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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\)\(A=\left\{x\in R|x< 3\right\}\Rightarrow A=\left(\text{ -∞;3}\right)\)
\(B=\left\{-1;0;1;2;3;4;5\right\}\)
\(\Rightarrow A\cap B=\left\{-1;0;1;2\right\}\)
\(b,x=-1\Rightarrow y=1-2\left(-1\right)+m=m+3\)
\(x=1\Rightarrow y=1-2+m=m-1\)
\(\Rightarrow C=(m-1;m+3]\subset A\)
\(\Rightarrow C\subset A\Leftrightarrow m+3< 3\Leftrightarrow m< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left\{-3;-2;-1;0;1;2;3;4\right\}\)
Để \(B\cap C=\varnothing\Leftrightarrow a\in D\)
Với \(D=\left\{x\in Z;x\le-4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}$
$A=\left\{1; -4\right\}$
$B=\left\{-1; 2\right\}$
Do đó:
$A\cup B = \left\{-4; -1; 1;2\right\}$
$C_E(A\cup B)=\left\{-5;-3;-2; 0;3;4;5\right\}$
$A\cap B = \varnothing$
$C_E(A\cap B)=E$
![](https://rs.olm.vn/images/avt/0.png?1311)
1: A={-3;-2;-1;0;1;2;3}
B={2;-2;4;-4}
A giao B={2;-2}
A hợp B={-3;-2;-1;0;1;2;3;4;-4}
2: x thuộc A giao B
=>\(x=\left\{2;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A\cap B=\left(-3;1\right)\)
\(A\cup B\)=[-5;4]
A\B=[1;4]
\(C_RA\)=R\A=(-∞;-3]\(\cap\)(4;+∞)
b: C={1;-1;5;-5}
\(B\cap C=\left\{-5;-1\right\}\)
Các tập con là ∅; {-5}; {-1}; {-5;-1}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left\{x\in Z,x^2< 4\right\}\)
\(\Rightarrow A=\left\{-1;0;1\right\}\)
\(B=\left\{x\in Z,\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\}\)\(\Rightarrow\left[{}\begin{matrix}5x-3x^2=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{3}\left(loai\right)\\x=0\\x=3\\x=-1\end{matrix}\right.\)
\(\Rightarrow B=\left\{0;-1;3\right\}\)
\(\Rightarrow A\cap B=\left\{0;-1\right\}\) \(A\cup B=\left\{0;-1;1;3\right\}\)
\(A\backslash B=\left\{1\right\}\) \(B\backslash A=\left\{3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A=\left\{1;-4\right\}\)
\(B=\left\{2;-1\right\}\)
a) Với mọi x thuộc A đều thuộc E \(\Rightarrow A\subset E\)
Với mọi x thuộc B đều thuộc E \(\Rightarrow B\subset E\)
b) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)
\(A\cap B=\left\{{}\begin{matrix}x>m\\x\le\dfrac{2m-1}{3}\end{matrix}\right.\left(1\right)\)
\(TH1:m< \dfrac{2m-1}{3}\)
\(\Leftrightarrow m-\dfrac{2m-1}{3}< 0\)
\(\Leftrightarrow\dfrac{m-1}{3}< 0\)
\(\Leftrightarrow m< 1\)
\(\left(1\right)\Leftrightarrow A\cap B=\left\{x\in Z|m< x\le\dfrac{2m-1}{3}\right\}\)
\(TH2:m>\dfrac{2m-1}{3}\)
\(\Leftrightarrow m-\dfrac{2m-1}{3}>0\)
\(\Leftrightarrow\dfrac{m-1}{3}>0\)
\(\Leftrightarrow m>1\)
\(\left(1\right)\Leftrightarrow A\cap B=\varnothing\)
nếu thế thì thừa TH1 nhỉ?