题目内容

(2007?宜兴市)
用递等式计算.
(1)0.48+10.8÷0.36×0.54

(2)[9.08-(0.61+1.75)]+3.2
(3)
13
14
-
15
28
÷
5
8
+
1
4
(4)
4
5
÷[(
3
4
-
2
3
5
8
]
分析:(1)先算除法和乘法,再算加法;
(2)(4)先算小括号内的,再算中括号内的,最后算括号外的;
(3)先算除法,再算减法和加法.
解答:解:(1)0.48+10.8÷0.36×0.54,
=0.48+30×0.54,
=0.48+16.2,
=16.68;

(2)[9.08-(0.61+1.75)]+3.2,
=[9.08-2.36]+3.2,
=6.72+3.2,
=9.92;

(3)
13
14
-
15
28
÷
5
8
+
1
4

=
13
14
-
15
28
×
8
5
+
1
4

=
13
14
-
6
7
+
1
4

=
1
14
+
1
4

=
9
28


(4)
4
5
÷[(
3
4
-
2
3
)÷
5
8
],
=
4
5
÷[
1
12
÷
5
8
],
=
4
5
÷[
1
12
×
8
5
],
=
4
5
÷
2
15

=
4
5
×
15
2

=6.
点评:用递等式计算,特别注意运算顺序和运算法则.
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