题目内容
乘积(1-
)(1-
)…(1-
)(1-
)应等于( )
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 92 |
| 1 |
| 102 |
分析:根据所给的式子利用平方差公式因式分解,再把所得的式子分子与分母进行约分,即可求出答案;
解答:解:(1-
)(1-
)…(1-
)(1-
)
=(1+
)(1-
)(1+
)(1-
)…(1+
)(1-
)(1+
)(1-
)
=
×
×
×
…
×
×
×
=
×
=
;
故选C.
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 92 |
| 1 |
| 102 |
=(1+
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 10 |
=
| 3 |
| 2 |
| 1 |
| 2 |
| 4 |
| 3 |
| 2 |
| 3 |
| 10 |
| 9 |
| 8 |
| 9 |
| 11 |
| 10 |
| 9 |
| 10 |
=
| 1 |
| 2 |
| 11 |
| 10 |
=
| 11 |
| 20 |
故选C.
点评:此题考查了因式分解;解题的关键是找出规律,再把分子与分母进行约分.
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