o i l O l i l G NANG lOi; SANG TAO CHO HOC SINH TRUNG HOG CO Sfl THDNG QOA KHAI THAC SOY LOAN LOGIC Tff CAC OAI TOAN HINH HOC PRO RfEN TRONG OAY HOC TOAN TS, PHAN A N H " Mpt trong nhdng muc tieu quan[.]
Trang 1oil O l i l G NANG lOi; SANG TAO CHO HOC SINH TRUNG HOG CO Sfl THDNG QOA KHAI THAC SOY LOAN LOGIC Tff CAC OAI TOAN HINH HOC
PRO RfEN TRONG OAY HOC TOAN
TS, P H A N A N H "
Mpt trong nhdng muc tieu quan trpng trong
dgy hpc loan la day eho hpe sinh (HS) biel
suy nghT, biet sang tao, nghTa la hinh thanh
eho cac em kha nang caibien tri thdc, bien ddi cac ddi
tupng toan hoc, lao nen san pham lu duy cd tlnh
mdi me va cd gia In ddi vcri ban than Phai then nang
luc sang tao eho HS bao ham nhieu van de, lien quan
den viee trang bi kien thde eho cac em dua tren dpng
CO, xue cam va tri lue, Trong pham vi bai bao nay,
ehung ldi trinh bay each thuc khai thac suy luan logic
ldmptsd/ja//oan/)/n/?/?oc(BTHH)phdbien dtrudng
tnjng hpc casd (THCS) nhim giup HS kham pha ra
cac BT mdi, gdp phan bdt dudng nang lue sang lao
ChoHS
1, Cau true logic cua mot s o ' B T H H p h d b i e n d
trudng THCS va cac djnh hudng khai thac nham
thiet lap bai toan (BT) m d l
CacBTHH dtrudng pho thong rat da dang va
phong phu ve the loai, c h i n g hgn nhu: BT linh
loan, BT chdng minh, BT tim cue tri htnh hpc, BT
quy tich, BT dung hinh, Viec dung logic menh de
de md ta chung bdi mdt cdng thdc chung la kha
phdc tgp Tuy nhien, eae BTHH dupe phat bieu
dudi dgng: "Chung minh menh deA => B trong dd:
A = Pt Ap2/\ /\p„;B = q, /\q2/\ Aq„," (*) la
kha phd bien tnsng chuong trinh THCS, an tarig trong
nhieu dang loan da dan ra dtren Can chu y ring: khi
chdng minh menh deA=^B nghTa la ehung minh
A^B\a menh de dung Neu A la menh de sai thi
hien nhien A=>B\a menh de dijng, Trong menh de
A^B.A duqc gpi la gia thiet, B dupc gpi la kel luan,
Cac BT dgng (') thudng cd mat trong saeh giao
khoa, saeh bai tap; ed thd khai thac suy lugn logic tu
cac BTHH phd bien dtrudng THCS de kham pha ra
cae BT mc^ theo eae hudng sau:
- Hu^g thtf nhat: [<gh\en euu kTqua trinh chung
minh BT(*) (chdng minh menh de,4=:>e),ngoaimuc
dieh hoan ihanh ldi giai edn cdthe rut ra dupe ket luan
B, nao nda khong? Neu cd, hay phalbieu BT: "Chung minh menh de:A => B/' Neu trong qua trinh suy luan
chua sildung het gia Ihiet A, can chinh lai gia thiet va phat bieu cho hpp li
• Wudnfir//?d'/?a/'Sddung ket qua Strong BT(*) (cd the kel hpp vdi A) de suy dien ra cae ketqua khae, gia sulhu dupc kel qua B^ Tudd, cho phep HS phat
bieu BT: VhdngminhmenhdeA=^B^"
- W[/d'/75r//ji^6a,-Nghien cuu menh deA (giathiet
c u a B T O ) , datra van deed the tim dupc menh de
A, sao cho ^ , zi> >1 la menh de dung hay khong"? Neu cd menh deA, thoa man dieu do thi cdthe phat
bieu BT: "Chung minh menh de: A ^^^ B" Khikhai
Ihac suy luan logic theo hudng nay, HS can chij y den cac van de sau: + Oa cd /4, =? ^ la menh de
dung, w'iAjt Aj nen ngoai ket quaA,=>B\a menh
de dung cdn cd the lim thay nhieu dieu thu vj khac, can triet de khai thac; + Cht phat hien ra dupc menh deA,suydienramplsdgiathietnaoddcuaA;ehang
han, 4 =^ p^ A;72'^ Aj[j,;.s<«lamenhdedung; khi do, cd the phat bieu BT: "Chung minh menh de
A A P X A p,.2" - A /7„ ^ s " hoge tiep lue tim kiem menh deAjsao cho A^_ ^ p^^, A/>^,3 A , A ; J „
la menh de dung de thiel lap BT: "Chting minh menh
de A^AA^^B "
- Huong thutu: Lat nguac van de: lieu menh de
S =>-4 cd phai la menh de dung hay khdng? Thue ra menh de S => ^ rat it khi la menh de dung; bdi vay, can cd su dieu chmh ca A va B de ed dupe menh de diJng, Tddd, phat bieu BT mdi
Bd'n djnh hudng tren la nhdng ytUOTiglhiellap eae
BT mdi dua tren cac suy lugn logic khi giai BT ('); dT nhien, ehung ta cung cdthesudung phuang thdc nay ddi vdi eae he qua cua nd
"Tnrdng fiai hpc Ha Tinh
Trang 2m d i dija tren iihai thac s u y luan logic t i j m o t so
BTHH phoblen d t r u d n g THCS
Omuc 1 da trinh bay mpt so djnh hudng chung ve
viec ithai thac suy luan logic doi vdi mot so BTHH d
THCS nhSm hinh thanh cho HS kha nSng sang t?io
ra BT mdi Tuy nhien, dehien thuc hda trong day hgc,
GV can luuycho HS mpt sd van desau: - Khdng ptiai
BTHH nao cd dang: "Chiang minh menh deA => B"
cungcdthekhaithacdi/pc theo cacdungysu pham
B(S vay, giao vien (GV) can iua chpn trudc cac BT cd
liem nang" va phu hpp vdi chuong trinh day hpc; •
iVIuc dich cudi Cling can dat duoc la hinh thanh cho
HS kha nang sang lao BT mdi thdng qua khai thac
cac suy luan logic tu BT cho trudc; do dd, hoal dong
nay can dupc lap di, lap lai nhieu Ian trong day hoc
GV can thuc hien theo mot ban ke hoach cd dung y su
pham da dupc xay dung tl mi Id trudc
Sau day la mpt sd vi du minh hpa viec hien thuc
hda cac dinb budng da trinh bay dtieu muc I vao day
hpc binh bpc Idp 7 GV can tan dung cp bpi cd the (tiet
luyen lap, gid ngoai khda, ), giao nhiem vu cbo HS
giai BT sau:
BT /.•Cbo AABC, cd2AB = BC Cac diem IVI, D
theo thd tu la trung diem cua cac doan IbSng BC va
IVIB Chdng minb rSng: 2AD = AC
BT 1 la dang loan ma HS chua co thuat giai That
vay, HS chl mdi dupc lam quen vdi vipc chung minb
bai doan thang b5ng nhau thong qua chdng minh
bai tam giac b l n g nhau Vi lido nay, HS mong mudn
tim ra mpt Ibu Ihuat mdi de giai quyet nhung BTHH
yeu cau chung minb tisddpdai bai doan tb5ng bSng
k, vdi k 7^ 1 BT da cho cung khong qua xa la ddi vdi
HS cac em can chuyen dupc ve dang quen thupc vi
da cp trai nghiem vdi tinb hudng <r= 1 Be thuc hien
dugcdungysupbam.GVyeu cau HSdpckTde toan,
viet gia thiet va ket luan mpt each rd rang, n g j n gpn
Myc dich cda boat dpng nay nh5m giup HS njm vuhg
cau true logic cda BT, ddng thdi, cac em ylbdc dupc
gia thiet nao da cbo, van de nao can tim bay nhiem vu
nao can giai quyet Dudi sudieu khien cua G V, HS cd
tbegiaiBTI nhusau:
Lay I la Irung diem cua AC, ndi I vdi M (hinh I), ta
BM Kel bop vdi (2), ta cd: IM=-BIH = DB
T0f(1)suyraZ/MC = Z D S A H o n n u a M C = B A
dpdd,A/WC=ADBAsuyra:>lD=:C/.Vay,^/lD=/1C
3 8 [ T a p chi Gido due so 315
Hinh I
dung cac ytudng
da trinh baydmuc
1 degiup HS phat hi§n ra BT mdf B Theo hudng thd nhat, sau khi HS
da giai quyet xong BT, G V yeu cau cac em xem xel liii mpt lan niJa, sau dd dat van de: trong qua trinh suy diin di tim ket qua, cd the phat hien ra dieu gi khac nua hay khdng? Tuy theojOmg dd'i tuong HS maGV
cd nhdng goi y cu the; chSng han: hay cho biet mdi quan bp gida dudng thing AM va doan thing Dl Lan lai each chung minh dtren, HS cd the suy diin dupc AMJa dudng trung true cua doan thing Dl GV hudng d i n HS thay ket luan cua BT ban dau de co dupc BT 2 sau:
BT2:Cho AABCco 2AB = BC Cac diem M, D, I
theo thdtu la trung diem cac doan thing BC, MB va
AC Chung minh r i n g : AM ia dudng thing tnJc cua doan thing Dl
Theo hudng hudng thu bai, dua theo ketqua cua
BT 1,GV dat van de: neu D'iampI diem tren c^nh AC sao cho4AD' = AC Ihiti sddd dai oua cac doan thing
D D ' v a B C b i n g bao nhieu? Giai quyet dupc van de nay tao dieu kipn cho HS phat bieu duoc BT 3:
S r 3 C h o A A B C c d 2 A B = BC.CacdiemDvaD' tbu tu n i m tren cac canh BC va AC sap cho:
BD AD' \ , 0 0 ' 3
— = — = - C h u n g m i n h r a n g : — = j Ddi vdi HS Idp 8, viec giai BT 3 la kha don gian, nhung VdiHS kjp 7cac em se gap nhieu khd khan khi giai bai toan nay
Theo budng thd ba, cho HS xem xet lai oSu tttic logic cua BT 1, sau dd GV dat van de: tam giac dac biet nao cd do dai mdt canh gap ddi dp dai ciia mdt c^nh khac? Cdthethay ddl gia thiet cua BT1 sao cho van suy dien ra ket iu?n cua BT do hay khdng? Til nhung tac dpng su pham nay, HS lien tudng Idi lam giac vudng cd mot gde b i n g 30° va day la yeu tdco the hoan ddl gia thiet de tliiet lap nen BT mdi GV luu yydi HS r i n g , suhoan ddl nay cdthe suy diln ra cac ket qua thu vi khac, ngoai ket luan 2AD = AC Gpi y nay cua GV la sugiao nhiem vu mpt each linh hoal cbo H S, tgo cd hoi cho cac em phat bilu dirpc BT 4
BT4:Cho&ABCc6 ^BAC = 90',ZACB= 30'
U D, I thu' t l / la tnjng diem cua cac do^n thing
BC, MB va AC Chiing minh r i n g : a) 2 A 0 = AC;
b) ADlBCc) MDI la tam giao d i u
Trang 3ttio^ Igt nguac van d l bSng cau hdi sau: Menh de:
"Neu tsABC cdcac diem M, D Ihdti/la trung diem eua
BCvaBM,ddng thdi 2,4D=/flCthi^>4SseC"c6 phai
la menh de dung hay khong? Oe HS lra Idi duac eau
hdi nay, G V ed the hudng d i n eho cac em thuc hien
cae hoat dpng sau: - Lay diem P tren tia AD sao clio
D la tnmg diem AP; gol I la tnjng diem eua AC, H la
giao didm cua AM va PC [hinh 2); - Thuc hien qua
trinh suy nguoc theo sa do:
2AB = BC*-MI = MD^ AMIA = AMDA *- ZPAH = Z.CAH *~
AH la trung luydn, ddng thdi la dudng eao eua
A / I P C ^ Mia trgng tam cua A^PC{c&n tgi A) (sddd
suy luan nguoc nay da duac gian \uqc)<^2AD=AC
Suy lugn ra M
la trong tam eua
A/lPCIadddang;
bdi vay, menh de
ma GV dua ra d ^ ^ / '^
tren la menh de l^^^^ '^ H
dung Den day, GV p Hinh 2
cho HS kdt hgp vdi
BT 1 de phat bieu BT md^ ed ket lugn manh hon
Trong quatrinh nay, GV can ehu y sda chda cac sai
lam trong hogt dong ngdn ngU cua HS de cac em
phat bleu duoe BT 5:
Sr5.-Cho ^ABC; M, D theo thd tu la trung diem
cua BC va BM Chdng minh ring: 2AB = BC khi va
ehikhi2AD = AC
mdilampttrong nhflng each day hpc giup HSblet suy nghT.bietsangtgo Phuang thdc nay Chung tdlmudn
GV trao ddi vdi HS khdng phai duac thuc hign mot sdm, mol ehieu ma la ca qua trinh dgy hpc, vdi mpt chudi cac hogt ddng cua ngudi hpc tuang thich vdl y
dd su phgm duac l§p di lap Igi nhieu lan Qua dd, HS hinh thanh kha nang khal thac cae BT ed "tiem nang"
de phat hien ra BT mdi; giup HS khdng ehi sang tgo
ra BT mdi macdn gdp phan vao viec hinh thanh dcac
em kha nang tu hpe, tu nghien cdu, nang dpng sang tgo tn:ng hoal ddng thue tien sau nay •
Tai lieu tham khao
1 TrSn Lu^n V^n di^ng tu tu&ng su phgm ciia G
Polya xdy dung ndi dung vd phucmg phdp d^y hpc trin CO sd cdc he thdng bdi tdp theo chii dS nhdm phdt huy ndng luc sdng t(io ciia hoc sinh chuySn todn cdp II LuandnPh6 titfn si Khoa hoc sir pham-Tam !i,
1996,
2 Nguyfin Canh To&n - Nguyln Van Le - Chau An
Khoi d^y ti^m n^ng s a n g t a o NXB Gido due,
H 2005
SUMMARY
In this paper, we Interpret tbe way to train stu-dents to invent new wathematica! problems via ex-ploiting logical inferences from typical problems in secondary school curriculum in order to develop leam ers' area tivity
Hirdng dan sinh vien tham gla
(Tiep theo trang 20)
Xh^c hign eae BT (chuan bj kithuat lien hanh ldng BT)
->_Kiem tra va tu van (phat hien ldi va giai dap nhdng
th^c m^c) ->Danh gia va sddung ket qua (danh gia
va tgo gla tri cho ket qua)
Hudng din S V Ihuc hanh nghien edu TLHS trong
day va hpe cac hpc phan Ihudc mdn TLH la mot viec
lam ddi hdi su chuyen tam cua ca thay va trd De
hogt dpng nay thuc hien ed hieu qua can luu y:
-Tang cudng thdi luang thue hanh trong dgy va hpc
cac hpe phan TLH - dae biet la THNC; - Lua chpn,
xay dung, xae dinh ro rang cac bai tgp THNC va sd
dung chting nhula nhflng phuong tign, eong cu cho
viec dgy va hpc trong cac co sd dao tgo giao vien;
- Hudng d i n S V THNC theo mdt quy trinh chat ehe
trong dgy va hpc •
(hi 1 8/2013)
-(1), (3) Hoang Phe (chu bifin), Tur dien Tieng Viet NXB £»d M n g , 2001
(2) Lt Tu- Thanh, Logic hoc v^ phuong phdp lu§n nghifin cihi Idioa hoc, NXB Tre, TP H6 Chi Minh, 1996
Tai Il$u tham khao
1, Michel Deveiay Mfit sd' vS'n de vS d a o t^o giao
vifin, NXB Gido due, H 1998
2 Weinert RE, (chu bien), S y p h a t trien nhSn thiic,
h9C t§p va giang day NXB Gido due, H, 1998
SUMMARY
Instruching the students how to practtve research-ing when teachresearch-ing and learnresearch-ing is best way to de-velop the positive and active attitude of learners the most as well as link teaching to the life reality and society's needs This article aims to introduce and make clear the process of Instruching the students how to practive researching In the teaching & learn-ing the module of Psychology
Tqp chi Gido due so 315 39