鸡兔同笼应用题解法

鸡兔同笼应用题解法 一、提出问题 大约在1500年前～《孙子算经》中记载了这样一个有趣的问题。书中说:“今有鸡兔同笼～上有三十五头～下有九十四足～问鸡兔各几何,”意思是:有若干只鸡兔同在一个笼子里～从上面数～有35个头,从下面数～有94只脚。问笼中鸡和兔各有几只, 这就是我们通常所说的鸡兔同笼问题～如何解决这个1500年前古人提出的数学问题～就是我们这节课要研究的内容。,板书课题, 二、解决问题 出示例1 :鸡兔同笼～有20个头～54条腿～鸡、兔各有几只, ,同时出示鸡兔同笼情境图, 师:想一想～如何来解决这个问题,请同学们把你的想法～你的 思考过程用你喜欢的方式达出来。 学生思考、分析、探索～接下来是讨论、交流、争辩。,老师参与其中～启发、点拔、引导适当～师生互动。, 10分钟后进入小组汇报、集体交流阶段。 师:谁能说一说你们小组探究的结果～鸡、兔各有几只,你们是怎样得出结论的, 学生汇报表达的方式: 生1:我们利用画图凑数的: ?先画10个头。 ?每个头下画上两条腿。 数一数～共有40条腿～比题中给出的腿数少54-20=14条腿。 ?给一些鸡添上两条腿～叫它变成兔(边添腿边数～凑够54条腿。 每把一只鸡添上两条腿～它就变成了兔～显然添14条腿就变出来7只兔(这样就得出答案～笼中有7只兔和13只鸡。 2(列表法: 生1:我们一个一个地试～把结果列成表格～最后得出7只鸡、3只兔。 头/个 鸡/只 兔/只 腿/条 20 1 19 78 20 2 18 76 20 3 17 74 20 4 16 72 … … … … 13 7 54 生2:我们组得出的结果也是只13鸡、7只兔～但我们不是一个一个地试～这样太麻烦了～我们是5个5个地试。 头/个 鸡/只 兔/只 腿/条 20 1 19 78 development of freezing plans and emergency precautions. 3, night-time setting the full-time attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible listens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified planting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABT-3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, improve the survival rate is obvious. (2) KD-1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature expansion 多了 20 5 15 70 20 10 10 60 20 15 5 50 20 14 6 52 20 13 7 54 生3:因为鸡、兔共20只～我们先假设鸡、兔各10只～这样共有60条腿～比54条腿多6条～说明假设的兔多了3只～鸡少了3只～于是兔只有7只～鸡有13只。 生4:我们是先按鸡兔各一半来算的。 头/个 鸡/只 兔/只 腿/条 20 10 10 60 20 12 8 56 20 13 7 54 师:同学们的探索精神和方法都很好～都能用自己的方法成功地解决“鸡兔同笼问题”。 师:谁还有其他的解法吗,, 老师让举手的其中三名学生上台板演, 生5:假设20只都是鸡～那么兔有:,54-20×2,?,4-2,=7,只,～鸡有20-7=13,只,。 生6:假设20只都是兔～那么鸡有:,4×20-54,?,4-2,=13,只,～兔有20-13=7,只,。 生7:设鸡有X只～那么兔有,20-X,只。 2X+4,20-X,=54～X=13～ 20-13=7,只, 即鸡有13只～兔有7只。 师:同学太聪明了～想出了这么多好办法～通过以上的学习～你有什么发现～有什么想法吗, 生:解决一个问题可以有不同的方法。 …… 三、想一想～做一做: 1(尝试解答课前提出的古代《孙子算经》中记载的鸡兔同笼问题。书中说:“今有鸡兔同笼～上有三十五头～下有九十四足～问鸡兔各几何, 2(完成书中练一练中的4道题 第4道题～ 小结: 师生共同总结～我们今天学习的鸡兔同笼问题～发现了可以用画图的方法解决、可以用列表的方式进行分析。还可以用假设的方法,亦可称作置换法,～可以先假设都是一种事物,换成同一种事物,～再根据题中给出的条件进行修正、推算。有的同学还用方程来解决这个问题～一个问题可以用多种方法来解决～真是条条大路通罗马呀:希望同学们今后在学习中也能象今天一样肯于动脑～勤于思考～使我们每一个同学都越学聪明。 of puffing chicken manure. High temperature expansion 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, night2 一,基本问题 "鸡兔同笼"是一类有名的中国古算题.最早出现在《孙子算经》中.许多算术应用题都可以转化成这类问题,或者用解它的典型解法--"假设法"来求解.因此很有必要学会它的解法和思路. 例1 有若干只鸡和兔子,它们共有88个头,244只脚,鸡和兔各有多少只 解:我们设想,每只鸡都是"金鸡独立",一只脚站着;而每只兔子都用两条后腿,像人一样用两只脚站着.现在,地面上出现脚的总数的一半,〃也就是 244?2=122(只). 在122这个数里,鸡的头数算了一次,兔子的头数相当于算了两次.因此从122减去总头数88,剩下的就是兔子头数 122-88=34, 有34只兔子.当然鸡就有54只. 答:有兔子34只,鸡54只. 上面的计算,可以归结为下面算式: 总脚数?2-总头数=兔子数. 上面的解法是《孙子算经》中记载的.做一次除法和一次减法,马上能求出兔子数,多简单!能够这样算,主要利用了兔和鸡的脚数分别是4和2,4又是2的2倍.可是,当其他问题转化成这类问题时,"脚数"就不一定是4和2,上面的计算方法就行不通.因此,我们对这类问题给出一种一般解法. 还说例1. 如果设想88只都是兔子,那么就有4×88只脚,比244只脚多了 88×4-244=108(只). 每只鸡比兔子少(4-2)只脚,所以共有鸡 (88×4-244)?(4-2)= 54(只). 说明我们设想的88只"兔子"中,有54只不是兔子.而是鸡.因此可以列出公式 鸡数=(兔脚数×总头数-总脚数)?(兔脚数-鸡脚数). 当然,我们也可以设想88只都是"鸡",那么共有脚2×88=176(只),比244只脚少了 244-176=68(只). 每只鸡比每只兔子少(4-2)只脚, 68?2=34(只). 说明设想中的"鸡",有34只是兔子,也可以列出公式 兔数=(总脚数-鸡脚数×总头数)?(兔脚数-鸡脚数). 上面两个公式不必都用,用其中一个算出兔数或鸡数,再用总头数去减,就知道另一个数. 假设全是鸡,或者全是兔,通常用这样的思路求解,有人称为"假设法". 现在,拿一个具体问题来试试上面的公式. 例2 红铅笔每支0.19元,蓝铅笔每支0.11元,两种铅笔共买了16支,花了2.80元.问 achi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABT3 红,蓝铅笔各买几支 解:以"分"作为钱的单位.我们设想,一种"鸡"有11只脚,一种"兔子"有19只脚,它们共有16个头,280只脚. 现在已经把买铅笔问题,转化成"鸡兔同笼"问题了.利用上面算兔数公式,就有 蓝笔数=(19×16-280)?(19-11) =24?8 =3(支). 红笔数=16-3=13(支). 答:买了13支红铅笔和3支蓝铅笔. 对于这类问题的计算,常常可以利用已知脚数的特殊性.例2中的"脚数"19与11之和是30.我们也可以设想16只中,8只是"兔子",8只是"鸡",根据这一设想,脚数是 8×(11+19)=240. 比280少40. 40?(19-11)=5. 就知道设想中的8只"鸡"应少5只,也就是"鸡"(蓝铅笔)数是3. 30×8比19×16或11×16要容易计算些.利用已知数的特殊性,靠心算来完成计算. 实际上,可以任意设想一个方便的兔数或鸡数.例如,设想16只中,"兔数"为10,"鸡数"为6,就有脚数 19×10+11×6=256. 比280少24. 24?(19-11)=3, 就知道设想6只"鸡",要少3只. 要使设想的数,能给计算带来方便,常常取决于你的心算本领. 下面再举四个稍有难度的例子. 例3 一份稿件,甲单独打字需6小时完成.乙单独打字需10小时完成,现在甲单独打若干小时后,因有事由乙接着打完,共用了7小时.甲打字用了多少小时 解:我们把这份稿件平均分成30份(30是6和10的最小公倍数),甲每小时打30?6=5(份),乙每小时打30?10=3(份). 现在把甲打字的时间看成"兔"头数,乙打字的时间看成"鸡"头数,总头数是7."兔"的脚数是5,"鸡"的脚数是3,总脚数是30,就把问题转化成"鸡兔同笼"问题了. 根据前面的公式 "兔"数=(30-3×7)?(5-3) =4.5, "鸡"数=7-4.5 =2.5, 也就是甲打字用了4.5小时,乙打字用了2.5小时. 答:甲打字用了4小时30分. rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, nightof puffing chicken manure. High temperature expansion 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-4 例4 今年是1998年,父母年龄(整数)和是78岁,兄弟的年龄和是17岁.四年后(2002年)父的年龄是弟的年龄的4倍,母的年龄是兄的年龄的3倍.那么当父的年龄是兄的年龄的3倍时,是公元哪一年 解:4年后,两人年龄和都要加8.此时兄弟年龄之和是17+8=25,父母年龄之和是78+8=86.我们可以把兄的年龄看作"鸡"头数,弟的年龄看作"兔"头数.25是"总头数".86是"总脚数".根据公式,兄的年龄是 (25×4-86)?(4-3)=14(岁). 1998年,兄年龄是 14-4=10(岁). 父年龄是 (25-14)×4-4=40(岁). 因此,当父的年龄是兄的年龄的3倍时,兄的年龄是 (40-10)?(3-1)=15(岁). 这是2003年. 答:公元2003年时,父年龄是兄年龄的3倍. 例5 蜘蛛有8条腿,蜻蜓有6条腿和2对翅膀,蝉有6条腿和1对翅膀.现在这三种小虫共18只,有118条腿和20对翅膀.每种小虫各几只 解:因为蜻蜓和蝉都有6条腿,所以从腿的数目来考虑,可以把小虫分成"8条腿"与"6条腿"两种.利用公式就可以算出8条腿的 蜘蛛数=(118-6×18)?(8-6) =5(只). 因此就知道6条腿的小虫共 18-5=13(只). 也就是蜻蜓和蝉共有13只,它们共有20对翅膀.再利用一次公式 蝉数=(13×2-20)?(2-1)=6(只). 因此蜻蜓数是13-6=7(只). 答:有5只蜘蛛,7只蜻蜓,6只蝉. 例6 某次数学考试考五道题,全班52人参加,共做对181道题,已知每人至少做对1道题,做对1道的有7人,5道全对的有6人,做对2道和3道的人数一样多,那么做对4道的人数有多少人 解:对2道,3道,4道题的人共有 52-7-6=39(人). 他们共做对 181-1×7-5×6=144(道). 由于对2道和3道题的人数一样多,我们就可以把他们看作是对2.5道题的人((2+3)?2=2.5).这样 兔脚数=4,鸡脚数=2.5, achi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABT5 总脚数=144,总头数=39. 对4道题的有 (144-2.5×39)?(4-1.5)=31(人). 答:做对4道题的有31人. 习题一 1.龟鹤共有100个头,350只脚.龟,鹤各多少只 2.学校有象棋,跳棋共26副,恰好可供120个学生同时进行活动.象棋2人下一副棋,跳棋6人下一副.象棋和跳棋各有几副 3.一些2分和5分的硬币,共值2.99元,其中2分硬币个数是5分硬币个数的4倍,问5分硬币有多少个 4.某人领得工资240元,有2元,5元,10元三种人民币,共50张,其中2元与5元的张数一样多.那么2元,5元,10元各有多少张 5.一件工程,甲单独做12天完成,乙单独做18天完成,现在甲做了若干天后,再由乙接着单独做完余下的部分,这样前后共用了16天.甲先做了多少天 6.摩托车赛全程长281千米,全程被划分成若干个阶段,每一阶段中,有的是由一段上坡路(3千米),一段平路(4千米),一段下坡路(2千米)和一段平路(4千米)组成的;有的是由一段上坡路(3千米),一段下坡路(2千米)和一段平路(4千米)组成的.已知摩托车跑完全程后,共跑了25段上坡路.全程中包含这两种阶段各几段 7.用1元钱买4分,8分,1角的邮票共15张,问最多可以买1角的邮票多少张 二,"两数之差"的问题 鸡兔同笼中的总头数是"两数之和",如果把条件换成"两数之差",又应该怎样去解呢 例7 买一些4分和8分的邮票,共花6元8角.已知8分的邮票比4分的邮票多40张,那么两种邮票各买了多少张 解一:如果拿出40张8分的邮票,余下的邮票中8分与4分的张数就一样多. (680-8×40)?(8+4)=30(张), 这就知道,余下的邮票中,8分和4分的各有30张. 因此8分邮票有 40+30=70(张). 答:买了8分的邮票70张,4分的邮票30张. 也可以用任意假设一个数的办法. 解二:譬如,假设有20张4分,根据条件"8分比4分多40张",那么应有60张8分.以"分"作为计算单位,此时邮票总值是 4×20+8×60=560. 比680少,因此还要增加邮票.为了保持"差"是40,每增加1张4分,就要增加1张8分,每种要增加的张数是 (680-4×20-8×60)?(4+8)=10(张). 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, nightof puffing chicken manure. High temperature expansion6 因此4分有20+10=30(张),8分有60+10=70(张). 例8 一项工程,如果全是晴天,15天可以完成.倘若下雨,雨天一天 工程要多少天才能完成 解:类似于例3,我们设工程的全部工作量是150份,晴天每天完成10份,雨天每天完成8份.用上一例题解一的方法,晴天有 (150-8×3)?(10+8)= 7(天). 雨天是7+3=10天,总共 7+10=17(天). 答:这项工程17天完成. 请注意,如果把"雨天比晴天多3天"去掉,而换成已知工程是17天完成,由此又回到上一节的问题.差是3,与和是17,知道其一,就能推算出另一个.这说明了例7,例8与上一节基本问题之间的关系. 总脚数是"两数之和",如果把条件换成"两数之差",又应该怎样去解呢 例9 鸡与兔共100只,鸡的脚数比兔的脚数少28.问鸡与兔各几只 解一:假如再补上28只鸡脚,也就是再有鸡28?2=14(只),鸡与兔脚数就相等,兔的脚是鸡的脚4?2=2(倍),于是鸡的只数是兔的只数的2倍.兔的只数是 (100+28?2)?(2+1)=38(只). 鸡是 100-38=62(只). 答:鸡62只,兔38只. 当然也可以去掉兔28?4=7(只).兔的只数是 (100-28?4)?(2+1)+7=38(只). 也可以用任意假设一个数的办法. 解二:假设有50只鸡,就有兔100-50=50(只).此时脚数之差是 4×50-2×50=100, 比28多了72.就说明假设的兔数多了(鸡数少了).为了保持总数是100,一只兔换成一只鸡,少了4只兔脚,多了2只鸡脚,相差为6只(千万注意,不是2).因此要减少的兔数是 (100-28)?(4+2)=12(只). 兔只数是 50-12=38(只). 另外,还存在下面这样的问题:总头数换成"两数之差",总脚数也换成"两数之差". 例10 古诗中,五言绝句是四句诗,每句都是五个字;七言绝句是四句诗,每句都是七个字.有一诗选集,其中五言绝句比七言绝句多13首,总字数却反而少了20个字.问两种诗各多少首. 解一:如果去掉13首五言绝句,两种诗首数就相等,此时字数相差 13×5×4+20=280(字). achi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABT7 每首字数相差 7×4-5×4=8(字). 因此,七言绝句有 28?(28-20)=35(首). 五言绝句有 35+13=48(首). 答:五言绝句48首,七言绝句35首. 解二:假设五言绝句是23首,那么根据相差13首,七言绝句是10首.字数分别是20×23=460(字),28×10=280(字),五言绝句的字数,反而多了 460-280=180(字). 与题目中"少20字"相差 180+20=200(字). 说明假设诗的首数少了.为了保持相差13首,增加一首五言绝句,也要增一首七言绝句,而字数相差增加8.因此五言绝句的首数要比假设增加 200?8=25(首). 五言绝句有 23+25=48(首). 七言绝句有 10+25=35(首). 在写出"鸡兔同笼"公式的时候,我们假设都是兔,或者都是鸡,对于例7,例9和例10三个问题,当然也可以这样假设.现在来具体做一下,把列出的计算式子与"鸡兔同笼"公式对照一下,就会发现非常有趣的事. 例7,假设都是8分邮票,4分邮票张数是 (680-8×40)?(8+4)=30(张). 例9,假设都是兔,鸡的只数是 (100×4-28)?(4+2)=62(只). 10,假设都是五言绝句,七言绝句的首数是 (20×13+20)?(28-20)=35(首). 首先,请读者先弄明白上面三个算式的由来,然后与"鸡兔同笼"公式比较,这三个算式只是有一处"-"成了"+".其奥妙何在呢 当你进入初中,有了负数的概念,并会列二元一次方程组,就会明白,从数学上说,这一讲前两节列举的所有例子都是同一件事. 例11 有一辆货车运输2000只玻璃瓶,运费按到达时完好的瓶子数目计算,每只2角,如有破损,破损瓶子不给运费,还要每只赔偿1元.结果得到运费379.6元,问这次搬运中玻璃瓶破损了几只 解:如果没有破损,运费应是400元.但破损一只要减少1+0.2=1.2(元).因此破损只数是 on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, nightof puffing chicken manure. High temperature expansion 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -8 (400-379.6)?(1+0.2)=17(只). 答:这次搬运中破损了17只玻璃瓶. 请你想一想,这是"鸡兔同笼"同一类型的问题吗 例12 有两次自然测验,第一次24道题,答对1题得5分,答错(包含不答)1题倒扣1分;第二次15道题,答对1题8分,答错或不答1题倒扣2分,小明两次测验共答对30道题,但第一次测验得分比第二次测验得分多10分,问小明两次测验各得多少分 解一:如果小明第一次测验24题全对,得5×24=120(分).那么第二次只做对30-24=6(题)得分是 8×6-2×(15-6)=30(分). 两次相差 120-30=90(分). 比题目中条件相差10分,多了80分.说明假设的第一次答对题数多了,要减少.第一次答对减少一题,少得5+1=6(分),而第二次答对增加一题不但不倒扣2分,还可得8分,因此增加8+2=10分.两者两差数就可减少 6+10=16(分). (90-10)?(6+10)=5(题). 因此,第一次答对题数要比假设(全对)减少5题,也就是第一次答对19题,第二次答对30-19=11(题). 第一次得分 5×19-1×(24- 9)=90. 第二次得分 8×11-2×(15-11)=80. 答:第一次得90分,第二次得80分. 解二:答对30题,也就是两次共答错 24+15-30=9(题). 第一次答错一题,要从满分中扣去5+1=6(分),第二次答错一题,要从满分中扣去8+2=10(分).答错题互换一下,两次得分要相差6+10=16(分). 如果答错9题都是第一次,要从满分中扣去6×9.但两次满分都是120分.比题目中条件"第一次得分多10分",要少了6×9+10.因此,第二次答错题数是 (6×9+10)?(6+10)=4(题)〃 第一次答错 9-4=5(题). 第一次得分 5×(24-5)-1×5=90(分). 第二次得分 8×(15-4)-2×4=80(分). 习题二 1.买语文书30本,数学书24本共花83.4元.每本语文书比每本数学书贵0.44元.每本语文书和数学书的价格各是多少 2.甲茶叶每千克132元,乙茶叶每千克96元,共买这两种茶叶12千克.甲茶叶所花的 achi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABT9 钱比乙茶叶所花钱少354元.问每种茶叶各买多少千克 3.一辆卡车运矿石,晴天每天可运16次,雨天每天只能运11次.一连运了若干天,有晴天,也有雨天.其中雨天比晴天多3天,但运的次数却比晴天运的次数少27次.问一连运了多少天 4.某次数学测验共20道题,做对一题得5分,做错一题倒扣1分,不做得0分.小华得了76分.问小华做对了几道题 5.甲,乙二人射击,若命中,甲得4分,乙得5分;若不中,甲失2分,乙失3分.每人各射10发,共命中14发.结算分数时,甲比乙多10分.问甲,乙各中几发 6.甲,乙两地相距12千米.小张从甲地到乙地,在停留半小时后,又从乙地返回甲地,小王从乙地到甲地,在甲地停留40分钟后,又从甲地返回乙地.已知两人同时分别从甲,乙两地出发,经过4小时后,他们在返回的途中相遇.如果小张速度比小王速度每小时多走1.5千米,求两人的速度. 三,从"三"到"二" "鸡"和"兔"是两种东西,实际上还有三种或者更多种东西的类似问题.在第一节例5和例6就都有三种东西.从这两个例子的解法,也可以看出,要把"三种"转化成"二种"来考虑.这一节要通过一些例题,告诉大家两类转化的方法. 例13 学校组织新年游艺晚会,用于奖品的铅笔,圆珠笔和钢笔共232支,共花了300元.其中铅笔数量是圆珠笔的4倍.已知铅笔每支0.60元,圆珠笔每支2.7元,钢笔每支6.3元.问三种笔各有多少支 解:从条件"铅笔数量是圆珠笔的4倍",这两种笔可并成一种笔,四支铅笔和一支圆珠笔成一组,这一组的笔,每支价格算作 (0.60×4+2.7)?5=1.02(元). 现在转化成价格为1.02和6.3两种笔.用"鸡兔同笼"公式可算出,钢笔支数是 (300-1.02×232)?(6.3-1.02)=12(支). 铅笔和圆珠笔共 232-12=220(支). 其中圆珠笔 220?(4+1)=44(支). 铅笔 220-44=176(支). 答:其中钢笔12支,圆珠笔44支,铅笔176支. 例14 商店出售大,中,小气球,大球每个3元,中球每个1.5元,小球每个1元.张老师用120元共买了55个球,其中买中球的钱与买小球的钱恰好一样多.问每种球各买几个 解:因为总钱数是整数,大,小球的价钱也都是整数,所以买中球的钱数是整数,而且还是3的整数倍.我们设想买中球,小球钱中各出3元.就可买2个中球,3个小球.因此,可以把这两种球看作一种,每个价钱是 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, nightof puffing chicken manure. High temperature expansion10 (1.5×2+1×3)?(2+3)=1.2(元). 从公式可算出,大球个数是 (120-1.2×55)?(3-1.2)=30(个). 买中,小球钱数各是 (120-30×3)?2=15(元). 可买10个中球,15个小球. 答:买大球30个,中球10个,小球15个. 例13是从两种东西的个数之间倍数关系,例14是从两种东西的总钱数之间相等关系(倍数关系也可用类似方法),把两种东西合井成一种考虑,实质上都是求两种东西的平均价,就把"三"转化成"二"了. 例15是为例16作准备. 例15 某人去时上坡速度为每小时走3千米,回来时下坡速度为每小时走6千米,求他的平均速度是多少 解:去和回来走的距离一样多.这是我们考虑问题的前提. 平均速度=所行距离?所用时间 去时走1千米,要用20分钟;回来时走1千米,要用10分钟.来回共走2千米,用了30分钟,即半小时,平均速度是每小时走4千米. 千万注意,平均速度不是两个速度的平均值:每小时走(6+3)?2=4.5千米. 例16 从甲地至乙地全长45千米,有上坡路,平路,下坡路.李强上坡速度是每小时3千米,平路上速度是每小时5千米,下坡速度是每小时6千米.从甲地到乙地,李强行走了10小时;从乙地到甲地,李强行走了11小时.问从甲地到乙地,各种路段分别是多少千米 解:把来回路程45×2=90(千米)算作全程.去时上坡,回来是下坡;去时下坡回来时上坡.把上坡和下坡合并成"一种"路程,根据例15,平均速度是每小时4千米.现在形成一个非常简单的"鸡兔同笼"问题.头数10+11=21,总脚数90,鸡,兔脚数分别是4和5.因此平路所用时间是 (90-4×21)?(5-4)=6(小时). 单程平路行走时间是6?2=3(小时). 从甲地至乙地,上坡和下坡用了10-3=7(小时)行走路程是 45-5×3=30(千米). 又是一个"鸡兔同笼"问题.从甲地至乙地,上坡行走的时间是 (6×7-30)?(6-3)=4(小时). 行走路程是3×4=12(千米). 下坡行走的时间是7-4=3(小时).行走路程是6×3=18(千米). 答:从甲地至乙地,上坡12千米,平路15千米,下坡18千米. 做两次"鸡兔同笼"的解法,也可以叫"两重鸡兔同笼问题".例16是非常典型的例题. 例17 某种考试已举行了24次,共出了426题.每次出的题数,有25题,或者16题,或者 3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABTachi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD11 20题.那么,其中考25题的有多少次 解:如果每次都考16题,16×24=384,比426少42道题. 每次考25道题,就要多25-16=9(道). 每次考20道题,就要多20-16=4(道). 就有 9×考25题的次数+4×考20题的次数=42. 请注意,4和42都是偶数,9×考25题次数也必须是偶数,因此,考25题的次数是偶数,由9×6=54比42大,考25题的次数,只能是0,2,4这三个数.由于42不能被4整除,0和4都不合适.只能是考25题有2次(考20题有6次). 答:其中考25题有2次. 例18 有50位同学前往参观,乘电车前往每人1.2元,乘小巴前往每人4元,乘地下铁路前往每人6元.这些同学共用了车费110元,问其中乘小巴的同学有多少位 解:由于总钱数110元是整数,小巴和地铁票也都是整数,因此乘电车前往的人数一定是5的整数倍. 如果有30人乘电车, 110-1.2×30=74(元). 还余下50-30=20(人)都乘小巴钱也不够.说明假设的乘电车人数少了. 如果有40人乘电车 110-1.2×40=62(元). 还余下50-40=10(人)都乘地下铁路前往,钱还有多(62>6×10).说明假设的乘电车人数又多了.30至40之间,只有35是5的整数倍. 现在又可以转化成"鸡兔同笼"了: 总头数 50-35=15, 总脚数 110-1.2×35=68. 因此,乘小巴前往的人数是 (6×15-68)?(6-4)=11. 答:乘小巴前往的同学有11位. 在"三"转化为"二"时,例13,例14,例16是一种类型.利用题目中数量比例关系,把两种东西合并组成一种.例17,例18是另一种类型.充分利用所求个数是整数,以及总量的限制,其中某一个数只能是几个数值.对几个数值逐一考虑是否符合题目的条件.确定了一个个数,也就变成"二"的问题了.在小学算术的范围内,学习这两种类型已足够了.更复杂的问题,只能借助中学的三元一次方程组等代数方法去求解. 习题三 1.有100枚硬币,把其中2分硬币全换成等值的5分硬币,硬币总数变成79个,然后又把其中的1分硬币换成等值的5分硬币,硬币总数变成63个.求原有2分及5分硬币共值多少钱 2."京剧公演"共出售750张票得22200元.甲票每张60元,乙票每张30元,丙票每张18 -rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, nightof puffing chicken manure. High temperature expansion 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use12 元.其中丙票张数是乙票张数的2倍.问其中甲票有多少张 3.小明参加数学竞赛,共做20题得67分.已知做一题得5分,不答得2分,做错一题倒扣3分.又知道他做错的题和没答的题一样多.问小明共做对几题 4.1分,2分和5分硬币共100枚,价值2元,如果其中2分硬币的价值比1分硬币的价值多13分.问三种硬币各多少枚 注:此题没有学过分数运算的同学可以不做. 5.甲地与乙地相距24千米.某人从甲地到乙地往返行走.上坡速度每小时4千米,走平路速度每小时5千米,下坡速度每小时6千米.去时行走了4小时50分,回来时用了5小时.问从甲地到乙地,上坡,平路,下坡各多少千米 6.某学校有12间宿舍,住着80个学生.宿舍的大小有三种:大的住8个学生,不大不小的住7个学生,小的住5人.其中不大不小的宿舍最多,问这样的宿舍有几间 测验题 1.松鼠妈妈采松籽,晴天每天可以采20个,雨天每天只能采12个. 它一连几天采了112个松籽,平均每天采14个. 问这几天当中有几天有雨 2.有一水池,只打开甲水龙头要24分钟注满水池,只打开乙水龙头要36分钟才注满水池.现在先打开甲水龙头几分钟,然后关掉甲,打开乙水龙头把水池注满.已知乙水龙头比甲水龙头多开26分钟.问注满水池总共用了多少分钟 3.某工程甲队独做50天可以完成,乙队独做75天可以完成.现在两队合做,但是中途乙队因另有任务调离了若干天.从开工后40天才把这项工程做完.问乙队中途离开了多少天 4.小华从家到学校,步行一段路后就跑步.他步行速度是每分钟600 ,跑步速度是每分钟140米.虽然步行时间比跑步时间多4分钟,但步行的距离却比跑步的距离少400米.问从家到学校多远 5.有16位教授,有人带1个研究生,有人带2个研究生,也有人带3个研究生.他们共带了27位研究生.其中带1个研究生的教授人数与带2,3个研究生的教授人数一样多.问带2个研究生的教授有几人 6.某商场为招揽顾客举办购物抽奖.奖金有三种:一等奖1000元,二等奖250元,三等奖50元.共有100人中奖,奖金总额为9500元.问二等奖有多少名 7.有一堆硬币,面值为1分,2分,5分三种,其中1分硬币个数是2分硬币个数的11倍.已知这堆硬币面值总和是1元,问5分的硬币有多少个 第三讲 答案 习题一 1.龟75只,鹤25只. 2.象棋9副,跳棋17副. 3.2分硬币92个,5分硬币23个. 应将总钱数2.99元分成2×4+5=13(份),其中2分钱数占2×4=8(份),5分钱数占5份. 4.2元与5元各20张,10元有10张. -improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABTachi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can13 2元与5元的张数之和是 (10×50-240)?[10-(2+5)?2]=40(张). 5.甲先做了4天. 提示:把这件工程设为36份,甲每天做3份,乙每天做2份. 6.第一种路段有14段,第二种路段有11段. 第一种路段全长13千米,第二种路段全长9千米,全赛程281千米,共25段,是标准的"鸡兔同笼". 7.最多可买1角邮票6张. 假设都买4分邮票,共用4×15=60(分),就多余100-60=40(分).买一张1角邮票,可以认为40分换1角,要多6分.40?6=6……4,最多买6张.最后多余4分,加在一张4分邮票上,恰好买一张8分邮票. 习题二 1.语文书1.74元,数学书1.30元. 设想语文书每本便宜0.44元,因此数学书的单价是 (83.4-0.44×30)?(30+24). 2.买甲茶3.5千克,乙茶8.5千克. 甲茶数=(96×12-354)?(132+96)=3.5(千克) 3.一连运了27天. 晴天数=(11×3+27)?(16-11)=12(天) 4.小华做对了16题. 76分比满分100分少24分.做错一题少6分,不做少5分.24分只能是6×4. 5.甲中8发,乙中6发. 假设甲中10发,乙就中14-10=4(发).甲得4×10=40(分),乙得5×4-3×6= 2(分).比题目条件"甲比乙多10分"相差(40-2)-10=28(分),甲少中1发,少4+2=6(分),乙可增5+3=8(分). 28?(6+8)=2. 甲中10-2=8(发). 6.小张速度每小时6千米,小王速度每小时4.5千米. 王的速度是每小时 注:为了避免分数运算,路程以米为单位,时间以分钟为单位,就可以达到目的. “鸡兔同笼”问题最早见于《孙子算经》～至今一直为人们所喜闻乐见。作为小学数学应用题中的一类重要问题～是智力训练的好问题～古今中外许多人都对它的解法作过研究～可以说它的解法已“箩成筐”了。 “鸡兔同笼”问题:今有鸡兔同笼～上有三十五头～下有九十四足。问鸡兔各几何, 1.鸡兔共35只～它们不可能全是鸡～因为那么一来它们将只有70只脚了～它们也不 3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, nightof puffing chicken manure. High temperature expansion 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-rvival rate is obvious. (2) KDimprove the su14 可能全是兔子～因为那样就将有140只脚。 但是它们应该恰好有94只脚。如果正好有20只鸡～15只兔子～ 那么它们就将有100只脚～列表如下: 鸡 兔 脚 35 0 70 0 35 140 20 15 100 如果把鸡的数目取小一些～那么必须把兔子的数目取大一些～而这就使得脚数增大了。反之～如果把鸡的数目增加一些～那么兔子的数目就减少一些～而这就使得脚数减小了。根据脚数随鸡数变化的规律～23只鸡和12只兔子～恰好有74只脚。 以“探索”为特征的这种逐次逼近的解法～由一系列的试探组成～其中每一次都企图纠正前面一次所带来的误差～整个说来～误差随着进一步的试探而减少～而依次进行的试探则越来越接近于所要求的结果。当然数字较大或较为复杂时～用这种方法求解～就需要实验多次。教师应该鼓励学生巧妙地应用逐次逼近这种基本的方法。 2.孙子的解法“上置三十五头～下置九十四足。半其足得四十七。以少减多～再命之～上三除下四～上五除下七。下有一除上三～下有二除上五～即得”。按其所述～在筹算板上的演示过程如下: 附图{图} 翻译成算术方法就是: 兔数 ,94?2,,35,12 鸡数 35,12,23 美国杰出数学教育家G 〃波利亚对这种解法创设了教学情景:意外地看见笼中的禽畜正在作一种古怪的姿式～每一只鸡都用一条腿站着～而每一只兔子都用其,两条,后腿站着～在这个不寻常的情况下～只用了半数的腿～即47条腿。在70这个数目中～鸡的头只计算了一次～而兔子的头则计算了两次～从47这个数减去所有头数35～就剩下兔子的头数了。当然～鸡的只数可立刻求出。 这种解法是巧妙的～但它需要清晰地掌握题中的数量关系～不是所有学生都能理解的。 3.第一种解法是假想这35只都是鸡或兔～思路虽然巧妙～却使学生想不通:明明有鸡有兔为什么假设只有一种呢,第二种解法巧妙而有趣～但其出发点仍似天外飞来～不易使全体学生掌握。张景中院士独具匠心～他从学生的常识出发～自然地引出了解答。 先问:“兔有四只脚～为什么鸡只有两只脚”这岂不是太不公平了吗,” 经过思考～学生会找到理由:“不是不公平～鸡还有两只翅膀呢:” 问:“如果翅膀也算脚～总共该有多少只脚,” 这容易回答:35×4,140～140只脚。 “但题中翅膀不算脚～只有94只脚～可见有多少翅膀呢,” anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientificto ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABTachi 15 “140,94,46～46只翅膀:” 于是学生兴奋地喊出来:“23只鸡:” 这种解法～每个学生都能立即理解～即使不再复习～半年后他们仍能回忆起来。同时这个例子告诉我们～要充分利用学生认知结构中已有的知识去创设问题情景～这样有利于学习中的正迁移的发生～能促进新知识的掌握、巩固和应用。 4.把数量关系问题和图形结合起来考虑～借助于图示的鲜明直观性～帮助学生理解题目中的数量关系～适合小学生的思维特点～是小学数学教学中的一贯作法: 根据题意作出图示: 附图{图} 由图示可以看出:兔数,,94,35×2,?2,12,只,～鸡数,35,12,23,只, 5.设笼中有x只鸡～则有,35,x,只兔～根据题意～得2x，4,35,x,,94解之～得x,23,只,～35,23,12,只,。 推广～若以h代替35～f代替94～得到这类问题的求解公式: f f兔数,??,h～鸡数,2h,??～即:兔子数等于脚数的一半减头数～ 2 2鸡数等于头数的两倍减去脚数的一半。 这种解法比较简单～小学五,或六,的学生都能掌握。 6.盈不足算法是我国古代解决算术应用题的基本方法。对于“鸡兔同笼”问题～可以通过两次假设～试算～化归为盈不足模式～然后套用现成的公式演算。 假设笼中有x[,1],20只鸡～则有兔15只～得脚100只～则知盈,多出脚数,y[,1],6,假设笼中有鸡x[,2],25只～则有兔10只～得脚90只～则知不足y[,2],4。 于是有盈不足算法,鸡,x[,1]y[,2]，x[,2]y[,1],?,y[,1]，y[,2],,,20×4 ，25×6,?,4，6,,23,只,或兔,15×4，10×6,?,4，6,,12,只,。 “盈不足术被阿拉伯人称为中国算法～它是以特定的数学模型来处理一大类应用问题的方法～在指导数学课外活动小组活动时～介绍这种方法～既开阔了学生的视野～又能增强学生的民族自尊心和自豪感～激发学生的爱国热情。” 7.假设有20只鸡～15只兔～那么就会有,20×2，15×4,,100 只脚～比实际多出了,100,94,,6只～这是因为把鸡看成兔～每只多算了2只脚～共把,6?2,,3只鸡看成兔～加上原来的20只鸡共有23只鸡～12只免。 这种先在假设的条件下推算～得出与已知数量不符～最后再适当调整的方法～是小学数学教材教法第一册第一章中整数四则应用题特殊解题思路部分介绍的～是小学数学教师应该了解的方法。 这个问题～是我国古代著名趣题之一。大约在1500年前～《孙子算经》中就记载了这个有趣的问题。书中是这样叙述的:“今有鸡兔同笼～上有三十五头～下有九十四足～问鸡兔各几何,这四句话的意思是:有若干只鸡兔同在一个笼子里～从上面数～有35个头,从下面数～有94只脚。求笼中各有几只鸡和兔, of puffing chicken manure. High temperature expansion 1 Super absorbent polymers. Insurance agent can enhance soil water storage, water retention and improve survival. (3) the use-rvival rate is obvious. (2) KDimprove the su3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -on of new technology to ensure quality: 1, the use of some new scientific achievements and improve survival. (1) ABTcatithe planting process except in strict accordance with the tender specified planting methods and technical measures, the appli new technology, new technology, new equipment, new material measures in order to ensure the survival of seedlings planted, in ion ofste water facilities, in order to ensure normal construction. The 12th chapter, resources planning and advancement, introductbefore construction began. 5, higher in the winter rain, prior to construction, priority project areas, and built a simple waments t finished filling the pit covered with a poncho, should be drained before the next fill clean water content meet the requireens to and published on the weather conditions. 4, should complete the construction of the winter rainy season Shi ... Has notime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible list-time setting the full-development of freezing plans and emergency precautions. 3, night16 解答思路是这样的:假如砍去每只鸡、每只兔一半的脚～则每只鸡就变成了“独角鸡”～每只兔就变成了“双脚兔”。这样～,1,鸡和兔的脚的总数就由94只变成了47只,,2,如果笼子里有一只兔子～则脚的总数就比头的总数多1。因此～脚的总只数47与总头数35的差～就是兔子的只数～即47,35,12,只,。显然～鸡的只数就是35,12,23,只,了。 这一思路新颖而奇特～其“砍足法”也令古今中外数学家赞叹不已。这种思维方法叫化归法。化归法就是在解决问题时～先不对问题采取直接的分析～而是将题中的条件或问题进行变形～使之转化～直到最终把它归成某个已经解决的问题。 to ensure the survival of seedlings planted, in the planting process except in strict accordance with the tender specified pl order , resources planning and advancement, introduction of new technology, new technology, new equipment, new material measures in, priority project areas, and built a simple waste water facilities, in order to ensure normal construction. The 12th chapterctionnext fill clean water content meet the requirements before construction began. 5, higher in the winter rain, prior to constru ction of the winter rainy season Shi ... Has not finished filling the pit covered with a poncho, should be drained before thens to and published on the weather conditions. 4, should complete the construtime attendant to ensure someone on duty and on duty day and night recording, while setting the Weatherman, responsible liste-time setting the full-emergency precautions. 3, nightdevelopment of freezing plans and ansionenhance soil water storage, water retention and improve survival. (3) the use of puffing chicken manure. High temperature exp 1 Super absorbent polymers. Insurance agent can-improve the survival rate is obvious. (2) KD3 rooting powder. This kind of rooting powder for Evergreen coniferous trees take root and valuable species of fast rooting, -evements and improve survival. (1) ABTachi anting methods and technical measures, the application of new technology to ensure quality: 1, the use of some new scientific17