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题目详情
(1)
-
-
-
-
-
-
(2)[1.25+(1
÷
-2.5÷3
)]÷25%
(3)
-
+
-
+
-
(4)81
×
+71
×
+61
×
+51
×
+41
×
+31
×
.
-
-
-
-
-
-
(2)[1.25+(1
÷
-2.5÷3
)]÷25%
(3)
-
+
-
+
-
(4)81
×
+71
×
+61
×
+51
×
+41
×
+31
×
.
1 1 2 2
-
-
-
-
-
(2)[1.25+(1
÷
-2.5÷3
)]÷25%
(3)
-
+
-
+
-
(4)81
×
+71
×
+61
×
+51
×
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×
+31
×
.
1 1 6 6
-
-
-
-
(2)[1.25+(1
÷
-2.5÷3
)]÷25%
(3)
-
+
-
+
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(4)81
×
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×
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×
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+41
×
+31
×
.
1 1 12 12
-
-
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(2)[1.25+(1
÷
-2.5÷3
)]÷25%
(3)
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+
-
+
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(4)81
×
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×
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×
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×
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+31
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.
1 1 20 20
-
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(2)[1.25+(1
÷
-2.5÷3
)]÷25%
(3)
-
+
-
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(4)81
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+31
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.
1 1 30 30
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(2)[1.25+(1
÷
-2.5÷3
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(3)
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+
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.
1 1 42 42
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÷
-2.5÷3
)]÷25%
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.
1 1 56 56
÷
-2.5÷3
)]÷25%
(3)
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.
1 1 4 4
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(3)
-
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2 2 3 3
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(3)
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.
1 1 3 3
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.
3 3 2 2
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-
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5 5 6 6
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7 7 12 12
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9 9 20 20
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11 11 30 30
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13 13 42 42
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1 1 7 7
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4 4 5 5
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1 1 3 3
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1 1 2 2
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2 2 3 3
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(3)
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(4)81
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(4)81
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▼优质解答
答案和解析
(1)
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=[
2 2 23 3 3-2.5÷3
)]÷25%,
=[
+(
×
-
×
)]÷
,
=[
1 1 13 3 3)]÷25%,
=[
+(
×
-
×
)]÷
,
=[
5 5 54 4 4+(
×
-
×
)]÷
,
=[
5 5 54 4 4×
-
×
)]÷
,
=[
3 3 32 2 2-
×
)]÷
,
=[
5 5 52 2 2×
)]÷
,
=[
3 3 310 10 10)]÷
,
=[
1 1 14 4 4,
=[
问题解析 问题解析
(1)通过观察,从第二个分数开始,每个分数都可以拆成两个分数相减的形式,然后通过分数加减相抵消的方法,求得结果;
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算. (1)通过观察,从第二个分数开始,每个分数都可以拆成两个分数相减的形式,然后通过分数加减相抵消的方法,求得结果;
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算.名师点评 名师点评
本题考点: 本题考点:
分数的巧算;整数、分数、小数、百分数四则混合运算. 分数的巧算;整数、分数、小数、百分数四则混合运算.
考点点评: 考点点评:
完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算. 完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算.
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userProvince = "\u56db\u5ddd",
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- 问题解析
- (1)通过观察,从第二个分数开始,每个分数都可以拆成两个分数相减的形式,然后通过分数加减相抵消的方法,求得结果;
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算.
- 名师点评
-
- 本题考点:
- 分数的巧算;整数、分数、小数、百分数四则混合运算.
-
- 考点点评:
- 完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算.
作业帮用户
2017-10-24
举报
- 问题解析
- (1)通过观察,从第二个分数开始,每个分数都可以拆成两个分数相减的形式,然后通过分数加减相抵消的方法,求得结果;
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算.
- 名师点评
-
- 本题考点:
- 分数的巧算;整数、分数、小数、百分数四则混合运算.
-
- 考点点评:
- 完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算.
作业帮用户
2017-10-24
举报
作业帮用户作业帮用户
2017-10-242017-10-24
举报
举报
- 问题解析
- (1)通过观察,从第二个分数开始,每个分数都可以拆成两个分数相减的形式,然后通过分数加减相抵消的方法,求得结果;
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算.
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算.
(2)把小数和百分数化为分数,先计算小括号内的除法,再计算小括号内的减法,然后算中括号内的加法,最后算括号外的除法;
(3)通过观察,把每个分数可以拆成两个分数相加的形式,然后通过分数加减相抵消的方法,求得结果;
(4)通过观察,把每个分数可以拆成整十数和假分数的形式,运用乘法分配律简算.
- 名师点评
-
- 本题考点:
- 分数的巧算;整数、分数、小数、百分数四则混合运算.
-
- 考点点评:
- 完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算.
- 本题考点:
- 分数的巧算;整数、分数、小数、百分数四则混合运算.
- 本题考点:
- 分数的巧算;整数、分数、小数、百分数四则混合运算.
- 考点点评:
- 完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算.
- 考点点评:
- 完成本题应注意分数的拆分,灵活运用运算技巧,进行简便计算.
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(1)835×2.7+6.2×1.4+3.5×8.6(2)11-180+543×374543×37 2020-04-07 …
大家帮我看看这个lingo小程序啊,总是说没有可行解data:a1,a2,b1,b2,c1,c2, 2020-05-13 …
帮我破解下这串数字什么意思9(2)4(3)6(2)2(1)4(3)3(1)4(3)9(1)6(3) 2020-05-17 …
为什么老是@bin那句出错model:sets:course/1..40/:a,b,c,d,t;t 2020-07-15 …
1.0.3:12=2.0.25:4=3.2.8:0.7=4.2.1:1.2=5.0.1:0.8=6 2020-07-17 …
[-|98|+76+(-87)]*23[56+(-75)-(7)]-(8+4+3)5+21*8/2 2020-07-17 …
直接写得数.13×2=25×4=18×5=20×45=0.7+1.5=420÷60=780÷6=1 2020-07-19 …
求助1/2*4+1/4*6+1/6*8+…+1/2010*2012=1/2(1/2-1/4+1/4- 2020-11-07 …
(2011•郑州模拟)计算,能简算的要简算:(1)1÷4÷8÷0.25÷12.5(2)9.75-5. 2020-11-12 …
6道有理数混合运算那位同学帮我算计算题?1.|-7/4-(-3/4)|-(|-5/4|-|-3/4| 2021-01-22 …