题目内容

(2011?昆明模拟)
计算.
1
2
×(
5
8
-
1
4
3.68-0.82-0.18 7.28-(1.28+0.25)
2
3
-
1
2
+
1
6
)÷
1
24
4
5
+
1
4
)÷
7
3
+
7
10
[1-(
1
4
+
3
8
)]÷
1
4
分析:(1)先算小括号里面的减法,再算括号外的除法,最后算乘法;
(2)(3)根据连续减去两个数等于减去这两个数的和简算;
(4)运用乘法分配律简算;
(5)先算括号里面的加法,再算括号外的除法,最后算加法;
(6)先算小括号里面的加法,再算中括号里面的减法,最后算括号外的除法.
解答:解:(1)4÷
1
2
×(
5
8
-
1
4
),
=4÷
1
2
×
3
8

=8×
3
8

=3;

(2)3.68-0.82-0.18,
=3.68-(0.82+0.18),
=3.68-1,
=2.68;

(3)7.28-(1.28+0.25),
=7.28-1.28-0.25,
=6-0.25,
=5.75;

(4)(
2
3
-
1
2
+
1
6
)÷
1
24

=(
2
3
-
1
2
+
1
6
)×24,
=
2
3
×24-
1
2
×24+
1
6
×24,
=16-12+4,
=8;

(5)(
4
5
+
1
4
)÷
7
3
+
7
10

=
21
20
×
3
7
+
7
10

=
9
20
+
7
10

=
23
20


(6)[1-(
1
4
+
3
8
)]÷
1
4

=[1-
5
8
1
4

=
3
8
÷
1
4

=
3
2
点评:这类型的题目先观察算式,看能不能运用简便运算的方法简算,若不能就要按照运算顺序逐步运算.
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