Đáp án:
ĐKXĐ: \[\left\{\begin{matrix}
x\ne 0 & \\
y\ne 0 & \\
z\ne 0 & \\
x+y+z\ne 0 &
\end{matrix}\right.\]
\[\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2\]
\[\Rightarrow \left\{\begin{matrix}
y+z+1=2x & \\
x+z+2=2y& \\
x+y-3=2z & \\
x+y+z=\dfrac{1}{2}&
\end{matrix}\right.\]
\[\Rightarrow \left\{\begin{matrix}
x=\dfrac{1}{2} & \\
y=\dfrac{5}{6}& \\
z=-\dfrac{5}{6}&
\end{matrix}\right.\]
