TlM Hifu MOTSOBRTOnN NH^M R€N UIV€N KHH NANG KHfll QUHT HOII CHO HOC SINH T1€U HOC TrUorng ce Hi i DUcfng ] Trong day hpc mdn Toin d eip t i l u hpe, ben canh vile trang bj cho hpe sinh (HS) nhQng cP[.]
TlM Hifu MOTSOBRTOnN NH^M R€N UIV€N KHH NANG KHfll QUHT HOII CHO HOC SINH T1€U HOC TrUorng c e Hii DUcfng ] T rong day hpc mdn Toin d eip t i l u hpe, ben canh vile trang bj cho hpe sinh (HS) nhQng cP sd v i quy t i c cda tu kinh nghiem, giio v i l n (GV) cin tQng budc trang bj eho HS tuduy li luin khoa hoe De hien thyc hda mdt edeh dung d i n xu hudng quan trpng ndy ddi hdi ed sy nghiin cQu riing td hop cac van de logic va tam li hpc, l i l n quan d i n bdn chat cae tri thUe kinh nghiem vd IT ludn, tuong quan giQa cac khia canh cCia hoat ddng nhdn thdc eua ngudi, nhU: c i m tinh vd li tinh, tryc quan vd trOu tUdng, cai cu t h i va cai khai quat Cd sd ben lien k i t cac khia canh dd la eac q u i trinh khai quat hda (KQH) t u va cac qua trinh cd liln quan chat e h l vdi ehung la hinh khdi n i l m - mdt dang hoat ddngtritueePbin Cdch 2:C6 sdu sd 10, sau s6 16, sdu sd 13, ta ed tdng sau: x + x + x = 60+ 96+ 78 = 234 Cdch3:Tdng (10+16) bing tdng (13 +13) N h u v i y c d 18 lan sd 13 dupe cdng lai vdi nhau, hay: 13x18 = 234 Ddi vdi dang bai toan t r i n , d i l u c i n t h i l t Id giup HS hudng d i n vile tim kilm mdt mo hinh hay mdt Idi giii mau Di thyc hiln d i l u nay, GV ehia bai toan t h i n h cac phin nhd, y l u c i u HS giai quyet ede phan nhd trudc Trong vi du t r i n , GV ed t h i yeu cau HS giii quylt ede bdi todn nhd hay cae phan nhd nhU: Trung binh cdng eua ede sd tren la bao nhieu? Mdt sd tdng t r i n xuat hien bao nhieu lan? Cd the tim dupe nhQng cap so nao ed tdng bang nhau? Ddi vdi HS tieu hpe, t u van cdn phu thudc nhieu vao edi eu the; toan hpe mang nang tinh trUu tupng, khdi q u i t cao, vay viec hinh thdnh khd ndng, nang lyc KQH eho HS tieu hpe Id rat can thiet Cung t u eua HS t i l u hpc la t u cy the n i n n h i l u bdi toan d tieu hoe mpt chQng myc ndo dd dd xdy dyng mau dang toan cd Idi giai thdng nhat theo cae bddc eu t h i , mot quy t i c chung n h i t ma chi thUe h i l n theo eac budc hoac cac quy t i c ta thu dugc ldi giai Vdi If dd, bdi v i l t ndy trinh bay mpt sd bai toan nham rdn luyen k h i ndng KQH eho HS tieu hpe vd van de p h i t t r i l n ngdn ngQ toan eho HS tieu hpe qua trinh hinh va phat trien k h i nang KQH Khigapmpt bai toan tUPngtyvdi bai t o i n nay, d i l u can phai cd ddi vdi moi HS la mgt sdng ;^ ndy sinh td tiem thdc Bii t o i n HS dd da gap bao gid chua? Cd the ap dung duoc gi tU bai toan tuong ty? TU dd hay tao co hpi cho HS t u minh tim tdi phuong phdp g i i i toan, trudc dUa hudng dan TU cde dac diem va cde d i u hieu don I I cua cde tri giac va b i l u tupng cd the rdt tinh khdi qudt ciia van de do, vd d ddy la hUdng giai quyet eho mdt loat cac bai todn tuong tu Bii toan {Bai t o i n v l tinh tong nhieu so): Hdy tinh tong cUa ede so sau ddy: Gidi: a Sd doan day thu duoc l i : 35:7 = (doan ddy) Bai toan 2: Cd mdt dogn ddy ddi 35em Ngudi ta cdt dogn ddy dd thdnh cdc dogn ddy nhd cd ddi 7cm Hdi: a NgUdl ta thu dugc bao nhiiu dogn ddy nhd?; b Ngudi ta edn edt bao nhiiu nhdt cdt? 10 + 16+13+10+16+13+10 + 16+13 + 10+16 + d day, HS ed the sii dung nhQng kien thUc ve p h i p chia de gidi 13 + 10+16+13+10+16+13 Gidi: Cdch /.Tong t r i n cd tong (10 + 16+ 13) vay ket q u i tim dupc l i : (10+16 +13) X = 234 b GV CO the hudng dan HS giii nhusau: N l u thu dupe doan ddy nhd thi can cat bao 29 iKliNTHffcl nhilu nhit? (vdi md hinh b i n , s6 n h i t eit Id nhdt eit v d i = - ) — I — Nlu thu dupc do^n d i y nhd thi e i n eit bao nhilu nhit? (bing md hinh b i n , s6 n h i t c i t l i —]—|—• nhitvi2 = 3-l) N l u thu dupe doan d i y nhd thi c i n c i t bao nhilu nhdt? (theo md hinh b i n , s6 n h i t c i t l i vd = 4-1) —I—I—I— Nlu thu duoc do^n d i y nhd th) c i n cdt bao nhilu nhdt? (cOng vdi gidi phdp md hinh sd nhdt c i t Id = - ) —I—I—I—I— Nhu v^y, vdi doan ddy ddi 35em, eit thdnh do?n nhd, mdi doan ddi 7cm, ta c i n c i t t i t ed nhdt D i n ddy GV eho HS nhin nhdn lai bdi todn t r i n vd dua v i n d l n h u Nlu sd do^n d i y nhd thu dupe l i ddng k l (50 doan, 100 do?n ) thi vl§c gidi quylt bdi todn b i n g md hinh nhu t r i n cdn g i i trj hQu hi|u hay khdng? Cd giii phdp ndo khde khdng? GV y l u eau HS nhin nhdn bdi todn t r i n mdt l i n nUa vd y l u eiu HS trd ldi ede edu hdi nhu Hay so sdnh s6 nhdt eat vd sd doan day cd dupc sau cde l i n eit? TU sd doan ddy Idm the ndo d l ed sd nhdt eit? TQ sd n h i t eit Idm the ndo de cd sd doan day thu dupe sau cdc l i n cit? Vilt menh di b i l u thj mdi quan h i giQa sd doan d i y vd sd nhdt eit? Cde edu hdi t r i n dupe GV lap di lap lai cd chd y hudng HS v i o cic d i l m gidng cCia cdc trudng hpp riing le Lde hoat ddng so sdnh chic chin se d i l n dau cua mdi HS hudng tdi tim nhQng dgediim gidng nhau, tiie Id hdnh ddng KQH dUpc thyc hiln Khi dd viec dua mpt quy t i c chung n h i t d l g i i i quylt dang bdi todn t r i n Id can thiet vd chinh xic, dd la: sd nhdt edt bdng sd dogn ddy cd dugc bdt di I NhU vdy KQH p h i i b i t nguon tU hoat ddng so sinh dac d i l m eua eie trudng hpp riing da xem xet dya t r i n ket q u i so sinh, quan sdt phdt h i l n cdi chung cua ede trudng hpp cy the da thyc h i l n Giii: Bing md hinh, HS quan s i t vd thyc h i l n p h i p d i m ta cd d i p sd (3 hinh tam giic) NhUng n l u sd do?n t h i n g tCr m^t dinh xudng mdt canh ddi dl$n md tdng I l n d i n g k l (7 doan, 10 doan, 100 do?n, ) thi vl^e xdc dinh sd tam gide cCia hinh t r i n Id mdt v i n d l khdng d l g i i i q u y l t Ta xem x l t vdi ede trudng hpp ey t h i cd t^o duoc edeh giii hQu hl$u khdng: N l u sd do^n t h i n g l i 2, sd tam gide Id tam gide vd ta cd = + + (thuc hi$n p h i p so sdnh vdi bilu thUe + + 3) N l u sd doan t h i n g Id 3, sd tam gide Id 10 tam giic vd taed 10= + + + { t h y c h i § n p h l p s o s d n h v d i b i l u thQc + + + 4) Ciing vdi sy so sdnh thi d ddy ed sy p h i n tieh giQa mdi tuong quan sd doan thing vd b i l u thUe l i tdng cua eie sd t y nhiln l i l n t i l p khde khdng VI v i y ed t h i d i n d i n : Nlu sd do^n t h i n g l i 4, sd tam gide tao thdnh se Id + + + + = 15 tam gide Khi dd vdi n doan t h i n g tU mdt dinh d i n mdt canh ddi d i l n thi sd tam giic thu dupc s e l i + + + + + + n.Din day bdi todn trd v l bdi todn tinh tdng eua day sd t y nhiln Jiln t i l p m i HS da cd len' giii B i i toan 4: Viit so thich hgp vdo chd chdm vd gidi thich cdch viit: a 1,3, 5, , b 1,2,3, 5,8 , e.3,7,11,15, , O l giii bdi todn niy, HS p h i i n l u dupc b i n chit (cdc quy luit) cda ede day sd a HS cd t h i nh§n t h i y d i y l i day so I I cich d i u , hoac sd hang l i l n sau hon so h^ng l i l n trudc nd don % Dya vdo k i t ludn bdn chat dd HS se thu duoc Idi gidi b.TQl v i Idm t h i ndo d l cd 3? + = 3.TQ2vd3 Idm the n i o d l cd 5? + = TQ va Idm t h i ndo d l ed 8? + = Bai t o i n 3: Cho tam gide ABC TCr dlnh A ki m^t dogn thdng xuong cgnh BC (hinh ve) Hdi ta cd bao nhiiu hinh Vdy, b i t d i u tU sd t h d ba trd di moi sd nhu the nao vdi tam gide? hai sd liln trUdc nd? HS nhin nhdn ede b i l u thde tren theo mdi quan h i tryc t i l p , nhan dae diem ehung n h i t se thu dUpc ket l u i n la quy luat cda day so, Id: b i t d i u tU sd thU ba, mdi so b i n g tdng hai sd lien trUde n6.ThUc hien theo quy t i c ta thu dUpe Idi giii c Ta cd - = Viy cieh may dPn vj? t h i m may thi ed 7? bdt di m i y thi cd 3? V d i v d 11,ta cung c o l l - = 4.Vdy 7cdch 11 miy 30 BAN TIN GlAO OgCTtlXAfi TAI CHlJC I TiMHiEuwiJMinay:^ dPn vi? t h i m may thi ed 11 ? 11 bdt di m i y thi ed 7? V|y day so tren Id d i y so each d i u miy? Bing mdt loat cae so sinh va n h i n dien cic die trung co b i n cCia cic sd (mdi lien h§ giu^ hai sd hang lien kl) day sd t r i n , hoat ddng KQH dUpc thyc hiln dua quy ludt eCia day s6: Day sd edeh d i u ddn vj HS thUe hien theo quy ludt ta cd ldi giii cda bai t o i n Moi mpt quy tic, tfnh chit todn hpc la k i t q u i cda hoat ddng KQH m i thdng thUdng ehung dupc bilu dien bdi mdt menh d l hay mdt cdng thQc, hdm ehUa nhQng die diem ehung n h i t cCia nhQng trUdng hpp riing dd dupc xem x l t trUde dd Ndi eich khic, q u i trinh KQH se dan d i n vile p h i t hiln va p h i t bilu quy tic, tinh chat dd b i n g nhieu hinh thUe khde mdt each sinh ddng \fiec khuyin khich HS phdt b i l u nhu vay la mdt khau cQa qud trinh day hoe ede quy tie, tinh chat can t h i l t cho viec hinh thdnh vd ren Iuyln ndng KQH cho HS, ddng thdi edn tao nhQng tinh hudng tdt d l thye hiln vile rIn Iuyln va phdt triln ngdn ngQ ndi ehung vd ngdn ngQ todn hpe ndi riing Trong giii quylt mdt v i n d l hpc tap, HS edn g i p nhOfng khd khdn v l hdnh ddng vd Idi ndi, nhdng khd khin t r i n phdOng diln hdnh ddng thi HS cd t h i vUpt qua, song t r i n phdong dl^n ldi ndi v i n cdn nhQng tdn tai cie em edn khd khdn v i n d l d i l n dat lai suy nghT cda minh Mdt nhQng y l u c i u ein dat dype sau mdi t i l t hoe Id gidp HS sUdung cic ngdn tU p h i t bilu nhQng k i t luin rut tQ bdi hoc hay k i t q u i eda KQH d dd tudi HS t i l u hoc, sU KQH duoc t i l n hdnh d blnh d i l n b i l u tupng Ndi dung cua nd mdc du van mang tinh c h i t b l ngodi, c i m tinh v l cdc tinh chat eua sU vdt nhung da ed t h i d i m b i o tuong ddi day du vd chlnh xdc vi?e sd dung KQH thyc hien cic bai t i p khdc nhau, ddi hdi phdi nhdn dang, phdn loai v i he thdng hda cie sy vdt O SdSS- 12/2012 31 ... khde mdt each sinh ddng \fiec khuyin khich HS phdt b i l u nhu vay la mdt khau cQa qud trinh day hoe ede quy tie, tinh chat can t h i l t cho viec hinh thdnh vd ren Iuyln ndng KQH cho HS, ddng... bdt di I NhU vdy KQH p h i i b i t nguon tU hoat ddng so sinh dac d i l m eua eie trudng hpp riing da xem xet dya t r i n ket q u i so sinh, quan sdt phdt h i l n cdi chung cua ede trudng hpp... 11 ? 11 bdt di m i y thi ed 7? V|y day so tren Id d i y so each d i u miy? Bing mdt loat cae so sinh va n h i n dien cic die trung co b i n cCia cic sd (mdi lien h§ giu^ hai sd hang lien kl) day