a) Q \(\cap\) I = \(\varnothing\)

b) R \(\cap\) I = I

a) Q \(\cap\) I = ∅

b) R \(\cap\) I = I

\(A\cup B=\left(-1;+\infty\right)\)

\(A\cap B=(2;5]\)

Bài 1: A={a;b;c;d}; B={a;b}

B⊂X⊂A

=>{a;b}⊂X⊂{a;b;c;d}

=>X={a;b;c}; X={a;b;d}; X={a;b;c;d}

Bài 2:

a; A={1;2;3;4;5}; B={2;4;6}; C={1;3;5}

A\(\cup\) B={1;2;3;4;5}\(\cup\) {2;4;6}

={1;2;3;4;5;6}

A\(\cap\) B={1;2;3;4;5}\(\cap\) {2;4;6}

={2;4}

B\(\cap\) C={2;4;6}\(\cap\) {1;3;5}

=∅

b: (A\(\cup\) B)\(\cap\) C={1;2;3;4;5;6}\(\cap\) {1;3;5}

={1;2;3;4;5;6}

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

={1;2;3;4;5}

\(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)

\(\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)=\left(-2;7\right)\)

\(\left(0;12\right)\cap[5;+\infty)=[5;12)\)

\(R\cap\left[-1;1\right]=\left[-1;1\right]\)

A

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=∅

Lời giải:
Theo đề thì: \(B\subset A\) nên \(A\cap B = B [-2;1)\)