| 1 |
| a+1 |
| a+3 |
| (a+1)(a−1) |
| (a−1)2 |
| (a+1)(a+3) |
=
| 1 |
| a+1 |
| a−1 |
| (a+1)2 |
=
| a+1−(a−1) |
| (a+1)2 |
=
| 2 |
| a2+2a+1 |
∵a2+2a-8=0，
∴a2+2a=8，
∴原式=
| 2 |
| 8+1 |
| 2 |
| 9 |
故答案为
| 2 |
| 9 |
| 1 |
| a+1 |
| a+3 |
| a2−1 |
| a2−2a+1 |
| (a+1)(a+3) |
| 1 |
| a+1 |
| a+3 |
| (a+1)(a−1) |
| (a−1)2 |
| (a+1)(a+3) |
| 1 |
| a+1 |
| a−1 |
| (a+1)2 |
| a+1−(a−1) |
| (a+1)2 |
| 2 |
| a2+2a+1 |
| 2 |
| 8+1 |
| 2 |
| 9 |
| 2 |
| 9 |