Question

[Nguyễn Thị Thảo]

cho\(\frac{xy+1}{y}=\frac{yz+1}{z}=\frac{zx+1}{x}\)chung minh\(x=y=z\)hoac \(x^2y^2z^2=1\)

Answer 1

[Trịnh Quỳnh Nhi]

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