题目内容
用简便方法计算 273-187-13 |
2000÷125÷8 | 99×38+38 |
88×125 | 404×25. |
分析:(1)根据减法的性质进行计算,
(2)根据除法的性质进行计算,
(3)根据乘法分配律进行计算,
(4)先把88化顾8×11,再根据乘法交换律进行计算,
(5)先把404看作(400+4),再根据乘法分配律进行计算.据此解答.
(2)根据除法的性质进行计算,
(3)根据乘法分配律进行计算,
(4)先把88化顾8×11,再根据乘法交换律进行计算,
(5)先把404看作(400+4),再根据乘法分配律进行计算.据此解答.
解答:解:(1)273-187-13,
=273-(187+13),
=273-200,
=73;
(2)2000÷125÷8,
=2000÷(125×8),
=2000÷1000,
=2;
(3)99×38+38,
=(99+1)×38,
=100×38,
=3800;
(4)88×125,
=11×88×125,
=11×(88×125),
=11×1000,
=11000;
(5)404×25,
=(400+4)×25,
=400×25+4×25,
=10000+100,
=10100.
=273-(187+13),
=273-200,
=73;
(2)2000÷125÷8,
=2000÷(125×8),
=2000÷1000,
=2;
(3)99×38+38,
=(99+1)×38,
=100×38,
=3800;
(4)88×125,
=11×88×125,
=11×(88×125),
=11×1000,
=11000;
(5)404×25,
=(400+4)×25,
=400×25+4×25,
=10000+100,
=10100.
点评:本题综合考查了学生在计算中合理运用简便算法的能力,注意要细心.
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