Bài 1:

a: A={1;2;3}; B={2;3;6;7}

A\(\cap\) B={1;2;3}\(\cap\) {2;3;6;7}

={2;3}

A\(\cup\) B={1;2;3}\(\cup\) {2;3;6;7}

={1;2;3;6;7}

A\B={1;2;3}\{2;3;6;7}

={1}

B\A={2;3;6;7}\{1;2;3}={6;7}

b: B={2;3;6;7}; C={3;4;5;8}

B\C={2;3;6;7}\{3;4;5;8}

={2;6;7}

A={1;2;3}; B\C={2;6;7}

A\(\cap\) (B\C)

={1;2;3}\(\cap\) {2;6;7}

={2}

A\(\cap\) B={2;3}; A\(\cap\) C={1;2;3}\(\cap\) {3;4;5;8}={3}

(A\(\cap\) B)\(A\(\cap\) C)

={2;3}\{3}

={2}

Do đó: A\(\cap\) (B\C)=(A\(\cap\) B)\(A\(\cap\) C)

Bài 2:

a: A\(\cap\)A=A

A\(\cup\)A=A

A\(\cap\) ∅=∅

A\(\cup\) ∅=A

b: A\A=∅

A\∅=A

∅\A=∅

\(E = \{ x \in \mathbb{N}|x < 8\}  = \{ 0;1;2;3;4;5;6;7\} \)

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=(-∞;2]\(\cup\) [2;+∞)

=(-∞;+∞)

A\(\cap\) B=(-∞;2]\(\cap\) [2;+∞)

={2}

B\(\cap\) C=[2;+∞)\(\cap\) (-2;2]

={2}

A\(\cap\) B\(\cap\) C=(-∞;2]\(\cap\) [2;+∞)\(\cap\) (-2;2]

={2}

A\B=(-∞;2]\[2;+∞)

=(-∞;2)

(A\(\cap\) B)\C={2}\(-2;2]

=∅

\(C_{R}A\) =R\A=R\(-∞;2]

=(2;+∞)

\(C_{R}B\) =R\B=R\[2;+∞)

=(-∞;2)

\(C_{R}C\) =R\C

=R\(-2;2]

=(-∞;-2]\(\cup\) (2;+∞)

a) \(\left(A\cap B\right)\cup A=A\)

b) \(\left(A\cup B\right)\cap B=B\)

c) (\(A\)\ \(B\)) \(\cup B=A\cup B\)

d) (\(A\)\ \(B\)) \(\cap\)(\(B\)\\(A\)) \(=\varnothing\)

\(A =  \left\{ {0;1;2;3;4;5;6} \right\}\)

\(\,B = \left\{ {1;2;3;6;7;8} \right\}\)

Vậy

\(A \cap B = \left\{ {1;2;3;6} \right\}\)

\(A \cup B = \left\{ {0;1;2;3;4;5;6;7;8} \right\}  = \left\{ {x \in \mathbb{N}|\;x < 9} \right\}\)

\(A\;{\rm{\backslash }}\;B = \left\{ {0;4;5} \right\}\)

\(A=\left\{1,2,3,6,9,18\right\}\)

\(B=\left\{1,2,3,5,6,10,15,30\right\}\)

\(A\cap B=\left\{1,2,3,6\right\}\)

\(A\cup B=\left\{1,2,3,5,6,10,15,18,30\right\}\)

A \ \(B=\left\{9,18\right\};B\)\\(A=\left\{5,10,15,30\right\}\)

a) \(A \cup B = \{ a;b;c;d;e;i;u\} \), \(A \cap B = \{ a;e\} \)

b) Phương trình \({x^2} + 2x - 3 = 0\) có hai nghiệm là 1 và -3, nên \(A = \{ 1; - 3\} \)

Phương trình \(B = \{ x \in \mathbb{R}|\;|x|\; = 1\} \) có hai nghiệm là 1 và -1, nên \(B = \{ 1; - 1\} \)

Từ đó, \(A \cup B = \{ 1; - 1; - 3\} \), \(A \cap B = \{ 1\} .\)

Tham khảo:

+) \(A \cap B = [0;3] \cap (2; + \infty ) = (2;3]\)

+) \(A \cup B = [0;3] \cup (2; + \infty ) = [0; + \infty )\)

+) \(A\,{\rm{\backslash }}\,B = [0;3]\,{\rm{\backslash }}\,(2; + \infty ) = [0;2]\)

+) \(B\,{\rm{\backslash }}\,A = (2; + \infty )\,{\rm{\backslash }}\,[0;3] = (3; + \infty )\)

+) \(\mathbb{R}\,{\rm{\backslash }}\,B = \mathbb{R}\,{\rm{\backslash }}\,(2; + \infty ) = ( - \infty ;2]\)

\(A\cap B=(2;3]\).

\(A\cup B=[0;+\infty)\)

\(\text{A \ B}=\left[0;2\right]\)

\(\text{B \ A}=\left(3;+\infty\right)\)

\(\text{R \ B}=(-\infty;2]\)