得a2+b2+c2+2ab+2bc+2ac=4，
將已知代入，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 |
故選D．