NGHIEN OAJ &U>JG DUNG KHfll THQC SiKU GlflO KHOfl TRONG DAY HOC BANG NHQN 2 TORN LOP 2 Sdch ^ido khoa (SGK) Id thiet bj dgy hgc quan trgng, dugc Bg Gido dye Ddo tgo ban hdnh vd cai cdch thudng xuyen,[.]
NGHIEN OAJ &U>JG DUNG KHfll THQC SiKU GlflO KHOfl TRONG DAY HOC BANG NHQN TORN LOP T h S P h a m Thj Kim Chdu Tru&ng Dgi hgc Dong Thdp S dch ^ido khoa (SGK) Id thiet bj dgy hgc quan trgng, dugc Bg Gido dye - Ddo tgo ban hdnh vd cai cdch thudng xuyen, klii dgy hgc GV phdi tudn theo SGK Todn ticu hgc cung cap cdc kidn thiic co ban thdng qua kenh chO' vd kenh hinh Ngoai kien thirc d kenh chO dudi dang cdc cdng thiic, qui tdc, , thi cdc kidn thiic todn cdn dugc the hign d kenh hinh rdt da dang, cd tinh tryc quan minh hga cao, ldm cho kidn thiic gidm trim tugng vd HS dd hieu dd tiep thu, kdnh hinh dugc xem Id diing day hgc hidu qua SGK ggi md cho GV cac y tudng de dinh hudng dung frong day hgc, chang hgn bdi hgc bdng nhan toan ldp 2, SGK trinh bdy nhu sau: (khi to chirc dgy hgc, GV thudng ggi hinh dnh ] • • Id tam bia cd cham trdn) Vdi cdch trinh bdy nhu the, GV can khai thdc nhu the ndo vd td chirc Idp hgc de ddm bdo hinh thdnh cho HS cdch xay dyng bdng bdng cdch Ihem (ket qud mOt cdng thirc dugc nham nhanh bdng cdch them vao ket qua cdng thiic lien trudc) Tir trudc ddn nay, hau het GV tidu hgc td chiic cho HS xdy dung bang nhdn ma van khdng sii dyng quy tdc them 2, cu the nhu sau: "GV ldy 01 tdm bia, hdi tdm bia duac ldy mdy ldn? (01 ldn) Ta co phep nhdn ndo? (phep nhdn x / Co tdt dugc l^y 11ln, ta vi^t: 2x1=2 2x1=2 2x2 = 2x3 = dugc lAy ldn, ta cd: 2x4 = 2x2=2+2=4 2x5 = V§y: X = 2x6 = 2x7 = du-gc ldy ldn, ta c6; 2x8 = 2x3=2+2+2=6 2x9 = Vdy : X = 2x10 = Ngdy nhgn bdi 29/8/2012, Ngdy duy?t ddng 25/10/2012 • TAP CHI THIET BI GIAO DUC-SO 87-11/2012 cd mdy chdm Iron? (dem chdm trdn) Vgy la co 2x1=? ^xl=2; GV lay 02 tdm bia, hdi tdm bia dug^c lay may ldn? (02 ldn) Ta co phep nhdn ndo? (phep nhdn 2x2/ Co tdt cd mdy chdm trdn? (dim chdm Iron) Vdy ta cd 2x2=? (2x2=4/ Tuang tu cho phep nhdn 2x3, vd tuang tu cho cdc phep nhdn cdn lgi bdng Den cdng thiic 2x10, GV phdi lay 10 tdm bia vd HS phdi dem tdt cd cdc chdm trdn tren 10 tdm bia" Vdi each DH nhu the, sau xdy dung xong mgt cdng thiic thi cat hdt cdc tam bia, mudn xay dung cdng thiic tiep theo GV phdi bdt dau Idy Iai cdc tdm bia tu ddu Trong bdng nhdn cd 10 cdng thiic thi phdi tidn hanh 10 quy trinh xay dyng dgc lgp Tidt day rat cdng kenh, mat nhidu thdi gian cho thao tac lay vd cat cac tdm bia, ddng thdi GV mat nhidu thdi gian chudn bi den 10 tdm bia Vigc xay dyng cdng thiic lien tidp hoan toan khdng cd mdi lidn he, vi§c tim kdt qud moi cdng thiic HS phdi ddm tii chdm frdn thut nhdt den chdm frdn cudi cung Vd mdt hinh thiic vdn hoan thdnh bdng nhan nhung ve bdn chat chua thd hidn quy tdc thdm Til cdng thiic 2x1 din cdng thiic 2x10 ddu phu thudc wsBmmmu vao cdc tam bia Van de ddt la: tir cdng thiic 2x4 frd di ndu ta khdng minh hga tren cdc tdm bia, vdy ta tim kdt qud 2x4 bdng cdch nao? Didu dd cho thdy cdch khai thac cac tam bia nhu trdn la chua hgp Iy, vdy chiing ta cdn nghien ciiu khai thac cac tdm bia nhu thd ndo cho hidu qud? Mgt bidn phap hidu qua md chung tdi dd cap d day Id: vdi cdng thlie ddu tien ciia bdng nhan (2x1, 2x2, 2x3), mdi cdng thirc ta chi can thao tac "Lay them mgt tdm bia" chir khdng nen "lay lgi tir dau cac tam bia" Vide lay them mdt tam bia se the hidn dugc y nghia cua viec chuyen tir phep nhan sang phep cdng vd the hien quy tac thdm 2, the hidn sy thdng nhat giua true quan "lay them" va su trim tugng "cdng them" Tu cdng thiic 2x4 frd di, chiing ta van dyng quy tdc thdm de xay dung Theo cdch ndy, GV chi can chudn hi 03 tdm bia, cy the: "GV ldy 01 tdm bia, hdi tdm bia dugc lay mdy ldn (mgt ldn) Ta co phep nhdn ndo? (phep nhdn x / Cd tdt cd mdy chdm trdn? (2 chdm trdn) Vdy ta CO 2x1=^1(2x1=2) GV lay them 01 tdm bia, hoi cdc tdm bia dugc ldy mdy ldn (hai ldn) Ta co phep nhdn ndo? (phep nhdn x / Cd tat cd mdy chdm trdn? (2 chdm trdn them chdm trdn ta cd chdm trdn) Vgy ta co 2x2=7 (2x2=2+2=4/ GV /4v them 01 tdm bia, hoi cdc tdm bia dugc ldy mdy idn (ba ldn) Ta cdphep nhdn ndo? (phep nhdn x / Co tdt cd mdy chdm Iron? (4 chdm trdn them chdm trdn la co chdm Iron) Vdy la co 2x3=? (2x3=2+2+2=4+2=6j Yeu cdu HS nhdn xel moi lien he ve kit qud giira cong thiic 2x1 vd 2x2, 2x2 vd 2x3? (kit qud cdng thirc 2x2 bdng kit qud cong thdc 2x them 2, kit qud cong thuc 2x3 bdng kit qud cong thirc 2x2 thim cdch khac thac nd nhu the nao cho higu qud vd chinh xdc SGK Todn Idp ludn frinh bay theo hudng md, tao didu kign cho GV tiep cgn theo nhieu cdch Ld GV chiing ta can nghien ciiu khai thdc SGK nhu the ndo cho dung vdi y nghTa ciia kien thiic, diing vdi y dd ciia tdc gid va ddm bao myc tieu dgy hgc de Sy khai thac triet de kien thiic d Vdy cdc em: Khi dd co kenh chQ va kenh hinh Iam 2x3=6, ta CO the tim dugc kit cho boat dgng nhgn thiic kien qud cdn^ thiic 2x4 bdng cdch thiic todn trd nen tryc quan ndo? (bdng cdch them vdo sinh ddng hem, giup HS kham ket qud cdng thiic x / M&i phd kien thiic dl dang ban, mgt bgn len viet (2x4=8/ luu giii kidn thirc ki ire Pay 2x5=? (2x5=10/ Em bdn chat hon, chdc chan ban tinh bang cdch ndo? (em them vdo kit qud cong thicc 2x4) Tdi ii^u tham khao Yeu cdu HS tu xdy dung hit Vu Qudc Chung (chu cdc cong thiic lgi bien) Phuang phdp dgy hgc bdng " Todn tiiu hgc NXB Gido dye Theo each ndy, cd cdng thuc ta cd the tim ket qua 2007 Trdn Bd Hodnh (chu cua cdng thiic kd tidp bang cdch cdng them Nhu vay bien) Ap dung day vd hgc nhung cdng thiic cdn Iai tich cue frong mdn Toan hgc bang HS de ddng xdy dung NXB Dai hgc Su phgm Hd tuang tu vd rat dd hgc thudc Ndi 2003 bang Khdng nhung thd, thay Bd Gido dye va Ddo vi phdi ehuan bi 10 tdm bia thi tao Toan (SGK, SGV) ta chi cdn 03 tdm bia NXB Giao due 2011 Kidn thlie Toan tieu hgc khdng nhidu khdng khd, day Summary tidu hgc khdng phdi nhdi cho Grade Mathematics HS ddy kidn thiic ma can day textbooks are presented in cdc em cdch hgc, cdch tu duy, the open, enabling teachers cdch khdm pha kien thiic Dgy to access in many ways With cdch hgc cho HS cang nhd cdng khd, ddi hdi GV phdi Table 2, the teacher should hidu tam ly HS vd hidu ngi be utilized textbooks how to dung chuang trinh Khi hidu form for buildmg two tables rd ndi dung chuang trinh, chi by adding two rules for the cdn nhin vdo SGK, GV se bidt pupils TAP CHI THIFT BI GIAO DUG - SO 87 -11/2012 • 35 ... co 2x3=? (2x3 =2+ 2 +2= 4 +2= 6j Yeu cdu HS nhdn xel moi lien he ve kit qud giira cong thiic 2x1 vd 2x2, 2x2 vd 2x3? (kit qud cdng thirc 2x2 bdng kit qud cong thdc 2x them 2, kit qud cong thuc 2x3... nhdn x / Cd tat cd mdy chdm trdn? (2 chdm trdn them chdm trdn ta cd chdm trdn) Vgy ta co 2x2=7 (2x2 =2+ 2=4/ GV /4v them 01 tdm bia, hoi cdc tdm bia dugc ldy mdy idn (ba ldn) Ta cdphep nhdn ndo?... chdm trdn? (2 chdm trdn) Vdy ta CO 2x1=^1(2x1 =2) GV lay them 01 tdm bia, hoi cdc tdm bia dugc ldy mdy ldn (hai ldn) Ta co phep nhdn ndo? (phep nhdn x / Cd tat cd mdy chdm trdn? (2 chdm trdn them