Giải:
`(x - 1)^2016 >= 0` ; `(2y - 1)^2016 >= 0` ; `|x + 2y - z|^2017 >= 0` `(∀x,y,z)`
Mà `(x - 1)^2016 + (2y - 1)^2016 + |x + 2y - z|^2017 = 0`
`->` $\left\{ \begin{array}{l}(x - 1)^(2016) = 0\\(2y - 1)^(2016) = 0\\|x + 2y - z|^(2017) = 0\end{array} \right.$
`->` $\left\{ \begin{array}{l}x - 1 = 0\\2y - 1 = 0\\x + 2y - z = 0\end{array} \right.$
`->` $\left\{ \begin{array}{l}x = 1\\y = \dfrac{1}{2}\\1 + 1 - z = 0\end{array} \right.$
`->` $\left\{ \begin{array}{l}x = 1\\y = \dfrac{1}{2}\\z = 2\end{array} \right.$
Vậy `x = 1, y = 1/2, z = 2`
