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 |