Áp dụng TCDTSBN ta có:
\(\dfrac{x+y+2017}{z}=\dfrac{y+z-2018}{x}=\dfrac{z+x+1}{y}=\dfrac{x+y+2017+y+z-2018+z+x+1}{z+x+y}=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\dfrac{z+x+1}{y}=\dfrac{2}{x+y+z};\dfrac{z+x+1}{y}=2\\ \Rightarrow\dfrac{2}{x+y+z}=2\\ \Rightarrow x+y+z=1\)
\(\left\{{}\begin{matrix}\dfrac{x+y+2017}{z}=2\\\dfrac{y+z-2018}{x}=2\\\dfrac{z+x+1}{y}=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y+2017=2z\\y+z-2018=2x\\z+x+1=2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y+z=3z-2017\\y+z+x=3x+2018\\z+x+y=3y-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3z-2017=1\\3x+2018=1\\3y-1=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3z=2018\\3x=-2017\\3y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}z=\dfrac{2018}{3}\\x=\dfrac{-2017}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{-2017}{3}\\y=\dfrac{2}{3}\\z=\dfrac{2018}{3}\end{matrix}\right.\)
