Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
{x^2} - 2x + 1 \ne 0\\
x \ne 0\\
x - 1 \ne 0\\
{x^2} - x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne 0
\end{array} \right.\\
P = \dfrac{{{x^2} + x}}{{{x^2} - 2x + 1}}:\left( {\dfrac{{x + 1}}{x} + \dfrac{1}{{x - 1}} + \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\left( {x + 1} \right)}}{{x - 1}}.\dfrac{x}{{x + 1}}\\
= \dfrac{{{x^2}}}{{x - 1}}\\
b)P = \dfrac{{ - 1}}{2}\\
\Leftrightarrow \dfrac{{{x^2}}}{{x - 1}} = - \dfrac{1}{2}\\
\Leftrightarrow 2{x^2} = 1 - x\\
\Leftrightarrow 2{x^2} + x - 1 = 0\\
\Leftrightarrow 2{x^2} + 2x - x - 1 = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\left( {tm} \right)\\
x = \dfrac{1}{2}\left( {tm} \right)
\end{array} \right.\\
Vay\,x = - 1;x = \dfrac{1}{2}\\
c)P = \dfrac{{{x^2}}}{{x - 1}} = \dfrac{{{x^2} - 1 + 1}}{{x - 1}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + 1}}{{x - 1}}\\
= x + 1 + \dfrac{1}{{x - 1}} \in Z\\
\Leftrightarrow \dfrac{1}{{x - 1}} \in Z\\
\Leftrightarrow \left( {x - 1} \right) \in \left\{ { - 1;1} \right\}\\
\Leftrightarrow x \in \left\{ {0;2} \right\}\\
Do:x \ne 0\\
\Leftrightarrow x = 2\\
Vay\,x = 2\\
d)x > 1\\
\Leftrightarrow x - 1 > 0\\
P = x + 1 + \dfrac{1}{{x - 1}}\\
= x - 1 + \dfrac{1}{{x - 1}} + 2\\
Theo\,Co - si:\\
\left( {x - 1} \right) + \dfrac{1}{{x - 1}} \ge 2\sqrt {\left( {x - 1} \right).\dfrac{1}{{x - 1}}} = 2\\
\Leftrightarrow \left( {x - 1} \right) + \dfrac{1}{{x - 1}} + 2 \ge 2 + 2\\
\Leftrightarrow P \ge 4\\
\Leftrightarrow GTNN:P = 4\\
Khi:\left( {x - 1} \right) = \dfrac{1}{{x - 1}}\\
\Leftrightarrow x - 1 = 1\\
\Leftrightarrow x = 2\\
Vay\,GTNN:P = 4\,khi:x = 2
\end{array}$
