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![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(A\cap B=\left[1;4\right]\); \(A\cap B=\left[-4;7\right]\); \(A\ B=[-4;1)\)
b/\(A\cap B=\varnothing\) ; \(A\cup B=\left[-4;-2\right]\cup\left(3;7\right)\) ; \(A\ B=A\)
c/ \(A\cap B=\varnothing\) ; \(A\cup B=(-\infty;-2]\cup[3;+\infty)\)
d/ \(A\cap B=[3;4)\) ; \(A\cap B=\left(0;+\infty\right)\); \(A\backslash B=[4;+\infty)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\cap B=\left[-1;3\right]\\ A\cup B=\left(-\infty;5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ A = (3;\(+\infty\)), B=[0;4]
A \(\cap\) B= (3;4)
A\(\cup\) B=[0;+\(\infty\))
A\B= (4;\(+\infty\))
B\A= [0;3]
b/ A=(\(-\infty\);4], B=(2;\(+\infty\))
A\(\cap\)B=(2;4]
A\(\cup\)B= R
A\B= (\(-\infty\);2]
B\A=(4;\(+\infty\))
c/ A=[0;4] , B=(\(-\infty\);2]
A\(\cap\)B= [0;2)
\(A\cup B\) = (\(-\infty\);4]
A\ B=[2;4]
B\A=(\(-\infty\);0)
![](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)
Dễ thấy nếu \(A\cap B=\varnothing\Rightarrow A\in[-3;3)\Rightarrow\left\{{}\begin{matrix}m-1\ge-3\\\dfrac{m+3}{2}< 3\end{matrix}\right.\)
\(\Leftrightarrow-2\le m< 3\)
Do đó để \(A\cap B\ne\varnothing\Rightarrow m\notin[-2;3)\Rightarrow\left[{}\begin{matrix}m< -2\\m\ge3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A\cap B=\varnothing\)
\(A\cup B=\left[-2;7\right]\)
A\B=[-2;0]
B\A=[1;7]
Lời giải:
a)
\(A\cap B=[1;4]\)
\(A\cup B=[-4;7]\)
\(A\setminus B=[-4;1); B\setminus A=(4;7]\)
b)
\(A\cap B=(0;4)\)
\(A\cup B=[-3;\infty)\)
\(A\setminus B=(-3;0]\cup [4;\infty)\)
\(B\setminus A=\varnothing\)