x'=x(y-z);y'=y(z-x);z'=z(x-y);x'=x(y-z);y'=y(z-x);z'=z(x-y);其中xz为t的函数，y,z;

Question

x'=x(y-z);y'=y(z-x);z'=z(x-y);
x'=x(y-z);y'=y(z-x);z'=z(x-y);其中x z为t的函数，y,z;

Answer 1

由题知：x/(y-z)+y/( z-x)+z/(x-y)=0
则分别将等式两边同乘以1/y-z ,1/z-x ,1/x-y
即得到三个等式：x/(y-z)^2+y/(z-x)(y-z)+z/(x-y)(y-z)=0 (1)
x/(y-z)(z-x)+y/(z-x)^2+z/(x-y)(z-x)=0 (2)
x/(y-z)(x-y)+y/( z-x)(x-y)+z/(x-y)^2=0 (3)
将（1）（2）（3）式相加得：x/(y-z)²+y/(z-x)²+z/(x-y)² +y/(z-x)(y-z)+z/(x-y)(y-z)+ x/(y-z)(z-x)+z/(x-y)(z-x)+x/(y-z)(x-y)+y/( z-x)(x-y)=0
经化简得：不加粗的可化为y(x-y)+z(z-x)+x(x-y)+z(y-z)+x(z-x)+y(y-z)除以（x-y）*（y-z）*(z-x)
经计算得分母为0,故不加粗的式子为0
因此 加粗的式子x/(y-z)²+y/(z-x)²+z/(x-y)² =0（得证）
设x+y-z/z=x-y+z/y=y+z-x/x=k
有x+y-z=kz
x-y+z=ky
y+z-x=kx
三式相加得x+y+z=k（x+y+z） k=1
得x+y=（k+1）z
x+z=（k+1）y
y+z=（k+1）x
(x+y)(y+z)(x+z)/xyz=（k+1）^3xyz/xyz=(k+1)^3=8
还漏了一种情况 -1
当x+y+z=0时,则直接可以知道x+y=-z,x+z=-y,y+z=-x,代入要求的式子中,可以求出值为-1.
所以是-1或8