题目内容

递等式计算.
[3.6-(
1
2
+2.8×
3
11
)]÷1.2   
4
1
2
÷[(
1
2
+0.75)×1
1
5
]
4.5÷[(5.7+9.38×
3
7
-
18
25
)×
1
20
]
[
5
6
+1
8
9
×(2
1
4
-1
1
8
)÷
5
24
].
分析:(1)先计算括号内部的,在计算括号外面的,数值较大,认真计算.
(2)、(3)先计算中括号内部的小括号,在计算中括号的,最后计算括号外面的.
(4)先计算中括号内部的再计算括号外面的,数值较大,认真计算.
解答:解:(1)[3.6-(
1
2
+2.8×
3
11
)]÷1.2,
=[3.6-(
1
2
+2.8×
3
11
)]÷
6
5

=[3.6-(
1
2
+
14
5
×
3
11
)]×
5
6

=[3.6-(
1
2
+
42
55
)]×
5
6

=[3.6-(
55
110
+
84
110
)]×
5
6

=[3.6-
139
110
5
6

=3.6×
5
6
-
139
110
×
5
6

=3-
139
132

=3-1
7
132

=1
125
132
;  

(2)4
1
2
÷[(
1
2
+0.75)×1
1
5
],
=
9
2
÷[(
2
4
+
3
4
)×
6
5
],
=
9
2
÷[
5
4
×
6
5
],
=
9
2
÷
3
2

=
9
2
×
2
3

=3;

(3)4.5÷[(5.7+9.38×
3
7
-
18
25
)×
1
20
],
=4.5÷[(5.7+9.38×
3
7
-
18
25
)×
1
20
],
=4.5÷[(5.7+4.02-0.72)×
1
20
],
=4.5÷[(9.72-0.72)×
1
20
],
=4.5÷[9×
1
20
],
=4.5÷
9
20

=4.5×
20
9

=10;

(4)[
5
6
+1
8
9
×(2
1
4
-1
1
8
)÷
5
24
],
=[
5
6
+
17
9
×(2
2
8
-
1
1
8
)×
24
5
],
=
5
6
+
17
9
×
9
8
×
24
5

=
5
6
+
51
5

=
25
30
+
306
30

=
331
30

=11
1
30
点评:本题主要考查了分数的四则混合运算的顺序,同时考查了分数的计算能力.
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