Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\rm{DK:}}\,\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
x \ne \pm 1
\end{array} \right.\\
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{{x + 1}}{x} + \dfrac{1}{{x - 1}} + \dfrac{{2 - {x^2}}}{{x\left( {x - 1} \right)}}} \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^2} - 1 + x + 2 - {x^2}}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}:\dfrac{{x + 1}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{x.\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{x.\left( {x - 1} \right)}}{{x + 1}}\\
= \dfrac{{{x^2}}}{{x - 1}}\\
b,\\
\left| {2x - 5} \right| = 3 \Leftrightarrow \left[ \begin{array}{l}
2x - 5 = 3\\
2x - 5 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 1
\end{array} \right.\\
x \ne 1 \Rightarrow x = 4 \Rightarrow A = \dfrac{{{4^2}}}{{4 - 1}} = \dfrac{{16}}{3}\\
c,\\
A = 4 \Leftrightarrow \dfrac{{{x^2}}}{{x - 1}} = 4 \Leftrightarrow {x^2} = 4x - 4 \Leftrightarrow {x^2} - 4x + 4 = 0 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
d,\\
A < 2 \Leftrightarrow \dfrac{{{x^2}}}{{x - 1}} < 2\\
\Leftrightarrow \dfrac{{{x^2}}}{{x - 1}} - 2 < 0\\
\Leftrightarrow \dfrac{{{x^2} - 2\left( {x - 1} \right)}}{{x - 1}} < 0\\
\Leftrightarrow \dfrac{{{x^2} - 2x + 2}}{{x - 1}} < 0\\
{x^2} - 2x + 2 = {\left( {x - 1} \right)^2} + 1 \ge 1 > 0,\,\,\forall x\\
\Rightarrow x - 1 < 0 \Leftrightarrow \left\{ \begin{array}{l}
x < 1\\
x \ne 0\\
x \ne - 1
\end{array} \right.\\
e,\\
A = \dfrac{{{x^2}}}{{x - 1}} = \dfrac{{\left( {{x^2} - 1} \right) + 1}}{{x - 1}} = \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 1}}{{x - 1}} = x + 1 + \dfrac{1}{{x - 1}}\\
A \in Z \Leftrightarrow \dfrac{1}{{x - 1}} \in Z \Leftrightarrow x - 1 \in \left\{ { - 1;1} \right\} \Rightarrow x \in \left\{ {0;2} \right\}\\
x \ne 0 \Rightarrow x = 2
\end{array}\)
