Giải thích các bước giải:
\(\begin{array}{l}
1,\\
A = \left( {\frac{{2 + x}}{{2 - x}} + \frac{{4{x^2}}}{{{x^2} - 4}} - \frac{{2 - x}}{{2 + x}}} \right):\frac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}\\
a,\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
x \ne 3\\
x \ne \pm 2
\end{array} \right.\\
b,\\
A = \left( {\frac{{2 + x}}{{2 - x}} + \frac{{4{x^2}}}{{{x^2} - 4}} - \frac{{2 - x}}{{2 + x}}} \right):\frac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}\\
= \left( {\frac{{2 + x}}{{2 - x}} + \frac{{4{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \frac{{2 - x}}{{2 + x}}} \right):\frac{{x\left( {x - 3} \right)}}{{{x^2}\left( {2 - x} \right)}}\\
= \left( {\frac{{{{\left( {2 + x} \right)}^2} + 4{x^2} - {{\left( {2 - x} \right)}^2}}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}} \right):\frac{{\left( {x - 3} \right)}}{{x\left( {2 - x} \right)}}\\
= \frac{{{x^2} + 4x + 4 + 4{x^2} - {x^2} + 4x - 4}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}:\frac{{x - 3}}{{x\left( {2 - x} \right)}}\\
= \frac{{4{x^2} + 8x}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\frac{{x\left( {2 - x} \right)}}{{x - 3}}\\
= \frac{{4x\left( {x + 2} \right).x.\left( {2 - x} \right)}}{{\left( {2 - x} \right)\left( {2 + x} \right)\left( {x - 3} \right)}}\\
= \frac{{4{x^2}}}{{x - 3}}\\
c,\\
\left| {x - 5} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 2\\
x - 5 = - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = 3\,\,\,\,\,\left( L \right)
\end{array} \right.\\
x = 7 \Rightarrow A = \frac{{{{4.7}^2}}}{{7 - 3}} = 49\\
2,\\
\frac{{2 - x}}{{2002}} - 1 = \frac{{1 - x}}{{2003}} - \frac{x}{{2004}}\\
\Leftrightarrow \frac{{2 - x}}{{2002}} + 1 = \frac{{1 - x}}{{2003}} + 1 - \left( {\frac{x}{{2004}} - 1} \right)\\
\Leftrightarrow \frac{{2 - x + 2002}}{{2002}} = \frac{{1 - x + 2003}}{{2003}} - \frac{{x - 2004}}{{2004}}\\
\Leftrightarrow \frac{{2004 - x}}{{2002}} = \frac{{2004 - x}}{{2003}} - \frac{{x - 2004}}{{2004}}\\
\Leftrightarrow \left( {2004 - x} \right).\left( {\frac{1}{{2002}} - \frac{1}{{2003}} - \frac{1}{{2004}}} \right) = 0\\
\Leftrightarrow 2004 - x = 0\\
\Leftrightarrow x = 2004
\end{array}\)
