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\(A\cap B=A\) ; \(B\cap C=B\)
\(\Rightarrow\left(A\cap B\right)\cup\left(B\cap C\right)=A\cup B=B\) (đáp án A đúng)
\(B\backslash C=\varnothing\Rightarrow A\cup\left(B\backslash C\right)=A\) (B cũng đúng)
\(A\backslash\left(B\cap C\right)=A\backslash B=\varnothing\) (C đúng)
Vậy D sai
\(\left(A\cap C\right)\cup B=A\cup B=B\) chứ ko phải C
E={0;1;2;3;4;5;6;7;8}
\(C_E^{A\cup B}=E\backslash\left(A\cup B\right)=E\backslash\left\{1;3;5;7;2;6\right\}=\left\{0;4\right\}\)
\(C_E^{A\cap B}=E\backslash\left\{1;3\right\}=\left\{0;2;4;5;6;7;8\right\}\)
=>\(C_E^{A\cup B}\subset C_E^{A\cap B}\)
Nguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh HằngRibi Nkok NgokMysterious PersonVõ Đông Anh TuấnPhương AnTrần Việt Linh
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
\(X=\left\{1;2;3;4;5;6;7;8;9\right\}\)
\(A\cap B=\left\{4;6;9\right\}\Rightarrow\left\{{}\begin{matrix}\left\{4;6;9\right\}\subset A\\\left\{4;6;9\right\}\subset B\end{matrix}\right.\)
\(A\cup\left\{3;4;5\right\}=\left\{1;3;4;5;6;8;9\right\}\Rightarrow\left\{1;4;6;8;9\right\}\subset A\)
\(B\cup\left\{4;8\right\}=\left\{2;3;4;5;6;7;8;9\right\}\Rightarrow\left\{2;3;4;5;6;7;9\right\}\subset B\)
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)
\(\Rightarrow\left\{{}\begin{matrix}1\notin B\\8\notin B\end{matrix}\right.\) \(\Rightarrow B=\left\{2;3;4;5;6;7;9\right\}\)
\(A=\left\{1;4;6;8;9\right\}\)
\(11-3x>0\Leftrightarrow x< \frac{11}{3}\Rightarrow A=\left\{0;1;2;3\right\}\)
\(B=\left\{-3;-2;-1;0;1;2;3\right\}\)
\(A\cup B=B=...\)
\(A\cap B=A=...\)
\(C_BA=\left\{-3;-2;-1\right\}\)
\(A\backslash B=\varnothing\)
\(B\backslash A=\left\{-3;-2;-1\right\}\)
\(X=A;\left\{-3;0;1;2;3\right\};\left\{-2;0;1;2;3\right\};\left\{-1;0;1;2;3\right\}\) ; \(\left\{-3;-2;0;1;2;3\right\};\left\{-3;-1;0;1;2;3\right\};\left\{-2;-1;0;1;2;3\right\};B\)
a/ \(\left\{1;2\right\};\left\{1;2;3\right\};\left\{1;2;4\right\};\left\{1;2;5\right\};\left\{1;2;3;4;5\right\}\)
b/ \(\left\{1;2;3;4\right\}\)