题目内容
0.13+0.23+0.33+…+0.93．
分析：从1开始的连续自然数的立方和公式为：13+23+33+43+…+n3=（1+2+3+4+…+n）2，据此利用公式计算比较简便．
解答：解：0.13+0.23+0.33+…+0.93，
=（1+2+3+…+9）2×
，
=2025×
，
=2.025．
=（1+2+3+…+9）2×
| 1 |
| 1000 |
=2025×
| 1 |
| 1000 |
=2.025．
点评：熟记从1开始连续自然数的立方和公式是解决此题的关键．
练习册系列答案
寒假大串联黄山书社系列答案
寒假创新型自主学习第三学期寒假衔接系列答案
相关题目
| 180÷60= | 12×25= | 0÷90= |
| 4×5÷4×5= | 0.25×4+4= | 131+218+269+182= |
| 25×6×4= | 420÷35= | 36×5×4= |
| 120÷5÷3= | 198-75-25= | 120-20÷5= |
| 59亿-58亿= | （1.3+3.5）÷0.6= | 75÷2.5= |
| 0.45÷1.25÷8= | 2.5×7.6×4= | 97.6÷0.02= |
| 8.8-（0.25+0.4）= | 0.25+87+0.75= | 0.8-0.35×2= |
| 0.1÷0.01×0.1= | 9.9+9.9×99= | 2.7×3.6+6.4×2.7= |
| 720÷8÷9= | 2.9+5.7+2.1= | 2-0.16-0.64= |
| 16.7+84.5-16.7= | 6.8-（3.8+2.9）= | 9×1.3+9×6.7= |
| 5.5-5.25+5.2= | 0.8×25-10= | 42.8×25×4= |
| 7.2+2.8= | 0.9÷0.01= | 3.5+7.6= |
| 3×0.2×0.5= | 0.96÷0.3= | 1.5×0.4= |
| 7÷0.25= | 2.5×0.7×0.8= | 8-2.5= |
| 16÷1.6= | 5.6×1.01= | 8×（2.5+0.25）= |
| 0.36+0.64= | 64.32÷16= | 72.8÷0.8= |
| 8.4×0.2+1.6×0.2= | 8.4÷4.2= | 0.05×0.8= |
| 2.56-0.37= | 1.23×3= | 25×0.8×0.5= |
| 12×2.5= | 3.2÷1.6= | 3.5+3.5×3= |
| 19.6÷2= | 8.8÷2.2= | 3.9+2.03= |
| 7.2×0.3+2.8×0.3= | 13.8+9.9= | 2.4×2.5= |
| 0.75×4= | （1.5+0.25）×4= | 4.2÷0.1= |
| 4.2÷3.5= | 2.14-0.9= | 27+456+73= |
| 12÷0.3= | 2.87÷0.7= | 13÷4= |
| 24÷4+56÷4= | 2.1×0.3÷2.1= | 2.3-0÷8.1= |
| 1.37-1.37÷1.37= | 4.4÷（5×0.2）= | 7×1.3= |
| 0.96÷0.06= | 0.36×0.3= | 0.35÷0.7= |
| 13.65÷13= | 4×0.8×0.25×1.25= | 13+0.7÷0.7= |
| 7.05×23.7+7.05×76.3= | 0.125×3.2×2.5= | （12+23.1÷0.75）×0.4-5.12= |
| 728+199= | 637-201= | 16×25= |
| 1.6×[（12-7.2）÷（0.84÷2.1）]= |