Question

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

Answer 1

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