兔的部首：小数的基本性质

2020年11月30日发(作者：冯德培)
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 relatonship 4, and using segment figure, and list method, method analysis number
四年级下册基础知识归纳
1、小数的意义：分母是10、100、1000……的分数，可以用小数来表示。
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110、1100、11000……
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：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： < br>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< br>按照从左往右的顺序依次计算。
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, aalysis, 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
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*(35)*(53)=2*[(35)*(53)]=2*1=2,
用字母表示a*b*c=(a*b)*c=a*(b*c)；
乘法分配律：
15*(13+25)=15*(13)+15*(25)=5+6=11,
a*(b+c)=a*b+a*c;
(16+115)*30=(16)*30+(115)*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 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
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 19 ，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14，小兰今年多少岁？
2）校园面积的35是空地，空地的23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12除以13表示：已知13与一个因数的积是12，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*(35)*(53)=2*[(35)*(53)]=2*1=2,
用字母表示a*b*c=(a*b)*c=a*(b*c)；
乘法分配律：
15*(13+25)=15*(13)+15*(25)=5+6=11,
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, aalysis, 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
a*(b+c)=a*b+a*c;
(16+115)*30=(16)*30+(115)*30=5+2=7,
(a+b)*c=a*c+b*c.
三、分数乘除法应用题：
关键：确定单位一，画线段图表示必须先画单位一的量
单位一已知，用乘法；单位一未知，用除法（或方程）。
1、分数乘法应用题：
确定单位一，单位一已知，求它的几分之几是多少，用乘法。
比单位一多，乘 1加几分之几
比单位一少，乘 1减几分之几
2、分数除法应用题：
确定单位一，单位一未知， 已知单位一（一个数）的几分之几是多少，求单
位一（这个数）用除法。
比单位一多，除以1加几分之几
比单位一少，除以1减几分之几
或按照分数乘法应用题的计算方法用方程解决。
四、百分数
意义：像22%、117.5% 这样的数叫做百分数，表示一个数是另一个数的百分之
几。
百分数也叫做百分率或百分比，只可 以表示分率，不能带单位。

百分数、分数、小数之间的互化：
小数与百分数的互化：
把小数化成百分数，只要把小数点向右移动两位，同时在后面添上百分 号；
把百分数化成小数，只要把百分号去掉，同时把小数点向左移两位。
分数与百分数的互化：
分数化成百分数：
把分数化成小数（除不尽时，通常保留三位小数），在写成百分数。
也可以把分子分母同乘一个数将其化成一百分之几的数，再写成百分数。
百分数应用题有下列三种计算问题：
① 求一个数是另一个数的百分之几，用除法。用一个数除以另一个数。
例：求45是225的百分之几，即 ＝20％．
② 求一个数的百分之几是多少．用乘法。
例：求 2.2的 75％是多少．即 2.2×75％＝1.65．
③ 已知一个数的百分之几是多少，求这个数，用除法．
例：已知一个数的75％是165，求这个数．即165÷75％＝220．
运算法则
1. 整数加法计算法则：
相同数位对齐，从低位加起，哪一位上的数相加满十，就向前一位进一。
2. 整数减法计算法则：
相同数位对齐，从低位加起，哪一位上的数不够减，就从它的前一位退一作十，
和本位上的数合并在一起，再减。
3. 整数乘法计算法则：
先用一个因数每 一位上的数分别去乘另一个因数各个数位上的数，用因数哪一位
上的数去乘，乘得的数的末尾就对齐哪一 位，然后把各次乘得的数加起来。
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
4. 整数除法计算法则：
先从被除数的高位除起，除数是几位数，就看被除数的前几位； 如果不够除，
就多看一位，除 到被除数的哪一位，商就写在哪一位的上面。如果哪一位上不够
商1，要补“0”占位。每次除得的余数 要小于除数。
5. 小数乘法法则：
先按照整数乘法的计算法则算出积，再看因数中共 有几位小数，就从积的右边起
数出几位，点上小数点；如果位数不够，就用“0”补足。
6. 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。
7. 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（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. 同分母分数加减法计算方法:
同分母分数相加减，只把分子相加减，分母不变。
9. 异分母分数加减法计算方法:
先通分，然后按照同分母分数加减法的的法则进行计算。
10. 带分数加减法的计算方法:
整数部分和分数部分分别相加减，再把所得的数合并起来。
11. 分数乘法的计算法则:
分数乘整数，用分数的分子和整数相乘的积作分子，分母不变；分数乘分数，用
分子相乘的积作 分子，分母相乘的积作分母。
12. 分数除法的计算法则:
甲数除以乙数（0除外），等于甲数乘乙数的倒数



















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, aalysis, 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
三年级下册
1、小数的意义：像 3.50 、1.06 、16.85 这样的数都是小数。
读数：16.85读作：十六点八五
写数：二十点零三写作：20.03
2、小数加减法的计算方法：
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。
3、两位数乘两位数：
1）末尾有0的数的计算方法：
1）先将0前面的数相乘，在数数两个因数后面共有几个0，就再乘的的积的
后面添上几个0.
2）列竖式时要将0前面的数字对齐。
3）两位数乘两位数：
相同数位对齐，先用 个位上的数去乘，积的末位与个位对齐，再用十位上的
数去乘，积的末位与十位对齐，（满十向前一位进 1）。然后把两次相乘的积相
加。
4因数与积的变化规律：
1）一个因数不变，另 一个因数乘或除以几（0除外），积也乘或除以几（0除外）
2）一个因数乘一个数（或除以一个数）， 另一个因数乘（或除以）另一个数，
积就乘（或除以）这两个数的积。
4分数的认识
1）.分数与分数单位的意义：
把单位‘1’平均分成若干份，表示这样一份或几份的数，叫 做分数。表示这样
一份的数，叫做分数单位。
2）.单位‘一’的意义：
一个物体 ，一个计量单位，或由许多物体组成的一个整体，都可以用自然数‘一’
来表示，通常我们把它叫做单位 ‘1’
3）.把单位平均分成若干份,表示这样的一份或几份的数叫做分数。分母表示
把一个 物体平均分成几份,分子表示取了其中的几份。
1 →分子
—→分数线
2 →分母 读作:二分之一
分数中间的一条横线叫做分数线，分数线上面的数叫做分子，分数线下面的数
叫做分母.
5、分数的大小比较
同分母分数相比较：分母相同的分数分子的的数比较大。
同分子分数相比较：分子相同的分数分母校的书比较大。
6、同分母分数相加减：
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
分母不变，只把分子向加减。

二年级下册
1、整数除法的意义：
平均分：把一个数平均分成若干份，求一份是多少。
如：20个苹果平均放在4个盘子里，每个盘子里放几个？
包含除：求一个数里面有几个几
如：每盘放5个苹果，20个苹果可以放几盘？
2、有余数除法：计算有余数除法时，要想除 数和几相乘的积比被除数小且最接
近被除数．（余数一定要比除数小）
3、混合运算：
在一个算式中，既有加减又有乘除，要先算乘除，再算加减。有小括号要先
算小括号里面的。
4、读数写数方法：
读法：读数时末尾的0一个也不读，中间有一个0或连续几个0只读一个0。
写法：从高位写起，这一位上是几就写几，哪一位上什么都没有就写0。
5、