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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left\{-2;-1;0;1;2;3;4\right\}\)
\(B=\varnothing\)
\(C=\left\{0;1;2;3;4;5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-3;5\right)\cap\left(2;4\right)=\left(-3;5\right)\)
\((-\infty;3]\cap\left[3;5\right]=(-\infty;5]\)
\(\left(-4;2\right)\cap[2;5)=\left(-4;5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A\cap B=\)[\(1;2\)) \(\cup\) (\(3;5\)]
b) \(A\cap B=\)\(\left(-1;0\right)\cup\left(4;5\right)\))
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (\(-\infty;0\)] \(\cup\left[1;2\right]\cup\) [\(3;+\infty\))
b) (\(-\infty;4\)] \(\cup\) [\(5;+\infty\))
c) \(\left(-2;1\right)\cup\left(3;7\right)\)
d) (\(-1;1\)] \(\cup\) [\(4;5\))
![](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)
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)
\(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)
\(\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)=\left(-2;7\right)\)
\(\left(0;12\right)\cap[5;+\infty)=[5;12)\)
\(R\cap\left[-1;1\right]=\left[-1;1\right]\)
a) \(\left\{-2;-1;0;1;2;3;4;5\right\}\)
b) \(\varnothing\)
c) \(\left\{2\right\}\)
d) \(\left\{0;1;2;3;4;5\right\}\)
là sao