Đáp án:
\(x \in \left( { - \infty ; - \dfrac{1}{2}} \right] \cup \left[ {3; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:2{x^2} - 3x - 2 \ge 0\\
\to \left( {x - 2} \right)\left( {2x + 1} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 2\\
x \le - \dfrac{1}{2}
\end{array} \right.\left( 1 \right)\\
\left( {{x^2} - 3x} \right)\sqrt {2{x^2} - 3x - 2} \ge 0\\
\to {x^2} - 3x \ge 0\\
Do:\sqrt {2{x^2} - 3x - 2} \ge 0\forall x \in \left( { - \infty ; - \dfrac{1}{2}} \right] \cup \left[ {2; + \infty } \right)\\
\to x\left( {x - 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 3\\
x \le 0
\end{array} \right.\left( 2 \right)\\
\to x \in \left( { - \infty ;0} \right] \cup \left[ {3; + \infty } \right)\\
\left( 1 \right);\left( 2 \right) \to x \in \left( { - \infty ; - \dfrac{1}{2}} \right] \cup \left[ {3; + \infty } \right)
\end{array}\)
