题目内容
用递等式计算.(能简算的要简算)
(1)4368÷42+58×37
(2)43×25+25×47
(3)65×(203-87÷29)
(4)(187-87)×450÷150.
(1)4368÷42+58×37
(2)43×25+25×47
(3)65×(203-87÷29)
(4)(187-87)×450÷150.
分析:(1)先分别算出乘法和除法,再算加法,
(2)运用乘法的分配律进行简算,
(3)先算小括号里的除法,再算小括号里的减法,最后算括号外的乘法,
(4)先算小括号里的减法,再从左向右进行计算.
(2)运用乘法的分配律进行简算,
(3)先算小括号里的除法,再算小括号里的减法,最后算括号外的乘法,
(4)先算小括号里的减法,再从左向右进行计算.
解答:解:(1)4368÷42+58×37,
=104+2146,
=2250;
(2)43×25+25×47,
=(43+47)×25,
=90×25,
=2250;
(3)65×(203-87÷29),
=65×(203-3),
=65×200,
=13000;
(4)(187-87)×450÷150,
=100×450÷150,
=45000÷150,
=300.
=104+2146,
=2250;
(2)43×25+25×47,
=(43+47)×25,
=90×25,
=2250;
(3)65×(203-87÷29),
=65×(203-3),
=65×200,
=13000;
(4)(187-87)×450÷150,
=100×450÷150,
=45000÷150,
=300.
点评:本题主要考查学生依据四则运算计算顺序进行计算,以及运用简便算法解决问题的能力.
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