已知ax/(a-b)=ay/(b-c)=az/(c-a)(a不得0),试求出x+y+z的值

题目

已知ax/(a-b)=ay/(b-c)=az/(c-a)(a不得0),试求出x+y+z的值

答案

设等式ax/(a-b)=ay/(b-c)=az/(c-a)=k
则有：ax=k(a-b);ay=k(b-c);az=k(c-a)
三个式子相加,可得a(x+y+z)=0;由于a不得0,所以x+y+z=0