Question

若x/(y-z)+y/(z-x)+z/(x-y)=0,求证x/(y-z)

过程清楚点，偶不算是个天才

Answer 1

题目应是：若x/(y-z)+y/(z-x)+z/(x-y)=0,求证x/(y-z)²+y/(z-x)²+z/(x-y)²=0
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：[ ]内可化为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 式子x/(y-z)²+y/(z-x)²+z/(x-y)² =0（得证）

Answer 2

你的意思是解出x/(y-z)等于多少么？？