Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{\left(y+z+x+z+x+y\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Khi đó:
\(\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\\dfrac{1}{x+y+z}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\2x+2y+2z=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y+2z+2=4x\\2x+2z+4=4y\\2x+2y-6=4z\\2x+2y+2z=1\end{matrix}\right.\)
\(\circledast\) Xét \(2x+2y+2z=1\Leftrightarrow2y+2z=1-2x\Leftrightarrow1-2x+2=4x\)
\(\Leftrightarrow3-2x=4x\Leftrightarrow6x=3\Leftrightarrow x=\dfrac{1}{2}\)
\(\circledast\)Xét \(2x+2y+2z=1\Leftrightarrow2x+2z=1-2y\Leftrightarrow1-2y+4=4y\)
\(\Leftrightarrow5-2y=4y\Leftrightarrow6y=5\Leftrightarrow y=\dfrac{5}{6}\)
\(\circledast\)Xét \(2x+2y+2z=1\Leftrightarrow2x+2y=1-2z\Leftrightarrow1-2z-6=4z\)
\(\Leftrightarrow-5-2z=4z\Leftrightarrow-5=6z\Leftrightarrow z=-\dfrac{5}{6}\)
