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

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