Lời giải:
Xét 2 TH sau:
TH1: \(x+y+z=0\)
\(\Rightarrow \frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z=0\)
\(\Rightarrow \left\{\begin{matrix} x=0\\ y=0\\ z=0\end{matrix}\right.\)
TH2: \(x+y+z\neq 0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(x+y+z=\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+x+z+1+x+y-2}\)
\(\Leftrightarrow x+y+z=\frac{x+y+z}{2(x+y+z)}=\frac{1}{2}\)
Cộng thêm 1 vào mỗi biểu thức:
\(\Rightarrow \frac{x+y+z+1}{y+z+1}=\frac{x+y+z+1}{x+z+1}=\frac{x+y+z-2}{x+y-2}=x+y+z+1\)
\(\Leftrightarrow \frac{3}{2(y+z+1)}=\frac{3}{2(x+z+1)}=\frac{-3}{2(x+y-2)}=\frac{3}{2}\)
\(\Leftrightarrow \left\{\begin{matrix} y+z+1=1\\ x+z+1=1\\ x+y-2=-1\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} y+z=0\\ x+z=0\\ x+y=1\end{matrix}\right.\Rightarrow x+y+z=\frac{1+0+0}{2}=\frac{1}{2}\)
Do đó:
\(\left\{\begin{matrix} x=\frac{1}{2}-0=\frac{1}{2}\\ y=\frac{1}{2}-0=\frac{1}{2}\\ z=\frac{1}{2}-1=\frac{-1}{2}\end{matrix}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Theo bài cho
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=2\)\(\Rightarrow y+z+1=2x;x+z+2=2y;x+y-3=2z;x+y+z=\dfrac{1}{2}\)từ đây bạn thay số vào nha
