祖国在我心中手抄报资料-电气自动化毕业论文

choose the correct meaning; (4) to correct the typos; (5) so the child write words (ABAB, and AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sentence types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sentence; (5) modified sentences. 2, knowledge classification (1) the common conjunctions coordinate: ... ... 一面…… ... 1, to examine the topic, dentify problems associated with two 2, analysis, alternative question two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unknown, column proportion type 4, and solutions proportion type 5, and test, wrote answer language plenary, and subject:
application problem (1)--simple application problem and composite application problem review content simple application problem composite application problem answers application problem of general steps 1, and figure out meaning --through examines the, find known conditions and by seeking problem 2, and analysis number relationship--analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine problem-solving method and problem-solving steps. 3, and column type calculation--lists formula, is out subdivisions 4, and test, and wrote answer--check, and checking, and wrote answers typical application problem 13, and subject: application problem (3)--column equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equivalent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according
to meaning find equivalent relationship of common method 1 , And according to common of number relationship type, established equivalent relationship 2, and according to has learn had of calculation formula, 3, and according to problem in the of focus described sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number
小学五年级下册数学知识点复习
主编人：
一、简便计算部分
加法结合律：（a+b）+c=a+(b+c) 减法的性质：a-b-c=a-(b+c) a-(b-c)=a-b+c
例：
5< br>711215
71222
20202
3

( )
18181717

17291729
414117
71115< br>
2
12

722

5
2020 2
3













18

17

18

17
29

1717

2 9
414117
311

11
2
2


2
17

二、计算部分
1、注意计算结果约分，尤其是分子和分母是3的倍数的分数。2、快速找到几
个分数的公分母。例：
1

35

1






7171

5115< br>3

412



2
9318 6

86412
2

95

3









< br>
661212

42424
14
31
18



1
248
18
612

1


2
三、解方程
等式的性质：a±c=b±c a÷c=b÷c a×c=b×c c≠0

53
x
35
113

x
x
74
812
124

5535
3353
113
解
：x

解
：x

解
：
xxx

7747
88128
124

1
19
311
x
x
x

28
24
412

1
x
四、长方体和正方体的计算
6

h
b
a a
长方体的棱长和=4a+4b+4h=4(a+b+h) 正方体的棱长和=12a （带
长度单位）
长方体的表面积= 2(ab+bh+ah) 正方体的
a
2
表面积=
（带面积单位）
长方体的体积= abh 正方体
a
3
的体积=
（带体积单位）五、知识点
1、几个最小：最小的自然数是0，最小的偶数是0，最小的奇数是1，最小的质
数是2，最小 的合数是4。
2、一个数的最大因数是它本身，最小因数是1；一个数的最小倍数是它身，没
有最大倍数。
一个数的最大因数等于它的最小倍数。

choose the correct meaning; (4) to correct the typos; (5) so the child write words (ABAB, and AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sentence types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sentence; (5) modified sentences. 2, knowledge classification (1) the common conjunctions coordinate: ... ... 一面…… ... 1, to examine the topic, identify problems associated with two 2, analysis, alternative question two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unknown, column proportion type 4, and solutions proportion type 5, and test, wrote answer language plenary, and subject:
application problem (1)--simple application problem and composite application problem review content simple application problem composite application problem answers application problem of general steps 1, and figure out meaning --through examines the, find known conditions and by seeking problem 2, and analysis number relationship--analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine problem-solving method and problem-solving steps. 3, and column type calculation--lists formula, is out subdivisions 4, and test, and wrote answer--check, and checking, and wrote answers typical application problem 13, and subject: application problem (3)--column equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equivalent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according
to meaning find equivalent relationship of common method 1 , And according to common of number relationship type, established equivalent relationship 2, and according to has learn had of calculation formula, 3, and according to problem in the of focus described sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number
3、图形的变换有：平移、对称、旋转、放大与缩小。
4、旋转的三要素：方向、角度、中心点（定点）。
5、长方形的对称轴有2条，正方形的对 称轴有4条，圆形有无数条对称轴，半
圆只有1条对称轴，扇形只有1条对称轴，等腰三角形只有1条对 称轴，等边三
角形有3条对称轴，
等腰梯形只有1条对称轴，菱形有2条对称轴。一般的平行四边形不是轴对称图
形。
6、长方体和正方体都有6个面，8个顶点，12条棱。长方体每个面一般都是长
方形，特殊情况有相对 的两个面是正方形，其余四个面都是面积相等的长方形。
长方体相对的棱长度相等，相对的面的面积相等 ，长方体有4条长，4条宽，4
条高。正方体也叫立方体，是长、宽、高都相等的特殊的长方体，正方体 每个面
都是正方形且面积都相等。
7、体积：物体所占空间的大小。常用的体积单
m
3
.dm
3
.cm
3
位有：
容积：容器、桶、仓库等所能容纳物体的体积。常用的容积单位有：l ml
体积与容积间的单位换
ldm
3
........lcm
3
算：
8、分数与除法的关系：分数的分子相当于除法里的被除数，分母相当于除法里
的除数，分数线 相当于除法里的除号，分数的大小（分数的值）相当于除法里的
a
商。区别：分数是一种数，除 法是一种运算。它的关系用字
ab

b0

母表示为：
b

9、分子比分母小的分数叫真分数，真分数小于1；分子比分母大（或相等）的< br>分数叫假分数，假分数大于或等于1。
10、分数的基本性质：分数的分子和分母同时乘以或除 以相同的数（0除外），
分数的
bbcbc
大小不变。
(a0,c0)

aacac
11、最简分数：分子和分母只有公因数1的分数叫最简分数。
12、同分数加减法的计算法则：分母不变，把分子相加减。
13、异分母加减法的计算法则：先通分，再按照同分母加减法的计算法则进行计
算。
14、众数：一组数据中出现次数最多的数据，它反映一组数据的集中情况。在一
组数据中可能没有众 数，也可能有多个众数。
15、常用的统计图：条形统计图，折线统计图，扇形统计图。
1 6、找次品的方法：一般把产品个数分成三份，其中两份数量要相等。利用最不
利原则去分析和考虑。2 -3个数量至少需要1次找到，4-9个数量至少要2次找
到，10-27个数量至少要3次找到，28 -81个数量至少4次找到。
17、奇数：不是2的倍数的数。偶数：是2的倍数的数。
1 8、质数：一个数除了1和它本身两个约数，没有别的约数的数。合数：一个数
除了1和它本身以外，还 有别的约数的数。1不是质数，也不是合数。
19、2的倍数的特点：个位上是0、2、4、6、8的 数。5的倍数的特点：个位上
是0或5的数。3的倍数的特点：一个数各位上的数字之和是3的倍数的数 。
20、互质数：只有公因数1的两个数。如：2和5，9和8，7和15，4和9。
六、解决问题
1、求一个量是另一个量的几分之几的？
方法：用一个量除以另一个量。注意：结果约成最简分数。

choose the correct meaning; (4) to correct the typos; (5) so the child write words (ABAB, and AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sentence types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sentence; (5) modified sentences. 2, knowledge classification (1) the common conjunctions coordinate: ... ... 一面…… ... 1, to examine the topic, identify probems associated with two 2, analysis, alternative question two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unknown, column proportion type 4, and solutions proportion type 5, and test, wrote answer language plenary, and subject:
application problem (1)--simple application problem and composite application problem review content simple application problem composite application problem answers application problem of general steps 1, and figure out meaning --through examines the, find known conditions and by seeking problem 2, and analysis number relationship--analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine problem-solving method and problem-solving steps. 3, and column type calculation--lists formula, is out subdivisions 4, and test, and wrote answer--check, and checking, and wrote answers typical application problem 13, and subject: application problem (3)--column equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equivalent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according
to meaning find equivalent relationship of common method 1 , And according to common of number relationship type, established equivalent relationship 2, and according to has learn had of calculation formula, 3, and according to problem in the of focus described sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number
例：把5克糖放入20克水中，糖的 重量占水的几分之几？糖的重量占糖水的几
分之几？
解答思路：第一问题是求糖的重量是水的 几分之几应该用糖的重量去除以水的重
量。而第二问题是求重量是糖水的重量的几分之几应该用糖的重量 去除以糖水的
重量。根据分析列式为：
51

520
204

51
5(205)
2、分数加减法应用题
255
3
22
例1：水果店里原有水果吨，卖出吨后又运进吨。水果店现在有水果
53
4多少吨？
解答思路：由于每个分数都带上了单位，所以每个分数表示具体的数量。应该用
我们以前学的整数应用题的解答方法进行解答。
32245244061

(吨
）

45360606060
21
例2： 五四班有45人，有的同学参加了语文兴趣小组，有的同学参加了
53
数学兴趣小组，其余的参 加了音、体、美兴趣小组。参加音、体、美兴趣小组的
同学占全班同学的几分之几？
解答思路 ：本题的每个分数没有带单位，它表示量与量之间的关系。因此本题应
把全班45人看作单位“1”进行 思考。

2115654
1

5315151515

3、长方体正方体表面积、体积的应用
方法：根据题意学会画图进行分析思考，抓住重点词句，利用好其计算公式。
例1：给一个无 盖长方体水缸抹水泥，从里面量得长8分米，宽4分米，深6分
米；抹水泥的面积是多少？
解 答思路：这是关于长方体的表面积的应用，从无盖和抹水泥的面积中可以看出。
在计算时，由于无盖只算 五个面。
8×4+8×6×2+4×6×2=176（平方分米）
例2：学校有一个长方体 沙坑，长2.4米，宽1.5米，深0.6米。如果每筐沙有
0.03立方米，填满这个沙坑要多少筐沙 ？
解答思路：根据每筐沙有0.03立方米，可以看出本题是与长方体的体积有关。
应先求长 方体沙坑的体积，看它的体积里面有多少个0.03立方米，就求出了问
题。
2.4×1.5×0.6÷0.03=72（筐）
例3：把一个长15厘米的长方体平均截成三个小正方体，表面积会增加多少平
方厘米？



解答思路：根据画图观察，我们以现平均截成三小正方体后，每个小正 方体的棱
长为（15÷3）厘米，而且表面积增加了4个小正方形的面积。

choose the correct meaning; (4) to correct the typos; (5) so the child write words (ABAB, and AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sentence types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sentence; (5) modified sentences. 2, knowledge classification (1) the common conjunctions coordinate: ... ... 一面…… ... 1, to examine the topic, identify problems associated with two 2, analysis, alternative question two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unknown, column proportion type 4, and solutions proportion type 5, and test, wrote answer language plenary, and subject:
application problem (1)--simple application problem and composite application problem review content simple application problem composite application problem answers application problem of general steps 1, and figure out meaning --through examines the, find known conditions and by seeking problem 2, and analysis number relationship--analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine problem-solving method and problem-solving steps. 3, and column type calculation--lists formula, is out subdivisions 4, and test, and wrote answer--check, and checking, and wrote answers typical application problem 13, and subject: application problem (3)--column equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equivalent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according
to meaning find equivalent relationship of common method 1 , And according to common of number relationship type, established equivalent relationship 2, and according to has learn had of calculation formula, 3, and according to problem in the of focus described sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number
15÷3=5（厘米） 5×5×4=100（平方厘米）
例4：把一个长20分米的长方体锯成5个同样大小的长方体，表面 积增加了160
平方分米，
原来这个长方体的体积是多少立方分米？

< br>解答思路：根据画图观察，我们发现锯成5个同样大小的正方体后增加了8个小
正方形的面积，所 每个小正方形的面积为160÷8，根据长方体的体积公式底面
积乘高。从而求出其体积。
160÷8×20=400（立方分米）
4、最大公因数和最小公倍数的应用
例1 ：五一班有48人，五二班有56人。如果把这两个班分成人数相等的小组，
每组最多几人？一共可分几 个小组？
解答思路：根据题意，要想两个班分成的人数相等，说明这个人数既是48的因
数， 也是56的因数，由于是求每组人数最多几人，所以是求它们的最大公因数。
48的因数有：1，2，3，4，6，8，12，16，24，48.
56的因数有：1，2，4，7，8，14，28，56。
48和56的最大公因数是8。所以每组人数最多是8人。
48÷8+56÷8=13（组）
例2：一个班有40多人，如果4个人一组或6个人一组都能刚好分完，这个班
有多少人？ < br>解答思路：根据题意，4人一组或多或6人一组都能刚好分完，所这个班的人数
既是4的倍数也是 6的倍数。所以是4和6的公倍数，并且是在40多的一个公
倍数。
4的倍数：4，8，12，16，20，24，28，32，36，40，44，48。
6的倍数：6，12，18，24，30，36，42，48。
4 和6的公倍数有：12，24，36，48。
所以这个班有48人。
5、找次品
有一批零件共15个，其中有一个比其它零件轻一些，你能用天平找出这个次品
来吗？至少要几次一定能 找到这个次品？
解答：15个零件（5，5，5）先天平各放5个，如果不平衡，将其中轻的5个零< br>件再分成（2，2，1），又将天平各放2个，如果不平衡，最后将轻的2个零件再
分面（1，1 ）。这样至少三次就可以找出这个较轻的零件了。
七、旋转 （顺时针旋转和逆时针旋转。）

choose the correct meaning; (4) to correct the typos; (5) so the child write words (ABAB, and AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sentence types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sentence; (5) modified sentences. 2, knowledge classification (1) the common conjunctions coordinate: ... ... 一面…… ... 1, to examine the topic, identify probems associated with two 2, analysis, alternative question two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unknown, column proportion type 4, and solutions proportion type 5, and test, wrote answer language plenary, and subject:
application problem (1)--simple application problem and composite application problem review content simple application problem composite application problem answers application problem of general steps 1, and figure out meaning --through examines the, find known conditions and by seeking problem 2, and analysis number relationship--analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine problem-solving method and problem-solving steps. 3, and column type calculation--lists formula, is out subdivisions 4, and test, and wrote answer--check, and checking, and wrote answers typical application problem 13, and subject: application problem (3)--column equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equivalent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according
to meaning find equivalent relationship of common method 1 , And according to common of number relationship type, established equivalent relationship 2, and according to has learn had of calculation formula, 3, and according to problem in the of focus described sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number











































































































































八、钟面上的旋转
每个大格是30度，每个小格是6度。
九、最大公因数和最小公倍数
方法：列举法 短除法 集合法 口算法
18和12（6）[24] 30和60（30）[60] 7和5（1）[35] 8、6和12（2）[24]

如果两个数是倍数关系，则它们的最大公因数是较小的数，最小公倍数是较大的
数。
如果两个数是互质数，则它们的最大公因数是1，最小公倍数是它们的乘积。
十、通分与约分
bbcbc
(a0,c0)
示： 依据：分数的基本性质 用字母表
aacac
例1：将下面的分数约成最简分数

3
181863
6623

8824242464

273

364

4

例2：将下面的各组分数进行通分
75
35

和
和

24

12
86
77
3339



2424
88324


20
510
554



1224
66424
十一、分数与小数的互化
小数化分数的方法：先将小数改写成分母是10、100、1000的分数，能约分的再
约分。
例
4441341742
0.043.4或3.433

100100425105105

分数化成小数的方法：一般根据分数与除法的关 系，用分子除以分母，除不尽的
保留一定的小数位数。
例
3325753
1
5
0.75或30.75560.83

4
0

.125

44251004
8
6

3
常用的分数与小数间的互化。
0.375
8
5
0. 625
8
7
0.875
8

choose the correct meaning; (4) to correct the typos; (5) so the child write words (ABAB, and AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sentence types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sentence; (5) modified sentences. 2, knowledge classification (1) the common conjunctions coordinate: ... ... 一面…… ... 1, to examine the topic, identify problems associated with two 2, analysis, alternative question two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unknown, column proportion type 4, and solutions proportion type 5, and test, wrote answer language plenary, and subject:
application problem (1)--simple application problem and composite application problem review content simple application problem composite application problem answers application problem of general steps 1, and figure out meaning --through examines the, find known conditions and by seeking problem 2, and analysis number relationship--analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine problem-solving method and problem-solving steps. 3, and column type calculation--lists formula, is out subdivisions 4, and test, and wrote answer--check, and checking, and wrote answers typical application problem 13, and subject: application problem (3)--column equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equivalent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according
to meaning find equivalent relationship of common method 1 , And according to common of number relationship type, established equivalent relationship 2, and according to has learn had of calculation formula, 3, and according to problem in the of focus described sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number
1

0.2
1
0.5

1
5
0.05
2
十
2
二、分解质因数
20
1
0.4
法：将合数写成几个质数
1
相乘的形式。
0.25
方
5
0.04
4
28、30、24、32、7 7、100
25
3
3
0.6
2×7
0.28=2
75
×
5
1
0.02
4
十三、分< br>4
数的意义
50
0.8
把单位“1”平
5
均分成 若干份，表示其中的一份或几份的数。
附：五年级下册数学精典习题选