`#tnvt`
Ta có: `x+y+z=0`
`=>{(x+y=-z),(x+z=-y),(y+z=-x):}`
`Q=(1+x/y).(1+y/z).(1+z/x)(x,y,z\ne0)`
`=(\frac{y+x}{y}).(\frac{z+y}{z}).(\frac{x+z}{x})`
`=\frac{-z}{y} .\frac{-x}{z}.\frac{-y}{x}`
`=\frac{-1}{1}.\frac{-1}{1}.\frac{-1}{1}`
`=-1`
Vậy tại `x+y+z=0`
`=>QQ=-1`