题目

1.已知abc=1,求a/ab+a+1 + b/bc+b+1 + c/ac+c+1的值
2.已知a2+b2+c2=1,a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3,求a+b+c的值

提问时间：2021-03-04

答案

1.原式=a/(ab+a+1)+ab/(abc+ab+a)+abc/(ac*ab+c*ab+ab)=a/(ab+a+1)+ab/(1+ab+a)+abc/(a+1+ab)=(a+ab+abc)/(ab+a+1)=(ab+a+1)/(ab+a+1)=1
2.a/b+a/c+b/a+b/c+c/a+c/b=-3
(b/a+c/a+a/a)+(a/b+c/b+b/b)+(b/c+a/c+c/c)=0
(a+b+c)/a+(a+b+c)/b+(a+b+c)/c=0
(a+b+c)*(1/a+1/b+1/c)=0
因为(a+b+c)2=a2+b2+c2+2ab+2bc+2ca,ab+bc+ca=0.5(a+b+c)2-0.5(a2+b2+c2)=0.5(a+b+c)2-0.5
1/a+1/b+1/c=[0.5(a+b+c)2-0.5]/abc
所以(a+b+c)*(1/a+1/b+1/c)=(a+b+c)*[(a+b+c)2-1]/2abc=0
所以（a+b+c)(a+b+c+1)(a+b+c-1)=0
所以a+b+c=0 或a+b+c+1=0 或a+b+c-1=0
所以a+b+c=0 或1 或-1