Ta có :
\(\left(x-1\right)^{2006}\ge0\)
\(\left(2y-1\right)^{2016}\ge0\)
\(\left(x+2y-z\right)^{2017}\ge0\)
Mà \(\left(x-1\right)^{2016}+\left(2y-1\right)^{2016}\)\(+|x+2y-z|^{2017}\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2006}=0\\\left(2x-1\right)^{2016}=0\\|x+2y-z|^{2017}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\2y-1=0\\x+2y-z=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=1\\2y=1\\1-1-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=2\end{cases}}}\)
Vậy ...
Ta có :
\(\left(x-1\right)^{2006}\ge0\)
\(\left(2y-1\right)^{2016}\ge0\)
\(\left|x+2y-z\right|^{2017}\ge0\)
Mà \(\left(x-1\right)^{2006}+\left(2x-1\right)^{2016}+\left|x+2y-z\right|^{2017}=0\)
Suy ra : \(\hept{\begin{cases}\left(x-1\right)^{2006}=0\\\left(2x-1\right)^{2016}=0\\\left|x+2y-z\right|^{2017}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-1=0\\2y-1=0\\x+2y-z=0\end{cases}}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=1\\2y=1\\1+1-z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=2\end{cases}}}\)
Vậy \(x=1\)\(;\)\(y=\frac{1}{2}\) và \(z=2\)
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