TRUQNG DAI H O C DONG THAP Tap Chi Khoa hoc so 27 (08 2017) MOT SO BIEN PHAP B 6 I D U ^ O N G NANG LlTC PHAN TICH TIM LC« GIAI " BAI TOAN CHO HOC SINH THONG QUA 0 A Y HQC GIAI BAI T A P CHU DE PHirON[.]
Trang 1M O T SO BIEN PHAP B 6 I D U ^ O N G N A N G LlTC P H A N TICH TIM LC« GIAI
" BAI T O A N C H O H O C SINH T H O N G QUA 0 A Y H Q C GIAI BAI T A P CHU DE PHirONG PHAP TOA DO T R O N G M A T P H A N G - H I N H HOC 10
• Nguyen Thj Xuan Mai'*', Nguyen Diidng Hoang^'J
Tdm tat
Bdi bdo trinh bdy tdng quan ve nang luc todn hgc, ndng luc gidi todn, ndng luc phdn tich tim lai giai bdi todn, tit dd di xudt mot sd bien phdp nham bdi duang ndng luc phdn tich tim ldi gidi bai todn cho hoc sinh thdng qua dgy hgc chu de "Phuang phdp tog do trong mat phang-Hinh hgc 10 " Tit khoa: Ndng luc phdn tich, bien phdp, hgc sinh, bdi tap, tga do trong mat phang
1 Dat van de
0 trudng phd thong, day toan la dgy hogt
ddng toan hpe cho hpe sinh (HS) [5, tt 206],
ttong dd gidi toan la hinh thiic chu yeu Do vay,
dgy hpe giai bai t ^ toan cd tam quan ttpng dac
biet va tir lau da la mpt van de ttpng tam cua
phuong phdp day hpc todn d trudng pho thdng
Theo Polia [2], ed 4 budc de di den Idi giai
bdi toan:
1) Tim hieu ndi dung bdi toan; 2) X ^ dpng
chuong trinh gidi; 3) Thue hien chuong trinh
giai; 4) Kiem tia nghien eiiu ldi gjai Nhu vay
ttong giai bdi toan, edng viec tim tdi ldi gidi bdi
toan la khau quyet dinh Dii cd ky thudt eao, cd
thdnh thao ttong vipc thyc hien cac thao tae va
eac phep tinh nhung khi chua ed phuang hudng
hogc phuong hudng chua tot thi chua the cd ldi
giai hope ldi giai ehua tot Tim tdi ldi gidi bdi
todn ciing chinh la co sd quan trong ttong vipc
ren li^en khd ndng lam viee ddc lap, sang tgo
cua HS Van de dat ra Id bdi dudng nhu the ndo
cd hieu qua de cho cac em cd nang lpc khi dimg
trudc mdt bdi todn cd tiie tp giai duoc mpt each
hop li Bai viet nay de xuat mdt so bipn phap de
boi duong nang lpe phan tich tim dudng loi giai
bai todn cho HS thdng qua dgy hpe gidi bai tgp
chii de "Phuang phdp tga do trong mdt
phang-Hinh koc 10"
2 Nang Inc phan tich tim Icri giai bai toan
2.L Nang Iprc toan hoc
Theo V A Krutecxki [6, tt 13], nang lpe
toan hpc dupe hieu theo 2 y nghia, 2 miic dp
Mdt la, theo y nghia nang lpe hpe t?^ (tai
tgo) tiic la nang lue doi vdi viec hpe Toan, doi
Hgc vien cao hpc, Tnrdiig Dai hgc D6i^ Thap
Tru6ng Dai hoc D6iig Thap
vdi viee ndm sach giao khoa mdn toan d tmdng pho thdng, nam mdt each nhanh va tdt cac kien
thiic, ky nang, ky xao tuong iing
Hai Id, theo y nghTa ndng luc sang tao (khoa
hoc), tuc la nang luc hopt dpng sang tgo Toan hpe, tao ra nhimg ket qua mdi, khach quan co gia tri doi vdi xa hpi loai ngudi
Ong dua ra dinh nghia "Ndng luc hpc t ^ toan hoc la ede dae diem tam ly ea nhdn (trudc het ia cae dgc diem boat ddng tri tue) dap ung yeu eau hoat ddng toan hoc va giiip cho viec nam giao trinh Toan mdt each sdng tao, giiip cho viec ndm mdt each tuong doi nhanh, de dang vd sau sdc kien thirc, ky nang vd ky xao toan hpc" [7, tt 14]
Theo Le Ngpc San [9, tt 21], neu coi qua trinh hpc t ^ Id qua trinh thu nhan va xii li thong tin thi ndng luc toan hpc ciia HS bao gom:
- Ndng lpe thu nhan thdng tin toan hoc: nang lpe tri giac hinh thiie hda tai lieu toan hoc, ndm eau tnie hinh thiie cua bai toan
- Nang lpc ehe bien thdng tin toan hpe: nang Ipc tu dio' logic ve quan he so luong va hinh dgng khdng gian bang ki hieu toan hoc; nang iuc khai quat hda eae doi tupng toan hpc, quan h^ toan hpe va eae phep toan; nang lyc tu di^ linh hogt bdng riit gpn qua tiinh suy luan va cau true toan hpc nit gpn; nang lpe chuyen hudng qua hinh tu duy
- Nang lue luu trii thdng tin todn hpc: nang luc ghi nhd (tri nhd khai quat, dac diem ve loai,
so do suy lugn va chimg minh, phuong phdp giai bai toan),
Ndi den HS cd ndng luc toan hpc Id ndi den
HS cd tri thdng minh tiong vipc hpe Toim, Tat ca moi HS deu ed kha nang vd phai ndm duoc
chuong trinh tnmg hpc, nhung cac khd ndng do
Trang 2khac nhau tii HS nay qua HS khdc Cac khd nang
ndy khdng phai co dinh, khdng thay doi Cdc
nang luc nay khdng phdi khdng the thay doi md
hinh thdnh vdjyhdt trien trong qud trinh hoc tap,
luyin tgp de nam dupe boat dpng tuong iing, vi
vay, can nghien cim de nam dupe ban ehat cua
ndng luc va cac con dudng hinh thanh, phat trien,
hoan thi^n nang lue
2.2 Nang luc giai toan
Ndng luc giai toan Id gi? Chung tdi quan
niem nhu sau: Nang lue gidi toan la mdt phan
cua nang lpc toan hpc, la to hop cac ky nang ddm
bao thuc hien cac hogt ddng giai toan mpt each
cd hieu qua cao sau mot so budc thuc hien
Mpt ngudi ed nang lpe gidi todn neu ngudi
do ndm vung tri thiic, ki nang, ki xdo cua hogt
dpng gidi toan vd dat duoc ket qua cao so vdi
trinh dd trung binh cua nhimg ngudi khdc cimg
tien hanh hogt ddng gidi toan do ttong eae dieu
kien tuong duong
Tir dgc diem boat dpng tri tue cua nhiing HS
cd nang lpe toan hpe vd khdi nipm ve nang lyc
giai toan, ed the rut ra mdt so dgc diem va cau
true cua nang lue giai toan nhu sau:
- Khd nang liiih hpi nhanh chdng qiQ' trinh
giai mpt bai toan va cac yeu cau ciia mdt ldi giai
dep va rd rdng
- Sp phat trien manh eua tu day logic, tu
duy sang tao the hien d khd ndng \^ lugn ehinh
xac, v4 quan he giiia cae du kipn eua bai toan,
- Cd nang Iuc phdn tich, tong hop ttong linh
vuc thao tdc vdi edc ki hipu, ngdn ngir toan hpe
Khd nang chuyen doi tir i^eu kien cua bai toan
sang ngdn ngu: ki hieu, quan h§, phep toan giua
cdc dai lupng da biet, ehua biet vd nguoc lai
- Cd tinh dpc lap va dpc ddo eao ttong khi
gidi toan va su phdt trien eua nang luc gidi quyet
van de
- Cd tinh tich cue, kien tri ve mat y ehi vd
khd ndng huy ddng tri dc cao ttong gjai toan
- Khd nang tim tdi nhieu ldi giai, huy ddng
nhieu kiln thue mot luc vao vipc giai bai tap, tir
do lua ehpn ldi gidi toi uu,
- Cd khd nang kiem tia eac ket qua da dgt
dupe va hinh thanh mdt so kien thiic mdi thdng
qua hoat dpng gidi todn, ttanh dupe nhung nham
lan ttong qua trinh giai toan
- Cd khd nang neu ra dupe mdt so bai tgp
tuong tp cimg vdi each gidi (cd the la ^ n h hudng
gidi, hodc quy trinh ed tinh tiiudt toan, hogc thuat toan de giai bai toan do)
- Cd kha nang khai quat hda tir bdi toan eu the den bai todn tong quat, tir bdi toan ed mdt so yeu to tong qudt den bai toan ed nhieu yeu to tong quat, nhd edc thao tac tri tup: phan tich, so sanh, tcing hpp, tuong tu, trim tupng, he thong hda, dac biet hda,
2.3 Nang lure phan tich tim Idi giai bai toan
Xet ve binh dien triet hpc thi "Phdn tich Id phuang phdp phdn chia cdi todn thi ra thdnh timg bg phdn, timg mat, timg yeu td de nghien citu vd hiiu duac edc bd phgn, mdt, yiu to dd"
[8, tt 86] Ngodi ra cdn ed nhieu dinh nghia khdc nhau ve phan tich Theo Nguyen Ba Kim [5]
"phdn tick id tdch (trong tu tudng) mdt he thdng thdnh nhihig vdt tdch mdt vdt thdnh nhirng bd phdn rieng le"; Theo Hodng Chung [1], "Phdn tich Id diing tri 6c chia cdi todn the thdnh edc thdnh phdn, hogc tdch ra timg thudc link hay khia cgnh riing biit nam trong edi todn thi do"
Tren co sd phan tich edc dinh nghia tien, ed
the quan mem ve phan tich nhu sau: "Phdn tich
Id dimg tri dc tdch doi tugng tu duy thdnh nhimg thugc tinh, nhimg bd phdn, cdc mdi lien he quan
he de nhdn thirc ddi tugng duge sdu sac han"
Cd hai hinh thiic phan tich' Thii nhat, do Id tach van de thanh eac bp phgn theo tieu chi, Chang han phan tich khai
niem so thdnh hai bp phgn: so chan vd so \i Viec
tdch nhu the nao tuy thudc tung dae diem, yeu cau, mue £eh bai toan
Thu hai, dd Id tach ra mot phan, tdp tnmg chii y vao thanh phan dd, thu thap cae thdng tin tir viec nghien cim thanh phan vira tach ra Chang han, ttong mpt phuang tiinh, tdch ve phai cua phucmg trinh, quan sat, xem xet cae phep todn, edc con so ttong bieu thirc ve phdi, tir do dua ra cdc thdng tin ve bieu thirc ndy Theo chung tdi quan ruem: Ndng luc phan tich tim ldi gidi bai toan la khau dau tien ttong qua tiinh gidi toan, ddi hdi ngudi gidi toan phai
ed kha ndng nhu dy doan, md mam, dac biet hda, khdi qudt hda, tuong tp hda, xdc dinh dupe the loai bai todn, vach phuong hudng gidi bai toan, tim duoc cac phuang phap va cdng cp thich hop
de giai bai toan
3 Mpt so bi^n phap boi du-ong nang lu-c phan tich tim ldi giai bai toan cho HS thong
Trang 3qua d^y hoc giai bai t^p chu de "Phuwng p h i p
toa dp trong m | t p h ^ g - H i n h h^c 10"
3.1 Thanh to nang lire giai toan cua chu
de "Phmmg phap tpa dp trong m^t
phang-Hinh hpc 10"
Ndi dung chii yeu cua ehu de "Phuang p h ^
tpa dp trong mgt phdng-Hnh hpe 10" bao gom:
Tpa dp diem, veeta; Phuang trinh dudng tiiang;
Vi tri tuong doi giua hai dudng thang; Gde;
Khoang each tir mdt dilm den dudng thdng;
phuong trinh dudng ttdn, vi tri tuong d6i giua
dudng ttdn va dudng thdng
Can eu vao dac diem ciia nang Ipc gidi toan,
can cii vdo ndi dung hp tiiong bai t ^ eua chu de,
chiing tdi xde $nh eae thanh to cua nang lpc giai
bai toan chu de "Phuong phap tpa dp ttong mat
phang-Hinh hoc 10":
1 Ndng luc huy ddng, van dyng edc tinh
chdt, cdng thuc, dinh ly vdo viec gidi nhanh vd
chinh xdc cdc bdi tgp
2 Ndng luc phdn tich, tdng hgp du kien vd
yeu cdu bdi todn de dinh hudng cdch gidi
3 Ndng luc tim ldi gidi bdi todn tga do
trong mat phdng bdng nhiiu cdch khdc nhau
4 Ndng luc trinh bdy ldi gidi mot cdch chdt
che vd cd ca sd
3.2 M$t so bi|n phap boi dirong
3.2.1 Bien phdp 1: Tap luyen cho HS biit
sir dung cdc thao tdc tu duy nhu dit dodn, md
mdm, phdn tich, tdng hap da Men vd yiu cdu bdi
todn de dinh hudng cdch gidi
Khi diing trudc mdt bdi toan HS can tp dgt
cho minh cdu hdi: De gidi bdi toan nay ta can
nhiing kien thiic nao? Tir do giiip hp lien tudng
den cac kien thiic hen quan de gidi bdi toan, Mgt
khac, gido vien (GV) can dua ra bai toan ngau
nhien, khong xep theo mpt trinh tp nao dgt trude
dl ren luypn eho HS nang luc huy dpng va vgn
dung kien thue vao gidi bdi todn
Vi dii 1: Trong mgt phang tpa dp Oxy, cho
hinh chu nhgt ABCD cd tam /
&•)
thiie da hpc (tinh ch4t cda hmh chii nhgt) lln kien thiie dang hpc (1§P phuang tiinh duong thdng), Nhimg neu ttong qua trinh gidng dgy (bai phuang trinh dudng tiidng), GV hp tiiong dupe kiln thiic (cac yeu to can thilt de 1 ^ phuong trinh dudng tiiang) va HS nam vung dupe cac kiln tiiuc ^ thi vipc gjai bai toan sS khdng ed gi khd khan
- Gia thilt: Biet hmh chii nhgt ABCD, suy raAB//CD, AB ±AD;AB LBC
Biet AB:x-2y + 2 = 0, suy ra vl ph^i
phuang trinh la w = ( l ; - 2 ) ; / —;0 Id tam ciia
cgnh AB
ed phuong trinh la x - 2 y + 2 = 0 va AB = 2CD
L ^ phuang trinh cae canh edn lai eiia hinh
chii nhat
Phan tich bai toan:
Bai todn ed gia tiuet kha phiic tap, ddi hdi
HS phdi huy dpng tat cd cdc Itien thiie: tir kien
hinh chu nhgt nen
d(I,AD) = 2d(I,AB)
d(J,AB) = d{I,CD),
Hinhl
- Ket l u ^ : l,ap phirong tiinh cac cgnh
cua hinh chii nhgt (cae dudng tiling
AD; BC; DC)
Tom tat each giai bai toan:
Tir kiln tiiiic da biet: Biet AB//CD, ma phuang trinh AB:x~2y + 2 = 0 sity ra phuong trinh CD cd dgng x-2y + c = 0(ci^2) Tuong tix, do AB±AD;AB1BC; suy ra phuong trinh AD va BC cddgng 2x+y+c'=0
- De tim dupe c, chung ta can lien he cong
thiie khoang each tir mpt cKem den mpt dudng thang Do biet tam 7 —;0 ciia hinh ehu nh§t
I ll 5 [c=2(i) nen d{I,AB) = d{I,CD) '^\c-\-\=-^ _'• Vgy phuang trinh canh CD •.x-2y-'i = ^
- Tuong ti; tim c: Do AB = 2AD suy ra d{I,AD) = 2d(I,AB) o | c ' + l | = 5 O _ • Vgy phuong ttinh AD vd BC lan lupt la
2:t + _V + 4 = 0 v d 2 j : + ; ' - 6 = 0
Trang 43.2.2 Bien phdp 2: Tap luyin cho HS
biit nhin nhgn tinh hudng dgt ra dudi nhieu
gdc do khdc nhau va biet gidi quyet van di
bang nhiiu phuang phdp khdc nhau, lua chgn
cdch gidi toi iru
Bien p h ^ nay giiip HS nhin nhgn bai toan
duoi nhieu hinh thiie, nhieu gde dp khac nhau,
Tir dd, HS se tim tdi ra dupe nhieu each gidi
khdc nhau cho mpt bdi toan va tim dupe phuang
phap giai dpe dao Dieu nay ren luyen tu diQ'
sang tao, dpc Igp giai quyet van de eho HS
Vi dy 2: Viet phuang trinh dudng ttdn npi
tiep tam giae ABC biet phuang trinh cac canh
AB:lx+Ay-6=(y, AC:Ax+3y-\=(i, BC:y=Q
Phan tich bai toan
- Hudng dan HS ve hinh
3x+4y-6=0 4x+3y-l==0
Htnb2
- Ddt eau hdi goi y cho HS tim phuang an
ttd ldi:
GV: Cd till tim dupe ylu to ndo khi biet
phuong trinh ba canh ciia tam giac?
HS: Cae dinh A, B, C lan lupt la giao diem
cua cae cgp canh AB va AC, AB va BC, AC vd
BC Ta tim dupe tpa dp ba dinh A, B, C
GV: Tam dudng ttdn npi tiep tam giac ABC
duoc xac dinh bang each ndo?
HS: Cdba hudng:
+ Tdm I la giao cua ba dudng phan gidc
ttong cua tam giae
+ Ap dung tinh chat khoang each tir tam I
den ba cgnh bang nhau
+ Gpi l(a;h) la tam dudng ttdn npi tiep
tam giac ABC Vi B, C nam tten trpc Ox spy ra
l{a,b),b>ii^l{a;r); S = p.r=> r = - Ap
dung cdng thiic khodng each dyI,ABj = r ta
tim duoc tpa dp tam /
GV: Hudng thit ba gpi l{a;b) thi xet xem
a, b phdi thda dilu kipn gi (dua vao tpa do ciia
A, B, cy
HS Do dudng ttdn ndi tiep tam giac ABC
nen t a m / ( a ; 6 ) phdi thda—<a < 2 ; 0 < 6 < 3
4
- Tir nhiing gpi y nay, HS ed the dl dang dinh hudng tim ra nhilu each giai cho bdi toan Tdm tat each giai bai toan:
Cach 1:
Tpa dp cua ^ la nghiem cua hp phuang ttinh
3x + 4 y - 6 = 0 [x = -2 ^
<=>J = > ^ ( - 2 ; 3 )
4x + 3 : v - l - 0 [>' = 3 ^ '
Tuong tp,tatim s ( 2 ; 0 ) , C - , 0 Phuong trinh cac dudng phdn giae ttong va
ngodi cua gdc A la 3x+Ay-6 _^Ax^3y-\ \x-y + 5^0 0) V3^+4' " U^+Z" [x + ; ^ - l - 0 (2)' Thay lan lupt toa dd ciia B, C vao v l tiai cua
(1), ta ehpn duoc (2) la phuong trinh dudng phan
giac ttong cua gde A
Phuong trinh cac dudng phan giae ttong va
ngoai ciia gdc B la 3JC + 4 ; ' - 6 \Zx-y-e = Q (3)
— ±—_ ^ + V o
^3^+4^ \x+3y-2 = Q (4) Thay lan lupt tpa dp ciia A, C vdo ve ttdi eua
(4), ta chpn duoc (4) la phuong ttinh dudng phan
giac ttong eiia gdc B
Gpi l{x;yj vd r Id tam va ban kinh dudng trdn npi tiep tam giae ABC
Khi dd tpa dp / la nghiem eua he phuang
'x + y-\ = Q
trinh J " ' -^ * " Cndi he ta tim dupe [x + 3^- - 2 = 0
= i h a y / [ - , - ] , S u y r a r = ^ ( / , S C ) = - ,
Trang 5Vgy phuong trinh duong tron ngi tiep tam
g i a c ^ C l a ( ^ x - i j +[y-^j = i
Cach 2:
Tim tga dp cua ^ ( - 2 ; 3 ) , B ( 2 ; 0 ) , d-.O
Ggi 7(a;6) la tam duong trgn ngi tiep tam
giac ABC voi -2<a<2;0<b<3 Khi do
ii{I;AB) = d{I,AC) = d(r,BC)
l3a + 4 4 - 6 | l4a + 3 i - l | lil
o ' =^ = ^-j= - n "
V 3 ' + 4 ' V4'+3^ VO' + l^
n3a + 4 6 - 6 | = |4a + 3 6 - l |
Vay phuang trinh duong tion n^i tiep tam
3a+ 4 6 - 6 = 5 *
1
Giai he ta dugc a=b = —, suy ra ban kinh
2
duong tron ngi tiep r = d\I;BCj = —
Vay phuong tiinh duong tron noi tiep tam
giac ABC :
Cach 3:
T i m t g a d 6 c u a 4 - 2 ; 3 ) , B(2A
"(-ij<'-0=i
2;0), c(i;o)
Ggi 7(a;6) voi - < a < 2 ; 0 < 6 <3 la tam
' 4
duong tron ngi tiep tam giac ABC Vi S va C
nam tren trye Ox suy ra l{a;r), AB = 5,
= - , A = 3 T a c o S = -BC.h = —
4 2 8
tia S = p.r ^ r =
P 2
Mat khac d(l,AB) = r^
3a + 4 i - 6
2 1
1 13
Suy ra a = — hoac a = — (loai)
2 " 6
giac ABC
»(-3<'-i]=J
3.2.3 Bien phdp 3: Khac phuc sai ldm cua
HS khi gidi todn qua cdc giai dogn
- Giai do^n 1: Sai lam chua xuat hien
Giai dogn nay nhdm phdng ttanh sai lam ciia HS, GV cd the dp dodn trude de "phdng ttanh" bdng each ttang bi cho HS nam vung cac kien thirc todn, kien thiic ve phuong p h ^ gi^i toan, nhan manh edc ehu y can thiet doi vdi HS
- Giai do^n 2: Sai lam xuat hien ttong
Idi giai Quy ttinh d giai doan nay Id GV theo doi thay sai lam -> GV gpi y de HS tp tim ra sai lam—>HS tp tim ra sai lam->GV goi y dieu chinh ldi gidi ->HS the hipn ldi giai diing->GV tong ket vd nhan manh sai lam HS dabi mdc phai
Tiiy theo mite do sai lam md GV chpn cac bien phap su phgm thieh hpp:
+ Dua ra ldi gidi dimg dl HS doi chilu + Chii dpng dua ra ldi gidi sai dl HS nhgn dang cdc dau hieu tim ra sai lam
+ Dua ra nhieu ldi gidi khac nhau (co tiie dung phuong ph^q) ttae nghiem) de HS phdn bi$t dung sai cua cac ldi gidi
- Giai do^n 3: Sai lam da duoc phdn tich va
sua ehua Mpt sai lam cua HS mac dii da dupe phan tich va sura chiia nhung vdn ed the tai dien De khdc phpc tinh tigng nay, GV can thir thach thudng xuyen HS qua edc bdi toan de din den cdc sai lam ma HS thudng mdc phdi Viee chia ba giai doan sai lam chi mang y nghTa nhan manh tiidi diem ciia sai lam, Trong dpy hpe gidi bai tgp toan, GV ed khi dong tiioi tae dpng den ea ba giai doan, bdi vi vira phong ttanh edc sai lam ehua xuat hipn, vira phan tich
va sira chiia cac sai lam dang xuat hi^n dong thoi xda nhimg sai lam da sira ehua
Vi dy 3: Trong mgt phang vdi he tpa do
Oxy, cho 4em ^ ( 2 ; 2) va cac dudng th§ng d^:x + y-2 = 0,d^:x + y-^ = 0 Tim tpa do edc diem 5 va C lan luot thupc d^ vd d^ sao cho tam giae ABC vudng can tai A
Trang 6Du- doan sai lam: HS se khdng nhgn ra vi
tri tuong doi cua rf, va d^ [d^ lld^)\ quen tinh
ehat eiia tam gidc vudng can dan tdi gidi bdi toan
sai hodc khdng tim dupe each giai Chinh vi vay,
trudc khi HS gidi bdi toan GV cho HS dn nhac
Igi cae kiln thiie ndy de ttanh sai lam
- Hudng dan HS ve binh:
(6-l)(c-4) = 2
V ^\ ' vadat x=h-\, y = c-4, [(b-lf-ic-4f=3'
ta cd hp •i Tir he nay ta tim duac x, y
- Khi HS giai xong, GV cho nhan xet vd nhan mgnh lai nhung chd HS sai lam GV cd thi eho bai t ^ tuong tp eho HS giai de khac sdu kiln thirc
Tom tat each giai:
Vi B€d^,CGd^ nen B(b;2-b\ C ( c ; 8 - c )
Tir gid thiet ta ed he:
[M^=0 f6c-46-c+2=0 f(6-])(c-4)-2
\AB = AC V - 2 6 = C ^ - 8 C + 1 8 \(p-V)^ -{c-Af =^
HinhS
Sai lam suat hien trong qua trinh giai toan:
GV: Do B^{d^),Ce{d^) tin S, C ta
chpn tpa dp nhu thi nao?
HS Chon B{x„y,)^(d,),C(x„y,)^(d,)
- Neu HS ehpn toa dp hai 4 e m B, C nhu
the se dan tdi sai lam la tim tdi bon an
Xg, yg, x^, y^ Khi dd GV gpi y dieu chinh
each chon hai dilm 5 (ft; 2 - ft), C{c; 8 - c)
GV: Hai diem S va C phai thda man dieu
kipn gi dl tam giae ABC vuong can tai Al
HS: ABLAC vaAB = AC
Sau dd lap he phuang ttinh:
\ABAC = 0 \bc-4b-c + 2^0
[AB =AC [ft^ - 2ft - C^ - 8C + 18
- Doi vdi he nay HS giai bdng phuang phdp
the se rat phiic tgp, cd the khdng tim dupe d ^ so
hope tim sai GV cd the gpi y biln doi he thdnh
Ddt x=ft-l, ;^'^c-4 tacdhe xy = 2
•y = 3
y = ~l hogc Gai hp ta duoc x
x = 2, y^l
Suy ra 5(-l,3), C(3;5) hogc 5(3;-l), C(5;3)
4 Ket l u ^ Nhiing van de dua ra ttong bai vilt nay da phan nao ldm ro tinh chat quylt dinh cua khau phan tich tim dutmg loi gjai bai toan ttong gidi toan Dong thdi bdi viet da dua ra dupe mpt so thdnh to ndng luc giai toan vd tii do dl xuat dupe
ba bien p h ^ boi dudng nang lue phan tich tim dudng loi gidi toan eho HS thdng qua day hpc gidi bdi t ^ chii de Phuang phdp tpa dp ttong mat phang De ren luyen ndng luc giai toan ndi rieng
va nang Ipc todn hpe cho HS ddi hdi su no luc quyet tam eua dpi ngu GV todn Chiing tdi hy vpng day Id mdt tai lieu tham khdo ed ieh eho
GV tiong viec ren luyen nang Ipc gidi todn cho
HS, ciing nhu eiing nhu ttong day hoc toan d trudng tnmg hoc pho thdng /
Tai lidu tham khao
[ 1 ] Hoang Chung (1979), Phuang phdp day hgc mdn todn d trudng trung hoe ea sd, NXB Giao
dpe Ha Npi
[2], G Polia (\997), Gidi mdt bdi todn nhu thi ndo^, NXB Gido due, Hd Ndi
[3] Tran Van Hao (Tong chu bien), Nguyen Mdng Hy (Chu bien), Nguyen Van Doanh, Tran
Dire Huyen, Hinh hgc 10, NXB Giao dpe Ha Npi
[4] Huynh Thanh Huong (2010), Phdt triin tu duy sdng tgo cho hgc sinh thdng qua viec dgy hoc chuang "Phuong phdp tga do trong mat phdng-Hinh hoc 10-Ndng cao ", Lugn van Thac sT giao due
hpc, Trudng Dai hpc Vinh, Vinh
Trang 7[5] Nguyen Ba Kim (Chij bien), Vii Duong Thuy (1992), Phitcmg phdp dgy hgc mdn Tom,
NXB Gao dpe Ha Ndi
[6] Kruchetxki V A (1973), Tdm li ndng luc todn hgc cua hgc sinh, NXB Gido dpe Ha Ndi,
[7], Le Ngoc Son (2015), "Dgy hpc todn ttong trudng pho thdng theo dinh hudng phdt trien nang lue", Tgp chi Todn hpe ttong nhd trudng, (so 1), tt 21-25
[8] Dao Tam (Chii bien), Le Hien Duong (2009), Cdc phuang phdp day hgc khdng truyen thong trgng day hoc todn a truang dai hoc vd trudng phd thdng, NXB Dgi hpe Su phgm Ha Ndi [9] Nguyen Dinh Trai (2001), Nang luc tu duy li lugn cho cdn bd gidng dgy li ludn Mdc - Lenin
d edc trudng chinh tri, Luan an Tien sT triet hpc, Vi§n Triet hoc - Hpc Vipn Chinh tri Quoc gia Ho
Chi Minh, Hd Npi
SOME METHODS IMPROVING STUDENTS' COMPETENCY OF ANALYZING FOR MATHS SOLUTIONS VIA TEACHING THE LESSON THEME
OF COORDINATE PLANE METHOD-GRADE 10 GEOMETRY
Summary This article reviews competencies of maths, analyzing matiis problems and finding solutions; thereby it suggests some methods improving students' competency of analyzing for finding solutions via teaching the lesson theme: "Coordinate Plane Metiiod-Grade 10 Geometry"
Key words: competency, analysis, method, students, exercise, coordinate plane
Ngay nhdn bar 12/5/2017; Ngdy nhgn lgi- 19/6/2017; Ngdy duyet ddng: 15/8/2017