题目内容
请用平方差公式计算:
(1-
)×(1-
)×(1-
)×…×(1-
)×(1-
).
(1-
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 992 |
| 1 |
| 1002 |
考点:乘方
专题:计算问题(巧算速算)
分析:把每个括号内的算式运用完全平方差公式展开,然后求出每个括号内的结果,约分计算即可.
解答:
解:(1-
)×(1-
)×(1-
)×…×(1-
)×(1-
)
=(1+
)×(1-
)×(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)×(1+
)×(1-
)
=
×
×
×
×
×
×…×
×
×
×
=
×
=
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| 992 |
| 1 |
| 1002 |
=(1+
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 99 |
| 1 |
| 99 |
| 1 |
| 100 |
| 1 |
| 100 |
=
| 3 |
| 2 |
| 1 |
| 2 |
| 4 |
| 3 |
| 2 |
| 3 |
| 5 |
| 4 |
| 3 |
| 4 |
| 100 |
| 99 |
| 98 |
| 99 |
| 101 |
| 100 |
| 99 |
| 100 |
=
| 1 |
| 2 |
| 101 |
| 100 |
=
| 101 |
| 200 |
点评:此题解答的关键在于灵活运用完全平方差公式以及约分计算.
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