题目内容
(1-| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 52 |
| 1 |
| 92 |
| 1 |
| 102 |
分析:1-
=(1+
)(1-
)=
×
;同理可得:1-
=
×
,…,1-
=
×
,根据此条件即可解答.
| 1 |
| 22 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
| 1 |
| 32 |
| 2 |
| 3 |
| 4 |
| 3 |
| 1 |
| 102 |
| 9 |
| 10 |
| 11 |
| 10 |
解答:解:原式=
×
×
×…×
×
=
.
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 9 |
| 10 |
| 11 |
| 10 |
| 11 |
| 20 |
点评:利用平方差公式把每个因式进行分解,注意到各个因式的特点是解决本题的关键.
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