题目内容
用递等式计算.
720÷(22-12)×2
[558-(195+38)]÷25
360×[480÷(236-116)].
720÷(22-12)×2
[558-(195+38)]÷25
360×[480÷(236-116)].
分析:(1)先算括号内的,再算括号外的除法和乘法;
(2)(3)先算小括号内的,再算中括号内的,最后算括号外的.
(2)(3)先算小括号内的,再算中括号内的,最后算括号外的.
解答:解:(1)720÷(22-12)×2,
=720÷10×2,
=72÷2,
=36;
(2)[558-(195+38)]÷25,
=[558-233]÷25,
=325÷25,
=13;
(3)360×[480÷(236-116)],
=360×[480÷120],
=360×4,
=1440.
=720÷10×2,
=72÷2,
=36;
(2)[558-(195+38)]÷25,
=[558-233]÷25,
=325÷25,
=13;
(3)360×[480÷(236-116)],
=360×[480÷120],
=360×4,
=1440.
点评:此题主要考查了学生对运算顺序的掌握与运用情况.
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