题目内容
(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)=
.
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| 3 |
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| 3 |
| 1 |
| 99 |
| 1 |
| 99 |
| 1 |
| 2 |
| 1 |
| 3 |
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| 3 |
| 1 |
| 99 |
| 1 |
| 99 |
| 50 |
| 99 |
| 50 |
| 99 |
分析:先调整一下数字顺序,把原式变为(1-
)×(1+
)×(1-
)×(1+
)×…×(1-
)(1+
),然后把每个括号内的结果求出来,通过前后分数的约分,最后剩下
×
,约分得出结果.
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| 2 |
| 1 |
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| 3 |
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| 99 |
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| 99 |
| 1 |
| 2 |
| 100 |
| 99 |
解答:解:(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)(1-
),
=(1-
)×(1+
)×(1-
)×(1+
)×…×(1-
)(1+
),
=
×
×
×
×…×
×
,
=
×
,
=
.
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| 1 |
| 3 |
| 1 |
| 99 |
| 1 |
| 99 |
=(1-
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| 1 |
| 2 |
| 1 |
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| 3 |
| 1 |
| 99 |
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| 99 |
=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 98 |
| 99 |
| 100 |
| 99 |
=
| 1 |
| 2 |
| 100 |
| 99 |
=
| 50 |
| 99 |
点评:对于这类问题应仔细审题,认真观察,找出简算的途径,由此解决问题.
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