题目内容
计算:
(1)
+
-
;
(2)
-x-1.
(1)
| m+2n |
| n-m |
| n |
| n-m |
| 2m |
| m-n |
(2)
| x2 |
| x-1 |
(1)原式=
+
+
=
=
;
(2)原式=
-
=
=
.
| m+2n |
| n-m |
| n |
| n-m |
| 2m |
| n-m |
=
| m+2n+n+2m |
| n-m |
=
| 3(n+m) |
| n-m |
(2)原式=
| x2 |
| x-1 |
| x2-1 |
| x-1 |
=
| x2-x2+1 |
| x-1 |
=
| 1 |
| x-1 |
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