TtiM NANG CUA CAC BAI TOAN KET THUC Mff IRONG VliC H i TRO HOC SINH PHAT TRIIN NANG IIIC SUY IDAN NGDAI SOY O TtiS, TRUONG THI KHANH PHI /ONG '''' Nhilu cdng trinh nghl3n cuu khoa hpc trong nhi/ng ndm gd[.]
TtiM NANG CUA CAC BAI TOAN KET THUC Mff IRONG VliC Hi TRO HOC SINH PHAT TRIIN NANG IIIC SUY IDAN NGDAI SOY O N h i l u cdng trinh nghl3n cuu khoa hpc nhi/ng ndm gdn ddy dd nhdn mpnh d i n tdm quan trpng cOa suy tudn ngopi suy (SLNS) vl§c thuc dd'y ngudi hinh thdnh y tudng sdng tpo vd khdm phd ll thuylt mdi, Trong day hpc todn d phd thdng, vdn d l ddt Id: hogt ddng ndo cd khd ndng rfiuc dd'y hqe sinh (HS) su dyng SLNS? cdc bdi todn kit t/it/c md (BTKTM) cd liSn hi nhu thi ndo din viec phdt triin ndng lye SLNS cho HS? Suy luqn ngogi stjy SLNS Id qud trinh suy ludn nhdm tim kiim nhi/ng gid thuyitphu hgp nhd't digldi thich cho mot kit qud quan sdt dugc (I) Ndi cdch khde, SLNS nhdm trd Idi cdc cdu hdi: Ddy Id trudng hgp cua quy tdc ndo^; Diiu gi dan din kit qud ndy?Vi dy kinh d i l n ve SINS todn hpc Id Goldboch quon sdt thdy: + = , + 17 = , + 17 = 30 vd nhdn xet rdng ede so 3, 7, 13, 17 Id ede sd nguyen td \k, h> dd dng d d phdf b i l u mpt gid thuylt ndi Hing: Moi so chdn Idn hon Id tdng cua hai sd nguyen td'le Qud trinh SLNS dt/pc dua hd\ J Josephson vd S Josephson (2) gdm cdc budc sau: - Cho D Id mot tdp ede du lieu (sy kien, quan sdt, edi d d cho); - Xudt hien gid thuylt H gidl thfch cho D; - Khdng ed gid thuyet ndo khde ed t h i gidl thich D tdt hon H; - H ed le Id dung Trong day hpc todn, SLNS cua HS thudng gdn l i l n vdl cdc hopt ddng hpc tap nhu: khdo sdt, gidi cdc BTKTM, tdng qudt hdo, phdt b i l u cdc bdi todn (BT) md rpng Nhi/ng suy ludn dien ro HS khdo sdt tren cdc md hinh todn hpc ddng nhdm phdt hien cdc djnh If, cdng thuc todn hpc, d y dodn so' hqng tdng qudt cua mdt d d y sd' b i l t tri/dc mot vdi sd hpng, Hm quy tich cua mdt d i l m , thdm chi dua ro cdc If thuyet d e g i d i thfch cho sy xud't hiSn eua mpt khdi niem mdi khdng phdi td h/ suy luqn suy dien, md ehu y l u Id SLNS TtiS, TRUONG THI KHANH P H I / O N G ' Bdi todn k i t thuc m d Tu nhimg ndm 1971 -197d, ede BTKTM dd Hd thdnh mdt cdng cy d l ddnh gid h/ bpc coo todn hpc Theo Pehkonen (3), BTddnglddgng BTcd cdu true hodn chinh, mot cdu hd Idi dung ludn dugc xdc dinh rd rdng tu nhimg gid thiit vua du dugc cho tinh hudng Trong dd, BTKTM eo «cdu &UC yiu'vdi rdt it dieu kien rdng huge cdc da Iliu dugc cung cap vd mang Igi cho hqc sinh nhieu edeh tiip can khde Thdng thudng, cdc BTKTM dua nhdng tinh hudng vd yeu cdu HS thSm gid thuyet vdo BTdimqt tinh chdt ndo dugc thda mdn, gidi thich cdc kit qud, tgo BT mdi cd liin quan hay tdng qudt hda BT Theo Foong (4), mdt BTKTM cdn neu dt/qc mdt sd dqc dllm sou: - Tqo eo hql cho HS t h i hien sy ndm vihig k i l n tht/c, soo cho tdt ed HS d I u cd khd ndng tim di/pc cdu Hd Idi; - Duo nhimg thdch thuc doi vdi qud Hinh tu vd suy luqn euo HS; - Cho phep HS dp dyng n h l l u cdch Hep cdn vd chiln li/qe khde d l di den Idi gidi BT Xudt phdt h/ nhi/ng quan d i l m Hen, Hong bdi v i l t ndy, chung tdi xet cdc BT ed du bo ddc tnmg sou dupe coi Id BTKTM: 1) Khdng cd phuemg phdp hoy thuqt todn c d djnh d l g i d l ; 2) Chd'p nhdn n h l l u ddp dn dung khde Cdc dap dn co t h i thay ddi cdc d i l u kien rdng budc thay d l i hodc cdc thdng tin tdt hem duoc cung cd'p; 3) Co n h i l u cdch tiep edn d y a H&n cdc muc khac nhau, tdt cd HS deu cd t h i chpn cho minh edeh h i p can phu hpp vd gidl thfch dupe If cho su lya ehpn d d Vf du sou ddy minh hoa mdt each ehuyln ddi h> BT ddng (BTl) song BTKIM (BT2) BT1: Mdt hinh chi/ nhdt ed c h l l u ddi Id 10m, ehilu rpng Id 5m Tim ehu vl vd dien tfch cOa hinh chi/ nhdt ndy BT2: N l u ehu vl mdt hinh chu nhdt Id 30m thi dien tfch eiJa nd cd t h i Id boo nhidu? * Tnrdng B9I h^c Y Dinrc -fiaihoc M Tap chi filao due s6 (kt a • la/aoiij Tuong ung vdl sy chuyin ddi tren, qud trinh suy luqn d l gidi quyet vd'n de euo HS cung dupe djch chuyin tu suy luqn suy d i l n vdi viec su dyng cdng thuc d d b i l t (cdng thuc tinh dien tfch vd chu vi hinh ehCr nhdt) song cdc hopt ddng de xud't gid thuylt vd If gidi (cd t h i ehd'p nhdn nhieu gid thuylt khde ve kfeh thudc hinh chu nhdt, mien id cd su gidi thfch hpp If) Id ede thdnh phdn eo bdn cua SLNS Khai thdc cdc BTKTM nhdm phdt trien ndng luc SLNS cho HS Nhimg phdn h'ch Hen cho thdy, d i l u kl§n thudn Ipi de SLNS dien Id HS dupe ddt mdt mdi tn/dng cd vdn d l vd k i t ludn di/o ro h/ nhCrng d i ; kien da cho vdn cdn Id d i l u nghi vd'n, ddi hdi HS h/ d l xud't cdc gid thuylt, Hm k i l m cdc cdch If gidi phu hqp Cdc BTKTM ddp ung tdt nhi/ng d i l u ndy Hon ni/o, n l u BTKTM tdn dyng tdt ede y l u td sou ddy se thuc ddy HS h l n hdnh SLNS d l gidi q u y l t vd'n de, cy t h i nhu sou: a) Thieu gid thiit Suy ludn suy dien chi dupe su dyng HS dd thu thdp mdt he thdng ddy du cdc gid thilt h> BT Vdl cdc BT t h i l u gid thilt, HS cdn dua kit ludn cd thi chd'p nhdn duqe dua tren mdt tap hqp cdc dd lieu khdng ddy du hoqe lya chqn gid thiit them vdo BT di cd thi dua nhung li gidi phu hqp nhdt Dd Id nhi/ng k i t ludn mong bdn ehd't cua SLNS Vi dy 1: V i l t mdt phuong Hinh cd t h i dt/pc dudi dpng g(x) = a(x - hf'+k cho hdm so g(x) ed dd thj nhu hinh Gidi thfch vi soo pht/ong Hinh cua bqn Iqi cd If Q u a n sdt qud trinh thdo ludn vd b d i Idm cOo HS, ehung tdi nhqn thdy HS d d tien hdnh SLNS n h d m tim k i l m ede quy tde «khdp" vdl trudng hpp rieng dong xet, Hinh I Idm eo sd d l dua k i t ludn Sou ddy Id tdng k i t mot qud Hinh SLNS H§u b i l u dupe HS thye hien: Ngoqi suy 1: Tren he true too chira cd ede d l l m chia don vj, ede khodng chia don vj tr&n hai trye tqa d d dupe xdc djnh b i l t vj trf diem { ; 1) Trudng hgp: Gpl d l l m M , N ed too dp tuong i/ng Id: ( ; 1) vd (2; 4) thi M , N thudc thj hdm so' f,(x} = ax? Tuy chung ta chua b i l t ey Tap chi Blao due so (ki a • la/aon) t h i d l l m ndy ndm d vj Hi ndo tr&n dd thi nhung to ed t h i xe djch ehung d i n hoi vj trf ndo dd tren d d thj hdm so f(x) (thudc gdc phdn tu thu I eua he true Oxy) cho hodnh dp vd tung dp eua M , N «khdp" vdl he trye tpa dp, nghTa Id vj tri d i l m ( ; 1) dupe xdc djnh K i t ludn: Cdc khodng chia don vj tren hai trye too dp diroc xdc djnh - Ngogi suy 2: Quy tdc: Dd thj hdm so g(x) = a(x - h)' + k Id tjnh tien euo dd thj hdm sd fjx) = ox^ song phdi mdt doqn bdng h don vj n l u h duong, song trdi h don vj n l u h dm, sou dich ISn tren mot doqn bang k don vj n l u k dtrong, xudng di/di k don vi n l u k dm Trudng hqp: Dd thj hdm g(x) Id phep tjnh h l n eua dd thj hdm sd fjx) = a x ' sang phdi mpt dopn bdng h don vi vd djch l6n tren mot doqn bdng k dem vj, Idn euo h, k dugc udc tupng dya vdo cdc khodng chio tren hai trye too dp Kit ludn: h xdp xi bdng 4, k xd'p xi bdng - Ngoqi suy 3: Quy tde: Porobol ed b l lom hudng ten tren a > hudng xud'ng dudi • < Trudng hgp: Parabol g(x) co b l Idm hudng xud'ng di/di K i t luqn: a < - Ngoqi suy 4: Quy tde: Hai nhdnh cua parabol F,(x) = o x ' cdng «dyng dung" len n l u gid trj hjyet ddi eua a cdng Idn, cdng «md rdng" ro n l u gid tri tuyet ddi cua a cdng nhd Trudng hqp: Parabol F,(x) = ox^ cd hai nhdnh md rdng ro hon so vdi dd thj hdm sd fj{x} = -x' (dd'i xung vdi f,(x) = x^ qua true hodnh) Kit ludn: a < -I b) Khdng ed quy trinh gidi cy the, cd nbiiu cdch tiip can dua tren cdc muc khde Cac BT ndy khdng co quy trinh cy the di dl din Idi gidi vd tieu chudn dixdc dinh ndo thi Idi gidi dugc chd'p nhdn (5), nen HS dt/pc de xud't cdc gid thuylt md cdc em cho Id hop If vd ddnh gid xem tdi gidi ndo Id tdi uu mpt sd Ht/dng hop Ddy Id cdc gioi dopn euo qud Hinh SLNS Vidy 2: Nhd An cdch trudng km, nhd Binh cdch Hudng km Theo em, nhd A n vd nhd Binh ed t h i cdch boo xa? Gidi thfch? (quy udc: nhd An Id d i l m A , nhd Binh Id diem B, Hudng hpc Id d i l m T) Sou ddy Id hai pht/ong dn chfnh dupe so HS t h i hien: Phuong dn I: Khi A , B, T thdng hdng thi AB = 7km hodc AB = 1km Khi A , B, T khdng thdng # hdng, chOng kio thdnh ba dinh eiio mdttom gide vdl AT = , BT = Phep ngopi suy giup HS llSn tudng d i n quy tdc lien quan d i n dp ddi bo eqnh cua mot tam gidc: | a - b | < c < a + b v d dua ro k i t luqn: i AB ^ (km) Phuemg dn 2: Vdl cdc khodng cdch khdng ddi AT = , BT = 3, phep ngoqi suy giup HS llSn tudng d i n cdc di/dng trdn (A; 4) vd (B; 3) De tdn tqi d l l m chung T cua hai dudng trdn ndy, HS duo ro gid thuylt (A; 4) v d (B; 3) phdl edt hodc h i p xuc nhau, don d i n rdng buqc: AB £ + = (km) vd AB a - = (km) Mdt sd HS khde edgdng t i l n hdnh SLNS nhdm trd Idl cdu hdi: N l u cd'd|nh trudc d l l m T thi A vd B ed t h i d nhi/ng vj trf ndo Hong mdt phdng? Dudi ddy Id nhCrng hinh ve bdi Idm cua mot HS so dd (hinh 2) Hinh c) Chua dung cdc tinh hud'ng thuc ti V d i ede BT thuc t l , HS phdi SLNS nhdm It/a chpn nhimg dir kien cdn thu thdp, kiin thuc ndo cdn dp dyng, cdc phuemg dn di gidi quyit vd'n de Qud Hinh chuyen h/ cdc Hnh hud'ng thuc t l sang md hinh todn hpc bao gdm ede hoqt ddng lien quan den SLNS nhu dgt gid thuyit, tdng qudt hda, tim kiim cdc quy ludt vd mdi lien he giua cdc yiu td Vidy 3: Hdi ddng thdnh p h d q u y l t djnh dyng mdt edy den dudng mdt cdng vien nhd hinh torn gide soo cho nd ehilu sdng todn bd cdng vien Ngudi ta nen dpt nd d ddu? SLNS dupe HS su dyng nhdm xdy d y n g md hinh todn hpc cho vd'n d l ndy: Cdng vien cd t h i dupe col nhu Id mot tom gidc, dnh sdng phdt ro tl/ bdng den nhu mot hinh edu m d bdng den Id tdm eua nd, d o d d edy den cdn ddt phfa cdng vien Nhu cdu duo ro mdt Idi gidi td'i uu hem thiie ddy HS nghT d i n vfee Idm t h i ndo de dnh sdng phdt d I u khdp cdng vign, sou d d dung SLNS ddnh gid cdc gid thuylt van edn nghi ngd: Vj trf ddt edy den ed t h i Id trpng tdm, tdm dudng trdn ndi t i l p , hay tdm dudng trdn ngopi Hep eua cdng vien hinh tam gidc? Vdi lya ehpn tdm dudng trdn ngopi t i l p , HS nhdn thd'y N e u mpt ba gdc cua cdng vien Id t u , thi Idl gidl trd nen khdng thyc t l vl cdy d6n ndm ngodi cdng vidn, HS H i p tyc ngoqi suy d l dua ro mpt lya chpn phu hpp theo cdch It gidi riSng eOa cdc em Hnh hudng ddc biSt ndy: vj tri ddt edy dkn ndy nSn Id trung d l l m cqnh Idn nhd't ciia cdng vien hinh tam gidc, Nht/ng phdn tfch v l mdt li t h u y l t vd k i t qud thye nghiSm cho thd'y: cdc BTKTM thuc ddy HS su d y n g SLNS q u d trinh khdm phd vd g i d l quyet vd'n d l TInh «cd'u true y l u " ciia BTKTM k h i l n HS g d p khd khdn mu6n si> d y n g suy ludn suy d i l n , ngupc Iql nd tqo co hdi d l ede em m d rdng k i l n thuc thdng qua v\%c di xudt gid t h u y l t mdi d d p u n g dupe cdc d l i u klSn r d n g budc cua BT M d t khde, d l gidl q u y l t cdc BTKTM, HS phdi suy ludn ngupc tu k i t q u d d l djnh hudng xem If t h u y l t hoy quy tdc ndo cdn vdn d y n g , phuong dn ndo cdn Hien khoi d i g l d l q u y l t vd'n d l Nhi/ng kTndng ndy dupe phdt t r i l n trSn n I n tdng SLNS V i vdy, cdn ddy mqnh hon nua viec su d y n g cdc BTKTM vdo q u d trinh d q y hqe todn nhdm phdt Hlln ndng lye SLNS cho HS p h d thdng G (1) Patokorpi, E Role of Abductive Reasoning in Digital Interaction Doctoral Dissertation Abo Akademi University, Finland, 2006 (2) Rivera F - Becker, J, Abduction in pattern generalization Proceedings of the 11" conference of the IMF Vol (4), pp 97-104, Seoul, Korea 2007 (3) Pehkonen, F Use of open-ended problems in mathematics classroom- Research Report 176 University of Helsmki Finland 1997 (4) Foong, P.Y Open-ended problemsfor higher-order thinking in mathemalic Source: Teaching and Learning, 20(2), pp 49-57, Institute of Education Singapore, 1996 (5) Chan, C, M E Engaging Students in Open-Ended Mathematics Problem Tasks: A Sharing on Teachers Production and Classroom Experience Nanyang Technological University, Singapore, 2005 SUMMARY Alxiuction IS the process of reasoning to find ttie most appropriate h^x>theses wNch explain some obsen/ations This article analyses the potential of open-ended problems which support students developing cAxiuctive reasoning competency Tap chi Biao due so (ki a - la/aoiij ...Tuong ung vdl sy chuyin ddi tren, qud trinh suy luqn d l gidi quyet vd''n de euo HS cung dupe djch chuyin tu suy luqn suy d i l n vdi viec su dyng cdng thuc d d b i l t (cdng thuc... thuc ddy HS h l n hdnh SLNS d l gidi q u y l t vd''n de, cy t h i nhu sou: a) Thieu gid thiit Suy ludn suy dien chi dupe su dyng HS dd thu thdp mdt he thdng ddy du cdc gid thilt h> BT Vdl cdc BT... xd''p xi bdng - Ngoqi suy 3: Quy tde: Porobol ed b l lom hudng ten tren a > hudng xud''ng dudi • < Trudng hgp: Parabol g(x) co b l Idm hudng xud''ng di/di K i t luqn: a < - Ngoqi suy 4: Quy tde: Hai