Question

tìm các số nguyên x,y,z thỏa mãn \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=x+y+z=3\)

Answer 1

\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\)

<=>x²z+y²x+z²y=x²y+y²z+z²x

<=>(x²z-x²y)+(y²x-z²x)+(z²y-y²z)=0

<=>x².(z-y)-x.(z-y)(z+y)+yz.(z-y)=0

<=>(z-y)(x²-xz-xy+yz)=0

<=>(z-y)(x-z)(x-y)=0

<=>x=y=z

Mà x+y+z=3

=>x=y=z=1

Answer 2

Có thể   \(x=y=z=1\)