已知a,b,c为有理数,满足ab+bc+ac不等于0,a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值.
问题描述：
已知a,b,c为有理数,满足ab+bc+ac不等于0,a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值.
答
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3a(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