Ta có
\(\frac{xy+1}{y}=\frac{yz+1}{z}=>x+\frac{1}{y}=y+\frac{1}{z}=>x-y=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}\left(1\right)\)
\(\frac{yz+1}{z}=\frac{zx+1}{x}=>y+\frac{1}{z}=z+\frac{1}{x}=>y-z=\frac{1}{x}-\frac{1}{z}=\frac{z-x}{xz}\left(2\right)\)
\(\frac{zx+1}{x}=\frac{xy+1}{y}=>z+\frac{1}{x}=x+\frac{1}{y}=>z-x=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}\left(3\right)\)
Nhân từng vế (1),(2),(3) ta có:
\(\left(x-y\right)\left(y-z\right)\left(z-x\right)=\frac{\left(y-z\right)\left(z-x\right)\left(x-y\right)}{x^2y^2z^2}\)
<=>\(x^2y^2z^2\left(x-y\right)\left(y-z\right)\left(z-x\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
<=>\(\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x^2y^2z^2-1\right)=0\)
=> (x-y)(y-z)(z-x)=0 hoặc x2y2z2-1=0
• (x-y)(y-z)(z-x)=0 => x=y=z
• x2y2z2-1=0 => x2y2z2=1
Vậy x=y=z hoặc x2y2z2=1
