Question

A{x∈Q/ (2x-x\(^2\))(2x\(^2\)-3x-2)=0}

Answer 1

\(\left(2x-x^2\right)\left(2x^2-3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x\left(2-x\right)=0\\\left(x-2\right)\left(2x+1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-\dfrac{1}{2};0;2\right\}\)

Answer 2

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