {\dfrac{{2 + x}}{{2 - x}} + \dfrac{{4{x^2}}}{{4 - {x^2}}} - \dfrac{{2 - x}}{{2 + x}}} \right):\left( {\dfrac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}} \right)\\
= \dfrac{{{{\left( {2 + x} \right)}^2} + 4{x^2} - {{\left( {2 - x} \right)}^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}:\dfrac{{x\left( {x - 3} \right)}}{{{x^2}\left( {2 - x} \right)}}\\
= \dfrac{{4 + 4x + {x^2} + 4{x^2} - 4 + 4x - {x^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\dfrac{{{x^2}\left( {2 - x} \right)}}{{x\left( {x - 3} \right)}}\\
= \dfrac{{4{x^2} + 8x}}{{2 + x}}.\dfrac{x}{{x - 3}}\\
= \dfrac{{4x\left( {x + 2} \right)}}{{2 + x}}.\dfrac{x}{{x - 3}}\\
= \dfrac{{4{x^2}}}{{x - 3}}\\
2)\left| {x - 7} \right| = 4\\
\Rightarrow \left[ \begin{array}{l}
x - 7 = 4\\
x - 7 = - 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 11\left( {tm} \right)\\
x = 3\left( {ktm} \right)
\end{array} \right.\\
Khi:x = 11\\
\Rightarrow A = \dfrac{{4{x^2}}}{{x - 3}}\\
= \dfrac{{{{4.11}^2}}}{{11 - 3}} = \dfrac{{4.121}}{8} = \dfrac{{121}}{2}
\end{array}$
