day hoc tich cuc

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TRNG I HC ANGIANGKHOA S PHM

KHA LUN TT NGHIPCHUYN NGNH PHNG PHP DY HC TON

CI TIN PHNG PHP DY HC VI YUCU TCH CC HA HOT NG HC TPTHEO HNG GIP HC SINH PHT HIN

V GII QUYT VN QUA VIC T CHCDY HC HM S LIN TC

SINH VIN : L QUANG VINH GVHD : Th.S NGUYN VN VNH

LP : DH5A2 NIN KHA : 2004 - 2008

An Giang , nm 2008


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MC LCTHAY LI TA

PHN M UI. L DO CHN TI.................................................................................................................. ...II. MC CH NGHIN CU ........................................................................................... ................1III. NHIM V NGHIN CU..........................................................................................................2IV. GI THUYT NGHIN CU .....................................................................................................2V. PHNG PHP NGHIN CU.................................................................................... ...............2VI. CU TRC LUN VN ....................................................................................... ........................PHN NI DUNGA. MT S VN V L LUN DY HCI. DY HC TCH CC HA HOT NG NHN THC CA HC SINH .........................3

1. Th no l tch cc..................................................................................................................2. Hot ng hc tp l mt qu trnh nhn thc tch cc .....................................................43. Dy hc tch cc ha hot ng nhn thc ca hc sinh l cch dy ph hp

vi quy lut nhn thc...........................................................................................................4. Nhng du hiu dc trng ca phng php dy hc tch cc ..........................................5

II. PHNG PHP DY HC PHT HUY TNH TCH CC CA HC SINH, XT THEOQUAN IM TM L HC .......................................................................................................6

III. DY HC PHT HIN V GII QUYT VN .............................................. ..................81. C s l lun ..........................................................................................................................

1.1 C s trit hc .............................................................................................................1.2 C s tm l hc ...........................................................................................................1.3 C s gio dc hc ......................................................................................................

2. Nhng khi nim c bn ........................................................................................................2.1 Vn .........................................................................................................................2.2 Tnh hung gi vn .................................................................................................

3. c trng ca dy hc pht hin v gii quyt vn ........................................................9

4. Nhng hnh thc dy hc pht hin v gii quyt vn ..................................................94.1 T nghin cu vn ..................................................................................................4.2 m thoi gii quyt vn ........................................................................................

4.3 Thuyt trnh gii quyt vn .................................................... ................................105. Nhng cch thng dng to tnh hung gi vn .......................... ..............................10


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Lun vn tt ngh ip G VHD: Ths.N gu y n Vn Vnh

PHN M U YZ

I. L DO CHN TIKin thc mnh mng nh mt i dng rng ln. S hiu bit ca con ngivchng th qu hn hp, do phi to hng th cho ngi nghin cu h m rngs hiu bit cho mnh v cho th gii ca chngta.Dy v hc l qu trnh em li kin thc mt cch sinh ng ca th h trctruynli cho th h sau. Khi vai tr ca ngi thy rt quan trng trong vic truynt,v ngi hc ng vai tr tip thu mt cch sng to nhng kin thcy.Do phng php gio dc ph thng phi pht huy tnh tch cc, t gic,chng sng to ca hc sinh, phi hp vi c im tng lp hc, mn hc, bidng phng php t hc, t nghin cu, rn luyn k nng vn dng kin thc vnhng

iu hc c vo thc tin, em li nim vui v hng th cho hcsinh.Lut gio dc nm 1998 (iu 24-2) vit: phng php gio dc phi pht huytnhtch cc, t gic, ch ng sng to ca hc sinh, ph hp vi c im ca tnglphc, mn hc, bi dng phng php t hc, rn luyn k nng vn dng kinthcvo thc tin, em li nim vui hng th cho hcsinh.Vn ct li ca vic i mi phng php dy hc mn Ton trng phthngl lm cho hc sinh hc tp vi thi tch cc, ch ng v sng to. Trongqutrnh gio dc, hc sinh ng vai tr l ch th ca hot ng nhn thc, hngvoci bin bn thn tch ly kin thc, k nng, k xo, dn dn pht trin t duyca bn thn Qu trnh ny ph thuc vo s hot ng ca mi hc sinh, khng aicth lm thay cho bn thn hc sinh. S tc ng ca hon cnh, mi trng c thl

s hng dn ca thy c, gip ca b bn, tp th ch l th yu, n ch htr cho qu trnh ny t kt qu tthn.Hot ng hc tp l hot ng trc tip hng vo vic tip thu, lnh hi tri thcvk nng. Dy hc mn ton v thc cht l hot ng ton hc m trc tin lhotng tduy.V vy ni dung ct li ca vic i mi phng php dy hc trng ph thngl pht huy tnh tch cc hc tp ca hc sinh vi tinh thn t gic. Phng phpdyhc pht hin v gii quyt vn l mt phng php c th gip cho hc sinhthchin c s t gic hc tp v tch cc mt cch sngto.Xut pht t nhng iu trn, chng ti nghin cu ti Ci tin phngphpdy hc vi yu cu tch cc ha hot ng hc tp theo hng gip hcsinhpht hin v gii quyt vn qua vic t chc dy hc hm s lintc.

II. MC CH NGHINCU1. H thng ha mt s vn v t tng tch cc ha hot ng nhnthc,

hc tp ca hc sinh nhm lm r kh nng tch cc ha hc sinh trong qu trnhtchc dy hc theo phng php dy hc pht hin v gii quyt vn.

Sinh vin : L Quang Vinh Trang 1


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2. Thit lp qui trnh dy hc, xy dng cc bin php s phm gp phngipcho gio vin t chc hp l qu trnh tm ti, pht hin v gii quyt vn cahcsinh.

3. Trn c s , p dng vo vic t chc v hot ng lnh hi tri thchms lin tcnhm nng cao cht lng dy hc ca trng phthng.

III. NHIM V NGHINCU Nghin cu ti ny chng ti thc hin cc nhim vsau:1. Nghin cu l lun v tm l hc dy hc lm c s cho nhng bin php

s phm nhm nng cao tnh tch cc ca hcsinh.2. Phn tch cc bc tng qut v hnh thc t chc ca dy hc pht hinv

gii quyt vn pht huy tnh tch cc trong hot ng nhn thc ca hcsinh.3. Xy dng qui trnh dy hc pht hin v gii quyt vn.4. Xy dng bin php s phm tng ng gip gio vin t hiu quging

dy trong qu trnh thchin.5. p dng vo vic dy hc hm s lintc.

IV. GI THUYT KHOAHCS dng bin php s phm ra c th gip hc sinh pht huy tnh tch cchotng nhn thc trong qu trnh dy hc hm s lintc.

V. PHNG PHP NGHINCU1. Nghin cu llun2. c kt kinhnghim

3. Thc nghim s phm4. nh gi, ktlun

VI. CU TRC LUNVNLun vn gm phn m u, phn ni dung, phn kt lun, phn ph lc v cctiliu thamkho.



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PHN NIDUNG

A. MT S VN L LUN DYHC Y

Z

I. DY HC TCH CC HA HOT NG NHN THC CA HCSINH:1. Th no l tnh tchcc

Tnh tch cc vn l mt phm cht vn c ca con ngi trong i sng x hi,conngi sn xut ra ca ci vt cht cn thit cho s tn ti, pht trin ca x hi,sngto ra nn vnha.Tnh tch cc biu hin trong hot ng. Hot ng l ch th ca la tui ihc.Tnh tch cc trong hot ng hc tp thc cht l tnh tch cc nhn thc, ctrng kht vng hiu bit, c gng tr tu v ngh lc cao trong qu trnh chim lnhtrithc. Qu trnh nhn thc trong hc tp nhm lnh hi nhng hiu bit m loingi tch ly c, hc sinh khm ph ra nhng ci mi i vi bn thn. Hc sinhsthng hiu, ghi nh nhng g nm c qua hot ng ch ng, n lc cachnhmnh.Tnh tch cc nhn thc trong hot ng hc tp lin quan trc ht vi ng c hctp. ng c ng to ra hng th. Hng th l tin ca t gic. Hai yu tnyto nn tnh tch cc. Tnh tch cc sn sinh np t duy c lp. Suy ngh c lplmm mng ca sngto.Tnh tch cc biu hin: hng hi tr li cu hi ca gio vin, b sung cu tr lica bn, thch pht biu kin ca mnh trc vn c nu ra, hay nu thc mc,ihi gii thch cn k vn cha r, ch ng vn dng kin thc k nng hcnhn thc vn mi, tp trung ch vo vn ang hc, kin tr hon thnhcc bi tp, khng nn trc tnh hungkhTnh tch cc hc tp t nhng cp t thp ncao:

+ Bt chc: gng sc lm theo cc mu hnh ng ca thy, ca bn+ Tm ti: c lp gii quyt vn nu ra, tm kim nhng cch giiquyt

khc nhau v mt vn+ Sng to: tm ra cch gii quyt mi, c o, huhiu

Cn theo t in ting Vit, tch cc l mt trng thi tinh thn c tc dngkhngnh v thc y s pht trin. Trong hot ng hc tp n din ra di nhiuhnhthc khc nhau : tri gic ti liu, thng hiu ti liu, ghi nh, vn dngv cth

hin nhiu hnh thc phong ph a dng khcnhau: Xc cm hc tp: th hin nim vui, hng hi thc hin yu cu cagio

vin. Ch : th hin vic lng nghe v theo di mi hnh ng ca giovin,

thc hin nhanh gn, chu o, chnh xc cc yu cu. S n lc ca ch : th hin s kin tr nhn ni vt kh khi giiquyt

nhim v nhn thc; khn trng khi thc hin cc hnh ng tduy.



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Kt qu lnh hi: khi cn th ti hin nhanh, ng v vn dng c khictnh hungmi.Khi ni v tnh tch cc, ngi ta thng nh gi n cp c nhn ngihctrong qu trnh thc hin mc ch hot ng dy hc chung. Tnh tch cctronghot ng nhn thc l trng thi hot ng ca hc sinh, c c trng bikhtvng hc tp, s c gng tr tu vi ngh lc cao trong qu trnh nm vng kinthccho chnh mnh ( theo I.F.Kharlamop) v theo G.I.Sukina chia tnh tch cc ra lm3cp:

9 Tnh tch cc bt chc, ti hin (xut hin do tc ng bnngoi).9 Tnh tch cc tm ti (i lin vi qu trnh hnh thnh khi nim, giiquyt

tnh hung nhnthc).9 Tnh tch cc sng to (th hin khi ch th tm ti kin thcmi).2. Hot ng hc tp l mt qu trnh nhn thc tchcc

Hc tp l hot ng nhn thc ca hc sinh, mun nm kin thc mt cch suscv vng chc, hc sinh phi thc hin y cc hot ng tr tu v theo ngconng nhn thc m Lnin vch ra: T trc quan sinh ng n t duytru

tng v t t duy tru tng n thc tin, l con ng bin chng nhnthc chn l, nhn thc hin thc khchquan.3. Dy hc tch cc ha hot ng nhn thc ca hc sinh l cch dyph

hp vi quy lut nhnthcThc cht dy hc tch cc ha hot ng nhn thc ca hc sinh l qu trnhtchc, hng dn hc sinh t tm hiu, pht hin v gii quyt vn trn c s tgic, c to kh nng v iu kin ch ng trong hot ng hc tp cahcsinh.Tch cc ha hot ng hc tp ca hc sinh l vic thc hin mt lot cc hotngnhm lm chuyn bin v tr t th ng sang ch ng, t i tng tip nhntrithc sang ch th tm kim tri thc nng cao hiu qu hctp.Phng php dy hc tch cc ha hot ng nhn thc ca hc sinh l phng phpdy hc trong gio vin t chc qu trnh dy hc da trn sc lc v tr tucahc sinh, mi hc sinh t nghin cu, thc hnh tm ra kin thc, hnh thnhk nng nhn thc. Phng php ny c cc ctrng:

Mi hc sinh u c tch cc ha hot ng t duy, c t lc tipcnkin thc mi cp khc nhau v gii quyt vn theo mt quytrnh.

Gio vin gi vai tr ch o, t chc cc tnh hung hc tp, hng dnhcsinh gii quyt vn , khng nh kin thc mi trong vn tri thc sn c camnh.Theo Vgotxki, vai tr ca dy hc trong s pht trin tr tu v phng din lchsl khng ngng tng ln v hin nay n mang tnh cht quyt nh. Dy hc phicnc vo dng tm l hot ng ca tr, tuy n cha phc tp nhng c xuthin.Trong khi thc hin cc hot ng vi s gip ca ngi ln, tr em chuyntvng pht trin gn nht ti vng pht trin tch cc. Trong vng ny tr c thtmnh thc hin li cc hot ng ny mt cch clp.



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Do qu trnh dy hc v pht trin tr tu c lin quan cht ch vi nhau.Thtvy, dy hc da trn trnh t c ca pht trin v to iu kin pht trinchotr, chuyn tr sang trnh pht trin tip theo, ti mt trnh caohn.

4. Nhng du hiu c trng ca phng php tchcc:(Theo p dng dy v hc tch cc trong Tm l- Gio dc hc ca GS TrnB

Honh- Ths L Trn nh Ts Ph c Ha-NXB I HC S PHM H NI).4.1 Dy v hc thng qua t chc cc hot ng ca hcsinh

Trong phng php tch cc, ngi hc - i tng ca hot ng dy, ngthil ch th ca hot ng hc c cun ht vo trong hot ng hc tp dogiovin t chc v ch o, thng qua t lc khm ph nhng iu mnh cha rchkhng phi th ng tip thu nhng tri thc c gio vin sp t. c tvotnh hung ca i sng thc t, ngi hc trc tip quan st, tho lun, lmthnghim, gii quyt vn t ra theo cch suy ngh ca mnh, t va nmckin thc k nng mi, va nm c phng php lm ra kin thc, k nngkhng rp theo nhng khun mu sn c, c bc l v pht huy tim nngsngto.

Dy theo cch ny th gio gio vin khng ch gin n truyn t tri thc mcnhng dn hnh ng. Chng trnh dy hc phi gip cho tng hc sinh bithnhng v tch cc tham gia cc chng trnh hnh ng ca cng ng.

4.2 Dy v hc ch trng rn luyn phng php t hc:Phng php tch cc xem vic rn luyn phng php hc tp cho hc sinhkhngch l mt bin php nng cao hiu qu dy hc m cn l mt mc tiu dy hc.Trong x hi hin i ang bin i nhanh - vi s bng n thng tin, khoa hc,k thut, cng ngh pht trin nh v bo th khng th nhi nht vo u c trkhilng kin thc ngy cng nhiu. Phi quan tm dy cho tr phng php hcngayt bc Tiu hc v cng ln bc hc cao hn cng phi c chtrng.

Trong cc phng php hc th ct li l phng php t hc. Nu rn luynchongi hc c c phng php, k nng, thi quen, ch t hc th s to chohlng ham hc, khi dy ni lc vn c trong mi ngi, kt qu hc tp scnhn ln gp bi. V vy, ngy nay ngi ta nhn mnh mt hot ng hc trongqutrnh dy hc, n lc to ra s chuyn bin t hc tp th ng sang t hc chng,t vn pht trin t hc ngay trong trng ph thng, khng ch t hc nhsau bi ln lp m t hc c trong tit hc c s hng dn ca giovin.

4.3 Tng cng hc tp c th, phi hp vi hc tp hptcTrong mt lp hc m trnh kin thc, t duy ca hc sinh khng th ngutuyt i th khi p dng phng php tch cc buc phi chp nhn s phn havcng , tin hon thnh nhim v hc tp, nht l khi bi hc c thitkthnh mt chui cng tc hc tp.p dng phng php tch cc trnh cng cao th s phn ha ngy cngln.Vic s dng cc phng tin cng ngh thng tin trong nh trng s p ngyucu c th ha hot ng hc tp theo nhu cu v kh nng ca mi hcsinh.Tuy nhin, trong hc tp, khng phi mi tri thc, k nng, thi u chnhthnh bng nhng hot ng c lp c nhn. Lp hc l mi trng giao tip thy-tr, tr - tr, to nn mi quan h hp tc gia cc c nhn trn con ng chimlnh



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II. PHNG PHP DY HC PHT HUY TNH TCH CC CAHCSINH XT THEO QUAN IM TM LHC:Mi phng php u c chc nng iu hnh ton b qu trnh dy hc, tc l nsquy nh cch thc chim lnh tri thc v kinh nghm hot ng ca hcsinh.Qu trnh hc tp ch nn din ra theo kiu tm kim, pht hin, khai thc, bin iv

ngi hc t kin to kin thc, k nng tng thch vi kinh nghim v bnchtngi hc, ngi hc tr thnh ch th tch cc hn. m bo gip hc sinhlnhhi y kin thc quy nh trong mt n v thi gian th khng th ch vndngmy mc mt cch dy hc m phi kt hp nhun nhuyn chng li vinhau. Nhng trong cch chim lnh tri thc bng cch nh hng n hot ng citotch cc dn n vic pht minh kin thcmi.Mc tiu gio dc o to phi hng vo o to nhng con ngi lao ngtch, sng to, c nng lc gii quyt nhng vn thng gp, qua gp phntchcc thc hin mc tiu ln ca t nc l dn giu, nc mnh, x hi cng bng,dn ch, vn minh (theo hi ngh ln VIII ca Ban chp hnh Trung ngngCng Sn Vit Nam).

V phng php gio dc, phi khuyn khch t hc, phi p dng nhng phng php gio dc hin i bi dng cho hc sinh nng lc t duy sng to, nnglcgii quyt vn . Phi i mi phng php gio dc o to, khc phc litruynth mt chiu, rn luyn thnh nt t duy sng to ca ngi hc. Tng bcpdng cc phng php tin tin v phng tin hin i vo qu trnh dy hc,m bo iu kin v thi gian t hc, t nghin cu cho hc sinh, c bit l sinhvinihc.Lut gio dc 1998 c vit: phng php go dc ph thng phi pht huy tnhtchcc, t gic, ch ng sng to ca hc sinh, ph hp vi c im ca tng lphc,mn hc, bi dng phng php t hc, rn luyn k nng vn dng kin thcvothc tin; tc ng n tnh cm em li nim vui hng th cho hcsinh.

Mt phng php dy hc ch c kh nng bi dng nhng phm cht ca tduykhi n thc hin s pht ng, thc y s suy ngh tch cc ca ngi hc vdndt s suy ngh y theo con ng ngn nht, hp l nht t ti kin thc vk nng. Phng php phi da vo thnh tu khoa hc nghin cu v tduy.Theo cc nh tm l hc, con ngi ch bt u t duy tch cc khi ny sinh nhucut duy tc ng trc mt tnh hung gi vn . Tnh hung gi vn l mttnhhung gi ra cho hc sinh nhng kh khn v l lun hay thc tin m h thycnthit v c kh nng vt qua, nhng khng phi ngay tc khc nh mt quy tcctnh cht thut gii m phi tri qua mt qu trnh tch cc suy ngh, hot ng bin i i tng hot ng hoc iu chnh kin thc sn c. Trc tnh hungcvn , ngi ta s bn khon suy ngh, tm cch gii quyt nhng bt u tu,theo phng hng no, but h phi thc c vn thng c th hin cht c cu hi hoc nu c thc mc, y l vn quan trng ca tchccha.Gio dc l thch ng a tr vo mi trng x hi ngi ln; l thch nghiconngi vo mi trng x hi xung quanh, o to c con ngi khi vo i lconngi t ch, nng ng v sng to th phng php gio dc cng phi hngvovic khi dy, rn luyn, pht trin kh nng ngh v lm mt cch t ch, nngngsng to trong hc tp v lao ng nh trng, nhm kch thch hc sinh tchcc



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suy ngh, ch ng tm ti sng to gii quyt vn , t ti kin thc mimtcch vng chc v susc. Nhn chung, mt phng php dy hc pht huy tnh tch cc ca hc sinh l phng php m trong ngi hc tr thnh trung tm, ch th c nh hng tmnh tm ra kin thc chn l bng hot ng ca bn thn. Gio vin gi mttrchnhim mi chun b cho hc sinh tht nhiu tnh hung phong ph to iu kinchohc sinh gii quyt vn ch khng nhi nht kin thc vo u hc sinh, lnginh hng cho hc sinh t khm ph ra chn l, t tm ra kin thc vi s hptcca tp th. Thy gio by gi tr thnh ngi nh hng, c vn cho hc sinhcamnh khm ph ra nhng iu mi, nhng kin thc mi, nhng chn l mi vishp tc ca cc ch th khc trong lphc.Phng php no m bo s kt hp nhun nhuyn hai hot ng ti hin vtmkim kin thc trong c hi v iu kin vic tm kim kin thc mi chimuth, kt hp hi ha vi tnh sn sng hc tp ca hc sinh th phng php ckh nng tch cc ha c qu trnh hc tp ca hc sinh, t hnh thnh phngthc hnh ng cng nh kinh nghim hot ng cho ccem.Tm li, dy hc theo phng php truyn thng cn bn ch thch hp cho ccthh qua, cho con ngi ch phong kin, n s kh p ng c mc tiuoto con ngi pht trin ton din ca ch ngha x hi. Cn dy hc theo kiutchcc, pht huy cao tnh tch cc hot ng ca tr, bin cc em thnh ch th,chng pht hin ra kin thc mi cn phi hc, n lun ph hp vi qu trnh phttrin ca x hi.

III. DY HC PHT HIN V GII QUYT VN1. C s l lun

1.1 C s trithc:

Trit hc duy vt bin chng cho thy nu ta gii quyt c mu thunths vt, hin tng s pht trin ngha l mu thun chnh l ng lc ca qutrnh pht trin. hc sinh mu thun gia yu cu nhim v nhn thc vi kin thcvkinh nghim snc.

1.2 C s tm lhc:Theo cc nh tm l hc khi ng trc mt kh khn v nhn thchay

mt tnh hung gi vn th con ngi bt u t duy tchcc.1.3 C s gio dc hc:Dy hc pht hin v gii quyt vn ph hp vi tnh tch cc, t gicv

n kch thch v to ng c cho ch th hot ng pht hin vn.

2. N h ng khi n i m c b n 2.1 Vn

Mt h thng cc mnh v cc cu hi to nn mt vn nu tha mn cciukin:

+ Cu hi cha c giip



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Trong :+ Mt y ca hnh hp (ABCD): ni dung dyhc+ Mt chnh din (AADD): qu trnh gingdy+ Mt bn (ABBA): qu trnh hctp

Gii thch:+ S Graph l m hnh tnh ca h dy hc. Mi mt thnh t b phn xut pht ban u ca h dy hc l mt b phn tim nng ca h Tonvn+ Khi s tng tc gia cc thnh phn ny xut hin th qu trnh dy hc c btu.+ Khi qu trnh dy hc bt u vn hnh, mi mt b phn tim nng tr thnhmt b phn c yu ca h Ton vn v l mt h con ca h Dyhc.

2.3.Cu trc ca cc hcon:2.3.1) N i dung dyh c :

c xc nh bi cc mc ch dy hc, c qui nh trong chng trnh mnhcv c th hin trong cc sch giokhoa.Cc nhim v dy hc tng hp vi ni dung dy hc, v cc giai on khcnhauca qu trnh dy hc l s m bo t c cc mc ch dyhc.Trong qu trnh dy hc, cc nhim v sau y c t ra v i hi giiquyt:

(N1): xut gy ng c nhnthc.

(N2): Tnh thi s ca kin thc, cch thc hot ng, bo lu skinmi.

(N3): Lnh hi cc ti liu hc tp, s khi qut ha cc ti liu hctp. (N

4): Cng c, hon thin kin thc, hnh thnh k nng, k xo.

(N5): Khi qut ha v h thng ha cc ti liu hc tp, vn dngkinthc, k nng, k xo, cch thc hot ng trong hon cnh cth.

(N6): Phn tch kt qu dy hc, s pht trin ca hc sinh (v tdaylogic, tr thng minh, kh nng sng to). Kim tra, nhgi.

N h nxt : Ni dung dy hc ch c hc sinh chp nhn nh mt i tng cahot

ng nhn thc nu n c th hin di dng mt h cc bi ton hc tp.Vicgii quyt cc bi ton hc tp trn y c t ra trong qu trnh hctp

VyN = { N1,N2,N3,N4,N5,N6 }Cc ni dung b phn c t ra v gii quyt tng hp vi cc nhim v trn.

2.3.2) Qu trnh g i ngd y :Theo Xibecnetic (iu khin hc): Qu trnh ging dy l s iu khin, dndt,kim tra s tng tc cah.



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Trong qu trnh dy hc, thy s dng mt cch c h thng, nh hng cc phng php ging dy trin hai ni dung dy hc, kch thch hot ng nhn thc cahcsinh, tp trung vo hai pha: Thng tin v iukhin.Qu trnh ging dy: l mt h con gm cc thnh t b phn

QTD = {PD1, PD2, PD3, PD4}

PD1: Phng php gii thch minhha PD2: Phng php trnh by nu vn PD3: Phng php tm ti tng phn PD4: Phng php nghincu

2.3.3) Qu trnh h c tp:Hot ng nhn thc tch cc ca hc sinh bao gm: tri gic, phn tch, suy on,ghinh ti liu hc tp, l thuyt, thc hnh ch bin cc thng tin thunhn.Qu trnh hc tp din ra cc trnh khc nhau, tng hp vi hot ngnhnthc ca hcsinh

PH1: Trnh tihin PH2: Trnh tm ti tng phn bng thcnghim PH3: Trnh tm ti tng phn bng suylun PH4: Trnh nghincu

VyPH = {PH1,PH2,PH3,PH4}Ch s tng th nhim v dy hc, tnh phc tp ca phng php truyn t viukhin hot ng nhn thctng.

2.3.4) Tnh h u ng d yh c :

Do cch xem xt cu trc cc h con ca qu trnh dy hc, c th trnh by s Graph chi tit ca H DYHC:

B C

A D

B C

A D

C tt c 6 x 4 x 4 = 96 khi hp ch nhtnh



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00

x x0 f ( x0)

Trng hp hms k ( x)

f ( x) =l ( x)

nunu

x x0 x < x0

Th thay v tatnh lim f ( x) th tatnh x x0

lim x x+

f ( x) v lim f ( x) x x



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x 2

k (2) = 2a + 1

Hm s lin tck ti im x0= 2 khi v ch khi:

limk ( x) = k (2) a =11

x 2 2

Vy gi tr cn tm la =11

.2



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y = f ( x) c lin tc ti trn mt khong( a; b )

Pht biu v t mc ch giiquyt:Chng ta cn tm cch gii tng qut v iu kin thc hin vic xt hms y = f ( x) cho c lin tc ti trn mt khong( a; b ) (mt on[a; b] ) noBc 2: Gii quyt vn

Phn tch vn , lm r mi quan h gia ci bit v ci phitm:Xt tnh lin tc ca hms y = f ( x) trn khong( a ; b )

Cu hi t ra : chng minh hm s lin tc trn mt khong( a; b ) no tacnlmg?

Tr li : Ta cn chng minh hm s lin tc ti tt c cc im trn khong( a ; b )



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Lun v n t t ng h ip G V HD: Ths .N g u y n V n V nh

Vn ny sinh: Khi hm s lin tc trn khong( a; b ) th ta cn thm iu king hm s lin tc trn on[a ; b] ?Tr l i:Da vo v d ta ch cn chng minh thm:

lim x a +

f ( x) = f (a ) v lim f ( x) = f (b) x b

xut hng gii quyt v thc hin gii quyt vn:

T trng hp trn, ta c th i n phng php chung xt tnh lin tc cahms y = f ( x) trn khong( a; b ) nh thno?

Ta chng minh vimi x0 thuc khong( a; b ) th hm s cho u lin tc ti x0

Phng php chung xt tnh lin tc ca hmsno?

y = f ( x) trn on[a ; b] nh th

Ta chng minh hm s lin tc trn khong( a; b ) r i chng minh thm:

lim x a +

f ( x) = f (a ) (1)

v lim f ( x) = f (b) x b

(2)

Vn ny sinh: Vy th hm s lin tc trn na khong( a ; b] , [a ;+ ) th sao?Gi hng gii quyt: Nu c c hai iu kin (1) v (2) th ta c hm s lintctrn on[a ; b] . By gi chng minh hm s lin tc trn na khong( a ; b] th tacn my iukin?Tr l i:Ta ch cn chng minh thm iu kin th(2)Phng php chung chng minh hm s lin tc trn na khong( a ; b]minh hm s lin tc trn khong( a; b ) v thm iu kin(2)

l chng

Cn mun chng minh hm s lin tc trn na on[a;+) th lm nh thno?Tr l i:Chng minh hm s lin tc trn khong( a ; + ) v chng minh thm iu kin(1)

Bi ton ktthc. Nh vy hm s lin tc khi biu din bng th c nt g c bit haykhng?

Gio vin giithiu:


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th hm s lin tc trn mt khong( a ; b ) l mt ng lin trn khong.



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a) f ( x) = 2 x3 + 5 x 7 lin tc tiim

x0

b) g ( x) = x + 2

x 1

lin tc ti mi im x0 1 v gin on tiim x0 = 1

c) h( x) = sin x

5lin tc vi mi x

Sau khi hc sinh lm xong cc bi ton trn th gio vin t vnYu cu hc sinh cho bit tn gi chung ca cc hms

Tr l i: f ( x) l hm athc

g ( x) l hm phn thc hut

f ( x); g ( x); h( x)

h( x) l hm lnggic

Vy ta c th kt lun v min lin tc ca cc hm s nykhng? Gii thch v chnh xc ha vn :

Ta cn tm ra min lin tc ca cc hm s: f ( x); g ( x); h( x) ; t tng qutln

tm min lin tc ca cc hm s : a thc, phn thc hu t; lnggic. Pht biu v t mc ch giiquyt:

Mc ch ca chng ta l tm ra min lin tc ca cc hm s c bit : a thc, phnthc hu t; lnggic.Bc 2: Gii quyt vn

Phn tch vn , lm r mi quan h gia ci bit v ci phitm:T v d c th ca hms f ( x) ta chng minh c hm s ny lin tc timi

im x0 . l iu bit

Gio vin yu cu hc sinh cho bit min xc nh ca hms

Tr li: Min xc nh ca hm s f l f ( x)

Gio vin dn : Vy i vi hm s a thc ny, min xc nh v min lin tcclin quan g vi nhau haykhng?Tr li: Chng trng nhau. Gio vin khng nh li vn : hm a thc lin tctrnmin xc nh can.T v d c th ca hm phn thc hu t ta chng minh c hm s lin tctimi im x

0 1 v gin on tiim


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f (a )

Vy bn no tr li ng, bn no tr lisai?

Hng dn hc sinh trl i:



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3) t f ( x) = x3 + x a

Tp xc nh ca hm s: D =

Hm s lin tc trn nn lin tc trn on[0; a ] .Ta c : f (0) = a

f (a ) = a 3

Suyra: f (0). f (a ) = a 4 < 0

Vy phngtrnh

T ta cpcm. f ( x) = 0 c t nht 1 nghim trn on[0; a ]


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C. THC NGHIM S PHM

I. GII THIU THC NGHIM S PHM kim nghim li l lun dy hc nu ra, chng ti tin hnh thc nghim trn2lp (1 lp trng THPT Thoi Ngc Hu, 1 lp ca trng THPT M Hi ng)c dy theo phng php tch cc (pht hin v gii quyt vn ). ng thi pht phiu thm d gio vin ca trng THPT M Hing

II. MC CH THCNGHIM:Thc nghim ln 1 nhm kim tra thc trng cng nh k nng gii ton ca hcsinh trng ph thng theo phng php truynthng.Thc nghim ln 2 nhm kim tra tnh kh thi ca ti. Tc l hng dy hcgiphc sinh pht hin v gii quyt vn qua vic dy hc hm s lin tc c nnpdng vo trng ph thng hay khng v n t kt qu rasao.Phiu thm d gio vin : Tm hiu kh khn gio vin khi tin hnh dy hcgiiton hm s lin tc, cc kin thc trng tm ca chng trnh ton lp 11.ngthi qua kim nghim thc t qu thy c nhng trng trn, chng ti munk imnghim xem hng dy hc gip hc sinh pht hin v gii quyt vn qua victchc dy hc hm s lin tc c nn a vo trng ph thng hay khng vhcsinh c tip thu tt haykhng?

III. HNH THC THC NGHIM:Thc nghim ln I: Hc sinh lm bi c nhn trong 30 phtThc nghim ln II: Cho hc sinh lm bi kim tra trong 45 pht. ng thi kimtracht lng gio n kim tra tnh kh thi ca ti.Thit lp phiu thm d gio vin gm 6 cu him.

IV. PHN TCH KT QU THCNGHIM:1.Thc nghim cho hcsinh

1.1.T h c ngh i m l n 1 : (Theo lun vn tt nghip ca T ThHongLan nm 2004 GVHD: thy Nguyn Vn Vnh. Lp kim tra: lp 11A3trngTHPT Nguyn Cng Tr - TP H Ch Minh)

1.1.1 N h nx t Cu 1 : Cc em ch quan tm n s lin tc ti cc im u mt. cu ny,tytheo gi tr ca a m kt lun tnh lin tc ca hm s nn trong trng hp lin tc,90% cc em u khng m ch kt lun lin tc trnkhong.



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Cu 2: Nhn vo ta c cm gic cu ny l mt cu d nhng khng phi emnocng tr li ng hon ton. Mc ch cu ny l kim tra s hiu khi nim lintcch c nh ngha trn tp xc nh ca hm s, v vy vic xc nh tp xcnhl mt khu rt quan trng. Cc em sailm:

+ Vic kh i nhn t chung t v mu thc lm rt nhiu em kt lun saivtp xc nh ca hm s, dn n kt lun sai v tnh lin tc ca hms.

+ Vic hiu khi nim ch mt chiu, ch nh n biu thc xc nh hmsm khng n iu kin thuc khong xc nh. C em xc nh ng tpxcnh nhng li kt lun hai im khng thuc tp xc nh l hai im ginon.Cu 3: lin quan n lng gic, kh hn nhng a s cc em u lmcCu 4 : Nhiu em cha tnh c giihnCu 5: Nhiu em qun mt iu kin lin tc, hu nh cc em bit chn hais a , b sao cho f (a). f (b) < 0

1.1.2 Kt qu kim tra trng Nguyn CngTr:

STT H vtn im1 Nguyn Ngc ThyAn 92 L NgcAnh 63 Trng Th LanAnh 104 Nguyn Trn MinhCnh 25 Phan Trn LanChi 86 Nguyn Th NgcChung 27 inh Th ThyDung 108 Nguyn Th KimDung 79 Nguyn Trn NgcDuy 1010 ng Th NgcH 611 V Th NgcH 412 Nguyn HngHnh 813 Mai Th ThanhHng 614 Lng ThanhHin 615 Bi Th NguytHng 916 Dng NgcHuyn 317 Mai Th ThuHuyn 10



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Lun v n t t ng h i p G V HD: Ths .N g u y n V n V nh

18 V Th LanHng 619 H Th ThanhHng 320 Th KimKhnh 621 Trn HongLan 6

22 Trn Th XunLoan 1023 Mai Th KimLin 624 Phan Th BchLin 825 Nguyn VnLnh 326 Trn Chu MaiLoan 527 Nguyn Th ThuLng 628 Trn AnhLng 729 L Th XunMai 630 L Ngc Kim Ngn 231 ng Th Bch Ngc 832 Nguyn Qunh Nh 333 Nguyn Ngc DimPhng 934 on DanhPhc 235 inh Th BchPhng 636 Nguyn Ngc BoQunh 437 Nguyn Th ThanhTm 438 Trn Ngc BngTm 439 Vng Th LanThanh 940 ng Th ThuTho 741 Nguyn Th ThanhTho 742 Nguyn Th ThuTho 543 V Th AnThun 344 Phm Th ThuThy 945 Trn L XunThy 946 Cao NhnTin 547 Bi KhnhTon 148 Kim HuynTrang 249 Nguyn HiT 950 ng DimUyn 2



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x n x x ( x x)2 n ( x x)2

01234567

8910

026554115

476

-5,91-4,91-3,91-2,91-1,92-0,910,091,09

2,093,094,09

34,9224,1015,288,463,640,830,011,19

4,379,5516,74

048,2091,6942,3118,223,310,095,95

17,4966,88100,41

394,55


51 L Th nh Vn 552 Nguyn Th Thanh Vn 753 Nguyn Thanh Vy 1

1.1.3 Ph54n tch k ut Dqu :ng Ngc Xun 10

55 Nguyn Hong Hi Yn 4

Phn phi ims:

imxi

0 1 2 3 4 5 6 7 8 9 10 Tngs

Shsni

0 2 6 5 5 4 11 5 4 7 6 55

Trung bnh cng:

x = = xi n

i = 5,

91n

Phng sai v lchchun:

i i i i i i

Phng sai: s2 = ni (

xi

x)2=

394,55 7, 31


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1

%

h c s

i n h


lch chun: s 2, 70

Tn sut:

xiS % hc sinh t im xi

0 1 2 3 4 5 6 7 8 9 10% 0 3,64 10,91 9,09 9,09 7,27 20 9,09 7,27 12,73 10,91

25

20

20

15

10

5 3.64

0

0

10.91

9.09 9.097.27

9.097.27

12.7310.91

1 2 3 4 5 6 7 8 9 10 11

im

1.2 Thc nghim dnh cho trng THPT Thoi Ngc Hu ( ln 2)1.2.1. S lc v trng THPT Thoi NgcHu:

Trng THPT Thoi Ngc Hu nm ngay trung tm thnh ph Long Xuyn AnGiang. L mt trong nhng trng gii nht ca tnh An Giang (so vi cctrng ph thng trong tnh). T l tt nghip trung hc ph thng cao ( trn 90%), vc bit l trng c t l hc sinh trng tuyn vo cc trng i hc Cao ngcngkh cao so vi cc trng THPT khc trongtnh.T khi thnh lp n nay trng lun t c nhiu thnh tch to ln: danhhiutrng tin tin, danh hiu trng tin tin xutsc.

1.2.2 Nhnxt:1:

o Bi 1:

Hu nh cc em bit cch xc nh tnh lin tc ca hm s i vi tng dng hms(hm phn thc hu t).Bit cch bin i biu thc hu t v dng khng cn v nh tnh giihnTnh ton kh chnh xc. Tuy nhin vn cn mt s em trnh by cha rrng.

o Bi 2: cu ny cc em hiu lm tm im gin on ca hm s l tm nhng imhms khng xcnh.



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x0


Thc cht bi ny yu cu xt tnh lin tc ca hm s, qua ch ra im ginon(nuc).Tuy nhin vn c mt b phn hc sinh hiu c yu cu bi ton, ch racnhng im gin on ca biton.

o Bi 3:

Mc ch bi ny l kim tra s hiu nh ngha hm s lin tc ca hcsinh.Hu nh cc em bit hms f ( x) lin tc ti mtim 0 khi lim f ( x) = f ( x ) x x 0

a s cc em cn cha ch ra min xc nh ca hms.im kh ca bi ny l vic bin i cng thc lnggic.

o Bi 4:Mc ch bi ny l kim tra mc vn dng nh l 3 chng minh phngtrnhc nghim.

Hu nh cc em bit vn dng nh l ny chng minh phng trnh cnghim.o Bi 5:

Tng t bi 4, mc ch ca bi ny cng l kim tra mc vn dng nh l 3chng minh phng trnh c nghim. Bi ny kh hn bi 4 ch n c s thamgiaca tham sa.a s cc em bit s dng nh l chng minh phng trnh c nghim vi mia. Nhn chung qua ny, cc em nm c kin thc c bn, bit c khinohm s lin tc ti mt im, trn mt khong, trn mton. Nm c nh l 2 bit c min lin tc ca cc hm s thnggp.

Bit vn dng nh l 3 chng minh phng trnh cnghim. 2:

Nhn chung cc em cng nm c kin thc nh cc em lm 1.1.2.3 K t qu k im tra tr ng THPT Tho i N gcHu



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Lun v n t t ng h i p G V HD: Ths .N g u y n V n V nh

STT H vtn im1 Hunh Th Kim Anh 82 Nguyn Vn Cng 23 T Th Ngc Dim 5

4 Th Thy Dung 55 Nguyn Th Ngc Hn 46 Nguyn Bo Hnh 107 H Khnh Ho 58 Trn Th Ngc Hng 99 M Khc Huy 810 L Th H Linh 811 Hng Loan 8

12 Nguyn Phng Mai 913 Phm Th Mai 914 Anh M 915 L Vit Bch Ngn 816 Mai Ngn 8

17 Bi Phm Nh Nguyt 918 Phm Ngc Dung Nhi 919 Chu Minh Quang 8

20 Ng ThKim QuynThanh

3

21 Dng Ngc 522 Lan Thanh

Thanh

9

23 Trn ThL 824 H Bch Thy 825 T Th Hong Tin 926 Nguyn Hong Ngc Trm 827 Trng Hoa Trm 7

28 Phm Th M Trn 829 Kiu Trang 830 Nguyn Th Bch Trang 831 Dng Th Tuyt Trinh 8



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% h

c s

i n h


10 2 2,47 6,10 12,2

124,94

Phng sai: s2 = ni (

xi

x)2=

124,94 3, 47

1

lch chun: s1 1, 86

Tn sut:

n 1 36


0 1 2 3 4 5 6 7 8 9 10

% 0 0 2,78 2,78 2,78 11,11 0 8,32 41,67 25,00 5,56

4541.67

40

35

30

25.00

25

20

15

10

5

0.00

0

2.78 2.78 2.78

11.11

0.00

8.32

5.56

1 2 3 4 5 6 7 8 9 10

im

1.2.5 So snh kt qu:Trung bnh cng:

= xini


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1o Ln I: x = n5, 91



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n ( x x)2 394,55n 1 55

n ( x x)2 124,94

2

1

s

x

2

Lun v n t t ng h ip G V HD: Ths .N g u y n V n V nh

= xi

ni o Ln II: x =n

7, 53

Phng sai v lchchun:o Phng sai:

LnI: s2 = i i = 7, 31

LnII: 2 =n 1 36

o lchchun:

3, 47

Ln I: s1 2, 70

Ln II: s2 1, 86

1.3 Thc nghim dnh cho trng THPT M Hing1.3.1 S lc v trng THPT M Hing:

Trng THPT M Hi ng l mt trong nhng trng cng lp ca huynCh Mi. Tuy mi thnh lp t nm 2000 nhng trng cng c nhng thnh tukh iu tt. Nm no cng c hc sinh gii cp tnh ( t nht 1 hc sinh). Mc ttnghip cc nm gn y trn 70%, cng l iu ng phn khi ca mttrng

cn nontr.1.3.2 Nhnxt:1: Bi 1:

a) Mc ch ca bi ny l n li kin thc v khi nim hm s lin tc timtim, xc nh min lin tc ca tng dng hm s thng gp (nh l2).a s hc sinh bit cch xt hm s cho c lin tc tiim 0 hay khng?

Trong qu trnh tnh gii hn, hc sinh bit cch phn tch a thc thnh nhn t tthc n gin vi mu thc trong vic tnh gii hn hms.Bit khi no tnh gii hn tri v gii hn phi.im chung ca cc em l thiu ch ra min xc nh ca hms. b) Tng t nh cu a, mc ch ca bi ny l kim tra mc nm khi nimhms lin tc, lin tc ti mt im, lin tc trn mt khong, lin tc trn mton.Bit c min lin tc ca hm s khi bit c dng hm s (phn thc hut).Bi ny c s tham gia ca biu thc lng gic thay v l phn thc hu t nh cuaa s cc em bit bin i biu thc lng gic v dng c th tnh c giihn( kh dng vnh)

Bi 2:


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8 Nguyn ThXun Diu 69 Nguyn ThThy Dung 610 onM Giang 511 NguynThnh Giang 512 Nguyn ThCm Giang 5



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x n x x ( x x)2 n ( x x)2

012345

678910

0000318

94710

-5,93-4,93-3,93-2,93-1,93-0,93

0,071,072,073,074,07

35,1624,3015,448,583,720,86

0,0051,144,289,4216,56

0000

11,1615,48

0,054,5629,969,42

0

70,63


imxi

0 1 2 3 4 5 6 7 8 9 10 Tngs

Shsni

0 0 0 0 3 18 9 4 7 1 0 42

Trung bnh cng:

x = = xi n

i

n5, 93

Phng sai v lchchun:

i i i i i i

Phng sai: s2 = ni (

xi

x)2=

70,63 1, 68

1n 1 42

lch chun: s1 1, 3

Tn sut:


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74/129

1

2

% h

c s

i n h



0 1 2 3 4 5 6 7 8 9 10

% 0 0 0 0 7,14 42,86 21,43 9,52 16,67 2,38 0

50

45

40

35

30

25

20

15

10

50 0 0

0

7.14

42.86

21.43

9.52

16.67

2.380

1 2 3 4 5 6 7 8 9 10

im

1.3.5. So snh ktqu:Trung bnh cng:

= xini

o Ln I: x =

o Ln II: x =

n

= xin

i

n

5, 91

5, 93

Phng sai v lchchun:Phng sai:

LnI: s2 = ni (

xi

x)2=

394,55 7, 31

1 n 1 55

LnII: s2

= ni ( xi


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x)2=

70, 63 1, 682n 1 42

lchchun:Ln I: s1 2, 70



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1


Ln II: s 1, 30

1.4 N h n x t c h un g:Qua kt qu trn, nhn chung c 2 lp khng chuyn Ton (1 lp chuyn vnthuc

trng Thoi Ngc Hu, 1 lp c bn thuc trng M Hi ng) t c ktnhtrn th c th phn no thy c hiu qu ca phng php dy hc tch ccnichung v phng php dy hc pht hin v gii quyt vn ni ring. iv ilp 11V trng Thoi Ngc Hu th t kt qu kh hn lp 11A4 trng THPTMHi ng. Kt qu c th chp nhn c bi Trng THPT M Hi nglmt trng mi thnh lp, cc em trng Thoi Ngc Hu hiu bi su hn,vndng linh hot hn cc em trng M Hi ng bi u vo hc sinh trngMHi ng tng i thp. Nhn chung, cc em:

Bit cch xt tnh lin tc ca hm s ti mt im, trn mtkhong,mton.

nghim.

Bit xc nh min lin tc ca hm s thng qua dng hms. Bit cch vn dng cc nh l chng minh phng trnhc

2. Trc nghim giovin2.1. Tin trnh thcnghim:

2.1.1 C Nguyn Th Ngc Yn (T trng t ton -TrngTHPTM Hing)Cu hi 1: C c nhn nhn g v trnh v kh nng tip cn b mn Tonca

hc sinh trong trng? c bit l bi Hm s lin tc ?T r l i : Nhn chung trnh v kh nng tip cn b mn Ton ca hc sinh trongtrngcn thp so vi yu cu ca bc trung hc ph thng, mt l do ch yu l hnhthcthi tt nghip trung hc c s cha tht s nghimtc.Mt khc l do cht lng u vo ca hc sinh cn cha t yu cu, phn nglt trung bnh tr xung, thi quen lm bi tp, c bit l bi tp v nh cn rt tnnk nng bin i gii ton cn sai, chm, thiu chnhxc.

+ V kh nng tip cn b mn Ton th gp nhiu kh khn trongcch bin i Tonhc

+ Tip thu cn yu cc khi nim c bn, php bini,. Ngoi ra cc em nm v hiu v cc phng php gii v hm s lin tc cn m h,cha r rng. Mt phn l do khng nm vng cc kin thc bi trc, trong c bi tnh gii hn hm s, dy s., mt phn l do tnh ton c qu nhiu sai lmvgii ton cn kh chm ( chng hn nh: khng tin tng vo bi lm ca mnh),mt phn l tip nhn tri thc khoa hc v phng php gii tng loi ton cn lnln,kiu bi ny li p dng cch gii ca kiu bikia.



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Ni tm li l rt km trong vic gii mt bi ton v hm s lin tc, c bitlkhu bin i tnh gii hn hm s.Cu hi 2: Cc phng php dy hc ca gio vin trong trng hin nayhuht l theo kiu truyn thng hay tchcc?T r l i :

Cc phng php dy hc ca gio vin b mn Ton trong trng thng sdngcc phng php dy hc c in l thuyt trnh, ging giiv c l y l phng php tch cc nht i vi i tng hc sinh trngny.Tuy nhin, mt s bi gio vin c s dng phng php tch cc ging dychohc sinh, chim 35% ni dung chng trnh ging dy. V trong tng lai, phng php tch cc c l s c pht huy hnna.Cu hi 3:C c nhn nhn g v kh nng tip thu kin thc mn Ton i vi tng kiu dy hc: truyn thng v tch cc?T r l i : Ni chung phng php no cng c u im v nhc im ring da trn cthca b mn i s v Hnh hc, cc i tng hc sinh. Nhng phn ln cngtythuc vo kin thc v nng khiu ging dy ca giovin.Dy theo kiu truyn thng, theo tin trnh suy din: Hc sinh tip thu kin thcm imt cch p t, khng t nhin, thng l dn n thng.Dy hc theo kiu tch cc, theo tin trnh quy np: Hc sinh hiu bi mt cchtnhin, tip thu bi d hn trong vic lnh hi tri thcmi.Cu hi 4: C c ngh rng chng ta nn thay i kiu dy hc ca giovintrong trng haykhng?T r l i :Theo kin ch quan ca ti, trong trng nn dn dn thay i kiu dy, nndytheo tin trnh quy np. Mt s khi nim n gin th dng suy din, tuy nhin philm sao cho hc sinh hiu c bn cht ca khi nim, v cn ch trng bcnhndng v th hin khinim.Cu hi 5:C c suy ngh g v kiu dy mi : Kiu dy hc pht huy tnhtchcc ca hc sinh ?T r l i :Kiu dy pht huy tnh tch cc ca hc sinh l mt phng php dy ttv:

+ Kin thc c tnh cht ktha+ Kin thc c tnh hthng

+ Pht huy c tnh tch cc ca hc sinh, ca cc i tng hcsinhtrong lp hc, hc sinh t pht hin kin thc mi nn s nh bi luhn.Mc du vy khi dy phng php ny cn ch:

+ Cu hi phi r rng v va sc vi tng i tng hc sinh : Gii Kh Trung bnh Yu.



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.+ H thng cu hi t ra phi c tnh gi m, tn ti tnh hung cvn

+ Ngi thy phi nhn tng qut v chng trnh mn hc, c thccp di v phi c thi gian chun b chu o v h thng cuhi.

2.1.2 Thy Trn Vn S (Gio vin ton-Trng THPT MHi

ng)Cu hi 1: Thy c nhn nhn g v trnh v kh nng tip cn b mnTonca hc sinh trong trng? c bit l bi Hm s lintc?T r l i: Theo kin ch quan ca ti, trnh v kh nng tip cn b mn Ton cahcsinh trong trng cn rt yu, a s cc em b mt cn bn t lp di nn victipthu kin thc mi rt vt v, dn n s chn nn trong tm l hc sinh khi hctithc mnTon.iu quan trng l cc em rt th ng, ch bit p dng nhng bi tp tng tmkhng t duy theo hng tch cc, nhng bi khng ging dng bi v d thkhng

chu suy ngh lm bi, khng t mnh tm hiu nghin cu gii quyt vn.V biHm S Lin Tccng khng ngoi l, hc sinh ch bit p dng li cc bitp khun mu, m khng nm vng v phng php chung gii cc bi tonhms lin tc, cc em khng bit bin i t dng cha bit v dng quenthuc. Ni vy, mt phn l do cht lng u vo cn qu thp, nhiu gio vin cnkhtr v cn nhiu hnch.V trnh : ni chung hc sinh trong trng hc cn yu v mn ton ni ringvcc mn t nhin nichung.Cu hi 2:Cc phng php dy hc ca gio vin trong trng hin nay huht l theo kiu truyn thng hay tch cc?

T r l i :Theo ti, phng php dy hc ca gio vin hin nay hu ht i theo xuhngtruyn thng (ch mi mt phn no theo hng tch cc). i khi trong mt bi phi kt hp c 2 phng php truyn thng ln tch cc h tr cho nhau. Vcn phi ty vo i tng hc sinh, ty vo trnh v kh nng tip thu ca hcsinh c cch dy cho phhp.Cu hi 3:Thy c nhn nhn g v kh nng tip thu kin thc mn Ton i vi tng kiu dy hc: truyn thng v tch cc?T r l i : Nu dy theo phng php truyn thng, tc l thy c, tr chp v hc theonhngg thy dy s lm cho hc sinh th ng, khng pht huy c tnh ch ng,tchcc, sng to ca hc sinh. iu lm cho hc sinh c tm l th ng, li,t iu tnh ti , nhiu khi cc em khng hiu, mun hi nhng li khng dmhi.iu ny lm cho cht lng ging dy ngy cng i xung mt cch vhnh.Cn dy theo phng php tch cc, v cc em hot ng l ch yu, cc em c thtmnh khm ph ra nhng kin thc mi thng qua nhng cu hi, gi hngdnca gio vin, t to nn s hng th, m rng kh nng t duy sng to chohc



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sinh, v nh th cht lng hc tp v ging dy s tt hn. i i vi phng phpny th gio vin cng phi ch n nhiu vn trong h thng cu hi phikch thch c hc sinh suy ngh, phi c tnh hung gi vn t ci c hcsinh c th khm ph racCu hi 4: Thy c ngh rng chng ta nn thay i kiu dy hc ca giovintrong trng haykhng?T r l i :Theo suy ngh ca ti, chng ta nn tng bc thay i kiu dy hc ca giovin,nn a ra phng php dy hc mi, c kh nng pht huy c tnh tch ccvtnh sng to ca hc sinh trong qu trnh tip nhn tri thc khoa hc mi. Tc llyngi hc lm trung tm gingdy.Theo ti phi i mi nhng phi i mi dn lng ghp mt s bi hc c bnvd suy din, khi hc sinh quen dn th ta i mi dn kiu dy l iu haynht. Nu ta thay i qu t ngt th cng lm cho hc sinh kh hiu hn, khngthchnghi kp phng php mi, iu hon ton khngnn.Cu hi 5: Thyc suy ngh g v kiu dy mi : Kiu dy hc pht huy tnhtchcc ca hc sinh ?T r l i :Theo nh gi ch quan ca ti, kiu dy hc mi pht huy tnh tch cc ca hcsinhl kiu dy hc rt tt, rt ph hp. Nhng phi ty vo trnh v kh nng tipthuca hc sinh tng lp m dy cho hpl.i vi hc sinh min nng thn (nh a phng ti ang dy) th ta khngnni mi tt c mt cch nhanh chng m phi lm dn dn m tt nht l i mitlp di, tp cho cc em c thi quen hc tch cc t trc. Nht l lp 10, tannlng ghp mt s bi dy c tnh tch cc v tng dn ln cho cc nm sau thhcsinh tip thu tt hn, v s t nh hng n tng i tng hc sinh. Bng cchno

lm cho mt bng hc sinh ngy cng ng u th vic s sng phng phpdyhc mi d pht huy tc dnghn.Cu hi 6: Theo kin ca thy th chng ta c nn s dng phng phpdyhc tch cc (chng hn nh phng php dy hc pht hin v gii quyt vn)ca hc sinh trong ging dy bi hm s lin tc hay khng? V n s emli tc dng nh th no i vi kh nng tip thu tri thc khoa hc ca hcsinh?Tr li:Ti ngh nn a phng php ging dy tch cc vo dy hc l rt hay, iv iton b chng trnh ton hc ni chung v i vi bi Hm s lin tcni ring. N s gip cho hc sinh hiu v nm vng hn v vic xt tnh lin tc ca hmskhi no xt trn khong, khi no xt trn on, khi no xt ti im v vn dngtnhlin tc trn on ca hm s chng minh phng trnh cnghim.

2.2 Phn tch trc nghim giovin Nhn chung cc gio vin tn thnh vic vn dng phng php dy hc nhm phthuy tnh tch cc ca hc sinh. Nhng tt c u c mt im chung l nn ch ntrnh ca hc sinh khi p dng phng php dy hcny.



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Bn cnh , cc gio vin cng nu ln nhng iu kh khn khi thc hin phng php ny, v n kh ph hp vi trnh hc sinh lp mnh trc tip gingdy.Tuy nhin, v vn c nn p dng hay khng vo vic a phng php dyhc pht huy tnh tch cc ca hc sinhvo ging dy mn Ton ni chung v biHmS Lin Tc(Chng trnh Gii Tch lp 11- SGK chnh l nm 2007) ni ringthhu ht cc gio vin u tn ng vic tn dng phng php ny, v cho rngnkh ph hp vi mi i tng, gp phn tch cc trong vic gip cho hc sinhhiu bi v nm r phng php gii cho tng kiu bi v tnh lin tc ca hm s,cngnh trong vic nhn dng v c cch gii ph hp gii quyt bi ton mtcchnhanh chng v chnhxc.V. Gio n gingdy:

GIO N

BI HM S LINTCi tng hc sinh:kh

Ngy son:10/03/2008Lp : 11

I. Mc c h ,yucu :1. V kinthc:

Gip hc sinh nm c khi nim hm s lin tc ti mt im, hm s lintctrn mt khong, cc nh l ca hm s lintc.Gip hc sinh bc u hnh thnh c biu tng trcquan

v hm s lin tc ti mt im, nm c phng php s dng hm s lin tcchng minh mt phng trnh cnghim.

2. V k nng v t tng:Gip hc sinh rn luyn k nng kho st s lin tc ca hm s ti mtim,

mt khong v chng minh phng trnh cnghim.Hc sinh rn luyn c tnh cn thn, tnh k lut trong tnhton.3. V t duy:Rn luyn t duy thut ton, t duy bin chng cho hcsinh.4. Trng tm cabi:Khi nim hm s lin tc ti mt im, hai nh l u v hm s lin tcv

ng dng hm s lin tc trong vic chng minh phng trnh cnghim



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II. P h ngphp , p h ng t i n d y h c,c h un b c a gio vin v h cs in h :Gio vin: ch yu l phng php vn p gi m c xen phng phpthuyt

trnh. Phng php vn p gi m l trngtm.Phng tin: thc thng, phn mu, bng ph, gion. Hcsinh:Xem li kin thc ca bi gii hn hm s v cch tnh gii hn hms.

III. Ti n trnh :



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Thilng Hot ng cathy Hot ng catr

Ni dung lubng

1 pht

10 pht

I.n nh l p

II.K i m tra bi c :

Hot ng 1:Gio vin chia bng lm4, gc bng bn phi, cho 3hms:

x ne u x > 2 y = f 1 ( x) = x + 2 ne u x 2(C 1)

x2 ne u x 2 y = f ( x) =

(C 2 ) y = f ( x) = x2 (C )3 3

Tnh cc gi tr sau (nuc): f = m f ( x) = ?

x

f = m f ( x) = ? x

Mt hc sinh ln bng invo: f 1(2) = 2, lim f 1( x) = lim x = 2

x 2 x 2m f ( x) = lim( x + 2) = 4

x

f 1 ( x) khng c gii hnkhi x 2

f (2) = 1

lim f ( x) = lim x2 = 4 x 2 2 x 2

f 3(2) = 4

lim f ( x) = lim x2 = 42 3 2 x x

Cu tr li mong c cahcsinh : thC3 l mt nglin.Hai th cn li khng lmt

(Phn th v cc thng snyhc sinh khng phi ghi vov)

yy=x +2

4

2

O 2 xy=x

f (2) = 2, lim f ( x) = lim x = 2 x 2 x 2

lim f 1( x) = lim( x + 2) = 4 x 2+ x 2+

2

Lun vn tt nghip GVHD : Ths Nguyn VnVnh

Tit 1

li 12+ x 2+

2

1 ne u x = 2

1(2) ?, li 12 1 1

3 (2) ?, li 32



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f = m f ( x) = ? x

Sau khi hc sinh tnh ccktqu, gio vin a ra 3 bng ph ca 3 hm strn.

Yu cu hc sinh quan st 3th v nn nhn xt v 3 thti im c honh x0 = 2(Nu hc sinh gp lngtng,tac th gi : th ca hms c lin ntkhng?...)

Gio vin ch tng th,davo kt qu hc sinh tnhc gii thch cho hc sinh bitv sao th lin nt hayngtqung ti im c honh bng2.

Sau gio vin a raktlun : trong trng hptrnngi ta ni rng hms f 1 ( x), f 2 ( x) khng lin tcti im x = 2, cn hm s f 3 (

x) l lin tc ti im x =2.Vy nu hm s f ( x) xcnhti x = 2 th hm s f ( x) lintc ti im x = 2 khino?

ng lin, b ngt qungti

im c honh x0 = 2 .

Cu tr li mong c cahcsinh: hm s lin tc tiim x = 2 khi lim f ( x) = f (2)

x 2

Hm s y = f ( x) lin tctimt im 2 thuc tp xcnhca hm s khi tac:lim f ( x) = f (a ) x a

f ( x) khng c gii hn khi x 2

y f 2(2) = 1 (C2)

lim f 2( x) 4 B (2;4) x 2= lim x2 = 4

x 2

1 A (2;1)

O 2 x

y(C3)

4 B (2;4) f 3(2) = 4lim f 3( x) x 2= lim x2 = 4 1 A (2;1)

x 2

O 2 x


2(2) ?, li 2 12



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7 pht

Mt cch tng qut, emhynu nh ngha mt hms y = f ( x) lin tc ti mtima thuc tp xc nh cahms?

III.Vo bi m i:

Hot ng 2:nh ngha khinim:Gio vin vit tm ttnhngha ln bng.

Vy mt hm s lin tc tintim x0 khi c cc iuking?

Suy ra hm s khng lintcti x0 khino?

Khi c lim f ( x) v x x0

lim f ( x) = f ( x0 ) x x0

Khi f ( x) khng c gii hnkhi x x0 hoc c f ( x0 ) vlim f ( x) nhng

x x0

lim f ( x) f ( x0) x x0

Bi 5: HM S LINTC

I.Hm s lin tc ti mtim:Cho hm s y = f ( x) xc nhtrnkhong (a ; b) , x0 ( a; b ) . Ta nhngha:a) Hm s f ( x) gi l lin tctiim x0nu : lim f ( x) = f ( x0) .

x x0

* f ( x) lin tc ti im x0Ton tai x x f ( x)lim

0lim f ( x) = f ( x0) x x0

Suy ra f ( x) khng lin tc tiim

x0 nu xy ra mt trong hai iukinsau:- Hm s f ( x) khng c gii hnkhi x x0 (C2)

-C f ( x0 ) v lim f ( x) nhng x x0

lim f ( x) f ( x0 ) (C3) x x0



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3 pht

5 pht

Hot ng 3: Cng ckhinim:

Vy hm s f ( x) =1

lin tc x

hay gin on ti x = 0?Hm s f ( x) = x c lintchay gin on ti 0 hay khng?

V d1:Trc khi hc sinh lm bi,gio vin t cu hi:vymunxt tnh lin tc cahm

s ti im, ta lm thno?

Cu tr li mong c lkhngth xt tnh lin tc ca hmsny ti 0 v 0 khng thuctpxc nh.Tuy nhin s rtnhiuhc sinh tr li khng lintc.

Cu tr li mong c lkhngtr li c v 0 khngthuckhong xc nh.

-Tnh f ( x0 )- Tnh lim f ( x)

x x0

- So snh f ( x0 ) v lim f ( x) x x0

b)Hm s khng lin tc ti x0 thc gi l gin on ti im , x0c gi l im ginon.

Ch : Hm s f ( x) = 1 , f ( x) = x x

khng c khi nim lin tc tiim

x = 0.



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Thilng Hot ng cathy Hot ng catr Ni dung lubng5 pht Cc nh l:

Trong bi gii hn hms,ta c cc nh l vgiihn ca tng, hiu,tch,thng ca cc hm scgii hn khi x x0

, tngt nh vy, em no cthd on v tnh lin tccatng, hiu, tch, thngcacc hm s lin tc tiim x = x ?

T nh l 1, ta c nh l2.Cc nh l ny ta thanhnm khng chngminh.Vn dng nhl

V da:

Tr li:Tng, hiu, tch, thngcacc hm s lin tc tiim

x0 cng lin tc tiim

x = x

IV.Cc nh l:Ta tha nhn khng chng minh cc nhlsau:

nh l1 :Tng, hiu, tch, thng (vi mu khc 0)canhng hm s lin tc ti mt im l lintcti im.

nh l 2 :Cc hm s a thc, hm s hu t, hmslng gic l lin tc trn tp xc nhcachng


Ti t 2:

0

0



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20 pht Hm s ny dngg?Vy trong hai nh l trn,tavn dng nh l no xttnh lin tc ca hms?Gio vin hng dnchohc sinhgii

Hm s ny c nhnghatrn hai min:(;1) ,(1; + ) v ti im x = 1.Mun xt tnh lin tccahm s ny, ta phi xttnhlin tc ca hm s trnmymin?Gio vin cho hc sinhngti ch xt tnh lin tccahm s trn(;1) , (1; + ) bng cch vn dng nhl2.Ti im x = 1, ta xttnhlin tc ca hm s nhthno?Gio vin cho hc sinhln bnglm.

Thng ca hai hm athcnh l2

Ta c hm s xc nhtrnhai min, nn ta xt tnhlintc trn tngmin.

Ta dng nh ngha hmslin tc ti mtim.

Vd:a)Xt tnh lin tc v tm im gin on(nuc) ca hm s sau:

x + 1 f ( x) =

x 2 + 2 x 3Gii:TX: D = \ {1; 3}

x ( ; 3), (3;1), (1;+ ) , f ( x) xcnh

f ( x) lin tc x \ {3;1}Kt lun: f ( x) lin tc trn \{3;1} ,khngc im ginon. b) Xt tnh lin tc ca hm ssau:

x 2 + x 2 x 1

( x > 1)

f ( x) = 2 x ( x < 1)3 ( x = 1)

Gii:TX: =

x 2 + x 2 x (1; + ) , f ( x) = xcnh x 1

f ( x) lin tc trn (1; + )



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15 pht

Qua hai v d trn ta cktlun chung g? (Giovinlu hc sinh l xt hmslin tc trnon)

Vy mun xt tnh lintcca hm s, ta lmnhng bc no?( gio vin niliv d 1 v v d 2 hcsinh tng qutln)

ng dng:Cho mt hm s c thnh hnh bn. Em cnhnxt g v tnh lin tcca

hm s trn on a; b ?

Yu cu hc sinh tm trnth im no c tung lnnht v nhnht?

Ngi ta chng minhcrng, nu mt hm slin

Ta tm tp xc nh cahmsTrn tng khong xc nhtadng nhlTi cc u mtcakhong , ta dng nhnghahm s lin tc ti mtimSau cng l kt lunchung

Hm s ny lin tctrn

on

im B c tung lnnht,im A c tung nhnht.

f (1) = 3 x 2 + x 2lim f ( x) = lim = lim( x + 2) = 3

x 1+ x 1+ x 1 x 1+

lim f ( x) = lim2 x = 2 x 1 x 1+

lim f ( x) = f (1) lim f ( x) x 1+ x 1Vy hm s khng lin tc ti1Ktlun:

Hm s lin tc tn( ;1

) v (1; + )

Phng php xt tnh lin tc ca hms:- Tm tp xc nh ca hms- Trn tng khong xc nh ta p dnghainh ltrn- Ti cc u mt ca khong, ta dngnhngha hm s lin tc ti mtim- Tng kt cc khong/on lintc

V.ng d ng c a hm s lin t c:



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1 2


tc trn mt on a; b

thn t gi tr ln nht vnhnht trn on .Tathanhn nh l sau mkhngchngminh.

Theo nh l trn, nu tac

y

BM

Trc honh ct th ti3

im

f (a). f (b) < 0 (nh hnhv

bn) ta c nhn xt g vvtr ca th v trchonh?

iu cng c nghal a O b phngtrnh f ( x) = 0 lun

c nghim.Ta c c h xqu . m

A

Theo h qu, nu hms f ( x) lin tc trn a; b v f (a). f (b) < 0 th chng nh l: SGK

minh c phngtrnh f ( x) = 0 c t nht1 Nu hm s y = f ( x) lin tc trn on a; b

nghim trn khong(a; b ) .Vy mun lm bi tontrn,

Hai ln trn hai khongrinhau

v f (a). f (b) < 0c (a; b ) saocho

th tn ti t nht mtim f (c) = 0 .

ta phi vn dung h quny

my ln v trn nhngon

x , x


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phng trnh cnghim.Bi tp v nh: bi tpschgio khoa v h thng bitpsau.

(2; 0) v 1 nghim thuc(0; 2) .Vy phng trnh lun c t nht 2nghimtrn (2; 2)



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Bi tp Mc ch

Bi 1:

Xt tnh lin tc ca cc hm s sau ti im x0 x 2 + 5 x 6

a) f ( x) = x 1 n eu x 1 ti x = 17 n eu x = 1 x

b) f ( x) = ti x0 = 0 x + 1 x + 1 neu x < 0

c) f ( x) = 0 neu x = 0 ti x = 00 x 2 + 1 neu x > 0

x 2 + 4 x 4 n eu x 1d) f ( x) = x

2 ne u 0 x < 1 ti x = 100 ne u x < 0

Hc sinh bit xt tnh lin tc

ca hm s ti mt im vcng c hai du hiu khi hms khng lin tc ti mt im.ng thi, cc bi tp cngtri rng dng hc sinhn tp cch tnh gii hn cahm s.

Bi 2:Xc nh hng s a cc hm s sau y lin tc tiim x0 :

ax2

+ 1 ne u x 2a) f ( x) = 3 ti x = 2 2 x + 4 2 0neu x > 2

x 2 4 x 2 + 2 x

b) f ( x) = x 2n eu x 0 ti x = 00

a n eu x = 0 sin x cos x

n eu x

c) f ( x) = 4 x 4 ta i x =

0 + a n eu x =

4

2 41 cos x.co2 x

neu x 0d) f ( x) = x 2 ti x = 00a n eu x = 0

Hc sinh luyn tp bi tonc cha tham s v tnh giihn hm s lng gic.

Lun vn tt nghip GVHD: Ths. Nguyn Vn Vnh

0



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Tm im gin on ca cc hm s sau:a. y = 4 x 2 + x 2

x 2 + 3 x + 5 b. y = x

3+ x 2

11 + 2 x 4 3 x + 2 x 2 n eu x > 1c. f ( x) = x

2 + 4 x + 3 3 ne u x 1

4

x 2 16d. f ( x) = x 4

n eu x 4

8 neu x = 4

nim im gin on v nh im gin on trc tin phi l mt im thuc tpxc nh ca hm s

Bi 6:Chng minh rng phng trnh :a. 4 x 4 + 2 x 2 x 3 = 0 c t nht hai nghim

phn

bit trn(1;1)

b. x 5

5 x 2

+ 4 x 1 = 0 c ng 5 nghim

c. 2 tan4 x + 2 x 3 = 0 lun c nghim trong0;5

Hc sinh luyn tp k nng vndng hm s lin tc chngminh phng trnh cnghim.Cu c mc ch chcho hc sinh thy sai lm nu

chn a = 0 v b =5

l sai6

v trn 0; 5

hm s

6

khng lin tc ti x =

2

Bi 7:Chng minh rng cc phng trnh sau lun cnghim vi mi m:a. m tan

4 x + 4 tan x m = 0

b.1

1

+ m = 0cos x sin x

c. x 7

+ 2 x m = 0

Bi 7, 8, 9 cho hc sinh thyc nhiu cch chn a, b vchng minh f (a). f (b) < 0hoc f (a). f (b) 0 sao choc th vn dng c h qu.


6



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Bi 8:Chng minh phng trnh

x 2 1 + m( x + 1)( x 2)( x 4) 2 = 0 lunc nghim trong(1; 4)

vi mi m.

Bi 9:Chng minh rng phng trnh sau c t nht 1nghim:ab ( x a)( x b) + ac( x a)( x c)

Bi 10:a.Chng minh rng nu f ( x) lin tc trn v phng trnh f ( x) = 0 v nghim th f ( x) khng th

i du trn . b.Cho f ( x) lin tc trn v phng trnh f ( x) = x v nghim.Chng minh rng phng trnh f f ( x ) = x cng v nghim

Bi 10 cho hc sinh thyc mt ng dng khc cahm s lin tc da trn cckin thc hc: nu phngtrnh f ( x) = 0 v nghimtrn 1 khong th hm s y = f ( x) mang mt du trnkhong .

Trn y ch l mt phng n dy hc hm s lin tc theo hng tch cc cho hc nhin trn c s chng ti va nu s cn nhiu phng n khc ty thuc vo ttng hc sinh. H thng bi tp trn c mt s cu kh, gio vin c th a vkhng a vo ty theo tng n v lp.



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o Bc 2: Gii quyt vn o Bc 3: Kim tra vn dng

6. Cc bin php s phm tng thch gip gio vin thc hin qui trnh nhtnh kh thi v hiu qu ging dy.

II. Cc kt qu thu nhn c cho php chng ti kt lun:1. C th tch cc ha hot ng nhn thc, bi dng nng lc pht hin v

quyt vn , ng thi pht huy tnh tch cc ca hc sinh.2. Nhng nghin cu v qu trnh tin hnh thc nghim cho thy gi thit k

hc ca lun vn l chp nhn c.3. Lun vn c th dng lm ti liu tham kho cho gio vin ging dy mn T

trng ph thng.



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PH LC(Cc kim tra v p n)

1. kim tra v p n ca lp 11A3 trng THPT Nguyn Cng Tr

1 :

Bi 1:Xt tnh lin tc ca hm s:

x2 + x 2 nu x >1

f ( x) = x3 1

Bi 2:ax +

1

nu x 1

Xt tnh lin tc ca hm s v tm im gin on nu c: x3 + 1

f ( x) =

Bi 3: x

2 x 2

Tm im gin on ca hm s:

x + 4 3 x + 10 nu x > 2

g ( x) = x2 5 x 6+ +

x + 1 nu x 2

Bi 4:Tm a hm s lin tc ti x = 01 cos x


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h( x) = x2a nu x 0

nu x = 0

Bi 5:Chng minh rng phng trnh :a cos4 x 2 x a + 1 = 0 lun c nghim vimi a.

p n v thang i m:Bi 1: (2,5 im)

x (;1) : f ( x) = ax + 1 xc nh f ( x) lin tc trn (;1)



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x (1;+ ) : f (

x) =

Ti x = 1: f (1)= a + 1

x2 + x 2

x3

1xc nh f ( x) lin tc trn (1;

+ )

0,5

0,25

lim f ( x) = lim x2 + x 2

= lim x + 2 = 1 0,5

x 1+ x 1+ x3 1 x 1+ x2 + x + 1

lim f ( x) = lim(ax + 1) = a +1

0,25

x 1

Bin lun: x 1

a = 0 : f (1)= lim f ( x) = lim f ( x)=

f (1)

x 1+ x 1

Hm s f ( x) lin tc trn 0,5

a 0: f (1)= lim f ( x) lim f ( x) x 1 x 1+

Hm s khng lin tc ti x = 1Kt lun:Hm s lin tc trn( ;1), (1; + ) 0,5

Bi 2 :(1,5 im)Tp xc nh: D = \ { 1; 2}

x3 + 1

0,5

x \ { 1; 2} : f ( x) =

x2 x 2lin tc 0,5


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0,75

0,5

2:Bi 1:

Xt tnh lin tc ca hm s:

x3 + x2 + x 14

nu x >2

f ( x) = x2 3 x 10

Bi 2:

+

7 x + 3 nu x 2

Tm im gin on ca hm s:



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a


1+ 3 x 2 x2 + 5 x 3 nu x >1

g ( x) =

8 x + 2 x

2 1

nu x 1

Bi 3:Xt tnh lin tc ca hm s v tm im gin on nu c:

x3 8 f ( x) =

Bi 4: x

2+ x 6

Tm a hm s h(x) lin tc ti x = 0:

cos2 x 3cos x +2

nu x 0

h( x) = 2 x2nu x = 0

Bi 5:

Chng minh phng trnhm sin 4 x 2 x m + 1 =0

lun c nghim vi mi m.

p n v thang d i m:Bi 1: (2,5 im)

x (; 2), f ( x) = 7 x + 3 xc nh

f ( x)

x3 + x2 + x 14

lin tc trn (; 2)

x (2; + ) : f (

x) =

Ti x = 2:

f

(2)=

17 x

2+ 3 x 10


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x (;1) : g ( x) = 8 x + 2 xc nh g ( x) lin tc trn

(;1)

xc nh

g ( x) lin tc trn (1;+ )Ti x = 1: g (1)= 10

0,5

0,25

lim g ( x) =lim

1 + 3 x 2 x2 + 5 x 3

= lim 2 x2 2 x + 4

x 1+

= lim

x 1+ x2 1

2 x 4 x 1+

( x2

1)( 1+

3 x +

= 3

2 x2

+ 5 x 3)

0,5

x 1+ ( x + 1)( 1+3 x +

2 x2 + 5 x 3) 4

lim g ( x) = lim(8 x + 2) = 10 lim g ( x)

0,25

x 1 x 1 x 1+

Vy hm s gin on ti x = 2. 0,5

Bi 3: ( 1,5 im)TX : D = \ { 3;

2} x3 8

0,5

x \ { 3; 2} : f (

x) = x2 + x 6

lin tc 0,5

Hm s khng c im gin on (do 3, 2 D ) 0,5


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Bi 4: (1,5 im)h(0)= a

cos2 x 3cos x + 2 (cos x 2)(cos x 1)

0,25

limh( x) =lim = lim

x 0 x 0 2 x2 x 0 2 x2

2 sin2 x (cos x

2)= lim 2 =

1

0,75

x 0 2 x2 4

Hm s h(x) lin tc ti im x = 0 limh( x) = h(0) x 0

a = 14

0,25

0,25

Bi 5: ( 2 im)



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a

nu x 1

Bi 3:Tm a hm s sau lin tc ti x = 0

h( x) =1 + xsin x cos

x x2

nu x 0

nu x = 0

Bi 4: Chng minh phng trnh tan4 x + x 2 = 0 lun c nghim trong (0;5

)6

Bi 5: Chng minh rng phng trnh sau lun c nghim vi mi a

x5 + x a = 0



111/129

a


2:Bi 1:Xt tnh lin tc ca hm s sau:

2 x3 x2 + 2 x

1

f ( x) = 2 x3 + x2 + x 1 5

nu

nu

x 12

x =1

9 2

Bi 2:Tm im gin on ca hm s sau:

1 1+ 2 x

4 3 x +2 x2

nu x >-1

+ g ( x)

=

Bi 3:

3

4

x2 x 2

nu x -1

Tm a hm s sau lin tc ti x = 0

1 tan x

1 + tan x

nu x 0

h( x) = sin 2 x nu x = 0

Bi 4: Chng minh rng phng trnh 2 tan4 x + 2 x 3 = 0 lun c nghim trong (0;5

)6

Bi 5: Chng minh rng phng trnh sau lun c nghim vi mi m

x5 + 2 x m = 0

p n v thang i m 1 :

Bi 1 : (2 im)


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x \ {2} : f (

x) =

x3 2 x2 + x

23 2

xc nh lin tc trn \

{ 2}

0,5

Ti x = 2: f (2)=

57

x x x 2

0,25



113/129


lim f ( x) = lim x2 + 1 5

= = f (2) 0,5

x 2 x 2 x2 + x + 1 7

Vy f(x) lin tc ti 2 0,5Kt lun: f(x) lin tc trnR 0,25

Bi 2: (2 im)

x > 1: g ( x)

=

2 + x + x2 x2 + x 2

3 +

x

xc nh g(x) lin tc x > 1

x < 1: g ( x) =13

lin tc g(x) lin tc x 03 3

f (0). f (

) 1a) Xt tnh lin tc ca hm s f ( x) = x

1ti x0 = 1

x2 + 2 x + 2 nu x 1

1 cos x.cos 2 x nu x 0

b) Xt tnh lin tc ca hm s g ( x) = x 5 nu x = 0

ti x0 = 0

4Bi 2:CMR cc phng trnh sau lun c nghim :

a) a cos4 x 2 x a + 1 =

0

2:Bi 1 :

b) x5

+ x a = 0

a) Xt tnh lin tc ca hm s

f ( x) =3 2cos x

x2 2 cos

x

nu x 0

nu x = 0 ti x0 = 0

b) C gi tr no ca a hm s

2 x3 x2 1


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p n v thang i m: 1:Bi 1:a) (2,5 im)

x2 + x 2 ( x 1)( x + 2)lim f ( x) = lim = lim = lim ( x + 2)= 3 0,5 x 1+ x 1+ x

1 x 1+ x

1 x 1+

lim f ( x) = lim( x2 + 2 x + 2) =5

0,5

x 1 x 1

lim f ( x) lim f ( x)

0,5

x 1+ x 1

f(x) khng c gii hn khi x 1 0,5

Vy hm s f(x) khng lin tc ti ti im x = 1. 0, b) (2,5 im)

lim g ( x) = lim1 cos x.cos 2 x

= lim1 cos x + cos x cos x. cos 2 x

x 0 x 0 x22 x

x 0 x22 x

= lim0

2 sin

x22 + lim

x 0

cos x(1 cos2 x)

x2= lim

x 0

2 sin2

x2+ lim

x 0

cos x.2 sin2 x x2

1,5

x

44

= 1 + 2 = 52 2

5= lim g ( x) g (0)=

50,5

2 x 0 4Vy hm s g(x) gin on ti 0 0,5


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Bi 2:a) (2 im)t f ( x) = acos4 x 2 x a +

1

0,5

f(x) lin tc trn 0,5 f (0)= 1 0,25

f ( ) = 2 + 1

f (0). f ( ) = 2 + 1 < 0,a

0,25

0,25

Vy phng trnh cho c nghim trn(0; ) , vi mi a. Suy ra pcm 0,25



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b) (2 im)t f ( x) = x5 + x

a

0,5

f(x) lin tc trnR 0,25f(0).f(a) = -a6,

a

0,75

Vy phng trnh cho c nghim vi mi a 0,

2;Bi 1:a) (2,5 im)

lim f ( x) = lim3 2cos x 2 cos x

= lim1 cos x

x 0 x 0 x2

2 sin2 x x 0

x2( 3 2cos x

+

2 cos x

)1,5

= lim 2 = 2 = 1

x 0 x2

4 ( 3 2cos x +4

2 cos x

)

4(1+ 1) 4

1= lim f ( x)

4 x 0 f (0)= 1 0,5

Vy hm s f(x)gin on ti 0 0,5 b) ( 2,5 im)g(1) = a 0,5

lim g ( x) =

lim


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2 x3 x2 1 ( x 1)(2 x2 + x + 1)= lim

x 1 x 1 x3 4 x2 + 5 x 2

x 1 ( x 1)( x2 3 x +2)

1,0

2 x2 + x + 1 2 x2 + x + 1= m = lim =

li1 x2 3 x + 2

1 ( x 1)( x 2)

x x

g(x) khng c gii hn khi x 1 0,5

Vy khng c gi tr thc no ca a hm s g(x) lin tc. 0

Bi 2:a) (2,5 im)t f ( x) = ms in4 x 2 x m +

1

0,5

f(x) lin tc trn R 0,5

f (

) = m m + 1 = 1

2

0,25



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f (

) = m + m + 1 = 1

+ 2

0,25

f ( ). f ( ) = 1 2 < 0 vi mi m. 0,52 2

Vy phng trnh cho lun c nghim trong

(

;

) vi mi m. 0,52 2

b)t f ( x) = x3 + 2 x

m

0,5

f(x) lin tc trnR 0,5 f (0)= m

m m3 f ( ) =2 8

m4

1,0

f (0). f (m ) = 0 ,vi mi m.2 8

Vy phng trnh cho c nghim vi mi m 0,5


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129/129


TI LIU THAM KHO

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#

Vn Thng : Gio trnh Tm L Hc i cng Gio trnh Tm L Hc S Phm v La Tui Phng php nghin cu khoa hc gio dc

Hong Chng _ Sng To Ton Hc NXB Gio dc 1998 L Thnh t _ Lun Vn Tt Nghip 2004 GVHD: Ths Nguyn Vn V Nguyn Hi Chu Nguyn Th Thch Phm c Quang Gii thiu n Ging dy Ton 11

Nguyn B Kim V Dng Thy Nhng Xu Hng Dy Hc KhTruyn Thng 1993

T Th Hong Lan _ Lun Vn Tt Nghip 2004 GVHD: Ths Nguyn VVnh

Trn B Honh Ths L Trn nh TS.Ph c Ha _ p Dng DyHc Tch Cc Trong Tm L Hc Gio Dc Hc

Trn Vn Ho (tng ch bin), V Tun (ch bin) _ i s v Gii 11.NXBGD 2007

Vn Nh cng Ng Thc Lanh Trn Vn Ho _ Ti liu hng ging dy Ton 11

Vng Vnh Pht _Gio trnh L Lun Dy Hc 2007

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