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E={0;1;2;3;4;5;6;7;8}
\(C_E^{A\cup B}=E\backslash\left(A\cup B\right)=E\backslash\left\{1;3;5;7;2;6\right\}=\left\{0;4\right\}\)
\(C_E^{A\cap B}=E\backslash\left\{1;3\right\}=\left\{0;2;4;5;6;7;8\right\}\)
=>\(C_E^{A\cup B}\subset C_E^{A\cap B}\)
\(E = \{ x \in \mathbb{N}|x < 8\} = \{ 0;1;2;3;4;5;6;7\} \)
a) Ta có: \(A\backslash B = \left\{ {0;1;2} \right\}\), \(B\backslash A = \left\{ 5 \right\},\)\((A\backslash B) \cap {\rm{(}}B\backslash A) = \emptyset \)
b) Ta có: \(A \cap B = \{ 3;4\} ,\;{C_E}(A \cap B) = \{ 0;1;2;5;6;7\} \)
\({C_E}A = \{ 5;6;7\} ,\;{C_E}B = \{ 0;1;2;6;7\} \Rightarrow ({C_E}A) \cap ({C_E}B) = \{ 6;7\} \)
c) Ta có: \(A \cup B = \{ 0;1;2;3;4;5\} ,\;{C_E}(A \cup B) = \{ 6;7\} \)
\({C_E}A = \{ 5;6;7\} ,\;{C_E}B = \{ 0;1;2;6;7\} \Rightarrow ({C_E}A) \cup ({C_E}B) = \{ 0;1;2;5;6;7\} \)
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
\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A=\left\{1;-4\right\}\)
\(B=\left\{2;-1\right\}\)
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) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)