题目内容
计算:(1-
)×(1-
)×…×(1-
)
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 20082 |
考点:平方差公式
专题:
分析:先根据平方差公式分解因式,再进行约分,即可得出答案.
解答:解:原式=(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)
=
×
×
×
×
×
×…×
×
=
×
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 2008 |
| 1 |
| 2008 |
=
| 3 |
| 2 |
| 1 |
| 2 |
| 4 |
| 3 |
| 2 |
| 3 |
| 5 |
| 4 |
| 3 |
| 4 |
| 2009 |
| 2008 |
| 2007 |
| 2008 |
=
| 1 |
| 2 |
| 2009 |
| 2008 |
=
| 2009 |
| 4016 |
点评:本题考查了约分和平方差公式的应用,主要考查学生的计算能力.
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