题目内容
已知
+(ab-2)2=0
观察:
=
-
,
=
-
,
=
-
…通过观察,求:
+
+
+…+
的值.
| a-1 |
观察:
| 1 |
| 1×2 |
| 1 |
| 1 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| ab |
| 1 |
| (a+1)(b+1) |
| 1 |
| (a+2)(b+2) |
| 1 |
| (a+2010)(b+2010) |
分析:先根据非负数的性质求出a、b的值,再把a、b的值代入代数式进行计算即可.
解答:解:∵
+(ab-2)2=0,
∴
,解得
,
∴原式=
+
+
+…+
=
+
-
+
-
+…+
-
=1-
=
.
| a-1 |
∴
|
|
∴原式=
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 2011×2012 |
=
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2011 |
| 1 |
| 2012 |
=1-
| 1 |
| 2012 |
=
| 2011 |
| 2012 |
点评:本题考查的是分式的化简求值,根据题意找出规律是解答此题的关键.
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