Question

Tìm x, y, z, biết nếu x + y + z = \(\dfrac{x}{y+z-2}=\dfrac{y}{z+x-3}=\dfrac{z}{x+y+5}\)

Answer 1

Đễ thấy \(x=y=z=0\) là bộ nghiệm cần tìm.

Xét \(x,y,z\ne0\)

\(x+y+z=\dfrac{x}{y+z-2}=\dfrac{y}{z+x-3}=\dfrac{z}{x+y+5}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=\dfrac{1}{2}\\\dfrac{x}{y+z-2}=\dfrac{1}{2}\\\dfrac{y}{z+x-3}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{5}{6}\\z=\dfrac{11}{6}\end{matrix}\right.\)

Vậy \(\left(x,y,z\right)=\left(0,0,0;-\dfrac{1}{2},-\dfrac{5}{6},\dfrac{11}{6}\right)\)