JOURNAL OF SCIENCE OF HNUE
Educational Sci„ 2009, Vol. 54, No. 8, pp. 3-13
T A P L U Y E N C H O H O C S I N H H O A T D O N G L I E N T U ' D N G VA H U Y D O N G K I E N T R U ' C T R O N G Q U A T R I N H
C H I E M L I N K T R I THLfC T O A N H O C
Nguyin HfJu Hau
Trudng THPT Dong San 2 - Thanh Hod
1. Md dau
Day Toan la day hoat dong Toan hpc [1;12], cho nen mot trong nhung yeu cau ciia day Toan la phai khai day dugc kha nang doc lap suy nghi va kham pha cua ngUdi hoc. Kha nang van dung tot nhiing kien thiic da dugc hoc de giai quyet nhQng vin de mdi la mot trong nhQng nhiem vu quan trong ciia viec hoc.
Lien tudng va huy dpng la nhQng kha nang rat quan trpng can phai ren luyen cho HS. Ngu cd kha nang lien tuong tdt thi khi dQng trUdc mot bai toan kho, hpc sinh se lien he vdi cac kien thQe lien quan co kha nang giai quyet van de. NgUpc lai, neu kha nang lien tudng kem thi khi gap mot van de nidi HS thudng khong biet dat nd trong moi lien he vdi cac kign thQe da biet, vi thg each nhin vS,n de mang tinh cue bp, rdi rac, trong khi Toan hpc la mot he thdng cac kign thiic cd lien he mat thiet vdi nhau.
2. N o i d u n g n g h i e n ciJu
2.1. K h a i n i e m h e n tu'dng va vai t r o ciia h e n tifdng diidi goc do t a m h
Theo tQ dign Tiing Viet, lign tudng cd nghia la: "Nhan su vat, hien tUpng nao dd ma nghi din sU vat hien tUpng khac c6 lien quan" [10; 568].
Ta CO thg higu khai quat vg thuyit lien tudng thong qua mot so luan digm chfnh sau:
- Tam If dupe cau thanh tQ cam giac. Cac cau thanh cao hpn nhu bigu tUpng, y nghi, tinh cam,... la cai thQ hai, xuat hien nhd lign tudng cac cam giac. Noi each khac, con dudng hinh thanh tam If ngudi la lien kgt cac cam giac va cac y tudng;
- Dien kien dg hinh thanh lien tUPng la sU gan gui ciia cac qua trinh tam li;
- Su lien ket cac cam giac va y tucfng dg hinh thanh y tudng mdi khong phai la su kit hpp gian don cac cam giac hoac cac y tudng da cd, ma la sU kit hpp phQc tap tao ra nhung y tudng mdi doi khi mang tinh nhay vpt;
- Cac lien tudng bi qui dinh bdi su linh hoat ciia cac cam giac va cac y tudng thanh phan dupc lign tuctng va t i n so nhdc lai ciia chiing trong kinh nghiem;
- Cac lien tudng dupc hinh thanh theo mot so qui luat: Qui luat tuong tu. Qui luat tuong can. Qui luat nhan qua [6;33,34].
Cac nha lien tuong hpc quan tam hon ca tdi toe dp va miic dp lien kit cac hinh anh, cac bigu tupng da cd. Theo hp, cd bon loai lien tuong la: Lign tudng giong nhau, Lign tudng tUdng phan, Lien tudng gkn ke nhau vi khong gian va thdi gian, va Lign tudng nhan qua, trong dd, Lign tudng nhan qua cd vai trd dac biet quan trpng [6;34l.
Theo Bill Van Hue, lign tudng dupc chia thanh bdn loai: lien tudng gdn nhau ve khong gian va thdi gian, lien tudng giong nhau vi hinh thii hoac noi dung, lign tudng trai ngupc nhau, lign tUPng nhan qua. Theo tac gia, lien tudng co vai trd rat cjuan trpng trong viec ghi nhd va nhd lai [3;69].
Nha tam If hpc P. A. Sgvarev da nghign cQu ti mi nhung moi lien tudng khai quat dpc dao va vai tro ciia chiing trong day hpc. Ong chi ra rang, nhung moi lign tudng khai quat bao gom ba kieu cP ban: nhung lien tudng dupc biin ddi mot nua, nhung lign tudng - biin thien va nhQng lien tUPng cu thg - biin thien [7; 136].
K. K. Plantondv xem tU duy la mot qua trinh gom nhieu giai doan ki tiip nhau, cu thg la: xudt hien cac lien tUdng, sang Ipc cac lien tudng va hinh thanh cac g i a t h u y i t [4;121].
Theo tac gia Nguyin Ba Kim va Vu Duong Thuy: "Trong day hpc, cdn chu y ren luyen cho HS ki nang biin doi xudi va ngUpc chiiu mOt each song song, nhdm giup cho viec hinh thanh cac lign tudng ngupc diin ra dong thdi vdi viec hinh thanh cac lien tudng thuan" [5; 174].
Nhu vay, co thg thay rang: Vai tro cua lign tUPng trong qua trinh t u duy rdt quan trpng, lien tuong ciing ddng vai tro quan trpng trong hoat dpng t u duy khi giai toan. Dao Van Trung cho rdng "Lign tudng la doi canh ciia tu duy, la cdu noi giai quyit vdn de" [9;73].
2.2. Lien tif dng va h u y d o n g kign thtJc t r o n g d a y h o c T o a n Khi d i cap din su phan loai cac bai toan, G. Polia quan niem: "Mot su phan loai tot phai chia cac bai toan thanh nhung loai (kiiu, dang) sao cho mdi bai toan xac dinh trudc mot phuPng phap giai". Dua vao muc dfeh cua bai toan, dng chia cac loai bai toan thanh hai loai: nhung bai toan tim tdi, nhung bai toan chQng minh.Vi mQc dp khd, d i cua bai toan, G. Folia cho rdng "Khong de dang xet doan vg mQc dp kho ciia bai toan, lai cang khd hon nua khi xac lap gia tri giao due cua nd" [7;132].
Giao vien (GV) nen nam dupc each phan loai dp khd ciia bai toan d i phuc vu cho viec giang day.
Trudc khi giai toan, khd khdng dinh dupc chdc chan rdng se dung nhung kiin thQe (dinh nghia, dinh If, menh de, qui tdc, cdng thQe,...) nao, trQ khi dd la bai toan da cd thuat giai. Trong qua trinh giai mot bai toan cu t h i nao dd, le duong nhign khong can huy dOng din mpi kiin thQe ma ngudi giai da thu thap, tfch luy dupc tQ trudc. Can huy dpng nhung kiin thQe nao, cdn xem xet din nhQng moi lien he nao phu thuOc vao kha nang chpn Ipc ciia ngudi giai toan. Ngudi giai toan da tfch luy nhung tri thQe ay, gid day riit ra va van dung mot each thfch hpp d i giai toan. G.
Tap luyen cho hoc sinh hoat dong lien tudng vd huy dong kien thiCc trong qud trinh...
Folia gpi viec nhd lai cd chpn Ipc cac tri thQe nhu vay la sU huy dpng, viec lam cho chung thich Qng vdi bai toan dang giai la sU td chQc [7;310, 311].
Trudc khi bdt tay vao giai mot bai toan cu the, ngudi giai da tich luy dupe rdt nhiiu kiin thQe, nhung dirng kiin thQe nao cho phii hpp thi that khong d i dang.
Tham chf niu cd kem theo chi d i n ve nhung kiin thiic cdn a[) dung thi bai toan ehua hdn da giai dupc ngay. bdi vi, tliQ nhdt, chua hdn HS da nhd dupc ngay noi dung kiin thQe dd, thQ hai, tlii nhd dupc noi dung kiin thQe thi viec ap dung no cung khdng han la d i dang.
Mat khac, nipt bai toan c6 chi ddn doi khi dua cac em vao trang thai bi dpng kiin thQe, cdn bai toan khdng dupc chi ddn lai dg cac em co t h i tu do lua chpn kiin thQe giiip giai quyit van de bdng nhiiu eon dudng khac nhau. Tuy nhien, kho va d i chi la nhung khai niem mang tinh chat tuPng doi, phu thupc vao nhung tinh hudng, thdi digm cu t h i .
Cac kiin thQe ma HS linh hoi dupc la san pham ciia hoat dpng, la mot bai toan ma muon chiim linh thi HS phai trai qua nhung hoat dpng tuPng Qng.
Kha nang huy dpng kiin tliQc khong phai la diiu b i t biin, nd thay doi theo lupng tfch luy tri thQe cua hpc sinh va mQc dp luyen tap thanh thao ciia cac em.
HS huy dong kien th'iic de gidi quyet tot vdn de cdn tuy thuoc vdo khd ndng Men doi vdn de, biSn doi bdi todn kho, m,di, la ve cdc bdi todn dcln gidn hon, quen thuoc Hon md cdc em dd tiing gidi quyet.
Chang han, xet bai toan: Tim m, de pInMng trinh sa,u co nghiem thuc:
3 \ / F ^ ^ + m. V-v~+T = 2 \fx'^~^~ 1. (2.1) Cac bai toan eo chQa tham s6 la mot dang toan tu'Piig doi kli6 vdi HS. Viec
chuyin bai todn da cho vi bai toan tUOng dUPiig khong dupc HS y thQe diy dii, co nhiiu sai 1dm khac vi mat tinh toan, van dung dinh li, qui tdc,... do kha nang khai quat van d i ciia cac em bi han ehi. Mot yiu to khac cung gay kho khan cho hpc sinh la khau kiim tra dap an tUPng doi phQc tap (nhiiu khi khdng thuc hien dupe).
0 day chung toi chi hudng d i n HS dinh hudng tim ldi giai bai toan 6 budc biin ddi tQ dd lign tudng, huy dOng kiin thQe d i dua vi bai toan mdi tuong dUPng vdi bai toan da cho giiip HS co diiu kien giai dung bai toan.
GV: Diiu kien xac dinh cua bai toan la gi? Cd t h i biin d6i dupc bai toan vi dang quen thupc hdn khdng?
HS vdi cau tra ldi mong dpi: Diiu kign xac dinh ciia phuong trinh la :c > 1, va phuong trinh tUPng ditpng vdi phUPng trinh:
cd nghiem .'t: > 1.
Nhin vao phupng trinh (2.2) da biit each giai bai toan chua?
HS vdi cau tra ldi mong dpi la dat an phu t = ^ - , vdi niQi .c > 1, thi dieu kien ciia t Ik 0 < t < 1, ngUpc lai vdi mdi t ma 0 < t < 1 thi diu eho .;; > 1. Va
bai toan da cho tuong duong vdi bai toan: Tim diiu kien cua m de phuang trinh:
rn = 2t-3t^ (2.3)
cd nghiem 0 < f < 1.
Cung phai ndi thgm rdng. 0 bai toan nay, viec tim dung diiu kien ciia / dg chuyin vi bai toan tUPng dUPng (2.3) la khdng di, b i n h i t HS chi tim dupe diiu kien t > 0 mh khong tim dupc diiu kien t < I. Di tim dung diiu kien ciia / lai doi hoi HS phai huy dpng kiin thQe biin ddi va danh gia. Ta co:
o < f = a- ^ < 1, V.c> 1.
V - r + l
GV hoi tiip: TQ phuong trinh (2.3) da biit each giai t i i p bai toan chua?
HS: So nghigm cua phUPng trinh (2.3) bdng s6 giao digm cua dd thi y = m.
(cung phuong vdi true O.r) va p h i n dd thi y = g{t) = 2t - 3t'^ xet tren 0 < f < 1.
TQ do HS giai dung bai todn: Gia tri cua m thoa man bai toan la --1 < rn < ~.
o Nhu vay, hoat dpng lien tUPng va huy dpng kiin thQe r i t quan trpng trong qua trinh giai toan. GV cin dac biet chu y phat triin hoat dpng nay cho HS, giup cac em cd k h i nang dpc lap giai quyit cac bai toan.
Hoat dong Hen tudng vd huy dong kiin thdc moi ngudi moi khac. Khd ndng lien tudng ndy phu thudc vdo su nha.y cdm d khdu phdt hien vdn de.
Khdng ed kha ndng lien tuPng va huy dpng kiin thQe thi nang luc giai todn bi han chi, cai nhin vi bai toan thudng cue bp, rdi rac. Tuy nhign, dQng trUdc mot bai toan cu thg, khdng nhdt thiit tdt cd kiin th-dc lien tudng vd huy ddng duac deu cd ich cho viec gidi bdi todn. C i n chpn Ipc thong qua cac phep thu - sai de tiin tdi mot su lign tUOng huy dpng phu hpp.
Chang han, xet bai toan: Chicng minh rdng niu o.,b.,c la do ddi cdc canh cua mot tam gide thi a^ -I- b^ -\- c^ < 2{ah + be + ca).
Ta nhan t h i y rdng bai toan cd de cap din mdi quan he giua cac canh ciia mot tam giac, ta hay huy dOng nhQng dinh If, tinh chit da biit vi quan he giQa cac canh cua mot tam giac:
a>b-c . (2.4) a <b + c ' (2.5)
a.'^ =6- + c^ -2bccosA (2.6)
2
a- ='-+ 2m-c - b'^ (2.7)
Dl chpn Ipc nhQng kien thQe thfch hpp, trudc h i t ta hay loai (2.6) va (2.7) vi chung d i cap den moi quan he "ddng thQe" chQ khdng phai " b i t ddng thQe" nhu diiu pjiai chQng minh.
Tap luyen cho hoc sinh hoat ddng lien tudng vd huy dong kien thilc trong qua trinh...
Hay cpian sat (2.4), hai vi eua nd diu la bae n h i t , trong khi do diiu cin phai cliQng minh lai lign c^uan den bae hai. Di lam x u i t hien bde hai, co t h i binh phuong hai vi hoac nhan ca hai vi vdi a.
Niu tiin hanh theo eon dudng binh phUPng hai vi, cin phai cdn than vi vi phai chua hdn la sd duong, tuy nhign cho du vi phai la am thi cung khong t h i cd a < —{b — c), do dd ta vin ed a'^ > {b — c)", tQc la a'^ > b'^ — 2bc -\- (?.
Do vai tro cua o, b, c la binh ddng nhu nhau, bdi vd}' sau khi co a^ > b'^ — 2bc.+c'^
cuiig cin phai rut ra kit qua tuong tU: 6^ > a^ — 2ac -I- c^, c^ > b'^ — 2ba + a^. Cong tQng vi cac BDT ta cd diiu phai chQng minh.
Bay gid thu quan sat (2.5), eung suy nghi nhu tren, d i lam x u i t hign a'^, eo t h i binh phuong hai ve hoac nhan hai vi vdi a.
Niu tiin hanh theo con dudng binh phUPng hai vi, ta cd a^ < b"^ -\- 2hc -I- c^, mot each tUdng tu cung cd 6^ < ci^ -\- 2ac -\- c^, c^ < a^ -I- 2ab + b~. Cong tQng vi ta suy ra a^ -I- 6" -f c" > -~2{a,b-\rbc + ca). Day la BDT hien nhign dung nhUng do khong phai la dieu cin phai chQng minh.
Niu tiin hanh theo eon dudng nhan ea hai vi vdi a, khi do a^ < ah -H ac.
Vi vai trd a, 6, c binh ddng nhu nhau ngn ta cung cin su dung not nhung b i t ddng thQe tuong tU: h^ < bc-\- ah, c^ < ca -^ cb. Cong tQng vi eac BDT nay ta co a^ + b'^ + c^ < 2{ab + bc + ca.).
2.3. N h i i n g q u a n d i g m chi d a o t r o n g v i e c t a p l u y e n cho H S h o a t d o n g lien tifdng va h u y d o n g k i e n thiJc
* Quan diem 1. Trong qud trinh truyen thu tri, thdc Todn hoc cho HS, GV cdn quan tdm tap luyen nhdn dang, phdt hien cdc the hien khde nhau, tii dd nhdn manh khd ndng icng dung cua nd bang viec lua chon he thdng bdi tap de HS thdy dugc moi lien he gvBa cdc ndi dung Todn hoc.
Trong day hpc thi viec truyin thu eho HS kiin thQe cP ban la r i t quan trpng, dd la nhung dinh nghia, dinh If hay cac qui tdc. Nhung diiu quan trpng la hpc sinh phai cd kha nang van dung chung trong qua trinh lam bai tap. Vi t h i GV cin minh hpa kiin thQe bdng cdc vi du cu t h i sinh dpng d i HS nhan biit each thQe Qng dung nhQng kiin thQe dd.
Chdng han, khi day hpc dinh If Lagrange "Cho ham lign tuc trgn doan [a, 6], co
dao ham trgn khoang (a, b). Ton tai mot so c G (a, b) sao cho /'(c) = ", GV se gpi y cho HS phat hien ra cac Qng dung khac nhau cua dinh li.
l}ng dung 1: Su dung dinh If d i chQng minh b i t ddng thQe. Ta c6, niu m <
F'ic) < M, Vc E (a, b) thi m < ^H^LzfM. < M ^ m.{b - a) < F{b) - F{a) <
b — a
M{b -a). Do dd, d i dp dung dupc kit qua trgn vao viec chQng minh BDT, dieu quan trpng la phai tun ra dupc ham s6 F{x).
Vf du: Chiing minh rang ln(x -^ 1) < x vdi moi x > 0.
Cho HS viit lai BDT trgn d i lam x u i t hien ham so F{x) : ln(x -|- 1) - In 1 <
fx + 1) - 1 o i ^ - t i l ^ ! ^ < 1. Khi do, HS se d i dang lien tudng d i n viec
^ ^ ^ (:r + 1) - 1
xet ham so F{t) = hit lign tuc va cd dao ham tren [l,.-?: + 1], {x > 0) theo dinh If Lagrange ludn tdn tai c G (1, .x -H 1) vdi a; > 0 sao cho:
,, , F ( . T + 1 ) - F ( 1 ) 1 l n ( : c + l ) , / , ^^ ^
^ ^ (x -h 1) - 1 c X c Ta c6 1 < c < .T - M suy ra ln(.i- + 1) = '-<-= x, dd la diiu phai chQng minh.
U'ng dung 2: SQ dung dinh li Lagrange d i chQng minh phuong trinh cd nghiem.
TQ dinh If, niu F{b)-F{a) = 0 thi tdn tai c G (a, b) sao cho F'{c) = 1^^^~ ^ ^ hay phuong trinh F'{x) = 0 cd nghigm thupc khoang (a, b).
Vdy, d i ap dung kit qua tren vao vige chQng minh phQOng trinh f{x) = 0 cd nghigm trong khoang (a, 6), diiu quan trpng la phat hien ra dupe ham so F{x) (la nguygn ham ciia ham sd f{x)).
Vi du: Chiing minh rang phuang trinh: acosx + 6COS2.T + ccosSx = {)(*) cd nghiem thudc khodng (0; vr) vdi moi a, 6, c.
Di phat hien ra ham so F{x) ta gpi y HS tim nguygn ham ciia ham so f{x) = acosx + 6cos2,T -F ccos3x, {F{x) = - a s i n . T - - sin 2x - -sin3.a:). Khi do ta thdy rdng F{x) lien tuc va cd dao ham trgn (0;7r) va F'{x) ^ a c o s x -+- 6cos2x -h ccos3x, F{TT) - F ( 0 ) = 0.
Khi dd theo kit qua trgn HS d i dang nhan thay tdn tai x„ G (0; TT) sao cho:
7-1/ \ TPfr\\
F'{xo) = —^ —^ tuong duong vdi acosxo-hbcos 2xo+ccos 3xo = 0 hay phuOng
TT — 0
trinh (*) cd nghiem XQ G (0;7r).
TQ cac vf du tren, GV cd t h i dua ra mot he thong cac bai tap sao cho HS co t h i van dung true tiip hoac gian tiip dinh If Lagrange d i giai. Thdng qua he thdng bai tap, HS se tiip thu kiin thQe va khde sdu kiin thQe vi dinh If Lagrange va Qng dung cua nd.
* Quan diem 2. Cdn chuyen tri thUc day hoc vi vimg phdt trien gdn nhdt trong khi tap luyen cho HS hoat ddng lien tudng vd huy ddng kiin thiic.
Thuc tiin su pham cho thiy, d i phat hien ra vin d i , phat hien ra cong cu va xac lap tiin trinh giai quyit vin dg-, HS thudng lien tudng tdi cac kiin thQe, kl ndng da CO. SQ lign tuong nay hinh thanh tren co sd so sanh, ddi chiiu cac dQ kien vd cac ygu ciu cua bai toan dat ra vdi cac kiin thQe If tliuyit va cac bai toan ma hp da luu giu trong trf nhd. Su xac lap cac lign tudng nay chfnh la xac lap moi lign he giQa tri thQe sdn cd vdi cac kham pha dpc lap cua HS trong cpia trinh hpc tap. Ndi each khac, theo quan diim eua Vugotxki, dd chfnh la qua trinh di chuyin tri thQe tQ "viing phat triin gin n h i t " d i n trinh dp hien tai.
Do vay, d i tao lap dQpc mdi tritdng thuc ddy boat dpng hpc tap cua HS, GV phai biit giao nhiem vu hpc tap gan vdi "vung phat trien g i n n h i t " cho cac ddi 8
Tap luyen cho hoc sinh hoat dong lien tudng vd huy dong kien tfiiic trong qua trinh...
tupng HS de tQ dd khuyen khfeh su "mao hiim" va kfch tliich nhu can cliiiin linh tri thQe cua hp. Viec day hpc cd higu qua thuc chit la qua trinh to eliQc va diiu khiin cua GV sao cho su "di chuA'in" tri tliQc trong hai vimg nay dien ra nipt each thuan ldi, dg tQ dd giup HS chiim linh tri thQe mot each tu giac, tfch ci.rc ddng thoi phdt trien kha nang tu hpc, kl ndng van dung tri tliQc mot each sang tao.
Vi du: Sau khi day bdi "Phuoiig trinJi luonq qidc ca bdn.'' GV ?/cw cdu HS qidi
TT TV '
phuang trinh: cos(.7- + 7 ) -I- sin(—h 2.r) -— 0.
o z
Hiin nhign ygu ciu nay khong qua xa ddi \di nhung kiin tliQc ma HS tich luy dupe sau khi hpc bai "FhuPng trinh lupng giac cP ban". HS biit rdtig edn phai thuc hign mot so phep biin ddi lupng giac de dua cluing vi mot trong cdc dang eiia phuong trinh lupng giac cd ban da hpc.
GV ed thg ngu cac can hoi:
- Phuong trinh lupng giac tren cd phai la phuong trinh lupng giac ed ban?
- Di dua phitdng trinh tren vi mot trong cae dang ciia phupiig trinh lupng giac CP ban ta cin thuc hien iihuiig phep biin doi nao?
- Trong cac phep biin ddi da hpc ta ciu lua chpn phep biin doi nao de co Ipi cho bai toan? (mdi cjuan he giua cae ham so lupng giac ciia hai cung phu nhau).
NhQng cdu hoi nhu vdy giup HS dg y din s i n ( - -I- 2.r) ^-^ (•()s2x.
GV cin biit Ipi dung su phdn bdc hoat ddng de diiu khigii qua trinh hpc tap.
Do do, GV cd thg dira vao su phan bae hoat dong de tudn tu ndng cao yeu cdu doi vdi HS. Viec lam do se phat huy dupc thih tieh cUc:, ehii dclug hpc tap \'a sU phat trien trf tue ciia HS. Chdng han, GV dua ra he thdng bai toan co eiing phitdng phaj) giai nhung mQe dp kho d i n . Con khi HS gap kho khan trong hoat dpng, ta eo tlie tam thdi ha thij) yeu ciu. Sau khi hp dat dupc nic tliip nay, yen cin lai dupc tudn tu ndng cao. Lam nhu vdy phu hpi) vdi Ly tlinyit \iigotxki vi vung [)hat trien gan nhit [5; 167].
Kiin thQe trung gian dong vai tro la "ciu ndi" giua kiin thuc ma HS dupe hpc vdi nhung bdi todn nang cao. Di cd thg giai dupc nhung bai toan dri. doi hoi HS phai cd kha nang lign tudng. huy dpng va van dung kiin thQe da biit. \'iee van dung kien thQe trung gian d i giai bdi toan se lam eho HS phat huy kha ndng du dodn ede van d i . Qua trinh do se ren luyen eho HS tinh thin hoai nghi khoa hpc, tfnh dpc lap, tfnh phg phan cua tU duy, gop phan ren luyen tfnh sang tao cho HS.
* Quan diem 3. Chu trong khd ndng gidi quyet vdn de trong qud trinh ren luyen khd ndng lien tudng, huy ddng kien thdc cua HS tren ca .td van dung cdc cdp do ciia dciy hoc Phdt hien vd gid.i quyit vdn de trong nhdng tinh /mdng phi). liOp.
Day hpc Phat hien giai quyit van d i "la hinh thQe cd hieu qua de to eliQe sir tim toi tri tue khi tiep thu tri tliQc thdng qua viec giai quyit cac vin d i " (2; 70].
Cd t h i ndi rdng, HS chua t h i dp dung ngay mot each diy dii phuong phap lam viec cua cac nha khoa hpc, GV chi c6 t h i lam eho lip bdt dau lam quen vdi phupng phap dd va tao dieu kien cho HS thuc hien mot so khau trong qua trinh tim tdi P nhQng mQc dp khac nhau. Trgn cP SP dd, cluing toi dua ra bon mQc dp thfch
hpp trong viec day cho HS lign tudng va huy dpng kiin thQe trong [5|.
Chdng han, xet vi du: Cho tam, gidc ABC, ttm, gid tri, nhd nhdt (GTNN) cua bieu thUc:
T = sill A + sin B + sin C -f- -—-• -I- -.—-- -f ~-^.
sm A sm B sm G
Thuyit trinh phat hien va giai quyit van d i cho HS bai toan trgn: GV dua ra nhan dinh, vi A, B, C la ba gde ciia tam giac ngn 0 < A, B, C < ISO" =>
sin/I, sin S , sin C dUOng. Biiu thQe T cho d dang tdng cac sd duong cd tfch khdng ddi, d i tim GTNN cua T ta phai danh gia theo chieu " > " nen mOt lien tudng gan n h i t cd t h i sU dung BDT Cauchy d i giai ta dupc: T > 6. T h i nhung, d i u " = " lai khdng t h i xdy ra {vi A = B = C = — man t h u i n vdi A -\- B + C = TT).
Vay hudng sQ dung BDT Cauchy cho 6 sd khong cli den kit qua, ta phai khai thac hudng khac. Ta xem thQ cd lien he giua sin yi, sin B , sin C hay khdng? Mot BDT quen thupc trong tam giac dupc lien tudng tdi la: sin A-F sin B -I-sin C < . Luc dd bai toan clua vi dang dai sd thuan tuy: "Cho ba sd dUPng a, 6, c thoa man
o / ^ 1 1 1
a-^b + c< -—-. Tim GTNN cua T = o + 6 -h c -h - + - + -". Dua vao vai trd binh 2 ^ a b c
ddng cua a, b, c trong bai toan va ddng thdi thu mot sd trudng hdp, ta dU doan rdng T dat GTNN khi a = b = c=—.
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Mat khac, d i u "=" trong BDT Cauchy xay ra khi cae hang tu bdng nhau.
m - - 1 » - - 7 V 3 1 1 1 2 ^ ,
l u cae du doan tren ta co: a = o = c = — , - = - — -=- ——. Dg cd cac so 2 a b c ^3
hang bang nhau trong BDT Cauchy, ta p h i i tim sd o sao cho aa = ob = ac = - =
r- "' 1 1 , \ / 3 2 4 • , . .i: :.
-r = - hay a— = --y= =» ft = - .
^ c •' 2 ^ 3 3 Tren cd sd dd, ta di din kit qua BT:
6 3 3 a b c 3 2 Trong vi du trgn, GV da thuyet trinh lai qua trinh tim kiim ldi giai BT. Thay biet ddt minh vao vi trf cua HS, hinh dung va binh luan cac sai l i m ma HS thudng mac phai, biit xoay chuyin hudng suy nghi khi gap khd khan, chQ khdng phai dot nhign dQa ra ngay mot ldi giai dung. Dd cung la yiu td lam nen uu diim cua phuong phap thuyit trinh phat hien va giai quyit v i n de, nhd nd HS dupc phuPng phap kiin tao tri thQe chQ khong phai chi t i i p nhan tri thQe.
Trong bai toan vi GTNN vQa xet tren day, niu sQ dung c i p dp ddm thoai phdt hien vd gidi quyit vdn di, GV cd t h i dua ra nhQng cau hdi nhu:
- Qui trinh giai bai toan tim GTNN? Mien gia tri cua sinA, sinB, sinC?
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Tap luyen cho hoc sinh hoat dong lien tudng vd huy dong kien thiic trong qud trinh...
- Theo em du doan T dat GTNN khi a, b, c nhdn nhung gia tri nao?
Nhin lai ldi giai cua bai toan trgn, chung ta t h i y cai "mit" chinh la P chd biit each phan tfch:
4 1 4 1 4 1 sin A = - shi A — ~ sin A, sin B = - shi B - - sin /?, sin C = -- sin C sin C.
3 3 3 3 3 3 Tai sao trong r i t nhiiu each phan tich, chung ta lai chpn each phan tfch do? Do la diiu ma GV phai lam eho HS hiiu. Khi GV neu cau hdi "Miin gia tri eua sin A, sin B, sin C?", chdc rdng nhiiu em se tra ldi la: "0 < sin A. sin B, sin C < 1", do khong phai la diiu ta mong dpi. Tuy nhign luc ay G \ ' eo t h i gpi y thgm "Co BDT lign quan nao trong tam giac ma cac em da biit hay khong?"
q /q
TQ BDT sin >l-|-sin i?-|-sin C < —;— va vai tro binh ddng cua sin A. sin B, sin C trong bai todn, GV t i i p tue gdi y d i HS du doan dupc dau "=" dat tai sin A = sin B = s i n C =: — , lam cP sp cho bude da ngu d phin trgn.
2 _ Nhu vay, trong c i p dp dam thoai phat hien giai quyit vin di, thiy sQ dung he thdng cdu hdi gpi y hpp 11 d i d i n ddt HS phat hien va giai quyit vin di.
Theo K. K. Plantondv thi tU duy la mc)t qua trinh gdm nhiiu giai doan ki tiip nhau, ma hai trong sd cae giai doan iy la sU x u i t hien lien tudng, sang loc lign tudng va hinh thanh gia thuyit. Chang han, trong vf du trgn, thu suy nghi:
Can danh gia theo chiiu ">"; A. B,C la ba gde cua mot tam giac nen cac hang tu trong bieu tliQc ludn duong. C i n danh gia tdng cac sd dupng theo chiiu " > " gpi cho cac em lign tUdng tdi BDT Cauchy. Tuy nhign, khi dp dung true tiip thi diu
" = " khdng xay ra. 0 day, cau hdi ciia GV nham giup HS huy dpng thgm kiin thQe
q /q
sin A -\- sin B -\- sin C < d i xac dinh hudng giai.
Trong gid hpc giai quyit vin di, cae can hdi nhdm vao viec gdi lai cac tri thQe CO lign quan trong vdn tri thQe da dupc linh hpi trudc day ciia HS. Cac can hdi cua GV cd tac dung thue ddy budc tim toi tri tliQc co lien quan de tim ra hudng giai quyit thfch hpp, loai trQ dupe nhung sai lech cd thg tren budc dudng giai quyit dung ddn khi HS dua diiu minh da biit vdo trong nhung mdi lign he thfch hpp.
* Quan diem 4- Can tao cho HS tlidi quen nhin nhdn mdt vdn de dudi nhieu goc do khac nhau trong qud trinh truyen thu tri thdc.
Mot bai toan eo t h i nhin dudi nhieu khfa canh khac nhau va Qng vdi mdi each nhin c6 thg cho ta mot ldi giai khae nhau. Hon nua mpi each giai diu dua vao mot sd dac diim nao do cua cae du kign, eho ngn viec tim nhiiu each giai luyen cho HS biit nhin nhan vin d i theo nhieu khfa canh khac nhau, diiu do r i t bd fch cho vige phdt triin ndng luc tu duy.
Mot ddc diim tam If cua HS trong ciua trinh giai todn la: Tu duy ludn cd siic i. Di ren luyen tu duy linh hoat, phd vd siic i cua tu duy ta cdn phdi tliuc liien cdc hoat ddng thdnh phdn sau:
-f Huy dpng kiin thQe lign quan din gia thiit va kit ludn cua bai toan theo 11
cac he thdng kien thQe lien (pian khac nhau;
-t Hoat dpng chuyin ddi ngdn ngQ trong mot iipi dung Toan hpc hoac ehuyen ddi ngdn ngQ nay sang ngon ngU khac trong khi day hpc cac tinh hudng dign hinh;
-^ Biin ddi bai toan thanh bai toan khac tUcJiig dudng (niQc dp tdng quat eo t h i khae nhau).
\'i du: Gidi phurJng trinh luang gidc:
eos'-^ X -h 2\/3 sin x cos:;; -|- 3 sin" x = 1 (2.8)
Ta CO t h i dinh hudng nhu sau:
ThQ n h i t : Bdi vi, eos" .r -f siii^ x = 1 nen cd t h i dua phUdng trinh (2.8) vi dang phuong trinh ddng c i p bdc hai ddi vdi sin x, cos .r.
ThQ hai: Do sd hang thQ hai eo chQa sin x. cos x nen ed t h i dua (2.8) vi phupng trinh bac n h i t ddi vdi sin2;r,cos2x nhd edng thQe ha bae.
ThQ ba: Xem cos".;; -I- 2 \ / 3 s i n x c o s x -I- 3 s i n - x = 1 la phuong trinh bac hai ddi vdi i n la cos:r. TQ phudng trinh dd ta co A' = 3siii'^,r -- 3sin".r -F 1 = 1.
Nhu vdy, khi phan tfch d i dinh hudng bai toan chung ta phai lien tudng giua eac pham vi khac nhau, lien tudng den tQng chi tiit trong bai toan. G. Polia cho r i n g "Nhd nghien cQu lien tiip tiUig chi tiet mot, bdng nhiiu each, cudi cung chung ta cung cci t h i nhin dupe toan bp vin d i dudi mot anh sang hoan toan khae trUdc va do do nit ra mot each cliQng minh nidi" |7; 94].
Hudng d i n cho HS co t h i van dung kiin thQe tdng hpp, kiin thQe lign mon, thiit lap su lien tudng giua eae pham vi dg tim nhiiu each giai eho mot bai todn.
Vf du: Cho ba sd duang a, b. c thoa mdn a > c. b > c. Cliiing minh rdng:
y ^ - r ) + V ^ ( 6 - C ) < v^6.
GV ed the dinh hudng giup HS giai bai toan nhu sau:
Dinh hudng 1: T h i y giao cd t h i thuyit trinh: Nhin vao c i u tao eac biiu thQe trong BDT ta t h i y
\fc+{a - c) =-- ^/d. s/c-{- [b -- c) •-- \/h.
Neu nhdn tQng vi eua ddng tliQc ta ditpe: y^c + (« — c).\/c f {b - c) = y/ab.
GV ed t h i ngn cho HS can hdi: Vdi phep biin ddi eiia eae ddng thQe tren gpi cho em lien tudng din BDT nao? Mot BDT r i t quen thupc? Chung ta mong dpi HS tra ldi rdng: Do la BDT Bunliiaeopxki.
Vdi lign tudng nhu trgn, HS ap dung BDT Bunhiaeopxki:
>/^-y/{^-- c) + V'cViJ^- c) < y/{c + (a - c)).{c f {b - c)) - v/^Ift.
Ta cd dieu i)liai chQng minh.
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Tap luyen cho hoc sinh hoat dong lien tudng vd huy dong kien thiic trong qua trinh....
Dinh hudng 2: Vdi vige dp dung BDT Biiiihiaeoi)xki ta co lien tudng din tich vd hudng cua hai vec to, khi biiu thQe tfch vo hudng cua hai vec td dupc biiu thi dudi dang tpa dp.
That vay, dat it =- {^/cT- c, ^/c). Ii = {s/c.\/h- c) --> fiv = \/c\fa - c ^-
\fc\fb — c va |v7||?7| =- ^/dh, mat khac ta ludn co u.v < \u\\f.'\. nen ta dupc y^cy^a - c-\- y/c\/b~— c < Vdh.
Tuy thupc vao each nhin vd kha nang biin ddi bai todn ciia HS, tQ d(3 G \ ' c6 thg dinh hudng eho HS bdng each ehuyen ddi sang ngdn iigu hinh lipe. ngon ngu lupng giac hoac cd t h i biin ddi tUPng dupiig, sU dung eong eu (hu) ham.
3. Ket luan
C i n thiit phai tap luyen cho HS kha nang lien tUPng va huy dpng kiin thii'c trong cpia trinh chiim linh tri thQe. Thue hien diiu do la goj) i)hin phat triin tu duy toan hpc cho HS. Tuy nhign, mudn dat dupe diiu nay cin phai lua chpn noi dung va phUPng phap day hpc mot each thich hpp.
TAI LIEU T H A M K H A O
[1] A. A. Stoliar, 1969. Gido due hoc Todn hoc. Nxb Giao due i\linsk(Tiing Nga).
[2] M. Alecxeep, V. Onhisuc, M. Gruglidc, V. Zabontin. X. Veexcle, 1976. Phdt trien tu duy hoc sinh. Nxb Giao due Ha Noi.
[3] Bui Vdn Hue, 2000. Gido trinh tdm li hoc. Nxb Dai hpc Qudc gia. Ha Noi.
>' [4] Pham Minh Hac, Pham Hoang Gia, Trin Trpng Thuy. NgUA'in Quang Udn, 1992. Tdm li hoc. NXIJ Gido d u e Hd Noi.
[5] Nguyin Ba Kim, Vu Dudng Thuy, 2001. Pliuang ptidp dq.y hoc man tod,n.
Nxb Giao due. Ha Ndi.
[6] Phan Trpng Ngp, 2005. Da,y hoc vd pliuang phdp day hoc trong nhd trudng, Nxb Dai hpc Su pham.
[7] A. V. Ptrdvxki, 1982. Tdm li hoc h'ia tudi vd td,m li hoc ,sii ph(im, tap 2.
Nxb Gido due. Ha Ndi.
[8] G Polia, 1995. Sdng tao todn hoc. Nxb Giao due, Ha Noi-
[9] Dao Vdn Trung, 2001. Ldm thi ndo de hoc tot todn phd tfwng. Nxb Dai hpc Qudc gia, Ha Noi
[10] Tic die'n Tiing Viet. 2005. Nxb Da Ndng vd Trung tdm TQ diin hpc, Ha Ndi - Da Ndng.
A B S T R A C T
Training s t u d e n t s associating activities and mobilizing knowledge during m a k i n g up and compiling m a t h e m a t i c s knowledge In this paper, we dealt with training students associating activities and mo- bilizing knowledge during making up and compiling preparing mathematics knowl- edge. The main result is to indicate the role of associating activies and mobilizing mathematical knowledge. The methodology of teaching and carrying out the main ideals to drill activities for students.
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