题目内容
0.7×1
+2
×l5+0.7×
+
×l5.
| 4 |
| 9 |
| 3 |
| 4 |
| 5 |
| 9 |
| 1 |
| 4 |
分析:此题先用加法交换律和结合律把原式变为(0.7×1
+0.7×
)+(2
×15+
×15),然后每个括号内用乘法分配律的逆运算进行计算,达到简算的目的,从而解决问题.
| 4 |
| 9 |
| 5 |
| 9 |
| 3 |
| 4 |
| 1 |
| 4 |
解答:解:0.7×1
+2
×l5+0.7×
+
×l5,
=(0.7×1
+0.7×
)+(2
×15+
×15),
=(1
+
)×0.7+(2
+
)×15,
=2×0.7+3×15,
=1.4+45,
=46.4.
| 4 |
| 9 |
| 3 |
| 4 |
| 5 |
| 9 |
| 1 |
| 4 |
=(0.7×1
| 4 |
| 9 |
| 5 |
| 9 |
| 3 |
| 4 |
| 1 |
| 4 |
=(1
| 4 |
| 9 |
| 5 |
| 9 |
| 3 |
| 4 |
| 1 |
| 4 |
=2×0.7+3×15,
=1.4+45,
=46.4.
点评:这道题主要考查对加法交换律和结合律以及乘法分配律的灵活运用能力.另外,0.7不要化为分数,到最后再做决定.
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