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![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A\subset B\Leftrightarrow\left\{{}\begin{matrix}m+3\ge5\\2m-1< -4\end{matrix}\right.\Rightarrow m\in\left\{\varnothing\right\}\)
b, \(B\subset A\Leftrightarrow\left\{{}\begin{matrix}m+3\le5\\2m-1>-4\end{matrix}\right.\Leftrightarrow-\dfrac{3}{2}< m\le2\)
c, \(A\cap B=\varnothing\Leftrightarrow\left[{}\begin{matrix}2m-1>5\\m+3\le-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>3\\m\le-7\end{matrix}\right.\)
d, \(A\cup B\) là một khoảng \(\Leftrightarrow\left\{{}\begin{matrix}m+3>5\\2m-1\le5\end{matrix}\right.\Leftrightarrow2< m\le3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(-3;-1\right)\cup\left(1;2\right)\)
\(B=\left(-1;+\infty\right)\)
\(C=\left(-\infty;2m\right)\)
\(A\cap B=\left(-3;-1\right)\)
Để \(A\cap B\cap C\ne\varnothing\Leftrightarrow2m\ge-1\)
\(\Leftrightarrow m\ge-\dfrac{1}{2}\)
Vậy \(m\ge-\dfrac{1}{2}\) thỏa đề bài
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\cap B\ne\varnothing\Leftrightarrow\left[{}\begin{matrix}m+1< 2m-1< m+3\\m+1< 2m< m+3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2< m< 4\\1< m< 3\end{matrix}\right.\) \(\Rightarrow1< m< 4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A\cap B\cap C=A\cap (B\cap C)$
Để tập hợp trên khác rỗng thì trước hết $B\cap C\neq \varnothing$
Điều này xảy ra khi $2m>m\Leftrightarrow m>0$
Khi đó: $B\cap C=(m; 2m)$
$\Rightarrow A\cap B\cap C=((-3;-1)\cup (1;2))\cap (m; 2m)$
$=((-3;-1)\cap (m;2m))\cup ((1;2)\cap (m; 2m))$
$=(1;2)\cap (m; 2m)$ (do $m>0$)
Để $(1;2)\cap (m; 2m)\neq \varnothing$ thì:
\(\left\{\begin{matrix} 2m>1\\ m< 2\end{matrix}\right.\Leftrightarrow m\in (\frac{1}{2};2)\)
Vậy...........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\cap B=\varnothing\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2m-3>m+1\\m+1\ge-1\\2m-3\le3\end{matrix}\right.\\\left\{{}\begin{matrix}2m-3>m+1\\m+1\ge5\\\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m>4\\m\ge-2\\m\le3\end{matrix}\right.\\\left\{{}\begin{matrix}m>4\\m\ge4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\cap B=\left\{{}\begin{matrix}x>m\\x\le\dfrac{2m-1}{3}\end{matrix}\right.\left(1\right)\)
\(TH1:m< \dfrac{2m-1}{3}\)
\(\Leftrightarrow m-\dfrac{2m-1}{3}< 0\)
\(\Leftrightarrow\dfrac{m-1}{3}< 0\)
\(\Leftrightarrow m< 1\)
\(\left(1\right)\Leftrightarrow A\cap B=\left\{x\in Z|m< x\le\dfrac{2m-1}{3}\right\}\)
\(TH2:m>\dfrac{2m-1}{3}\)
\(\Leftrightarrow m-\dfrac{2m-1}{3}>0\)
\(\Leftrightarrow\dfrac{m-1}{3}>0\)
\(\Leftrightarrow m>1\)
\(\left(1\right)\Leftrightarrow A\cap B=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) để \(A\subset B\Leftrightarrow\left\{{}\begin{matrix}2m-1\ge-4\\m+3\le5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ge\dfrac{-3}{2}\\m\le2\end{matrix}\right.\Leftrightarrow\dfrac{-3}{2}\le m\le2\)
b) để \(B\subset A\Leftrightarrow\left\{{}\begin{matrix}2m-1\le-4\\m+3\ge5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{-3}{2}\\m\ge2\end{matrix}\right.\Rightarrow m\in\varnothing\)
c) để \(A\cap B=\varnothing\Leftrightarrow\left[{}\begin{matrix}m+3< 4\\5< 2m-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m< 1\\m>3\end{matrix}\right.\)
\(\Rightarrow m\in\left(-\infty;1\right)\cup\left(3;+\infty\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A\cap B=\varnothing\)
\(\Rightarrow\left[{}\begin{matrix}5-2m< -5\\1-2m\ge3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2m< -10\\-2m\ge2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m>5\\m\le-1\end{matrix}\right.\)
Vậy: ...
Bài 1:
Để A giao B bằng rỗng thì \(\left[{}\begin{matrix}m+3< -3\\2m-1>6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m< -6\\m>\dfrac{7}{2}\end{matrix}\right.\)