Ta có: \(\left(x-1\right)^{2018}>=0\forall x\)
\(\left(2y-1\right)^{2018}>=0\forall y\)
\(\left|x+2y-z\right|^{2019}>=0\forall x,y,z\)
Do đó: \(\left(x-1\right)^{2018}+\left(2y-1\right)^{2018}+\left|x+2y-z\right|^{2019}>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\2y-1=0\\x+2y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\\z=x+2y=1+2\cdot\dfrac{1}{2}=1+1=2\end{matrix}\right.\)
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