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\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A=\left\{1;-4\right\}\)
\(B=\left\{2;-1\right\}\)
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) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)
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
=>\(x=\left\{2;-2\right\}\)
\(a,\)\(A=\left\{x\in R|x< 3\right\}\Rightarrow A=\left(\text{ -∞;3}\right)\)
\(B=\left\{-1;0;1;2;3;4;5\right\}\)
\(\Rightarrow A\cap B=\left\{-1;0;1;2\right\}\)
\(b,x=-1\Rightarrow y=1-2\left(-1\right)+m=m+3\)
\(x=1\Rightarrow y=1-2+m=m-1\)
\(\Rightarrow C=(m-1;m+3]\subset A\)
\(\Rightarrow C\subset A\Leftrightarrow m+3< 3\Leftrightarrow m< 0\)
\(X=\left\{1;2;3;4;5;6;7;8;9\right\}\)
\(A\cap B=\left\{4;6;9\right\}\Rightarrow\left\{{}\begin{matrix}\left\{4;6;9\right\}\subset A\\\left\{4;6;9\right\}\subset B\end{matrix}\right.\)
\(A\cup\left\{3;4;5\right\}=\left\{1;3;4;5;6;8;9\right\}\Rightarrow\left\{1;4;6;8;9\right\}\subset A\)
\(B\cup\left\{4;8\right\}=\left\{2;3;4;5;6;7;8;9\right\}\Rightarrow\left\{2;3;4;5;6;7;9\right\}\subset B\)
Nếu \(\left[{}\begin{matrix}1\in B\\8\in B\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\in A\cap B\\8\in A\cap B\end{matrix}\right.\) (ktm)
\(\Rightarrow\left\{{}\begin{matrix}1\notin B\\8\notin B\end{matrix}\right.\) \(\Rightarrow B=\left\{2;3;4;5;6;7;9\right\}\)
\(A=\left\{1;4;6;8;9\right\}\)
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a, \(X\in\left\{a;b\right\},\left\{a;b;c\right\},\left\{a;b;d\right\},\left\{a;b;e\right\},\left\{a;c;d\right\},\left\{a;c;e\right\},\left\{a;d;e\right\},\left\{a;b;c;d\right\},\left\{a;b;c;e\right\},\left\{a;c;d;e\right\},\left\{a;b;c;d;e\right\}\)
b,
\(X=\left\{3;4;5\right\}\)
c,đề có sai hay sao ý ạ
\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)
Giải phương trình sau :
\(\left(x-2x^2\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow x\left(1-2x\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-2x=0\\x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\\x=2\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)
\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)
Giải bất phương trình sau :
\(3< n\left(n+1\right)< 31\)
\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)
Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)
\(\Rightarrow A\cap B=\left\{2\right\}\)
\(11-3x>0\Leftrightarrow x< \frac{11}{3}\Rightarrow A=\left\{0;1;2;3\right\}\)
\(B=\left\{-3;-2;-1;0;1;2;3\right\}\)
\(A\cup B=B=...\)
\(A\cap B=A=...\)
\(C_BA=\left\{-3;-2;-1\right\}\)
\(A\backslash B=\varnothing\)
\(B\backslash A=\left\{-3;-2;-1\right\}\)
\(X=A;\left\{-3;0;1;2;3\right\};\left\{-2;0;1;2;3\right\};\left\{-1;0;1;2;3\right\}\) ; \(\left\{-3;-2;0;1;2;3\right\};\left\{-3;-1;0;1;2;3\right\};\left\{-2;-1;0;1;2;3\right\};B\)