问题补充：
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y 求(x+y)(y+z)(z+x)/xyz(x+y-z)/z=(y+z-x)/x=(z+x-y)/y 求(x+y)(y+z)(z+x)/xyz
答案：
设：(x+y-z)/z=(y+z-x)/x=(z+x-y)/y =k
{x+y-z=kz (1)
{y+z-x=kx (2)
{z+x-y=ky (3)
(1)+(2)+(3)得:
(x+y+z)=k(x+y+z)
(x+y+z)(k-1)=0
k=1或x+y+z=0
当k=1时{x+y-z=z
{y+z-x=x
{z+x-y=y
解得x=y=z (x+y)(y+z)(z+x)/xyz=8
当x+y+z=0时
(x+y)(y+z)(z+x)/xyz=(-z)(-x)(-y)/xyz=-1
======以下答案可供参考======
供参考答案1:
(x+y-z)/z=（x+y）/z-1=(y+z-x)/x=(y+z)/x-1=(z+x-y)/y=(z+x)/y-1
(x+y)/z=(y+z)/x=(z+x)/y
所以(x+y)(y+z)(z+x)/xyz=1
供参考答案2:
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y
等价于(x+y)/z-1=(y+z)/x-1=(z+x)/y-1
即(x+y)/z=(y+z)/x=(z+x)/y
所以(x+y)(y+z)(z+x)/xyz =[(x+y)/z][(y+z)/x][=(z+x)/y]=[(z+x)/y]^3
供参考答案3:
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y
=kx+y-z=kz
y+z-x=kx
z+x-y=ky三个式子相加
得k=1所以x=y=z(x+y)(y+z)(z+x)/xyz=8
供参考答案4:
。。首先x+y+z=0，大家都推出来了，然后(x+y)(y+z)(z+x)/xyz分解后=（x+y）/z +(y+z)/x +(x+z)/y +2 =(x+y+z)/z -1 +(x+y+z)/x -1 +(x+y+z)/y -1 +2 = -1
供参考答案5:
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y
(x+y)/z-1=(y+z)/x-1=(z+x)/y-1
x/z+y/z=y/x+z/x=z/y+x/y
x/z+y/z=y/x+z/x
y=(z/x-x/z)/(1/z-1/x)
=(z^2-x^2)/(x-z)
=-(x+z)
同理，x=-(y+z),z=-(x+y)
(x+y)(y+z)(z+x)/xyz=-1
供参考答案6:
解法一：设：(x+y-z)/z=(y+z-x)/x=(z+x-y)/y =k{x+y-z=kz (1){y+z-x=kx (2){z+x-y=ky (3) (1)+(2)+(3)得:
(x+y+z)=k(x+y+z)
(x+y+z)(k-1)=0
k=1或x+y+z=0
当k=1时{x+y-z=z
{y+z-x=x
{z+x-y=y
解得x=y=z(x+y)(y+z)(z+x)/xyz=8 当x+y+z=0时
(x+y)(y+z)(z+x)/xyz=(-z)(-x)(-y)/xyz=-1
解法二：(x+y-z)/z=(y+z-x)/x=(z+x-y)/y
[x+y]/z-1=[y+z]/x-1=[z+x]/y-1 [x+y]/z=[y+z]/x=[z+x]/y 设[x+y]/z=[y+z]/x=[z
