Sửa đề:
Lời giải:
\(\dfrac{x}{y+z}=\dfrac{y}{x+z}=\dfrac{z}{x+y}=\dfrac{1}{x+y+z}\)(nghĩ vậy,vì đề bạn thiếu)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z}=\dfrac{y}{x+z}=\dfrac{z}{x+y}=\dfrac{x+y+z}{y+z+x+z+x+y}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)Suy ra: \(\left\{{}\begin{matrix}\dfrac{x}{y+z}=\dfrac{1}{2}\\\dfrac{y}{x+z}=\dfrac{1}{2}\\\dfrac{z}{x+y}=\dfrac{1}{2}\\\dfrac{1}{x+y+z}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\x+z=2y\\x+y=2z\\x+y+z=2\end{matrix}\right.\)
\(\circledast\)Từ \(x+y+z=2\Leftrightarrow y+z=2-x\)
Nên \(2-x=2x\Leftrightarrow3x=2\Leftrightarrow x=\dfrac{2}{3}\)
\(\circledast\)Từ \(x+y+z=2\Leftrightarrow x+z=2-y\)
Nên \(2-y=2y\Leftrightarrow3y=2\Leftrightarrow y=\dfrac{2}{3}\)
\(\circledast\)Từ \(x+y+z=2\Leftrightarrow x+y=2-z\)
Nên \(2-z=2z\Leftrightarrow3z=2\Leftrightarrow z=\dfrac{2}{3}\)
Vậy \(x=y=z=\dfrac{2}{3}\)
