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

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

Vậy

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

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

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

\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)

Giải phương trình sau :

 \(\left(x-2x^2\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow x\left(1-2x\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-2x=0\\x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\\x=2\end{matrix}\right.\)

\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)

\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)

Giải bất phương trình sau :

\(3< n\left(n+1\right)< 31\)

\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)

\(\Rightarrow A\cap B=\left\{2\right\}\)

(2x-x^2)(2x^3-3x-2)=0

=>x(2-x)(2x^3-3x-2)=0

=>x=0 hoặc 2-x=0 hoặc 2x^3-3x-2=0

=>\(x\in\left\{0;2;1,48\right\}\)

=>\(A=\left\{0;2;1,48\right\}\)

3<n^2<30

mà \(n\in Z^+\)

nên \(n\in\left\{2;3;4;5\right\}\)

=>B={2;3;4;5}

=>A giao B={2}

=>Chọn B

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, \(A\cup B=(-5;2]\cup(3;6]\)

\(A\cap B=\left\{0\right\}\cup[4;5)\)

\(A\backslash B=(0;2]\cup\left[-5;6\right]\)

\(B\backslash A=[-5;0)\cup\left(3;4\right)\)

a: A=(-7/4; -1/2]

\(B=\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\)

\(C=\left(\dfrac{2}{3};+\infty\right)\)

b: \(\left(A\cap B\right)\cap C=\varnothing\)

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

\(=(-\dfrac{7}{4};-\dfrac{1}{2}]\cup\left(\dfrac{2}{3};+\infty\right)\cap\left[\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\right]\)

\(=\left(4;\dfrac{9}{2}\right)\)

\(3k-1=5m-2\)

\(\Leftrightarrow3k-9=5m-10\)

\(\Leftrightarrow3\left(k-3\right)=5\left(m-2\right)\)

Do 3 và 5 nguyên tố cùng nhau \(\Rightarrow k-3⋮5\Rightarrow k=5n+3\) với \(n\in Z\)

Vậy \(A\cap B=\left\{5n+3|n\in Z\right\}\)

Ta có M = (18-5), M = (18-9), M = (12-5), M = (12 - 9), M = (81 - 5), M = (81-9)

Vì M được xác định là a - b

Do đó có tất cả 6 tập hợp 

Ta có 

M=(18-5);M=(18-9);M=(12-5);M=(12-9);M=(81-5);M=(81-9)

Vì M xác định là 

a-b

nên :  có tất cả 6 tập hợp 

Ta cóM=(18-5);M=(18-9);M=(12-5);M=(12-9);M=(81-5);M=(81-9)

Vì M xác định làa-b

nên : có tất cả 6 tập hợp

hc tốt

1: A={-3;-2;-1;0;1;2;3}

B={2;-2;4;-4}

A giao B={2;-2}

A hợp B={-3;-2;-1;0;1;2;3;4;-4}

2: x thuộc A giao B

=>\(x=\left\{2;-2\right\}\)

\(C\cap B=[-5;a]\)

mà \(B=\left\{x\in R|-5\le x\le5\right\}\) có độ dài là \(\left|-5\right|+\left|5\right|=10\)

\(\Rightarrow C\cap B=[-5;a]\) có độ dài là \(5\) thì \(a=10:2-5=0\)

\(D\cap B=[b;5]\) có độ dài là 9 thì \(b=10:2-9=-4\)