Câu hỏi của Đinh Doãn Nam

Question

Tìm $A\cap B;A\cup B;A/B;B/A$
$A=\left\{x\in Z|x^2< 4\right\};B=\left\{x\in Z|\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\} $

Hồng Quang · Accepted Answer

$A=\left\{x\in Z,x^2< 4\right\}$

$\Rightarrow A=\left\{-1;0;1\right\}$

$B=\left\{x\in Z,\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\}$$\Rightarrow\left[{}\begin{matrix}5x-3x^2=0\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{3}\left(loai\right)\x=0\x=3\x=-1\end{matrix}\right.$

$\Rightarrow B=\left\{0;-1;3\right\}$

$\Rightarrow A\cap B=\left\{0;-1\right\}$ $A\cup B=\left\{0;-1;1;3\right\}$

$A\backslash B=\left\{1\right\}$ $B\backslash A=\left\{3\right\}$

Câu trả lời:

$A=\left\{x\in Z,x^2< 4\right\}$

$\Rightarrow A=\left\{-1;0;1\right\}$

$B=\left\{x\in Z,\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\}$$\Rightarrow\left[{}\begin{matrix}5x-3x^2=0\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{3}\left(loai\right)\x=0\x=3\x=-1\end{matrix}\right.$

$\Rightarrow B=\left\{0;-1;3\right\}$

$\Rightarrow A\cap B=\left\{0;-1\right\}$ $A\cup B=\left\{0;-1;1;3\right\}$

$A\backslash B=\left\{1\right\}$ $B\backslash A=\left\{3\right\}$

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