题目内容
观察下列分母有理化的运算.1 | ||
|
2 |
1 | ||||
|
3 |
2 |
1 | ||||
|
4 |
3 |
…
1 | ||||
|
2006 |
2005 |
1 | ||||
|
2007 |
2006 |
利用上面的规律计算:
(
1 | ||
|
1 | ||||
|
1 | ||||
|
1 | ||||
|
1 | ||||
|
2007 |
分析:根据给出的分母有理化的规律对所求式子进行分母有理化,寻找抵消规律,再运用平方差公式计算.
解答:解:(
+
+
+…+
+
)(1+
)
=(
-1+
-
+
-
+…+
-
)(1+
)
=(
-1)(1+
)
=2007-1
=2006.
1 | ||
|
1 | ||||
|
1 | ||||
|
1 | ||||
|
1 | ||||
|
2007 |
=(
2 |
3 |
2 |
3 |
2 |
2007 |
2006 |
2007 |
=(
2007 |
2007 |
=2007-1
=2006.
点评:本题考查了分母的有理化,题目较为新颖,注意发现抵消规律.
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