题目内容
(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 100 |
| 1 |
| 100 |
分析:首先把括号内的加减运算完成,然后前后分子分母约分,即可得解.
解答:解:(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)
=
×
×
×
×
×
×…×
×
=
×
×
×
×
×
×…×
×
=
×
=
.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 100 |
| 1 |
| 100 |
=
| 3 |
| 2 |
| 1 |
| 2 |
| 4 |
| 3 |
| 2 |
| 3 |
| 5 |
| 4 |
| 3 |
| 4 |
| 101 |
| 100 |
| 99 |
| 100 |
=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 3 |
| 4 |
| 5 |
| 4 |
| 99 |
| 100 |
| 101 |
| 100 |
=
| 1 |
| 2 |
| 101 |
| 100 |
=
| 101 |
| 200 |
点评:利用分数的分数的四则混合运算,首先计算括号内的,然后约分是解决此题的关键.
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