Đáp án:
$\begin{array}{l}
a)Dkxd:{x_0};x \ne 1\\
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}}}:\dfrac{{\left( {x + 1} \right).\left( {x - 1} \right) + x + 2 - {x^2}}}{{x\left( {x - 1} \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^2}.\left( {x + 1} \right)}}{{\left( {x - 1} \right).\left( {x + 1} \right)}}\\
= \dfrac{{{x^2}}}{{x - 1}}\\
b)\left| {2x - 5} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
2x - 5 = 3\\
2x - 5 = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
x = 1\left( {ktm} \right)
\end{array} \right.\\
Thay\,x = 4\\
\Rightarrow A = \dfrac{{{x^2}}}{{x - 1}} = \dfrac{{16}}{3}\\
c)A = 4\\
\Rightarrow \dfrac{{{x^2}}}{{x - 1}} = 4\\
\Rightarrow {x^2} = 4x - 4\\
\Rightarrow {x^2} - 4x + 4 = 0\\
\Rightarrow {\left( {x - 2} \right)^2} = 0\\
\Rightarrow x = 2\left( {tmdk} \right)\\
d)A < 2\\
\Rightarrow \dfrac{{{x^2}}}{{x - 1}} < 2\\
\Rightarrow \dfrac{{{x^2} - 2x + 2}}{{x - 1}} < 0\\
\Rightarrow x - 1 < 0\left( {do:{x^2} - 2x + 2 \ge 1} \right)\\
\Rightarrow x < 1\\
Vay\,x < 1;x \ne 0\\
2)\\
a)Dkxd:x \ne 2;x \ne - 2;x \ne 0\\
A = \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)}^2} + 4{x^2} - {{\left( {2 - x} \right)}^2}}}{{\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} - 4 - {x^2} + 4x}}{{2 + x}}.\dfrac{x}{{x - 3}}\\
= \dfrac{{4{x^2} + 8x}}{{2 + x}}.\dfrac{x}{{x - 3}}\\
= \dfrac{{4{x^2}}}{{x - 3}}\\
b)\left| {x - 5} \right| = 2\\
\Rightarrow \left[ \begin{array}{l}
x - 5 = 2\\
x - 5 = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 7\left( {tm} \right)\\
x = 3\left( {ktm} \right)
\end{array} \right.\\
Thay\,x = 7\\
\Rightarrow A = \dfrac{{4{x^2}}}{{x - 3}} = \dfrac{{{{4.7}^2}}}{{7 - 3}} = 49
\end{array}$
