题目内容
计算:
(1)
÷
(2)(1+
)÷
(3)1÷
•(m2-1)
(4)(x-
)÷(1-
)
(1)
| a2-b2 |
| a2+2ab+b2 |
| 2a-2b |
| a+b |
(2)(1+
| 1 |
| x |
| x2-1 |
| x |
(3)1÷
| 1+m |
| 1-m |
(4)(x-
| 2x-1 |
| x |
| 1 |
| x |
分析:(1)先将除法转化为乘法,因式分解后通分即可;
(2)先将括号内的部分通分,再将除法转化为乘法,然后约分;
(3)将除法转化为乘法,因式分解并约分;
(4)先将括号内的部分通分相减,再将除法转化为乘法,约分即可.
(2)先将括号内的部分通分,再将除法转化为乘法,然后约分;
(3)将除法转化为乘法,因式分解并约分;
(4)先将括号内的部分通分相减,再将除法转化为乘法,约分即可.
解答:解:(1)原式=
•
=
;
(2)原式=(
+
)•
=
•
=
;
(3)原式=
•(m-1)(m+1)
=-m2+2m-1;
(4)原式=
÷
=
•
=x-1.
| (a-b)(a+b) |
| (a+b)2 |
| a+b |
| 2(a-b) |
=
| 1 |
| 2 |
(2)原式=(
| x |
| x |
| 1 |
| x |
| x |
| (x-1)(x+1) |
=
| x+1 |
| x |
| x |
| (x-1)(x+1) |
=
| 1 |
| x-1 |
(3)原式=
| 1-m |
| 1+m |
=-m2+2m-1;
(4)原式=
| x2-2x+1 |
| x |
| x-1 |
| x |
=
| (x-1)2 |
| x |
| x |
| x-1 |
=x-1.
点评:本题考查了分式的混合运算,将除法转化为乘法并进行因式分解是解题的关键.
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