Đáp án:
$\begin{array}{l}
Dkxd:x \ne 0;x \ne 1;x \ne - 1\\
P\left( x \right) = \dfrac{{{x^2} + x}}{{{x^2} - 2x + 1}}:\left( {\dfrac{{x + 1}}{x} + \dfrac{1}{{x - 1}} + \dfrac{{2 - {x^2}}}{{{x^2} - x}}} \right)\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}:\left( {\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + x + 2 - {x^2}}}{{x\left( {x - 1} \right)}}} \right)\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{x\left( {x - 1} \right)}}{{{x^2} - 1 + x + 2 - {x^2}}}\\
= \dfrac{{x\left( {x + 1} \right)}}{{x - 1}}.\dfrac{x}{{x + 1}}\\
= \dfrac{{{x^2}}}{{x - 1}}\\
P = \dfrac{{{x^2}}}{{x - 1}}\\
\Rightarrow {x^2} = P.x - P\\
\Rightarrow {x^2} - P.x + P = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta \ge 0\\
P \ne 0\\
{\left( { - 1} \right)^2} - P.\left( { - 1} \right) + P \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{P^2} - 4P \ge 0\\
P \ne 0\\
P \ne - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
P\left( {P - 4} \right) \ge 0\\
P \ne 0\\
P \ne - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
P \ge 4\\
P < 0
\end{array} \right.\\
\Rightarrow P \ge 4\\
\Rightarrow GTNN:P = 4 \Leftrightarrow x = 2
\end{array}$
