题目内容
脱式计算 5845÷7+1020×5= |
2150÷(104-99)= |
1200-200÷5+28= | 250+255+260+265+270= |
分析:(1)先分别求出除法算式和乘法算式,再算加法;
(2)先算小括号里的减法,再算括号外的除法;
(3)先算除法,再从左向右进行计算;
(4)运用加法的交换律、结合律进行解答即可.
(2)先算小括号里的减法,再算括号外的除法;
(3)先算除法,再从左向右进行计算;
(4)运用加法的交换律、结合律进行解答即可.
解答:解:(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.
=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.
点评:题主要考查整数的四则混合运算的运算顺序以及计算能力.
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