Đáp án:
$\begin{array}{l}
A = \left\{ {x \in R/\dfrac{{x + 2}}{{x + 5}} \le 0} \right\}\\
\dfrac{{x + 2}}{{x + 5}} \le 0\\
\Rightarrow - 5 < x \le - 2\\
Do:x \in R\\
\Rightarrow A = \left\{ { - 4; - 3; - 2} \right\}\\
B = \left\{ {x \in R/\left( {3x + 2} \right)\left( {x - 1} \right) > 0} \right\}\\
\left( {3x + 2} \right)\left( {x - 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 1\\
x < - \dfrac{2}{3}
\end{array} \right.\\
\Rightarrow B = R/\left\{ 0 \right\}
\end{array}$
