题目内容
请先阅读下列一段内容,然后解答问题:因为:
| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 9×10 |
| 1 |
| 9 |
| 1 |
| 10 |
所以:
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 9×10 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 9 |
| 1 |
| 10 |
| 1 |
| 10 |
| 9 |
| 10 |
计算:
(1)
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 2007×2008 |
(2)
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 49×51 |
分析:此类分数的加法计算要熟练运用拆分的方法达到抵消的目的,进行简便计算.
解答:解:(1)原式=1-
+
-
+…+
-
=1-
=
;
(2)∵
=
=
(1-
),
=
=
×(
-
),…
=
×(
-
),
∴原式=
(1-
+
-
+…+
-
)=
(1-
)=
.
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2007 |
| 1 |
| 2008 |
| 1 |
| 2008 |
| 2007 |
| 2008 |
(2)∵
| 1 |
| 1×3 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×5 |
| 1 |
| 15 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 49×51 |
| 1 |
| 2 |
| 1 |
| 49 |
| 1 |
| 51 |
∴原式=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 49 |
| 1 |
| 51 |
| 1 |
| 2 |
| 1 |
| 51 |
| 25 |
| 51 |
点评:本题是一道找规律的题目,要求学生通过观察,分析、归纳发现其中的规律,并应用发现的规律解决问题.熟练掌握分数的拆分计算.
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