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=∅

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

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

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

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

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

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: \(A \cup B = \{ 0;1;2;3;4;5\} ,\;{C_E}(A \cup B) = \{ 6;7\} \)

\({C_E}A = \{ 5;6;7\} ,\;{C_E}B = \{ 0;1;2;6;7\}  \Rightarrow ({C_E}A) \cup ({C_E}B) = \{ 0;1;2;5;6;7\} \)

Giao của hai tập hợp là \([ - 2;3] \cap (1; + \infty ) = (1;3]\)

Hiệu của \(B \backslash A \) là \( (1; + \infty ) \backslash [ - 2;3] = (3; + \infty )\)

Phần bù của B trong \(\mathbb{R}\) là: \({C_\mathbb{R}}\;B = \mathbb{R}{\rm{\backslash }}\;(1; + \infty ) = ( - \infty ;1]\)

a) (0, 7)

b) (2, 5)

c) [3, +∞)

a)(0,7).

b)(2,5).

c)(3,\(+\infty\)).