Ta có: \(\left(x^2+y^2+2xy+2yz+2xz\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=3\)

\(\Rightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=3\)

\(\Rightarrow\left(x+y+z\right)^2\le3\)

Dấu "=" xảy ra <=> x=y=z

Do đó \(-\sqrt{3}\le x+y+z\le\sqrt{3}\)

\(\Rightarrow-\sqrt{3}\le A\le\sqrt{3}\)

=> \(\hept{\begin{cases}Min_A=-\sqrt{3}\Leftrightarrow x=y=z=\frac{-\sqrt{3}}{3}\\Max_A=\sqrt{3}\Leftrightarrow x=y=z=\frac{\sqrt{3}}{3}\end{cases}}\)

Lời giải:
Ta có:

\(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)

\(\Leftrightarrow (4x^2-4xy+y^2)+2z^2+y^2-2z(2x-y)-6y-10z+34=0\)

\(\Leftrightarrow (2x-y)^2-2z(2x-y)+z^2+(y^2-6y+9)+(z^2-10z+25)=0\)

\(\Leftrightarrow (2x-y-z)^2+(y-3)^2+(z-5)^2=0\)

Vì \((2x-y-z)^2; (y-3)^2; (z-5)^2\geq 0, \forall x,y,z\). Do đó để \((2x-y-z)^2+(y-3)^2+(z-5)^2=0\) thì:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\)

\(\Rightarrow \left\{\begin{matrix} x=4\\ y=3\\ z=5\end{matrix}\right.\)

Khi đó:

\(S=(4-4)^{2018}+(3-4)^{2019}+(5-4)^{2020}=0+(-1)+1=0\)