Đáp án:
$\begin{array}{l}
a)\left| {2 - 3x} \right| \le \left| {{x^2} - 2} \right|\\
\Rightarrow \left[ \begin{array}{l}
2 - 3x \le {x^2} - 2\\
2 - 3x \ge 2 - {x^2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 3x - 4 \ge 0\\
{x^2} - 3x \ge 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x + 4} \right) \ge 0\\
x\left( {x - 3} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le - 4\\
x \ge 3\\
x \le 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 4
\end{array} \right.\\
Vậy\,x \le - 4\,hoặc\,x \ge 3\\
b)\sqrt {{x^2} - 3x + 2} \ge \sqrt {2 - 5x} \\
\Rightarrow \left\{ \begin{array}{l}
{x^2} - 3x + 2 \ge 0\\
2 - 5x \ge 0\\
{x^2} - 3x + 2 \ge 2 - 5x
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 2} \right) \ge 0\\
x \le \dfrac{2}{5}\\
{x^2} + 2x \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
x \le \dfrac{2}{5}\\
\left[ \begin{array}{l}
x \ge 0\\
x \le - 2
\end{array} \right.
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \le \dfrac{2}{5}\\
\left[ \begin{array}{l}
x \ge 0\\
x \le - 2
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \le - 2\\
0 \le x \le \dfrac{2}{5}
\end{array} \right.\\
Vậy\,x \le - 2\,hoặc\,0 \le x \le \dfrac{2}{5}
\end{array}$
