ĐK: x>0.
Pt\(\left[{}\begin{matrix}x-2016=2016x\\x-2016=-2016x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2015}{2016}\left(l\right)\\x=\frac{2016}{2017}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{2016}{2017}\right\}\)
Ta có: \(\left|x-2016\right|=2016x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2016=2016x\\x-2016=-2016x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2016-2016x=0\\x-2016+2016x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2015x=2016\\2017x=2016\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2016}{2015}\\x=\frac{2016}{2017}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{2016}{2015};\frac{2016}{2017}\right\}\)
