Sửa lại đề bài :
\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=x+y+z\)
\(\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\Rightarrow\dfrac{x}{\dfrac{1}{2}-x+1}=\dfrac{1}{2}\Rightarrow2x=\dfrac{3}{2}-x\Rightarrow x=\dfrac{1}{2}\)
\(\dfrac{y}{z+x+1}=\dfrac{1}{2}\Rightarrow\dfrac{y}{\dfrac{1}{2}-y+1}=\dfrac{1}{2}\Rightarrow2y=\dfrac{3}{2}-y\Rightarrow y=\dfrac{1}{2}\)
\(x+y+z=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}-\left(x+y\right)=\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)=-\dfrac{1}{2}\)
Vậy \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{2};-\dfrac{1}{2}\right)\)
