题目内容
[2.1+7÷(3
-1.625)]×1
.
| 1 |
| 12 |
| 1 |
| 3 |
分析:根据四则混合运算的顺序,先算乘除后算加减,有括号的先算括号里的,计算过程中注意分数与小数的互化.
解答:解:[2.1+7÷(3
-1.625)]×1
,
=[2.1+7÷(3
- 1
)]×1
,
=(2.1+7÷
)×1
,
=(2.1+7×
)×1
,
=(2.1+
)×
,
=2.1×
+
×
,
=2.8+6.4,
=9.2.
| 1 |
| 12 |
| 1 |
| 3 |
=[2.1+7÷(3
| 1 |
| 12 |
| 5 |
| 8 |
| 1 |
| 3 |
=(2.1+7÷
| 35 |
| 24 |
| 1 |
| 3 |
=(2.1+7×
| 24 |
| 35 |
| 1 |
| 3 |
=(2.1+
| 24 |
| 5 |
| 4 |
| 3 |
=2.1×
| 4 |
| 3 |
| 24 |
| 5 |
| 4 |
| 3 |
=2.8+6.4,
=9.2.
点评:此题考查了按四则混合运算的顺序计算.
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