题目内容
30×(
+
+
+
+
+
)
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
| 1 |
| 143 |
| 1 |
| 195 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:通过观察,括号内可变成(
+
+
+
+
+
,分母中的两个因数相差2,因此提出2,把每个分数进行拆分,通过加减相互抵消,解决问题.
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 7×9 |
| 1 |
| 9×11 |
| 1 |
| 11×13 |
| 1 |
| 13×15 |
解答:
解:30×(
+
+
+
+
+
)
=30×(
+
+
+
+
+
)
=30×
×(
-
+
-
+
-
+
-
+
-
+
-
)
=15×(
-
)
=15×
=4
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
| 1 |
| 143 |
| 1 |
| 195 |
=30×(
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 7×9 |
| 1 |
| 9×11 |
| 1 |
| 11×13 |
| 1 |
| 13×15 |
=30×
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 11 |
| 1 |
| 13 |
| 1 |
| 13 |
| 1 |
| 15 |
=15×(
| 1 |
| 3 |
| 1 |
| 15 |
=15×
| 4 |
| 15 |
=4
点评:此题采用了裂项消项法,先进行分数裂项,然后通过加减相互抵消,求出结果.
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