题目内容
计算:(1-
)(1-
)…(1-
).
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| 22 |
| 1 |
| 32 |
| 1 |
| 102 |
考点:因式分解的应用
专题:
分析:根据所给代数式的结构特点,运用因式分解法来变形、化简,即可解决问题.
解答:解:原式
=(1-
)(1+
)(1-
)(1+
)(1-
)(1+
)…(1-
)(1+
)
=
×
×
×
×
×
…×
×
=
×
=
.
=(1-
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| 2 |
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| 3 |
| 1 |
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| 1 |
| 4 |
| 1 |
| 10 |
| 1 |
| 10 |
=
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| 2 |
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| 3 |
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=
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| 11 |
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=
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点评:该题主要考查了因式分解及其应用问题;解题的关键是灵活运用因式分解来变形、化简、运算、求值.
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