题目内容
1-
-
-…-
= .
| 2 |
| 1×(1+2) |
| 3 |
| (1+2)×(1+2+3) |
| 10 |
| (1+2+…+9)×(1+2+…+10) |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:通过观察,每个分数可拆成两个分数相减的形式,然后通过加减相互抵消,解决问题.
解答:
解:1-
-
-…-
=1-2×(
-
+
-
+…+
-
)
=1-2×(
-
)
=1-2×
=1-
=
故答案为:
.
| 2 |
| 1×(1+2) |
| 3 |
| (1+2)×(1+2+3) |
| 10 |
| (1+2+…+9)×(1+2+…+10) |
=1-2×(
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 9×10 |
| 1 |
| 10×11 |
=1-2×(
| 1 |
| 2 |
| 1 |
| 10×11 |
=1-2×
| 54 |
| 110 |
=1-
| 54 |
| 55 |
=
| 1 |
| 55 |
故答案为:
| 1 |
| 55 |
点评:此题解答的关键在于根据数据特点,进行分数拆分,灵活简算.
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