令(y+z)/x=(z+x)/y=(x+y)z=k
则:1式:y+z=kx
2式:z+x=ky
3式:x+y=kz
1式+2式+3式得:2(x+y+z)=k(x+y+z)
∵xyz≠0,∴x,y,z均不为0
由此得:k=2
(z+y)(x+z)(y+z)/[xyz]=[y+z)/x]×[(z+x)/y]×[(x+y)z]=k×k×k=2×2×2
=8
令(y+z)/x=(z+x)/y=(x+y)z=k
则:1式:y+z=kx
2式:z+x=ky
3式:x+y=kz
1式+2式+3式得:2(x+y+z)=k(x+y+z)
∵xyz≠0,∴x,y,z均不为0
由此得:k=2
(z+y)(x+z)(y+z)/[xyz]=[y+z)/x]×[(z+x)/y]×[(x+y)z]=k×k×k=2×2×2
=8