题目内容
(
+
+…+
)+(
+…+
)+…+(
+
)+
= .
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 20 |
| 2 |
| 3 |
| 2 |
| 20 |
| 18 |
| 19 |
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| 20 |
| 19 |
| 20 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:根据数字特点,把分母相同的分数相加,原式变为
+(
+
)+(
+
+
)+…+(
+
+…+
)+(
+
+…+
),然后把每个括号内的数字算出来,在运用乘法分配律以及高斯求和公式简算即可.
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| 19 |
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| 19 |
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| 19 |
| 1 |
| 20 |
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| 20 |
| 19 |
| 20 |
解答:
解:(
+
+…+
)+(
+…+
)+…+(
+
)+
=
+(
+
)+(
+
+
)+…+(
+
+…+
)+(
+
+…+
)
=
+1+1
+…+9+9
=
×(1+2+3+…+19)
=
×[(1+19)×19÷2]
=
×190
=95??
故答案为:95.?
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| 19 |
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| 19 |
| 20 |
=
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| 1 |
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| 19 |
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| 19 |
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| 19 |
| 1 |
| 20 |
| 2 |
| 20 |
| 19 |
| 20 |
=
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| 1 |
| 2 |
| 1 |
| 2 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=95??
故答案为:95.?
点评:此题主要考查学生能否根据数字特点,巧妙灵活地运用运算定律或运算技巧,使复杂的问题简单化.
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