题目内容
用递等式计算.
49200÷24-1898
3.5×1.6+4.65÷3.1
+
×(
-
)
303÷[(72-37)×
].
49200÷24-1898
3.5×1.6+4.65÷3.1
5 |
6 |
8 |
15 |
2 |
3 |
1 |
4 |
303÷[(72-37)×
6 |
7 |
考点:整数、分数、小数、百分数四则混合运算
专题:运算顺序及法则
分析:(1)先算除法,再算减法.
(2)先算乘除,再算加法.
(3)先算括号内的,再算括号外的乘法,最后算加法.
(4)先算小括号内的,再算中括号内的,最后算括号外的.
(2)先算乘除,再算加法.
(3)先算括号内的,再算括号外的乘法,最后算加法.
(4)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)49200÷24-1898
=2050-1898
=153
(2)3.5×1.6+4.65÷3.1
=5.6+1.5
=7.1
(3)
+
×(
-
)
=
+
×
=
+
=
(4)303÷[(72-37)×
]
=303÷[35×
]
=303÷30
=10.1
=2050-1898
=153
(2)3.5×1.6+4.65÷3.1
=5.6+1.5
=7.1
(3)
5 |
6 |
8 |
15 |
2 |
3 |
1 |
4 |
=
5 |
6 |
8 |
15 |
5 |
12 |
=
5 |
6 |
2 |
9 |
=
19 |
18 |
(4)303÷[(72-37)×
6 |
7 |
=303÷[35×
6 |
7 |
=303÷30
=10.1
点评:此题主要考查了整数、小数、分数的四则混合运算,注意运算顺序和运算法则.
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