已知abc=1,a+b+c=2,a²+b²+c²=3,则1/(ab+c-1)+1/(bc+a-1)+1/(ac+b-1)的值是（ ）
问题描述：
已知abc=1,a+b+c=2,a²+b²+c²=3,则1/(ab+c-1)+1/(bc+a-1)+1/(ac+b-1)的值是（ ）
题是没有错的哦!我看见人家解答出来了,但是由于太过复杂...所以希望有人能有更简单的方法~~
答
a+b+c=2
ab+c-1=ab+2-a-b-1=ab-a-b+1=(a-1)(b-1)
1/(ab+c-1)=1/(a-1)(b-1))
1/(ab+c-1)+1/(bc+a-1)+1/(ac+b-1)
=1/(a-1)(b-1))+1/(c-1)(b-1))+1/(a-1)(c-1))
=(a+b+c-3)/((a-1)(b-1)(c-1))
=-1/((a-1)(b-1)(c-1))
(a-1)(b-1)(c-1)
=abc+a+b+c-1-ab-ac-bc
=1+2-1-ab-ac-bc
=2-ab-ac-bc
(a+b+c)^2=a²+b²+c²+2ab+2ac+2bc
2^2=3+2(ab+ac+bc)
ab+ac+bc=1/2
(a-1)(b-1)(c-1)=2-1/2=3/2
1/(ab+c-1)+1/(bc+a-1)+1/(ac+b-1)
=-1/((a-1)(b-1)(c-1))
=-2/3