Có :
$x^3+y^2+z^3=3xyz$
$\to (x+y+z).\dfrac{1}{2}.[(x-y)^2+(y-z)^2+(z-x)^2] = 0 $
$\to x=y=z$ ( Do $x+y+z \neq 0 $)
Khi đó : \(C=\dfrac{3\cdot x^{2019}}{\left(3x\right)^{2019}}=\dfrac{1}{3^{2018}}\)
$⇔x^3+y^3+z^3-3xyz=0$
$⇔(x+y)^3-3xy(x+y)-3xyz+z^3=0$
$⇔(x+y+z)((x+y)^2-(x+y)z+z^2)-3xy(x+y+z)=0$
$⇔(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$
$⇔x^2+y^2+z^2-xy-yz-zx=0$
$⇔(x-y)^2+(y-z)^2+(z-x)^2=0$
Mà $(x-y)^2+(y-z)^2+(z-x)^2≥0∀x;y;z$
suy ra $\begin{cases}(x-y)^2=0\\(y-z)^2=0\\(z-x)^2=0\end{cases}$
$⇔\begin{cases}(x-y)=0\\(y-z)=0\\(z-x)=0\end{cases}$
$⇔x=y=z$
Nên $C=\dfrac{3.x^{2019}}{(9.x^{2019}}=\dfrac{1}{3}$
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