题目内容
计算
(1)
•
;
(2)(
+
)÷
.
(1)
| x |
| x2-1 |
| x+x2 |
| x2 |
(2)(
| a+2 |
| a-2 |
| 4 |
| a2-4a+4 |
| a |
| a-2 |
(1)原式=
×
=
;
(2)原式=[
+
]•
=
•
=
.
| x |
| (x+1)(x-1) |
| x(x+1) |
| x2 |
=
| 1 |
| x-1 |
(2)原式=[
| ( a+2 )( a-2 ) |
| ( a-2 )2 |
| 4 |
| ( a-2 )2 |
| a-2 |
| a |
=
| a2 |
| ( a-2 )2 |
| a-2 |
| a |
=
| a |
| a-2 |
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