题目内容
(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
)等于
.
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
99 |
1 |
99 |
50 |
99 |
50 |
99 |
分析:此题如果按部就班地进行计算,计算量可想而知,所以要寻求巧算的方法,此题可利用乘法结合律进行简算.
解答:解:(1+
)×(1-
)×(1+
)×(1-
)×…×(1+
)×(1-
),
=[(1+
)×(1+
)×…×(1+
)]×[(1-
)×(1-
)×…×(1-
)],
=[
×
×…×
]×[
×
×…×
],
=
×
,
=
.
故答案为:
.
1 |
2 |
1 |
2 |
1 |
3 |
1 |
3 |
1 |
99 |
1 |
99 |
=[(1+
1 |
2 |
1 |
3 |
1 |
99 |
1 |
2 |
1 |
3 |
1 |
99 |
=[
3 |
2 |
4 |
3 |
100 |
99 |
1 |
2 |
2 |
3 |
98 |
99 |
=
100 |
2 |
1 |
99 |
=
50 |
99 |
故答案为:
50 |
99 |
点评:此题考查了学生乘法结合律的知识,以及巧算的能力.
练习册系列答案
相关题目