题目内容
| 3 |
| 2×5 |
| 3 |
| 5×8 |
| 3 |
| 8×11 |
| 3 |
| 1991×1994 |
| 3 |
| 1994×1997 |
考点:分数的拆项
专题:计算问题(巧算速算)
分析:把原式拆分为:
-
+
-
+
-
+…+
-
+
-
,然后通过加减相互抵消,即可得出答案.
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 8 |
| 1 |
| 8 |
| 1 |
| 11 |
| 1 |
| 1991 |
| 1 |
| 1994 |
| 1 |
| 1994 |
| 1 |
| 1997 |
解答:
解:
+
+
+…+
+
,
=
-
+
-
+
-
+…+
-
+
-
,
=
-
,
=
;
故答案为:
.
| 3 |
| 2×5 |
| 3 |
| 5×8 |
| 3 |
| 8×11 |
| 3 |
| 1991×1994 |
| 3 |
| 1994×1997 |
=
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 8 |
| 1 |
| 8 |
| 1 |
| 11 |
| 1 |
| 1991 |
| 1 |
| 1994 |
| 1 |
| 1994 |
| 1 |
| 1997 |
=
| 1 |
| 2 |
| 1 |
| 1997 |
=
| 1995 |
| 3994 |
故答案为:
| 1995 |
| 3994 |
点评:本题考查了分数拆项方法的灵活应用,拆项公式是:
=
-
.
| a |
| n(n+a) |
| 1 |
| n |
| 1 |
| n+a |
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