题目内容
[(1-
)×(1-
)×(1-
)×…×(1-
)]×(1+
)×(1+
)×(1+
)×…×(1+
)]= .
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考点:分数的巧算
专题:计算问题(巧算速算)
分析:通过观察,此题可先求出每个小括号内的结果,然后运用乘法交换律与结合律简算.
解答:
解:[(1-
)×(1-
)×(1-
)×…×(1-
)]×(1+
)×(1+
)×(1+
)×…×(1+
)]
=[
×
×
×…×
]×
×
×
×…×
=
×(
×
)×(
×
)×…×(
×
)×
=
×
=
故答案为:
.
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=[
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| 2013 |
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=
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=
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| 2013 |
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=
| 2013 |
| 4024 |
故答案为:
| 2013 |
| 4024 |
点评:注意观察题目中数字构成的特点和规律,善于灵活运用运算定律或运算技巧,巧妙解答.
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