\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{1}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{1}=\dfrac{x+y+z}{\left(y+z+1\right)+\left(x+z+1\right)+\left(x+y-2\right)}\)
\(=\dfrac{x+y+z}{2x+2y+2z}\)
\(TH1:x+y+z=0\)
⇒ \(\dfrac{x+y+z}{1}=0\)
⇒ \(x=y=z=0\)(loại vì trái với điều kiện đề bài )
\(TH2:z+y+z\)≠ 0
⇒ \(\dfrac{x+y+z}{2x+2y+2z}=\dfrac{x+y+z}{2.\left(x+y+z\right)}=\dfrac{1}{2}\)
Vậy \(x+y+z=\dfrac{1}{2}\)
\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\) ⇒ \(2x=y+z+1\)⇒\(2x=y+z+2\left(x+y+z\right)=2x+3y+3z\)
⇒ \(3y+3z=0\) ⇒ \(y+z=0\) ⇒ \(2x=1\) ⇒ \(x=\dfrac{1}{2}\)
\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\) ⇒ \(2y=x+z+1=x+z+2\left(x+y+z\right)=2y+3x+3z\)
⇒ \(3x+3z=0\) ⇒ \(x+z=0\) ⇒ \(2y=1\) ⇒ \(y=\dfrac{1}{2}\)
\(x+z=0\) ; \(x=\dfrac{1}{2}\)
⇒ \(z=0-\dfrac{1}{2}=\dfrac{-1}{2}\)
Vậy \(x=\dfrac{1}{2}\) ; \(y=\dfrac{1}{2}\) ; \(z=\dfrac{-1}{2}\)
