Ta có: $3(x^2+y^2+z^2)=(x+y+z)^2$
⇔$2x^2+2y^2+2z^2=2xy+2yz+2zx$
⇔$(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=0$
⇔$(x-y)^2+(y-z)^2+(z-x)^2=0$
Do: $(x-y)^2≥0; (y-z)^2≥0; (z-x)^2≥0$ nên pt⇔$(x-y)^2=0; (y-z)^2=0; (z-x)^2=0$
⇒$x=y=z$
Suy ra: $\frac{x+2y-4z}{3}=\frac{-x}{3}$
Lại có: $x^{2018}+y^{2018}+z^{2018}=3.x^{2018} = 27^{478}=3^{1434}$
$⇔x^{2018}=3^{1433}$
TH1: x=$\sqrt[2018]{3^{1433}}$
⇒$\frac{-x}{3}=\frac{-\sqrt[2018]{3^{1433}}}{3}=\frac{-1}{\sqrt[2018]{3^{585}}}$
TH2: x=$-\sqrt[2018]{3^{1433}}$
⇒$\frac{-x}{3}=\frac{\sqrt[2018]{3^{1433}}}{3}=\frac{1}{\sqrt[2018]{3^{585}}}$
