PHNG PHP HC TPNHANH GII TON CN LM VIC KHCTi sao ni : Nhanh gii ton cn lm vic khc?Chng ta cng bit ch mt s lng % rt nh nhng ngi yu ton theo con ng ton hc vi am m v coi l mc ch cuc i. Hu ht chng ta chn ng khc. Vic khc y cn hiu theo ngha rng : Bao nhiu lnh vc ph hp hn cho chng ta la chn. Nhng ai cng bit ton v suy ngh Ton hc l mt cng c khng th thiu chim lnh hu ht cc nghnh khoa hc v kinh t khc, m ra tng lai y v nhiu mu sc hn. T khc y hiu theo ngha gii cng c ton lm tt cng vic chnh ca mi ngi la chn.Mun hc v lm ton hiu qu khng th ly chm ch b thng minh, phi lun lun sng to v mnh dn tm cch khc! Thng qua mt bi ton vi 3 cch gii khc nhau, chng ta tm hiu s chuyn bin t thay i chin thut gii ton sang kin to chin lc lm ton. Nu chng ta lun c phong cch gii v nhn bi ton t nhiu gc , cng sc lao ng Ton hc ca chng ta s hiu qu hn v nhiu kt qu khng ng xut hin.Bi ton: Cc im P, Q, R ln lt l cc im gia ca cc cnh BC, CA, AB cu tam gic ABC. B ba im L, M, N v L,M, N cng xoay theo chiu thun sao cho L i xng vi L qua P v nm trn BC, M i xng vi M qua Q v nm trn CA, N i xng vi N qua P v nm trn AB ( hnh v). Chng minh rng din tch hai tam gic LMN v LMN bng nhau.

Chng minh. K hiu x=BL= CL, y=CM=AM , z=AN=BN; dt(XYZ)= din tch tam gic XYZ .Cch 1 (p n ly t mt thi HSG nc ngoi 2009 - 2010) dt(LMN) = dt(ABC)-dt(AMN)-dt(BNL)-dt(CLM) Tng t ta c: dt(LMN) = dt(ABC)-dt(AMN)-dt(BNL)-dt(CLM)Do vic chng minh dt(LMN)=dt(LMN) tng ng vi chng minhdt(AMN)+dt(BNL)+dt(CLM) = dt(AMN)+dt(BNL)+dt(CLM) (1) p dng nh l Sin bin i c hai v ca (1) ta c: z(b-y)sin + x(c-z)sin + y(a-x)sin y(c-z)sin +z(a-x)sin + x(b-y)sin (2)ph ngoc v n gin c hai v ta c (2) tng ng vi zb.sin + xc.sin + ya.sin yc.sin +za.sin + xb.sin (3)S dng kt qu: 2T=ab. sin=bc.sin=ca.sin ta nhn c: z. + x. + y. = y. + z. + x. ( dpcm).

Cch 2. (Da trn li gii ca Nguyn L Phc )Ta s dng mt b n gin: Hai tam gic ABC (cnh a, b, c ) v ABC ( cnh a, b, c) c gc A bng gc A. Khi = .

Chng minh iu ny ch cn s dng tnh cht ng dng ca hai hnh tam gic vung ABH v ABH. Quay tr li chng minh bi ton ban u. K hiu u=x/a, v=y/b, w=z/c. = w(1-v) + v(1-u) + u(1-w) = u + v + w vw uv uw Tng t c: =(1-w)v + (1-v)u + (1-u)w = u + v + w vw uv uwsuy ra dt(LMN) = dt(LMN) (pcm) Cch 3. Ta chng minh dt(LMN)=dt(LMN) tng ng vi dt(PMN)=dt(PMN) ( y cng chnh l trng hp M v M trng nhau v trng vi P) (4)

Tht vy: dt(LMN) dt(PMN) = dt(PLN) dt(PLM) = dt(LMN) dt(PMN) = dt(PLN) dt(PLM) = PL.d/2 = Qua lp lun (4) suy ra ch cn chng minh trong cc trng hp N trng vi N, M trng vi M v trng vi im gia tng ng ca cc cnh. Khi ny tam gic LMN v tam gic LMN trng nhau. Hin nhin pcm.Nhn xt : Chng ta va tham kho ba li gii tng i c trng ca bi ton.- Vi li gii th nht bi ton xng ng c c mt trong mt cuc thi hc sinh gii cp THPT. V l bi luyn tp tt cho cc lp chuyn.- Chuyn sang li gii 2 bi ton c a v cho cc bn hc sinh cp THCS, thm ch li gii cn n gin v thanh thot hn ca li gii u tin.- n li gii th 3 vi mt k nng bt ng: bi ton c chuyn v khng phi gii v hin nhin v ch cn k nng lp 6 - 7.S hnh thnh v pht trin ca h thng cc cch gii ca bi ton ny phn nh tng i y tinh thn v ni dung ca phng php. Chng ta tm gi l: PHNG PHP 6D - c th gi gn bng khu hiu: n gin hn na v hy nhn theo cch khc. Cng vic n gin hn na chnh l ng lc, a cc vn l thuyt n mi trng ng dng - cng vic sng cn ca cng ngh trong thi bui cnh tranh. Vic hy nhn theo nhng cch khc nghe th hay nhng cha bit phi lm sao? Lnh vc ny i hi mt s u t tr tu nghim tc ca cc nh khoa hc v gio dc, ca cc thy c gio hng ngy bn cnh hc sinh. Nhng cng trnh khoa hc sng gi c mt c im chung l: cch tip cn chng ging ai v v th kt qu c bit him hoi - chng ging ai - ch c th ny sinh t nhng ci u dm nhn bng cch khc. Mun c tc phong hc tp sng to ny th tt nht l c luyn tp pht huy ngay t nh. Phi chng y cng l cha kha th h tr c th ui kp cc dn tc tin tin nhanh nht v bn vng nht. S thay i c l phi bt u t mt nn gio dc cht lng.Mt v d c th hi t, nhng cng nu mt hng i tinh gin:n gin hn na v hy nhn theo cch khc.Thnh cng ca mi ngi s mang li ngh ca phng php ny!