题目内容
(1-2-
)-1(1-2-
)(1+2-
)(1+2-
)(1+2-
)(1+2-
)=( )
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| 32 |
| 1 |
| 16 |
| 1 |
| 16 |
| 1 |
| 8 |
| 1 |
| 4 |
| 1 |
| 2 |
A、
| ||||
B、
| ||||
C、
| ||||
D、(1-2-
|
分析:如果把每一个括号看作一个数,本题是求5个数的积.如果从左往右计算,则非常麻烦.观察发现(1-2-
)(1+2-
)符合平方差公式结构,利用平方差公式首先计算这两个数的积,所得结果再与(1+2-
)相乘,也符合平方差公式结构,依次计算即可.
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| 16 |
| 1 |
| 16 |
| 1 |
| 8 |
解答:解:原式=(1-2-
)-1[(1-2-
)(1+2-
)](1+2-
)(1+2-
)(1+2-
)
=(1-2-
)-1(1-2-
)(1+2-
)(1+2-
)(1+2-
)
=(1-2-
)-1(1-2-
)(1+2-
)(1+2-
)
=(1-2-
)-1(1-2-
)(1+2-
)
=(1-2-
)-1(1-2-1)
=
(1-2-
)-1.
故选A.
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| 16 |
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| 8 |
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=(1-2-
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| 32 |
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| 8 |
| 1 |
| 8 |
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=(1-2-
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| 32 |
| 1 |
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| 1 |
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=(1-2-
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| 32 |
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=(1-2-
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| 32 |
=
| 1 |
| 2 |
| 1 |
| 32 |
故选A.
点评:本题主要考查了平方差公式及幂的乘方运算性质.牢固掌握平方差公式的结构特征是解题的关键,难度中等.
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