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(1)12-16-112-120-130-142-156(2)[1.25+(114÷23-2.5÷313)]÷25%(3)32-56+712-920+1130-1342(4)8117×78+7116×67+6115×56+5114×45+4113×14+3112×23.
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
×
+41
×
+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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(4)81
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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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(4)81
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▼优质解答
答案和解析
(1)
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)]÷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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2017-10-24
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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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口算22×40=9+14=120÷6=300+400=56÷8+3=56÷7=46+24=15+1 2020-04-07 …
(1)61*60分之59+122又120分之1除以11 (2)4*5分之1+5*6分之1+6*7分 2020-05-15 …
口算22×40=9+14=120÷6=300+400=56÷7=46+24=15+15=60×30 2020-07-09 …
直接写出得数920×2=700×7=96÷3=430×1=280÷4=1.8-1.6=100×0= 2020-07-09 …
用简便方法计算92×6+×9(920)(16.5×5+7.5×5)÷6(20)3/5÷6[(7/9 2020-07-09 …
数字推理1)3,4,6,12,36()2)-2,-1,1,5,(),293)215,124,63, 2020-07-17 …
3÷〔9×(127−136)〕×1293.6÷〔(6-2.88)×(578-1.875)〕7.05 2020-07-17 …
700-8×5×4=540840÷6÷7+630=650530+120÷4×9=800315-60 2020-07-19 …
=1,=1*2,=1*2*3,······那么(1)5!;(2)(4!+8!)/5!(1)5!=1 2020-07-19 …