题目内容
计算(1)| a2 |
| a+1 |
| 1 |
| a+1 |
| a-1 |
| a2-4a+4 |
| a2-4 |
| 2a-2 |
(3)m+2-
| 4 |
| 2-m |
| x |
| x+y |
| 2y |
| x+y |
| xy |
| x+2y |
| 1 |
| x |
| 1 |
| y |
分析:(1)(3)直接通分,因式分解后约分;
(2)直接因式分解后约分;
(4)把括号里的通分,除法化为乘法,再约分.
(2)直接因式分解后约分;
(4)把括号里的通分,除法化为乘法,再约分.
解答:解:(1)
-
=
=
=a-1;
(2)
×
=
×
=
;
(3)m+2-
=
=
;
(4)(
+
)•
÷(
+
)=
•
•
=
.
| a2 |
| a+1 |
| 1 |
| a+1 |
| a2-1 |
| a+1 |
| (a+1)(a-1) |
| a+1 |
(2)
| a-1 |
| a2-4a+4 |
| a2-4 |
| 2a-2 |
| a-1 |
| (a-2)2 |
| (a+2)(a-2) |
| 2(a-1) |
| a+2 |
| 2(a-2) |
(3)m+2-
| 4 |
| 2-m |
| m2-4+4 |
| m-2 |
| m2 |
| m-2 |
(4)(
| x |
| x+y |
| 2y |
| x+y |
| xy |
| x+2y |
| 1 |
| x |
| 1 |
| y |
| x+2y |
| x+y |
| xy |
| x+2y |
| xy |
| x+y |
| x2y2 |
| (x+y)2 |
点评:本题主要考查分式的混合运算,通分、因式分解和约分是解答的关键.
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