Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x \ne \pm 2\\
x \ne 1
\end{array} \right.\\
A = \left( {\frac{{2 + x}}{{2 - x}} - \frac{{4{x^2}}}{{{x^2} - 4}} - \frac{{2 - x}}{{2 + x}}} \right).\left( {\frac{{6 - x - {x^2}}}{{4\left( {{x^2} - 2x + 1} \right)}}} \right)\\
= \left( {\frac{{{{\left( {2 + x} \right)}^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}} - \frac{{4{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \frac{{{{\left( {2 - x} \right)}^2}}}{{\left( {2 + x} \right)\left( {2 - x} \right)}}} \right).\frac{{\left( {2 - x} \right)\left( {x + 3} \right)}}{{4{{\left( {x - 1} \right)}^2}}}\\
= \frac{{{{\left( {2 + x} \right)}^2} + 4{x^2} - {{\left( {2 - x} \right)}^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\frac{{\left( {2 - x} \right)\left( {x + 3} \right)}}{{4{{\left( {x - 1} \right)}^2}}}\\
= \frac{{{x^2} + 4x + 4 + 4{x^2} - 4 + 4x - {x^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\frac{{\left( {2 - x} \right)\left( {x + 3} \right)}}{{4{{\left( {x - 1} \right)}^2}}}\\
= \frac{{4{x^2} + 8x}}{{\left( {2 + x} \right)}}.\frac{{x + 3}}{{4{{\left( {x - 1} \right)}^2}}}\\
= \frac{{4x\left( {x + 2} \right)\left( {x + 3} \right)}}{{\left( {x + 2} \right).4.{{\left( {x - 1} \right)}^2}}}\\
= \frac{{x\left( {x + 3} \right)}}{{{{\left( {x - 1} \right)}^2}}}
\end{array}\)
