得a2+b2+c2+2ab+2bc+2ac=4,
将已知代入,得ab+bc+ac=
1 |
2 |
由a+b+c=2得:c-1=1-a-b,
∴ab+c-1=ab+1-a-b=(a-1)(b-1),
同理,得bc+a-1=(b-1)(c-1),
ca+b-1=(c-1)(a-1),
∴原式=
1 |
(a−1)(b−1) |
1 |
(b−1)(c−1) |
1 |
(c−1)(a−1) |
=
c−1+a−1+b−1 |
(a−1)(b−1)(c−1) |
=
−1 |
(ab−a−b+1)(c−1) |
=
−1 |
abc−ac−bc+c−ab+a+b−1 |
=
−1 | ||
1−
|
2 |
3 |
故选D.