题目内容
计算:
(1)
-
;
(2)
-
.
(1)
| 2m |
| 5n2p |
| 3n |
| 4mp2 |
(2)
| 2x |
| x2-64y2 |
| 1 |
| x-8y |
考点:分式的加减法
专题:
分析:先把分式进行通分,再相加减即可.
解答:解:(1)
-
=
-
,
=
,
(2)
-
=
-
,
=
-
,
=
,
=
.
| 2m |
| 5n2p |
| 3n |
| 4mp2 |
=
| 2m•4mp |
| 20n2p2m |
| 3n•5n2 |
| 20n2p2m |
=
| 8m2p-15n3 |
| 20n2p2m |
(2)
| 2x |
| x2-64y2 |
| 1 |
| x-8y |
=
| 2x |
| (x+8y)(x-8y) |
| 1 |
| x-8y |
=
| 2x |
| (x+8y)(x-8y) |
| x+8y |
| (x+8y)(x-8y) |
=
| x-8y |
| (x+8y)(x-8y) |
=
| 1 |
| x+8y |
点评:本题主要考查了分式的加减法,解题的关键是分式的通分.
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