\(A=\left\{-1;0;1\right\}\\ \left(5x-3x^2\right)\left(x^2-2x-3\right)=0\\ \Leftrightarrow x\left(5-3x\right)\left(x-3\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\left(ktm\right)\\x=3\\x=-1\end{matrix}\right.\\ \Leftrightarrow B=\left\{-1;0;3\right\}\)
Do đó \(A\cap B=\left\{-1;0\right\};A\cup B=\left\{-1;0;1;3\right\};A\B=\left\{1\right\};B\A=\left\{3\right\}\)
Đúng 1
Bình luận (0)
