题目内容
计算:
(1)(
-
)÷
(2)
÷
+
.
(1)(
| x2 |
| x+2 |
| 4 |
| x+2 |
| x-2 |
| x |
(2)
| x-1 |
| x+2 |
| x2-2x+1 |
| x2-4 |
| 1 |
| x-1 |
(1)原式=
•
=
•
=x;
(2)原式=
•
+
=
+
=
=1.
(2)原式=
•
+
=
•
+
=
+
=
=
=1.
| x2-4 |
| x+2 |
| x |
| x-2 |
=
| (x+2)(x-2) |
| x+2 |
| x |
| x-2 |
=x;
(2)原式=
| x-1 |
| x+2 |
| (x+2)(x-2) |
| (x-1)2 |
| 1 |
| x-1 |
=
| x-2 |
| x-1 |
| 1 |
| x-1 |
=
| x-2+1 |
| x-1 |
=1.
(2)原式=
| x-1 |
| x+2 |
| x2-4 |
| x2-2x+1 |
| 1 |
| x-1 |
| x-1 |
| x+2 |
| (x+2)(x-2) |
| (x-1)2 |
| 1 |
| x-1 |
| x-2 |
| x-1 |
| 1 |
| x-1 |
| x-2+1 |
| x-1 |
| x-1 |
| x-1 |
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