Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Theo bài ra: \(\dfrac{x+y-3}{z}=\dfrac{x+z+2}{y}=\dfrac{y+z+1}{x}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{x+y-3}{z}=\dfrac{x+z+2}{y}=\dfrac{y+z+1}{x}=\dfrac{1}{x+y+z}=2\)
\(\Rightarrow x+y-3=2z\left(1\right);x+z+2=2y\left(2\right);y+z+1=2x\left(3\right);x+y+z=\dfrac{1}{2}\)
\(+)x+y+z=\dfrac{1}{2}\Leftrightarrow y+x=\dfrac{1}{2}-z\). Thay vào \(\left(3\right)\), ta được\(\dfrac{1}{2}-x+1=2x\Rightarrow\dfrac{3}{2}=3x\Rightarrow x=\dfrac{1}{2}\)
\(+)x+y+z=\dfrac{1}{2}\Leftrightarrow x+z=\dfrac{1}{2}-y\). Thay vào \(\left(2\right)\), ta được \(\dfrac{1}{2}-y+2=2y\Rightarrow\dfrac{5}{2}=3y\Rightarrow y=\dfrac{5}{6}\)
\(\Rightarrow x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+z\Leftrightarrow z=-\dfrac{5}{6}\)
Vậy \(x=\dfrac{1}{2};y=\dfrac{5}{6};z=-\dfrac{5}{6}\)
