Đáp án: $\left[ \begin{array}{l}
x \ge \dfrac{{ - 3 + \sqrt {105} }}{6}\\
\dfrac{{ - 3 - \sqrt {105} }}{6} \le x <  - 2
\end{array} \right.$

Giải thích các bước giải:

$\begin{array}{l}
Dkxd:x\# 1\\
\dfrac{2}{{{x^3} - 3x + 2}} \ge \dfrac{3}{{x - 1}}\\
 \Leftrightarrow \dfrac{2}{{{x^3} - {x^2} + {x^2} - x - 2x + 2}} - \dfrac{3}{{x - 1}} \ge 0\\
 \Leftrightarrow \dfrac{2}{{\left( {x - 1} \right)\left( {{x^2} + x - 2} \right)}} - \dfrac{3}{{x - 1}} \ge 0\\
 \Leftrightarrow \dfrac{2}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}} - \dfrac{3}{{x - 1}} \ge 0\\
 \Leftrightarrow \dfrac{{2 - 3\left( {x - 1} \right)\left( {x + 2} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}} \ge 0\\
 \Leftrightarrow \dfrac{{2 - 3{x^2} - 3x + 6}}{{\left( {x + 2} \right)}} \ge 0\left( {do:{{\left( {x - 1} \right)}^2} > 0} \right)\\
 \Leftrightarrow \dfrac{{3{x^2} + 3x - 8}}{{x + 2}} \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3{x^2} + 3x - 8 \ge 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3{x^2} + 3x - 8 \le 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{{ - 3 + \sqrt {105} }}{6}\\
x \le \dfrac{{ - 3 - \sqrt {105} }}{6}
\end{array} \right.\\
x >  - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{{ - 3 - \sqrt {105} }}{6} \le x \le \dfrac{{ - 3 + \sqrt {105} }}{6}\\
x <  - 2
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{{ - 3 + \sqrt {105} }}{6}\\
\dfrac{{ - 3 - \sqrt {105} }}{6} \le x <  - 2
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x \ge \dfrac{{ - 3 + \sqrt {105} }}{6}\\
\dfrac{{ - 3 - \sqrt {105} }}{6} \le x <  - 2
\end{array} \right.
\end{array}$