`(x-y-z)/x=(-x+y-z)/y=(-x-y+z)/z`
- Xét tử `=0`
$⇒\begin{cases}x-y-z=0\\y-x-z=0\\z-x-y=0\end{cases}$
`⇒x=y=z=0` (loại)
- Xét tử `\ne 0`
`⇒(x-y-z)/x=(-x+y-z)/y=(-x-y+z)/z=(x-y-z-x+y-z-x-y+z)/(x+y+z)=-1`
$⇒\begin{cases}x-y-z=-x\\-x+y-z=-y\\-x-y+z=-z\end{cases}$
$⇒\begin{cases}y+z=2x\\x+z=2y\\x+y=2z\end{cases}$
`⇒x=y=z`
Ta có: `A=(1+y/x)(1+z/y)(1+x/z)`
`A=2.2.2=8`
Vậy `A=8`