小学数学基础知识基本概念大全之一

1、choose the corre ct meani ng; 4 to correct the typos; 5 so the child write w ords a bab, and aabb; 6 in accordance wit h written w ords; 7 the complete w ord, a nd explai n the me aning of the word; 8 collocation; 9 make sentences w ith the wor d; 10 the written lang uage as r equired. c t he main s

2、ente nce types 1 complete sentences; 2 write dow n the meani ng of a sente nce or ex pressi onof thoughts and feeling s; 3 write sentences as re quired; 4 finish malalignme nt of the se ntence; 5 modified se ntences. 2, k nowle dge classifi cati on 1 the common conjunctions coordi nate: . . 一 面 . 1,

3、 to examine the topi c, i dentify probl ems associated wit h two 2, a nalysi s, alternative question two is i n dire ct pr oportion to t he amount of the a ssociated r elationshi p is inversely proporti onal r elationshi p. 3, and set unk nown, colum n proportion type 4, and soluti ons pr oportion t

4、ype 5, and test, wrote a nswer la nguag e ple nary, and subject:appli cation problem 1-simple a ppli cation problem and composite applicati on probl em review content simple a ppli cation problem composite applicati on probl em answ ers applicati on probl em of general ste ps 1, a nd figure out mea

5、ning -t hroug h examines t he, find k nown conditions a nd by seeki ng problem 2, a nd a nalysis number relationshi p-a nalysis k now n conditions zhijian, a nd conditi ons a nd problem zhijia n of relationship, determine probl em-solving method a nd problem-solvi ng steps. 3, and colum n type calcu

6、lation-lists formula , is out subdivi sions 4, a nd test, a nd wrote answer-che ck, and checki ng, a nd wr ote answ ers typi cal applicati on probl em 13, a nd subject: a ppli cation problem 3-col umn equati on soluti ons applicati on pr oblem revie w content ov erview pr oblem -solving ste ps 1, a

7、nd figure out mea ning, find by seeking of unk now n and x said 2, and accor ding to mea ning find equivalent relati onship, lists equation 3, a nd soluti ons equation 4, and test, and wrote a nswer s accordi ng to meaning find equivalent relati onship of common method 1 , and accor ding to common o

8、f num ber relationshi p type, e stabli she d equivalent relati onship 2, and accordi ng to ha s learn had of ca lculation form ula, 3, a nd a ccording to problem in the of focus described se ntence from overall sha ng determine basi c of equivalent relati onshi p 4, and usi ng segme nt figure, and l

9、ist method, method a nalysis number学校数学基础学问基本概念大全之一自然数 :用来表示物体个数的0 、1 、2 、3 、4 、5 、6 、7 、8叫做自然数；整数:自然数都是整数，整数不都是自然数；小数: 小数是特别形式的分数；但是不能说小数就是分数；带小数 : 小数的整数部分不为零的小数叫带小数；纯小数 : 小数的整数部分为零的小数，叫做纯小数；有限小数 : 小数的小数部分只有有限个数字的小数（不全为零）叫做有限小数；无限小数 :小数的小数部分有很多个数字（不包含全为零） 的小数，叫做无限小数；循环小数都是无限小数，无限小数不肯定都是循环小数；例如，圆周率

10、也是无限小数；分数: 表示把一个 “单位 1”平均分成如干份，取其中的一份或几份的数， 叫做分数；（分成0 份在此不争论）真分数 :分子比分母小的分数叫真分数；假分数 : 分子比分母大，或者分子等于分母的分数叫做假分数；（分母、分子为零在此不争论）choose the corre ct meani ng; 4 to correct the typos; 5 so the child write w ords a bab, and aabb; 6 in accordance wit h written w ords; 7 the complete w ord, a nd explai n the

11、 me aning of the word; 8 collocation; 9 make sentences w ith the wor d; 10 the written lang uage as r equired. c t he main sente nce types 1 complete sentences; 2 write down the meani ng of a sente nce or ex pression of thoughts and feeling s; 3 write sentences as re quired; 4 finish malalignme nt o

12、f the se ntence; 5 modified se ntences. 2, k nowle dge classifi cati on 1 the common conjunctions coordi nate: . . 一 面 . 1, to examine the topi c, i dentify probl ems associated wit h two 2, a nalysi s, alternative question two is i n dire ct p oportion to t he amount of the a ssociated r elationshi

13、 p is inversely proporti onal r elationshi p. 3, and set unk nown, colum n proportion type 4, and soluti ons pr oportion type 5, and test, wrote a nswer la nguag e ple nary, and subject:appli cation problem 1-simple a ppli cation problem and composite applicati on probl em review content simple a pp

14、li cation problem composite applicati on probl em answ ers applicati on probl em of general ste ps 1, a nd figure out mea ning -t hroug h examines t he, find k nown conditions a nd by seeki ng problem 2, a nd a nalysis number relationshi p-a nalysis k now n conditions zhijian, a nd conditi ons a nd

15、problem zhijia n of relationship, determine probl em -solving method a nd problem-solvi ng steps. 3, and colum n type calculation-lists formula , is out subdivi sions 4, a nd test, a nd wrote answer-che ck, and checki ng, a nd wr ote answ ers typi cal applicati on probl em 13, a nd subject: a ppli c

16、ation problem 3 -col umn equati on soluti ons applicati on pr oblem revie w content ov erview pr oblem -solving ste ps 1, a nd figure out mea ning, find by seeking of unk now n and x said 2, and accor ding to mea ning find equivalent relati onship, lists equation 3, a nd soluti ons equation 4, and t

17、est, and wrote a nswer s accordi ng to meaning find equivalent relati onship of common method 1 , and accor ding to common of num ber relationshi p type, e stabli she d equivalent relati onship 2, and accordi ng to ha s learn had of ca lculation form ula, 3, a nd a ccording to problem in the of focu

18、s described se ntence from overall sha ng determine basi c of equivalent relati onshi p 4, and usi ng segme nt figure, and list method, method a nalysis number带分数: 一个整数（零除外）和一个真分数组合在一起的数，叫做带分数；带分数也是假分数的另一种表示形式，相互之间可以互化；关于 n表示自然数 是否是分数 :是分数，但不能用分数的意义去说明它，它既不属于真分数，也不属于假分数，而是一个特别分数，叫零分数；数是由数字和数位组成；0 的意

19、义 :0 既可以表示 “没有 ”，也可以作为某些数量的界限；如温度等；0 是一个完全有确定意义的数； 0 是一个数；0 是一个偶数；0 是任何自然数 0除外 的倍数； 0 有占位的作用； 0 不能作除数； 0 是中性数；十进制 : 十进制计数法是世界各国常用的一种记数方法； 特点是相邻两个单位之间的进率都是十； 10 个较低的单位等于 1 个相邻的较高单位；常说“满十进一 ”，这种以 “十”为基数的进位制，叫做十进制；加法: 把两个数合并成一个数的运算，叫做加法，其中两个数都叫“加数”，结果叫 “和”；减法: 已知两个加数的和与其中一个加数，求另一个加数的运算，叫做减法；减法是加法的逆运算；其

20、中“和”叫“被减数 ”，已知的加数叫 “减数”，求出的另一个加数叫“差”；乘法: 求 n 个相同加数的和的简便运算，叫做乘法；其中相同的这个数及 n 个这样的数都叫 “因数”，结果叫 “积”；除法: 已知两个因数的积与其中一个因数，求另一个因数的运算，叫做除法；除法是乘法的逆运算；其中“积”叫做 “被除数 ”，已知的一个因数choose the corre ct meani ng; 4 to correct the typos; 5 so the child write w ords a bab, and aabb; 6 in accordance wit h written w ords;

21、7 the complete w ord, a nd explai n the me aning of the word; 8 collocation; 9 make sentences w ith the wor d; 10 the written lang uage as r equired. c t he main sente nce types 1 complete sentences; 2 write down the meani ng of a sente nce or ex pression of thoughts and feeling s; 3 write sentences

22、 as re quired; 4 finish malalignme nt of the se ntence; 5 modified se ntences. 2, k nowle dge classifi cati on 1 the common conjunctions coordi nate: . . 一 面 . 1, to examine the topi c, i dentify probl ems associated wit h two 2, a nalysi s, alternative question two is i n dire ct pr oportion to t h

23、e amount of the a ssociated r elationshi p is inversely proporti onal r elationshi p. 3, and set unk nown, colum n proportion type 4, and soluti ons pr oportion type 5, and test, wrote a nswer la nguag e ple nary, and subject:appli cation problem 1-simple a ppli cation problem and composite applicat

24、i on probl em review content simple a ppli cation problem composite applicati on probl em answ ers applicati on probl em of general ste ps 1, a nd figure out mea ning -t hroug h examines t he, find k nown conditions a nd by seeki ng problem 2, a nd a nalysis number relationshi p-a nalysis k now n co

25、nditions zhijian, a nd conditi ons a nd problem zhijia n of relationship, determine probl em -solving method a nd problem-solvi ng steps. 3, and colum n type calculation-lists formula , is out subdivi sions 4, a nd test, a nd wrote answer-che ck, and checki ng, a nd wr ote answ ers typi cal applicat

26、i on probl em 13, a nd subject: a ppli cation problem 3 -col umn equati on soluti ons applicati on pr oblem revie w content ov erview pr oblem -solving ste ps 1, a nd figure out mea ning, find by seeking of unk now n and x said 2, and accor ding to mea ning find equivalent relati onship, lists equat

27、ion 3, a nd soluti ons equation 4, and test, and wrote a nswer s accordi ng to meaning find equivalent relati onship of common method 1 , and accor ding to common of num ber relationshi p type, e stabli she d equivalent relati onship 2, and accordi ng to ha s learn had of ca lculation form ula, 3, a

28、 nd a ccording to problem in the of focus described se ntence from overall sha ng determine basi c of equivalent relati onshi p 4, and usi ng segme nt figure, and list method, method a nalysis num ber叫做“除数 ”，求出来的另一个因数叫做“商”；加、减法的运算定律加法交换律： 两个数相加，交换两个加数的位置，和不变，叫做加法交换律；加法结合律： 三个数相加，先把前二个数相加，再加第三个数，或者，先

29、把后二个数相加，再加上第一个数，其和不变；这叫做加法结合律；在减法中，被减数、减数同时加上或者减去一个数，差不变；在减法中，被减数增加多少或者削减多少，减数不变，差随着增加或者削减多少；反之，减数增加多少或者削减多少，被减数不变，差随着削减或者增加多少；在减法中，被减数减去如干个减数，可以把这些减数先加，差不变；乘、除法运算定律乘法的交换律： 两个数相乘，交换两个因数的位置，积不变；这叫做乘法的交换律；乘法的结合律 ：三个数相乘，先把前两个数相乘，再乘以第三个数，或者，先把后两个数相乘，再和第一个数相乘，积不变；这叫做乘法结合律；乘法安排律：两个数的和（或差）与一个数相乘，等于把这两个数分别与

30、这个数相乘，再把两个积相加（或相减）；这叫做乘法安排律；乘法的其他运算定律一个因数扩大如干倍，必需把另一个因数缩小相同的倍数，其积不变；除法的运算定律 - 商不变性质choose the corre ct meani ng; 4 to correct the typos; 5 so the child write w ords a bab, and aabb; 6 in accordance wit h written w ords; 7 the complete w ord, a nd explai n the me aning of the word; 8 collocation; 9 m

31、ake sentences w ith the wor d; 10 the written lang uage as r equired. c t he main sente nce types 1 complete sentences; 2 write down the meani ng of a sente nce or ex pression of thoughts and feeling s; 3 write sentences as re quired; 4 finish malalignme nt of the se ntence; 5 modified se ntences. 2

32、, k nowle dge classifi cati on 1 the common conjunctions coordi nate: . . 一 面 . 1, to examine the topi c, i dentify probl ems associated wit h two 2, a nalysi s, alternative question two is i n dire ct p oportion to t he amount of the a ssociated r elationshi p is inversely proporti onal r elationsh

33、i p. 3, and set unk nown, colum n proportion type 4, and soluti ons pr oportion type 5, and test, wrote a nswer la nguag e ple nary, and subject:appli cation problem 1-simple a ppli cation problem and composite applicati on probl em review content simple a ppli cation problem composite applicati on

34、probl em answ ers applicati on probl em of general ste ps 1, a nd figure out mea ning -t hroug h examines t he, find k nown conditions a nd by seeki ng problem 2, a nd a nalysis number relationshi p-a nalysis k now n conditions zhijian, a nd conditi ons a nd problem zhijia n of relationship, determi

35、ne probl em -solving method a nd problem-solvi ng steps. 3, and colum n type calculation-lists formula , is out subdivi sions 4, a nd test, a nd wrote answer-che ck, and checki ng, a nd wr ote answ ers typi cal applicati on probl em 13, a nd subject: a ppli cation problem 3 -col umn equati on soluti

36、 ons applicati on pr oblem revie w content ov erview pr oblem -solving ste ps 1, a nd figure out mea ning, find by seeking of unk now n and x said 2, and accor ding to mea ning find equivalent relati onship, lists equation 3, a nd soluti ons equation 4, and test, and wrote a nswer s accordi ng to me

37、aning find equivalent relati onship of common method 1 , and accor ding to common of num ber relationshi p type, e stabli she d equivalent relati onship 2, and accordi ng to ha s learn had of ca lculation form ula, 3, a nd a ccording to problem in the of focus described se ntence from overall sha ng determine basi c of equivalent relati onshi p 4, and usi ng segme nt figure, and list method, method a nalysis number两个数相除， 被除数和除数同时扩大或者缩小相同的一个数（0 除外），商的大小不变；乘法的意义 : 一道乘法算式一般有下面几个意义：一、求几个相同加数的和是多少？例如：27× 13 ，表示求 13个 27 的和是多少？也可以表示求27 的 13倍是