TAO DDNG CO VA D O N G TDD DOC TAP CDO DOC SIND NHAM PDAT DOY KDA NANG CDO D Q N G ' C D I E M LlNH TDI THiJC TRONG DAY HOC T O A N 6 P H 6 THONG
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rong qua trinh day hpc, dua vao dae diem tam li Ida tuoi ctja hpe sinh (HS), giao vien (GV) can tao ra mdt mdi tn/timg hoc tap (HT) tich cue de d i n d i t eae em tham gia vao cac hoat ddng hpe tap (HDHT) chiem ^nh tri thiJfc mdi, tang hieu qua cua hoat dpng day hpc (HODH) theo djnh hucflig doi mdi phuangphapdayhoc(PPDH).Mdttrong nhung myc tieu day hpc la hinh thanh va thiJc day dpng eo hpc tap (DCHT) ben trong cho H S d e cae em hiing thii hpe tap (HTHT).1. OCHT la khong co san, cung khdng theap dat ma phai dupe hinh thanh dan dan trong qua trinh HS HT chiem finh tri thue qua tung tiet hpc, dudi su hudng dan,toehue cua GV.Cdthehieu,OCHT eua HS ehinh la ket qua HT ma eae em dat dugc de thoa man nhu cau eua minh.DeedDCHT, tn/dc het phai cddoitugng dben ngoai chuthe.cd gia tri dd'i vdi chu the va lam nay sinh nhu cau c ^ chiem linh nd. Khi nhu cau chiem ITnh tri thdc dU(K ea nhan y thdc. se trd thanh ddng co thiic day, djnh hudng va duy tri hanh dpng. Ddng co ludn gan vdi nhu eau, mong mud'n cua ca nhan. Hay ndi each khae, nhu eau, mong mud'n la nhung yeu to ben trong quan trong nha't de hinh thanh ddng ea.
Nhieu cdng trinh nghien cdu cua cae nha tam li hgc da eoi HTHT la mpt bieu hien dae biet eua nhan thdc. Theo A. G. Kovaliov: HTHT chinh la thaiddh/a ehgn dacbletcua chij^edolvadoitugngcua HDHT, vi su thu hut ve mat tinh cam vay nghia thuc tien eua nd trong ddi song eua ea nhan (2). HTHT cua HS se thuc day tinh tich cue cua tri tue, su nd luc cua y chi trong HD nhan thuc. HTHT la ddng ca chinh, kich thich HS chiem ^nh tri thue mdi mot each tich cue va ben bi. Dudi anh hudng cua HTHT. HS tien hanh eae hoat dgng tri tue khae nhau de di sau vao ban chait cua cac dd'i h/gng va hien tugng dang xet.
Giao dye hung thtj g i n vcri doi mdi PPDH. Tn/de het, giao due hdng thu la tien de cho day each phat
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hien van de. HS se khdng phat hien dugc van de neu khdng tieh cue HD va tuduy. Tiep den, giao dye hdng thti gan lien vdi day hpc theo quan diem^HD; khi cd hung thtJ. HS nd luc bien ddi dd'i tugng de xam nhap vao doi fe/gng. Cd hai loai hdng thii nhan thdc, dd la:
hung thd dcae ket qua cQa nhan thiic va hdng thu a nhung phuang phap ctia HD nhan thue. Hdng thu d cac ket qua thudng dugc bieu hien d ehd: HS cam thay HTHT neu eae em dat ket q ua tot, dutjc G V khen nggi va dat die'm eao. Hdng thii phuong phap dugc bieu hien d: HS cam thay hdng thu vdi cae kien thdc mcri thii vj, Idi giai hay, ngan ggn, vdi each thdc giai quyet bat ngd, ludn tim ra nhdng phuong phap mdi de giai quyet van de. Nhuvay, GV can biet dieu khien (ke ca dieu khien ve mat tam li, bao gom su ddng vien, hudng dan, dieu khien qua trinh day hpc b^ng he thd'ng cau hdi, dan d I t HS, tao eae tinh hud'ng dgy hgc) qua trinh day hgc sao cho tao ra cho eac em nhung xiie cam tieh ei/c khi tim hieu mdn hgc, giiip cac em td ehd tim hdng thii ben ngoai den viec tim hdng thu ben trong.
Hunq thunhust/thue day ben trong lam giam sif cang thang met nhge va mdra con dudtig din den st/
hieu biet, nim vdng tri thue mgt each dSdang va hieu qua han trong qua trinh nhan thtJee. Hdng thu va dgng ed CO quan he vdi nhau. Ddng ca tao ra ht/ng thu, hdng thu la tien decua ti/glae. Htmg thu va tuglae la yeutd'tam litao nen tinh tieh cue, tuduy ddc lap, sang tao (3). Hung thii kich thieh, thiic day eon ngudi tich cue HD disau nghien cdu doitugng. Do vay, HTHT la mpt dang bieu hien cua DCHT.
2. Tao dgng co va HTHT cho HS Trong day hgc loan dphd'thdng, ddng cova HTHT cua HS eoquan he bien chdng, hdu eavdi nhau; GV can giiip cac em y thdc dugc md'i quan he mat thiet
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giua HS va doi tugng hpc. Oe tao ddng ca va HTHT choHS,GVed thethyc hien thdng qua nhung phuang phapsau:
1) Thiet ke, toehue, hudng din HS thuc hien cac HBHT ydi hinh thuc da dang, phong phd, cd st/c hap dan, phu hgp vdi dae trung bai hgc, dae dim, trinh do ci/a HS va dieu kien Idp hgc. Hay noi each khac, deday hgc cd hieu qua, dieu quan trgng laGV tao ra dugc tinh hud'ng day hoc cd y nghia dd'i vdiHS. Nghe thuat cua ngudi day la dua ra vS'n de ma nguiji hpc muon tim ra each giai quyet nd. Chang hgn, khi day hpe eae khai niem va djnh li toan hpe, de HS cd hdng thii, GV can tao ra cac tinh h u o n g g a y s U c h u y v d i c a c e m . D o d d , khiday hpc khai mem va dmh li toan hgc, G V can quan tam den kha nang ung dung, tri thue phuang phap va xay dung quy trinh giai.
Vidu /.-Sau khi hpc xong dinh li Viet, G V cd the' lat nguge vain d l : Cho hai so a va b thoa man:
{a + h = S
U , = p ' ' ' . Hay neu caeh xac dinh a vab?
GV mong dgi HS lap luan:
\iih =P [tH,S-ti) = P [a--«S + /' = 0 Vgy.athoa man (*) thi a thoa man phuong trinh:
A^-SX+P=0. M5t khae, tuhe(*), neu ta thay abdii) va /) bdi a thi h$ tren khdng thay doi, suy ra b ciJng thoa man phyang trinh: X^ - SX+P=0. Vay, a va b la hai s6'thoa man (*) khi va ehi khi a va b la hai nghiem cua phuang trinh X'-SX-1-9 = 0.
Neu HS gap khd khan, GV ed the d i n dat: Khi giai hS phuong trinh hai an, ta thudng s u dyng nhung phuang phap nao? Em ed nhan xet gi ve md'i lien he gida a va b trong hg {*)? Sau khi HS giai quyet xong vl'n de, GV can nhan manh: trong tht/c tiln giai toan, ed nhdng he phuang trinh ma khi giai, ta phai dua ve
GV dSt ra cac cau hdi: he phuang trinh (*) cd nghiem khi nao, ed hai nghiem phan biet khi nao? Ta mong dgi HS suy luan: x, y la hai so thoa man (*) khi va ehl khi x. y la hai nghiem ctia phuang trinh X^ - SX-i-P=0, maphuang trinh A ' - S X + P ^ O cd nghiem khivachikhiA = 5 - " - 4 / ' a o » i - £ 4 / ' , V a y . h e ( * } e d nghipm khi va ehi khi 5 ' £4/>; neu (x; y) la nghiem thi (y; x) cung la nghiem. do dd neu he (*) cd hai
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nghiem thi x * y . Mat khac, haisd'x, y thoa man (') khi va ehl khi nd la hai nghiem ctia phuang trinh A^ - SX+P = 0, nen he (*)cd hai nghiem khi va ehl khi
^ = sl-AP>0»,?' >4/'.De'cunge6kienthuc.GV ed the cho HS giai BT sau: Cho he phuang trinh:
| ( I + v) n> = /» • ^^ Gial he PT khi m = 2; b) Tim m dehe phuang trinh ed nghiem.
Sau do, G V hudng dan HS cac budc giai he phuong trinh CO chda bieu thdc dd'i xting x + y va xy nhu sau:
- Budc ?;Nhan dang he phuang trinh; - Budc2:Ba\ x +y = S;xy = P, vcri dieu kien-^'s 4/';-Si/dcS.'Chuyen he PT cd chua an x, y ve he phuang trinh ed chda an S. P;-Sf/dc4.Giaihephuangtrinh v d i a n S . P . S o sanh vdi dieu kien a budc 2, neu thoa man dieu kien thi ehuyen sang budc 5, neu khdng thoa man thi ket luan he phuong trinh vd nghiem; - 6£/dc 5; Sudyng dinh li Viet, eoi x,y la nghiem eua PT:A'^-SA'+P=0;
-et/dcfi.'Ketluan.
2) Tap luy^n eho HS HD ehuyen do) BT de giai quyet eac van debang nhieu each khac nhau, nhin nhan van de toan hoc theo cac khia canh khac nhau
Vidu2:Tm m dephuong trinh x^ + 2x + m +1 = 0 v6 nghiem.
GV e d t h e d i n d^t HS tim dugc ba each diln dat khac nhau eua BT, do la: - Vdi gia trj nao cua m thi bieu thdc x^ + 2x + m + 1 ludn duong; - Vdi gia tri nao eua m thi parabol y = x^ + 2x + m + 1 khdng elt taic hoanh?; • Vdi gia tri nao cua m, thi parabol y = x^+1 n i m phia tren dudng thang y = - 2x - m?
Ba phuong thdc dien dat dtren da ehuyen van de vd nghiem eua phuang trinh ve BT tim m detam thdc khdng dd'i dau. each thd 2 dua ve BT xac dmh m de do thi ham so bae hai nIm tren taic Ox, each thd 3 dua ve viee xet vj tri tuong dd'i giua do thj ham so y = x^+1 vdi dudng thing y = -2x - m. Nhu vay, khdng nhung edsu g i n ket vdi nhung kien thdc da hgc trudc day, ma con khic sau va he thd'ng hda kien thdc. Vdi quan diem suy nghT chi xuat hien trong nhung hoan canh cy the, cdn niem tin la tu tudng hinh thanh tusu trai nghiem, G V nen dat eau hoi, tao tinh huong, hoan canh khac nhau, yeu cau HS thuc hanh,tratnghlem hoae giao nhiem vu cho HS de cac em dugc suy nghT, lam vipevaHD.
Nhuehiing ta da biet, hieu sau vain de can giai quyet la mau chd't khi giai quyet vain de. Do sau
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cua su hieu biet nay ehu yeu the hien d viec n I m vung ban chat cua vain de ya bieu dgt nd dudi nhung dang khac nhau. GV can^ ren luyen eho HS dao sau suy nghT sau khi giai mdi BT va ti/ dat ra cac eau hoi: BT cdn ed each giai nao khac nua hay khdng? Cd Idi giai nao tot han khdng? Cd the phat trien dugc BTnua hay khdng?.., Khi HS tral(Snhi>ig eau hdi dd se dan den nhu cau xem xet BT theo nhieu each khae nhau. Nhd the, GV ed the ren luyen eho HS each nhin BT mpt each toan dien, da dang, khai thac dugc eae thude tinh, md'i lien he gida eae ddkien; tao cho cac em suhdng thu trong HT chiem Unh tri thuc.
3) Khai thac cai hay, cai dep hoae nhung chi tiet, stf kien li thij lien quan den ngi dung day hgc nhim tao an tugng eho HS. Cai dep cua nhung Idi giai hay, ngan ggn, doe dao edn thehien d eae tinh hud'ng nhu; mpt BT rat phdc tap nhung lai cd mdt each giai n g i n ggn, ddi khi chi qua phep bien dd'i kTthuat; Chang han, vdi BT hinh hgc chi can ke them dudng phu, hoae mdt BT dai so chi can them bdt yeu td' nao dd; cung cd the la mdt each giai ddc dao. GV c^n phat huy n i l m say me HT eua HS, hudng eac em t d s i / y§u thich den chu dgng tim tdi, sang tao de chiem linh tri thuc.
V/du5.Chdng minh r i n g : (1 +a)(1 +b)(1 +e){1 + d)(1 + e ) > 1 + q + b + e + d + e,trongdda,b,c, d, e Cling dau va deu Idn han - 1 .
Viec chdng m inh tn/c tiep bat d i n g thdc nay tuang dd'i khd khan ve mat phuong phap vi viec khai trien hay nhdm vetrai la bat kha thi. GVed the ggiyehoHS phat bieu BT dudi dang td'ng quat. Vdi su ggi y nhu vay, HS cd thephatbleu BT td'ng quatva ehung minh mgt each d l dang han neu sd dung phuong phap quy nap toan hgc. BT td'ng quat la: Chdng minh ring (i + a,)(l + o,)...(i + o j £ i + 0| + Oj + ...+a„, trong dd ff,. t ; , . . . . o „ la eae so eung dau va deu Idn h a n - 1 . Dung y cua BT dua ra la giup HS thay dugc ddi khi chdng minh tn/e tiep mpt BT cu the nao dd cdn khd han khi ta tong quat hda BT, sau dd chdng minh bang mdt cdng ey khae se d l dang han so vdl BT ban dau. Bdi HS thudng cd suy nghT la chdng minh BT twig quat rait khd, ta phai ditdnhungBT cuthe nhung dday thi nguac lai, HS thay hdng thii han trong qua trinh giai toan.
Vi du 4: Chdng minh ring neu phuang hinh i'+OT'+&t=+aT+1 = 0 0 ) CO nghiem thi c ' + i f t - s i ' ^ j -
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Nhin vao BT. dudng nhu gida gia thiet va ket luan cd rait it md'i lien he vdi nhau, GV cd the gay dong co, hiing thii cho HS b i n g vi$c hucffig d i n cae em khai thac triet de gia thiet cua BT vdi mye dich quy ^ ^'^
quen. Vdi gia thiet eua BT, ta can tim dieu kien de khi phuang trinh (1) cd nghiem se cd md'i quan he gi vcri dieu phai chdng minh. Vdi each dinh hudng nhuvay, HS nhan ra la PT bae bon day du ed dang dae biet (phuong trinh hoi quy).
Giasdphuongtrinh(1)ednghiem x„ ^O.khidd, ta cd dang thdc diing sau: x^ + axl + 6x,; + or,, + i = o.
Chia ea halve cho x,;tadugc:-^J+A-+o(^*.+-j+6=o.
Dat "«*—='., vdi \i,\^2, ta thu dugc d i n g thdc:
/,; + ai„ + 6 - 2 = 0. Sau khi bien dd'i, H S ehuyen BT da eho thanh tim dieu kien de phuang trinh:
(• + a/„ + A - 2 = 0 ed nghiem [(J > 2, vdi each bien dd'i nhuvay. viec giai quyet BT trdnen khd khan va cd the HS se gap be tac, GV can dinh hudng de' chuyen hudng suy nghT eua HS.
Vdi yeu cau eua BT cd xua't hi§n a, (b • 2) nen td(1). la bien dd'i thanh /.' = - K + 6 - 2 ) ( 2 ) . Bieu thdc a^ + f/j - 2 / l a tong binh phuang cua cac he so cua nhj thue ar+ i? - 2. G V dat cau hoi eho HS: Oe lam xuat hien bieu thdc dd ed the lien tudng den bat d i n g thdc nao quen thupc hay khdng? HS de dang lien tudng den Bat d i n g thdc Bunhiaeopxki va tiep tuc bien doi (2) thanh i* = (at„ + b-2f va (a/„ + 6-2)'<[a'+(6-c)-](/,; + l). T d dO thu dUOC ( i ' + ( f t - 2 ) - a p i ^ . D e n d a y , H S d a t h a y d u g c m a q u a n he giua gia thiet va ket lugn. Van de edn lai la danh gia bieu t h u c ; j ^ . Vdi dieu kien r,;s 4 n § n r r i ^ ; y ^ ' i ' * T . Nhu vay, b i n g each phan tich, bien dd'i BT v l dang quen thude va ea ban, GV da hudtig cho HS nhin mdt van de ludn trong trang thai van ddng, cac em cd tlie se tim dugc m dt each giai quyet van de ddc dao. Thdng qua cae HO giai toan va s u d i n dat hgp li ctia GVdetao dugc subat ngdva gay HTHT cho HS.
Ngoai ra, de tao ddng ea va HTHT eho HS, GV can chii y: - Dgng vien, khuyen khieh. tao ea hoi va dieu kien cho HS tham gia mgt each tich cue, chu (Xem t^'p trang 26)
