解三元一次方程式

bc=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+ca)
4 = 3 + 2 (ab+bc+ca)
ab+bc+ca = 1/2

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

I think this Q worth 20 point...

很明顯這是根與係數的問題
不過只有五點~算了~~