NGHIEN cUu & UNG DUNG
B6I DUOPG nAno LUC PHAT HIEO VA GIAI OUVET
VAn OE TRono HOC ToAn voi sir HO TRQ CUA cnii
Le Van 'Hiyen - Sd Gido dtic vd Ddo tgo Tuyen Quang TS. Tran Vift Cwdng - Truang DHSP - DH Thdi Nguyen nhftng dieu kien ndy sinh van de va |MFi - MF2I = 2a.
1. Nang l\|rc phdt hifn vk gidi quyit van de
Nang luc phdt hifn va giai quy8t vdn de (trong hpc tgp) la mdt hf thong cac thupc tinh cua cd nhdn con ngudi the hifn d cdc kha nang (tu duy va hanh dpng) ttong hoat dgng hgc tgp nhdm phdt hifn va giai quyit c6 hifu qua cac van de, nhifm vu ttong hogt dpng do.
Bieu hien cua nang luc phat hifn va giai quyit ttong hpc tap todn dugc the hifn d cac mat sau:
Biet huy dong dugc kien thitc Toan hgc lien quan tdi hoat dgng gidi quyet mgt ngi dung Toan hgc cu the; Co kl nang tien hanh dugc cac hoat dpng: giai bai todn, xay dung va nam vung khai niem Toan hgc va chiing minh djnh li...; Dat dugc ket qua phii hop vdi muc dich yeu cau; Biet van dung sang tao va co ket qua ttong cac tinh huong cua bai toan khac; The hifn dugc thai dg, tinh cam cua minh vdi nhiing ldi giai bai todn: Phat hifn sai lam va sira sai, thay dugc cai hay, sau sac ttong moi each giai...
Nang luc phat hien va giai quyet van de co the dugc chia theo cac miic do sau:
- O mice do thd nhdt: HS dap ling dugc nhung yeu cau ca ban phat hifn va giai quyet van de khi van dfi da dugc GV dat ra mgt each tuong dii rd rang.
- O muc do thu hai: HS nhan ra dugc van de do GV dua ra;
biit hodn tdt vifc phat hifn va giai quyet vdn di dudi su ggfi y, ddn ddt cua GV.
- Cfmuc dp diu ba: HS chu dgng phdt hifn dugc van de, dir doan
nhan x6t cdch thuc tiep can de phat hifn vd giai quyit van de.
Bai bao nay, chiing toi gidi thifu khd nang boi duong nang lire phdt hifn va gidi quyet van de cho HS vdi su hd ttg cua cong nghf thong tin thong qua day hgc ngi dung cac dudng Conic trong chuang trinh Hinh hgc ldp 10 Trung hgc pho thong. De mo ta cho dieu do, chiing toi minh hoa bang mgt so vi du sau:
2. M0t so vi dy ve viec su dung cong nghf thdng tin gop phan boi dir&ng nang luc phat hien va giai quyet van de cho HS Vl dy 1. Trong gio day bai phuang trinh dudng Hypebol, GV CO the khai thac sir dung phan mem Cabri Geometry ttong day hgc nhu sau:
• Hinh thdnh hinh dnh Hyberbol
- Lay hai diem F,, F^ co dinh va mgt doan thang AB khdng doi CO do dai bang 2a nho hon F^F^.
Lay diem C bat ky thuoc doan thdng AB.
- Ldn lugt dung dudng tron tam Fj CO ban kinh bing dg dai cua doan thang AC va dudng tron tam F, c6 ban kinh bang dg dai cua doan thang CB.
- Lay mgt diem A bat ky tren dudng trdn (F,; CB).
- Dyng dudng tron (A; AC).
- Dung dudng thang AF,.
- Gpi giao diem cua dudng thang AFi va dudng tron (A; AC) laF'2.
- Dung duong trung tryc F2F'2.
- Gpi M la giao diem ciia FiF'2 va F2F'2 Ro rang ta thdy
Cho diem A thay doi tten dudng tron (F,; CB) ta nhan dugc hinh anh tryc quan vi tgp hgp cac diem M, day chinh la hlnh dang ciia Hyberbol (Hinh 1).
Hinhi
Mat khac, neu sii dung each minh hoa nhu sach giao khoa da trinh bay thi chi mdi diing d miic do minh hoa hmh anh Hyberbol, con tai sao dp dai cua sgi day lai phai nho hon F1F2 thi HS chua biet. Khi su dung phan mem Cabri Geometry thi van de nay rat dan gian, ta su dung chugt cho diem C thay doi tren AB HS quan sat hinh anh quy tich diem M di tu minh giai quyet van di nay.
• Phucmg trinh chinh tdc cua Hyberbol
Ve phucmg trinh chinh tac ciia Hyberbol, sach giao khoa sau khi dua ra viec chgn true toa do Oxy sao cho F, = (-c; 0) va Fj = (c; 0) da dua ra ket luan M(x; y) e (H)
<^|5-^=1(]) trong do b^ = a^ + cl HS hoan toan co the dat cau hoi, tap hgp cac diem M(x; y) co toa dp thoa man (1) lieu cd hinh dang diing nhu Hyberbol ta vira biet hay khong?
Ta cd the giai quyet van de tren bang each sir dung chirc
NoAv nhAn hAi 7(1/01 non. Nc^ duvet dane 25/01/2013
TAP CHI THIEF BI GIAO DUC-5691-03/2013 • 9
Ill
NGHIEN CUU & UNG DUNG ndng "Ham bieu dien dudng vd tog dp diim" chi vdo dudng quy tich diim M khi dd ta se cd ket qud nhu ttong hlnh (Hlnh 2)Hinh 2 Vf dy 2. X6t bdi todn: Trong mgt phdng tog dq Oxy cho diem M(x, y) di dgng co log dg luon thod man ^Z%smV• 'rongdo t Id tham so. Hdy cho biet diem M di dgng tren dudng ndo?
GV cd the to chiic cho HS tham gia cac hoat dgng vdi phan mem Cabri Geometry nhu sau.
- Hoat dgng 1. GV sir dyng phan mem Cabri Geometry, cho hifn hf tryc toa dg va dung diim M(5cost; 4sint) len hf true toa dg (Hinh 3).
Hinh 3
Resiit -2 20 Result 3 59
+ Chgn G] Point on Object Lay diim X (t; 0) bdt ky tten true Ox.
+ Chpn ES Eguation and Coordinates: Chi vao diem X de hien toa do cua diem X ra man hinh.
+ Chgn K Calculate: Nhap bieu thiic tinh gia tri 5*cost va 4*sint, ttong do t la hoanh dp diem X.
f Chgn [^ Measurement Transfer: Lan lugt bam chpn gia tri vira tinh dupc sau dd chi vao
tryc tunc Ox vd Oy. Ta xdc djnh dupc diem A thupc Ox vd diim B thugc Oy.
+ Chgn 13 Perpendicular Line:
Lan Iugt dyng cdc dudng vudng gde vdi tryc Ox tgi diem A, vudng gde vdi Oy tgi diem B.
+ Chgn Kl Intersection Points:
Xdc giao diem M ciia hai dudng thane vudng gde vira dyng. M se Id diem c6 tog dg (5cost; 4sint).
- Hogt dgng 2. Du dodn guy tich cUa diem M. GV sii dyng chiic nang cua phan mem di lgi dau vit cho diem M vd cho diem X thay ddi de HS quan sdt:
+ Chgn S Trace On/Ojf. Gan thugc tinh de Iai vet cho diem M.
+ Cho diim X thay doi khi d6 vet de lai ciia diem M se cho ta hinh dnh ciia diem M.
Khi do, HS se thdy dupc hinh anh ciia quy tich la Blip (Hinh 4).
dnh cua quy tich ciJa diem M Id Elip (Hmh 5) vd phuong trinh ciia Blip chinh Id phuong trinh (•).
Hinh 4 - Ho^t dpng 3. Chirng minh bdi todn. Dya vao hogt dgng 2 va sir dyng suy luan logic, HS di dang chiing minh dupc fj+7^ = ' (*)hay M di dgng tten elip (E) cd phuong trinh (*).
- Hogt dfng 4. Minh hoa kel gud bdi todn. GV sir dyng chiic nang cua phdn mem de xdc dinh quy tich ciia diem M ciing nhu phuong trinh cua dudng quy tich de HS quan sdt:
+ Chpn 13 Locus: Ve quy tich ciia diem M khi diim X diay dii.
+ Chpn S Eguation and Coordinates: Xdc dinh phucmg trinh ciia dudng quy tich.
Khi dd, HS se thdy dugc hinh
Hinh 5 Vi dy 3. X6t bdi toan: Cho dudng trdn C/F^; 2a) cd dinh va mgt diem F^ co dinh ndm trong (C). Dudng trong (C) bdt ki co tdm M luon di gua diem F^ vd tiep xiic vol du&ng tron (CJ. Hay chung to Mdi dgng tren mgt elip.
GV su dyng phdn mim Cabri Geometry giup do HS giai quyk bai toan nhu sau:
- Hoat dgng 1. 77m hieu bdi todn. Md file "VD3.fig" di HS quan sdt hinh ve, ve hinh va xac dinh yeu td cd dinh, yeu td di dpUE... (Hinh 6).
Hinh 6 - Hoat dgng 2: 77m hirang chung minh bdi todn.
GV: Cho diim N diay dii, yeu cau HS dy dodn mdi quan hf gi&B 3 diem F,, M va N, tii do tinh dO dai doan MFj.
HS: Vi (C) tiip xiic vcd (C,}
tai N nen ta cd F,, M vd N thdng hdng. Do do, ta c6 NF, - F,M + MN = 2a hay MF, = 2a - R (R la bdn kinh ciia (C)).
GV: Nhan xet dp dai hai do?ii thdng MN va MF^.
HS: Do N va F^ tiiupc (C) nfin 1 0 • TAP CHI THIET BI GIAO DUC - SO 91 - 03/2013
NGHIEN cUu & UNG DUNG ta cd MFj = MN = R.
GV: Nhan xet moi quan hf giua MF,, MFj va 2a, tir do nhdn xetvi diimM.
HS: MF, + MF^ = 2a. Do do M thugc Blip cd tieu diem la F, vaFj.
- Ho^t dfng 3: Chicng minh bdi todn. Dya vao ho^t dfng 2 vd sii dyng suy lu^n logic, HS di dang chiing minh dupc MF, + MF^ = 2a hay M di dgng tren elip (E) cd hai tieu diim Id F, vd F^ va tryc ldn la 2a.
- Hoat dpng 4: Minh hog ket gud bdi todn. Sau khi HS thyc hifn xong boat ddng 3, GV cho diem N chay tten dudng ttdn (€,) va dk l^i ddu vit cua diSm M. HS
Hinh?
se dugc quan sdt each tryc quan quy tich cua diem M (Hinh 7).
Quy tich M tten man hinh hoan todn triing vdi kit qua ma HS da chl ra dugc ttong chiing minh.
- Hoat dgng 5. Md rgng bdi todn. Ddi vdi dii tuong HS hgc ban nang cao, GV cd the khai thdc phan mem de md rgng bai toan nhu sau:
+ GV cho diem F^ gdn sat vdi diem F, va ydu cau HS quan sat hinh dnh cua quy tich. HS dl dang nhan ra rang: Khi F, cdng gan F^
thi Elip cang gdn vdi dudng ttdn va khi F, triing vdi F^ thi Eli^ la dudng ttdn tdm F,, ban kinh bang a hay M ndm tten dudng ttdn (F,, a) (Hmh 8).
+ GV cho diim F^ di chuyin ra ngodi dudng trdn (C,) vd yeu cau
Hinh 8 HS quan sat hlnhLdnh ciia quy tich.
HS dl dang nhan thdy: Khi F^ nam ngoai dudng trong (Cj) thi quy tich
Hinh 9 ciia M la mdt Hypebol hay M nam tren mdt dudng Hypebol (Hinh 9).
Tuong ty, vdi vi du tten, GV cd the diing phan mem Cabri Geometry de to chiic cho HS giai quyet dugc bai toan: Cho hai du&ng tron C/F,; R) va C/F^;
RJ. Du&ng tron (C) ndm trong du&ng tron (CJ vd F^ khdc F^.
Ggi M Id tdm cita du&ng tron (C) thay doi nhtmg luon tiep xdc ngodi v&i (CJ vd tiep xdc trong v&i (CJ. Chiing to diem M luon di dgng trin mgt elip.
3. Kitlu^n
Vifc bdi dudng cho HS nang luc phat hien va giai quyet van de ttong hgc todn se giup cho viec hgc tap mdn toan frd nen hifu qua, giiip HS se ndm vung cdc tri thiic, phat trien tu duy, hinh thanh cac ki nang, ki xao cdn thiit cho ban than. Nd gop phdn hinh thanh cdc phdm chat tri tuf ciia con ngudi ndi chirng, ciia HS phd thdng ndi rieng. Bdi dudng
nang tyc phdt hifn vd gidi quyet vdn d8 Id gdp phan phdt huy tinh tich cyc ciia HS ttong hgc tap, t^o cho HS cd nhung nhan td, nhiing dieu kifn frong pham chdt nhan cdch di tham gia vdo cupc sing vdi nhimg yeu cau cua con ngudi mdi ttong nen kinh ti md cura, hfi nh^p, ttong sy phdt triin da dang cua ddi song xa hpi.
Tdi lifu tham khdo [ 1 ] Tir Dlic Thao, Boi du&ng ndng luc phdt Men vd gidi quyit vdn de cho hgc sinh trung hgc pho thong trong dgy hoc Hinh hoc, Luan an Tien sy giao dye hgc.
(2012).
[2] Doan Quynh (ting chii bien). Van Nhu Cuong (chii bien), Pham Via Khue, Biii Van Nghi, Hinh hgc ndng cao 10, NXB Giao due. (2006).
[3] Nguyen Mgng Hy (chit bien), Nguyin Van Doanh, Trdn Dlic Huyen: Bdi tdp Hinh hgc 10. NXB Giao due (2006).
Summary If students have the capacity to detect and solve problems will help the process of mathematical convenience, thereby developing the intellectual capacity for themselves. This paper, we present some examples in the direction of fostering the capacity to detect and solve problems for students with the support of information technology.
Keyword. Students, capacity, information technology
TAP CHI THIET BI GIAO DUC-SO 91-03/2013 • 1 1