题目内容
计算:
(1)125×888=
(2)1
+3
+5
+7
+9
=
(1)125×888=
111000
111000
.(2)1
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
25
| 5 |
| 6 |
25
.| 5 |
| 6 |
分析:(1)125×888,将原式转化为:125×8×111,然后运用乘法结合律进行简算.
(2)1
+3
+5
+7
+9
,首先将原式转化为:(1+3+5+7+9)+
+
+
+
+
,把各个分数分别进行拆分,有规律地达到抵消的目的,依此计算即可.
(2)1
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
解答:解(1)125×888=,
=(125×8)×111,
=1000×111,
=111000;
(2)1
+3
+5
+7
+9
,
=(1+3+5+7+9)+
+
+
+
+
,
=25+
+
-
+
-
+
-
+
-
,
=25+1-
,
=25
;
故答案为:11100,25
.
=(125×8)×111,
=1000×111,
=111000;
(2)1
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
=(1+3+5+7+9)+
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 20 |
| 1 |
| 30 |
=25+
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 6 |
=25+1-
| 1 |
| 6 |
=25
| 5 |
| 6 |
故答案为:11100,25
| 5 |
| 6 |
点评:此题考查了分数的巧算,关键是对各个分数进行拆分,从而达到抵消的目的.
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