题目内容
计算:(1-
)(1-
)(1-
)…(1-
)(n是正整数).
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| n2 |
考点:因式分解-运用公式法
专题:
分析:直接利用平方差公式分解因式,进而化简求出即可.
解答:解:(1-
)(1-
)(1-
)…(1-
)
=(1+
)(1-
)(1+
)(1-
)(1+
)(1-
)…(1+
)(1-
)
=
×
×
×
×
×
×
×
…
×
=
×
=
.
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| n2 |
=(1+
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n |
=
| 3 |
| 2 |
| 1 |
| 2 |
| 4 |
| 3 |
| 2 |
| 3 |
| 5 |
| 4 |
| 3 |
| 4 |
| 6 |
| 5 |
| 4 |
| 5 |
| n+1 |
| n |
| n-1 |
| n |
=
| 1 |
| 2 |
| n+1 |
| n |
=
| n+1 |
| 2n |
点评:此题主要考查了公式法分解因式,熟练应用平方差公式是解题关键.
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