脱式计算5845÷7+1020×5=2150÷(104-99)=1200-200÷5+28=250+255+260+265+270=-数学

题目简介

脱式计算5845÷7+1020×5=2150÷(104-99)=1200-200÷5+28=250+255+260+265+270=-数学

题目详情

脱式计算
5845÷7+1020×5=
2150÷(104-99)=
1200-200÷5+28=250+255+260+265+270=
题型:解答题难度:中档来源:不详

答案

(1)5845÷7+1020×5,
=835+5100,
=5935;

(2)2150÷(104-99),
=2150÷5,
=430;

(3)1200-200÷5+28,
=1200-40+28,
=1160+28,
=1188;

(4)250+255+260+265+270,
=250+(255+265)+(260+270),
=250+520+530,
=1300.

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