题目内容
| 22 |
| 1×3 |
| 42 |
| 3×5 |
| 62 |
| 5×7 |
| 82 |
| 7×9 |
| 102 |
| 9×11 |
| 122 |
| 11×13 |
考点:分数的拆项
专题:计算问题(巧算速算)
分析:根据观察可知:每个分子比分母大1,据此可写成带分数的形式,再根据分数的折项进行解答.
解答:
解:
+
+
+
+
+
=
+
+
+
+
+
=1
+1
+1
+1
+1
+1
=(1+1+1+1+1+1)+(
+
+
+
+
+
)
=6+(1-
+
-
+
-
+
-
+
-
+
-
)×
=6+
×
=6+
=6
故答案为:6
.
| 22 |
| 1×3 |
| 42 |
| 3×5 |
| 62 |
| 5×7 |
| 82 |
| 7×9 |
| 102 |
| 9×11 |
| 122 |
| 11×13 |
=
| 4 |
| 3 |
| 16 |
| 15 |
| 36 |
| 35 |
| 64 |
| 63 |
| 100 |
| 99 |
| 144 |
| 143 |
=1
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
| 1 |
| 143 |
=(1+1+1+1+1+1)+(
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 7×9 |
| 1 |
| 9×11 |
| 1 |
| 11×13 |
=6+(1-
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 11 |
| 1 |
| 13 |
| 1 |
| 2 |
=6+
| 12 |
| 13 |
| 1 |
| 2 |
=6+
| 6 |
| 13 |
=6
| 6 |
| 13 |
故答案为:6
| 6 |
| 13 |
点评:本题的关键是找出原式中分子和分母的特点,再进行计算.
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