\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)
Mà \(xy+yz+zx=0\)(theo đề) nên \(2\left(xy+yz+zx\right)=0\)
\(\Rightarrow x^2+y^2+z^2=0\)
Vì \(\hept{\begin{cases}x^2\ge0\\y^2\ge0\\z^2\ge0\end{cases}}\) (với mọi x;y;z) nên \(x^2+y^2+z^2\ge0\) (với mọi x;y;z)
Để \(x^2+y^2+z^2=0\) \(\Leftrightarrow\) \(\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}\Leftrightarrow}x=y=z=0\)
Vậy \(A=\left(0-1\right)^{2016}+0^{2017}+\left(0+1\right)^{2018}=\left(-1\right)^{2016}+0+1^{2018}=2\)