题目：
先化简，再求代数式（1-| 3 |
| x+2 |
| x2-1 |
| x+2 |
答案
解：原式=| x-1 |
| x+2 |
| x+2 |
| (x+1)(x-1) |
=
| 1 |
| x+1 |
∵x=2cos30°-tan45°，
∴x=2×
| ||
| 2 |
| 3 |
∴原式=
| 1 | ||
|
| ||
| 3 |
解：原式=
| x-1 |
| x+2 |
| x+2 |
| (x+1)(x-1) |
=
| 1 |
| x+1 |
∵x=2cos30°-tan45°，
∴x=2×
| ||
| 2 |
| 3 |
∴原式=
| 1 | ||
|
| ||
| 3 |
| 3 |
| x+2 |
| x2-1 |
| x+2 |
| x-1 |
| x+2 |
| x+2 |
| (x+1)(x-1) |
| 1 |
| x+1 |
| ||
| 2 |
| 3 |
| 1 | ||
|
| ||
| 3 |
| x-1 |
| x+2 |
| x+2 |
| (x+1)(x-1) |
| 1 |
| x+1 |
| ||
| 2 |
| 3 |
| 1 | ||
|
| ||
| 3 |
（2009·三明）如图，△ABC是边长为2的等边三角形，将△ABC沿射线BC向右平移得到△DCE，连接AD、BD，下列结论错误的是（ ）