Question

已知abc=1,a+b+c=2,a^2+b^2+c^2=3,求(1/ab+c-1)+(1/bc+a-1)+(1/ac+b-1)的值。

Answer 1

∵abc = 1
a+b+c=2
a^2 + b^2 + c^2 =3
∴(a+b+c)^2-(a^2+b^2+c^2)
=2(ab+bc+ac) =4-3=1 ∴ab+bc+ac=1/2
又因为 a+b+c=2
c-1=1-a-b
ab+c-1=ab+1-a-b=(a-1)(b-1) 同理
bc+a-1=(b-1)(c-1) 1/ac+b-1=(c-1)(a-1)
原式=[1/(ab+c-1)]+[1/(bc+a-1)]+[1/(ca+b-1)]
=1/[(a-1)(b-1)]+1/[(b-1)(c-1)]+1/[(c-1)(a-1)]
=[(a-1)+(b-1)+(c-1)]/[(a-1)(b-1)(c-1)]
=[a+b+c-3]/[abc-(ab+bc+ac)+(a+b+c)-1]
=(-1)/[1-1/2+2-1]
=(-1)/(3/2)
=-2/3