已知a,b,c为有理数,满足ab+ac+bc不等于0,a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=

a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
a(1/a+1/b+1/c)+b(1/b+1/c+1/a)+c(1/c+1/a+1/b)=0
(a+b+c)(1/a+1/b+1/c)=0
(a+b+c)(ab+bc+ca)/abc=0
ab+ac+bc不等于0
所以,a+b+c=0