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[.]
Trang 1TtiM NANG CUA CAC BAI TOAN KET THUC Mff
IRONG VliC Hi TRO HOC SINH PHAT TRIIN NANG IIIC SUY IDAN NGDAI SOY
O TtiS, T R U O N G THI KHANH P H I / O N G '
Nh i l u cdng trinh nghl3n cuu khoa hpc trong
nhi/ng ndm gdn ddy dd nhdn mpnh d i n
tdm quan trpng cOa suy tudn ngopi suy
(SLNS) trong vl§c thuc dd'y con ngudi hinh thdnh
y tudng sdng tpo vd khdm phd ll t h u y l t mdi,
Trong day hpc todn d phd thdng, vdn d l ddt ra
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?
1 Suy luqn n g o g i 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 trong todn hpc Id
khi Goldboch quon sdt thdy: 3 + 7 = 1 0 , 3 + 17
= 2 0 , 1 3 + 17 = 30 vd nhdn xet rdng ede so 3, 7,
13, 17 Id ede sd nguyen t d \k, h> dd dng d d phdf
b i l u mpt gid thuylt ndi H i n g : Moi so chdn Idn
hon 2 Id tdng cua hai sd nguyen td'le
Q u d trinh SLNS dt/pc dua ra 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 khi HS khdo sdt tren cdc md hinh todn hpc
d d n g nhdm phdt hien cdc djnh If, cdng thuc todn
hpc, d y dodn so' hqng tdng qudt cua mdt d d y
sd' khi b i l t tri/dc mot vdi sd hpng, Hm quy tich
cua mdt d i l m , thdm chi d u a 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
2 Bdi t o d n k i t thuc m d
Tu nhimg ndm 1971 -197d, ede BTKTM dd Hd
thdnh mdt cdng cy d l ddnh gid h/ duy bpc coo
trong 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 trong tinh hudng Trong khi dd, BTKTM eo
«cdu &UC yiu'vdi rdt it dieu kien rdng huge trong cdc da Iliu dugc cung cap vd mang Igi cho hqc sinh nhieu edeh tiip can khde nhau Thdng thudng, cdc BTKTM dua ra nhdng tinh hudng vd yeu cdu
HS thSm gid thuyet vdo BTdimqt tinh chdt ndo do dugc thda mdn, gidi thich cdc kit qud, tgo ra 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 ra nhimg thdch thuc doi vdi qud Hinh tu duy vd suy luqn euo HS; - Cho phep HS dp dyng n h l l u cdch Hep cdn vd chiln li/qe khde nhau 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 nhau Cdc dap dn co
t h i thay ddi khi 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 do 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 do cho su lya ehpn d d
Vf du sou ddy minh hoa mdt each ehuyln ddi
h> BT d d n g (BTl) song BTKIM (BT2) BT1: Mdt hinh chi/ nhdt ed c h l l u ddi Id 10m,
e h i l u 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 - fiai hoc M
Tap chi filao due s6 2 7 6 (kt a • la/aoiij
Trang 2suy luqn d l gidi quyet vd'n de euo HS cung dupe
djch c h u y i n t u 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 nhau ve kfeh thudc hinh chu nhdt, mien id
cd su gidi thfch hpp If) Id ede thdnh phdn eo bdn
cua SLNS
3 Khai thdc cdc BTKTM n h d m phdt trien
ndng luc SLNS c h o HS
Nhimg phdn h'ch Hen cho thdy, d i l u kl§n thudn
Ipi de SLNS dien ra Id khi HS dupe ddt trong 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 t d
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 khi HS dd thu thdp mdt he thdng ddy du
cdc gid t h i l t h> BT Vdl cdc BT t h i l u gid thilt, HS
cdn dua ra 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 ra
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 1 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 h d n h SLNS n h d m tim
k i l m ede quy tde
«khdp" vdl trudng hpp rieng dong xet, Idm eo sd d l dua
ra 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 do 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 khi b i l t vj trf diem
{ 1 ; 1) Trudng hgp: G p l 2 d l l m M , N ed too dp
tuong i/ng Id: ( 1 ; 1) vd (2; 4) thi M , N thudc do
thj hdm so' f,(x} = ax? Tuy chung ta chua b i l t ey
Tap chi Blao due so 2 7 6 (ki a • la/aon)
Hinh I
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) sao cho khi hodnh dp vd tung
dp eua M , N «khdp" vdl nhau trong he trye tpa
dp, nghTa Id vj tri d i l m ( 1 ; 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 f j x ) = 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 do 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 f j x ) = a x ' sang phdi mpt dopn bdng h don vi vd djch l6n tren mot doqn bdng k
dem vj, do 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 1
- Ngoqi suy 3: Quy tde: Porobol ed b l lom
hudng ten tren khi a > 0 hudng xud'ng dudi khi •
< 0 Trudng hgp: Parabol g(x) co b l Idm hudng
xud'ng di/di K i t luqn: a < 0
- 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 do khde nhau Cac BT ndy khdng co quy trinh cy the di dl din Idi gidi vd tieu chudn dixdc dinh khi 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 trong mpt sd Ht/dng
hop Ddy Id cdc gioi dopn euo qud Hinh SLNS Vidy 2: Nhd An cdch trudng 4 km, nhd Binh
cdch Hudng 3 km Theo em, nhd A n vd nhd Binh
ed t h i cdch nhau boo xa? Gidi thfch? (quy udc: nhd A n 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 do 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
#
Trang 3vdl AT = 4 , BT = 3 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: 1 i AB ^ 7 (km)
Phuemg dn 2: Vdl cdc khodng cdch khdng ddi
AT = 4 , 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 nhau
hodc h i p xuc nhau, don d i n rdng buqc: AB £ 3
+ 4 = 7 (km) vd AB a 4 - 3 = 1 (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 trong bdi Idm cua
mot HS trong so d d (hinh 2)
Hinh 2
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 Q u d 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 trong mdt cdng vien nhd hinh
torn gide soo cho n d ehilu sdng todn b d cdng
vien Ngudi ta nen dpt nd d ddu?
SLNS dupe HS su d y n g 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 n d , d o d d edy den cdn d d t phfa trong
cdng vien Nhu cdu d u o ro mdt Idi g i d i td'i uu
hem thiie ddy HS nghT d i n vfee Idm t h i ndo de
dnh sdng phdt ra d I u khdp cdng vign, sou d d
dung SLNS ddnh gid cdc gid thuylt van edn nghi
n g d : Vj trf ddt edy den ed t h i Id trpng tdm, tdm
dudng trdn ndi t i l p , hay tdm d u d n g trdn ngopi
Hep eua cdng vien hinh tam gidc? V d i lya ehpn
tdm dudng trdn ngopi t i l p , HS nhdn thd'y N e u
mpt trong ba gdc cua cdng vien Id t u , thi Idl
cdng vidn, HS H i p tyc ngoqi suy d l d u a r o mpt lya chpn phu h p p theo cdch It g i d i riSng eOa cdc em trong 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 trong 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 khi mu6n si>
d y n g suy ludn suy d i l n , ngupc Iql n d 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 p h d i 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 N h i / n g 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 problems for 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