giải giùm m cau a …………………………………………………………………. October 19, 2020 by Calantha giải giùm m cau a ………………………………………………………………….
Đáp án: $\begin{array}{l}A = \left\{ {x \in R/\dfrac{{x + 2}}{{x + 5}} \le 0} \right\}\\\dfrac{{x + 2}}{{x + 5}} \le 0\\ \Rightarrow – 5 < x \le – 2\\Do:x \in R\\ \Rightarrow A = \left\{ { – 4; – 3; – 2} \right\}\\B = \left\{ {x \in R/\left( {3x + 2} \right)\left( {x – 1} \right) > 0} \right\}\\\left( {3x + 2} \right)\left( {x – 1} \right) > 0\\ \Rightarrow \left[ \begin{array}{l}x > 1\\x < – \dfrac{2}{3}\end{array} \right.\\ \Rightarrow B = R/\left\{ 0 \right\}\end{array}$ Reply
Đáp án:
$\begin{array}{l}
A = \left\{ {x \in R/\dfrac{{x + 2}}{{x + 5}} \le 0} \right\}\\
\dfrac{{x + 2}}{{x + 5}} \le 0\\
\Rightarrow – 5 < x \le – 2\\
Do:x \in R\\
\Rightarrow A = \left\{ { – 4; – 3; – 2} \right\}\\
B = \left\{ {x \in R/\left( {3x + 2} \right)\left( {x – 1} \right) > 0} \right\}\\
\left( {3x + 2} \right)\left( {x – 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 1\\
x < – \dfrac{2}{3}
\end{array} \right.\\
\Rightarrow B = R/\left\{ 0 \right\}
\end{array}$