Đáp án: $x \ge 2\,hoặc\,x \le 0$
Giải thích các bước giải:
$\begin{array}{l}
Dk{\rm{xd}}:x\# 1\\
f\left( x \right) = \dfrac{{1 - 3x + {x^2}}}{{x - 1}}\\
\Leftrightarrow f'\left( x \right) = \dfrac{{\left( {2x - 3} \right).\left( {x - 1} \right) - 1 + 3x - {x^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{2{x^2} - 5x + 3 - 1 + 3x - {x^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{{x^2} - 2x + 2}}{{{{\left( {x - 1} \right)}^2}}}\\
Khi:f'\left( x \right) \le 2\\
\Leftrightarrow \dfrac{{{x^2} - 2x + 2}}{{{{\left( {x - 1} \right)}^2}}} \le 2\\
\Leftrightarrow \dfrac{{{x^2} - 2x + 2 - 2{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}} \le 0\\
\Leftrightarrow \dfrac{{ - {x^2} + 2x}}{{{{\left( {x - 1} \right)}^2}}} \le 0\\
\Leftrightarrow - {x^2} + 2x \le 0\\
\Leftrightarrow {x^2} - 2x \ge 0\\
\Leftrightarrow x\left( {x - 2} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.\\
Vậy\,x \ge 2\,hoặc\,x \le 0
\end{array}$
