Question

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~~

Answer 1

∵a+b+c=2 a²+b²+c²=3
∴(a+b+c)^2=a²+b²+c²+2(ab+bc+ac)=3+2(ab+bc+ac)=4
∴ab+bc+ac=1/2

∵a+b+c=2
∴c-1=1-a-b
∴ab+c-1=ab+1-a-b=(ab-a)-(b-1)=a(b-1)-(b-1)=(a-1)(b-1)
同理可得： bc+a-1=(b-1)(c-1) ac+b-1=(a-1)(c-1)

1/(ab+c-1)+1/(bc+a-1)+1/(ac+b-1)
=1/(a-1)(b-1)+1/(b-1)(c-1)+1/(a-1)(c-1)
=[(c-1)+(a-1)+(b-1)]/(a-1)(b-1)(c-1)
=(a+b+c-3)/(abc+a+b+c-ab-ac-bc-1)
=(2-3)/(1+2-1/2-1)
=-2/3

Answer 2

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

Answer 3

我注意到 A+B+C大于等于3