题目内容
用递等式计算:(能简便的就用简便方法计算,并写出主要计算过程)
(1)38.78-19.59+41.22-30.41
(2)2.74×4.86+4.86×6.26+4.86
(3)0.08×(12.5+2.5)×0.1
(4)25.92+(10.1-14.5×0.2)
(5)0.487×12.5÷48.7
(6)(8.16+2.09)÷[(10.3-9.3)÷0.4].
解:
(1)38.78-19.59+41.22-30.41,
=38.78+41.22-19.59-30.41,
=(38.78+41.22)-(19.59+30.41),
=80-50,
=30;
(2)2.74×4.86+4.86×6.26+4.86,
=(2.74+6.26+1)×4.86,
=10×4.86,
=48.6;
(3)0.08×(12.5+2.5)×0.1,
=(0.08×12.5+0.08×2.5)×0.1,
=(1+0.2)×0.1,
=1.2×0.1,
=0.12;
(4)25.92+(10.1-14.5×0.2),
=25.92+(10.1-2.9),
=25.92+7.2,
=33.12;
(5)0.487×12.5÷48.7,
=0.487÷48.7×12.5,
=0.01×12.5,
=0.125;
(6)(8.16+2.09)÷[(10.3-9.3)÷0.4],
=10.25÷[1÷0.4],
=10.25÷2.5,
=4.1.
分析:(1)根据假发交换律和结合律以及连减的性质进行计算;
(2)根据乘法分配律进行计算;
(3)根据乘法分配律进行计算;
(4)先算乘法,再算减法,最后算加法;
(5)根据乘法交换律进行计算;
(6)先算小括号里面的加法和减法,再算中括号里面的除法,最后算括号外面的除法.
点评:四则运算,先弄清运算顺序,然后再进一步计算,能简算的要简算.
(1)38.78-19.59+41.22-30.41,
=38.78+41.22-19.59-30.41,
=(38.78+41.22)-(19.59+30.41),
=80-50,
=30;
(2)2.74×4.86+4.86×6.26+4.86,
=(2.74+6.26+1)×4.86,
=10×4.86,
=48.6;
(3)0.08×(12.5+2.5)×0.1,
=(0.08×12.5+0.08×2.5)×0.1,
=(1+0.2)×0.1,
=1.2×0.1,
=0.12;
(4)25.92+(10.1-14.5×0.2),
=25.92+(10.1-2.9),
=25.92+7.2,
=33.12;
(5)0.487×12.5÷48.7,
=0.487÷48.7×12.5,
=0.01×12.5,
=0.125;
(6)(8.16+2.09)÷[(10.3-9.3)÷0.4],
=10.25÷[1÷0.4],
=10.25÷2.5,
=4.1.
分析:(1)根据假发交换律和结合律以及连减的性质进行计算;
(2)根据乘法分配律进行计算;
(3)根据乘法分配律进行计算;
(4)先算乘法,再算减法,最后算加法;
(5)根据乘法交换律进行计算;
(6)先算小括号里面的加法和减法,再算中括号里面的除法,最后算括号外面的除法.
点评:四则运算,先弄清运算顺序,然后再进一步计算,能简算的要简算.
练习册系列答案
相关题目