语感启蒙2004打印文本

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少儿语感启蒙【三篇】

少儿语感启蒙【三篇】导读：本文少儿语感启蒙【三篇】，仅供参考，如果觉得很不错，欢迎点评和分享。

少儿语感启蒙（1）【手指游戏】the eentsy weentsy spider went up the water spout 小小蜘蛛，爬上了水管down came the rain, and washed spider out 雨滴落下，冲走了蜘蛛out came the sun and driedup all the rain 太阳出来，蒸发了水滴and the eentsy weentsy spider went up the spout again 小小蜘蛛，再爬上水管【读一读】the eentsy weentsy spider went up the water spoutout came the sun and dried up all the rain【词义解析】1.spider 蜘蛛2.water spout 水管3.rain 雨4.sun 太阳5.shoulder 肩膀例句：and the eentsy weentsy spider went up the spout again 小小蜘蛛，再爬上水管少儿语感启蒙（2）【手指游戏】Three little monkeys jumping on the bed, one fell off and bumped his head. 三只小猴子跳到床上，一只摔了下去，把他的头撞肿了。

Mama called the doctor,and the doctor said, "No more little monkeys jumping on the bed!" 妈妈给医生打电话，医生说："猴子不要再在床上跳了！" Two little monkeys jumping on the bed, one fell off and bumped his head. 两只小猴子跳到床上，一只摔了下去，把他的头撞肿了。

六年级上册语文2004电子书

六年级上册语文2004电子书1、1介绍工艺流程一般按照工序的先后顺序逐一介绍，突出每个步骤的操作要领。

[判断题] *对错(正确答案)2、下列词语中，加着重号字的注音正确的一项是（）[单选题] *A、外甥（shēn）窘迫（jiǒng）刮痧（shā）秩序（zhì）B、筹划（chóu）供给（gěi）家谱（pǔ）惦记（diàn）C、蛮横（hèng）拥挤不堪（kān）发愣（lèng）济南（jǐ）(正确答案)D、私塾（shú）廿三（niàn）丧事（sāng）撮土（chuō）3、下列选项中加着重号字注音正确的一项是（）[单选题] *A、嗜好shì厌恶èB、崇高chóng 熔化róng(正确答案)C、补偿chǎng结缘yuánD、包揽lǎn 苦难nán4、下列选项中加着重号字注音有错误的一项是（）[单选题] *A、尘嚣xiāo荫庇bì栽种zāi 埋骨máiB、稀疏shū栅栏zhà禁锢gù嬉戏xīC、飒飒sà心弦xuán 挖空kōng 奢华shē(正确答案)D、灵寝qǐn 墓冢zhǒng朝拜cháo 灌木guàn5、下列选项中加着重号字注音正确的一项是（）[单选题] *A、粗糙cāo 红缯zēng 乳酪lào(正确答案)B、背负bèi 树冠guān 萌蘖nièC、龟裂guī宋徽宗huī贮藏zhùD、谚语yàn 紫绡qiāo 果梗gěng6、1柳永《雨霖铃》是豪放词的典型代表。

[判断题] *对(正确答案)错7、下列关于《红楼梦》理解正确的一项是（）[单选题] *A.《红楼梦》里大量诗文判词戏曲歌词，充满了譬喻，往往是对人物关系和命运际遇的暗示。

《枉凝眉》曰：“一个是阆苑仙葩，一个是美玉无瑕”，其中“阆苑仙葩”指的是贾宝玉，“美玉无瑕”指的是林黛玉。

清华版语感启蒙文本含翻译文字版

清华版语感启蒙文本(含翻译)－文字版————————————————————————————————作者: ————————————————————————————————日期：ﻩＰart Ｏne１.Liｔtle Bｅｅ小蜜蜂Liｔtle bee, ｌiｔtle bee,rounｄrounｄrouｎｄ.Little bee, ｌｉtｔｌe bee,ｓoundｓｏｕnd sounｄ.Ｂｚzzzzzz小蜜蜂,小蜜蜂，飞飞飞。

小蜜蜂,小蜜蜂,嗡嗡嗡。

嗡……2.Ｐｅek－a-boｏ躲猫猫Peeｋ，peek,ｐeeｋ-a-boo. Ｐeek, pｅeｋ, I see ｙｏu. 猫儿,猫儿,躲猫猫。

猫儿,猫儿，看到你啦。

3．Ｏpeｎ，Sｈut 张，合Open，ｓｈｕｔ,open, shuｔ.Give a lｉtｔｌｅcｌap.Creｅp, creep, creｅｐ, cｒeｅｐ. Ｇiｖe a lｉttｌｅｆｌaｐ.张、合、张、合，啪啪啪啪拍拍手。

爬、爬、爬、爬,啪啦啪啦飞得快!4．ＨerｅIs The Bｅｅｈivｅ小小蜂窝Herｅiｓthe bｅｅ- ｈiｖｅ,Whｅre ａre the ｂeｅs？Hid- deｎａ－ｗay ｗheｒe ｎo- bｏdy sｅes．Ｗａtch aｎd you’ｌl seｅｔｈem coｍeｏｕt of tｈehiｖe.Ｏne, ｔwｏ,thｒee，four,five. Bzzzｚzｚzzzzｚzzzzzzzｚｚz蜂窝在这里，蜜蜂在哪里？藏起来,找不着。

快来看,飞出来，一、二、三、四、五，嗡……５.Ten Lｉtｔle Fingeｒs十个小手指Oｎe ｌitｔlｅ, ｔwo little, three liｔtle fin-gers．Ｆour litｔle, fivｅlｉｔtle, six ｌｉttlｅfin－geｒs．Sｅveｎlittle, ｅiｇht liｔｔｌe,niｎe ｌiｔtle fｉn－ｇe ｒs.Ｔen fｉｎgersｏn ｍy hａnds.小小手指，一二三;小小手指,四五六;小小手指，七八九；十根手指，是一家。

GALOIS MODULE STRUCTURE OF p TH-POWER CLASSES OF CYCLIC EXTENSIONS OF DEGREE p n J´AN MIN´AˇC,ANDREW SCHULTZ,AND JOHN SWALLOWDedicated to the memory of Walter FeitAbstract.In the mid-1960s Boreviˇc and Faddeev initiated thestudy of the Galois module structure of groups of p th-power classesof cyclic extensions K/F of p th-power degree.They determinedthe structure of these modules in the case when F is a localﬁeld.In this paper we determine these Galois modules for all baseﬁeldsF.ContentsIntroduction and Main Theorems2 1.Notation and Lemmas8 1.1.F p[G]-modules8 1.2.Kummer Subﬁelds of K/F and Exceptional Elements10 1.3.The Fixed Submodule J G of J12 1.4.F p[G]-Submodules of J141.5.Fixed Submodules of Cyclic Submodules of J162.Base Cases213.Free Submodules and Proof of Theorem1284.Exceptional Elements325.Proof of Theorem2356.Proofs of Corollaries407.Proof of Theorem341 References46 Date:October30,2004.2000Mathematics Subject Classiﬁcation.Primary12F10;Secondary16D70.Key words and phrases.cyclic extension,Galois module,Hilbert90,Kummer theory,multiplicative group of aﬁeld,norm.J´a n Min´aˇc was supported in part by the Natural Sciences and Engineering Re-search Council of Canada,by the special Dean of Science Fund at the University of Western Ontario,and by the Mathematical Sciences Research Institute,Berkeley. John Swallow was supported in part by National Security Agency grant MDA904-02-1-0061.12J´AN MIN´AˇC,ANDREW SCHULTZ,AND JOHN SWALLOWIntroduction and Main TheoremsIn1947ˇSafareviˇc initiated the study of Galois groups of maximal p-extensions ofﬁelds with the case of localﬁelds[ˇSa47],and this study has grown into what is both an elegant theory as well as an eﬃcient tool in the arithmetic ofﬁelds.From the very beginning it became clear that the groups of p th-power classes of the variousﬁeld extensions of a base ﬁeld encode basic information about the structure of the Galois groups of maximal p-extensions.(See[Ko02,Se02].)Such groups of p th-power classes arise naturally in studies in arithmetic algebraic geometry,as for example in studies of elliptic curves.In1960Faddeev began to study the Galois module structure of p th-power classes of cyclic p-extensions,again in the case of localﬁelds,and during the mid-1960s he and Boreviˇc established the structure of these Galois modules using basic arithmetic invariants attached to Galois extensions.(See[Fa60,Bo65].)In2003two of the authors ascertained the Galois module structure of p th-power classes in the case of cyclic extensions of degree p over all baseﬁelds F containing a primitive p th root of unity[MS03].Very recently,this work paved the way for the determination of the entire Galois cohomology as a Galois module in the case of a cyclic extension of degree p of a baseﬁeld containing a primitive p th root of unity,using Voevodsky’s recent work on Galois cohomology([LMS];see[Vo03,Vo]).In this paper we extend the results obtained in[MS03]in two di-rections.First,our results hold for cyclic extensions of any p th-power degree,rather than just p,and,furthermore,we no longer require that the baseﬁeld contain a primitive p th root of unity.Thus our results provide a complete determination of p th-power classes as Galois mod-ules for all cyclic extensions of p th-power degree.We expect that,just as the results and techniques in[MS03]helped to determine the entire Milnor K-theory modulo p as a Galois module in the case of cyclic extensions of degree p,so will the results and methods developed in this paper lead to the determination of the entire Milnor K-theory modulo p as a Galois module in the case of cyclic extensions of p th-power degree.In fact,precisely such a generalization has already taken place in the case of characteristic p[BLMS].Similarly,in the same way as the results and techniques developed in[MS03]led in[MS]to the solution of Galois embedding problemsGALOIS MODULE STRUCTURE OF p TH-POWER CLASSES 3and the discovery of a new automatic realization of Galois groups,it is clear that the results in this paper will also have such Galois-theoretic applications.In a subsequent paper [MSSa]we will consider some of these applications.Our basic approach to the problem is induction,and some of the results in [MS03]handle the base case.In the end,however,neither the results nor the techniques employed are straightforward generalizations of the work in [MS03].First,the possible generalization of the innocent summand of dimension 1or 2considered in [MS03]turned out to be rather subtle to handle.These new summands of dimension p i +1for some i ∈N are very interesting invariants of cyclic extensions of p th-power degree.Another substantial challenge was to generate enough norms,and the resolution involves several thorny induction arguments.Finally,the case p =2presented a new problem for quartic extensions,and this problem is taken care of as a separate base induction case.Fundamentally,the classiﬁcation of p th-power classes as Galois mod-ules depends upon arithmetic invariants,all of which originate from the images of the norms of the intermediate ﬁelds of K/F .The classiﬁca-tion,in short,has the ﬂavor of local class ﬁeld theory,and although the arguments underlying the classiﬁcation are not straightforward,the ﬁnal results,just as in local class ﬁeld theory,have a rather simple and elegant form,which we now describe.Let p be a prime number,n ≥1an integer,F an arbitrary ﬁeld,and K a Galois extension of F with group G = σ cyclic of order p n .Let F ×denote the multiplicative group of nonzero elements of F .Let J =J (K )=K ×/K ×p be the F p [G ]-module of p th-power classes,denoted by [γ]for γ∈K ×.Similarly,let J (F )=F ×/F ×p be the F p -module of p th-power classes of F ×,denoted by [f ]F for f ∈F ×.Let N K/F :K →F be the norm map,and write N :K ×/K ×p →F ×/F ×p for the map induced by N K/F .Also by abuse of notation we use the same symbol N to denote the endomorphism N :K ×/K ×p →K ×/K ×p induced by N :K ×/K ×p →F ×/F ×p deﬁned above,followed by the map induced by the inclusion map ǫ=ǫK :F ×→K ×.Further let K i ,i =0,...,n ,be the intermediate ﬁeld of K/F such that [K i :F ]=p i .Denote by H i the Galois group Gal(K/K i )⊂G .Let [K ×i ]denote the submodule of J which is the image of the map induced by the inclusion map K ×i →K ×: K ×i =K ×i K ×p /K ×p .Similarly,for other G -submodules A ⊂K ×,such as A =N K i /F (K ×i ),let [A ]=AK ×/K ×p .4J´AN MIN´AˇC,ANDREW SCHULTZ,AND JOHN SWALLOW Theorem1.Suppose that either•ξp∈F,or•p=2,n=1,and−1∈N K/F(K×).Then the F p[G]-module J decomposes asJ=Y n⊕Y n−1⊕···⊕Y0,where Y i is a direct sum of cyclic F p[G]-modules of dimension p i and[K×i]=J H i,0≤i≤n.It is easy to show that this decomposition of J is unique.(In fact this also follows from a well-known result of Azumaya.See[AnFu73, page144].)In the following corollary we determine the sizes of the modules Y i in terms of norms.Observe that direct sums of cyclic F p[G]-modules of dimension p i are free F p[G/H i]-modules.Lete i=dim Fp N K i/F K×i / N K i+1/F K×i+1 ,0≤i<n,[N K/F(K×)].and let e n=dim FpCorollary1.For each0≤i≤n,[N K(K×i)]=(Y i+Y i+1+···+Y n)G,i/FandY i=e i.rank Fp[G/H i]For K/F not satisfying the conditions of the theorem above,we adopt the conventions K×−∞={1}and p−∞=0and make the following deﬁnition.Deﬁnition(Exceptional Element).Suppose thatξp∈F and,if p=2, that either n>1or−1∈N K/F(K×).We seti(K/F):=min{i∈{−∞,0,1,...,n}|∃δ∈K×such that[N K/F(δ)]F=[1]F and[δ]τ−1∈[K×i]∀τ∈Gal(K/F)}.We say thatδ∈K×is an exceptional element of K/F if[N K/F(δ)]F= [1]F and[δ]τ−1∈[K×i(K/F)]for allτ∈Gal(K/F).Elements of K×that are not exceptional are said to be unexceptional.For simplicity,we often write m instead of i(K/F).GALOIS MODULE STRUCTURE OF p TH-POWER CLASSES5 Observe that[δ](τ−1)∈[K×i]for allτ∈G if and only if[δ](σ−1)∈[K×i]for aﬁxed generatorσ∈G.In what follows we will use this formulation for our given generatorσ.Note that ifδis an exceptional element then m=i(K/F)=−∞if and only if[δ]σ=[δ]and[N K/F(δ)]F=[1]F.Because the exceptionality of an elementγ∈K×is independent of the particular representativeγof[γ],we deﬁne[γ]to be exceptional ifγis exceptional.It is also useful to observe that if an F p[G]-generator[γ] of a module Mγ⊂J is exceptional,then so is any other F p[G]-generator [ω]of Mγ.Indeed,using additive notation for J for the moment,any such generator[ω]has the form[ω]=c0[γ]+c1(σ−1)[γ]+c2(σ−1)2[γ]+...,c0,c1,···∈F p,c0=0.Then[N K/F(ω)]F=[N K/F(γ)]c0F=[1]and[ω]σ−1∈[K×m].In Proposition2we show that exceptional elements always exist for K/F satisfying the hypothesis in the Deﬁnition above,and in Propo-sition7we show that,in fact,m≤n−1.Finally,note that since N K/F(K×n−1)⊂F×p,each exceptional elementδ∈K×n\K×n−1.Moreover,for these K/F,we have Kummer theory,becauseξp∈F.√a)for some a∈F.In section4we prove some Hence K1=F(pmore speciﬁc results about exceptional elements in terms of a:excep-tional elements satisfy[N K/F(δ)]F=[a]s F for s≡0mod p and that for all K/F as above,an exceptional elementδ∈K×exists satisfying [N K/F(δ)]F=[a]F.Theorem2.Suppose thatξp∈F and,if p=2,that either n>1or −1∈N K/F(K×).Letδ∈K×be any exceptional element of K/F.Then the F p[G]-module J decomposes asJ=X⊕Y,Y=Y n⊕Y n−1⊕···⊕Y0,where(1)X is the cyclic F p[G]-module generated by[δ],with dimensionp m+1;(2)Y i is a direct sum of cyclic F p[G]-modules of dimension p i;and6J´AN MIN´AˇC,ANDREW SCHULTZ,AND JOHN SWALLOW(3)for i∈{0,...,n},[K×i]= X(σ−1)⊕Y H i,m≤i;X(σ−1)(σp i−1)p m−i−1⊕Y H i,i<m.(Here X(σ−1)and X(σ−1)(σ−1)p m−i−1denote images of X under the action of(σ−1)and(σ−1)(σp i−1)p m−i−1respectively.)As before,lete i=dim Fp N K i/F K×i / N K i+1/F K×i+1 ,0≤i<n,[N K/F(K×)].and let e n=dim FpCorollary2.For each m<i≤n,(K×i)]=(Y i+Y i+1+···+Y n)G,[N Ki/Fand,if m≥0,for each0≤i≤m,[N K(K×i)]=(X+Y i+Y i+1+···+Y n)G.i/FFor i=m,rank FY i=e i,p[G/H i]while if m≥0,Y m=e m.1+rank Fp[G/H m]Finally,we present several interesting conditions equivalent to m= i(K/F)being a particular element of the subset ofﬁeld indices E= {−∞,0,...,n−1}.To express these conditions,we deﬁne−∞∔1=0 and,for e∈E with e≥0,we deﬁne e∔1=e+1.We also set (K×−∞)to be{1}.N Kn−1/FTheorem3.Suppose thatξp∈F and,if p>2,that either n>1or −1∈N K/F(K×).Theni(K/F)=min s|ξp∈N K/F(K×)N K n−1/F(K×s)=min s|ξp∈N K/K s∔1(K×)=min s|∃[δ]∈J H s∔1,[N K/K s∔1δ]K s∔1=[1]K s∔1 .GALOIS MODULE STRUCTURE OF p TH-POWER CLASSES7 One can also connect these equalities with the existence of solutions of particular Galois embedding problems.This connection will be pur-sued in a forthcoming paper.(See[MSSa].)X summands also lead naturally to the investigation of some cyclotomic cyclic algebras over F which,in turn,allow us to constructﬁelds with prescribed X sum-mands.This topic will also be pursued in a subsequent paper.(See [MSSb].)The proofs of Theorems1and2are inductive,resting on the base case n=1for Theorem1and two base cases n=1and p=2,n=2 for Theorem2.In these base cases as well as the inductive proof,we employ lemmas which establish the structure of theﬁxed submodule J G of J—in particular,whether thisﬁxed submodule is no more than the image of the p th-power classes of the baseﬁeld F—and specify which of these elements are norms.In fact,these lemmas reﬂect what has emerged,both in this work as well as in the work on determining the entire Milnor K-theory modulo p as a Galois module(see[BLMS]and[LMS]),as two essential foun-dational ingredients in the proof.Theﬁrst is Hilbert’s Theorem90, which in our situation may be viewed as a principle saying that we have enough norms.Indeed,Hilbert90tells us that the kernel of the norm map is as small as possible.In order to use Hilbert90eﬀectively,we need again and again the technical reﬁnements of this principle telling us that certain elements in a group of p th-power classes are norms.In this work these reﬁnements,for example,begin with Lemmas10,11, and12(identifying someﬁxed elements as norms),and are completed in the full proofs of Theorems1and2.The second essential ingredient is control of the image of p th-power classes of the baseﬁeld in the group of p th-power classes of ourﬁeld extension,which in this work is obtained from Lemma6(the Exact Sequence Lemma)and its technical relative Lemma5(the Fixed Sub-module Lemma).In this paper,both of these principles are elementary, but they are more sophisticated in the higher Milnor K-theory case. It is remarkable that one requires only repetitions of these two prin-ciples in order to determine fully the Galois module structure of the modules in question.Drawing out the structure from only these two ﬁrst principles,however,does not come without cost,and a number of technical observations turn out to be necessary for us toﬁt the puzzle pieces together.8J´AN MIN´AˇC,ANDREW SCHULTZ,AND JOHN SWALLOW Whenξp∈F,we need additional information to determine when an element[γ]∈J H i lies in[K×i]or is instead an exceptional element.We begin by standardizing choices of the a i in the presentations of subﬁelds K i+1=K i(√a i)in section1.2.Then,in section1.4,we collect several results used in identifying elements of[K×i].These are the Submodule-Subﬁeld Lemma(7)for free components,the Norm Lemma(8)for comparisons among norms from K to various K i(in order to determine when an exceptional element for K/F is an exceptional element for K/K i),and the Proper Subﬁeld Lemma(9)for elements that generate suﬃciently small cyclic submodules.In section1.1,we present lemmas which we use to manipulate F p[G]-representations formally:the Inclusion Lemma(1),the Exclusion Lem-ma(2),and the Free Complement Lemma(3).We begin the proof by proving the base cases for an induction in section2.Our inductive strategy isﬁrst to show that J contains a suf-ﬁciently large direct sum of F p[G]-submodules of p th-power dimensions. We do so in section3in Proposition6,the result of which is already enough to prove Theorem1.Whenξp∈F and,if p=2,n>1,we also need to establish the dimension of the X component and connect notions of exceptional elements for subextensions K/K i.We do so in section4.In section5,we prove an analogue of Proposition6which establishes Theorem2without the independence of X and Y,and then we prove Theorem2fully.Finally,in section7,we prove Theorem3. For the reader’s convenience,we have made our paper self-contained; in particular,it is independent from[MS03].1.Notation and Lemmas1.1.F p[G]-modules.Let G be a cyclic group of order p n with generatorσ.For an F p[G]-module U,let U G denote the submodule of Uﬁxed by G,and for an arbitrary element u∈U,let l(u)denote the dimension of the F p[G]-submodule of U generated by u.Denote by N the operator(σ−1)p n−1 acting on U.For an F p[G]-module V and an elementγ∈V,let γ denote the F p-subspace of V spanned byγ,and let Mγdenote the cyclic F p[G]-module generated byγ.If[γ]is an element of K×/K×p represented byγ∈K×,we write Mγinstead of M[γ].GALOIS MODULE STRUCTURE OF p TH-POWER CLASSES9 We will usually use additive notation for general F p[G]-modules, switching to multiplicative notation when considering the speciﬁc mod-ule J=K×/K×p.However,occasionally even in this case we employ additive notation,in particular writing{0}to denote{[1]}.Lemma1(Inclusion Lemma).Let U and V be F p[G]-modules con-tained in an F p[G]-module W.Suppose that(U+V)G⊂U and for each w∈(U+V)\(U+V)G there exists u∈U such that(σ−1)l(w)−1(w)=N(u).Then V⊂U.Proof.Let{T i}s i=1be the socle series of U+V:T1=(U+V)G and T i+1/T i=((U+V)/T i)G,and let s be the least natural number such that T s=U+V.Observe that since(σ−1)p n=0,we have s≤p n. We prove the lemma by induction on the socle series.By hypothesis,T1⊂U.Assume now that T i⊂U for some i<s. Then for each w∈T i+1\T i we have l(w)=i+1and(σ−1)l(w)−1(w)= N(u)=(σ−1)p n−1(u)for some u∈U.Therefore(σ−1)l(w)−1 w−(σ−1)p n−l(w)(u) =0. Therefore w−(σ−1)p n−l(w)(u)∈T i⊂U.Hence w∈U and T i+1⊂U. Therefore U+V=U and V⊂U as required. Lemma2(Exclusion Lemma).Let U and V be F p[G]-modules con-tained in an F p[G]-module W.Suppose that U G∩V G={0}.Then U+V=U⊕V.Proof.Let Z=U∩V and suppose that y∈Z\{0}.Letz=(σ−1)l(y)−1(y)=0.Then z∈U G∩V G,a contradiction.Hence U∩V={0}and U+V= U⊕V.The following lemma follows from the fact that each free F p[G]-module is injective.(See[Ca96,Theorem11.2].)We shall,however, provide a direct proof.Lemma3(Free Complement Lemma).Let V⊂U be free F p[G]-modules.Then there exists a free F p[G]-submodule˜V of U such that V⊕˜V=U.10J´AN MIN´AˇC,ANDREW SCHULTZ,AND JOHN SWALLOWProof.Let Z be a complement of V G in U G as F p-vector spaces,andlet Z be an F p-base of Z.For each z∈Z,there exists u(z)suchthat z=N(u(z)).Let M(z)be the F p[G]-submodule of U generatedby u(z).Then M(z)is a free F p[G]-submodule.Moreover,itsﬁxed submodule M(z)G is the F p-vector subspace generated by z.We claim that the M(z),z∈Z,are independent.First we showby induction on the number of modules that aﬁnite set of mod-ules M(z)is independent.The base case is trivial.Now let W=M(z)∩ z′=z M(z′).Now by the inductive assumption on indepen-dence,( z′=z M(z′))G= z′=z M(z′)G,and for each z,M(z)G= z . Since the z form an F p-base for Z,we obtain W G={0}.The ExclusionLemma(2)then gives that M(z)+ z′=z M(z′)=M(z)⊕ z′=z M(z′). The case of an inﬁnite sum follows from the same argument,since the fact that m∈M(z)G∩ z′=z M(z′)G forces m to be aﬁnite sum of elements m(z′).Hence the M(z),z∈Z,are independent.Set˜V:=⊕z∈Z M(z).Then˜V is a free F p[G]-submodule of U and ˜V G=Z.By the Exclusion Lemma(2),we have that V+˜V=V⊕˜V and(V⊕˜V)G=V G⊕˜V G=U G.Now let u∈U be arbitrary and let M be the cyclic F p[G]-submodule of U generated by u.Then(M+V+˜V)G⊂U G⊂V+˜V.Moreover, for any m∈(M+V+˜V)\(M+V+˜V)G,(σ−1)l(m)−1(m)∈(M+V+˜V)G⊂U G=(V+˜V)G=N(V+˜V) by the freeness of V and˜V.By the Inclusion Lemma(1),then,M⊂V+˜V.Hence U=V⊕˜V. Remark.At several points later,we use the same argument as that contained in the proof above to show that a possibly inﬁnite set of modules is independent,and we use the Exclusion Lemma(2)as an abbreviation for this argument.1.2.Kummer Subﬁelds of K/F and Exceptional Elements. Suppose thatξp∈F.In this case we have Kummer theory and may organize presentations of the extensions K i+1/K i as follows. Proposition1(Subﬁeld Generators).We may choose a i∈K×i,0≤i<n such thatGALOIS MODULE STRUCTURE OF p TH-POWER CLASSES11•K i+1=K i(p√a i)and•N K i/K j a i=a j for all0≤j<i<n.In what follows we will assume that the choices of a i have been made according to Proposition1,and we set a=a0.We prove this result by means of the followingLemma4.Suppose thatξp∈K and let L′/K be a cyclic extension ofdegree p2with L/K the intermediate extension of degree p.Then,for√,we have L=K(p N L/K(b)).every b∈L with L′=L(pProof.Letσbe a generator of Gal(L′/K).For each i∈{1,2,...,p−1}, we have p√b σi=p√bσifor a suitable choice of a p th root of bσi.Hencep√b 1+σ+···+σp−1=p√b1+σ+···+σp−1=p N L/K(b)∈L′for a suitable choice of a p th root of N L/K(b).Observe that sinceξp∈K the equalityp√b (1+σ+···+σp−1)(σ−1)=p√bσp−1=p N L/K(b)σ−1√) is independent of the choice of p th roots.Moreover,since L′=L(p√bσp−1=1.Hence we con-andσp generates Gal(L′/L),we see that pclude that L=K(p N L/K(b)).Proof of Proposition1.By Kummer theory,there exists a n−1∈K×n−1 such that K n=K n−1(p√a n−1).Then inductively deﬁne(a n−i+1)a n−i=N Kn−i+1/K n−ifor i∈{2,...,n}.Applying the lemma to extensions K n−i+2/K n−i,we have the results.Our deﬁnition of exceptional elements makes use of a subset of the set{δ∈K×|[N K/F(δ)]F=[1]F}.In general,however,this latter set may be empty.Consider,for example,the extension C/R,for which N C/R(C×)⊂R×2.The next proposition shows that under the conditions we require in the deﬁnition of exceptional elements,this set is never empty and therefore exceptional elements exist.12J ´ANMIN ´A ˇC,ANDREW SCHULTZ,AND JOHN SWALLOW Proposition 2.Let ξp ∈F and,if p =2,that n >1or −1∈N K/F (K ×).Then an exceptional element δexists.Proof.Consider δ=p √a n −1.If p >2then N K/K n −1(δ)=a n −1andhence N K/F (δ)=a 0=a .Now if p =2then N K/K n −1(δ)=−a n −1and for n >1we similarly have N K/F (δ)=a 0=a .If p =2and n =1,then −a =N K/F (√a )and hence −1∈N K/F (K ×)if and onlyif a ∈N K/F (K ×).Consequently,under our hypothesis,exceptional elements always exist.1.3.The Fixed Submodule J G of J .Recall that we write [F ×]for F ×K ×p /K ×p ⊂J .The following lemmas generalize [MS03,Lemma 2and Remark 2]:Lemma 5(Fixed Submodule Lemma).(1)If ξp ∈N K/F (K ×),J G =[F ×].(2)If ξp ∈N K/F (K ×),J G = [δ] ⊕[F ×],where δ∈K ×with δσ−1=λp ,N K/F (λ)is a primitive p th root of unity,and [N K/F (δ)]F =[a ]F .In particular,δis an exceptional element of K/F .Proof.Suppose that θ∈K ×such that [θ]∈J G .Then θσ−1=λp for some λ∈K ×,and hence N K/F (λ)p =1.Therefore N K/F (λ)is a p th root of unity.Now consider the ﬁrst case,ξp ∈N K/F (K ×).Then N K/F (λ)=1,because otherwise ξp would be the norm of a suitable power of λ.From Hilbert 90we see that θσ−1=(k p )σ−1for some k ∈K ×.We conclude that θ/k p ∈F ×and hence [θ]=[f ]for some f ∈F ×.Therefore if ξp ∈N K/F (K ×)then J G =[F ×]as required.Now assume that ξp ∈N K/F (K ×).Then ξp =N K/F (λ)for some λ∈K ×and by Hilbert 90there exists an element δ∈K ×such that δσ−1=λp .Then the F p [G ]-submodule of J generated by [δ]and ǫ(F ×)is isomorphic to [F ×]⊕ [δ] .GALOIS MODULE STRUCTURE OF p TH-POWER CLASSES13√)is a cyclic extension of F of degree By[Al35,Theorem3],K(pp n+1.Then repeated application of Lemma4gives thatK n−i=K n−i−1 p N K n/K n−i−1(δ)for i∈{0,1,...,n−1}.Hence K1=F(p N K/F(δ)).By Kummer theory, [N K/F(δ)]F = [a]F as subgroups of F×/F×p.By replacing δwith another power if necessary,then,[N K/F(δ)]F=[a]F andδσ−1=λp,where N K/F(λ)is a primitive p th root of unity.We have that [δ](σ−1)=[1]and so by deﬁnitionδis exceptional for K/F.Now for each[θ]∈J G,θσ−1=νp with N K/F(ν)=N K/F(λ)c for some c∈Z.Then we have(θδ−c)σ−1=νpλ−pc.Because N(νλ−c)=1,from Hilbert90we see that there existsω∈K×such thatωσ−1=νλ−c. Hence(θδ−c)σ−1=(ωp)σ−1and we see that[θ]∈[F×]+[δ]c.Hence J G∼=[F×]⊕ [δ] ,as required.Lemma6(Exact Sequence Lemma).There is an exact sequence1→A→F×/F×pǫ−→J G N−→Awhere A=(F×∩K×p)/F×p,ǫis the natural homomorphism induced by the inclusion F×→K×,and N is the homomorphism induced by the norm map N K/F:K×→F×.•Ifξp∈F,then A=1.•Ifξp∈F,A= [a] ,in which case the map N is surjective ifand only ifξp∈N K/F(K×).Proof.Ifξp∈F,then Kummer theory implies that theﬁrst occurrence of A in the exact sequence above is equal to A= [a]F .Otherwise, suppose thatξp/∈F.If char(F)=p then no primitive p th root of unity lies in the algebraic closure of F,whenceξp/∈K.If char(F)=p,then since2≤[F(ξp):F]≤p−1and[K:F]=p n,we similarly obtain ξp/∈K.In any case,then,ξp/∈K.Assume that k p=f∈F×.Then (k p)σ−1=(kσ−1)p=1,whence kσ−1is a p th root of unity,which must be1.Hence kσ−1=1,and we deduce k∈F and f∈F×p.Therefore A=1.The Fixed Submodule Lemma(5)then gives exactness at J G and that N is surjective if and only if eitherξp∈F orξp∈N K/F(K×). Exactness at F×/F×p follows from Kummer theory.14J ´ANMIN ´A ˇC,ANDREW SCHULTZ,AND JOHN SWALLOW 1.4.F p [G ]-Submodules of J .Lemma 7(Submodule-Subﬁeld Lemma).Let U be a free F p [G ]-sub-module of J and i ∈{0,1,...,n }.ThenU H i =U (σ−1)p n −p i =U ∩[N K n /K i K ×n ]=U ∩[K ×i ].Proof.Suppose [u ]∈U H i .Then [u ](σp i −1)=[u ](σ−1)p i =[1],so l (u )≤p i .Since U is free,[u ]=[˜u ](σ−1)p n −l (u )for some [˜u ]∈U .In particular,[u ]=([˜u ](σ−1)p i −l (u ))(σ−1)p n −p i .Hence U H i ⊂U (σ−1)p n −p i .Now suppose [u ]=[˜u ](σ−1)p n −p i .Then since [˜u ](σ−1)p n −p i =[N K n /K i (˜u )],U (σ−1)p n −p i ⊂U ∩[N K n /K i K ×n ]⊂U ∩[K ×i ].Finally suppose that [u ]∈U ∩[K ×i ].Then [u ]∈U H i and we see that all of our inclusions above are actually equalities. Remark.If U is a free F p [G ]-module,then U is also a free F p [H i ]-module.But then H 2(H i ,U )={0}.Hence U H i =N i (U ):=the imageof the norm operator N i .Thus U H i =U (σ−1)p n −p i as required.Just as with F =K 0,denote elements of the F p [G/H i ]-module J (K i )=K ×i /K ×p i by [γ]K i ,γ∈K ×i .Lemma 8(Norm Lemma).For all elements [γ]∈J with l (γ)<p n ,[N K/F (γ)]F ∈ [a ]F .Now suppose additionally that l (γ)≤p n −p i for some 0≤i <n .Then [N K/K i (γ)]K i ∈ [a i ]K i ,and [N K/F (γ)]F =[a ]s F if and only if [N K/K i (γ)]K i =[a i ]s K i .Proof.For the ﬁrst statement,observe that (1+σ+···+σp n−1)≡(σ−1)p n −1on J ,and hence [N K/F (γ)]=[γ](σ−1)p n −1.Since l (γ)<p n ,[N K/F (γ)]=[1].Therefore N K/F (γ)∈F ×∩K ×p ,which by Kummertheory is the union ∪p −1j =0a j F×p .We obtain [N K/F (γ)]F ∈ [a ]F .For the second statement,observe ﬁrst that if [γ]=[1]then the lemma is trivial.Otherwise,consider J as an F p [H i ]-module and letτ=σp i .The F p [H i ]-module generated by [γ]has dimension equal toGALOIS MODULE STRUCTURE OF p TH-POWER CLASSES15t ,where [γ](τ−1)t =[1]and [γ](τ−1)t −1=[1].Since τ−1≡(σ−1)pi on J ,this condition is equivalent to (t −1)p i <l (γ)≤tp i .Since l (γ)≤(p n −i −1)p i ,the dimension t is strictly less than p n −i .Hence (τ−1)p n −i −1annihilates the cyclic F p [H i ]-module generated by γ,and so its length,as an F p [H i ]-module,is less than p n −i .Applying the ﬁrst statement in the case of the cyclic extension K/K i ,we have [N K/K i (γ)]K i ∈ [a i ]K i .Now because N K i /F (a i )=a and[N K/F (γ)]F =N K i /F ([N K/K i (γ)]K i ),we have [N K/K i (γ)]K i =[a i ]s K i if and only if [N K/F (γ)]F =[a ]s F .Remark.Occasionally,we will cite the Norm Lemma (8)as an abbre-viation of the simple argument,at the end of the lemma’s proof,which shows that[N K/F (γ)]F =[a ]s F if and only if [N K/K i (γ)]K i =[a i ]s K i .Lemma 9(Proper Subﬁeld Lemma).Let [z ]∈J H i ,i <n .Then[z ]∈[K ×i ]if and only if [N K/F (z )]F =[1]F .Proof.If [z ]∈J H i ,then [z ](σp i −1)=[1].Since (σp i −1)≡(σ−1)p i on J ,l (z )≤p i .Consider J as an F p [H i ]-module.Then from the Fixed Submodule Lemma (5)applied to the ﬁeld extension K/K i ,we see that[z ]∈[K ×i ]or [z ]∈ [δ] ⊕[K ×i ]according to whetherξp /∈N K/K i (K ×)or ξp ∈N K/K i (K ×).(Here δ∈K ×with δσ−1=λp ,N K/K i (λ)is a primitive p th root of unity,and [N K/K i (δ)]K i =[a i ]K i .)Therefore if [z ]/∈[K ×i ]then [N K/K i (z )]K i =[a i ]c K i for c ≡0mod p ,and by the Norm Lemma (8),[N K/F (z )]F =[a ]c F ,which contradicts our hypothesis.Hence if [z ]∈J H i and [N K/F (z )]F =[1]F ,then [z ]∈[K ×i ].Conversely,if [z ]∈[K ×i ]then [z ]∈J H i and[N K/F (z )]F =[N K i /F (z )]pn −i F=[1]F ,since n >i .。

四五快读1-4册精心排版A4打印

第一课人口大中小哭笑一上下爸妈天太月大人小人大哭大笑大口小口第二课地阳亮星云火水三爸爸妈妈上天天上太大太小一天一月二天二月上上下下第三课土山石木我好有田天地大地太阳月亮星星天亮大火大水火星水星三天三月下地地上地下第四课牛羊聪耳目心和四土地大山小山土山石山火山土星木星好人田地水田我有我爸我妈我哭我笑好山好水我有好爸爸、好妈妈。

天上有太阳、月亮、星星。

地上有土、石、山、水田。

爸爸上山，妈妈下山。

第五课明头眉鼻手花树五大牛小牛水牛山羊小羊小心中心心中四月四天地上有小草和大树。

草地上有水牛和小山羊。

爸爸和妈妈心中有我。

我心中有爸爸和妈妈。

第六课草叶日风雨的孩六聪明明亮明天明月眉头鼻头石头木头心头小手小花大树小树树木五月五天手心我有小手。

头上有眉、目、耳、鼻和口。

眉下有目，鼻下有口。

树上有花和叶。

第七课白红是家多唱子七花草小草草地树叶一日大风大雨小雨下雨风雨雨水我的六月六日六天天上有明亮的太阳和月亮。

我有爸爸和妈妈。

明日有大风，下大雨。

石头上有小草和小花。

第八课爱爷奶少歌不朋入白云白天明白红花红日火红火花是的我是大家我家人家鼻子叶子七月七日七天红太阳我是爸爸、妈妈的好孩子。

天是太阳、月亮、星星的家。

白天，天上有红太阳；地上有红花、白花和小草。

爷爷、奶奶家有大水牛和小山羊。

第九课宝在学书游友儿九爷爷奶奶多少唱歌爱笑爱哭爱唱不爱不哭不笑不唱不是是不是好不好不好八日八月八天明天多云，有小雨。

聪聪爱唱歌。

我家有爷爷、奶奶、爸爸、妈妈和我。

爷爷爱我，奶奶爱我，爸爸爱我，妈妈爱我。

我爱爷爷，我爱奶奶，我爱爸爸，我爱妈妈。

第十课贝生习看戏字气十宝宝在家不在小人书游水朋友友人花儿儿子儿歌歌儿九日九月天我是宝宝，我在家爱唱儿歌。

我不爱哭，我爱笑。

我的好朋友是聪聪。

小头爸爸和大头儿子玩游戏。

第十一课见早雪鸡绿黄青鱼做飞跑要吃鸟他看见会见学会早上大雪小雪雪花白雪雪人雪山雪地下雪天小鸡大鸡绿草绿叶黄花黄牛青草我学会好多生字。

树上有爱唱歌的小黄鸟。

语感启蒙清华幼儿英语2004

语感启蒙清华幼儿英语2004《语感启蒙清华幼儿英语2004》是一本为幼儿提供英语学习启蒙的教材。

这本教材以培养孩子们对英语的语感为主要目标，通过丰富多样的活动和练习，帮助幼儿在轻松愉快的环境中建立起对英语的兴趣和自信。

这本教材强调的不仅仅是单词和句子的学习，更重要的是通过情境和故事情节来激发孩子们对英语的兴趣。

教材中设计了丰富的角色和故事，通过这些有趣的角色和情节，帮助幼儿在玩耍和娱乐中学习英语。

这种趣味性的设计能够让孩子们在学习中快乐地体验到英语的魅力，进而激发他们的学习热情。

此外，《语感启蒙清华幼儿英语2004》强调的是学习英语的综合能力的培养。

除了注重口语的训练外，教材还通过游戏和歌曲来锻炼幼儿的听力、阅读和写作能力。

通过这些多样化的学习方式，幼儿的英语学习能力得到了全面的提升。

这本教材还注重培养幼儿的交际能力。

在教学中，教材设计了各种角色扮演活动和小组合作训练，让幼儿在实际运用中学习如何与他人进行交流。

这样的设计不仅帮助幼儿提高了口语表达能力，也培养了他们的合作意识和团队精神。

总之，《语感启蒙清华幼儿英语2004》是一本以培养孩子们对英语兴趣和语感为目标的教材。

通过丰富多样的活动和练习，教材帮助幼儿在愉快的学习环境中提高他们的英语学习能力。

通过学习角色扮演、游戏和歌曲，幼儿不仅能够提高英语口语表达能力，还能够培养他们的综合能力和交际能力。

这本教材无疑是幼儿英语学习的良好启蒙之书。

《汉语语言文字启蒙1》教案一教程文件

《汉语语言文字启蒙I》教案教材简介：《汉语语言文字启蒙》（华语教学出版社出版；1997年第一版；2011年第七次印刷）作者是法国著名汉学家白乐桑先生，现为巴黎第七大学副教授以及法国汉语教师协会主席。

先后出版过《汉字的表意王国》《说字解词词典》等专著十余部。

本书中国作者张朋朋先生现就职于北京语言文化大学，具有多年对外汉语教学经验。

曾出版过《口语速成》（华语教学出版社2001年）《部首三字经》（北京语言大学出版社2002年）等著作。

本书属于汉语国际教育教材。

适用于在国外学习汉语并且注重汉字学习的零起点欧美国家的外国人。

分为上下两册。

上册共19课，由四部分组成。

第一部分为预备课。

之后每部分由六篇课文组成。

上册涉及400个最常用的汉字，约1500个词汇以及54个语法点。

下册大约涉及500个常用汉字，下册课文设计内容丰富，题材广泛。

本书始终贯彻“字本位”原则，将“字”作为汉语教学基本单位，“以字构词”，弥补了汉语国际教学中“汉字”和“口语”严重脱节的不足。

同时结合文化传播原则、循序渐进原则，紧密结合汉语国际教学实际。

第一课您贵姓？一．教学目标1.掌握代词“你”、“他、她”的用法。

并能够区分“你”和“您”交际运用的不同。

2.熟练并能够得体地询问别人姓名：“您贵姓？”“免贵x。

”“你叫什么名字？”“我叫XXX。

”3.熟悉另一种正反疑问句：“V+不+V”（叫不叫；去不去）“adj.+_不+adj”(好不好；大不大）4.能够熟悉本课生词，正确听辨并且书写。

5.对中国的姓氏文化有一定的了解。

二．教学重点和难点（一）语音：儿化音;“不”的变调；（二）词汇1.代词：您；你；他/她2.名词：名字；立阳；月文3.动词：姓；叫；去4.表示询问：什么；贵姓；哪儿（难点）（三）汉字1.独体字：立；月；文；么；去2.合体字：贵；阳；叫；字（难点）（四）语法：“V+不+V”（叫不叫；去不去）“adj.+_不+adj”(好不好；大不大）三．教学方法（1）变调练唱法。

史上全面英语启蒙资料

亲爱哒，感谢你的关注！我们辛苦收集整理了超多超好的学习资源在这里，欢迎下载。

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2004年高考语文试题及答案(北京春卷)

2004年普通高等学校春季招生考试语文（北京卷）一、（18分，每小题3分）1．下列各组词语中，加点的字读音全都不同的一组是（）A．边陲．锤．炼捶．打唾．手可得B．彼．此山坡．波．浪疲．惫不堪C．颠．沛嗔．责缜．密瞋．目而视D．屹．立起讫．呓．语自古迄．今2．下列各组词语中，没有错别字的一组是（）A．题词题纲题写文不对题B．连续连截连手藕断丝连C．挤兑兑奖汇兑兑换现金D．决断决胜决计决无此意3．依次填入下面这段话横线处的词语，最恰当的一组是（）阅读优秀作品，其语言，感受其思想、艺术魅力，不仅可以大自然和人生的多姿多彩，还可以激发自然、热爱生活的感情。

A．品味体味珍爱B．品评体验珍惜C．品评体验珍爱D．品味体味珍惜4．下列句子中，加点的成语使用不恰当的一句是（）A．17年卧薪尝胆．．．．，2003年中国女排终以11战全胜的成绩夺回世界杯赛冠军的称号。

B．也许是大家都知道巴金老人对玫瑰情有独钟．．．．，一束束象征热情与朝气的红玫瑰将冬日里巴老的病房装点得春意盎然。

C．1998年初，国际足联秘书长布拉特宣布参加国际足联主席的竞选，欧洲足联主席约翰松也积极参与竞选，一时间国际足联主席一职炙手可热．．．．。

D．他多次在千钧一发之际逃过仇敌追杀，但百密一疏．．．．，一年前不慎泄露行踪，最终未能幸免于难。

5．下列句子中，没有语病的一句是（）A．鸦片战争以来的中国近代史，对于大多数中学生是比较熟悉的，重大的历史事件都能说得一清二楚。

B．为保持北京的古都风貌，在北京市旧城区改造中，新的建筑应以故宫和皇城为中心向外分七个层次逐步提高。

C．初涉文坛，她的第一部处女作就是这样一部意味深长的巨著，不能不令人刮目相看。

D．“神舟”五号载人飞船把我国首位航天员成功送入浩瀚的太空并安全返回，这是我国航天事业取得的又一举世瞩目的成就。

6．把后面的句子依次填到横线上，衔接恰当的一项是（）上海交响乐迷中近六成的人收入并非十分丰厚，。

在防洪抢险的战斗中，，终于保住了大坝，战胜了洪水。

巧虎全套内容打印A4

巧虎全套内容打印A4巧虎贝贝版乐智小天地贝贝版01- 我爱爸爸妈妈 01 小小世界我的家人 02 和巧虎一起玩躲猫猫 03 动物乐园小狗汪汪 04 生活习惯我会应答 05 小手哆来咪亲亲我 06 故事小屋动物摸摸书 07 情绪表达喜欢、讨厌 08 唱游巧虎之歌09 亲子玩具巧虎小木车 10 重点词汇 11 巧虎日记 12 预告乐智小天地贝贝版02- 出门手拉手 01 和巧虎一起玩巧虎的一天 02 生活习惯出门手拉手 03 小手哆来咪小花笑哈哈 04 动物乐园小猫喵喵 05 唱游小花猫06 小小世界我的身体 07 故事小屋哈哈洞洞书 08 亲子玩具爸爸妈妈手偶 09 情绪表达开心伤心 10 重点词江 11 巧虎日记 12 预告乐智小天地贝贝版03- 小手洗一洗 01 生活习惯小手洗一洗 02 唱游两只小象 03 动物乐园大象04 小小世界家里有什么 05 亲子玩具缤纷玩水组 06 小手哆睐咪拳头歌 07 和巧虎一起玩捉迷藏 08 情绪表达期待失望如愿 09 故事小屋生活游戏书 10 重点词汇11 巧虎日记 12 预告13 商品使用说明巧虎宝宝版乐智小天地宝宝版01- 我会打招呼 01 小小主角我的新朋友--巧虎 02 唱游巧虎之歌03 动物世界可爱的小狗 04 宝宝健康操（一）伦敦桥 05 生活习惯我会打招呼 06 我和巧虎你好，再见07 宝宝健康操（二）体操大集合 08 亲子玩具指尖益智玩具 09 小手哆啦咪小手叠叠叠 10 玩具预告乐智小天地宝宝版02- 安全亲子乐 01 小小主角巧虎的一天 02 生活习惯出门手拉手 03 动物世界小猫咪04 宝宝健康操（一）爬小山 05 唱游小花猫之歌 06 我和巧虎我会手拉手 07 故事小屋小狗汪汪找妈妈 08 亲子玩具动物立体电子书 09 小手哆啦咪握紧拳头 10 宝宝健康操（二）爬爬爬 11 玩具预告缤纷玩水组乐智小天地宝宝版03- 吃饭真开心 01 小小主角和巧虎一起捉迷藏 02 唱游兔子跳跳跳 03 动物世界小兔子04 宝宝健康操（一）小车子 05 生活习惯吃饭真开心 06 我和巧虎宝宝吃饭喽 07 小手哆啦咪炒萝卜08 故事小屋小勺子和它的朋友们 09 亲子玩具缤纷玩水组10 宝宝健康操（二）看信号，做动作11 玩具预告生活认知布书乐智小天地宝宝版04- 我要上厕所 01 主题童话谁在上厕所 02 生活习惯我要上厕所 03 唱游我喜欢大象04 宝宝健康操（一）小车子 05 动物世界大象106 小手哆啦咪伊比呀呀 07 生活认知香蕉08 亲子玩具生活认知布书 09 故事小屋谁来了10 宝宝健康操（二）平衡游戏 11 玩具预告认知系列迷你书乐智小天地宝宝版05- 宝贝睡觉喽 01 生活习惯我会说“早”、“晚安” 02 唱游狮子LION 03 动物世界狮子04 宝宝健康操（一）节奏小球 05 故事小屋森林里的饼屋 06 小手哆啦咪饼干歌07 宝宝健康操（二）球球模仿秀 08 生活认知草莓09 亲子玩具认知系列迷你书 10 主题童话大家都睡觉了 11 预告乐智小天地宝宝版06- 健康小勇士 01 宝宝健康操（一）健康好宝宝 02 生活习惯巧虎生病了 03 主题童话忙碌的河马医生 04 动物世界绵羊 05 唱游小羊咩咩咩 06 生活认知包子07 小手哆啦咪拍拍小手 08 故事小屋小小音乐家09 宝宝健康操（二）企鹅走走走 10 预告乐智小天地宝宝版07- 我也来帮忙 01 唱游圣诞老公公02 故事小屋圣诞老公公来了 03 生活习惯我也来帮忙 04 主题童话谁来擦一擦05 动物世界长颈鹿06 宝宝健康操（一）叮叮当 07 小手哆啦咪炒饭喽 08 生活认知超市09 亲子玩具超市购物组10 宝宝健康操（二）小兔蹦蹦跳 11 预告乐智小天地宝宝版08- 牙齿亮晶晶01 主题童话牙齿亮晶晶 02 生活习惯刷牙刷刷刷 03 动物世界河马 04 唱游河马 HIPPO 05 故事小屋大手套 06 生活认知蔬菜07 宝宝健康操（一）好吃的蔬菜色拉 08 语言表达我的身体 09 小手哆啦咪碰碰碰 10 亲子玩具大嘴刷牙狗11 宝宝健康操（二）猩猩抬身体 12 预告乐智小天地宝宝版09- 整整齐齐过新年01 唱游新年快乐02 生活习惯一起收玩具 03 主题童话我的巧虎呢 04 语言表达恭喜恭喜 05 生活认知过年喽06 故事小屋“年”的故事 07 动物世界可爱的猪08 宝宝健康操（一）小猪打滚 09 小手哆咪咪做点心10 亲子玩具城市环游列车组 11 宝宝健康操（二）猴子扔扔扔 12 预告乐智小天地宝宝版10- 我们来分享 01 宝宝健康操（一）一起去郊游 02 主题童话你一半我一半 03 生活习惯给你吃谢谢 04 动物世界大熊猫 05 唱游提灯笼 06 语言表达接电话07 小手哆睐咪电话铃响了 08 亲子玩具小厨师切切组 09 故事小屋在哪里 10 生活认知橘子11 宝宝健康操（二）鸡妈妈护蛋 12 预告乐智小天地宝宝版11- 开心小伙伴 01 唱游小雨滴 02 生活认知公园03 生活习惯小猫对不起 04 主题童话拔萝卜205 宝宝健康操（一）小皮球 06 语言表达去哪里 07 HAPPY ENGLISH BINGO 08 动物世界乌龟09 宝宝健康操（二）小乌龟爬地 10 亲子玩具扮演游戏组 11 预告乐智小天地宝宝版12- 一起来洗手 01 生活习惯我会自己洗手 02 主题童话手脏了怎么办 03 唱游蜗牛歌04 动物世界青蛙和蜗牛 05 语言表达吃什么呢06 生活认知巧虎生日快乐 07 宝宝健康操（一）大鼓小鼓 08 小手哆来咪幸福拍手歌09 亲子玩具音乐节奏电子鼓（一） 10 故事小屋不爱洗澡的小羊11 HAPPY ENGLISH ROW ROW ROW YOUR BOAT 12 亲子玩具音乐节奏电子鼓（二） 13 宝宝健康操（二）扭扭小金鱼 14 预告乐智小天地宝宝版13- 和形状一起做游戏 01 语言表达圆形三角形方形 02 主题童话饼干宝宝在哪里 03 唱游企鹅舞 04 动物世界企鹅05 生活习惯关心小弟弟小妹妹 06 宝宝健康操（一）老鹰抓小鸡 07 生活认知推土机挖土机卡车 08 故事小屋这是什么车呢09 HAPPY ENGLISH THE FINGER FAMILY 10 宝宝健康操（二）模仿游戏 11 亲子玩具形状颜色积木 12 预告乐智小天地宝宝版14- 颜色真奇妙 01 语言表达红色蓝色黄色 02 唱游红黄蓝舞曲03 生活习惯一个一个轮流玩 04 动物世界动物模仿 05 生活认知西瓜06 主题童话黑色的水果07 宝宝健康操（一）彩色气球08 HAPPY ENGLISH MARY HAO A LITTLE LAMB 09 亲子玩具颜色认知海绵拼图 10 宝宝健康操（二）海豚救生圈 11 预告乐智小天地宝宝版15- 我会自己穿鞋 01 生活习惯我会自己穿鞋 02 主题童话小老鼠跌倒了 03 唱游顽皮熊04 动物世界动物亲子乐园 05 语言表达看看大家在干什么 06 故事小屋大家都来穿鞋子 07 宝宝健康操（一）娃娃兵 08 生活认知葡萄09 HAPPY ENGLISH TEN LITTLE KITTIES 10 亲子玩具穿线游戏书11 宝宝健康操（二）小羊钻山洞 12 预告乐智小天地宝宝版16- 去上厕所喽 01 唱游碰碰身体02 生活习惯去上厕所喽 03 成长小故事要便便啦 04 动物乐园狮子05 小小世界西瓜、香蕉、苹果 06 小手哆来咪西瓜 07 故事小屋水果拼拼乐 08 亲子玩具生活认知布书 09 重点词汇 10 巧虎日记乐智小天地宝宝版17- 见面问个好 01 生活习惯见面问个好 02 成长小故事你好你好 03 小手哆来咪炒萝卜 04 小小世界好吃的食物 05 动物乐园鸡的一家 06 唱游大树下07 亲子玩具动物发声电子书 08 故事小屋宝宝贴贴书 09 重点词汇 10 巧虎日记乐智小天地宝宝版18- 我会乖乖睡觉 01 唱游小羊咩咩咩 02 动物乐园小绵羊303 生活习惯做好准备睡觉喽 04 小手哆来咪伊比呀呀 05 故事小屋森林里的饼屋 06 亲子玩具好宝宝习惯组 07 成长小故事宝贝晚安 08 小小世界白天晚上09 重点词汇 10 巧虎日记乐智小天地宝宝版19- 自己吃饭真开心 01 唱游叮叮当02 生活习惯自己吃饭真开心 03 动物乐园长颈鹿 04 小手哆来咪饼干歌 05 成长小故事小勺子 06 小小世界逛超市喽 07 说说看谢谢谢谢08 故事小屋圣诞老公公来了 09 重点词汇 10 巧虎日记乐智小天地宝宝版20- 我喜欢刷牙 01 唱游河马HIPPO 02 动物乐园河马03 生活习惯妈妈帮我刷刷牙 04 成长小故事牙齿亮晶晶 05 小小世界车06 小手哆��咪大家来刷牙 07 故事小屋大手套 09 亲子玩具大嘴刷牙狗 10 说说看拜拜 09 重点词汇 10 巧虎日记乐智小天地宝宝版21- 交个朋友吧 01 唱游新年快乐02 生活习惯交个朋友吧 03 说说看恭喜恭喜 04 动物乐园大熊猫05 成长小故事有个朋友真好 06 小小世界冬天来了 07 小手哆��咪幸福拍手歌 08 故事小屋小火车旅行记 09 亲子玩具欢乐火车组 10 重点词汇 11 巧虎日记乐智小天地宝宝版22- 你一半我一半 01 唱游宝宝健康操 02 生活习惯大家分着吃 03 动物乐园金鱼04 小小世界（一）梨葡萄猕猴桃 05 说说看请进06 成长小故事你一半我一半 07 小小世界（二）想一想是哪个 08 小手哆��咪线团咕噜咕噜 09 故事小屋谁来了 10 亲子玩具欢乐火车组 10 重点词汇11 巧虎日记乐智小天地宝宝版23- 宝宝身体棒 01 唱游花开了02 生活习惯生病了我不怕 03 成长小故事忙碌的河马医生 04 小小世界春天05 动物乐园猴子06 故事小屋宝贝儿别怕 07 小手哆��咪电话铃铃铃 08 亲子玩具扮演游戏组 10 说说看接电话 09 重点词汇 10 巧虎日记乐智小天地宝宝版24- 颜色小精灵 01 唱游红黄蓝跳舞曲02 生活习惯 (一) 一起来收拾 03 开动脑筋红色蓝色黄色 04 动物乐园乌龟05小小世界巧虎生日快乐 06 故事小屋气球变变变07生活习惯 (二) 和巧虎一起收拾 08说说看好妈妈09成长小故事小老鼠跌倒了 10小手哆来咪小小乌龟 11亲子玩具拔萝卜布书12重点语汇 13巧虎日记 14预告乐智小天地宝宝版25- 有趣的形状 01 开动脑筋圆形三角形方形402 生活习惯对不起没关系 03 动物乐园青蛙04 小小世界工地上的车 05 唱游玩皮熊06 成长小故事奇妙的 \对不起\07小手哆来咪拍拍小手 08 故事小屋这是什么车呢 09 说说看碰碰车10 亲子玩具形状颜色积木 11 重点语汇 12 巧虎日记 13 预告乐智小天地宝宝版26- 爱心小宝贝01 唱游小雨滴02 开动脑筋黑色白色 03 生活习惯爱心小宝贝 04 小小世界下雨啦 05 小手哆来咪谁会这样 06 动物乐园是谁的声音呢 07 成长小故事宝宝当哥哥 08 说说看夏天09 亲子玩具颜色认知海绵拼图 10 重点语汇 11 巧虎日记 12 预告乐智小天地宝宝版27- 自己来穿鞋01 动物亲子乐园02 生活习惯自己来穿鞋 03 唱游模仿游戏04 小小世界神奇的裁缝店 05 成长小故事大家来穿鞋子 06 开动脑筋 123 07 说说看理发08 小手哆来咪奇妙的口袋 09 故事小屋小客人 10 重点语汇 11 巧虎日记 12 预告巧虎幼幼版乐智小天地幼幼版01- 1和许多 01 快乐的巧虎一家 02 欢迎歌03 概念认知 1和许多04 生活习惯吃饭时我不玩 05 唱游吃饭真快乐06 童话故事不爱吃饭的冬冬 07 小小世界动物大集合 08 巧虎小故事我是小帮手09 我会小心吃东西的安全10 HAPPY ENGLISH BAA BAA BLACK SHEEP 11 巧虎智力玩具多少认知购物组 12 预告乐智小天地幼幼版02- 我会刷牙喽 01 童话故事牙刷火车来啦 02 生活习惯我会刷牙喽03 巧虎智力玩具巧虎生活习惯组 04 唱游（一）医生超人舞 05 小小世界各种用途的车 06 我会小心坐车的时候要小心07 HAPPY ENGLISH HUSH LITTLE BABY 08 概念认知长和短09 巧虎小故事小火车快点好起来 10 唱游（二）晚安歌 11 预告乐智小天地幼幼版03- 比一比大中小 01 唱游（一）大小咚咚啪 02 紧急认知比一比大中小 03 小小世界要用什么东西呢 04 生活习惯我把小手洗干净 05 我会小心咬人的电器 06 童话故事肥皂飞走了 07 唱游（二）点点踏踏舞 08 巧虎小故事妈妈怎么了09 HAPPY ENGLISH THIS LITTLE PIG WENT TO HARKET10 巧虎智力玩具大小认知海绵动物组 11 预告乐智小天地幼幼版04- 形状配配对 01 唱游（一）圣诞快乐 02 生活习惯一次只买一个 03 概念认知形状配配对 04 唱游（二）水果歌 05 小小世界水果和蔬菜06 HAPPY ENGLISH WE WISH YOU A MERRY CHRIRSTMAS5感谢您的阅读，祝您生活愉快。

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精心整理PartOne1.LittleBee小蜜蜂Littlebee,littlebee,roundroundround.Littlebee,littlebee,soundsoundsound.Bzzzzzzz小蜜蜂，小蜜蜂，飞飞飞。

小蜜蜂，小蜜蜂，嗡嗡嗡。

嗡……One,two,three,four,five.Bzzzzzzzzzzzzzzzzzzzzz蜂窝在这里，蜜蜂在哪里？藏起来，找不着。

快来看，飞出来，一、二、三、四、五，嗡……5.TenLittleFingers十个小手指Onelittle,twolittle,threelittlefin-gers.精心整理Fourlittle,fivelittle,sixlittlefin-gers. Sevenlittle,eightlittle,ninelittlefin-gers. Tenfingersonmyhands.小小手指，一二三；小小手指，四五六；小小手指，七八九；十根手指，是一家。

6.DanceYourFingers跳动手指杰克飞走了，吉尔飞走了；杰克飞回来，吉尔飞回来。

8.JackandJill杰克和吉尔JackandJillwentupthehilltofetchapailofwa-ter. Jackfelldownandbrokehiscrown, andJillcametumbingaf-ter.精心整理杰克和吉尔，上山去打水。

杰克滑倒了，脑袋摔着了，吉尔跟着滑倒了。

9.Thechimney烟囱Hereisthechimney,Hereisthetop.O-penthelid,outsantawillpop.一个烟囱，一个盖儿。

揭开这个盖儿，圣诞老人冒出来！10.EentsyWeentsySpide小蜘蛛五只小鸟没有巣，五棵小树排成行。

快来树上搭个窝，随着树儿摇一摇。

2.FiveRabbits五只兔子Fiverabbitswerehid-inginthewoodsoneday.精心整理One,two,three,four,fivejumpedout,andthentheyrana-w ay.有一天，树丛里，五只兔，把身藏。

一、二、三、四、五，五只兔子跳出来，接着往外跑。

3.FiveLittleSausages五根小香肠Thislittlepigwenttomarket.Thislittlepigstayedhome. Thislittlepighadroastbeef.Thislittlepighadnone. Andthislittlepigcried“weeweewee”allthewayhome.这只小猪上集市，这只小猪在家里，这只小猪有烤肉，这只小猪都没有，这只小猪哭呀哭，呜呜呜呜回家去。

精心整理5.ThereLittleMonkeys三只小猴Threelittlemonkeysjumpingonthebed. Onefelloffandbumpedhishead. Mommycalledthedoctorandthedoc-torsaid “Nomorelittlemon-keysjumpingonthebed.”妈妈给医生打电话，医生说：“小猴不要再在床上跳。

”6.HickoryDickoryDock滴答滴答Hickorydickorydock,①themouseranuptheclock.Theclockstruckone.★②Themouserandown.Hickorydickorydock.Hickorydickorydock,themouseranuptheclock.Theclockstrucktwo.精心整理★★Themousesaid“Boo”.Hickorydickorydock.Hickorydickorydock,themouseranuptheclock.Theclockstruckthree.★★★Themousesaid“Wee”.Hickorydickorydock.Hickorydickorydock,themouseranuptheclock.Theclockstruckfour.★★★★Themousesaid“Nomore”.Hickorydickorydock.滴答滴答滴，老鼠爬上钟，挂钟敲一（当），老鼠跑下去。

滴答滴答滴。

Howareyoutodaysir?VerywellIthankyou.Runa-way.Runa-way. WhereisPinky?WhereisTPinky?HereIam,HereIam.Howareyoutodaysir?VerywellIthankyou.Runa-way.Runa-way.拇指在哪里？拇指在哪里？我在这里，我在这里。

今天你好吗？我很好，谢谢。

我走了，我走了。

精心整理食指在哪里……中指在哪里……无名指在哪里……小指在哪里……8.YouAreMySunshine你是我的阳光Youaremysunshine,myonlysunshine. Youmakemeha-ppy,whenskiesaregray.我爱你在清晨呀，唉你在午后！爱你在夜晚呀，爱在月光里！噢，呜啦啦呜啦啦呜啦啦啦，我爱你。

10.ClapYourHands拍手歌Clap,clap,clapyourhandsasslowlyasyoucan. Clap,clap,clapyourhandsasquicklyasyoucan.精心整理Shake,shake,shakeyourhandsasslowlyasyoucan. Shake,shake,shakeyourhandsasquicklyasyoucan. Roll,roll,rollyourhandsasslowlyasyoucan.Roll,roll,rollyourhandsasquicklyasyoucan.Rub,rub,rubyourhandsasslowlyasyoucan.Gobble,gobble,gobble,quack,quack,quack. Aturkeysaysgobble.Aducksaysquack.咯咯咯，嘎嘎嘎。

火鸡叫，“咯咯咯”。

鸭子叫，“嘎嘎嘎”。

2.OldMacdonald老麦克唐纳OldMac-donaldhadafarm,EIEIO.精心整理Andonhisfarmhehadaduck,EIEIO. Withaquackquackhere,andaquackquackthere,Herequack,therequack,every-wherequack,quack.OldMac-donaldhadafarm,EIEIO. OldMacdonaldhadafarm,EIEIO.OldMacdonaldhadafarm,EIEIO.老麦克唐纳有个农场呦，依呀依呀哟，农场上有鸭子呦，依呀依呀哟，这儿嘎嘎，那儿嘎嘎，嘎嘎，嘎嘎，到处都嘎嘎，老麦克唐纳有个农场呦，依呀依呀哟。

…小猪…哼哼……奶牛…哞哞……火鸡…咯咯…精心整理3.WalkingWalking走走停停Walking,walking.Walking,walking. Jump,jump.Jump,jump,jump.Jump. Running,running,running.Running,running,running. Nowlet’sstop.Nowlet’sstop.走走走走，走走走走，跳跳跳，跳跳跳。

Tiptoea-roundthero-sy.Apocket-fulofpo-sies. A-shes,a-shes,weallfalldown.围着玫瑰转啊转，口袋满是花瓣；花瓣飘，花瓣飘，我们都跌倒。

…跑啊跑……走啊走……单脚跳……双脚跳……躲猫猫…5.TeddyBear麦迪熊TeddyBear,TeddyBear,turna-round. TeddyBear,TeddyBear,touchtheground. TeddyBear,TeddyBear,turna-round. TeddyBear,TeddyBear,touchtheground. TeddyBear,TeddyBear,showyourshoe.Ifyou’rehappyandyouknowit,stompyourfeet.△△Ifyou’reha ppyandyouknowit,shouthurray.Hur-ray. Ifyou’rehappyandyouknowit,shout hurray.Hur-ray. Ifyou’rehappyandyouknowit,thenyourfacewillsurelyshowit. Ifyou’rehappyandyouknowit,shouthurray.Hur-ray..Ifyou’rehappyandyouknowit,doallthree.○○△△Hurray. Ifyou’rehappyandyou knowit,doallthree.○○△△Hurray. Ifyou’rehappyandyouknowit,thenyourfacewillsurelyshowit. Ifyou’rehappyandyouknowit,doallthree.○○△△Hurray. 如果感到幸福你就拍拍手。

如果感到幸福你就拍拍手。

如果感到幸福你就来表示吧。

如果感到幸福你就拍拍手。

⑵Youputyourrighthandin. Youputyourrighthandout. Youputyourrighthandin,andyoushakeitalla-bout. Youdotheho-keypokey,andyouturnyour-selfa-round. That’swhatit’salla-bout.⑵lefthand⑶rightfoot⑷leftfoot⑸rightshoulder⑹leftshoulder⑺righthip⑻lefthip⑼head⑽wholeself 把右手放进去，把右手拿出来，把右手放进去呀，甩呀甩呀甩，一起来玩游戏呀，转呀转个圈，就这样玩起来。

…左手……右脚……左脚……右肩……左肩……右胯……左胯……头……全身…Thumbsup.Thumbsup.Thumbsup.Thumbsup.A-tu-li-ta.A-tu-li-ta.A-tu-li-ta-ta.A-tu-li-ta.A-tu-li-ta.A-tu-li-ta-ta.⑵Elbowsback.Elbowsback.Elbowsback.Elbowsback.⑶Kneestogether.Kneestogether.Kneestogether.Kneestogether.⑷Turnaround.Turnaround.Turnaround.Turnaround.阿图丽塔，阿图丽塔，阿图丽塔，阿图丽塔，阿图丽塔，阿图丽塔，拇指向上，拇指向上，拇指向上，拇指向上。