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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) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)
Phương trình \({x^2} - 5x - 6 = 0\) có hai nghiệm là -1 và 6, nên \(A = \{ - 1;6\} \)
Phương trình \({x^2} = 1\) có hai nghiệm là 1 và -1, nên \(B = \{ - 1;1\} \)
Do đó
\(\begin{array}{l}A \cap B = \{ - 1\} ,\\A \cup B = \{ - 1;1;6\} ,\\A\backslash B = \{ 6\} ,\\B\backslash A = \{ 1\} ,\end{array}\)
Tham khảo:
+) \(A \cap B = [0;3] \cap (2; + \infty ) = (2;3]\)
+) \(A \cup B = [0;3] \cup (2; + \infty ) = [0; + \infty )\)
+) \(A\,{\rm{\backslash }}\,B = [0;3]\,{\rm{\backslash }}\,(2; + \infty ) = [0;2]\)
+) \(B\,{\rm{\backslash }}\,A = (2; + \infty )\,{\rm{\backslash }}\,[0;3] = (3; + \infty )\)
+) \(\mathbb{R}\,{\rm{\backslash }}\,B = \mathbb{R}\,{\rm{\backslash }}\,(2; + \infty ) = ( - \infty ;2]\)
\(E = \{ x \in \mathbb{N}|x < 8\} = \{ 0;1;2;3;4;5;6;7\} \)
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, \(A\cup B=(-4;5]\)
\(A\cap B=[-3;4)\)
\(A\backslash B=\left[4;5\right]\)
\(B\backslash A=\left(-4;-3\right)\)
b, \(A\cup B=\left(-3;7\right)\)
\(A\cap B=[1;2)\cup(3;5]\)
\(A\backslash B=\left[2;3\right]\)
\(B\backslash A=\left(-3;1\right)\cup\left(5;7\right)\)
c, \(A\cup B=\left[\dfrac{1}{2};3\right]\)
\(A\cap B=\left[1;\dfrac{3}{2}\right]\)
\(A\backslash B=[\dfrac{1}{2};1)\)
\(B\backslash A=(\dfrac{3}{2};3]\)
d, \(A\cup B=(-5;2]\cup(3;6]\)
\(A\cap B=\left\{0\right\}\cup[4;5)\)
\(A\backslash B=(0;2]\cup\left[-5;6\right]\)
\(B\backslash A=[-5;0)\cup\left(3;4\right)\)
\(E = \{ x \in \mathbb{N}|x < 10\} = \{ 0;1;2;3;4;5;6;7;8;9\} \)
\(A = \{ x \in E|x\) là bội của 3\(\} \)\( = \{ 0;3;6;9\} \)
\(B = \{ x \in E|x\) là ước của 6\(\} \)\( = \{1;2;3;6\} \)
Ta có: \(A\backslash B = \left\{ {0;9} \right\}\), \(B\backslash A = \left\{ {1;2} \right\}\)
\({C_E}A = \{ 1;2;4;5;7;8\} ,\;{C_E}B = \{ 0;4;5;7;8;9\} \)
\(A \cap B = \{ 3;6\} \Rightarrow {C_E}(A \cap B) = {C_E}B = \{0;1;2;4;5;7;8;9\} \)
\(A \cup B = \{ 0;1;2;3;6;9\} \Rightarrow {C_E}(A \cup B) = {C_E}A = \{ 4;5;7;8\} \)
\(A = \left\{ {0;1;2;3;4;5;6} \right\}\)
\(\,B = \left\{ {1;2;3;6;7;8} \right\}\)
Vậy
\(A \cap B = \left\{ {1;2;3;6} \right\}\)
\(A \cup B = \left\{ {0;1;2;3;4;5;6;7;8} \right\} = \left\{ {x \in \mathbb{N}|\;x < 9} \right\}\)
\(A\;{\rm{\backslash }}\;B = \left\{ {0;4;5} \right\}\)
a/ A = (3;\(+\infty\)), B=[0;4]
A \(\cap\) B= (3;4)
A\(\cup\) B=[0;+\(\infty\))
A\B= (4;\(+\infty\))
B\A= [0;3]
b/ A=(\(-\infty\);4], B=(2;\(+\infty\))
A\(\cap\)B=(2;4]
A\(\cup\)B= R
A\B= (\(-\infty\);2]
B\A=(4;\(+\infty\))
c/ A=[0;4] , B=(\(-\infty\);2]
A\(\cap\)B= [0;2)
\(A\cup B\) = (\(-\infty\);4]
A\ B=[2;4]
B\A=(\(-\infty\);0)
