Ta có: A = (2; 5] = {x \(\in\) R | 2 < x \(\le\) 5}
B = (m - 1; m + 3) = {x \(\in\) R | m - 1 < x < m + 3}
a, A \(\subset\) B \(\Leftrightarrow\) (\(\forall\)x : x \(\in\) A \(\Rightarrow\) x \(\in\) B)
\(\Rightarrow\) m - 1 < 2 < x \(\le\) 5 < m + 3
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}m-1< 2\\m+3>5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}m< 3\\m>2\end{matrix}\right.\)
\(\Leftrightarrow\) 2 < m < 3 (TM)
Vậy 2 < m < 3
b, B \(\subset\) A \(\Leftrightarrow\) (\(\forall\)x : x \(\in\) B \(\Rightarrow\) x \(\in\) A)
\(\Rightarrow\) 2 < m - 1 < x < m + 3 \(\le\) 5
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2< m-1\\m+3\le5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}m>3\\m\le2\end{matrix}\right.\)
\(\Leftrightarrow\) 2 \(\ge\) m > 3 (KTM)
Vậy m \(\in\) \(\varnothing\)
c, Ta có: A \(\cap\) B = \(\varnothing\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}m-1>5\\m+3\le2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}m>6\\m\le-1\end{matrix}\right.\)
Vậy m > 6 hoặc m \(\le\) -1
d, A \(\cup\) B là một khoảng \(\Leftrightarrow\) A \(\cap\) B = \(\varnothing\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}m>6\\m\le-1\end{matrix}\right.\)
Vậy ...
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