Li cam n Trong suot 1 thang tin hanh lam d tai , chung toi cacthanh vin nhom 13, d thuc su gp rat nhiu kho khn. Truoc ht, vi la lan dau tin duoc thuchanh lam tron ven mot d tai khoa hoc, hon na thoi gian lam d tai tuong doi ngn nn cac thanh vin trong nhom chua thuc su khai thac duoc mot cach trit d cac noi dung ma d tai nghin cuu. Tuv vav, duoc suhuongdanvachidavcuagiangvinVDgiangvinhocphan cuakhoatoantruongHSPHu,nhomdbit duoc phan nao cach hoan thin tron ven cau truc cua mot d tai khoa hoc.ong thoi, nhomcngxingiloicamonchanthanhdncacgiaovinvaHSlop11A2truong THPT Quoc Hoc d giup d nhom trong vic diu tra thuc nghim. c bit, nhng loi nhan xet, tu van cua cac sinh vin khoa toan truong HSP Hu, thuc su la nhng dong gop quv bau giup ca nhom hoan thin d tai mot cach hoan chinh nhat. Khi tin hanh thuc hin d tai khoa hoc nav, nhom d tham khao kha nhiu tai liu, trong do co luan vn ' Ren luvn nng luc giai toan cho hoc sinh PTTH thong qua vic phan tich va sa cha cac sai lam cua hoc sinh khi giai toan` do Tin si L Thong Nhat nghin cuu. Co th noi dav la mot tai liu ' xuong cot` cua nhom. Ji vav nhom cng xin duoc gi loi cam on sau sc dn Tin si vi nhng nghin cuu nav.Ja nhom cng xin gi loi cam on dn thanh vin cac din dan: www.tailieu.vn , www.baigiang.violet.vn , www.mathgroup.org , www.thucvientoanhoc.net .... d cho nhom nhng v kin thuc su hu ich. Cuoi cung, nhu d noi tu dau, d tai duoc thuc hin trong mot thoi gian ngn voi nhng thanh vin van chua thuc su co kinh nghim nn kho tranh khoi cac sai xot. Ji vav nhom thuc su mong cac doc gia thong cam va thng thn gi cac v kin dong gop toi cac thanh vin cua nhom 13. ST.01649821769 Mail: Ledethuong.tt@gmail.com. rST.01649826097 Mail: Tungocsp510@gmail.com. r ST.01674718379 Mail. Tranquangsp@gmail.com. u 6011 Nhom 1 : 2011 2 Muc luc I. do chnti............. 5 Muc ch nghien cu........... 7 II.Nhim vu nghien cu............7 III.hch th v i tng nghien cu...................................................7 IV.i thut ho hc...........7 V.hng php nghien cu...........8 VI.u trcti..................8 VII. hng I: Nghien cu v cc si lm ph bin cu S hi gii ton A. c si lm ph bin A. 1. Silm hi bin i cng thc...9 A. 2. Silm hi gii phng trinh.......................................................................10 A. 3. Silm hi chng minh BT11 A. 4. Silm hi tim gi tri Mx Min1 A. 5. Silm hi gii tm thc bc hi.........................13 A. 6. Silm hi hi gii h pt..................................14 A. 7. Silm hi tnh gii han............................................14 A. 8. Silm hi gii ton lien qun n ao hm..................................15 A. 9. Silm hi xet bi ton tip xc v tip tun15 A. 10. Silm hi xet cc ng tim cn16 A. 11. Silm hi gii ton nguen hm v tch phn17 B. hn tch nguen nhn dn n cc si lm cu S hi gii ton B. 1. Nguen nhn 1: iu hng u v chnh xc17 B. 2. Nguyn nhn 2: hng nm vng cu trc lgic.....21 B. 3. Nguen nhn :Thiu in thc cn thit v lgic....24 B. 4. Nguen nhn 4: S hng nm vng phng php gii.............26 hng II: c bin php ren lun nng lc gii ton thng qu vic phn tch v s ch cc si lm cu S TT hi gii ton A.s l lun A. 1.lun v phng php da hc9 : 2011 3 A. 2. Nhng vnc bn cu vntm l da hc1 B. B phng chm chi ao B. 1. hng chm 1: Tnh ip thi B. 2. hng chm : Tnh chnh xc4 B. 3. hng chm : Tnh gio duc5 C. Bn bin php s pham chu u C. 1. Bin php 1: Trng bi u chnh xc6 C. 2. Bin php : Trng bi cc in thc47 C. 3. Bin php : S c th thch5 C. 4. Bin php 4: Theo doi thng xuen s xo bo...............53 D. c eu cu i vi S v V D. 1. Ren lunthc vch56 D. 2. inh thnh hoat ng hc57 D. 3. X dng u tn GV........................................................................................................57 hng III: Thc nghim s pham 1. Muc ch thc nghim60 2. Ni dung thc nghim60 3. T chc thc nghim............60 4. hng php tin hnh......................................................................................60 5. t lun thc nghim62 6. nghi mt s hiu bit qun trng63 VIII. Ti liu thm ho....................................................................65 IX. hu luc.........................66 1. hu luc 1: hiu iu tr 66 : 2011 4 Bnghiu vit tt !ii qut xong vn . ?Si lm. o ng ai hc Sai hc s pham. GDio duc GDPTio duc ph thng. GVio vien. HSc sinh. KTin thc. NTp s t nhien NCKHNghien cu ho hc. NTNh trng. PPhng php. PTh thng. RTp s thc SLSi lm. SPS pham. TLTm l. THPTTrung hc ph thng. ZTp s nguen : 2011 5 I.L DO CHN TI. Trong qu trnh thc hin bi tp gia k cho hc phn P pp i u khoa hc, chng ti suy ngh rt nhiu v vic la chn mt ti thc s thit thc, ph hp vi kh nng ca c nhm v c bit l hu ch cho cc bn sinh vin khoa Ton trng i hc S Phm Hu. Sau mt qu trnh tho lun y nghim tc, chng ti thng nht theo cc quan im sau: -Ton hc l mt b mn khoa hc quan trng, c nhiu ng dng thc t trong cc nghnh khoa hc k thut. Cng ging nh ccmn th thao tr tu khc, Ton hcgipchngtanhiutrongvicrnluynphngphpsuyngh,phngphp suy lun, phng php hc tp, phng php gii quyt cc vn , gip chng ta rn luyn tr thng minh sng to. N cn gip chng ta rn luyn nhiu c tnh qu bu khc nh cn c v nhn ni, t lc gnh sinh, ch vt kh, yu thch chnh xc, ham chung nh l. D bn phc v ngnh no, trong cng tc no th kin thc v phngphptonhccngrtcnchoccbn.chnhlldochngtrnh GDPT hin nay lun xem ton hc l mt trong cc mn hc chnh, khng th thay th. Cc trng THPT cng rt xem trng b mn ny, c bit l i vi khi 12 - khihcchunbbcvokthittnghipcmntonlcnh.Tuynhin, khostthctindytonnctatrongnhiunmquacththyrngcht lngdytontrngphthngcnchatt,thhinnnglcgiitonca hcsinhcnhnchdohcsinhcnviphmnhiusailmvkinthc,phng php ton hc. -Gio vin dy ton chnh l cc hun luyn vin trong mn th thao tr tu ny. Cng vic dy ton ca chng ta nhm rn luyn cho hc sinh t duy ton hc cng nhng phm cht ca con ngi lao ng mi cc em vng vng tr thnh nhng chnhntnglaicatnc.Dovy,sinhvinsphmchngtacnthc csmnhcaocnykhngngngphnuhctp,rnluynpng yu cu ca ngh nghip. Tuy nhin, nhiu gio vin vn cha thc s lm tt chc nng s phm ca mnh, trong nhiu gio vin cn t kinh ngim trong cc vic: pht hin sai lm ca hc sinh khi gii ton, tm ra nhng nguyn nhn ca nhng sai lm v nhng bin php hn ch, sa cha chng, thm ch l sai lm khi khng ch n cc sai lm ca cc em v khng a ra c bin php ng n, kp thi. : 2011 6 DnnhiuquGDkhngcao.VnnycccnhtmlvGDhc quan tmn.Vd:J.A.Komenskykhng nh:Btkmtsailmnocng c th lm cho hc sinh hc km i nu nh GV khng ch ngay ti sai lm , bng cch hng dn HS t nhn ra v sa cha, khc phc sai lm. A.A.Stoliar cn nhn mnh: Khng c tic thi gian phn tch trn lp nhng sai lm ca hc sinh. -Hin nay, nhiu hc sinh c cm gic mt gc ton trm trng, dn n ccem ngi hc mn ton, khng c ch hc tp. Ngc li, nhiu em l hc sinh kh gii, thmchlxutscnhngvnmcccsailmkhcbn,thmchlphbin. B.V.Gownhenvencokhinura5phmchttonhcthcniti3phmcht lin quan ti vic trnh cc sai lm khi gii ton: -Nng lc nhn thy c tnh khng r rng ca suy lun; thy s thiu cc mc xch cn thit ca chng minh. -C thi quen l gii lgic mtcch y . -S chnh xc ca l lun. -CctiliunghincuvsailmcaHSTHPTckhnhiu,gmctiliu trong v ngoi nc. Nhng cc ti liu vn cha thc s ph bin v thit thc cho c HS v SV khoa ton chng ta. -ChngtichnitnglhcsinhTHPTvbchcnycnhimvhon chnhGDPT, chun b cho HS ra cuc sng v mt b phn ln hc bc Trung cp chuyn nghip, Cao ng, i Hc. Do vy, nu HS bc hc ny mc sai lm th s i n nhng hu qu kh nghim trng. T vic nht qun cc quan im trn, chng ti i n thng nht la chn ti: PHN TCH V SA CHA CC SAI LM CA HC SINH PH THNG KHI GII TON : 2011 7 II.M NIN U. Nghincuccsai lmph bincaHSTHPTkhi gii ton,ngthixut cc gii php s phm hn ch v sa cha cc sai lm ny, nm ch yu qua phn mn I S - GII TCH nhm rn luyn nng lc gii ton cho HS v gp phn nng cao cht lng dy hc mn ton trong cc trng THPT. III.NHIM V NGHIN CU. Nhim v nghin cu ca ti bao gm: iu tra cc sai lm ph bin ca hc sinh THPT khi gii ton. Phn tch cc nguyn nhn sai lm ca hc sinh khi gii ton. xut cc bin php s phmvi cc tnh hung in hnh hn ch, sa cha cc sai lm ca HS THPT khi gii ton. Thcnghimsphmxemxttnhkhthivtnhhiuqucaccbin php c xut. IV.KHCH TH V I TNG NGHIN CU. Khch th v i tng nghin cu ca ti bao gm: Hc sinh THPT ca mt s trng cp III trn a bn thnh ph Hu. GiovindytonTHPTcamtstrngcpIIItrnabnthnhph Hu. Mi trng s phm ca mt s trng cp III trn a bn thnh ph Hu, c bit l trong cc gi hc ton. V.GI THUYT KHOA HC. Nu cc GV ton trng THPT nm bt c cc sai lm ph bin ca hc sinh khi gii ton, ng thi bit cch phn tch v s dng cc bin php dy hc thch hp hn ch, sa cha cc sai lm ny th nng lc gii ton ca hc sinh s c nng cao hn, t cht lng gio dc ton hc s tt hn. : 2011 8 VI.N NIN U. 1.Nghin cu l lun:C s l lun v tm l hc, gio dc hc, l lun dy hc mn ton, iu khin hc, thng tin hc phn tch cc nguyn nhn v xy dng cc bin php dy hc nhm hn ch, sa cha cc sai lm ca hc sinh THPT khi gii ton. 2.iu tra tm hiu: Tin hnh tm hiu v cc sai lm thng qua cc GV ton trn a bn thnh ph Hu, thng qua bi kim tra trc tip HS cc trngTHPT. 3.Thc nghim s pham: TinhnhiutravnhgimcmcsailmcaHSlp11A2trng THPT Quc hc. Qua nhn thc c vai tr ca ti v xut mt s kin i vi SV khoa tonchng ta. VII.CU TRC TI Ngoi phn m u, kt lun v ti liu tham kho, ti chng ti thc hin gm 3 chng: - : Nghin cu cc sai lm ph bin ca HS THPT khi gii ton. - 2: Cc bin php rn luyn nng lc gii ton cho HS THPT thng qua phn tch v sa cha sai lm. - : Thc nghim s phm. Ngoi ra ti cn c 2 bng, 8 s v 1 ph lc. : 2011 9 hng I Nghin cu v cc sai lm ph bin cua hc sinh ph thng trung hc khi gii ton. Theo t in ting vit th: Sailm:ltriviyucukhchquanhoclphi,dnnhuqukhng hay. Phbin: l c tnh cht chung, c th p dng cho c mt tp hp hin tng, s vt. Vi cch hiu trn, chng ti nghin cu cc sai lm ph bin ca HS THPT khi gii ton. Hc sinh THPT hin nay vn mc nhiu sai lm khi gii ton v mi i tng hc sinh u c th mc sai lm khi gii ton. Mt s nguyn nhn ni tri: -Khng hiu khi nim, ni dung, tnh ton nhm ln. -Xt thiu trng hp, khng logic trong suy din . -Hiu sai ton, thiu iu kin, qun xt iu kin -Nh sai cng thc, tnh cht, din t km.... T vic iu tra, nghin cumt s lp hc trn a bn thnh ph Hu cng nh thng qua cc k thi, chng ti i n kt qu sau: Hc sinh cn mc nhiu sai lm khi gii ton, k c hc sinh kh gii cc lp chuyn. Di y l nhng sai lm ph bin m hc sinh kh gii thng mc phi.y l nhng sai lm c tn xut cao trong cc li gii ton ca hc sinh.Nh ni, cc sai lm ny nm ch yu b mn i s - Gii tch ca ph thng trung hc. A. Mt s sai lm thng gp. A. 1.Sai lm khi bin i cng thc. -Nhng sai lm khi bin i cng thc thng mc khi s dng cc ng thc m khng phi l hng ng thc, l cc ng thc- cha ng vi iu kin : 2011 10 no . i khi sai lm xut hin do hiu nhm cng thc, s dng cng thc m qun mt iu kin rng buc . -Cc v d: Saing log

log

log

log

||

2.2x = 4x2.2x = 21+x A. 2.Sai lm khi gii phng trnh, bt phng trnh. -NhngsailmkhigiiphngtrnhthngmckhiHSviphmquytcbin i phng trnh, bt phng trnh tng ng. t tha hay thiu cc iu kin u dn n nhng sai lm, thm ch sai n mc khng gii c na! Mt sai lm cn do hu qu ca vic bin i cng thc khng ng ( Xem mc VII. A. 1). -Cc v d: Saing 3 2 3 22 23 6 9 9( 2 3)3 ( 2 3) 9( 2 3)3 93x x x x xx x x x xxx = = == 3 2 2 22 2223 6 9 9( 2 3)3 ( 2 3) 9( 2 3)(3 9)( 2 3) 03 9 02 3 031331x x x x xx x x x xx x xxx xxxxxx = = = =

==

=

==

= : 2011 11 ( )( )( )( )( )2222lg x2mx lg x 10lg x2mxlg x 1x2mx x 1x 2m 1 x 1 0+ = + = + = + + = Phuong trinh co nghim duy nhat khi: 22(2 1) 4 04 4 3 03212mm mmmA = = =

=

=

( )( )( )( )2222lg x2mx lg x 10lg x2mxlg x 11x2mx x 11x(2m-1)x +1 0(*)xx+ = + = > + = > + = Phuong trinh co nghim duy nhat khi: 22(2 1) 4 04 4 3 03212mm mmmA = = =

=

=

Khi do. -m = 3/2: Pt(*) tr thnh

1 1(loai) -m = -1/2: Pt(*) tr thnh

1 1(loai) Vay khng ton tai m d pt d cho co mot nghim duy nhat. A. 3.Sai lm khi khi chng minh bt ng thc. -Cc sai lm thng bt ngun khi vn dng cc bt ng thc c in m khng n iu kin btngthc ng,sdngsai stccquytc suylun khi t bt ng thc ny suy ra bt ng thc kia. -Cc v d: : 2011 12 VD: So snh 1

Gii: p dng BDT Cauchy cho 2 s x v

ta c: 1

1

ng thc xy ra khi : x =

hay x2=1 hay x=1 Sai lm: Hc sinh mc sai lm v khng iu kin ca cc s a, b trong bt ng thc Cauchy:

Vi a , b

A. 4.Sai lm khi tm gi tr nh nht, ln nht. -Nhng sai lm khi tm gi tr ln nht v gi tr nh nht ca hm s hay ca biu thc nhiu n thng do vi phm quy tc suy lun lgic: - Nu{fxmxx

fx

mth min

fxm - Nu{fxMxx

fx

Mth mx

fxM -i vi biu thc nhiu n cng c quy tc tng t. -Cc v d: VD: Tm gi tr nh nht ca : F(x,y) = (x+y)2 + (x+1)2 + (y+1)2 Gii: Vi mi x, yR th (x+y)2

(x+1)2

(y+1)2

Vy F(x,y)hay min

x : 2011 13 Sai lm :HS khng ch ra cc gi tr ca x, y F(x,y)=0. Nhrng:F(x,y)vnutnti

saochoF(

)=0thmi ktlunmin

x.ivibinythkhngtnti

F(

)=0. Sa li: p dng bt ng thc Bunhiacopski ta c: 1= |x x 1| x1 x1 x

xng thc xy ra khi:

=> {1 { 45 KL: Min

x 1 { 45

A. 5.Sai lm khi gii cc bi ton tam thc bc hai. -Khi gii ton tam thc bc hai, cc sai lm xut hin do khng ch n gi thit ca cc nh l m vi vng p dng hoc l lm dng suy din nhng mnh khng ng hoc xt thiu trng hp cn bin lun. -Cc v d: VD: Tm m sao cho:

1(*)

1 : 2011 14 1Sailm:khinhnhaivca(*)vi

khichabitducabiu thc ny. A. 6. Sai lm khi gii h phng trnh, bt phng trnh. -Sai lm khi xt cc loi h phng trnh thng xut pht t nguyn nhn khng nmvngccphpbinitngnghockhngbinluncc trng hp xy ra. -Cc v d: VD: Gii h phng trnh:{

1

4

Gii: Tr tng v ca hai phng trnh ta c:

[

Vy h c nghim x= -1 hoc x=2. Sai lm: R rng x=-1 khng phi l nghim ca h?? Cn lu rng: ,

,

Li gii trn vi phm tnh tng ng v hiu rng: ,

A-B =0 Trong khi ta ch c: ,

A-B =0. Li gii ng l: H tng ng vi: {

1

T (**) ta c [

. V c hai gi tr ny u khng tha mnnn h cho v nghim. A. 7. Sai lm khi tnh gii hn. -Tipxcviccbitontnhgiihn,HSbctvngthuhnsang vng t v hn vi nhng i lng v cng b, v cng ln nn rt d mc sailm.Ccsailmcadngtonnythngbtnguntvickhngnm vng cc quy tc vn dng cc nh l v gii hn, c bit l phm vi c hiu lc ca nh l. : 2011 15 -Cc v d: VD: Tnh: L = lim

Gii: Ta c: L= lim

+lim

+..+lim

= 0 + 0 + +0 = 0 Sailm: V hc sinh khng nm vng kin thc: cc php ton gii hn ch p dng cho hu hn s hng, dn n sai lm trn. Li gii ng: 1

1

1

1 Do

1

m lim

1 Theo nh l kp th L=1. A. 8.Sai lm khi gii ton lin quan ti o hm: -Cc sai lm lin quan ti khi nim o hm thng gp khi tnh o hm v khi vn dng o hm gii ton. -Cc v d: VD: Cho f(x) ={

Tnh f `(0)? Gii: V f(0) = 0 = const => f `(0)=0. Sai lm : sai lm ca li gii trn l khi thay x=0 vo f(x) ri mi tnh o hm? Nu c nh vy th o hm ca f(x) tai mi x u bng 0.

Li gii ng: Theo nh ngha ta c: F(0) = lim

= lim

= *lim

+ = 1. : 2011 16 A. 9.Sai lm khi xt bi ton v tip xc v tip tuyn -Cc sai lm khi xt bi ton loi ny xut pht t vic khng nm vng thut ng hoc khng hiu ng s tip xc ca hai th l g? -Cc v d: VD: Cho hm s y = x

- 3x + 1 Vit phng trnh tip tuyn k t im A(3;19) ti th. Gii:Ta thy f(3) = 19A thuc th. Vy phng trnh tip tuyn cn xc nh l: y = f(3) = f(3)(x - 3) y = 24x 53. Sai lm:Phng trnh tip tuyn y = 24x 53 l tip tuyn ti A (nhn A lm tip im) ttnhinlktA.NhngvncthtiptuyniquaAmAkhngphiltip im. Kt qu ng l: C 2 tip tuyn tha mn bi ton : y = 24(x - 3) + 19. y =

(x - 3) + 19. A. 10.Sai lm khi xt cc ng tim cn ca th. -Khi nim v ng tim cn ca th quan h cht ch ti php tnh gii hn (k c php tnh gii hn mt pha). Nhiu hc sinh khng nm c nh ngha mchnhnvohnhthccahmsvsuyonmymcnndnnsai lm. Tt nhin vic tnh cc gii hn sai cng dn n sai lm khi tm cc ng tim cn. -Cc v d: VD: Tm ng tim cn ca ng y =

Gii: V lim

=nn th c hai ng tim cn ng l x = 1. V tp xc nh ca hm s l (-1; 1) nn lim

khng tn ti. Suy ra th khng c ng tim cn ngang.(?) Sailm: V tp xc nh ca hm sl (-1; 1) nn ch c lim

v lim

.Do khng vit lim

. : 2011 17 Cn lu thm th cng khng c tim cn xin v tp xc nh ca hm s l (-1; 1). A. 11.Sai lm khi gii ton nguyn hm, tch phn -Nhng sai lm loi ny lin quan ti s hiu bit khng ngcc khi nim v vn dng sai cc nh l, quy tc. -Cc v d: VD: Tnh x 1

dx. Gii:Ta c x 1

dx =

c Sai lm:Li gii trn vn dng cng thc : x

dx

c vi n1 y phi t u = 2x + 1dudx dx

c li gii ng. B. Phn tch cc nguyn nhn dn ti sai lm cua hc sinh ph thng trung hc khi gii ton. B. 1. Nguynnhn1: Hi k v chnh xc cc thuc tnh ca cc khi nim ton hc. Chngtabitrng:khinimlmttrongccsnphmcatduytonhc. Mi khi nim u c ni hm v ngoi din. Tp hp cc du hiu c trng cho bn cht ca cc i tng c phn nh trong cc khi nim chnh l ni hm ca cckhi nim. Tp hp cc i tng c cha cc du hiu trn l chnh l ngoi din ca khi nim s dn hc sinh ti s hiu khng trn vn, thm ch sai lch bn chtca khi nim.T, cc sai lmkhigii ton sxut hin.Mt khc nhiu khi nim trong ton hc l m rng hoc thu hp ca mt khi nim trc . Vic hc sinh khng nm vng khi nim ny s dn ti vic khng hiu v khng th c biu tng v khi nim khc. Mi quan h gia cc khi nim trong ton hc c tnh lin kt lgic. Nhiu khi nim kh trong ton hc mi c a vo chng trnhPTTHnh:vect,binhnh,nguynhm,tchphnNuchngtakhng : 2011 18 kp thi c nhng c gng hon thin mi v phng php ging dy cc khi nim th hc sinh th hc sinh s rt kh khn trong vic lnh hi cc khi nim . Nhiu ngi hay ni ti s mt gc ca hc sinh v kin thc th trc ht cn hiu rng l s mt gc v cc khi nim. Khng hiu s m rng khi nim gc hnh hc sang khi nim gc lng gic th hc sinh gp ngay kh khn trong vic nm vng khi nim v cc hm lng gic v t vic biu din gc lng gic, vicgiiccphngtrnh,cbitvicgiiccbtphngtrnhlnggics khng trnh khi cc sai lm. Nhiu hc sinh vit: sinx < 1 x < /2 +k2, hay khi gii phng trnh lng gic th cc s nguyn khc nhau u c k hiu l k v dn ti s thu hp tp nghim. Ngay hai n v o gc lng gic l v radian m hc sinh cng khng hiu c y l hai n v o khc nhau dn n sai lm 6

?. Hc sinh khng nm c khi nim gii hn ca dy s s dn ti mt lot s khng hiu cc khi nim tip theo: gii hn hm s, tnh lin tc, o hm, nguyn hm, tim cn cc ng cong. Thm ch nhiu em cn hiul cc s, nnsnsngvit ,0.=0, 1

=1Mtsemnghrng hai ng timcnnhauthkhngctnhau.Thmchnhiuemkhnghnhdungrac khi nim tip xc ca hai ng dn n sai lm ltip tuyn ti im un ca ng cong bc 3 khng tip xc vi ng bc 3 (ch v thy tip tuyn c bit ny i xuyn th). Hc sinh khng hiu v cn thc nn vit

hay

, t dn ti sai lm khi gii phng trnh v khi bin i cc biu thc. Hc sinh khng hiu cc khi nim v cc tr hm s nn khng phn bit c khi nim ny vi khi nim v gi tr ln nht v gi tr nh nht ca hm s, c bit s lm dng cc k hiu ymax v ymin. Hc sinh khng hiu cc khi nim v nguyn hm, dn ti vic chng minh h thcbngcchchngminhohmhaivbngnhau.Lraphihiurng nguyn hm ca mt hm s f(x) l mt tp hp cc hm F(x) sao cho F(x) = f(x) : 2011 19 nnchngminhhainguynhmbngnhautheonguyntcchngminhhaitp hp bng nhau. HS khng nm vng khi nim v h ta Decart vung gc, nn chn n v trn hai trc ox, oy khc nhau d v th ca mt hm no . HSkhngnmvngkhinimvparabolnnnhmlnkhigitnmts ngcdnghigingparabol,chnghnngy=x4.HSkhngnmvng khi nim qu tch nn nhiu khi mi lm xong phn thun vi vng kt lun qu tch cc im tha mn tnh cht ca bi ton l ng Hc sinh c khi cn nhm ln khi nim vi nh l, chng hn v khng nm ckhinimsm0calythannchngminh20 =1.HSkhngnm c khi nim s m thc vi khi nim cn thc nn c tng

vi mi x thuc R, t dn ti cc sai lm khi gii phng trnh

v a v phng trnh

1 Hc sinh khng nm vng khi nim v nghim ca h phng trnh nn nhiu khi kt lun h c hai n c hai nghim l x= v y= Nhvyquaccdnchngchothyvickhngnmvngcckhinimhc sinh c th b dn ti sai lm trong li gii. Chng ti xin lu ti nguyn nhn ny v gio vin nu khng c bin php s phm kp thi th chnh t s gy ra hu qu ln cho hc sinh th hin qua s sau: : 2011 20 Khng nm vng ni hm Khng nm vng cc thuc tnh khi nim Khng nm vng ngoi din Hc sinh Nhn dng sai Bin i sai K hiu sai Chng minh sai V hnh sai Din t sai Khng phn tch Khng cng c Khng phn loi Khng pht hin Gio vin TH HINSAI Hnh 1: Sai lm do khng nm hiu khng y v chnh xc cc thuc tnh ca cc khi nim ton hc. : 2011 21 B. 2. Nguyn nhn 2 : Khng nm vng cu trc lgic c nh l. nhllmtmnhckhngnhng.Cutrcthngthngca nhlcdngAB.TrongcutrccanhlABthAlgithuytca nh l v cho chng ta bit phm vi s dng c ca nh l. Ngi ta cn ni A ldiukincB.Nhngkhnhiuhcsinhkhngnmvnghoccoi thng gi thuyt A nn dn ti sai lm. NhiuhcsinhnhmgithuytAcanhlcngliukincnckt lun B nn mc sai lm. Nhiu hc sinh nhm gi thit A ca nh l l iu kin cn c kt lun B nn mc sai lm. Khi hc nh l v chiu bin thin ca hm s Nu f(x) > 0 vi mi xthuc(a;b)thhmsy=f(x)ngbintrn(a;b),khnhiuHSnghyl iu kin cn v hm s y= f(x) ng bin trn (a;b). Thc ra y ch l iu kin ( hm s y = x3 l hm s ng bin trn R. T , HS s dng nh l ny xcnhthamssaochohmsthamnyucubiton.Khicnhl: Nu f(x0) = 0 v f(x0)> 0 ( f(x0) < 0) th hm s t cc tiu (cc i) ti x = x0, HS cng mc sai lm khi gp tnh hung f(x0) = 0 v f( x0) = 0 li kt lun hm s khng c cc i, cc tiu. Tnh hung ny ch dn n mt suy ngh hp l l tr v qui tc 1 xt cc tr hm s nh o hm bc nht. phn th d cho sai lm ca HS l y=x4. Khi hc nh l Weiersrtrass v s tn ti gii hn dy, nhiu HS cng tng iu kin dy n iu l iu kin cn v l lun sai lm dy khng n iu nn khng c gii hn. KhngnmvnggithitcanhlnnHScthpdngnhlrangoi phm vica gi thit. Chng hn, hc quy tc tnh o hm ca hm s y = xn, HS khnglurngsmphilhngsnnpdngquytctrntnho hm ca hm s y = xx. Ngay HS PTTH m cn ni rng ( )( )2x 1xx 1++ +l s nguyn khi v ch khix+1 chia ht cho x2 + x + 1 mc d x thuc tp s thc R. iu trn ch ni c nux + 1vx2 + x + 1 nhn gi tr thuc tp s nguyn Z. Khi hc vbtngthcCauchy,HSkhngtigithitchpdngbtngthc : 2011 22 cho cc s khng m nn khi gp bi ton so snh x + 1/x vi s 2 p dng ngay c sai lm x + 1x> 2 vi x 1 v x + 1x= 2 vi x = 1.Nhiu HS lp 12 vn dng nh l Newton-Leibnitz tnh tch phn1221x} mc d hm s khng xc nh v lin tc ti x = 0 thuc [-2;1] c p s sai l -1,5, thc ra tch phn ny khng tn ti. nh l v cc php ton ca gii hn dy ch pht biu cho gii hn ca mt tng hu hn dy v cc dy ny phi tn ti gii hn, nhng nhiu khi HS vn p dng nh l cho tng v hn, thm ch khng gii hn ca tng dy c tn ti hay khng? Nguyn nhn ny cn dn n sai lm l nhiu HS s dng cc hng ng thc khi gii ton, chng hn: 2 2 2.log ( ) log ( ) log ( ).log .log log .a b aab a bxy x yb c c=== Tm li, vic khng nm vng cu trc logic ca nh l s dn HS ti nhiu sai lm trong khi hc ton v gii ton. Chng ti xin lu s sau: : 2011 23 nh l : AB Khng nm vng AKhng nm vng B Khng c A vn suy ra B Khng c Asuy ra khng c B S dng nh l tng t cha ng S dng B m khng nh A C B suy ra A C Anhng suy ra khng phiB Li gii sai Hc sinhGio vin Hnh 2: Sai lm do khng nm vng cu trc lgic ca nh l.. : 2011 24 B. 3. Nguyn nhn 3 : Thiu cc kin thc cn thit v lgic. Suylunlmthotngtrtucbitcaphnonmttrongcchnh thccatduy.Hotdngsuylungiitondatrncscalgichc.HS thiu cc kin thc cn thit v lgic s mc sai lm trong suy lun v t dn n cc sai lm khi gii ton. Vic khng c thc v php tuyn v php hi gy cho HS kh khn ngay c vic lnh hi cc kha nim, cc nh l. Nhiu nh l c gi thit v kt lun mang cu trc tuyn hoc hi. Nhiu tnh cht c trng ca mt khi nim cng c cc kiucu trc ny.Chng hnnh lNuhmstcctrti x=x*th hms khng c o hm ti x=x* hoc o hm ti trit tiu. Nhiu HS khng hiu c t suy ra mt khng nh Nu hm s c o hm ti im x=x* th o hm ti bng 0. Php ton ko theo ca lgic l php ton rt quan trng trong vic pht biu cc nhl,khinimvtronglpluncaligii.Chngtiphntchnguyn nhn cc HS khng nm vng cu trc lgic ca nh l nn dn ti cc sai lm khi gii ton. Nhng s thiu hiu bit v lgic , m c bit l php ton ko theo li l nguyn nhn ca nguyn nhn dn n cc sai lm. Nhiu HS khng hiu u l iu kin cn, iu kin v thm ch u l iu kin cn, u l iu kin , HS cng kh tr li. HS cn thiu nhng hiu bit v cc quy tc suy lun nn dn ti nhiu sai lm khi thc hin cc php tnh chng minh. Phn tch cc suy lun trong chng minh tonhctathymichngminhbaogmccbccbn,mmibcc thc hin theo nhng quy tc nht nh gi l cc quy tc suy lun. HS nhiu khi nhm php suy ngc tin l mt php chng minh. Chng hn, chng minh vi mi a, b, c ta c bt ng thc:2 2 2 23( ) ( ) a b c a b c + + > + + C HS gii nh sau: : 2011 25 2 2 2 22 2 22 2 23( ) ( )2 2 2 2 2 2 0( ) ( ) ( ) 0a b c a b ca b c ab ac bca b a c b c+ + > ++ + + > + + > Dobtngthccuicnghinnhinngnnbtngthccnchng minh cng ng ? -HS cn cha hiu thc cht ca php quy np ton hc, nhiu khi dng phptngtthayphpchngminhbngquynptonhc.Chnghn, tnhohmbcncahmsy=e2x,HSlnlttnhy=2e2x,y=4e2x, y=8e2x v tng t suy ra y(n)=2 e2x m khng chng minh g thm. - Khng nm vng cc php ton i s mnh : ph nh, ko theo, hi, tuyn, tng ng, khng nm vng thuc tnh ca cc lng t mi, tn ti cng nh v, hoc... Khng nm vng cc quy tc suy lun c bn: -Quy tc kt lun(Modus ponens): p, p qq -Quy tc suy lun bc cu: p q, q rp r -Quy tc suy lun phn chng:q p.pq q, p q

p hoc p qq p : 2011 26 - Quy tc la chn:p q, p

qvhocp q, qpv -Quy tc tch hi: p qp. hoc p q

q. -Quy tc nhp hi: p, qp q . -Quy tc nhp tuyn:

p

p q v hoc q

p q v -Quy tc hon v tin : p (q r)q (p r) -Quy tc hi tin : p (q r)p q r . B. 4.Nguyn nhn 4: Hc sinh khng nm v p pp ii bi to bn. -Khng nm vng phng php gii cc bi ton c bn, HS khng ngh c trng hp cn xt v dn n iu kin sai.-Khng nm vng phng php gii cc bi ton, HS s khng bin lun cc trng hp xy ra ca bi ton. : 2011 27 -Khng nm vng phng php gii cc bi ton, HS s khng p dng ng phm vi v dn n b tc, khng i n li gii. -Khng nm vng phng php gii cc bi ton, HS s b qua nhng bc quan trng v i ngay ti kt lun. -Khng nmvngphng php giicacngmtloi ton,HSkhng tmraphng php gii ti u cho mt bi ton c th. -Khng nm vng phng php gii, li gii ca hc sinh s khng c trnh t lgic v s khng bit khi no kt thc li gii. VD: Gii cc phng trnh: a)

4

5

b)1

5

1

c)4

6

9

Phng php gii: HS d dng nhn ra phng php gii c bn ca pt dng:

vi { *

V phng trnh c mt nghim bit

nh sau: -a phng trnh v dng (

)

(

)

1 -Ch ra x =

l nghim. -Nu

.

1 thf(x)= (

)

(

)

ng bin nn x=

l nghim duy nht. -Nu

.

1 thf(x)= (

)

(

)

nghch bin nn x=

l nghim duy nht. Vi phng php ny, HS s thnh cng khi gii phng trnh a, b, nhng s tht bi viptc.Phngtrnhnykhngdngphngphpgiitrnvkhngnhmra nghim

mcdvnchngminhc

lnghimduynht.Vkhngnmr cch gii phng trnh loi ny nn hc sinh i ngay n kt lun sai lm. Trong khi ta c th gii phng trnh c theo phng php: -Cho phng trnh A(a2)t(x) + B(ab)t(x) + C (b2)t(x) = 0 -t t =(

)

-Ta c phng trnh:At2 +Bt +C =0 -Gii phng trnh c bn trn tm ra

. : 2011 28 Kt lun:Tt c kt qu nghin cu chng ny, cho php chng ti khng nh: -HS cn mc nhiu sai lm khi gii ton. -Nhng sai lm ca HS c th h thng li gip GV d pht hin trong li gii ca HS. -Nhng sai lm khi gii ton ca HS xut pht t nhiu nguyn nhn v kin thc. -T nhng nghin cu ny, chng ti c c s thc tin v l lun ngh cc bin php hiu qu nhm phn tch, sa cha v hn ch cc sai lm ca HS khi gii ton. T , gp phn hon thin l lun dy hc mn ton v rn luyn nng lc gii ton cho HS THPT. : 2011 29 hng II Cc bin php rn luyn nng lc gii ton cho hc sinh ph thng trung hc qua phn tch v sa cha sai lm. A. s l lun Trong qu trnh nghin cu xy dngcc bin php hn ch v sa cha cc sai lm ca HS PTTH khi gii ton, chng ti da trn nhng c s l lun sau y: A. 1. L lun v phng php day hc. Cn c ny l tin b sung vo phng php dy hc mn ton nhm t mc ch ra l hn ch, sa cha cc sai lm ca HS khi gii ton. Phng php dy hc l cch thc lm vic ca thy c v ca tr trong s phi hp thng nht v di s ch o ca thy, nhm lm cho tr t gic, tch cc, t lc t ti mc ch dy hc. Phng php dy phi c hai chc nng l truyn t v ch o. Phng php cng c 2 chc nng l tip thu v t ch o. Phngphpkhoahctonhcvphngphpdyhctonhclngcu, nhng khng ng nht. Ngi HS ch ch ng sng to trong khun kh ca s ch o s phm ca GV, ca chng trnh o to, pht hin li chn l mi cho bn . Ccnhtmlhckhngnhrngmitrembnhthngkhngcbnhttg uckhnngtchcvntonhcphthng,cbnduchochngtrnh ton hin i ha . Nh vy c th thy rng cc sai lm ca HS khi gii ton l c th khc phc c. Gio dc hc mn ton lin h khng kht vi mt s khoa hc khc : khoa hc duy vtbinchngvduyvtlchs,tonhc,giodchc,tmlhc,lgichc,iu khin hc v l thuyt thng tin. Cc bin php sa cha sai lm cho HS khi gii ton cng phi da trn mi lin h hu c ca cc b mn khoa hc trn. : 2011 30 Cc bin php sa cha sai lm cho HS, cng nh phng php dy hc ni chung phi phn nh c : cu trc bn ngoi v cu trc bn trong, c bit i vi cu trc bn trong phi ch ra c cc thao tc tr tu, cch thc t chc lgic ca s nhn thc v lnh hi ca HS. i vi vic ch racc sai lm ca HS khi gii ton cng c nhiu quan im khc nhau trn th gii. Na th k sau ca th k XIX,mt s nh gio dc hc ngi cm tiu biu l Aphogut Lai cho rng: vic ch ti cc sai lm ca HS trong gi hc cnh hng xu n vic tip thu bi ging. c bit quan im ny ngh khng vit li li gii sai ln bng v iu ny lm cng c thm sai lm trong thc HS. y l mt quan nim c tnh cht my mc gio iu, khng da trn qui lut tip thu tri thc mt cch c thc ca HS. Chng ti ng nht quan im vi R.A.Axanop : Vic tip thu tri thc mt cch c thc c kch thch bi vic t HS phn tch mt cch c suy ngh ni dung ca tng sai lm m HS phm phi, gii thch ngun gc ca cc sai lm ny v t duy, l lun v bn cht ca cc sai lm . Chnh A.A.Stoliar cng t ra mt s bi ton phng php ging dy m trong lin quan ti cc tnh hung HS mc sai lm khi gii ton v khng nh cn phi c binphpnhmdyhcmntondatrnccsailm,khiccsailmcaHSxut hin. Mtkhc,ngoiccphngphpdyhctruynthng,ccnhnghincuv phng php dyhca ramts phng phpmimtnhhungmcsailm ca HS to iu kin pht huy u im ca phng php ny. Chng ti xin minh ha r trn bng quan im c th di y. Phng php dy hc gii quyt vn da trn tnh hung c vn trong dy hc. Khi HS mc sai lm li gii l xut hin tnh hung c vn , khng phi do GV ra theo mnh m t n ny sinh t lgic bn trong ca vic gii ton. Sai lm ca HS to ra mu thun v mu thun ny chnh l ng lc thc y qu trnh nhn thc ca HS. Sai lm ca HS lm ny sinh nhu cu cho t duy m t duy sng to lun bt u bng mt tnh hung gi vn (Rubinstein ). : 2011 31 SailmcaHSxuthinthskhugichotnghctpmHSsc hngch,gingctmrasailmvitiligiing.Tmracisaica chnh minh hay ca bn mnh u l s khm ph. T s khm ph ny, HS chim lnh c kin thc mt cch trn vn hn. Tuy nhin cn gy nim tin cho HS l bn thn mnh c th tm ra c sai lm trong mt li gii no . HS c th t suy ngh hoc trao i tm ra cc sai lm. TrongtnhtrngphncctrnhcaHSnhhinnay(ngaytrongmtlp)th phng php dy hc phn ha c tc dng rt bt dn s phn cc. GV c th i x cbit trong nhng pha dy hc ng lot nhm hn ch v sa cha cc sai lm ca HS khi gii ton. S phn ha trong nh thng qua nhng mc by sai lm khc nhau cho tng i tng HS, th hin ngay vic GV giao bi tp trn lp hoc bi tp v nh. S phn ha ngoi nh thng qua cc cng vic t chc hc tp theo nhm, t v ph o ring cho nhng HS mc nhiu sai lm trm trng. Tuynhin,chngtakhngcquntndngccuimcaccphngphp dy hc truyn thng vo mc ch m chng ta ang hng ti. Ccbinphpcxutudatrnquanimhotngtrongphngphp day hoc voi cac tu tung chu dao d duoc G.S Nguyn Ba Kim duc kt nhu sau. a)Cho HS thuc hin v tap luyn nhng hoat dong v hoat dong thnh phan tuong thch vi ni dung v mc ch dy hc. b)Gy ng c hot ng v tin hnh hot ng. c)Truyn th tri thc, c bit l tri thc phng php, nh phng tin v kt qu ca hot ng. d)Phn bc hot ng lm ch dc cho vic iu khin qu trnh dy hc. A. 2. Nhng vnc bn cua tm l day hc NgaytrongccnguynnhndntisailmcaHSkhigiitoncccnguyn nhn v tm l ca HS. Chnh v vy khi a ra cc bin php s phm, chng ti ly cc qui lut ca tm l hc dy hc lm c s l lun. : 2011 32 Nh l thuyt hot ng m ngnh tm l hc s phm nghin cu c nhiu kt qu do thc tin dy v hc t ra. Chng ti rt tn thnh quan im Suy cho cng gio dc l qu trnh bin nng lc caloingithnhnnglccamitrem.lmccngvicnycnphi thngquahotngdyhc.Ttnhin,hotngdykhngthtchrihotng hc. Cht lng hot ng hc ph thuc vo trnh iu khin v t chc ( trnh ngh nghip) ca thy, kt tinh trnh pht trin nhng hnh ng hc tp tch cc ca HS . Chng ti rt quan tm ti bn cht ca hot ng hc ca HS c khng nh: a)Tri thc v nhng k nng, k xo tng ng vi tri thc y l i tng ca hot ng hc. Vic lnh hi tri thc, k nng, k xo ca x hi s khng th thc hin c nu ngi hc l khch th b ng ca nhng tc ng s phm. b)Hot ng hc lm cho chnh ch th hot ng ny thay i v pht trin. Ch c thngquangihcmiginhcnhngkhnngkhchquanngy cng t hon thin chnh mnh. c)Hot ng hc cn lm cho HS c c nhng tri thc v hot ng hc m chng ta thng gi l phng php hc tp. Cc bin php sa cha sai lm cho HS khi gii ton phi tc ng v nhm ch vo hot ng ca HS. Trc ht cn to ra ng c hc tp sa cha cc sai lm. HS phi thy vic sa cha cc sai lm khi gii ton l mt nhu cu v cn phi tham gia nh mtchthmtcchtnguyn,saymhohng.HSphiccngchon thin tri thc. Cn ly hot ng hc tp ca HS lm c s cho qu trnh lnh hi tri thc. HotnghccaHSphithngquacchnhngcth:hnhngphntch, hnh ng c th ha. Cncvonhngktqunghincuvtmldyhc,chngtithycnhnh thnh HS nhng nng lc to ra nng lc, m trong bn thn nng lc tm ra cc sai lm khi gii ton s to ra nng lc gii ton cho HS. T HS t tin sa cha cc sai lm. : 2011 33 TmlhckhngnhmunhnhthnhkhinimHSphilyhnhngca ccemlmcs.NutchchnhngchoHSkhngttthHSkhngthnm vng cc thuc tnh ca khi nim v nguyn nhn gy ra sai lm s xut hin. Hn na, cc bin php phi tp trung vo pht trin hot ng, rn luyn cc k nng hc tp ca HS ( k nng nhn thc, k nng thc hnh, k nng t chc hot ng, k nng t kim tra, nh gi) . B. B phng chm chi ao s dung cc bin php s pham nhm han ch v sa cha cc sai lm cua hc sinh khi gii ton B. 1. hng chm 1: Tnh kp thi Cc bin php phi ch thch ng vi thi im thch hp. Bin php ch huy hiu qu nu c p dng ng lc. Khng thty tin trong vic phn tch vsa cha, cng nh hn ch cc sai lm ca HS. c bit, thi gian m GV tip xc trc tip vi HS l c hn.S khng kp thi s gy lng ph thi gian v GV s kh c iu kin ly li thi gian mt. Tnh kp thi ca cc bin php i hi s nhanh nhy ca GV trc cc tnh hung in hnh, nhm tc ng ng hot ng hc ca HS. Tnh kp thi i hi s tch cc ha hot ng nhn thc ca c GV v HS. TnhkpthiihiGVphinghincuvdoncccsailmcaHS tng thi im ca nm hc, tng gi ln lp. Tnh kp thi i hi GV lun t th thng trc vi mc tiu dy hc nhm hn ch v sa cha sai lm ca HS khi gii ton. Sai lm cng sa mun bao nhiu th s vt v ca thy v tr cng tng by nhiu. Tnh kp thi i hi GV phi vng vng v tm l ngh nghip, bit ch ng trong thi , bit kim ch khi kh chu v bit ng cm vi mi iu sai, ng ca HS. Tnh kp thi i hi GV phi tranh th giao tip vi HS, khng ch trn lp m cn trong nhiu hon cnh khc tn dng c hi thc hin cc bin php dy hc. Tnh kp thi i hi GV phi tm cch hn ch cc nguyn nhn sai lm ca HS k c khi cc sai lm cha xut hin. : 2011 34 Tnh kp thi cn i hi GV phi cng c thng xuyn cc sai lm sa cha cho HS, nhm khng cc sai lm ti din. B. 2. hng chm : Tnh chnh xc S chnh xc trong li gii l i hi ca ton hc, cng l s i hi ca nhim v dy hc mn ton trong nh trng ph thng o to c cht lng nhng ngi lao ng mi. Cc bin php xut phi i ti mc tiu lm cho li gii ca HS bo m chnh xc cao. Tnh chnh xc i hi GV phi din t chnh xc, t ngn ng thng thng n ngn ng ton hc. GV phi l mu mc v phng php t duy, t duy chnh xc, v li gii chnh xc cho cc bi ton. Tnh chnh xc i hi GV phi ch ra chnh xc nguyn nhn sai lm ca HS trong li gii. GV khng c ph nh li gii sai ca HS mt cch chung chung. Tnh chnh xc i hi cc bi ton ca GV a ra khng c sai lm. i vi HS gii th c th s sai lm ca bi ton s c HS pht hin, nhng i vi HS yu hoc trung bnh th bi ton d gy hoang mang v mt nim tin vo GV. Tnhchnh xcihi snh gichnhxcmcsailmcaHS.Chng hn, khiHSvit

thi thng thng cc V cho l mtsailmnghim trng v kin thc c bn. Tuy nhin, i vi mt s HS c th th sai lm ny c khi ch do s v thc gy nn. TnhchnhxcihiGVnhgibigiicaHSquaimsmtcchcng bng. Tnh chnh xc i hi GV phi bit hng dn iu chnh, sa cha mt li gii sai Hs t tm ra mt li gii ng. TnhchnhxcihiGVphilachnngbinphptiutrongtngtnh hung in hnh. : 2011 35 B. 3. hng chm: Tnh gio dcTnh gio dc i hi GV phi ly s pht trin nhn cch ca HS lm mc tiu cho cc bin php. Tnh gio dc gip cho HS thy c tm quan trng ca s chnh xc trong li gii. Tnh gio dc gip cho HS trnh c cc sai lm khi sai lm cha xut hin. Tnh gio dc gip cho HS xc nh c ng c hc tp mn ton.Tnh gio dc i hi GV phi c phm cht v nng lc xng ng l ngi thy. Tnh gio dc i hi GV khng lm cho HS b xc phm v nhn cch khi mc sai lm trong li gii. TnhgiodclmchoHScchtronghcton,giiton.HSkhngngikh, bitkintrvcnthnitiligiing.TnhgiodcgipchoHScnhng thi quen tt, nh bit t kim tra vic lm ca mnh, bit ph nh sai lm ca chnh mnh v bit gip bn nhn ra sai lm. TnhgiodcgipchoHSkhnggiudt,dmhikhikhnghiu,khngbit, trnh gian ln quay cp mong li gii ng. Tnh gio dc gip cho HS tch cc suy ngh, tng ci hot ng a n s ham m chim lnh kin thc chun xc. Tnh chnh xc i hi GV phi bit khen ngi, khch k HS khi sa cha c sai lm. Tnh gio dc lm cho HS thy c mi sai lm c th sa cha c nu ra tm ra nguyn nhn v c ch khc phc. TnhgiodclmchoHSbitcuimcatrcgiclcthgipnghra, kim tra li gii nhng cng chnh trc gic c th a HS n cc sai lm. Tnh gio dc i hi GV m nhn ra sai lm camnh trong li gii, trong cch nh gi HS. : 2011 36 TnhgiodcihiGVkhngnngvitrongvicthchinccbinphp mongmunchmdtsailmcaHS.CnhngsailmihiGVphihuyng nhiu bin php ng b v qua mt thi gian di mi khc phc ni. Tnh gio dc i hi cc bin php phi da trn tnh thng yu hc sinh, mong HS tin b v tuyt i khng xc phm hay quy kt sai nguyn nhn sai lm ca HS. Baphngchmtrnhtr,bsungchonhaulmchoccbinphpthchin ng mc ch v kt qu. Tnh kp thi lm cho tnh gio dc t c nhanh hn v ngc li tnh gio dc gip cho cc bin php thc hin c kp thi, thun li hn. Tnh chnh xc cng c cho tnh gio dc v to iu kin cho tnh kp thi. Ngc li, tnh kp thi l chun b iu kin th hin tnh chnh xc. Mtbinphp,mthotngcaGVhayHSnhiulcthhincbaphng chm ch o quan trng trn. Chng hn s tch cc ha trong vic nhn thc cc khi nim va c tnh kp thi phng cc sai lm, va c tnh chnh xc t c s hiu bit su sc khi nim v c tnh gio dc trong vic gip HS ch ng chim lnh cc kin thc chun C.Bn bin php s pham chu yu nhm han ch v sa cha sai lm cho hc sinh C. 1.Bin php 1: Trang b y , chnh xc cc kin thc v b mn ton. Bin php ny gii quyt bn tnh hung c th sau y: -Tnhhung1:DaykhinimTonhocnhuthnaodtrnhsailamchohoc sinh khi giai Ton? Ngoi cc hot ng dy hc khi nim c trnh by chng ti cn mun nhn mnh v hon thin thm mt s bin php. Giovincndontrc(bngkinhnghimbnthnhoctraoiving nghip), cc kh nng khng hiu ht cc thuc tnh ca khi nim. Chng hn i vi khi nimhmsngcth hc sinhc khnngno khng hiu ht cc thuctnh ca khi nim? D xy dng qua khi nim song nh i chng na, cui cng i vi hc sinh trung hc ph thng, chng ta u da vo phng trnh f(x) = y vi y thuc tp gi tr ca hm f cho trc. Nu phng trnh ny c nghim duy nht th chng ta c th xy dng c hm g sao cho nu f(x) = y th g(y) = x v g gi l hm ngc : 2011 37 ca hm f. Khng nhng th, ngi ta c th thu hp tp xc nh ca f tn ti g. Nhiuhcsinhkhinitihmfchobimtbiuthcgiitch,mkhngti hm f cho bi nhiu biu thc gii tch, thm ch cho bi cc cch khc nh bng gi tr, th vv Nhiu hc sinh khng ti tp xc nh ca f. Chng hn y = f(x) = x2 l khngn iutrnR nhngn iu trnR+.Hc sinhcoihmy=f(x)=x2 trn R v trn R+ l nh nhau v cng mt cch tng ng mi x vi y = x2. T , hc sinh hot ng nhn dang v thc hin d mc sai lm. Mt s hc sinh cn ni hm y =arcsinxlhmngccahmy=sinx,chkhngnhnmnhlhmngcca hm y=sinxvix[

;

].Hcsinhcnnghrnghaihmngcnhaufvgl khc nhau! Do khi tmhmngcca hmy=x hayhmy=

;y=

hcsinhng ngng khng dm kt lun hm ngc ca f chnh l f. Khi khng c phng trnh f(x) = y th hc sinh khng bit tm hm ngc nh th no. Chng hn hm s cho bi bng gi tr tng ng: x12345678 y87654321 Th hc sinh khng tm ra c hm s ngc. C khi hc sinh khng nm c nu hm g l hm ngc ca hm f th hmcng l hm ngc ca hm g. Nudoncccsailmthchcchngiovinschunbbigingca mnh phng trc sai lm cho hc sinh. S ch ng phng sai lm bao gi cngtchcchnllosachasauny.Nhngsailmcahcsinhvkhinim Ton hc mang du n kh phai v rt mt cng chnh li cho chnh xc. y cng lu phn bit cha hiu ht v hiu sai. C nhng khi nim kh, hc sinh khng hiu ht cc thuc tnh ngay mt lc m phi qua cc hot ng nhn dng v ci thin mi i ti s trn vn. chnh vic cha hiu ht cc thuc tnh s rt d dn nvichiusaikhinim.Docnhngsailmcahcsinhphilmchohc sinh hiu ht cc thuc tnh ca khi nim th mi mong hc sinh ht hiu sai. Chng hn khi dng k hiu ymax, ymin t lp 10, lp 11 hc sinh coi l k hiu cho gi tr : 2011 38 ln nht v gi tr nh nht ca hm s , nhng lc hc sinh cha hiu ht ci sai m lp 12 th hc sinh mi c gii thch tha ng. Trong l thuyt thng tin, khng nhiu thng tin chng ti c ra phng php m c th vn dng vo dy hc. Hc sinh c nhim v gii m thng tin m gio vin a n. Lm sao va sc gii m ca hc sinh. Hay i m hc sinh d gii m hn. Cc bin php gio dc trc quan c nhiu tc gi nghin cu, y chng ti cp ti mt bin php i m cho hc sinh bng cch nng gi mang ca thng tin. Chng hn khi dy hm s lin tc, chng ta c nh ngha: Hm s y = f(x) c gi l lin tc ti m x = x0 nu: 1)x0 l mt im thuc tp xc nh ca hm s, 2)

= f(x0) Mt nh ngha khc: Mthmsf(x)xcnhtrntpsD,gillintctiimxD,nu lim

= f(x0) Thc ra khi vit f(x0) mang thng tin x0 thuc tp xc nh, nhng cc nh ngha trn u nhnmnh ring yu cu ny chnh l to iu kin cho hc sinh gii m tt hn. Thm ch theo chng ti l rt tt. Chng ti cn lu ngay di nh ngha: Nh vy mt hm s lin tc ti x0 nu v ch nu 3 iu kin sau: 1)f(x) xc nh ti x = x0, 2)lim

tn ti, 3)lim

= f(x0) Vi lu ny chng ti to iu kin gii m cho hc sinh . Hy hnh dung l ra ch cn 3 nhng thm 1, v cui cng thm c 2. Hiu c dng s phm v ngha thng tin ny, gip vin s t chc hot ng nhn dng tt hn. Trong hot ng nhn dng th cc phn v d rt quan trng trong vic trnh cc sai lm ca hc sinh khi lnh hi khi nim. chng hn, hm s y =

khng lin tc ti x= -1 (vi phm 1), y ={

1 1 khng lin tc ti x= -1 : 2011 39 ( vi phm 3); y={

khng lin tc ti x=0 ( vi phm 2). Khi hc sinh i qua mt lot khi nim bng s rt d gy du n cho hc sinh. Chng hn mi quan h gia 3 khi nim quan trng ca gii tch: Ngayvicphnloikhinim,giovincngcnchrasphttrintheocon ng ton hc i hc sinh thy c hc sinh thy c ni hm v ngoi din khi nim. c th gio vin cn phi bit bc tranh ton cnh v cc khi nim quan trng trong chng trnh i s-gii tch trung hc ph thng. -Tnh hung 2: Dav cac dnh l Ton hoc nhu th nao d hoc sinh trnh cc sai lam khi giai Ton? Ni ti nh l Ton hc l ni ti mt khng nh ng. Tuy nhin vic quan trng m gio vin cn quan tm u tin l cu trc logic ca nh l. Nh chng ti phn tchvickhngnmvngcccutrcnhlsdntisailmkhihcsinhgii ton.cc nh l ton hc thng c din t theo cu trc A=> B. Ai cng bit A l gi thuyt B l khng nh,kt lun ca nh l. Nhng chng ti xin lu thm: A cho bit dng nh l khi no v B s cho bit kt lun suy ra c g khi c A. Dy nh l ton hc c th i theo 2 con ng: con ng suy din v con ng c khu suy on. Nhm hn ch v phng sai lm ca hc sinh khi gii ton chng ti thy cn phi phn tch r gi thit ca nh l.Hc sinh nhiu khi khng quan tm n gi thit nh l m ch quan tm ti kt lun nh l dn ti sai lm. Gio vin cn nhn mnh gi thit ca nh l c cu trc hi hay tuyn. Chng hn nh l Viete: Nu phng trnh bc hai ax2+ bx+ c = 0 (a0) c nghim x1, x2 th tng v tch ca nghim n l: f(x) kh vi trn (a;b) f(x) lin tc trn (a;b) f(x) kh tch trn (a;b) : 2011 40 {

Cu trc ca gi thit c cu trc hi : { a0}v { 0}. Trc khi dng nh l ny phi kim tra hoc t iu kin bi ton tha mn ng thi 2 iu kin ca gi thit. Hc sinh rt hay qun iu kin a0. Nhiu hc sinh vn tnh tng v tch nghim ca phng trnh x2+x+1=0 mc d phng trnh v nghim. Gio vin cn to ra cc th d m cc iu kin ca gi thit cha tha mn hon ton hc sinh thy rng mi iu kin ca gi thit l khng th thiu c. Th d hay nh l: Nu hm s f(x) c o hm f(x)= 0trn khong (a;b) th y l mt hm hng trn khong tc l f(x)=c vi mix (a;b). Nhiu hc sinh ch ti f(x)=0 m khng ti (a;b) nn dn ti sai lm. Khi nh lccu trcABthA liu kin cBchchachcliu kin cn. Gio vin cng cn nu ra th d thuyt phc, ch khng ch dng li vic nhc nh. Cc th d m c bit l cc phn v d bao gi cng to n tng su sc i vi hc sinh. Chng hn khi dy nh l: Cho hm s y = f(x) c o hm f(x)trn khong (a ; b. Nu f(x) > 0 vi mi x (a ; b) th f(x) ng bin ( tc l tng) trn khong . Gio vincnchra hmsy=x3 thc stng trnR, nhngy=3x2 vn trit tiu ti x = 0 , thm ch y = thc s tng trn [ 0 ; + ) nhng y =

khng xc nh ti x = 0, iu ny chng t gi thit f(x) > 0 vi mi x (a ; b) ch l iu kin ch khng phi l iu kin cn. khcsu nh l, gio vin cnch nhl nylkhiqut chonh l nom hcsinh hc trc.Chnghn nhlhmcosin khiqutchonh lPytago. nh l hm sin lm cho hc sinh nhn li nh l hm qu tch cung cha gc. : 2011 41 Khi dy nh l cn ch cho hc sinh cc ng dng ca nh l to s nhy cm cho hc sinh khi ng trc mt bi ton bit ngh ti vn dng nh l no. Chng hn nhlLagrangechngtrnhgiitchlp12,giovincnchra3hngng dng ca nh l ny:Chng minh h thc hoc rt gn biu thc, chng minh phng trnhcnghim,chngminhbtngthc.Chngtixindnra3vdcho3ng dng ny: ng 1: Cho m > 0 v a,b,c l cc s tha mn

+

= 0 Chngminh: phng trnh ax2 + bx +c = 0 c nghim thuc (0 ; 1).Bi ton ny hc sinh lp 10 gii nh nh l o v du ca tam thc bc hai, Hc sinh lp 11 da vo nh l gi tr trung gian ca hm s lin tc, c hai li gii u kh phc tp, hc sinh lp 12 c th da vo nh l Lagrange. Quabitonnygiovincngiphcsinhthycmtbihcthv:khi c cung cp thm cc kin thc lp trn th cng c nhng cch nhn mi v cng mt bi ton t c nhng li gii mi. Hc sinh thy rng s pht trin ca kin thc chnh l m rng tm nhn, ch khng phi mang thm gnh nng trong tr c. ng 2: Chng minh rng 0 < a < b th

>

, t kt qu trn suy ra iu phi chng minh. : 2011 42 iucbitcnchkhidytonhcchohcsinhl:Giovinchohcsinh thy r phng php phn tch chng nh l. Chnh bin php ny gip hc sinh d itichngminhngtronggiitonsauny.Dynhlchnhnhmmcch truyn th nhng tri thc phng php lin quan ti php chng minh. -Tnhhung 3:Cung cap cc kin thuc v logicnhu th nao d hoc sinh trnh sai lam khi giai ton? Nhm gii quyt tnh hung ny, chng trnh ton PTTH chuyn ban chnh thc a mt s hnh thc suy lun ton hc vo phn mn i s lp 10. Gio vin khi dy chng trnh ci cch gio dc cn tham kho vn ny tin hnh bin php cung cp thm cc kin thc v logic cho hc sinh ngay t u THPH. Trchtcnlu timnh vcc phptonmnh nhph nh, tuyn. hi, ko theo, tng ng. Theo thc nghim ca chng ti vic a ra cc th d theo ngn ng t nhin cn itrcccthdtheongnngtonhc.ychnhlconngittrcquan sinhngntduytrutngcanhnthc.Chnghn,cththeomnh A={trinng};B={im}ththngthnghcsinhcnhcnhNutri nng th i m nn hc sinh d nhn thc ra ngha ca php ko theoAB. A l c B nhng lu nhiu hc sinh vn i m khi tri khng nng, ngha l A cha phi l iu kin cn c B. cbitnuA BngthylthdnhnmnhmnhoB A khng ng. Hc sinh c th thy ngay vic i m khng lm cho tri nng! Mt th d khc, ph nh mnh do mt bn nu ra: T thng xuyn tp th dc bui sng th ch cn ch ra mt bui sng m bn y khng tp. T ch ra mi quan h gia hai lng t vi mi v tn ti. T ngn ng thng thng, gio vin bt u s dng cc khi nim, tnh cht,nh l ton hc m hc sinh bit phn tch chn l ca cc mnh hi, tuyn, ph nh, ko theo, tng ng ca cc mnh cho trc. Chng hn, nuA = {s t nhin c tn cng bng 0}; B = {s t nhin c tn cng bng 5}; : 2011 43 C = {s t nhin chia ht cho 5}th(AB)C ng thi C (A B), do (A B) C l tiu chun chia ht cho 5 ca s t nhin. Khi kim tra mt s chia ht cho 5 hay khng ch cn kim tra A hoc B. T ph nh mnh ny c( ) A B C . , qua y hc sinh thy mi quan h ca cc php tuyn, hi, ph nh, tng ng. Mtsailmcntrnhlsdngcckhiulogicmtcchtytin.nhiuhc sinh vit A B m A khng phi l iu kin c B. Chng hn c hc sinh vitx 1x = 1, l ra vit x 1 v x 1 x = 1 Khiukotheocncvittrongcuvn:onxei mtmbim mt. Vic trang b l thuyt khng ch dng li mt s bin php suy lun ton hc trong phn u chng trnh i s lp 10 m cn c thng xuyn cng c. Php quy np ton hc c nhiu dp gio vin nhc r cho hc sinh, Chng hn khi chng minh h thc: 1 + 2 + 3 +.+ n =

Thngoicchcschgiokhoatrnhby,GVcnyucuthmhcsinh dngthmcchth2nhphngphpquynptonhc.Khidyhcsinh tnh o hm bc cao, Gio vin limt ln na dng phng php ny gii ton.Cn phn bit suy on vi suy din. Chng hn tnh o hm bc n ca hm s y =

, hc sinh c th suy on y(n) =2ne2x nhng phi chng minh tt nht y l phng php quy np ton hc. GVkhngcntngminhdychohcsinhphptamonlunkhngnh, ph nh, la chn, bc cu. GV c th ch ng a ra cc suy lun sai hc sinh phn tch v trnh vp phi sau ny. cbit,cnlmchohcsinhnmcphngphpphntchiln,phn tch,tng hp, phn chng, quy np. : 2011 44 GV cn tn dng bt c c hi no, min l hp l, khc su kin thc logic cho hc sinh. Chng hn lp 10 i vi h phng trnh {

Th vic phn tch hai yu cu sau y l khc nhau chnh l tng cng kin thc logic: -Tm a sao cho vi mi b lun tn ti c h c nghim. -Tm a sao cho tn ti c h c nghim vi mi b. Hc sinh nm vng cc kin thc v logic s hn ch c sai lm khi gii ton. - Tnhhung4:Trangb phuong phap giai cacbaitoan coban nhuth nao d trnh sai lam hoc sinh khi giai ton? C th ni rng cc loi ton c bn trong chng trnh i s - Gii tch PTTH u cphng phpgii.Victrangbcc phng php gii nychnhlmchohc sinh nm vng cc loi ton c bn. Phng php gii h {

ccptrongchngtrnhphthnghailncungcptibncchgii: php th, php cng i s, phng php th v phngphp nh thc. Nhiu hc sinh thng tnh ngay D = |

| khi bt u li gii. Chng ti c s sau: : 2011 45

Hnh 3: thut ton gii h hai phng trnh bc nht hai n. T s trn hc sinh cn thy c Nu a1, a2, b1, b2 khng ng thi bng 0 v D = 0 th h c nghim khi ch khi Dx = Dy = 0 . y l kin thc quan trng hc sinh trnh sai lm khi gp bi ton Tm a sao cho vi mi b lun tn ti c h sau c nghim: 2

bx y ax by c c+= + = + Trong chng trnh i s 10, Hc sinh cn nm vng phng php gii so snh cc nghim ca tam thc bc hai f(x) = ax2 + bx + c ( a 0) vi mt s o . Gio vin cn tng kt theo s sau: V nghim V s nghim ph thuc vo 1 phng trnh. V nghim D =0 0 0 =0 =00 =0 0 : 2011 46 Hnh 4: Thut ton so snh cc nghim ca tam thc bc hai f(x) = ax2 + bx + c ( a 0) vi mt s o . Vic rn luyn cho hc sinh lp cc s trn va lm hc sinh nm vng phng phpgii,vaphttrintduychohctpnichungvhcbmnTonhcni ring. T hc sinh c th trnh sai lm khi gii ton. Tuy nhin cng c th lu hc sinh l vi mt loi ton c th c nhiu phng php gii khc nhau, hc sinh cn bit la chn phng php gii ti u gii quyt bi ton c th. Ccphngphpgiithngxuynccngchcsinhcnmvng. Chng hn bi ton so snh nghim ca tam thc bc hai vi cc s (lp 10) cn c s dng xt phng trnh lng gic, phng trnh m, bt phng trnh m, phng trnh logarit, bt phng trnh logarit, xt chiu bin thin v cc tr hm s. VN - - - - + + + : 2011 47 T li gii mt bi ton c th, GV cn gi cho hc sinh tm ra phng php li gii cho mt lp bi ton. Bin php ny gip hc sinh hiu bn cht li gii c th v t duy khi qut c pht trin. Trnh tnh trng lm bi no bit bi y; thy cy mchngthyrng.Lmciunygiovingiphcsinhhcmtbit mi. Th d khi hc sinh gii bi ton Chng minh sin20o >

th da vo sin60o = 3sin20o 4sin320o, hc sinh a ra f(x) = 4x3 3x +

nhn sin20o lm nghim, sau nhn thy sin20o,

thuc (-

;

) l min nghch bin f(x) nn sin20o >

f(sin20o) < f(

)0 <

-

v t d dng gii xong. Khng dng li gio vin cho tip cc bi tp ( c th l v nh): so snh cos10o vi

;

+9

vv hc sinh tng kt thnh phng php gii lp bi ton. Vic tng kt v h thng li cc phng php gii s gip hc sinh bt lng tng, vp phi sai lm ng tic khi gii ton. C. 2. Bin php 2 : Trang b cc kin th bn v p pp ii to, c bit l vic t kim tra pht hin ra cc sai lm trong cc li gii. Nhng kin thc cn thit v phng php gii ton c nhiu tc gi trong v ngoi nc nghin cu kh su sc. Ni bt l nhng nghin cu ca G.Poolya. Ngoi ra, cc tc gi Nguyn B Kim, VDng Thy, Nguyn Thi He, Hong Chng cng nhn mnh v c th ha cc t tng ca G.Poolya. Cc tc gi i su vo vic tr li cu hi : Tm ti li gii bi ton nh th no: -Tm hiu ni dung bi ton -Xy dng chng trnh gii -Thc hin chng trnh gii -Kim tra v nghin cu li gii. : 2011 48 Khngnhngth,G.Poolyacnaramtbngicthrtcchchomi ngi gii ton. ChngtiluthmnhngkincaL.MPhoritmanE.N.TuretkiV.la. Xtetxenc v s cc bc tm kim li gii ca cc bi ton nh sau: -Trong khi c k bi ton, cn phi c gng xc nh c bi ton thuc dng no -Nu cc bn nhn c trong bi ton chun ca dng quen thuc, th hy vn dng qui tc bit gii. -Nu bi ton l khng chun th cn phi hnh ng theo hai hng: tch t t bitonrahocchianhbitonrathnhnhngbitonnhcdngchun(th php chia nh) hoc din t li bi ton theo mt cch khc, dn bi ton n mt bi ton c dng chun ( th php m hnh ha) . Tuynhincchngdntrncngchlconngchung,chaphilcon ng ring hiu qu cho mi bi ton c th. Ngay G.Poolya cng tha nhn: .. t c kt qu thc s th anh ta cng cn hc tp c cch suy lun c l, l suy lun m ton b hot ng sng to ca anh ta s ph thuc vo . . p dng mt cch c hiu qu cc suy lun c l nh l mt k nng thc hnh v k nng cng nh mi k nng thc hnh khc u hc c bng cn ng bt chc v thc hnh. Ti d nh lm tt c nhng g ti c th lm c gip bn c ham mun hc thng tho cch suy lun c l, song tt c nhng g ti c th ngh th ch l nhng th d mu v kh nng thc hnh chu o . Vi mc ch hn ch v sa cha cc sai lm ca HS khi gii ton, chng ti ch ti hai bc cui cng trong bn bc m G.Poolya nu ra: -Thc hin chng trnh gii vi cc gi nhm trnh sai lm : hy kim tra li tng bc, c th chng minh c tnh ng n ca tng bc lm hay khng? -Tr li cch gii vi cc gi kim tra: c th kim tra li kt qu? C th kim tra li ton b qu trnh gii bi ton khng? Nh thc hin bin php 1, trong c vic trang b cc kin thc v lgic cho HS m vic thc hin kim tra s c l ca tng bc suy lun thc hin c thun li. : 2011 49 Li gii th d 1, mc 1.4 trang 9 li mc sai lm bc cui cng khi tng rng t F(x,y)>0 vi mi x,y thuc R th min F(x;y) = 0. Hs s trnh c sai lm nu tm cchchrax,ysaochoF(x;y)=0.Rrng,khngcgitrnoca x,ythamn iu ,nn Hs s thc hin con ng khc gii bi ton. thc hin bin php 2, chng ti s bn k ti ni dung sau: GV cn trang b cho HS phng php nhn bit ra li gii sai( mt s du hiu dn ti phn th d, tuy nhin i vi nhiu HS th ngay vic hiu nh th no l mt phn th d cng ang cn kh khn).Ti thiu, GV cn trang b cho HS cc du hiu quan trng sau y. -Duhiuthnht:Ktqualoigiaimuthuanvoiktquatrongtruonghop ring. Mt lot th d m chng ti trnh by chng 1 minh ha cho du hiu quan trng ny.-Du hiu th hai : - Truong hop ring kt qua khng thoa mn bi ton Chng hn vi ch ny HS c th thy cc nghim ngoi lai ca phng trnh , bt phngtrnhhocccloi h.Ccnghimngoilai xut hinchngtvicbin i khng tng ng hoc cn thiu mt bc quan trng trong li gii. -Du hiu th ba: Kt qua loi giai bi ton khng chua kt qua trong truong hop ring. i khi du hiu ny th hin nh du hiu th nht. Khi c bitha bi ton s dn timtktqu c thmktqucth nyphi thuc trong kt qu ca ton b bi ton. Nu iu ny khng xy ra th li gii c sai lm. Chnghn,khigiibitonTmmthy=x4 mx3+x2ctrcixng thng ng c th thy rng trong trng hp c bit m=0 th hm s tr thnh hm chn nn c trc i xng x=0 tha mn bi ton. Chng t trong cc gi tr cn tm ca m phi ti thiu c m=0. Nu kt qu li gii thiu m=0 th chc chn li gii c sai lm. : 2011 50 -Duhiuth:Ktquabitoncthkhcktquabitontngquatd bit. Nhiu bi ton m HS gii c th tng qut v c kt qu tng qut. Nu kt qu ca bi ton c th mu thun vi kt qu ca bi ton tng qut th r rng li gii c sai lm. Chng hn, ng hypepol, 2ax' 'bx cya x b+ +=+c hai tip tuyn vung gc khi v ch khicimccivcctiu.lktqutngqut.HStmmth my xx= +chaitiptuynvunggcthdtkhotpsphilm>0(tmm hm s c cc i, cc tiu l bi ton n gin hn), nu p s khc th chc chn li gii c sai lm. Mt th d khc, qu tch cc im k c hai tip tuyn vung gc vi nhau ti parabol y= ax2 +bx + c l ng thng cng phng vi trc honh (c th vit c phng trnh tng qut). Vy khi HS tm qu tch ny i vi hm bc hai c th gp kt qu mu thun th li gii c sai lm. Hay l, hm s 2ax' 'bx cya x b+ +=+ nu y c hai nghim phn bit x1,x2 th1 22x x ba+ =NuHStnhohmcahms 212x xyx+ +=+ cychainghimkhngtha mn 1 222x x += th php tnh o hm y hoc php tm nghim ca y b sai lm. Do trong nhng loi ton c th c, GV nn cung cp cho HS kt qu tng qut bng cch lm bi ton tng qut hoc dng li mc thng bo kt qu HS c vn kin thc kim tra li li gii c th. -Du hiu th : Kt qua tim duoc mu thuan voi thuc t. Tt nhin, nhiu khi du hiu ny xut hin do chnh bi ton ban u mu thun vi thc t. y, gi s rng bi ton cho ph hp vi thc t m kt qu mu thun vi thc t th li gii mc sai lm. : 2011 51 LigiingybinnitingchngminhlcsAsinchythuamtconra, khng bao gi ui kp con ra nu con ra chy trc l mu thun vi thc t, do sai lm. Tt nhin t vic bit ti li gii sai n vic ch ra c li sai, nguyn nhn sai l mt qu trnh rn luyn tch cc mi t c. Thctligiimuthunnhiukhilthctcathcnghimtrnmhnh. Tonhcchyulkhoahcsuydinvquynp.ThnhnggpnhngHS thcnghimtrnmhnhdonktquvkimtraligiicamnh.Khi chng ti cho bi ton Cho hai im A v B chuyn ng trn ng parabol y =x2 sao cho AB =1, tm qu tch trung im I ca AB, c HS ly on thc k cho 2 u trt trn parabol nhn nh kt qu ca bi ton l ng thng hay ng cong l sai ( tuy HS khng gii c nhng vn d on ng nhng kt qu khng th c).Trong cc bi ton tnh s ngi, tnh s cc kh nng m kt qu khng phi s nguyn dng th chc chn li gii sai lm. -Du hiu th su: Kt luan khong binh dng gia cc yu to binh dng cua gia thit. Vic phn tch cc gi thit, cc iu kin ca bi ton v c kt qu ca n gip cho ngi gii ton thy r qu trnh xy ra c quy lut ca mi bi ton. Ni c th hn l ngi gii ton s bit c vi cc gi thit, cc iu kin cho nh vy th tt yu kt qu phi din ra nh th no?. y chng ti nhn mnh s bnh ng ca cc yu t gi thit bi ton s tt yuthhintiptcsbnhngktlun.Khngxyraiunythchc chn li gii sai. - Duhiuth by: Kt qua cua loi giai ny khc kt qua cua loi giai khc, m loi giai sau c hnh anh tin cay. Nhiu khi kim tra li gii, ngi ta gii bi ton theo mt cch khc. Nu kt quca haicch giimuthunthct nhtmttronghailigiilsai.Khi : 2011 52 phi tin hnh kim tra tng bc thao tc ca mi li gii hoc dng cc du hiu trnh by trn xem xt li gii no sai( c khi c hai li gi u sai). Thng thng ngi ta hay kim tra li gii bng mt cch nhn khc, chng hn bng hnh v, tuy c th cch nhn ny khng cho kt qu c th cho li gii, nhng c th bo hiu cho chng ta mt du hiu li gii sai. -Du hiu th tm: on v do hai v cua mot dng thuc khc nhau( sai lam v thu nguyn). Trong cc bi ton c ni dung tnh ton mt i lng khc, cn lu ti n v o ca cc i lng ny. C. 3.Binphp3 HS c th t tng xuyn vi nhng bi ton d dn n sai lm khi gii ton. ylbinphpthngtrc,kckhisailmnocphntchvsa cha cho HS. thchinbinphpny,giovinphibittccbitoncchacc by. Chngtixinnhnghadidngmtkhinimbynhmtheongha tchcc vc tnhsphm.Chngti khngthin vsnhhc tr.Bis nh qu mc s dn HS n b tc ch khng sa by, khi cc bin php c ch nh v s phm s khng thc hin c. Chng ti cho rng by phi lm cho bi ton c tnh hp dn. chnh c trng ny lm cho HS tch cc tham gia hot ng gii ton. HS ch quan ngh rng bi ton khng c g un khc v d dng a ra li gii ca mnh cng vi sai lm do gio vin d kin trc. Chng ti thc nghim v by trong mt t iu tra 43 hc sinh lp 11A2 trng THPT Quc Hc. Vi bi ton Chng minh vi mi a, b,c th : (a2 + b2)(b2 + c2)(c2 + b2) 8a2b2c2 li cun 98,5% HS tham gia gii v c li gii. Hu ht HS lp 11 u nhn xt l bi ton d, ch lm cha n 10 pht. Nhng c ti 42% (195 HS) b chung mt sai lm khi nhn v theo v ca 3 bt ng thc: a2 + b2 2ab; b2 + c2 2bc; c2 + a2 2ca : 2011 53 ciuphichngminh,mcd2ab,2bc,2cachaphingthilccs khng m.Khi bit mnh b sai lm do li ny, nhiu HS rt thm tha v mt quy tc suy lun sai lm khi chng minh bt ng thc, mc d hu ht HS l HS cc lp chuyn. Nhvy,tmcchsphmthbyphilmchobitonctnhth thch o vng vng v nhng kin thc c th ca HS. C nhng bi ton c ci t lin tip cc by. HS vt qua c tt c cc by i n li gii chnh xc, chnh l qua mt hnh thc o kin thc. Bi ton tuy nhiu by nhng HS kh nhn ra. Chng hn, bi ton Gii bt phng trnh

+ 3(

> 12 lin tip c ci cc by sau y : - HS t t=(

> 0 a v gii t2 + t 12 > 0 v vit t1 = 3; t2 = -4 (loi). y iu kin t> 0l iukincanghimbtphng trnhchkhngphiiu kincho nghim tam thc bc nn khng th ni n vic loi gi trt2 = -4. - HS vit (

1 b sai lm v khng ch n c s

(0;1). - HS vit

< -11< -x b sai lm v khng th nhn hai v ca bt phng trnh vi x khi cha bit du ca x. Bitontrnhontonkhngqukh nhngkimtracnhiukinthc ca HS. Vy cui cng, by trong cc bi ton l cc kin thc mHS d b sai lm mt bc no ca li gii, cc kin thc ny c s chun b c ch nh ca GVnhmtc tnhhpdn cng vitnhththch iviHS.Torac by trong mt bi ton chnh l mt ngh thut s phm ca GV. Vic to cc by HS mc sai lm chnh l s phng trnh ch ng cc sai lm c th xut hin. Cc by ny cn cng c li, nhm xa hn nhng sai lm ca HS c sa cha trc . Tuy nhin cn s dng cc by c mc . S lm dng qu bin php ny s dn n s nh ch khng phi th thch HS v mc ch s phm. C. 4.Bin php 4: Theo di mt sai lm ca HS khi gii to q ii on. : 2011 54 tng cng hiu qu ca cc bin php trn, GV phi nhn thc c cc giai on c th ca mt sai lm no . i vi mt sai lm (GV c th d on trc) th tnh giai on th hin kh r. -G n 1: Sai lam chua xuat hin. giai on ny, cc bin php c huy ng nhm phng trnh sai lm xut hin. Khng c thc v vic ny chng ta d thiu tch cc trong giai on 1. Bin php s dngchyutrong giiai on nyl trangb ttkinthccho b mn ton (bin php 1), kin thc v phng php gii ton (bin php 2). Mt iu cn lu giai on ny l GV c th d bo trc cc sai lm v th hin cc ch i vi HS. Chng hn, GV c th ch bt ng thc Cauchy ch c p dng cho cc s khngm,vvychngminha(1a)

bngcchpdngbtngthc Cauchy cho hai s a v 1 a l sai lm. Tt nhin, d bo tt, GV phi c trang bhiubitvccsailmcaHSkhigiitonvphicnnglcchuynmn, kinh nghim s phm. -Giai n 2: Sai lam xuat hin trong loi giai cua HS. y l giai on i hi GV phi kt hp c ba nguyn tc kp thi, chnh xc vgiodc,cngvistchcchacaHSvndngcchiubitvvic kimtraligii(binphp2)nhmtmrasailm,phntchnguynnhnvsa cha li gii. Quy trnh giai on ny l GV theo di thy sai lmGV gi HS t tm ra sai lmHS t tm ra sai lmGV gi iu chnh li giiHS th hin li gii ngGV tng kt v nhn mnh sai lm b mc. Nhiu sai lm ca HS kh tinh vi, c khi GV khng pht hin kp thi. Giai onny i hi GV phi c thi i x kho lo s phm tng hiu gio dc. : 2011 55 Ty theo mc sai lm m GV quyt nh s dng cc bin php s phm thch hp. C khi GVcn a ra li gii ng HS i chiu v tm ra sai lm ca li gii sai, y cng l mt cch gi HS nhn ra sai lm. C khi GV ch ng a ra li gii sai HS nhn dng cc du hiu tm ra sai lm. C khi GV a ra nhiu li gii khc nhau HS phn bit s ng sai ca cc li gii, c th s dng phng php trc nghim mi HS u phi suy ngh v c kin. Giai on ny, GV d to ra c nhng tnh hung th v, c th ht huy u dim ca nhiu phng php dy hc nh : dy hc gii quyt vn , dy hc phn ha, dyhc theo l thuyttnh hung, dyhcmthoi, trn quan imhotng ca qu trnh dy hc. Ngc li, nu giai on ny GV khng kp thi phn tch v sa cha cc sai lm ca HS khi gi ton th cc sai lm s ngy cng trm trng , GV khng hon thnh nhim v dy hc, HS s st km v kt qu. -G n 3 : Sai lam d duoc phn tch v sa cha. MtsailmcaHStuycGVphntchvsacha,vncncthti din. Chng ta lu tnh ca t duy. c bit l cc sai lm gy ra t cc thi quen khng tt.Vicdtb thi quen khng n ginv thi quen nmtrong np sngca mt con ngi. Song song vi vic dt b mt thi quen, GV cn xa b hn sai lm ca HS. Vicchiabagiaionivimtsailmchcnghanhnmnhthiim ca sai lm. Trong mt thi im dy hc GV c khi ng thi tc ng n cba giai on, bi v va phng trnh cc sai lm cha xut hin, va lo phn tch v sachaccsailmangxuthin,ngthiloxahnnhngsailmsa cha. S sau ch r s kin tr xa b mt sai lm ca HS: : 2011 56 Hnh 5: Qu trnh xa b mt sai lm ca HS. D. Cc yu cu i vi hc sinh v gio vin Cc bin php s phm c xut ch t c hiu qu khi c s m bo ca cc yu cu v HS v GV. y l hai i tng ch yu tham gia thc hin cc bin php s phm. GV cn lu cc yu cu sau: D. 1.Rn luyn thc v ch hc tp cho hc sinh. Vic xc nh mc ch hc tp mn ton s nh hng rt ln ti iu kin c thc hc tp mn ton ca HS. Mc ch chung nht ca vic hc tp chnh l ng c hc tp. Nhiu HS cha xc nh c ng c cng nh mc ch ca vic hc ton. ThmchcHSchquannimtonlmnhcthicnnbtbucphihc.HS khng bit rng ton l mn hc gip rn luyn tr tu n v hiu qu nht. S n iu v phng php ging dy ca GV cng lm cho HS coi mn ton l mt gnh nng trong vic hc tp ph thng. Sai lm cha xut hin Sai lm xut hin Phong trnh Phn tch sa cha Cng c th thch Sai lm c xa b : 2011 57 HSchathccrngsaukhittnghipTHPT,dlmnghgtrongxhi cng cn n hc vn ton ph thng ( kin thc, phng php, k nng) . Theolthuytthngtinthmttrong3nguynnhndnntnhtrngmt thngtinlthckmcangithunhnthngtin(hainguynnhncnlilt chcthngtinvkthutthngtin).ChnhtHSnhnthcbigingcaGV khng y , khng chnh xc v dn n ti sai lm khi gii ton. Thiu thc hc tp, HS khng c k hoch thng xuyn n tp li cc kin thc chc.DocckinthccbndndnmnhttrongnhnthccaHS, nhiu l hng v kin thc xut hin, thm ch ch hc tp cng dn st km. Nhiu HSkhigiitonkhngctinhthnvtkh,khngchutnhtoncnthn,khng chu kim tra li kt qu lm v t dn n sai lm. D. 2.Hnh thnh hot ng hc cho hc sinh. Dy hc tp trung vo HS, ngha l ly HS lm mc tiu phn u ca s nghip GD v ly HS lm ng lc chnh tin hnh ton b qu trnh dy hc. T quan im dy hc ny nn c phng php dy hc pht huy tnh tch cc ca HS. Chnh v dy hc tp trung vo HS v nhm khc phc cc nguyn nhn sai lm ca HS khi gii ton, chng ti rt ch n hnh thnh hot ng hc cho HS theo mc ch ny. Chng ta bit rng hot ng hc ca HS bao gm 3 yu t ch yu l hot ng hc tp, mc ch hc tp v hnh ng hc tp. GV nn thy rng ng c hc tp ca HS khng th p t t bn ngoi m phi di s hng dn ca ngi thy. GV cn lm cho HS thy c s hp dn ca mn ton. S hp dn ny kch thch s mong mun chim lnh kin thc. D. 3.Xy dng uy tn ca gio vin da trn s b d c chuyn mn v phm cht ngh nghip. Chng ta bit rng: Nhn cch ca ngi GV l nhn t c ngha to ln i vi cht lng GD : 2011 58 Nhn cch gio vin c cu trc nh sau: Hnh 6: Cu trc ca nhn cch gio vin. GVkhngtthngxuynrnluynnnglcchuynmncamnhnndyHS khhiu,thmchnhiuGVilcgiisaiccbiton.GVbtpdng dy r c cu HS phn c m nt. C nhng GV vng kin thc chuyn mn nhng li yu v mt nng lc s phm, nn cng dy, HS cng kh hiu, cng thy vn phc tp v dn HS n cc sai lm. Nhiu GV khi dy nh l, khi nim, quy tc cha c kh nng phn tch r cu trc ton hc r rng v cha d kin c cc sai lm d mc phi ca HS. Ktlun:Bn bin php v ba phng chm ch o nhm mc ch lm cho HS c c kin thc chun xc th hin qua s : Nhn cch Xu hng T tng Kin thc c bn Tnh cht kh cht Phng php dy hc Hiu hc sinh T chc Giao tip Nng lcPhm cht : 2011 59 Hnh 7: Qu trnh HS c c kin thc chun. Xy dng uy tn thc hin bin php s phm Rn luyn thc, ch. Hnh thnh hot ng hc. B sung kin thc. Tm ra, sa cha sai lm Kin thc chun Gio vinHc sinh Phm cht v nng lc Yu t tm l v kin thc cha tt Truyn th kin thc v phng php Lnh hi kin thc v phng php Hc sinh sai lm khi gii ton : 2011 60 hng III THC NGHIMS M 1.Muc ch thc nghim -Sng t thm sai lm ca hc sinh khi gii ton l tnh trng ph bin hin nay, k c hc sinh chuyn Ton( nhng hc sinh c coi l c trnh kh gii v ton). -Thcnghimcnnhmthnghimccbinphpdyhcthchhppht hin, phn tch, hn ch v sa cha cc sai lm ca hc sinh khi gii ton. T c th xem xt tnh kh thi v tnh c hiu qu ca cc bin php xut. 2.Ni dung thc nghim -Sng t thm sai lm ca hc sinh khi gii ton l tnh trng ph bin hin nay, k c hc sinh chuyn Ton (nhng hc sinh c coi l c trnh kh gii v ton). -Trang b cho hc sinh cc kin thc v phng php gii ton, c bit l cc du hiu pht hin li gii gii sai, to cc by trong cc bi ton,nhm rn luyn cho hc sinh hn ch cc sai lm khi gii ton. 3.T chc thc nghim csngca ccBan gimhiu nh trng,chng ti chn itng thc nghiml hc sinh lp 11A2 chuyn Tontrng THPT chuyn Quc Hc.cimcaitngthcnghim:lhcsinhlpchuynToncamtngi trng t chun quc gia, kinh nghim dy mn chuyn rt dy dn, c th ni l mtmi trng o to rt tt cho hc sinh chuyn. Chng ti tin hnh iu tra cht lng u ca 43 hc sinh trong lp ny bng cng mt phiu iu tra ( Ph lc 1). 4.hng php tin hnh Phng php iu tra: Lp phiu iu tra gm 10 cu hi c gii sai hon ton. Pht phiu cho 43 hc sinh trong lp 11A2. Yu cu cc em nh gi bi lm trong phiu iu tra thng qua vic chm im (1im/cu). Thu phiu iu tra v nh gi mc nhn thc cc sai lm ca hc sinh theo cch thc: -Hc sinh cho im cng cao chng t cc em vn b nhm ln kh cao. : 2011 61 -Hc sinh cho im cng thp th mc nhn thc sai lm ca cc em cng tt. 5.t lun thc nghim -Tuy l cc hc sinh u t, c thnh tch hc tp tt nhng cc em vn mc cc sai lm kh ph bin. -Mt s em vn b nhm ln kh trm trng. -Theo di s thu c t cuc kho st: S liu thng k s lng hc sinh cho im bi lm trong phiu iu tra.

im 0 1 2 3 4 5 6 7 8 9 10 S hc sinh cho im ngi) 30649874110 Ti l (%)70149.320.918.616.39.32.32.30 : 2011 62 2: T l cho im bi lm trong phiu iu tra. (%) 6. nghi v mt s hiu bit qun trng cu sinh vien s pham ton Qua nghin cu v quan st ca mnh, chng ti c mt kin ngh tha thit sau y:Nhngsailmcahcsinhkhigiitonlmthiubitquantrng cm . Tht vy, nhn li chng trnh hc ca cc sinh vin s phm ton, chng ti thyccmnctnhchtotonghdyhcnhTmlhc,Giodchc c ging cha su sc v t c th d vn dng vo vic ging dy ton. T sinh vin khng thch hc hoc hc i ph vi mn ny. im 0im1im 2im 3im 4 im 5im 6im 7im 8im 9 : 2011 63 Nhng kin thc v cc sai lm ca hc sinh khi gii ton, thc s l mt hiu bit c tnh ngh nghip cn c trang b cho sinh vin s phm ton. Chng ti xut 3 con ng a hiu bit ny n sinh vin s phm ton. -ngthnht:Ccnghincu,aratrongticbsung xen vo trong chng trnh cc mn hc lin quan ti chng trnh ton PTTH. Ging vin cc mn ny c trang b cc ti liu cn thit kho lo a vo chng trnh dy cc thi dim thch hp. C th a nhng hiu bit ny vo trong chng trnh ngoi kha. -ngthhai:Binthnhmtchuyntrongchngtrnhrn luyn nghip v s phm cho sinh vin nm th ba. -ngthba:tvnchosinhvinlmcciutravnhng dngsailmcahcsinhtrunghcphthngkhigiitonthng dn sinh vin lm nin lun, kha lun v cc vn t m hn. y l con ng gn sinh vin vi thc t v tp dc cho sinh vin nghin cu khoa hc gio dc. T chng ti s gn cht thm ba nh ca tam gic o to s phm : Phn nh thc tin o to i ng Hc ngh s phm Thc Tp S Phm Trang b kin thc Thc tin Hnh 8: Tam gic o to s phm. Trng S PhmTrng Ph Thng Sinh Vin : 2011 64 VIII.Ti liu tham kho. [1]Nguyn Ngc Bo, Mot vai suv nghi v tnh tch cuc, tinh doc lap nhn thuc v moi lin h gia chng, Hoi thao: 'i moi giang day, nghin cuu tm l hoc v gio dc hoc` i hc Quc gia H Ni, Trng i Hc S Phm, Khoa Tm L Gio Dc, H.1995. [2]Nguyn Thanh Bnh, Kha nng giao tip cua sinh vin su pham trong thuc tap tot nghip, NCGD, 12-1994, tr23. [3]Phm Thanh Bnh, i moi manh m phuong phap dav hoc truong ph thng- yu cau cap bch cua gio dc hin nay, NCGD, 3-1995, tr10. [4]Nguyn Mnh Cng, V Dng Thy, Nghin cuu xy dung h thong bi tap va phuong phap day hoc ton, NCGD, 11-1986, tr. 9-11 [5]Nguyn Vnh Cn, L Thng Nht, Phan Thanh Quang, Sai lam ph bin khi giai ton, NXB Gio dc, H 1996.[6]Phan Hu Chu, Trn Lm Hch, Nhap mn l thuyt tap hop v logic. NXB Gio dc, H.1997. [7]Hong Chng, Phuong phap day hoc ton hoc, NXB Gio dc, H.1978. [8]Hong Chng, Rn luyn kha nng sang tao Ton hoc ph thng, NXB Gio dc, H. 1969. [9]Nguyn Ngha Dn, Phuong phap giao dc tch cuc voi mc tiu nhn cch sng tao, NCGD, 8-1996, tr5. [10]Phm Tt Dong, Can nang cao nng luc ki thuat cho hoc sinh ph thng,NCGD, 11-1977, tr 27-30. [11]Phm vn ng, Phuong phap day hoc pht huy tnh tch cuc mot phuong php v cng qu bu, NCGD, 12-1994, tr 1-2. [12]Phm Minh Hc, Phuong phap tip can hoat dong nhn cch- mot co s l luan cua phuong phap day hoc hin dai, Thng tin KHGD, tr 7-10. [13]Phm Vn Hon, Trn Thc Trnh, Rn luyn con nguoi qua day ton, NCGD, 10-1975, tr17-25. [14]Trn B Honh, Bn tip v day hoc lay hoc sinh lm trung tm. Thng tin KHGD 49 (1995), tr22-27. [15]Nguyn Thi He, Rn luyn tu duv qua giai bi tap ton, NXB Gio dc, H.1995. [16]Nguyn B Kim, Chnh xc ha mot so khi nim lin quan dn day hoc giai quyt van d, NCGD, 9-1991, tr2-3. : 2011 65 IX.hu luc. PI PHIU IU TRA NGHIN CU V SA CHA SAI LM CHO HS THPT KHI GII TON H v tn:Lp: Nhn xt v cho im tng cu(1im/cu) x x 16Bi 1: Gii Phng trnh:Gii: K: 164

6456 4

65 59 *

Bi 2: Gii phng trnh:

1 1 1 Gii: K: 1 Lc ta c: 1 1 1 Vi1 th 1 1 Vy pt v nghim.

1

Bi 3: Gii bt phng trnh:Tng im: : 2011 66 Bi 4: Chng minh rng vi a, b, c >0 th:

Gii: p dng BDT cauchy ta c:

V cc v u dng nn:

Bi 5: Tm gi tr nh nht ca : F(x,y) = (x+y)2 + (x+1)2 + (y+1)2 Gii: Vi mi x, yR th: (x+y)2

(x+1)2

(y+1)2

Vy F(x,y)hay min

x Bi 6: Tnh: L = lim

Gii: Ta c: lim

=0, lim

=0 . lim

Nn L = 0 Bi 7:Tm ng tim cn ca ng

1 1

t f(x) =

. Khi f(x) ng bin trn R nn vi x>-1 th: Vy nghim ca BPT l 1 : 2011 67 y =

Gii: V lim

= nn th chai ng tim cn ng l x = 1. V tp xc nh ca hm s l (-1; 1) nnlim khng tn ti.Suy ra th khng c ng tim cn ngang. Bi 8: Mt d tic c 10 nam v 6 n u khiu v gii. Ngi ta chn 3 nam 3 n ghp thnh 3 cp nhy. Hi c bao nhiu cch ghp 3 cp nhy ? Gii: Mi cch ghp 3 bn nam trong 10 bnnam l mt chnh hp 3 ca 10 nn s cchchn 3 bn nam c th t l

8917cch. Tng t s cch chn 3 bn n c th t l

4561 cch. Vy s cch b tr 3 cp nhy l:

7 1864Bi 9:Tnh gii hn:L = lim

Gii: L = lim

= lim

= 1. Bi 10: Tnh tch phn:I =

Gii: I =

=

|

=-4/3. Nhn xt: : 2011 68