Đáp án:

$\begin{array}{l}
A = \dfrac{{{x^2} + x}}{{{x^2} - 2x + 1}}:\left( {\dfrac{{x + 1}}{x} - \dfrac{1}{{1 - x}} + \dfrac{{2 - {x^2}}}{{{x^2} - x}}} \right)\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}:\left( {\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + x + 2 - {x^2}}}{{x\left( {x - 1} \right)}}} \right)\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{x\left( {x - 1} \right)}}{{{x^2} - 1 + x + 2 - {x^2}}}\\
 = \dfrac{{x\left( {x + 1} \right)}}{{x - 1}}.\dfrac{x}{{1 + x}}\\
 = \dfrac{{{x^2}}}{{x - 1}}\\
B = \left( {\dfrac{{2 + x}}{{2 - x}} - \dfrac{{4{x^2}}}{{{x^2} - 4}} - \dfrac{{2 - x}}{{2 + x}}} \right):\dfrac{{{x^2} - 3x}}{{2{x^2} - {x^3}}}\\
 = \dfrac{{\left( {2 + x} \right)\left( {2 + x} \right) + 4{x^2} - \left( {2 - x} \right)\left( {2 - x} \right)}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}.\dfrac{{{x^2}\left( {2 - x} \right)}}{{x\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} + 4x + 4 + 4{x^2} - {x^2} + 4x - 4}}{{2 + x}}.\dfrac{x}{{x - 3}}\\
 = \dfrac{{4{x^2} + 8x}}{{2 + x}}.\dfrac{x}{{x - 3}}\\
 = \dfrac{{4{x^2}}}{{x - 3}}\\
C = \left( {\dfrac{{2x}}{{2{x^2} - 5x + 3}} - \dfrac{5}{{2x - 3}}} \right):\left( {3 + \dfrac{2}{{1 - x}}} \right)\\
 = \dfrac{{2x - 5\left( {x - 1} \right)}}{{\left( {2x - 3} \right)\left( {x - 1} \right)}}:\dfrac{{3\left( {x - 1} \right) - 2}}{{x - 1}}\\
 = \dfrac{{2x - 5x + 5}}{{\left( {2x - 3} \right)\left( {x - 1} \right)}}.\dfrac{{x - 1}}{{3x - 3 - 2}}\\
 = \dfrac{{ - 3x + 5}}{{2x - 3}}.\dfrac{1}{{3x - 5}}\\
 = \dfrac{1}{{3 - 2x}}\\
D = \left( {\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}}} \right):\dfrac{{2x}}{{5x - 5}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{5\left( {x - 1} \right)}}{{2x}}\\
 = \dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1}}{{x + 1}}.\dfrac{5}{{2x}}\\
 = \dfrac{{4x}}{{x + 1}}.\dfrac{5}{{2x}}\\
 = \dfrac{{10}}{{x + 1}}
\end{array}$

Đáp án:

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