题目内容
用合理的方法计算.
1
|
52.4×
|
(8.6-7
| ||||||||||||||||||
6.15÷2
|
(4
|
10
|
考点:整数、分数、小数、百分数四则混合运算,运算定律与简便运算
专题:运算顺序及法则,运算定律及简算
分析:(1)先算除法,再算减法.
(2)通过数字与乘除法转换,运用乘法分配律简算.
(3)先算括号内的,再算括号外的除法,最后算加法.
(4)解答此题应注意数字转化,先算除法和括号内的,再算括号外的乘法,最后算减法.
(5)先算括号内的乘法,再算括号内的减法,最后算括号外的除法.
(6)运用减法的性质及加法交换律与结合律简算.
(2)通过数字与乘除法转换,运用乘法分配律简算.
(3)先算括号内的,再算括号外的除法,最后算加法.
(4)解答此题应注意数字转化,先算除法和括号内的,再算括号外的乘法,最后算减法.
(5)先算括号内的乘法,再算括号内的减法,最后算括号外的除法.
(6)运用减法的性质及加法交换律与结合律简算.
解答:
解:(1)1
-0.58÷5
=1
-0.58×
=1
-0.1
=
-
=
;
(2)52.4×
+45.6×0.75+2÷1
=52.4×0.75+45.6×0.75+2×
=52.4×0.75+45.6×0.75+2×0.75
=(52.4+45.6+2)×0.75
=100×0.75
=75;
(3)(8.6-7
)÷2
+7.64
=(8
-7
)÷2
+7.64
=
×
+7.64
=
+7.64
=0.36+7.64
=8;
(4)6.15÷2
-
×(0.05+4
)
=6.15×
-
×(0.05+4.36)
=2.46-
×4.41
=2.46-0.63
=1.83;
(5)(4
-1.6×
)÷1
=(4.8-1.2)÷1
=3.6×
=3;
(6)10
-(8.8-
)+
=10.8-8.8+
+
=(10.8-8.8)+(
+
)
=2+1
=3.
| 1 |
| 3 |
| 4 |
| 5 |
=1
| 1 |
| 3 |
| 5 |
| 29 |
=1
| 1 |
| 3 |
=
| 4 |
| 3 |
| 1 |
| 10 |
=
| 37 |
| 30 |
(2)52.4×
| 3 |
| 4 |
| 1 |
| 3 |
=52.4×0.75+45.6×0.75+2×
| 3 |
| 4 |
=52.4×0.75+45.6×0.75+2×0.75
=(52.4+45.6+2)×0.75
=100×0.75
=75;
(3)(8.6-7
| 4 |
| 7 |
| 6 |
| 7 |
=(8
| 3 |
| 5 |
| 4 |
| 7 |
| 6 |
| 7 |
=
| 36 |
| 35 |
| 7 |
| 20 |
=
| 9 |
| 25 |
=0.36+7.64
=8;
(4)6.15÷2
| 1 |
| 2 |
| 1 |
| 7 |
| 9 |
| 25 |
=6.15×
| 2 |
| 5 |
| 1 |
| 7 |
=2.46-
| 1 |
| 7 |
=2.46-0.63
=1.83;
(5)(4
| 4 |
| 5 |
| 3 |
| 4 |
| 1 |
| 5 |
=(4.8-1.2)÷1
| 1 |
| 5 |
=3.6×
| 5 |
| 6 |
=3;
(6)10
| 4 |
| 5 |
| 5 |
| 7 |
| 2 |
| 7 |
=10.8-8.8+
| 5 |
| 7 |
| 2 |
| 7 |
=(10.8-8.8)+(
| 5 |
| 7 |
| 2 |
| 7 |
=2+1
=3.
点评:此题考查了分数的四则混合运算,注意运算顺序和运算法则,灵活运用所学的运算律简便计算.
练习册系列答案
相关题目
一路车和2路车从早上8时起第一次同时发车,1路车和2路车第二次同时发车是在( )

| A、8:03 | B、8:05 |
| C、8:15 | D、8:30 |
要使得数为0,则135连续减去9的个数是( )
| A、15个 | B、10个 | C、12个 |