题目内容
运用平方差公式进行化简下式:(1-
)(1-
)(1-
)…(1-
).
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| 42 |
| 1 |
| n2 |
分析:根据平方差公式分解因式,再求出每个括号内的值,最后约分,即可求出答案.
解答:解:原式=(1-
)(1+
)(1-
)(1+
)(1-
)(1+
)…(1-
)(1+
)
=
×
×
×
×
×
×…
×
=
×
=
.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n |
=
| 1 |
| 2 |
| 3 |
| 2 |
| 2 |
| 3 |
| 4 |
| 3 |
| 3 |
| 4 |
| 5 |
| 4 |
| n-1 |
| n |
| n+1 |
| n |
=
| 1 |
| 2 |
| n+1 |
| n |
=
| n+1 |
| 2n |
点评:本题考查了平方差公式的应用,注意:平方差公式是:(a+b)(a-b)=a2-b2.
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