8x(x-2017)-2x+4034=0
\(\Leftrightarrow\)8x(x-2017)-2(x-2017)=0
\(\Leftrightarrow\)(x-2017)(8x-2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2017=0\\8x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2017\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy x\(\in\left\{2017;\frac{1}{4}\right\}\)