已知abc不等于0,且满足a+b+c=0 求a(b/1+c/1)+c(a/1+b/1)+b(a/1+c/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)+c(a/1+b/1)+b(a/1+c/1)的值
答
a(b/1+c/1)+c(a/1+b/1)+b(a/1+c/1)
=a(a/1+b/1+c/1)+c(c/1+a/1+b/1)+b(b/1+a/1+c/1)-a*a/1-b*b/1-c*c/1
=(a+b+c)(a/1+b/1+c/1))-a*a/1-b*b/1-c*c/1 因为a+b+c=0
=-a*a/1-b*b/1-c*c/1
=-3
答
b/1不是等于b吗，直接写b不就得了？还是你写错了？
答
a(b/1+c/1)+c(a/1+b/1)+b(a/1+c/1)=a(a/1+b/1+c/1)+c(c/1+a/1+b/1)+b(b/1+a/1+c/1)-a*a/1-b*b/1-c*c/1=(a+b+c)(a/1+b/1+c/1))-a*a/1-b*b/1-c*c/1 因为a+b+c=0=-a*a/1-b*b/1-c*c/1=-3
答
a(1/b+1/c)+c(1/a+1/b)+b(1/a+1/c)
=(a+b)/c+(a+c)/b+(b+c)/a
=-c/c-b/b-a/a
=-1-1-1
=-3