已知abc不等于0,且满足a+b+c=0.求a(b/1+c/1)+b(a/1+b/1)的值.
问题描述：
已知abc不等于0,且满足a+b+c=0.求a(b/1+c/1)+b(a/1+b/1)的值.
答
a+b+c=0.
所以b+c=-a
a+b=-c
a+c=-b
题目写错了,应该是
a(b/1+c/1)+c(a/1+b/1)+b(a/1+c/1)
a(b/1+c/1)+c(a/1+b/1)+b(a/1+c/1)
=b/a+c/a+a/c+b/c+a/b+c/b
=b/(a+c)+c/(a+b)+a/(b+c)
=b/(-b)+c/(-c)+a/(-a)
=-1-1-1
=-3