题目内容
计算(1-
)(1-
)…(1-
)(1-
)=( )
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 92 |
| 1 |
| 102 |
A、
| ||
B、
| ||
C、
| ||
D、
|
分析:先利用平方差公式把原式展开,得到原式=(1-
)(1+
)×(1-
)(1+
)×…×(1-
)×(1+
)×(1-
)×(1+
),然后算出括号里的数,再依次相乘即可得到答案.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 10 |
解答:解:原式=(1-
)(1+
)×(1-
)(1+
)×…×(1-
)×(1+
)×(1-
)×(1+
),
=
×
×
×
×…×
×
×
×
,
=
.
故选D.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 10 |
=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 8 |
| 9 |
| 10 |
| 9 |
| 9 |
| 10 |
| 11 |
| 10 |
=
| 11 |
| 20 |
故选D.
点评:本题考查有理数的乘方以及平方差公式,是各地中考题中常见的计算题型.解题的关键是利用平方差公式把原式展开再进行约分从而得出答案.
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