Day hpc yeu td hmh hpc d tieu hpc theo hUdng phat trien nang \\ic tir duy va lap iuan Toan hpc cho hpc sinh
Le Thj e l m Nhung TnJflng Cao ding Si/ phgm ThSi Nguyen Phi/dng Thinh OSn, Biinh phfl Th* Nguygn, tlnti That Nguygn, Vi$t Nam Email' nhungltc@tce,edu vn
TOM TAT: Day hgc theo hddng phit trien nang life ciia hoc sinh la yeu cau cOa ChUdng trinh Giao due phS thdng mdi. Ning luc Toin hgc li mdt loai hinh ning luc chuyen mdn, gan lien vdi mdn hgc. Mdt thinh td cua ning luc Toan hgc la ning luc tu duy va lap luan Toan hge. Day Toan li phii day tu duy. Viec day hgc theo hUdng ren luyen nang luc tu duy va lgp luin Toin hgc cho hgc sinh se gdp phin phit triin tu duy cho hgc sinh. Bii viit trinh bay quan niem, bign phap dgy hgc yiu to hinh hgc d tieu hgc theo hUdng phat triin nang lUc tu duy va lip luan Toan hgc cho hgc sinh.
J\i KHOA: TU duy; nang lUc Toan hgc; nang life tif duy va lap l u ^ n Toan hoc; day hoc yeu 15 hinh hoc.
-> Nhan bai 04/4/2020 -> Nhan bai aa chinh sila 09/5/2020 -» Duyet tfang 15/6/2020,
1. Dat van de
Ngay 26 thatig 12 nam 2018, Bp Giao due va Dao tao da ban hanh Chuang trinh Giao due ph6 th6ng {CTGDPT) theo dinh huong phat tri6n phim chdt va nang liic (NL) ciia nguoi hoc. CTGDPT mon Toan 2018 xac dinh eae th^h t6 c6t 16i eua NL Toan hpc la: NL tu duy va ISp luan Toan hpc (TD&LLTH), NL mo hinh hoa Toan hpc, NL giai quyet van de Toan hpe, NL giao tiep Toan hpe, NL su dung c6ng cu, phuong tien hpe Toan. Hinh thanh, phat triin NL TD&LLTH se giup hpc sinh (HS) khong chi hpc Toan tot ma con giup HS phat trien NL tu duy (TD), giai quyet tot cae van de eua eac mon hpc khae va trong cugc s6ng. Ci cAp Ti8u hpc, m6n Toan khong phan chia thanh eae phan mon. Cac y6u t6 hinh hoe (YTHH) gan ket chat ehe cung vai cac kien thuc ve Do luong de cau thinh mach n6i dimg Hinh hpc - Do luong, d6ng thai ciing vdi cac mach kiSn thuc khac tao nen mot mon Toan thong nhat. Ngoai ra, vi6c day hpc YTHH a tieu hpe con g6p phan phat trien tri tucmg tugng khong gian va tinh trgc giic cho HS. C( bai viet nay, chiing toi trinh bay mgt so quan ni?m vk NL TD&LLTH 6 ti6u hpc, bien phap dgiy hpc YTHH theo hucmg phat triSn NL TD&LLTH cua HS gop phin th\rc hi?n t6t CTGDPT moi.
2. N$l dung nghidn cuTu 2.1. Mot so khai niem 2.1,1. TU duy va nang luc Toan hpc
Trong th6 gicri hi?n thyc co rat nhi6u cai con nguoi chua nh^ thuc dugc. Nhigm vy cua euoc s6ng luon doi hoi con nguoi phai hieu thau nhung eai chua biet do, phai vach ra dugc cai ban chat va nhung quy M t tac dgng cua chung.
Qua trinh nhan thijre do ggi la TD. Theo tam h hpc, TD la thuOc tinh d§c bi?t eua vilt chat co t6 chue cao - bp nao ngudi. TD phan anh thi eidi vat chit dudi dang cae hinh
anh li tudng: "TD phan anh nhung thugc tinh ben trong, ban ehat, nhung moi lien he co tinh quy luat cua su vat, hien tugng ma trudc do ta chua biet" [1]. Qua trinh phan anh nay la qua trinh gian tiep, dgc lap va mang tinh khai quat, duae nay sinh tren ca sd hoat dgng thuc tien, tii su nhan thuc cam tinh nhung vupl xa cac gidi han ciia nhan thuc cam tinh. Theo Art Costa, tac gia eua nhieu cudn sach v6 TD: "TD la su cam nhan ciia chung ta khi chung ta nhan duac nhihig du kien, nhiing thong tin dien ra trong cac moi quan he". Ndi mpt each ngan gpn la: "Chiing ta suy nghT' [2]. Theo tam li hpc, TD diln ra thdng qua eac thao tac hi tue: phan tich - tdng hgp, so sanh - tuong tu, truu tupng hoa - khai quat hoa [ 1 ]:
A.M Phndman cho ring: "TD Toan hgc la tu duy li thuylt triiu tugng cao nhat, cac ddi tugng ciia nd ed the duge mo hinh hda, vut bd tdt ea cac tinh vat chdt va chi gill lai nhirng quan he da eho giua chiing". I.A.Khin chin neu ra 4 tinh chdt dac trung cua TD Toan hgc la: Suy luan theo sa dd logic chilm uu thi; Tinh nit ggn cua qua trinh suy luan; Tinh phan chia ro rang cua qua trinh suy luan;
Tinh hit sue chinh xac cua cac ki hieu duae su dung ttong qua trinh suy luan. TD Toan hgc co tinh khai quat va tinh logic.TInh khai quat thi hien b vile luon ludn tim hieu dac tmh ciia sir vat, hien tupng mpt each ban chat, cdt loi tu dd tim ra nhung quy luat hit sue tdng quat;
Tinh logic ciia tu duy Toan hpc thi hien d cho dua tren nhirag quy luat logic.
2.1.2. Nang lUc tU duy va lap luan Toan hoc d tieu hpc Trong qua trinh hgc Toan, cac hoat dgng TD & LLTH giu vai trd then chdt, vua la phuang tien nhan thiic vira gSn liin vdi muc tilu phat Oiln NL tu duy cho HS.
Tac gia Tran Nggc Bich da tiep can khia canh ngon ngii Toan hgc cua TD & LLTH khi nghien ciru giai phap giup
Sfi 32 thing 8/2020 "
NGHlENCUruULUAN
HS tieu hgc sir dvmg hilu qua ngdn ngu Toan hpc [4].
Cung theo huong nay, de xay di;ng bien phap ren luyen kl n5ng su dung ngdn ngQ Toan hgc trong day hgc mdn Toan cho HS ldp 4, ldp 5, tac gia Thai Huy Vmh da lam ro vai trd ngdn ngu Toan hgc vai tu duy Toan hpc [5].
CTGDPT mdn loan 2018 [6] da cu thi hda nhung biiu hien va kit qua hoat ddng TD & LLTH ddi vdi HS nhu sau (xem Bang 1). Nhu vay, cd the noi, cac chi bao quan trpng dk xac dinh NL TD&LLTH cua HS Tiiu hpc thi hien nhu sau:
-Th\rc hien dupe cac thao tac tu duy, biit quan sat, tim kilm su tuong dong va khae bilt, va biit khSng dinh kit qua ciia viec quan sat
- Neu dugc chiing cii, li le va biit lap luan hop li khi giai quyet van de.
Dudi day, chung tdi de xuat mgt sd bi|n phap day hpc YTHH nhdm gdp phdn phat hiin NL TD&LLTH cho HS Tieu hpc.
2.2. Bien ph^p day hpc yeu to hinh hpc theo hUdng phat trien ndng luc tu duy va ldp luan To^n hpc cho hpc sinh 2.2.1. Thi^t kg tinh hudng gpi van de de hpc sinh dUpc tiep xuc trUc quan, tham gia trdi nghiem va tich cue suy nghi Vi HS chi tu duy khi "cd vdn di" nln GV cSn xay dung cac tinh hudng cd van dl dl HS dupe "trai nghiem", duge huy dpng cac kien thiic va kinh nghilm thuc tien de suy nghi, biin d6i ddi tugng hoat ddng, tim ra hudng giai quyit vdn di, thdi thiic HS kham pha tun hiiu kiin thiic mdi.
Vi diji 1: Khi day hinh thanh cdng thiic tinh "Dien tich hinh chu nhat", GV giao phieu hge tap va td chirc cho HS iam viec ca nhan hoac theo nhdm ddi: So sanh di?n tieh cua ba hinh sau, biet mdi d vudng cd diln tich 1 cm^
(xem Hinh t):
(Dim theo hang ho$c cot), tir do HS cd thi nit ra nh|n xet: Hinh A cd chiiu rpng la 2 cm, chiiu dai la 4 cm, dm tich la 8 cm^. Hinh C cd chiiu rpng la 2 cm, chieu dii U 3 cm, di?n tich la 6 cm^. _ ,
Sau khi HS da so sanh dien tich cua ba hinh, cd the h6i HS: Hinh chu nhat cd chiiu rgng la 2 cm, chieu dai ia 7 cm thi diln tich cua nd la bao nhieu ? HS cd the tim ra dap an bdng each ghep hai hinh chO nhat A va C de ra kit qua. HS cd thi ve hinh, dim d vudng tim ra ket qui. C6 HS sg dv doan ngay kit qua la 2x7 = 14 cm^
Vf dy 2: Sau khi HS hgc each tinh do dai dudng gdp khilc, GV cd thi cho HS giai bai Toan: "Trong Hinh 2, dudng di cua Cao mau xanh, dudng di cua Nhim mau d6, dudng di cua Thd mau vang. Hdi dudng di ciia ai ngan nhdt?"
Hinh 2: Minh hga do dai duang gdp khtic Vi dy 3: Cd the td chiic cho HS trai nghiem vdi BT
"Cung xay tam cap" [7]: Mpt b§c tam cdp di len dupe lam b^ng nhung khdi da l^p phucmg xep nhu hinh vS (xem Hinh 3). Neu ta can lam m$t b^c thang nhu thi vak chilu cao bdng 9 idn canh ciia khdi l^p phucmg, ta can phai cd bao nhieu khdi l§p phucmg?
tffl ii
Hinh 1: Ba hinh so sdnh dien tieh
HS se thuc hien theo cac cdch khac nhau. So sanh bang cdch dim hmh, tach hinh roi so sanh theo phdn da tach.
GV hudng dan HS xac dinh chilu dai, chieu rdng cua hinh chu nhat va nlu each dem so d vudng trong hinh Bdng 1: Bi^u hien vd y^u cSu cOa NL TD & LLTH cAa HS tieu hpc
Hinh 3: Minh hpa bai "eiing xay tam cdp "
HS se thdng qua trai nghi?m hoac tich c\rc suy n ^ di tim Icri giai: Vi chieu cao ciia bac thang bdng 9 \h canh eua khdi lap phuong nen bac tltang s6 cd 9 b|c.
Khi dd, hang dudi ciing ciia bac thang sfi cd 9 khoi % phuang. Hang thii hai sg cd 8 khdi lap phuong. Cii
- Tlii/c hi^n diiOc cdc hdnh dOng:
So Seinh, phSn tich, tSng hdp, d^c bi^t h6a, kliai q u ^ hPa. Ufdng tu.
quy n^p, di^n djch,
• Chi ra duoc chCIng cii. li le Vii bigt l^p iu^n hgp II tnfdc khi k i t lu$n.
- GlctJ thich hogc d i l u chinh cSch ttiUc gitij quygt v i n de ve phuong dign Toiln hpc
-ThUc hign flUijc cdc thao t i c t u duy {d mOc diJ ddn giin) cl3c b l f l b * quan s i t , tim kifim sU tuang d6ng vJ k h i c biet trong nhOng tinh hu&ig quen thuOc vd bift khang djnh k f t qud ctia vi$c quan s i t - Nfiu UUOc chUng cit, li le va b<ft ldp ludn bop li tnidc khi ket ludn - Biet ddt va tid ldi cau hdi khi ldp luan, gidi quyft v i n ag Buflc ddu WA chi ra chifng cfl v i ldp l u i n c6 cd sd. oil II Ifi tniUc khi ket lu^n
TAP CHi KHOA HOC GIAO DUC VIET NAM
giam ddn nhu v^y thi din h^ng tren ciing sg cd 1 khdi ISp phuong. Vay sd khdi lap phucmg cdn phai dung de xip bac thang la: 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 - 4 5 (khdi). Khai quat hda cho bai Toan xay n bac tam cdp kiiu nhu trin hit sd khdi l§p phuang la n + n - 1 +....+
3 + 2 + 1 (khdi).
2.2.2. Tap luyen cho hpc sinh sif dung linh hoat cac thao tac tu duy tren cd sd tiep xuc, quan sat trUc quan
a. Ren luyen thao tde so sdnh - tirang ttf: Giiip HS tim thdy sy gidng va khac nhau trong cac hinh, phat hien ra nhihig ddu hi?u ban chdt ciia su vat. Ren luyen kha nang quan sat.
Vi du 3: 6 ldp 1, di HS ren luy?n kha nang nhan dang hinh vudng, hinh trdn, hinh tam giac, hinh chu nhat. GV CO thi cho HS l^m bai tap sau: "So sanh cac hinh" (xem Hinh 4).
• • • • •
Hinh 4: Minh hga bdi tgp "So sdnh cdc hinh "
Vi du 4: So sanh va tuong tu la hai thao tde tu duy gan liin nhau. Khi cho HS lam bai tap: "Hinh tiep theo se nhu thi nao?" (xem Hinh 5).
•Eai D
Hinh 5: Minh hga bai tap "Hinh tiep theo se the nao? "
HS phai so sanh cac hinh va nit ra nhan xet bdng thao tde tuong t\r dk tao hinh. Hoac: HS phai thao tac tu duy
"tuang ty" khi lam bai tap: "Ve theo mau rdi td mau"
(xem Hmh 6).
in<E>n<i>n<r>
Hinh 6: Minh hga bdi tgp " Ve theo mau rdi td mau "
b. Ren luypn thao tde phdn tich - tdng hgp Vi dy 5: Ci ldp 1, d i HS ren luy?n kha nang nhan dang hinh vudng, hinh trdn, hinh tam giac, hinh chii nhat va mdu sde cac hinh. GV cd thi cho HS lam bdi t^p sau:
"Nhd d d cudi eimg la nha nhu thi nao d i tdt ca nha d cac hang, cpt la khac nhau" (xem Hinh 7).
Hinh 7: Minh hoa vi du 5
HS muon tra ldi duac eau hdi cua bai Toan bat budc phai so sanh cac hang, cac cdt dl tim ra sir gidng va khac nhau cua cac hang: Mdi hang, mdi cot phai cd ba mau, phai cd ba cai nha, mai phai la cac hinh khac nhau, trang tri tren mai khae nhau. Dl giai dupe bai Toan, HS khdng chi dimg d vile so sanh, HS phai kit hpp ca phan tich cac yeu td cua hang, cdt, ngdi nha mdi cd thi so sanh, phai tdng hgp cae yiu td khae nhau mdi tim ra kit qua:
Ngdi nha d d can tim mau xanh, mai tdidng phai la hinh tam giac, hinh chii nhat, trang tri trin mai la hinh trdn.
Khi tra loi HS phai biet lap luan di chi ra each minh tim dung ket qua.
Vi dy 6: C) ldp 3, d i HS bilt dpc ki hieu hinh trin hang, cot, biit kit hgp hinh theo quy luat, cho HS lam bai tap:
"Cac hinh eon thiiu d cac d trdng nhu thi nao?"
•
A
• •
•
•
• 0 1 \A<o<l
B^ F.
Hinh 8: Minh hga bdi tim cdc hinh edn thieu lap 3 Vi du 7: Khi HS da hgc each tinh dien tich hinh chu nhat, cho HS lam bai tap: Tinh diln tich hinh bin:
Hinh 9: Hinh tinh dien tieh a vi du 7 Dl giai dupe bai tap, HS phai phan tich eac hinh dl nhan dang roi tdng hop lai dl dugc kit qua.
e. Ren luyen thao tde triru tucmg hda
HS Tiiu hgc nhdt la cac ldp ddu cap thudng tu duy cu thi, cdn cd dd dimg true quan hd tia cho tu duy, d ldp 1, ldp 2 cho HS lam cac bai tap dan gian di budc ddu hinh thanh kha nang trim tugng hda.
Vi dy 8: Lam thi nao dl biet cac hinh trong d vudng cudi? (xem Hmh 10).
So 32 thing 8/2020 39
NGHIEN C C T U U L U A N
• • m_9_
A 1=1
•
[=Z)
•
•
each xoay quanh 1 diem (mau dd) d 4 vi tri, ve lai ta dugc hai hinh ve (xem Hinh 15).
vatdmdu
Hinh 10: Minh hga vi dti 8
Bk ren luy|n cho HS tu duy khdng gian, kha ndng trim tugng hda cac bieu tugng khdng gian va mdi quan hg giira HS su chuyen ddng ciia mpt vat quanh mot diim (xoay), chuyen ddng theo mpt dudng thdng (dorgt - tmh tiin), ddi xiing (true, tam) GV cd thi cho HS lam eac vi du sau:
Vi du 9: GV gidi thilu vdi HS tu chir F ban ddu (xem Hmh 11):
Xoay Ddi xiing true Trugt
Hinh 11: Minh hga vidii 9
Sau do, cho HS chi ra ten cac chuyin dpng ciia hinh d cac trudng hpp sau (xem Hinh 12):
' X ,
Hinh IS: Minh hga vi du 10 Tucmg to, em hay xay dung cac hinh vS m « to moi htah trong Hinh 16 duoi day:
Hinh 16
Vi dii 11: Sir d\mg chuyfin dOng cua ducmg tr6n, c6 the tao ra cac mlu Ichac nhau. HSy quan sat each tjio mlu va ve cac hinh tocmg ty vao va ciia em. TCr cac each vg nay, em tim cich tao cac hinh vS Ithic nhau (xem Htah 17).
ir
Hinh 12: Chuyen ddng cua cdc hinh Hoac cho HS lam: Vdi each thuc hien tuang ty nhu vdi chu- P, em hay vl hinh vdi chii R va chu L (xem Hmh 13):
Hinh 17: Minh hga viduU Vi dy 12: Cat cac hinh tii gidy. Thyc hi?n c4c chuyin dgng tuang tu vdi chiing nhu trong Hinh 18 dudi dSy. VS vao vd hinh nhin dugc tucmg iing vdi timg chuyin dgng.
^ ^m WIP
Hinh 13: Minh hpa hdi ve hinh cha R va ehu L Hoac: "Tii hinh I ) cae ban ve theo thii ty nao?"
D ( • ( I D
Vi dy 9: Hdi tuang tu vdi hinh sau (xem Hinh 14):
® ®
Hinh 14: Minh hga vidg 9
VI dy 10: Hinh cd thi dugc di chuyen bdng each xoay nd theo cdc huong khic nhau xung quanh mpt diem cd dinh. Bdng each vg len cac hinh thu dupe theo each nay.
Cd thi "t^o" dugc eac hinh ve - nhung tac pham khac nhau. Chdng han, d Hinh 15 dudi day, di chuyen nd bing
Hinh 18: Minh hga vi dv 12 d. Ren luyen thao lac dgc bipt hda - khdi qudt hda - Khi day hpc each tinh chu vi, di?n tich ciia cic hinh, tir cac vi dy ey the HS trai nghi|m d i tim ra cich tinh, sau dd GV cd thi din ddt de HS tim ra quy tic tinh, Uiii quat thanh each tinh hoae cdng thiic tinh. Ching h^n, 6 ldp 3, HS da biet tinh ehu vi, di?n tich hmh vudng, hUi ehiJ nhit. Din ldp 4, HS dupe tiip tuc biit cich tinh chu vi, di|n tich hinh binh hanh, hinh thoi. Han nOa, cic quy tde tinh chu vi, di|n tich cac hinh dugc neu dudi d ^ khii quit bing cic cong thiic tinh bing chii, chdng h?D:
* P = a X 4; S = a X a (vdi a la dp dai canh hinh vudoft P la chu vi hinh vudng, S li di?n tich hinh vudng). * P = (a + b) X 2; S = a x b (P la chu vi, S la di|n tich hinh chO nh^t; a la chieu dii. b ta chieu rgng hinh chQ nh^t). • S = a X h (S la di?n tich hinh binh hanh, a la dp dai day h li chiiu cao). * S = mxn (S la di?n tich hinh ihoi; m. n la d?
TAP CHI KHOA HQC GIAO DIJC VlfT NAM
dai cua hai dudng cheo).
- Cling cd thi qua eic bai tap HS dugc hinh thanh thao tac quy nap
Vi du 13: CJ ldp 4, eho HS lam bai tap "Ve vao vd dudng trdn ed tam d O va bin kinh OA. Ve dudng kinh AB cua dudng trdn va dinh ddu cac dilm C, D, E nhu trong hinh(xem Hinh 19). Cau ndi sau ed diing khdng:
"Cac gdc ACB, ADB, AEB la eac gdc vudng"? Lay mdt diim F tren dudng trdn va khdng nam trin dudng kinh AB, xac dinh dang ciia gdc AFB. Em cd kit luan gi?"
chudn bj nh&ng tdm van (xem Hinh 20). Mdi ban chudn bi mpt tam, Tdm van ciia Sde ngin han tdm van cua Thd va tam van ciia Nhim ngdn han tdm van eua Sde. Hdi ehilu dai cua tdm van do mdi ban Sde, Thd, Nhim chudn bi cd mau gi?
2/ O ldp 2, cho HS lam bai tap: Lua chpn dap an diing danh cho Hinh 21:
Nlu hinh ve mau vang thi dd la hinh tron.
Tat ca cac tam giac diu cd mau xanh.
Nlu hinh ve ed mau xanh thi dd la hinh tii giac.
Hoan thanh cau sau dl nd la cau diing: "Nlu hinh vg la tir giac thi.,..".
Hinh 19: Minh hga vi du 13
Vf dy 14: Cho 5 diim A, B, C, D, E. Hdi khi ndi chiing l^i ta dugc bao nhilu do^n thing? Neu cd 2019 dilm thi khi ndi chiing l?i dugc bao nhilu doan thang? Cd nhilu each giai bii Toan niy nhung cd thi cho HS md mam:
- Niu cd 2 diem thi sd d o ^ thing ndi 2 diim la I = 0 +1 (doan thing).
- Nlu cd 3 diem thi sd dogn thing ndi 3 dilm la 3 = 0 + 1+2 (dogn thing).
V^y, nit ra quy lu§t d diy la: Niu cd n diim thi khi ndi chung lai ta dugc sd doan thing:
0 + 1 + 2 +...+ (n - 1) = n X (n - 1) : 2 (doan thing) Ap dyng quy luat trin niu cd 5 diem thi ndi ehiing lai ta sS dugc &h dogn thing Ii:
5 X (5 - I) : 2 = 10 (doan thing) Tirong ty, HS cd thi tim ra sd doan thing khi ndi 2019 diem.
Hoic giai theo cich: Ndi diim A vdi mdi diem cdn l?i ta se dugc 4 doan thing. Nhu vay, khi ndi 5 dilm dd vdi nhau ta sl dugc 4 x 5 = 20 (doan thing). Nhung mdi dogn thing dugc tinh 2 idn. Vi vay, sd doan thing dim dugc khi ndi 5 diim d i cho vdi nhau la: 20 : 2 = 10 (doan thing). Neu cd n doan diem thi ndi vdi n - 1 diem cdn lai, tinh tuong ty sd dogn thing ndi dugc la:
n X (n - 1 ) ; 2 (dogn thing)
Tii dd, cd the tinh dugc sd dogn thing khi ndi 2019 diem hay bdt ky' sd diim nio khic.
e. Ren luypn suy lugn logic, lap lugn Todn hge Ban ddu, cd thi cho HS lim cic bai tgp don gian, cic budc suy lugn ngin.
Vf dy 15: 1/ 0 ldp 1, cd thi cho HS lam bai tgp: Ba HS ciia ngdi dirdng trong rOmg la Sde, Thd va Nhim d3
Z^ Z^
O
Hinh 20: Tam van ciia Hinh 21: Minh hga vi di4 Soc, Thd, Nhim 15, lua ehgn ddp dn dung 3/ CJ ldp 2, cho HS lam bai Toan: "HS ciia mdt trudng trong khu rimg gom Sde, Thd, Nhim, Cio da ve 4 hinh, mdi ban vS mdt hinh. Nhim khdng ve da giic, Thd khdng vg tam giic, Cao da ve mdt hinh chir nhat. Hdi S6c vg hinh gi?" (xem Hmh 22).
Hinh 22: Hinh ve cua Soc, Thd, Nhim, Cdo Vi dy 16: d ldp 3, Cho HS lam bai tap: Chgn tit ea cac ciu diing cho inh nay (xem Hinh 23):
1/ Tdt c i cic cd dd cd dang hinh vudng.
2/ Neu cd cd hinh tam giac, thi nd cd miu xanh.
3/ Niu cd cd ba miu thi nd cd dang hinh chii nhat.
Hoan thinh cac cau sau d l trd thanh cau diing.
Neu cd cd mau vang thi nd Ii ...
Tat ea cic cd eo miu dd deu la...
M l
Hinh 23: Minh hga vi du 16
Vi dy 17: CJ ldp 3, dk HS biit phan logi hinh, bilt nhgn xet khai quit theo nhdm hinh, cho HS lim bai tap: "Cac hinh gi dugc trinh bay trong hinh v6 bin (xem Hinh 24) Cd thi chia cac hmh ra hai nhdm nhu thi n i o ' Hiy dua ra hai cich chia nhdm."
S<f 32 0)^8/2020 41
N G H I E N curu Li LUAN
a
Hinh 24: Minh hoa vidu 17 Vi dy 18: Sau khi HS hgc dien tich hinh vudng, hinh tam giac, d i HS ren luyen NL lap luan Toan hgc, biit chi ra chiing cii, li Ig va biit lap luan hgp li trudc khi kit luan, cho HS lam BT: "Chiing td rang, d mdi hinh trong Hinh 25, tdng dien tich phan mau vang bdng tdng dien tich phan miu xanh".Hinh 25: Minh hga vi du 18
Vi dy 19: CJ ldp 4, Idp 5, cd the cho HS bai tap "Diing, Minh va Anh ve cac hinh (xem Hinh 26). Biet Anh khdng ve hinh mau xanh. Dung khdng vg hinh mau xanh va mau vang, Hdi mdi ngudi da ve hinh mau gi?" Em hay ve cac hinh gidng nhu vay?
Hinh 26: Minh hga vi du 19 Thdng qua giai quyit bai tap, HS vira dugc ren luy^
NL TD&LLTH vira dugc ren luyen NL ve hmh, nhgn dang hinh,
3. Ket luan
NL TD&LLTH cd vai trd quan trgng trong ddi sdng va hgc tap cua HS. ^k ren luyen, phat triin NL TD&LLTH cho HS n-ong day hpc YTHH d tiiu hpc, can tgo ra cac tmh hudng cd vdn dk va tip luyen cho HS sir dyng linh hoat cac thao tac hr duy tten ca sd tiip xiic, quan sat Uyc quan tim cac bai tap, hoat dpng d i HS giai quyit thong qua ten luyen eac thao tic tu duy nhu so sanh, hrang ty, phan tich, tdng hgp, trim tugng hda, khii quit hda, suy luan logic. Bai viit da trinh biy mpt sd bai tgp cd dung y su pham d i GV sii dung giiip HS ren luy|n, phat hien TD&LLTH. Cac bien phip dgy hpc theo hudng phii triin TD&LLTH gdp phan hinh thinh, phat triin li luan ve day hpc YTHH theo hudng phit trien NL cho HS.
TM li^u tham khdo
[1] Pham Minh Hac, (1998), Gido trinh Tdm li hpc, NXB Giao due, Ha Noi,
[2] Anne J Udall and Joan E. Darnels, (1991), Creating Active Thinkers. Nine Strategies For a Thoughtful Classroom. Zephyr Press, the USA.
[3] Nguyin V5n Loc, (1995), Hinh thdnh kinang lgp lugn co cdn cu cho hoc sinh ede Idp ddu cdp pho thong co sd Viet Nam thdng qua dgy hoc hinh hpc, Luan an Tien sT Giao due hpc, Tmong Dai hpc Vinh.
[4] Tran Ngpc Bich, (2013), Mgt so biin phdp giiip hoc smh cdc lop ddu cdp Tieu hgc su dung hipu qud ngon ngir [7]
Todn hgc, Luan an Tien si Khoa hpc Giio dye, Vi?n khca hpc Giao dye Vi?t Nam.
Thai Huy Vinh, (2014), Ren luyen kT ndng sir dung ngon ngir Todn hpc trong dgy hgc mon Todn lop 4, ldp 5 a truang Tieu hoc, Luan an Tien sT Khoa hpc GiSo d^c.
Truong D^i hpe Vinh.
0 6 Due Thai (Chu bien) - D6 Tiin D?t - Nguyen Hoii Anh - Trin Ngpe Bieh - Do Diic Binh - Hoing Mai U - Tran Thuy Nga, (2018), Dgy hpc phdt triin ndng life mon Todn a Tieu hpc, NXB Dai hpc Su phgm. Ha NOi- T?ip chi Toan tu6i tha 1, s6 127 -128, thang 5-6/2011.
TEACHING GEOMETRY ELEMENTS IN ELEMENTARY SCHOOLS BASED ON DEVELOPING MATHEMATICAL THINKING AND REASONING COMPETENCE FOR STUDENTS
Le Thi Cam Nhung Thai Nguyen College of Education Thinh Dan ward. Thai Nguyen city, Thai Nguyen province, Vielnam Email: [email protected]
ABSTRACT: Teaching based on competence approach is a requirement of ttie new general education curriculum. Math competence is a type of profession^
competence, closely linked to the subjects. An element of mathematk:^
competence is the capacity of mathematical thinking and reasoning. Teaching math is to leach thinking. Teaching based on training students' mathemaSc^
thinking and reasoning skills will contribute to developing the students' thinking. This paper presents the concepts and the methods of teaching geometric elements in elementary schools based on developing mathematical thinking and reasoning competence for students.
KEYWORDS- Thinking; mathematical competence; thinking and mathematical reasoning teaching geometric elements.
42 TAP CHI KHOA HQC GIAO DUC VlfT NAM
