(1)(
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x-4 |
| x |
(2)
| x+1 |
| x-1 |
| 4 |
| x2-1 |
答案:
(1)原式=[| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x |
| x-4 |
=
| (x+2)(x-2)-x(x-1) |
| x(x-2)2 |
| x |
| x-4 |
=
| -3x |
| x(x-2)2 |
| x |
| x-4 |
=
| -3x |
| (x-2)2(x-4) |
即x2+2x+1-4=x2-1,
则2x-3=-1,
解得:x=1,
检验:当x=1时,(x+1)(x-1)=0,则x=1不是方程的解.
方程无解.

| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| x-4 |
| x |
| x+1 |
| x-1 |
| 4 |
| x2-1 |
答案:
(1)原式=[| x+2 |
| x(x-2) |
| x-1 |
| (x-2)2 |
| x |
| x-4 |
| (x+2)(x-2)-x(x-1) |
| x(x-2)2 |
| x |
| x-4 |
| -3x |
| x(x-2)2 |
| x |
| x-4 |
| -3x |
| (x-2)2(x-4) |