| x2-1 |
| x2-2x+1 |
| 1-x |
| x+1 |
| x |
| x-1 |
答案:
原式=[| (x+1)(x-1) |
| (x-1)2 |
| x-1 |
| x+1 |
| x-1 |
| x |
=
| x+1 |
| x |
| (x-1)2 |
| x(x+1) |
=
| (x+1)2-(x-1)2 |
| x(x+1) |
=
| 4x |
| x(x+1) |
=
| 4 |
| x+1 |
故答案为:
| 4 |
| x+1 |

| x2-1 |
| x2-2x+1 |
| 1-x |
| x+1 |
| x |
| x-1 |
答案:
原式=[| (x+1)(x-1) |
| (x-1)2 |
| x-1 |
| x+1 |
| x-1 |
| x |
| x+1 |
| x |
| (x-1)2 |
| x(x+1) |
| (x+1)2-(x-1)2 |
| x(x+1) |
| 4x |
| x(x+1) |
| 4 |
| x+1 |
| 4 |
| x+1 |