CONG NGHE DAY HOC PHAT TRIEN KHA NANG DAC BIET VA KHAI QUAT HOA CHO HOC SINH PHO THdNG V6l SlT HO TRO CUA CNH TS TrJnVÎ Cilttng Tru&ng DHSP Dgi hgc Thdi Nguyin SUMMARY With activefeatures and conserv[.]
CONG NGHE DAY HOC PHAT TRIEN KHA NANG DAC BIET VA KHAI QUAT HOA CHO HOC SINH PHO THdNG V6l SlT HO TRO CUA C N H TS.TrJnVI^Cilttng Tru&ng DHSP - Dgi hgc Thdi Nguyin SUMMARY With activefeatures and conservation of the structure of the mathematical model which desire by the teaching sofhvare, we can easily specialize or expand a convenient for the original objec for students The purpose is to observe and see the corresponding changes of the remaining objec in the mathematical model, which discovered the relationship between the objects, the quanti in the model to propose the comments and predictions In this paper, the relationship betwee specialize and generalize in mathematics teaching in high school with the support of informati technology is studied Nh^nbAi ngiy 10/11/2013 ngiky duyit dang25/12/2013 D$t vin dl Khdi qudt hod va d$c bi|t hod Id cdc dgng hogt dgng tri tu$ chung ciia tu Hogt dgng khdi qudt hod vd d$c bi^ hod Id cdc thao tdc tri tuf cd lien quan vdi khd ndng phan tich, so sdnh, tudng tugng vd hf thong hod Khdi qudt hod vd d$c bigt hod Id m$t nhihig hogt dgng co bdn cua tu sdng tgo Theo G.Polya "Bdn thdn sy khdi qudt hod, sy d$c bi$t hod la nhihig ngu6n g6c vi dgi cua sy phdt minh Trong qud trinh dgy hgc, vi^c tdng cudng ren luy^n khd ndng d$c bif t hod vd khdi qudt hod cho hgc sinh Id nhi^m vy vd myc tieu cua dgy hgc Ngay nay, CNTT ndi chung va PMDH ndi rieng ngdy mgt phdt triln Nhd tinh d0ng va bdo todn cdc cau true cua cdc mo hinh dgy hgc todn cho HS se giup cho qud trinh d$c bi?t hod cdc d6i tugng ho$c cho thay ddi cdc doi tugng mgt each thu|in Igi dk HS cd the khai qudt hod cdc d6i tugng dd mgt cdch thugn Igi Bdi vilt ndy, chung tdi nghien cdu theo hudng phdt trien khd ndng d$c bi^t hod vd khdi qudt hod cho hgc sinh thdng vdi sy ho trg cua CNTT Phdt trien kha ndng d$c bi$t hod vd khdi qudt hod cho hgc sinh thdng vdi sy h3 trg cua CNTT 44 • TAP CHt Ttil^BIGlAO DUC-sd 101-1/2014 a) Theo G.Polya "Khdi qudt hod Id chuyen tu vi^c nghien cdu mgt t|ip hgp doi tugng dd cho den vi^c nghi6n cdu mgt tap hgp Idn hon, bao gom ca tgp hgp ban dau" Nhu v|iy, khdi qudt hod cd tdc dyng dinh hudng din tdi gid thuyet khoa hgc Nhd tinh ndng ddng vd bdo todn ciu trdc cda PMDH, GV cho thay doi m$t vai thugc tinh cda diu bdi de HS quan sdt thiy dugc sy thay d6i tuang dng cua cdc doi tugng cdn Igi md hinh toan hgc, tir dd phdt hi^n m6i tuang quan giiia cdc dii tugng, cdc dgi lugng md hinh dl dua cdc nhdn xet va dy dodn Tiep theo, GV sd dyng cdc chdc ndng kilm tra cua phin mem de kilm thd cdc dy dodn ma HS dua Tir kit qud xd ly cda phin mIm, HS logi bd ho$c rim cdch chdng minh Vi?c GV hudng din HS chdng minh dugc cdc dy dodn dd da gidp HS tgo dugc cdc bdi toan mdi theo hudng khdi qudt hod cdc bdi todn ban ddu Vi dv Tim diem co djnh ciia hg dudng cong (dm):y = x^ + (m-l)x^-2(m+l)x + m-2 GV cd thi khai thdc phan mem Geometry Cabri dl hudng din HS gidi bdi toan nhu sau: Hogt dgng Dy dodn diem c6 dinh cda hg dudng cong CONG NGHE DAY HOC - Chon Laal Show Axes: De cho hien he tnic toa d0 Oxy - Chon L i l Point on Object: L^y cac di^m X (x; 0), JVtta: 0) bdt IcJ- tren triic Ox - Chpn H I £jaofton and Coordinates: Cho hi$n toj cua hai di^m X, M man hinh - Chpn IMl Calculate: Tinh gia trj cua him so x 14 hoJnh dp di^m X, m la hoanh dO cua diem M RI - Chpn U a Measurement Transfer: Lan lupt bam chpn gid trj vCra tinh sau chi vao tr\ic tung Oy Ta xic dinh dupe diem Y thupc Oy - Chpn L i l Perpendicular Line: Lin lupt dvng cac ducrng vuong goc vai tryc Ox tai diem X, vuong g6c vdi Oy tji diem Y - Chpn t i l Intersection Points: Xac giao diem N cua hai duong thing vuong goc vua dpig N sa la diem c6 to(i d0 (x; f(x)) - Chpn IMl Locus: L4n lupt chl vao di^m N vi diem X de Cabri Geometry dua thj cua ^^ ham so - Chpn Trace On/Offtoi bim vao ih thj de dp thupc tinh de l^i v^t cho thi - Chpn LAI Pointer: Cho di chuyen di^m m tren tryc hoanh Ket qui trvrc quan cho thdy: (dm) luon di qua diem co djnh c6 toa dp (1; - 4) (Hinhl) VdisifhSirffcua md hinh dpng, GVcho HS nghfin ciat bai todn md r^ng tren nhir sau: - Buoc 1: GV dua tinh huong co vdn dS; Voi a li mpt s6 th\rc bdt Icy (a ^ 1) thi ii?u rSng hp duong cong (dJ: y = ax' + (m -1):^ - 2(m +l}x + m- CO di qua diem co djnh nao khong? Noi chung diy sS li vdn de rdt kho doi voi HS Tuy nhiSn vdn de si trd nen "vita siic" hon d6i vdi HS neu c6 s\r h3 trp ciia phdn mem Geometry Cabri GV CO the cho HS thirc hi?n cac npi dung Hinh3 T4P CHl THifr B| GlAO DVC - Sfi 101 -1/2014 • 45 CONG NGHE DAY HOC - Vg dd thj cda hg dudng cong ( d J vdi gid tri a cu thi, chdng hgn vdi a =2 Cho m thay doi vd quan sdt hinh anh d6 thi ham so (d^^^) trSn mdn hinh Hinh dnh tryc quan cho thiy hg cdc dudng cong tuang dng V(5i gid tri a = ludn di qua diem CO dinh cd tog dd (1; -3) (hinh 2) - Cho thay dii gid tri a, gid sd a = Cho m thay dii vd quan sdt hinh anh dd thi bam so (d^) trSn mdn hinh Hinh dnh tryc quan cho thiy hg cdc dudng cong tuang dng vdi gid tri a = ludn dl qua diem c6 dinh cd tog dO (1; -2) (Hinh 3) - Bu&c 2: GV tilp tyc Igi neu tinh huong cd van de: Td cdc trudng hgp tren, hay nhdn xet ve moi quan h? cua cdc diem cd dinh cua hg dudng cong di qua dng vdi mSi gid tri cu thi cda a Td ket qud tryc quan vin cho thay "hinh nhu" cdc dilm co dinh ndm trin cung mgt dudng thing c6 dinh Den day HS dua "dy dodn": T§p hgp cdc dilm co dinh cua hg dudng cong ludn ndm tren mgt dudng thing c6 dinh x=l - Bu&c 3: Ldm sdng td dy dodn GV hudng dan HS su dyng lap lugn Id gie dl Idm sdng td dy dodn cda HS Kit qud chimg minh cho thay: Cdc diem co dinh cua hg d u ^ g cong (d ) cd tog dO (I; a - 5) Do dd, t^p hgp cdc diem CO dinh ciia hg dudng cong ludn ndm tren mdt dudng thing co dinh la x = I b) Theo G.Polya: D?ic bi?t hod Id chuyin td vi^c nghiSn cuu mgt t$p hgp doi tugng dd cho sang nghien cuu mgt tap hgp nhd hon chda t^p hgp dd cho D§c bigt hod gidi todn Id vi$c thu ggn viec nghien cdu tren nhdng tgp cd nhdng d|ic diem rieng noi b^t so vdi tap hgp ban dau Vdi su ho trg cda PMDH n6i rieng vd cua CNTT ndi chung, GV hudng dan HS tim toi ldi gidi cdc bdi toan Tiep dd, nhd tinh ndng dgng vd bdo todn cau true cua PMDH, GV cho dac bi$t hod mOt so thugc tinh cua diu bai, cho m$t s6 doi tugng tdi mgt so vi hi d$c bi?t dl HS quan sat thay dugc sy thay dii tuang iing cua cdc doi tugng cdn Igi cdc md hinh todn hgc, td dd phdt hign mli tuong quan gida cac doi tugng, cdc dgi lugng md hinh dl dua cdc nhgn xet vd d\[ dodn Vi?c GV hudng dan HS chdng minh dugc cac dy dodn dd gidp HS tgo dugc cdc bai toan mdi theo theo hudng d§c bi|t hod bai todn ban dau VI dij Cho hinh vudng ABCD Ldy mgt diim M bdt ki hinh vuong do, Dir&ng thdng d'quaMcdtABtgiE, c&tCDtgiG, du&ngthdng d" qua M vudng ^c v&i d'cdt BC tgi F, cdt AD tgi H Chimg minh rdng: EG = FH De giup HS gidi quylt bdi todn ndy, GV sd dyng PM Cabri dl hudng dan HS bing cdch thyc hi^n cdc hogt dgng sau: Ngi dung Hogt dgng cua gido vi€n Hogt dgng cua bgc sinh HD 1: Tim hilu bai todn - Mdfile "VD3.fig" {hinh 4) - Hay xdc dinh gid thilt, kit lu$n cua bai todn Quan sat hinh v5 tren man hinh, xdc dinh gid thil^ kit ludn ciia bdi toan V€ hinh, ghi gid diiet, kit lugn vdo vd HD2: Xay dyng chuong trmh gidi bdi todn - Cho d' thay doi den vi tri d$c bi?t ? Hay quan sat vd dy dodn d$ dai hai dogn EG - Khi dV/ BC thi d" // AB nen EG = FH (vi ldc dd H = T vd G = S, dd T la hinh chieu cua F ISn AD, S Id hinh chieu cda E len DC vd ES = B C = A B = FT) - AGES = AHFT - E G = FH HD3: Chdng minh Nhic la cdc phuang phdp chdng minh hai dogn thing bdng vkVli(hinh5) ? Cd nhgn xet gi ve hai tam gidc AGES va AHFT? ? Hay so sdnh dd ddi cda hai dogn thdng EG vdFH? ii ' W CHt THiflr B| GlAO DMC- SA101 -1/2014 HgSElDCFTlAD Khi ta cd AGES = AHFT Suy EG = BH CONG NGHE DAY HOC N$i dung Ho^t d$ng cua gido vien HD4: Khdo sdt ldi gidi dd tim dugc - Cho diem M trung vdi (tdm cua hinh vuong) [hinh 6) ? Cdc tam gidc AOEB, AOFC, AOGD, AOAH cd dac diem gi? ? Cdc tam gidc AOEA, AOFB, AOGC, AOHD cd d$c dilm gi? ? Cho biet td gidc EHGF Id hinh gi? GV kit ludn: Td suy Iu$n tren ta co bdi toan: Cho hinh vudng ABCD, hai du&ng thdng vudng gdc d', d" qua tdm cua hinh vuong cdt cdc cgnh hinh vudng tgi cdc diem tuong img E.G,F,H (hinh 6) Chung minh rdng: ^^ %B^a^ Hogt d$Dg cua hgc sinh - Quan sat hinh ve - AOEB = AOFC = AOGD = AOAH - AOEA = AOFB = AOGC = AOHD - EHGF la hinh vuong (vi SQg^= SoHDC=SoGCF=SoFBEVdOE = O H = OG=OF) + Sd dyng suy lu$n Id gie dl chdng minh bdi todn ^OGC " ^OFBE ^ '-^^ ^ABCD- b) EFGHld hinh vudng - Cho M trung I (I Id tmng dilm cua BC) va dudng thing d" qua A (/KK/I 7), ? So sanh dd ddi cdc dogn ME, MG va HM (hinh 7) ? Cho biet vi tri cua diem E? GV kit lu|in: AHME dyng dugc, diem B xdc dinh dugc (B la chdn dudng cao hg td M xuong HE) Vgy hinh vuong ABCD dyng dugc Td ta cd bai todn sau: "Dgng hinh vuong ABCD biet dinh A vd trung diem Mcua BC" - Quan sat hinh vS - M E = M G = HM/2 - E thugc d' vd cdch M m^t khodng bangHM/2 + Sd dyng suy lugn Id gie dl chimg minh bdi todn Ket lu^n Tvt mpt vii vi dy tren cho thdy, viec khai thac m6i quan h$ giiia d$c bi|t hoi va khii quit hoi qui trinh day hpc cho HS thong vai s\r hS trp cua CNTT giup HS phit trien dupe ning tu va nSng sing t^o cho bin than, T A I ISgV THAM KHAO Trjnh Thanh Hii, Trin Vift Cirong, Trjnh Thj Phuong Thio , Giio trinh Ung dyng tin hpc d^y hpc toin, hiXB Giio dye Vi?t Nam.(2013) Nguyin Bi Kim, Phucmg phip djy hpc mon toin, NXB Dji hpc Su ph?m.(2005) Hinh TAP CHl THift- B| GlAO DgC - Sfi 101 -1/2014 • 41 ... DAY HOC - Chon Laal Show Axes: De cho hien he tnic toa d0 Oxy - Chon L i l Point on Object: L^y cac di^m X (x; 0), JVtta: 0) bdt IcJ- tren triic Ox - Chpn H I £jaofton and Coordinates: Cho hi$n... v^t cho thi - Chpn LAI Pointer: Cho di chuyen di^m m tren tryc hoanh Ket qui trvrc quan cho thdy: (dm) luon di qua diem co djnh c6 toa dp (1; - 4) (Hinhl) VdisifhSirffcua md hinh dpng, GVcho... dy tren cho thdy, viec khai thac m6i quan h$ giiia d$c bi|t hoi va khii quit hoi qui trinh day hpc cho HS thong vai s\r hS trp cua CNTT giup HS phit trien dupe ning tu va nSng sing t^o cho bin