2022年小数的意义和性质单元备课

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1、choose the corre ct meaning; (4) to correct the typos; 5) so the child write words (ABAB, a nd 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 ma

2、in sentence ty pes (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 sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 .

3、 1, to examine the topi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and te

4、st, wrote answer language pl enary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the,

5、find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4,

6、 and test, and wr ote answer-check, and checking, and wrote answers typical appli cation 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, an

7、d according to meaning find equival ent 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 accordi

8、ng 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第一课时小数的产生和意义教学目标：( 一) 在学生初步认识分数和小数的基础上，进一步理解小数的意义。( 二) 使学生理解和掌握

9、小数的计数单位及相邻两个单位间的进率。( 三) 培养学生的观察、分析、推理能力。教学重点、难点：在学生初步认识一位和两位小数的基础上，进一步把认数范围扩展到三位小数，使学生明确小数表示的是分母是10，100，1000，的分数，并理解小数的计数单位及单位间的进率，既是本课的重点，也是本课的难点教具：直尺 2米彩带 0.9米彩带课件教学流程：一、情景导入教师：同学们喜欢做游戏吗？今天老师带大家做一个游戏，游戏的名字叫“猜一猜，测一测。”师出示 2米的彩带，同学们猜一猜有多长，指名回答后让学生测量验证。师再出示0.9 米的彩带，让学生猜测，然后测量出结果是9 分米。提问： 9 分米如果用米做单位用分

10、数做单位是多少米？（9/10 米）用小数表示是多少米？（0.9 米）二、探究新知1、教学例1 （1） 课件出示例一图（1）教师引导学生观察讨论，然后回答图下面的问题。教师引导小结：我们把分母是10 的分数写成的是一位小数。（2）课件出示例一图（2）教师引导学生观察讨论，然后回答图下面的问题。教师引导小结：我们把分母是100 的分数写成的是两位小数。（3）课件出示例一图（3）教师引导学生观察讨论，然后回答图下面的问题。教师引导小结：我们把分母是 1000 的分数写成的是三位小数。（4）教师：我们把1 米平均分成10、100、1000 份，用分数、小数都会说了，如果老师再把1 米平均分成 1000

11、0 份，这样的几份写成小数是几位小数；那么份呢？（5）教师引导学生观察这些分数和小数，然后讨论：分数和小数之间有什么联系呢？学生回答后教师小结：分母是10、100、1000的分数可以用小数表示这就是小数的意义。（教师板书）2、完成教材第51 页做一做。让学生独立完成，教师提醒学生要先看一看每一幅图平均分成了多少份？然后教师讲评。3、课件出示智慧关第一关0.3 里面有（）个 1/10 0.07里面有（）个 1/100 0.5 里面有（）个 1/10 0.9里面有（）个 1/100 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 1 页，共 16 页c

12、hoose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd 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 ma

13、in sente nce 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 sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面

14、. 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and

15、 test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the

16、, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions

17、4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem (3)-colum n equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2,

18、 and according to meaning find equival ent 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 acco

19、rding 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教师：学生讨论完成，并说一说为什么这样想？教师指名回答后小结：像0.3 、0.5 这样的一位小数，我们都可以看成有许多个

20、1/10 组成的，那么我们就说十分之一是一位小数的计数单位，写作0.1 。同理，像 0.07 、0.09 这样的两位小数，可以看成有许多个1/100 组成的，那么我们就说百分之一是两位小数的计数单位，写作0.01 。教师：同学们猜一猜三为小数的计数单位是什么？写作_ 4、课件出示智慧关第二关0.1 米里面有（）个 0.01 米 0.01米里面有（）个 0.001 米教师：同学们可以结合板书讨论完成。然后指名回答，并说出理由。教师小结：每相邻两个计数单位之间的进率是10。5、课件出示智慧关第三关0.8 里面有（）个 0.1 0.32里面有（）个 1/100 三、教师小结：同学们顺利的闯过了关，在

21、这节课上有什么收获？第二课时小数的读法和写法教学目的：1理解小数的计数单位，掌握小数的读写方法2培养学生类比、迁移和归纳总结的能力教学重点：理解小数的计数单位，掌握小数的读写方法教学难点：能熟练、正确地读写小数教学过程一、复习引入1读出下面各数：234 7093 31 10000 38950 0.7 2回忆一下：你是怎样读出这些数的？整数的数位顺序是什么？（个位、十位、百位、千位）整数的计数单位依次是什么？（一个、十、百、千）3试着读出下面的数：2.78 55.2 0.46 35.9 你是怎样读的？4导入：小数和整数一样，也有计数单位，也按照一定的顺序排列起来，这节课我们就来研究一下小数的数位

22、顺序二、探究新知1教学小数的数位顺序表教师提问：看整数数位顺序表，你认为小数的数位应在什么位置上？（在整数部分的右侧）中间怎样区分呢？（用小数点隔开）教师提问：我们知道小数的计数单位有十分之一、百分之一、千分之一，那么，十分之一表示小数精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 2 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with

23、 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 ty pes (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings;

24、(3) write sentences as required; (4) finish malalignment of the sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the topi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct roport

25、ion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite applicatio

26、n problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and cond

27、itions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application

28、 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 equival ent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers

29、 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

30、 determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number部分有几位小数？（有一位小数）所以十分之一就在小数点后的第几位？（第一位）它所占的位置叫做十分位，计数单位就是十分之一引导学生思考：你能推测小数部分的第二位是什么吗？为什么？小数部分的第二位是百分位，因为它表示有两位小数，计数单位是百分之一）教师提问：谁能依次说出小数部分后面几位的数位顺序以及相应的计数单位？（千分位千分之一；万分位万分之一；十万分位十万分之一；百万分位百万

31、分之一） 出示下表，提问：为什么后面用省略号？（表示后面还有很多数位）2学习小数的读法教师出例2：读出下面的小数： 6 5 004 0073 教师提问：你能读出下面这些小数吗？教师总结：整数部分是“0”的就读作“零” ；整数部分不是“0”的按照整数读法来读，小数点读作“点”，小数部分是几就依次读出来即可将你读出来的内容用汉字写下来，就可以了（板书）分组讨论：每个小数分别表示什么吗？教师讲解：六点五表示六十五个十分之一；零点零四表示四个百分之一；零点零七三表示七十三个千分之一3学习小数的写法出示例 3：写出下面小数：四点三九零点四零八三十点零一五教师提问：写小数时应该怎样写？教师说明：写小数的时

32、候，整数部分按照整数的写法来写（整数部分是零的写作“ 0” ）小数点写在个位右下角点，小数部分顺次写出每一个数位上的数字四点三九写作： 4 39 零点四零八写作： 0 408 三十点零一五写作： 30015 学生讨论：每个小数分别表示什么吗？439，表示四百三十九个百分之一；0408，表示四百零八个千分之一；3015，表示三千零一十五个千分之一三、巩固练习1填空：小数点右边第二位是（）位，第四位是（）位，第一位是（）位，第三位是（）位2读出下面的句子 （1）南京长江大桥全长6772 千米 （2）土星绕太阳一周需要2946 年（3） 1 千瓦时的电量可以使电车行驶084 千米3写出下面横线上的数

33、（1）我国科学工作者和登山运动员，精确测得珠穆朗玛峰的高度是海拔八千八百四十八点一三米（2）地球赤道的周长是四万零七十五点六九千米（3）非洲大甲虫长十四点八五九厘米，重九十九点七九克精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with written words; (7) the com

34、plete 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 sente nce types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as requ

35、ired; (4) finish malalignment of the sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to the amount of the

36、 associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content

37、 simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhiji

38、an of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem (3)-colum n equ

39、ation 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 equival ent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according to meaning f

40、ind 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 equ

41、ivalent relationship 4, and using segment figure, and list method, method analysis number4判断（1） 373 读作：三点七十三（ ） （2）零点三零七写作：0307 （ ）（3）五十点二零八写作：5 208 （ ）5提高题（1）有一个数，十位和百分位上都是6，个位和十分位上都是0，这个数写作（） ，读作（） （2）小华在读一个小数时，把小数点丢了，结果读成了四万五千零一原来的小数读出来只读一个零，原来的小数是多少？四、课堂小结这节课我们学习了小数的读法与写法，五、布置作业1读出下面横线上的数（1）南京长江

42、大桥全长6.772 千米（2）土星绕太阳转一周需要29.46 年（3） 1 千瓦时的电量可以使电车行驶0.84 千米2写出下面横线上的数（1）我国科学荏复登山运动员, 精确测得珠穆朗玛峰的高度是海拔八千八百四十八点一三米（2）地球赤道的周长是四万零七十五点六九千米（3）非洲大甲虫长十四点八五九厘米，重九十九点七九克自我问答 : 本节课的设计力求符合学生的认知特点，想方设法创造生动活泼的教学情景，使学生始终处于好奇好学的良好氛围中，让每一位学生学有多得，都能体会学习的成功和喜悦。第三课时小数的性质教学目标：1、初步理解小数的基本性质，并应用性质化简和改写小数。2、运用猜测、操作、检验、观察、对比

43、等方法，探索并发现小数的性质，养成探求新知的良好品质。3、感受透过现象看本质的过程以及数学在实际生活中的重要作用，体验问题解决的情趣。教学重点：让学生理解并掌握小数的性质。教学难点：能应用小数的性质解决实际问题教学过程：一、谈话导入、课前质疑我有个邻居小明的爸爸下岗了，最近他开了个便民小超市，想请大家帮忙给设计个标价牌，大家能帮这个忙吗 ?( 出示手套和毛巾图) 手套每副2 元 5 角、毛巾每条3 元，标价牌该怎么填呢？学生自由创作。（选择几种写法引起争论）板书:2.5 2.50 3 3.00 我们在商店里看到的标价一般是这样的：2.50 元 3.00元（课件演示）2.5 和 2.50 都表示

44、 2 元 5 角吗？ 3 和 3.00 相等吗？精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 4 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the word; (8) collocation; (

45、9) make sentences with the word; (10) the written language as required. (C) the main sentence ty pes (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 sente nce; (5) modified senten

46、ce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the topi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, an

47、d set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli

48、 cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-so

49、lving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem (3)-column equation solutions application problem review content overview problem-sol

50、ving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equival ent 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 commo

51、n 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, meth

52、od analysis number为什么会相等呢？学会今天的这节课你就明白了。今天学“小数的性质”（板书课题）二、探究新知、课中释疑1. 教学例 1。（1）课件出示1 分米、 10 厘米、 100 毫米的线段图请比较一下它们的大小。学生略加思考后马上提问，要求说说你是怎么知道的。（即想的过程）演示：重合法比较1 分米、 10 厘米、 100 毫米的大小。板书并演示：1 分米 =10 厘米 =100 毫米（2）导入例1：你能把它们都写成用米做单位的小数的形式吗？必须体现它们的原先单位。导：分米和米有什么关系？厘米、毫米呢？根据学生回答归纳演示： 1分米是 1/10 米，写成 0.1 米 10厘

53、米是 10 个 1/100 米，写成0.10 米100 毫米是 100 个 1/1000 米，写成 0.100 米并板书： 01 米 0.10米 0.100米那 0.1 米、0.10 米、0.100 米之间大小有什么关系呢？学生很快回答后课件演示。并在他们之间加上等号。我们还可以用重合法比较一下。（课件演示）（3）指导看黑板：1 分米 = 10厘米 = 100毫米 0.1米 = 0.10米 = 0.100米提问 : 这说明了什么问题? 请大家仔细观察这个等式，可以从左往右看，再从右往左看，什么变了？什么没变？在什么地方多（少）0？在这个小数的什么位置？多（少）0 还可以怎么说？根据学生回答逐一

54、板书：小数 0.1米=0.10 米 =0.100 米大小不变小数的末尾添上0 大小不变，去掉 0大小也不变。 是不是所有的小数都有这个性质呢？这是不是一个特例？我们还需再验证一下。2. 教学例 2。（1）出示例2: 比较 0.30 和 0.3 的大小。请同桌之间拿出印好的大小完全相等的两个正方形，用你喜欢的颜色分别表示出030 和 03。导：想想0.30 表示什么意思？0.3 呢？应该涂多少格？学生涂完色问：你为什么这样涂？之后演示涂色过程。（2）同桌商量比较，汇报结论。问：谁涂的面积大？0.30 和.0.3的大小怎样？你是怎么知道的？精选学习资料 - - - - - - - - - 名师归纳

55、总结 - - - - - - -第 5 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd 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 lan

56、guage as required. (C) the main sente nce 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 sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common con

57、junctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solu

58、tions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out

59、 meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists

60、formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem (3)-colum n equation solutions application problem review content overview problem-solving steps 1, and figure out meaning, find by see

61、king of unknown and x said 2, and according to meaning find equival ent 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 equiva

62、lent 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直观比较法：看上去都一样大；理论推导法：0.30 是 30 个

63、 1/100 ，也是 3 个 1/10 ；0.3 是 3 个 1/10 。课件演示重合图形。( 在原板书下再板书:0.30 0.3) （3）观察思考观察板书0.30=0.3 这个例子说明了什么？看来不仅仅是个特例，再次验证我们的猜测。3. 讨论归纳教师指着板书说：你能把上面的研究结论归纳成为一句话吗？4 人小组之间讨论一下，想想该怎么说才比较完整？教师提问几个小组代表让其归纳，不够完整可以由其他小组代表补充。得出小数的性质：在小数的末尾添上“ 0”或者去掉“0” ，小数的大小不变这叫做小数的性质（课件展示）4、指导阅读。讲述 : 书上也证实了我们的研究，并把它称为 小数的性质 。齐读小数的性质

64、。5、质疑问难： （判断）你们对这句话理解的够不够透彻呢？挑战一下你们。（以下题目陆续出现）（1）一个数的末尾添上0 或去掉 0 ，这个数的大小不变。举例说明后返回小数的性质，红字强调“小数”。（2）小数点的后面添上0 或去掉 0 ，小数的大小不变。举例说明后返回小数小性质，红字强调“末尾”。（3） 10.50=10.5=10.500 判断后返回小数小性质强调“大小不变”。三、巩固运用、交流反思小数的性质有什么作用呢? 强调：我们如果遇到小数末尾有“0”的时候，一般可以去掉末尾的“0” ，把小数化简l. 出示例 3: 把 0.70 和 105.0900 化简。思考：哪些“0”可以去掉，哪些“0

65、”不能去掉？（1）提问： 0.70 你认为可以怎么化简才能大小不变？（2）学生自己完成。指名回答，让其说说这样做的根据是什么？（3）为什么105.0900 的 5左边的 0 不能去掉呢 ?( 强调小数的性质中 小数的末尾的0 。) （4）练习 :下面的数，哪些0 可以去掉 ?哪些 0不能去掉 ? 0.40 1.820 2.900 0.080 12.000 问 12 应该去掉 0 后是多少？还可以怎样表示？精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 6 页，共 16 页choose the corre ct meaning; (4) to cor

66、rect the typos; (5) so the child write words (ABAB, a nd 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 ty pes (1) complete senten

67、ces; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the topi c, identify pr

68、oblems associated with two 2, a nalysis, alternative que stion two is in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and

69、 subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking pr

70、obl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and

71、 checking, and wrote answers typical appli cation 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 equival ent

72、 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 form

73、ula, 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强调： 12 去掉 0 后，小数部分没有数，可以把小数点也去掉。过渡：同样，应用小数的性质，我们还可以根据需要，把一个数改写成含有指定小数位数的小数2. 出示例 4: 。不改变数的大小，把0.2 、4.0

74、8 、 3 改写成小数部分是三位的小数。想想可以怎么做？（1）学生自己完成。（2）大家这样做的根据是什么?3 能不能直接在后面添0? （3）练习 :下列数如果末尾添0 ，哪些数的大小不变，哪些数的大小有变化? 3.4 18 0.06 700 3.0 4.90 如果整数想改成大小不变的小数，必须先做什么？（先添上小数点，再添0）四、数字游戏、拓展升华：摆数游戏1. 出示 5 张卡片： 2、 5、0、0、和“”并说明游戏要求：每小组利用老师发给的数字卡片按要求摆数。3 个人摆数，一个人做记录。2. 动手摆（ 1）用五张卡片摆一个数，这个数中的两个“”都能去掉（2）用五张卡片摆一个数，这个数中的两个

75、“”一个能去掉，一个不能去掉想一想：怎样摆才能既不重复又不遗漏3. 请小组板演汇报，不完全的可以补充。4. 说说这样摆的技巧，你为什么这样摆？五、课堂小结1. 这节课你学到了哪些知识？有哪些收获? 2. 现在知道刚才的标价为什么相等吗？我们帮助了小明的爸爸，我代他谢谢你们。同样，你们也得谢谢小明的爸爸，通过标价我们知道了小数的性质。第四课时小数的大小比较教学目标：（1）理解并掌握比较两个小数大小的方法，会正确比较两个小数的大小。（2）引导学生经历观察、比较、抽象和概括等数学思维活动，归纳出小数大小比较的方法。（3）让学生感受到数学知识来源于生活实际并运用于生活实际，培养学生解决实际问题意识和能

76、力。教学重点 : 引导学生经历观察、比较、抽象和概括等数学思维活动，归纳出小数大小比较的方法。掌握比较两个小数大小的方法。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 7 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with written words; (7) the complete word, and explain the me

77、aning of the word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sente nce 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

78、of the sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to the amount of the associated relationship is in

79、versely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem co

80、mposite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine

81、pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem (3)-colum n equation solutions application pr

82、oblem 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 equival ent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according to meaning find equivalent relationship of

83、 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 us

84、ing segment figure, and list method, method analysis number教学难点：比较位数不同的小数的大小。教学准备：多媒体课件、练习纸、小黑板教学过程：一、复习引入：多媒体出示校园运动会场景图1、多媒体出示对话场景：小亚说 : 在规定的比赛时间内我的跳绳成绩是159 次。小巧说：我的成绩是93 次。师提出问题：谁的成绩好？你是怎样比较这两个数的大小的? 结论 1：比较两个整数的大小，位数多的整数就大。师：老师还搜集了小胖和小丁丁参加运动会比赛的成绩表2、多媒体出示：校园运动会成绩表注：比赛时间相同踢毽子小胖236 小丁丁257 师提出问题：踢毽子

85、比赛中谁的成绩好？你是怎样比较这两个整数的大小的？结论 2：两个整数位数相同时，先从最高位比起，按数位顺序一位一位地去比，哪一位上的数大，那个数就大。二、揭示课题：多媒体出示问题1：在跳高比赛中，小林和小云的成绩分别是1.1 米和 0.9 米。谁的成绩好？要知道谁的成绩好？我们该怎么办? 引导学生把实际问题转化为数学问题：我们只要比较1.1 和 0.9 这两个小数的大小就可以知道谁的成绩好。教师揭示课题并板书：这节课我们就来学习小数的大小比较。二、分层探究：项目次数姓名精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 8 页，共 16 页choose

86、 the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd 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 se

87、ntence ty pes (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 sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, t

88、o examine the topi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, w

89、rote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find k

90、nown conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and t

91、est, and wr ote answer-check, and checking, and wrote answers typical appli cation 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 acco

92、rding to meaning find equival ent 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

93、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（一）解决问题1：在跳高比赛中，小林和小云的成绩分别是1.1 米和 0.9 米。谁的成绩好？师：接下来就请你们做个小裁判，比一比谁的成绩好

94、？说说理由。1、独立思考。2、反馈交流：师：谁的成绩好？你是怎样比较的？预设可能情况：生 1：我在数射线上找出1.1 和 0.9 对应的位置，很明显1.1 比 0.9 大。生 2: ：因为 1.1 里有 11 个 0.1, 而 0.9 里有 9 个 0.1, 所以 1.1 米0.9 米。生 3：因为 1.1=11/10 0.9=9/10 11/109/10所以 1.1 比 0.9 大。生 4：1.1 中的 1 比 0.9 中的 0 大, 所以 1.10.9 。( 教师要适时引导学生并板书结论) 3、比较感悟：你们认为哪一种方法最简便？在学生充分发表意见后，教师引导学生概括：比较两个小数的大小，

95、先看它们的整数部分，整数部分大的那个数就大。板书：看整数部分。（二）多媒体出示问题2：师：在激动人心的100 米男子短跑比赛中，谁会成为胜者？下面我们就将视线转向短跑比赛区。在 100 米男子短跑比赛中，小丁丁和小胖的成绩分别是17.26 秒和 17.84 秒。谁跑得快？1、在学生独立思考的基础上进行组内交流。2、全班反馈交流：师：小丁丁跑得快，为什么？你是怎样比较的？教师引导学生概括：当整数部分相同时，看十分位，十分位上的数大的那个数就大。板书：看十分位。（三）多媒体出示问题3：在跳远比赛中，小明和小军的成绩分别是1.284 米和 1.26 米，谁是胜者？1、学生自己尝试比较。2、反馈交流师

96、：小明是胜利者，为什么？你是如何比较它们的大小的。师：通过这个练习，你又能得出什么结论？教师引导学生明确：整数部分和十分位上的数都相同，要看百分位上的数，百分位上数大的那个数就大。板书：看百分位。( 四) 归纳方法：精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 9 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with written w

97、ords; (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 sente nce types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write

98、sentences as required; (4) finish malalignment of the sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to

99、the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application probl

100、em review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions

101、and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application proble

102、m (3)-colum n 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 equival ent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers accor

103、ding 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 deter

104、mine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number师：刚才我们讨论了各种情况的小数比较大小的方法，谁能完整地归纳概括一下？先自己说，再小组讨论后，总结出：比较两个小数的大小，先比较整数部分，整数部分大的那个数就大；整数部分相同的，再比较十分位上的数, 十分位上的数大的那个数就大；十分位上的数也相同的，再比较百分位上数, 百分位上的数大的那个数就大；( 教师可继续追问省略号的含义，然后用小黑板将这一结论呈现出来。) 师: 学习到这里，老

105、师要检验一下你们对小数的大小比较方法掌握的怎么样？四、巩固应用1、比较下面每组中两个小数的大小。3144.13 5.1925.129 12.001 12.01 0.473 0.46 10.34710.343 7.281 8.001 2、运动会跳远成绩公布出来了，你能给这三名同学排出名次吗？单位：米姓名小胖小丁丁小明成绩154 169 165 名次3、2004 年雅典奥运会男子110 米决赛中：加西亚的成绩是13.20 秒，刘翔的成绩是12.91 秒，特拉梅尔的成绩是13.18 秒。你能给他们排出名次吗？名次第一名第二名第三名姓名4、在运动会的跳高比赛中，小强、小军、小明的成绩和名次如下表：姓名

106、小强小军小明成绩1.20 米？1.10 米名次第一名第二名第三名小军的成绩可能是多少？五、课堂总结: 师：这节课你学到了什么？比较小数的大小时，你有什么要提醒同学的？第五课时小数点移动教学目标：1知识与技能：通过一组数的比较，观察各数之间的相同点和不同点，引导学生发现小数点位置的移动引起小数大小的变化规律，并应用这一规律计算有关的乘、除法。2过程与方法：通过操作、观察、归纳、概括等数学活动，发展数学思维能力。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 10 页，共 16 页choose the corre ct meaning; (4) to

107、correct the typos; (5) so the child write words (ABAB, a nd 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 ty pes (1) complete sen

108、tences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the topi c, identify

109、 problems associated with two 2, a nalysis, alternative que stion two is in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary,

110、and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking

111、 probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check,

112、and checking, and wrote answers typical appli cation 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 equival

113、ent 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 f

114、ormula, 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 number3. 情感态度价值观：培养学生的合作意识及知识迁移和推理能力。重点难点：重点：小数点位置移动引起小数大小变化规律的应探索及掌握。难点：小数点位置移动引起小数大小变化规律的理解及灵活应用。教学准备：小

115、黑板教学挂图（小数点移动）教学过程（一）复习准备1、提问。（1）把 5 米分别扩大10 倍、 100 倍、 1000 倍，各是多少米？（2）把 5000 厘米分别缩小10 倍、100 倍、 1000 倍，各是多少厘米？2、按从大到小的顺序排列。0.004 0.4 0.04 （二）导入新课1. 师： 出示小黑板 下面是四年级三位同学的身高纪录。请大家看一看，这些数据对不对？（小明 14.5 米，小红1.38 米，小李0.14 米）2. 师：你们笑什么呀？生： * 身高不对。 14.5 米太高了。生： 用手比 0.14米只有这么高师：两个错的数据错在哪里？小数点写错了位置。师：是啊，在小数点的末尾

116、添上0 或者去掉0不改变小数的大小，但是小数点的位置移动直接引起小数的大小发生变化。今天我们就一起来学习小数点移动的知识。 板书课题：小数点移动 （三）探究规律1、. 出示情景出示 ( 例 5 教学挂图）：教师便叙述边板书0.009 米-0.9米 0.9 米-9米同学们都看过西游记吧，齐天大圣孙悟空的“金箍棒”平时放在耳朵里，长只有0.009 米，遇到妖怪的时候，才亮出来，由小变大，0.009 米、 0.09 米、 0.9 米、 9 米、 90 米师：观察这组数和金箍棒的变化，你有什么发现？（从上往下观察小数点是怎样移动的？数的大小有什么变化吗？从下往上观察小数点是怎样移动的？数的大小有什么变

117、化？）小结：看来小数点向后移动，原来的数就扩大；小数点向左移动，原来的数就缩小。板书：右移扩左移缩2、合作探究（1）提问：从上往下观察它们都是把小数点向右移动，却得到了三个不同的数，对吗？看来小数点移动的位数不一样，原数大小的变化也就不一样。数的大小的变化既与小数点移动的方向有关，还与小数点移精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 11 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd

118、 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 sente nce types (1) complete sentences; (2) write down the meaning of a sentence or express

119、ion of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative q

120、ue stion two is in direct proportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application

121、 problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis k

122、nown conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation pro

123、blem 13, and subject: application problem (3)-colum n 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 equival ent relationship, lists Equation 3, and solutions equat

124、ion 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 des

125、cribed sentence from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number动位数的多少有关。（2）合作探究：究竟有怎样的关系呢？我们来继续深入研究。各组有这样一张表格和一张小数数位表，请你们小组选择其中的一种方法进行研究。先吧空白处填写完整，再观察小数点移动的位数与原来小数的大小变化。小数点可以向左移动，也可以向右移动。方法 1：表格小数点移动的位数（）米 =（）毫米 小数的大小

126、变化从（）往（）观察小数点向（）移动移动（）位（）米 =（）毫米移动（）位（）米 =（）毫米移动（）位（）米 =（）毫米方法 2： （学具中的数位表）（3）交流汇报谁来说一说，你们是选择哪种方法研究的？你们发现了什么？能概括地说一说我们发现的这个规律吗？ 指名学生对照板书说明小数向右移动引起小数扩大的规律 悟空打完妖怪，金箍棒要放回去了，谁来说一说这个时候金箍棒怎么变的？（从下到上观察）（四）实际应用1. 明确数的变化的方法我们大家研究得出这个规律有什么作用呢？1. 如果要吧一个小数扩大10 倍、 100 倍、 1000 倍可以怎么办？如果要缩小为1/10 、1/100 、1/1000呢？2.

127、 集体交流根据小数点移动的变化规律，如果要吧一个数扩大到它的10 倍、 100 倍、 1000 倍，只要把小数点向右移动一位、两位、三位就行了。要把一个数缩小到它的1/10 、1/100 、1/1000 ，只要把小数点向左移动一位、两位、三位。3. 强化去 0、添 0 的问题出示例 6、7 把 0.01 扩大到它的10 倍、 100 倍、 1000 倍，各是多少？精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 12 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5)

128、so the child write words (ABAB, a nd 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 ty pes (1) complete sentences; (2) write down

129、the meaning of a sentence or expression of thoughts and feelings; (3) write sentences as required; (4) finish malalignment of the sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the topi c, identify problems associated wi

130、th two 2, a nalysis, alternative que stion two is in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cati

131、on problem (1)-simple application problem and composite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analys

132、is number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote

133、 answers typical appli cation 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 equival ent relationship, lists

134、 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 accordin

135、g 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把 1 缩小到它的1/10 、1/100 、1/1000 ，各是多少？遇到位数不够怎么解决？小数点向左移动时，如果整数数位不够则要在数的左边用“0”补足。整百、整千的数，小数点向左移动后，小数末尾的“0”要去掉。4. 填空：把 2.3 的小数点

136、向右移动一位，就（）到原数（）倍。把 0.375 扩大到原数100 倍，小数点向（）移动（）位。把 0.73 的小数点向（）移动（）位，就缩小到原数的1/1000 。把 30 的小数点向（）移动（）位，原数变成0.003 。5. 把 1.8 改写成下面各数，它的大小有什么变化？0.018 180 0.0018 1.80 （五）总结本节知识，畅谈收获。第六课时生活中的小数教学目标1认识名数、单名数、复名数，掌握低级单位的名数与高级单位名数互化的方法，能正确进行低级单位的名数与高级单位的名数的互化。2在探索过程中培养分析、比较、归纳和概括的能力，提高解决实际问题的能力。3体会数学与实际生活的密切联

137、系，培养探索意识与合作能力。教学过程一、创设情境，提出问题1. 这段时间我们大家一起学习了小数，然后好多同学发现原来小数是我们生活中形影不离的朋友，那么今天我们再来更多地认识这位好朋友，好吗？（板书课题）先看课件的图片，请学生说说图片上小数表示的意义：（1）运动会上老师为同学们记录了成绩“我的 50 米赛跑成绩是7.98 秒。 ”（2）体育老师给大家测量身高体重“我的身高是1.41 米。 ” 、 “我的体重是39.4 千克。”（3）医生替小红量体温“你的体温是28.2 ，发烧了。 ”（4）面包的价格0.90 元、火腿的价格2.85 元、牛奶的价格5.98 元。2. 你还能说出一些生活中的小数吗

138、？（学生自由回答，把课前收集的或观察的小数进行汇报。）二、探究新知，解决实际问题1. 看看刚才老师出示的这些数：50 米、 7.98 秒、 1.41 米等等，还有大家收集的这些数后面好像都带着个小尾巴，是什么呀？精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 13 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with written wo

139、rds; (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 sente nce types (1) complete sentences; (2) write down the meaning of a sentence or expression of thoughts and feelings; (3) write s

140、entences as required; (4) finish malalignment of the sente nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to t

141、he amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application proble

142、m review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions a

143、nd problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem

144、 (3)-colum n 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 equival ent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers accord

145、ing 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 determ

146、ine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number我们把这些量得的数和单位名称合起来的数叫做名数。（出示具体概念，请学生默读。）2. 课件出示四名同学，先请学生们猜一猜他们的身高各是多少；然后给出确切的身高。这些数可以叫做什么？那么看1 米 45 厘米与其他三个名数不同在哪里？带有两个或两个以上单位名称的数叫做复名数。（出示具体概念，请学生默读。）设计意图：培养学生观察数据的能力，找出不同点与相同点，在一点一滴中体会学习；大胆鼓励学生猜

147、测身高，既锻炼了学生对实际数据掌握的能力，又激发了学数学、爱数学的情感。3. 这些数据太乱了，怎么比呢？改写成相同计量单位的数。都改成以米为单位的数。（1）请学生说出自己的办法。80 厘米 = 米=0.80 米。（2）填空后讨论第二种方法： 1000厘米 =（）米 200厘米 =（）米 3000厘米 =（）米 560厘米 =（）米集体讨论： 1 米=100 厘米，进率是100。1000 厘米里有 10 个 100 厘米，所以是10 米； 200 厘米里包含了2个 100厘米，所以是 2 米； 3000 厘米包含了30 个 100厘米，所以是 30 米； 所以都要除以100， 560100=5.

148、60（米）；80 厘米 =（80100）米 =0.80 米（板书）。（3）用自己喜欢的方法讨论1 米 45 厘米是多少米？小结：由低级单位向高级单位转化要除以它们之间的进率。（板书）（4）做一做： 23 分米 =（ ）米 1350克=（ ）千克 7450米=（ ）千米9020 千克 =（ ）吨（5）给刚才这四名小朋友按高矮顺序排队。（学生练习，巡视辅导） 4. 把这些数改写成以厘米为单位的数。1 米=100 厘米，进率是100。 0.95米 =0.95 100=95 厘米（板书）小结：由高级单位向低级单位转化要乘以它们之间的进率。（板书）三、巩固练习0.3 千克 =（ ）克 0.86平方米 =

149、（ ）平方分米 2.63千米 =（ ）米 3.7吨=（ ）千克四、小结这节课你有什么收获？还有什么问题？五、布置作业数学书 71 页 4 题、 5 题。第七课时求一个小数的近似数（1）教学目标 : 1. 理解和掌握用“四舍五入法”求一个小数的近似数; 理解精确度的意义。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 14 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6)

150、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 ty pes (1) complete sentences; (2) write down the meaning of a sentence or expression of thoug

151、hts and feelings; (3) write sentences as required; (4) finish malalignment of the sente nce; (5) modified sentence s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the topi c, identify problems associated with two 2, a nalysis, alternative que stion two i

152、s in direct roportion to the amount of the associated relationship is inversely proportional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and co

153、mposite application problem review content simple application problem composite application problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions

154、 Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solvi ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and su

155、bject: 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 equival ent relationship, lists Equation 3, and solutions equation 4, and test,

156、 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

157、from overall Shang determine basic of equivalent relationship 4, and using segment figure, and list method, method analysis number2. 经历求小数的近似数的过程，体验利用旧知迁移学习的方法。3感受数学知识与日常生活的密切联系，激发学习数学的兴趣, 培养学生的数感和数学意识。教学过程 : 一、师生对话, 迁移引入1. 师生谈话。师：猜猜李老师的体重和身高？2. 出示 : 李老师的体重大约是65 千克 , 身高是 170 厘米。3. 复习准备（1）教师引导比较，引出准确

158、数和近似数。（2）出示世博会主题馆的数据并求出近似数。（3）揭示“四舍五入法”方法进行求近似数。4. 揭示课题师: 生活中有很多求整数近似数的例子, 今天我们就来学习求一个小数的近似数。二、自主探究，方法获得1. 出示李老师女儿上幼儿大班时的身高0.984 米。师 : 你们说说她那时候的身高大约是多少米？2. 引导学生估计生1：小豆豆的身高大约是1 米。生 2： 豆豆的身高也可看作0.98 米。3. 探究学习（ 1）独立思考：他们是怎样得出这个身高的近似数呢?（2）在小组内说说，你是怎么想的？（3）试着归纳保留整数数、两位小数的方法。（4）拓展：师：还可以保留成几位小数呢？（5）理解近似数 1

159、.0 并强调注意点。 （6）比较三个近似数之间的相同点和不同点。4. 汇报交流：求小数近似数的方法“四舍五入法”。5. 小结评价。6. 感受精确度：师:0.98 、1.0 、1 这三个近似数那个最精确？7. 走进生活：图片展示刘翔、郑幸娟的破纪录数据，深刻理解精确度。三、练习巩固，提高升华1. “做一做”：保留两位小数，精确到十分位。2. 教材 P75第二题。下面各小数在哪两个相邻的整数之间？它们各近似于哪个整数？3. 取近似数1 的一位小数有哪些？4. 取近似数1.0 的两位位小数有哪些？四、全课总结。本节课是在学生学习了求整数的近似数的基础上进行教学的，目的是让学生学会用四舍五入法求小数的

160、近似数，并通过学生的自主探索、合作交流，培养学生的探究能力。第八课时求一个小数的近似数（2 ）精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 15 页，共 16 页choose the corre ct meaning; (4) to correct the typos; (5) so the child write words (ABAB, a nd AABB); (6) in accordance with written words; (7) the complete word, and explain the meaning of the

161、word; (8) collocation; (9) make sentences with the word; (10) the written language as required. (C) the main sente nce 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 sente

162、nce; (5) modified sente nce s. 2, knowledge classification (1) the common conjunctions coordinate: . . 一面 . 1, to examine the t opi c, identify problems associated with two 2, a nalysis, alternative que stion two is in direct proportion to the amount of the associated relationship is inversely propo

163、rtional relationship. 3, and set unk nown, column proportion type 4, and solutions proporti on type 5, and test, wrote answer language plenary, and subject: appli cation problem (1)-simple application problem and composite application problem review content simple application problem composite appli

164、cation problemanswers appli cation problem of general steps 1, and figure out meaning-through examines the, find known conditions and by seeking probl em 2, and analysis number relationship-analysis known conditions Zhijian, and conditions and problem Zhijian of relationship, determine pr oblem-solv

165、i ng method and probl em-solving steps. 3, and column type calculation-lists formula, is out subdivisions 4, and test, and wr ote answer-check, and checking, and wrote answers typical appli cation problem 13, and subject: application problem (3)-colum n equation solutions application problem review

166、content overview problem-solving steps 1, and figure out meaning, find by seeking of unknown and x said 2, and according to meaning find equival ent relationship, lists Equation 3, and solutions equation 4, and test, and wrote answers according to meaning find equivalent relationship of common metho

167、d 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 f

168、igure, and list method, method analysis number教学目标：1、 学生能够根据要求，把较大的数改写成“万”、 “亿”作单位的数。2、进一步掌握用“四舍五入法”求一个小数的近似数重点： 1、根据要求把较大的数改写成“万”、 “亿”作单位的数。 2 、用“四舍五入法”求一个小数的近似数。难点：把较大的数改写成“万”、 “亿”作单位的数与求一个的近似数混淆。教具、学具准备：师生分别准备木条( 或硬纸条 ) 钉成的三角形。教学过程：一、复习引入：1、求下面小数的近似数：（1）保留两位小数：10.0946 6.29647 （2）精确到十分位： 2.5464 0.

169、3496 2、把下面的数改写成“万”、 “亿”作单位的数。 0 3、过渡引入；把改写成“万”、作单位的数。二、学习新知： 1 、出示例2： 幻灯片（1）学生理解题意； （2）问：人改写成用“万人”作单位的数，就看里有多少个万，可以除以多少？把17250 缩小多少倍？怎样移动这个数的小数点？（3）说明：为了简便，改写时只在万位后面点上小数点。去掉小数末尾的0，写成 17.25 万人（板书：人=17.25 万人。 ）2、练习：把下面各数改写成用万作单位的数。3、过渡；我们会改写用万单位的数，会不会改写用亿单位的数。4、出示例3：（1）学生试做。（2）你是怎样改写的？（3）保留一位小数，求出它的近似

170、数。（板书： 0元=7697.82 亿元 7697.8 亿元）三、对比练习：（1）把改写成万作单位的数；（2）把改写成亿作单位的数。四、课堂总结1、小结：结合例2、3，常常把较大的数改写成用“万”或“亿”作单位的数。改写时，只要在“万”位或“亿”位的右边点上小数点，再在数的后面添写“万”或“亿”字。 2 、改写成用“万”或“亿”作单位的数和用“万”“亿”为单位求近似数时，要注意什么？（1）改写后的计数单位“万”或“亿”不能漏写，如有单位名称，也不能丢掉。（2）准确数与近似数不能混淆，既准确数要使用“=”近似数则使用“” 。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 16 页，共 16 页