Có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}.\)
\(\Rightarrow\dfrac{z}{x+y-2}=\dfrac{1}{2}.\)
\(\Rightarrow2z=x+y-2.\)
\(\Rightarrow2z+2=x+y.\) (quy tắc "chuyển vế").
Lại có:
\(x+y+z=\dfrac{1}{2}.\)
\(\Leftrightarrow2z+2+z=\dfrac{1}{2}.\)
\(\Leftrightarrow3z+2=\dfrac{1}{2}.\)
\(\Leftrightarrow3z=\dfrac{1}{2}-2=-\dfrac{3}{2}.\)
\(\Leftrightarrow z=-\dfrac{1}{2}.\)
Từ đó ta tìm được \(x=y=\dfrac{1}{2}.\)
Vậy \(\left\{x;y;z\right\}=\left\{\dfrac{1}{2};\dfrac{1}{2};-\dfrac{1}{2}\right\}.\)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
\(x+y+z=\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{2(x+y+z)}=\frac{1}{2}\)
\(\Rightarrow \frac{x+y+z+1}{y+z+1}=\frac{x+y+z+1}{x+z+1}=\frac{x+y+z-2}{x+y-2}=\frac{1}{2}+1=\frac{3}{2}\)
Thay \(x+y+z=\frac{1}{2}\) thu được:
\(\frac{\frac{3}{2}}{y+z+1}=\frac{\frac{3}{2}}{x+z+1}=\frac{\frac{-3}{2}}{x+y-2}=\frac{3}{2}\)
\(\Rightarrow \left\{\begin{matrix} y+z+1=1\\ x+z+1=1\\ x+y-2=-1\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} y+z=0\\ x+z=0\\ x+y=1\end{matrix}\right.\)
\( \Rightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{1}{2}\\ z=\frac{-1}{2}\end{matrix}\right.\)
