题目内容
递等式计算(能简便的要简便)
| ①( | ②2.5×13×0.42 | ③5.3×12-33.46÷0.35 |
| ④ | ⑤9.34-(2.34+0.49) | ⑥ |
解:(1)(
+
)÷
,
=(
+
)×24,
=
×24+
×24,
=9+20,
=29;
(2)2.5×13×0.42,
=32.5×0.42,
=16.65;
(3)5.3×12-33.46÷0.35,
=63.6-95.6,
=-32;
(4)
÷[(
+
)×
],
=
÷[
×
],
=
,
=
;
(5)9.34-(2.34+0.49),
=9.34-2.34-0.49,
=7-0.49,
=6.51;
(6)
+
×19,
=
×(1+19),
=
×20,
=16.
分析:(1)把除法转化为乘法,再根据乘法分配律进行计算;
(2)从左到右依次计算;
(3)先算乘除,再算减法;
(4)先算括号里面的,再算中括号里的,最后算除法;
(5)根据减法的性质进行计算;
(6)根据乘法分配律进行计算.
点评:本题主要考查了学生四则运算的计算能力,注意在计算中灵活运用简便算法的能力.
=(
=
=9+20,
=29;
(2)2.5×13×0.42,
=32.5×0.42,
=16.65;
(3)5.3×12-33.46÷0.35,
=63.6-95.6,
=-32;
(4)
=
=
=
(5)9.34-(2.34+0.49),
=9.34-2.34-0.49,
=7-0.49,
=6.51;
(6)
=
=
=16.
分析:(1)把除法转化为乘法,再根据乘法分配律进行计算;
(2)从左到右依次计算;
(3)先算乘除,再算减法;
(4)先算括号里面的,再算中括号里的,最后算除法;
(5)根据减法的性质进行计算;
(6)根据乘法分配律进行计算.
点评:本题主要考查了学生四则运算的计算能力,注意在计算中灵活运用简便算法的能力.
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