§2. Tập hợp

A={0;1;-1}

B={0;3;-1}

A giao B={-1;0}

A hợp B={0;1;-1;3}

A\B={1}

B\A={3}

A giao B={1;2;3}

A\B={4;5}

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

B\A={6;9}; A giao B={1;2;3}

=>B={1;2;3;6;9}

a: A giao B={1;2;3;4;5;7}

A giao C={1;3;6;7}

B giao C={1;3;7}

A\B={6}

B\C={-1;2;4;5;9}

A hợp C={1;2;3;4;5;6;7;-2}

A hợp C={1;2;3;4;5;6;-1;9}

b: (A giao B) hợp (A\B) hợp (B\A)

=(A\B) hợp (B\A)

=A hợp B

\(\left[{}\begin{matrix}\left[{}\begin{matrix}A\cap B=\left\{1;2;3;5;6;7;-1;4;9\right\}\\A\backslash B=\varnothing\end{matrix}\right.\\\left[{}\begin{matrix}A\backslash C=\left\{2;5\right\}\\A\cap C=\left\{1;2;3;5;6;7;-2\right\}\end{matrix}\right.\\\left[{}\begin{matrix}B\cap C=\left\{-2;-1;1;2;3;4;5;6;7;9\right\}\\A\cup C=\left\{1;3;6;7\right\}\end{matrix}\right.\\A\cup B=\left\{1;2;3;5;6;7\right\}\end{matrix}\right.\)

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

\(\)\(VT=VP=\left\{1;2;3;5;6;7;-1;4;9\right\}\)

CMR: \(\left(A\cup B\right)\cap C=\left(A\cap C\right)\cup\left(B\cap C\right)\)

\(VT=VP=\left\{1;3;6;2;7;5;-2\right\}\)

2:
Số bạn chỉ thích Văn là 15-8=7 bạn

Số bạn chỉ thích Toán là 20-8=12 bạn

Số học sinh tổng là;

7+12+8+10=37(bạn)

x+3<2x+4

=>2x+4>x+3

=>x>1

Vậy: \(A=\left(1;+\infty\right)\)

5x-3<4x-1

=>5x-4x<-1+3

=>x<2

\(B=\left(-\infty;2\right)\)

\(A\cup B=R\)

Chọn A

a: A={x|m>x>2m+1}

Để A là tập rỗng thì m<=2m+1

=>-m<=1

hay m>=-1

b: Để A là tập rỗng thì \(\left(-2\right)^2-4\cdot1\cdot m< 0\)

=>4-4m<0

=>4m>4

hay m>1

a) \(2x^2-5x+3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\) mà \(x\in Z\Rightarrow x=1\)

Vậy \(D=\left\{1\right\}\)

b) \(x\in Z\Rightarrow x+2\in Z\Rightarrow\left|x+2\right|\in Z\)

Mà \(\left|x+2\right|\le1\Rightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x+2=0\\x+2=1\\x+2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-1\\x=-3\end{matrix}\right.\)

Vậy \(F=\left\{-3;-2;-1\right\}\)

c) \(G=\left\{x\in N|x< 5\right\}=\left\{0;1;2;3;4\right\}\)

d) \(x^2+x+3=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{11}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\)

Vậy \(H=\varnothing\)

e) Ta có \(\left\{{}\begin{matrix}x+3< 4+2x\\5x-3< 4x-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\) mà \(x\in N\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy \(E=\left\{0;1\right\}\)