KHAI THAC TiHH TRlTC QUAN CUA GAC PHAN NIEM TROHG DHV HOC tollH (T TRinrHB PH0 THOHG
O TS. TRINH THANH H A I *
H
ien nay, viee khai thdc tinh trwc quan cuo cdc phdn mem hd trg dgy hge (DH) mdn Todn dugc cdc trudng phd thdng chu y. Tuy nhien, thuc te ed mdt sd gido vien (GV) qud «lqm dyng" hodc «le thudc" vdo phdn mem dgy hgc (PMDH) nen hieu qud DH chua coo vd thdm ehi cd ngudi cdn ddt cdu hdi «Lieu tinh true quan cua cdc PMDH cd Idm gidm khd ndng tw duy cua hgc sinh (HS) trong DH todn hay khdng?". Trong phqm vi bdi viet ndy, chung tdi hi vgng Idm sdng td vd duo ro edu trd ldi cho cdu hdi tren.1. Md'i quan he giua tinh true quan cua cdc PMDH vdi viec ren luyen khd ndng t u duy cho HS trong DH todn
Viee cho phep tgo ra cdc md hinh todn hge true quan Id mdt the mgnh vugt trdi ciia cdc PMDH so vdi cdc phuong tien, dd dung DH todn truyen thdng. Vd'n de Id d chd to khai thdc tinh true quan ciio PMDH trong DH todn nhu the ndo?
Qua thuc te trien khoi viee nghien eiiu ling dyng cdc PMDH, theo chung tdi de ren luyen vd phdt trien khd ndng tu duy cho HS trong DH todn, ta tdp trung vdo khai thdc tinh true quan cuo PMDH theo cdc hudng sou:
- Giup HS phdt hien ra vdn de bdng cdch quan sdt cdc md hinh dugc thiet ke bdi PMDH.
Khi edn nghien cifu, khdm phd mdt vdn de todn hge ndo dd, trudc het ta se sii dyng PMDH de thiet ke cdc md hinh true quan, ed the thay ddi ede gde dd tie'p cdn de HS phdt hien nhanh nhd quan sdt bdng mdt cdc md'i quan he todn hge cung nhu su thay ddi cua cdc ddi tugng todn hgc. Hon nifa, to ed the cho thay ddi mdt vdi thdnh phdn trong md hinh, HS se tu quan sdt su thay ddi tuong ting eua ede thdnh phdn edn Iqi, quo dd se phdt hien ro md'i tuong quan giifa cdc ddi tugng, ede dgi lugng trong md hinh.
- Giup HS kiem nghiem cdc du dodn trong qud trinh tim tdi ldi gidi. Trudc het, nhd PMDH, HS se duo ro cdc du dodn cuo minh. Tie'p sou dd, HS se tuong tdc vdi PMDH bdng cdch ddt ra cdc edu hdi md PMDH Id edng ey de trd ldi. Neu
Tap chi Giao due so 2 4 2 (ki 2 • 7/20101
du dodn eua HS dugc cung cd niem tin hi cdu
«trd ldi" cua PMDH, HS se tim cdch Idm sdng td vdn de. Quy trinh ndy cd the md td bdi luge dd:
Viec dimg PMDH de duo ro, kiem nghiem cdc du dodn vd viec sii dyng ldp ludn suy d i l n de Idm sdng td vdn de cd tdc dyng thuc ddy hd h-g cho nhau trong sudt qud h-inh HS tim tdi,ldi gidi.
2. V i du minh hda
Vidu i: '
Cho tom g i d c ABC vudng tqi A, dudng eao AH,HC-HB
=AB. Chiing minh rdng BC = 2AB.
Xet b d i t o d n t r e n , vd'n de Id d c h d : Ldm
sao cd the ve dwgc tam gidc ABC nhu gid thiet dd chd?
To sii dyng Cobri Geometry, ve mdt tom gidc vudng bdt ki rdi tim cdch thay ddi cdc ye'u td sao cho nd t h i hien dung dugc cdc tinh chdt cho bdi gid thie't, rdi sou dd tim ro nhifng yeu td tiem dn.
Budc 1: Ve hinh: Ve tam gidc vudng ABC vudng t q i A bdt ki. Dung
dudng cao A H . Ldy diem E thudc BC sao choHB = HE,vdyHC- HB = HC - HE = EC.
Tren BC Idy diem D sao cho CD = AS, (hinh 1).
V d i cdch d u n g
hinh nhu tren, ndi chung hoi diem D, E Id phdn biet, h i c l d H C - H B = C E ^ A B = CD.
Bwdc 2: Thay ddi hinh ve dequan sdt. GV eho thay ddi tam gidc vudng ABC soo cho HC - HB =
Hinh 1
' Tnrdng Dai hoc su pham - Dai hoc Thai Nguyen
Hinh 2
AB hay CE = CD tiic Id thay ddi tam gidc soo cho cho hoi diem E, D triing nhau (hinh 2)va yeu cdu HS quan sdt vd dua ra cdc du dodn:
- GV: Diem D ed gi dqc biet?
- HS: True gide cho thdy D «cd ve" Id h-ung diem cua BC. Dimg Cabri Geometry kiem tra cho thdy du dodn Id chinh xdc.
- GV: Hdy ndi A vdi D vd xet xem A A B D cd gi
ddc biet?
- HS: Bdng true quan cho thdy AABD deu vd BC
= 2DB = 2 A B . D u n g Cobri Geometry de kiem tra, ke't qud cho nhdn xet true quan tren Id dung.
Hodn todn tuong tu, HS phdt hien dugc A A D C Id tam gide cdn. Nhu vdy nhd Geomehy Cabri, HS phdt hien ra neu tam gidc vudng ABC thod mdn gid thuyet thi diem D phdi Id trung diem eua BC.
Budc 3: Gidi quyet bdi todn. Iu phdt hien tren, HS di den cdch gidi quyet bdi todn bdng cdch tqo them yeu to phy Id Idy diem D thudc HC soo cho HD = HB. Tam gidc ABD cd AH vifa Id dudng cao vira Id dudng trung tuye'n nen Id tom gide cdn, suy ra ZADB = ZB vd AB = AD (*).
Mdt khde DC = HC - HD = HC - HB = AB = AD hay tam gidc ADC Id tom gidc cdn suy ro Z D A C
= Z C , suy ra ZDAB = ZB (eung bdng 90° - ZC) (**). Tif (*) vd ( " ) suy ro ZADB = ZB = ZDAB hay tam gide ABD Id tom gide deu. Vdy to cd:
AB = BD = AD = DC hay BC = 2AB.
Vidu 2;Cho hg parobol (P^): / = x^-^ (2m+l)x + nr^- 1. Chifng minh rdng (P^) ludn tiep xuc vdi dudng thdng y = x - 1.
Thue te eho t h ^ da so GV, HS deu dimg Igi sou khi chimg minh dugc (Pm) ludn hep xuc vdi dudng thdng y = x- 1. Tuy nhien, neu khai thdc su hd tro ciia phdn mem Geomehy Cobri, GV cd the'hep hjc khai thdc bdi todn, tqo mdi h-udng phdt frien tu duy cua HS bdng cdch td chirc ede hoqt ddng sou:
Hogt ddng 1: Minh hga ket qud bdi todn a dqng dgng. Neu sii dyng cdc PMDH cd chue ndng ve dd thi hdm so nhu Graph, Maple... thi rdt khd minh hga cho HS tfidy dugc (P^) thay ddi theo tham so m nhu the ndo vd frong qud frinh thay ddi dd vdn ludn tiep xiic vdi dudng thdng y = x - /. Vd'n de ndy se don gidn hon khi ta sii dyng Geomehy Cobri, ey the:
- Budc 1: Ve do thi cua (PJ wng vdi mgt gid tri cu the cua m:
GV ve hode hudng ddn HS ve dd thj cuo hdm so dudi dqng quy tich ddng bdng cdch: - Dimg
chiie ndng Point on Object Idy hoi diem X(x; 0), M(m; 0) bdt ki tren tryc Ox; - Dimg chifc ndng Equation and Coordinates de xdc djnh tog do cua hai diem X, M ; - Dimg chire ndng Calculate tinh gid tri cuo hdm so y = f(x,m) trong dd x Id hodnh do diem X, m Id hodnh do ciia diem M;
- Dung chife ndng Measurement Transfer, Perpendicular Line dung diem M(x; y=f(x,m));
- Dung chiic ndng Locus dung dd thj ciio hdm so.
- Budc 2: Minh hga ket qud bdi todn mgt cdch true quan. Sou khi dd dung duge dd thj, to minh hga ke't qud bdi todn bang cdch dimg chifc ndng Trace O n / O f f gdn thudc tinh de Igi vet khi chuyen ddng cho dd thj, sou dd eho tham so m thay ddi bdng cdch di chuyen diem M tren tryc hodnh. Ke't qud cho to thdy rd (P ) ludn tiep xuc vdi dudng thdng y = x - 1 (hinh 3).
Hogt dgng 2: Xet bdi todn md rdng.
- Btrdc / ; GV dua ra tinh hudng cd vdn de:
Vdi a Id mdt so thuc bdt ki {a^^) thi tinh chdt (P^):
/ = ax^+ (2m+l)x + m^-1 ludn tiep xiic vdi dudng thdng cd dugc bdo todn? Ddy se Id vdn de rdt khd ddi vdi HS. Tuy nhien vd'n de se trd nen «vifa
sifc" hon ddi vdi HS neu cd , su hd t r g cua Geometry
Cobri. GV ed the cho HS thuc hien cdc ndi dung sou: Ve dd thj cuo (P ) vdi gid trj a cy the, chdng hgn vdi a = 2. Cho m thay ddi vd auon sdt. Hinh dnh h-yc quan cho thdy hg cdc
dudng eong tuong ling vdi gid trj o = 2 khdng cdn ludn tiep xuc vdi dudng thdng y = x - 1.
- Budc 2: GV tiep tue Iqi neu finh hudng cd vdn de: Vdi trudng hgp ey the a = 2, hinh dnh true quan eho thdy rd tinh chdt ludn tiep xuc vdi dudng thdng ciio hg (P^) khdng edn dung nifa, nhung lieu (P^) ed the ludn tiep xuc vdi mdt dudng ndo knde khdng? GV cho HS tie'p hJC thii nghiem vdi mdt vdi gid tri khde cuo a {a=^]). Ket qud true quan cho thdy hg ede dudng cong ndy dudng nhu ludn tiep xuc vdi mdt porabol. Den ddy HS dua ra du dodn: dd thj ciia hg porabol (P^):
y = ax^ + (2m + l)x + nf-1 vdi a Id mdt so thuc bdt ki (a ;t 1) ludn tie'p xue vdi mdt porabol.
- Bwdc 3: Ldm sdng fd du dodn. Xud't phdt tif gid thuye't (P^) ludn hep xuc vdi porabol, ddn de'n bdi todn md rdng: Tim dieu kien eiia cdc he sd b.
Hinh 3
bx'
+ CX + I#
c, d de porabol cd phuong trinh / = ludn tie'p xiie vdi (P_^).
Viec di xdc djnh cdc he so ddn de'n viec HS gidi he phuong h-inh vd ket qud HS dd ehi ra dugc frong
(Xem tiep trang 47)
Tap chi Giao due s6 2 4 2 (ki 2 - 7/201 o)
AE / / O C ) , OE vd OB ngugc h u d n g , nen OE = -|sMLOB = -|iOB. Tuong h/: 0F = - 5 ^ d c = A o c -
s, Vdy OA = -|ioB-5ioc, hoy s,OA + S,OB + S,OC = 0.
Ddy Id trong trudng hgp diem O ndm trong tam gidc ABC, edn trudng hgp O ndm ngodi tam gidc ABC thi ddng thife SiOA-i-SjOB-fSjOC = 0 cdn dung khdng?
Gid sii O ndm trong mien tgo bdi hoi tia CA vd CB, to dung hinh binh hdnh O M C N vdi M , N ndm tren cdc tia O A , OB (hinh 4). Theo quy tdc hinh binh hdnh to ed:
OC = OM + ON = — O A + — O B OA O B
Q . Q . •
^M)BtOA + ^^ec^OB ( d o O N / / M C , O M / / N q
= -^OA-i-^OB . "'• ,, . S3 S,
S u y r a : S , O A - f S 2 O B - S 3 O C = 0 .
Tif ddy, to cd bdi todn: Ggi O Id diem ndm ngodi tam gidc ABC, thudc mien goc tqo bdi hai tia CA vd CB, ggi S,, S^, S^ Idn lugt Id dien tich
cua cdc tam gidc AOBC, AOAC, AOAB. Chung minh rdng:
S,OA-i-S,OB •S3OC = 0 .
Vdn dyng quan diem bien . »,, Hinh 4 chifng ve md'i quan he giira edi
chung vd edi rieng vdo dqy hgc todn cd tdc dyng phdt trien tu duy cuo HS, dgc biet Id cdc logi hinh phdt trien tri tue nhu: phdn tich, tdng hgp, khdi qudt hod, tdng qudt hod vd dd Id tien de cho viec phdt trien tu duy sdng tgo.
Vdn dyng theo djnh hudng ndy giup HS hge tdp chu ddng, tich cue vd ndng cao hieu qud trong qud trinh dqy hgc mdn Todn, gdp phdn ddi mdi phuong phdp dgy hgc hien nay. G
Tai lieu tham khao
1. David Tall. Advanced Mathematical Thinking.
Kluwer Academic Publishers, Dordrecht/ Boston/
London, 1991.
2. G.I. Rudavin - A. Nuxanbaep - G. Sliakhin. M d t s o quan diem triet hoc trong toan hoc. NXB Gido due, H. 1979.
3. Nguyen Ba Kim. Phuong phap day hoc m6n Toaa NXB Dqi hgc suphqm Hd Ngi, 2009.
4. Nguy6n Canh Toan. Phucmg phap luan duy vSt bien chung vol viec hoc, day, nghien cihi Toan. NXB Dqi hge qudc gia Hd Ngi, 1997.
Khai thac tinh tnrc quan...
I (Tiep theo trang 44)
trudng hgp a^], (P^) ludn hep xuc vdi porabol ed phuongtrinh l d / = ^ o - l)x^+x- / ( l ) v d hinh dnh true quan mdt Idn nifa minh hga eho ke't qud cua bdi todn mdt each sinh ddng (hinh 4). Tie'p hjc khoi thdc h'nh h-yc quan cuo Geomehy Cabri, HS hep hjc thu dugc nhifng ket qud thu vj: - Khi a > 2: Hg (P^) ludn hep xuc trong (hinh 4); - Khi a < 2: Hg (P^) ludn Hep xuc ngodi (hinh 5\.
3. Ket lugn
Viec sii dyng PMDH de tgo ro cdc ddi tugng todn hgc sou dd tim tdi
khdm phd cdc thudc tinh dn chiia ben trong ddi tugng dd dd tgo mgt mdi trudng thudn Igi de HS phdt trien khd ndng suy ludn todn hgc vd tu duy logic, dgc biet Id ndng
Hinh 5
lue quan sdt, md t d , phdn tich so sdnh...
Tinh true quan cuo cdc PMDH khdng nhung khdng Idm g i d m khd ndng t u duy cuo HS md trdi Iqi dd hd t r g ta dot duoc muc dich hinh thdnh kien thiic, ren luyen k i n d n g vd phdt t r i e n t u d u y cho HS, g d p p h d n d d i m d i phuong phdp OH nhdm ndng coo chdt lugng DH todn. •
Hinh 4
Tai lieu tham khao
1. John Olive. Implications of Using Dynamic Geometry Technology for Teaching and Learning.
Portugal, 2000.
2. Michael D. De Villiers. Rethinking Proot with The Geometer's Skecchpad. Key Curriculum Press, CA, 1999.
3. Tringa, P.&Lipitakis. A Study of Teaching Mathematical Concepts with Computer. Mathematical Education in Science and Teachnology, 1993.
Tap chi Giao due so 2 4 2 (ki 2 • 7/201 o)
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