Ta có: \(x^2-2017x+2016=0\)
\(\Rightarrow x^2-x-2016x+2016=0\)
\(\Rightarrow x\left(x-1\right)-2016\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2016\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=2016\end{cases}}}\)
Vậy \(x=\left\{1;2016\right\}\)