giải giùm m cau a …………………………………………………………………. Question giải giùm m cau a …………………………………………………………………. in progress 0 Môn Toán Calantha 1 year 2020-10-19T00:36:21+00:00 2020-10-19T00:36:21+00:00 1 Answers 72 views 0
Answers ( )
Đáp án:
$\begin{array}{l}
A = \left\{ {x \in R/\dfrac{{x + 2}}{{x + 5}} \le 0} \right\}\\
\dfrac{{x + 2}}{{x + 5}} \le 0\\
\Rightarrow – 5 < x \le – 2\\
Do:x \in R\\
\Rightarrow A = \left\{ { – 4; – 3; – 2} \right\}\\
B = \left\{ {x \in R/\left( {3x + 2} \right)\left( {x – 1} \right) > 0} \right\}\\
\left( {3x + 2} \right)\left( {x – 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 1\\
x < – \dfrac{2}{3}
\end{array} \right.\\
\Rightarrow B = R/\left\{ 0 \right\}
\end{array}$