Cho 2 tập hợp A=(2m-1;m+3), B=(-4;5) Tìm m để a) A con B b) B con A c) A giao B = giỗng d) A hợp B là 1 khoảng 18/07/2021 Bởi Genesis Cho 2 tập hợp A=(2m-1;m+3), B=(-4;5) Tìm m để a) A con B b) B con A c) A giao B = giỗng d) A hợp B là 1 khoảng
Đáp án: $\begin{array}{l}A = \left( {2m – 1;m + 3} \right);\\B = \left( { – 4;5} \right)\\a)A \subset B\\ \Rightarrow \left( {2m – 1;m + 3} \right) \subset \left( { – 4;5} \right)\\ \Rightarrow – 4 \le 2m – 1 < m + 3 \le 5\\ \Rightarrow \left\{ \begin{array}{l}2m – 1 \ge – 4\\2m – 1 < m + 3\\m + 3 \le 5\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}m \ge – \dfrac{3}{2}\\m < 4\\m \le 2\end{array} \right.\\ \Rightarrow – \dfrac{3}{2} \le m \le 2\\b)B \subset A\\ \Rightarrow \left( { – 4;5} \right) \subset \left( {2m – 1;m + 3} \right)\\ \Rightarrow 2m – 1 \le – 4 < 5 \le m + 3\\ \Rightarrow \left\{ \begin{array}{l}2m – 1 \le – 4\\5 \le m + 3\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}m \le – \dfrac{3}{2}\\m \ge 2\end{array} \right.\left( {ktm} \right)\\ \Rightarrow m \in \emptyset \\c)A \cap B = \emptyset \\ \Rightarrow \left[ \begin{array}{l}m + 3 < – 4\\5 < 2m – 1\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}m < – 7\\m > 3\end{array} \right.\\Vay\,m < – 7\,hoac\,m > 3\\d)\exists A \cap B\\ \Rightarrow \left[ \begin{array}{l}2m – 1 \le – 4 < m + 3 \le 5\\ – 4 \le 2m – 1 < 5 \le m + 3\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}m \le – \dfrac{3}{2}\\m > – 7\\m \le 2\end{array} \right.\\\left\{ \begin{array}{l}m \ge – \dfrac{3}{2}\\m < 3\\m \ge 2\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l} – 7 < m \le \dfrac{3}{2}\\2 \le m < 3\end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A = \left( {2m – 1;m + 3} \right);\\
B = \left( { – 4;5} \right)\\
a)A \subset B\\
\Rightarrow \left( {2m – 1;m + 3} \right) \subset \left( { – 4;5} \right)\\
\Rightarrow – 4 \le 2m – 1 < m + 3 \le 5\\
\Rightarrow \left\{ \begin{array}{l}
2m – 1 \ge – 4\\
2m – 1 < m + 3\\
m + 3 \le 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ge – \dfrac{3}{2}\\
m < 4\\
m \le 2
\end{array} \right.\\
\Rightarrow – \dfrac{3}{2} \le m \le 2\\
b)B \subset A\\
\Rightarrow \left( { – 4;5} \right) \subset \left( {2m – 1;m + 3} \right)\\
\Rightarrow 2m – 1 \le – 4 < 5 \le m + 3\\
\Rightarrow \left\{ \begin{array}{l}
2m – 1 \le – 4\\
5 \le m + 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \le – \dfrac{3}{2}\\
m \ge 2
\end{array} \right.\left( {ktm} \right)\\
\Rightarrow m \in \emptyset \\
c)A \cap B = \emptyset \\
\Rightarrow \left[ \begin{array}{l}
m + 3 < – 4\\
5 < 2m – 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m < – 7\\
m > 3
\end{array} \right.\\
Vay\,m < – 7\,hoac\,m > 3\\
d)\exists A \cap B\\
\Rightarrow \left[ \begin{array}{l}
2m – 1 \le – 4 < m + 3 \le 5\\
– 4 \le 2m – 1 < 5 \le m + 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \le – \dfrac{3}{2}\\
m > – 7\\
m \le 2
\end{array} \right.\\
\left\{ \begin{array}{l}
m \ge – \dfrac{3}{2}\\
m < 3\\
m \ge 2
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
– 7 < m \le \dfrac{3}{2}\\
2 \le m < 3
\end{array} \right.
\end{array}$