Sửa đề: \(2x^2-3x+1=0\)
Ta có: \(2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{1;\frac{1}{2}\right\}\)