\(\frac{x-y\sqrt{2019}}{y-z\sqrt{2019}}=m\Rightarrow x-y\sqrt{2019}=my-mz\sqrt{2019}\)
\(\Leftrightarrow\left(mz-y\right)\sqrt{2019}=my-x\)
\(\Rightarrow\left\{{}\begin{matrix}mz-y=0\\my-x=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m=\frac{y}{z}\\m=\frac{x}{y}\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}=\frac{y}{z}\Rightarrow y^2=zx\)
\(\Rightarrow x^2+y^2+z^2=x^2+z^2+2zx-2zx+y^2\)
\(=\left(x+z\right)^2-2y^2+y^2=\left(x+z\right)^2-y^2=\left(x+y+z\right)\left(x+z-y\right)\)
Do \(x^2+y^2+z^2\) là SNT \(\Rightarrow x+z-y=1\Rightarrow y=x+z-1\)
Mặt khác \(y^2=zx\le\frac{\left(x+z\right)^2}{4}\Rightarrow y\le\frac{x+z}{2}\)
\(\Rightarrow x+z-1\le\frac{x+z}{2}\Rightarrow x+z\le2\)
\(\Rightarrow x=z=1\Rightarrow y=1\)
Vậy có duy nhất \(\left(x;y;z\right)=\left(1;1;1\right)\) thỏa mãn
