小数的基本性质
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2021年01月24日 05:32
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绩效奖励制度-我是一颗小草
choose
the
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(4) to correct the typos;
(5) so the child write w
ords (A
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(6)
in accordance wit
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ords;
(7) the complete w
ord, a
nd explai
n the me
aning
of the
wor
d;
(8) collocation;
(9) make sentences w
ith the wor
d;
(10) the written lang
uage as r
equired. (C) t
he main
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types (1) complete
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ntences; (2) write
dow
n the meani
ng of a
sente
nce or ex
pressi
on
of 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
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cati
on (1) the
common conjunctions
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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 pr
oportion 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
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,
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
catio
n problem (3)--col
umn e
quati
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 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
nd a
ccor
ding to
problem in the
of focus described se
ntence from overall Sha
ng determine basi
c of equivalent relat
onshi
p 4, and usi
ng segme
nt figure, and list method, method a
nalysis
number
四年级下册基础知识归纳
1
、小数的意义：分母是< br>10
、
100
、
1000……
的分数，可以用小数来表示。< br>
2
、小数的性质：小数的末尾添上
0
或者去掉
0
， 小数的大小不变。
3
、小数点的移动：
1
）小数点向右移动一位、 两位、三位
……
小数相应扩大到
时原小数的
10
倍、
100
倍、
1000
倍
……
2
）小数点向左移动一位、两位、三位
……
小数相应缩 小到时
原小数的
1/10
、
1/100
、
1/1000……
4
、比较小数大小的方法：先看两个小数的整数部分，整数部分大的那个小数就大；整数部分相同，就比较两个小数的十分位，十分位大的那个小数就大；十分
位上相同，
就比较两个小数的百分位
……
继续下去，
一直到比较出两个小数的大
小为止。
5
、小数加、减法的意义和计算法则：
加法意义：
小数 加法的意义与整数加法的意义相同，
也是把两个数合并成一个数
的运算。
减法意义：是已知和与一个加数，求另一个加数的运算
1
）
计算小 数加、
减法，
先把各数的小数点对齐
（也就是把相同数位上的数对齐）
，
2
）再按照整数加、减法的法则进行计算，最后在得数里对齐横线上的小数点 点
上小数点。
（得数的小数部分末尾有
0
，一般要把
0
去掉。）
6
、小数乘法意义和计算法则
意义：
1
）小数乘整数：与 整数乘法的意义相同，就是求几个相同加数的和的简
便运算。例如
:2.5×
6 ，表示
6
个
2.5
求和或
2.5
的
6
倍是多少。
（
2
）
一个数乘小数的意义：
与整 数乘法的意义有所不同，
它是整数乘法意
义的进一步扩展。它可以理解为是求这个数的十分之几 、百分之几、
千分之几
……
是多少。例如，
2.5 ×
0.6
表示
2.5
的十分之六是多少，
2.5 ×
0.98
表示
2.5
的百分之九十八是多少。
法则：< br>先按照整数的计算方法算出乘积，
再看因数中一共有几位小数，
就从积的
个位起 数出几位，点上小数点。
7
、小数除法意义和计算法则：
意义：
与整数除法的意义相同，
都是已知两个因数的积和一其中的一个因数，
求
另一 个因数的运算。
1
）除数是整数的小数除法计算法则：
先按照整数除法的法则去除，商的小数点要和被除数的小数点对齐；如
果除到被除数的末尾仍有余数， 就在余数后面添
“0”
，再继续除。
2
）除数是小数的除法计算法则：
先移动除数的小数点，
使它变成整数，
除数的小数点也向右移动几位，
被
除数的小数点也要向右移动几位（ 位数不够的补
“0”
），然后按照除数是整数
的除法法则进行计算。
8
小数的四则混合运算
顺序：同整数的运算顺序相同，有括号的，先算括号 里面的，要按照小括号，
中括号，大括号的运算顺序，然后再乘除，最后加减。没有括号的，要
按照从左往右的顺序依次计算。
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
wor
d;
(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
se
ntences; (2) write
dow
n the meani
ng of a
sente
nce or ex
pressi
on
of 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, to examine the topi
c, i
dentify
probl
ems associated wit
h two 2, a
alysi
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 pr
oportion 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
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,
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
cation problem (3)
--col
umn e
quati
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 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
nd a
ccor
ding 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
加法交换律：
运算性质：
2.5+1.7=1.7+2.5
用字母表示
a+b=b+a
；
加法结合律：
1.3+2.7+7.3=1.3+(2.7+7.3)=1.3+10=11.3
用字母表示
a+b+c=(a+b)+c=a+(b+c)
；
乘法交换律：
0.2*3=3*0.2=0.6
用字母表示
a*b=b*a
；
乘法结合律：
2*(3/5)*(5/3)=2*[(3/5)*(5/3)]=2*1=2,
用字母表示
a*b*c=(a*b)*c=a*(b*c)
；
乘法分配律：
15*(1/3+2/5)=15*(1/3)+15*(2/5)=5+6=11,
a*(b+c)=a*b+a*c;
(1/6+1/15)*30=(1/6)*30+(1/15)*30=5+2=7,
(a+b)*c=a*c+b*c.
8
、方程：
意义：含有
未知数
的
等式
叫
方程
。
方程的解：使方程左右两边相等的未知数的值叫做方程的解。
解方程：求方程的解的过程叫做解方程。
等式的性质：
等式两边 同时加
(
减
)
同一个数或同一个代数式，所得的结果仍是等式。
用字母表示为：若
a
＝
b
，
c
为一个数或一个代数式。则：
(1)a+c
＝
b+c
(2)a-c
＝
b-c
等式的两边同时乘或除以同一个不为０的数所得的结果仍是等式。
(3)
若
a=b,
则
b=a
（等式的对称性）。
(4)
若
a=b,b=c
则
a=c
（等式的传递性）。
用字母表示为：若
a
＝
b
，
c
为一个数或 一个代数式（不为０）。则：
a×c=b×c
a÷c＝b÷c
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
wor
d;
(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
se
ntences; (2) write
dow
n the meani
ng of a
sente
nce or ex
pressi
on
of 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, 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 pr
oportion 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
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,
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
cation problem (3)
--col
umn e
quati
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 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
nd a
ccor
ding to
problem in the
of focus described se
ntence from overall Sha
ng determine basi
c of equivalent relati
onshi
p 4, an
d usi
ng segme
nt figure, and list method, method a
nalysis
number
五年级下册基本知识归纳
一、分数乘法的意义和法则：
1
、分数乘整数：
意义：与整数乘法的意义相同，都是求几个相同加数和的简便运算
如：拖拉机耕一块地，每时耕这块地的
1/9
，一天工作
8
时，耕了这
块地的几分之几？
方法：分数乘整数，分子和整数相乘的积做分子，分母不变
2
、一个数乘分数：一个数乘分数的意义就是求这个数的几分之几是多少
包括整数乘分数和分数乘分数
如：
1
）叔叔今年
36
岁，小兰的年龄是叔叔的
1/4
，小兰今年多少岁？
2
）
校园面积的
3/5
是空地，
空地的
2/3是草坪，
草坪的面积占校园总面积
的几分之几？
法则：分数乘法先用分子乘以分子的积做分子
,
分母乘以分母的积做分母。
二、分数除法的意义及法则：
意义：
与整数除法的意义相同，
都是 已知两个因数的积与其中一个因数，
求另一
个因数的运算．如
1/2
除以1/3
表示：已知
1/3
与一个因数的积是
1/2
，求另一个因数是多少
.
法则：甲数除以乙数（
0
除外），等于甲数乘乙数的倒数。
当除数 小于
1
，商大于被除数；当除数等于
1
，商等于被除数；当除数大
于
1
，商小于被除数。
分数除法法则：甲数除以乙数（
0
除外），等于
甲数乘乙数的倒数。
只要是分数除法应用题，就先找单位
1.
单位
1
找
到了，方法也就出来了。
分数除法应用题：乙数的几分之几是甲数，
求乙数，就用甲数除以乙数。
不知道
单位
1
有就用除法
分数混合运算：
顺序：同整数的运算顺序相同，有括号的，先算括号里面的，要按照 小括号，中
括号，大括号的运算顺序，然后再乘除，最后加减。没有括号的，要按照从左往
右的 顺序依次计算。
加法交换律：
运算性质：
25+17=17+25
用字母表示
a+b=b+a
；
加法结合律：
13+27+73=13+(27+73)=13+100=113
用字母表示
a+b+c=(a+b)+c=a+(b+c)
；
乘法交换律：
2*3=3*2=6
用字母表示
a*b=b*a
；
乘法结合律：
2*(3/5)*(5/3)=2*[(3/5)*(5/3)]=2*1=2,
用字母表示
a*b*c=(a*b)*c=a*(b*c)
；
乘法分配律：
15*(1/3+2/5)=15*(1/3)+15*(2/5)=5+6=11,