初二年级数学分式题！急急急！

(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x
(x+y)/z-1=(x+z)/y-1=(y+z)/x-1
(x+y)/z=(x+z)/y=(y+z)/x

设(x+y)/z=(x+z)/y=(y+z)/x=k
x+y=kz
x+z=ky
两式相减得
y-z=k(z-y)

如果y≠z
k=-1
(x+y)(y+z)(z+x)/xyz=(x+y)/z*(x+z)/y*(y+z)/x=k*k*k=-1

如果y=z
由(y+z)=kx,得x=2y/k带回x+y=kz得2/k*y+y=ky
解得k=2或-1
(x+y)(y+z)(z+x)/xyz=(x+y)/z*(x+z)/y*(y+z)/x=k*k*k=8或-1

综上原式=8或-1

设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k,
则有
x+y-z=kz,(1)
x-y+z=ky,(2)
-x+y+z=kx,(3)
(1)+(2)+(3)得
x+y+z=k(x+y+z),
所以(x+y+z)-k(x+y+z)=0,
所以(x+y+z)(1-k)=0,
所以x+y+z=0或k=1,
1)当k=1时,
(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=1,
所以x+y-z=z,x-y+z=y,-x+y+z=x,
所以x+y=2z,x+z=2y,y+z=2x,
所以(x+y)(y+z)(z+x)/xyz=2z*2x*2y/xyz=8;
2)当x+y+z=0时,
x+y=-z,y+z=-x,z+x=-y,
所以(x+y)(y+z)(z+x)/xyz=-z*(-x)*(-y)/xyz=-1.

若x+y+z=0
则x+y=-z,y+z=-x,z+x=-y
所以(x+y)(y+z)(z+x)/xyz=-xyz/xyz=-1

若x+y+z不等于0
则用