`ĐK: m – 1 < (m + 1)/2` `<=> m < 3` `a)` Để `A ⊂ B` `<=>` \(\left[ \begin{array}{l}\dfrac{m + 1}{2} < -2\\m – 1 ≥ 2\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}m < -5\\m ≥ 3 (l)\end{array} \right.\) `=> m ∈ (-∞; -5)` `b)` Để `A ∩ B = ∅` `<=>` \(\left\{ \begin{array}{l}m – 1 ≥ -2\\\dfrac{m + 1}{2} < 2\end{array} \right.\) `<=>` \(\left\{ \begin{array}{l}m ≥ -1\\m < 3\end{array} \right.\) `=> -1 ≤ m < 3` `=> m ∈ [-1; 3)` Reply
Đáp án: $\begin{array}{l}A = \left[ {m – 1;\dfrac{{m + 1}}{2}} \right]\\Dk:\dfrac{{m + 1}}{2} \ge m – 1\\ \Rightarrow m + 1 \ge 2m – 1\\ \Rightarrow m \le 2\\B = \left( { – \infty ; – 2} \right) \cup \left[ {2; + \infty } \right)\\a)A \subset B\\ \Rightarrow \left[ \begin{array}{l}\left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \subset \left( { – \infty ; – 2} \right)\\\left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \subset \left[ {2; + \infty } \right)\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}\dfrac{{m + 1}}{2} < – 2\\m – 1 \ge 2\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}m < – 5\\m \ge 3\end{array} \right.\\Ket\,hop:m \le 2\\ \Rightarrow m < – 5\\Vay\,m < – 5\\b)A \cap B = \emptyset \\ \Rightarrow \left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \cap \left( { – \infty ; – 2} \right) \cup \left[ {2; + \infty } \right) = \emptyset \\ \Rightarrow \left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \subset \left[ { – 2;2} \right)\\ \Rightarrow – 2 \le m – 1 \le \dfrac{{m + 1}}{2} < 2\\ \Rightarrow \left\{ \begin{array}{l}m \ge – 1\\m < 3\end{array} \right.\\ \Rightarrow – 1 \le m < 3\\Ket\,hop:m \le 2\\Vay\, – 1 \le m \le 2\end{array}$ Reply
`ĐK: m – 1 < (m + 1)/2`
`<=> m < 3`
`a)` Để `A ⊂ B`
`<=>` \(\left[ \begin{array}{l}\dfrac{m + 1}{2} < -2\\m – 1 ≥ 2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}m < -5\\m ≥ 3 (l)\end{array} \right.\)
`=> m ∈ (-∞; -5)`
`b)` Để `A ∩ B = ∅`
`<=>` \(\left\{ \begin{array}{l}m – 1 ≥ -2\\\dfrac{m + 1}{2} < 2\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}m ≥ -1\\m < 3\end{array} \right.\)
`=> -1 ≤ m < 3`
`=> m ∈ [-1; 3)`
Đáp án:
$\begin{array}{l}
A = \left[ {m – 1;\dfrac{{m + 1}}{2}} \right]\\
Dk:\dfrac{{m + 1}}{2} \ge m – 1\\
\Rightarrow m + 1 \ge 2m – 1\\
\Rightarrow m \le 2\\
B = \left( { – \infty ; – 2} \right) \cup \left[ {2; + \infty } \right)\\
a)A \subset B\\
\Rightarrow \left[ \begin{array}{l}
\left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \subset \left( { – \infty ; – 2} \right)\\
\left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \subset \left[ {2; + \infty } \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{m + 1}}{2} < – 2\\
m – 1 \ge 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m < – 5\\
m \ge 3
\end{array} \right.\\
Ket\,hop:m \le 2\\
\Rightarrow m < – 5\\
Vay\,m < – 5\\
b)A \cap B = \emptyset \\
\Rightarrow \left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \cap \left( { – \infty ; – 2} \right) \cup \left[ {2; + \infty } \right) = \emptyset \\
\Rightarrow \left[ {m – 1;\dfrac{{m + 1}}{2}} \right] \subset \left[ { – 2;2} \right)\\
\Rightarrow – 2 \le m – 1 \le \dfrac{{m + 1}}{2} < 2\\
\Rightarrow \left\{ \begin{array}{l}
m \ge – 1\\
m < 3
\end{array} \right.\\
\Rightarrow – 1 \le m < 3\\
Ket\,hop:m \le 2\\
Vay\, – 1 \le m \le 2
\end{array}$