Đáp án: $x \in \left( { - \dfrac{3}{2}; - \dfrac{1}{2}} \right] \cup \left[ {2; + \infty } \right)$
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{2{x^2} - 3x - 2}}{{2x + 3}} \ge 0\\
\Rightarrow \dfrac{{\left( {2x + 1} \right)\left( {x - 2} \right)}}{{\left( {2x + 3} \right)}} \ge 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left( {2x + 1} \right)\left( {x - 2} \right) \ge 0\\
2x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\left( {2x + 1} \right)\left( {x - 2} \right) \le 0\\
2x + 3 < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 2/x \le - \dfrac{1}{2}\\
x > - \dfrac{3}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
- \dfrac{1}{2} \le x \le 2\\
x < - \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 2\\
- \dfrac{3}{2} < x \le \dfrac{{ - 1}}{2}
\end{array} \right.
\end{array}$
Vậy $x \in \left( { - \dfrac{3}{2}; - \dfrac{1}{2}} \right] \cup \left[ {2; + \infty } \right)$
