Nhm bin san lp 10 Tan 1

122131xOCABMPN Gio vin hng dn: KIM SN Nhm bin san lp 10 Tan 2

Trong hot ng ca mnh, con ngi lun lun i mt vi mt cu hi tm gi tr cc i hoc cc tiu ca mt i tng hnh hc no v di, din tch, b mt hoc th tch, Ngay trong t nhin, nhng hnh c dng u, chng mang nhng tnh cht rt c bit, trong n cha n nhng tnh cht cc tr m cc hnh khc khng c c nh tam gic u, hnh vung, lc gic u hoc hnh trn, khi cu,. Ngy nay nhng bi ton cc tr vn c quan tm v nghin cu. Nhng phng php gii v cc dng bi tp ny trong hnh hc rt c trng v bt ngun t l thuyt c bn ca ton hc. ta, nhng loi sch tng kt li nhng bi ton cc tr trong hnh hc cn him, nht l khng h thng phng php gii v a ra mt cch nhn mi trong hc tp, rt nhiu cun bi tp ch mang tnh cht lit k khng lm ni bt nhng tng ca ton v cc phng php tip cn gii ton Thi gian qua, nh s hng dn ca gio vin b mn, chng em xin gii thiu chuyn Mt s phng php tm cc tr trong hnh hc. Chuyn ny ch gii thiu v mt s phng php tm cc tr c bn thng gp trong hnh hc phng v hnh hc vect. Trong mi phng php s c cc v d minh ha. V cui cng l phn bi tp tng hp vi cc bi tp gii bng nhng phng php khc nhau.Trong qu trnh bin son, su tm v tp hp cc phng php cng nhng v d, bi tp, tuy chng em c gng rt nhiu nhng thiu st l iu kh trnh. V vy, chng em mong thy v cc bn thng cm. Xin chn thnh cm n! Nhm bin tp Phm Ngc Xun o Nguyn Th M Huyn V Th Dim Ph L Th Thu Tho Nguyn Hang Anh Th Trng Thanh Th Nhm bin san lp 10 Tan 3 Tm tt kin thc : 1)Cc tr hnh hc : Cho biu thc f ph thuc im X bin thin trn min D. Ta ni :

0 0( ) ,: ( )max ( )f X M X DX D f X Mf X Ms e- e ==

0 0( ) ,: ( )min ( )f X m X DX D f X mf X m> e- e == 2)Php ton vector:Php cng vector:* Quy tc 3 im:AB BC AC + =uuur uuur uuur * Quy tc hnh bnh hnh: nu ABCD l hnh bnh hnh thAB AD AC + =uuur uuur uuur Php tr vector: * Quy tc:AC AB BC =uuur uuur uuur Tch vector vi 1 s:Cho s k 0 v0 a =r r. Tch vector a vi s k l mt vector k hiukar, cng hng vector a nu k > 0 v ngc hng vector a nu k < 0 v c di bng| || | k ar Tch v hng ca hai vector :Cho, a br r khc vector 0 . Ta c :. | | . | | .cos( , ) a b a b a b =r r r r r r Mt s k hiu dng trong ti liu , , a b c : di cc cnh BC, CA, AB ca ABC A . , ,a b ch h h : di cc ng cao xut pht t cc nh A, B, C ca ABC A . , ,a b cm m m : di cc ng trung tuyn xut pht t cc nh A, B, C ca ABC A . , R r : bn knh ng trn ngoi tip, ni tipABC A . , ,a b cr r r : bn knh cc ng trn bng tip gc A, B, C ca ABC A . Mt s im c bit trong tam gic im Lemoine: Nhm bin san lp 10 Tan 4 nh ngha: Trn cc cnh BC, CA, AB caABC Aly cc im 1 1 1, , ABCtng ng sao cho 2 2 21 1 12 2 21 1 1, ,AC BA CB b c aC B a AC b B A c= = = (cc ng 1 1 1, , AABBCCl cc ng i trung). Khi cc ng thng1 1 1, , AABBCCng quy ti im L gi l im Lemoine. Tnh cht: Cho ABC A , L l im trong tam gic. Gi H, K, N theo th t l hnh chiu ca L trn BC, CA, AB. Khi L l im Lemoine caABC Akhi v ch khi L l trng tm ca HKN Akhi v ch khi2 2 20 aLA bLB cLC + + =uur uuur uuur r

im Toricelli: ChoABC Ac cc gc u nh hn 0120 . Khi tn ti duy nht im T c tnh cht cng nhn cc cnh BC, CA, AB di cc gc 0120 . im T nh vy gi l im Toricelli caABC A . im Gergone: ng trn ni tipABC Atip xc vi cc cnh BC, CA, AB ln lt ti 1 1 1, , ABC . Khi cc ng thng 1 1 1, , AABBCC ng quy ti im J gi l im Gergone. im Naghen: Cc ng trn bng ca ABC Atip xc vi cc cnh BC, CA, AB ln lt ti 1 1 1, , ABC . Khi cc ng thng 1 1 1, , AABBCC ng quy ti im N gi l im Naghen. Nhm bin san lp 10 Tan 5

Nhm bin san lp 10 Tan 6 Phng php 1: Vn dng quan h gia ng xin v ng vung gc. V d 1.1: Cho hnh vung ABCD. Trn cc cnh AB, BC, CD, DA ta ly theo th t cc im E, F, G, H sao cho AE= BF= CG= DH. Xc nh v tr ca cc im E, F, G, H sao cho t gic EFGH c chu vi nh nht. Gii:HAE= EBF(c-g-c)HE= EF. OCABDEHFG Tng t ta c: HE= EF= FG= GH nn t gic EFGH l hnh thoi. HAE= EBF cn suy ra AHE BEF =Ta li c 0 090 90 AHE AEH nnBEF AEH + = + =Do : 090 HEF = . Nh vy hnh thoi EFGH l hnh vung. Gi O l giao im ca AC v EG. T gic AECG c AE= CG, AE// CG nn l hnh bnh hnh, suy ra O l trung im ca AC v ca EG, do Ol tm ca c hai hnh vung ABCD v EFGH. HOE vung cn: 2 22. . 2 HE OE HE OE = =Chu vi EFGH= 4.HE=4 2 .OE. Do chu vi EFGH nh nht OE nh nht. KOK AB . Theo quan h gia ng vung gc v ng xin: OE > OK( di OK khng i) nn OE= OK E K Do min OE= OK Nh vy, chu vi t gic EFGH nh nht khi v ch khi E, F, G, H l trung im ca AB, BC, CD,DA. Nhn xt v phng php gii: trong cch gii trn c cc bin i tng ng sau: Chu vi EFGH nh nht HE nh nht OE nh nht. Quan h OEs OK (OK khng i) cho php ta xc nh v tr ca im E OE c di nh nht . V d 1.2: Cho on thng AB c di 2a. V v mt pha ca AB cc tia Ax v By vung gc vi AB. Qua trung im M ca AB c hai ng thng thay i lun vung gc vi nhau Nhm bin san lp 10 Tan 7 v ct Ax, By theo th t C, D. Xx nh v tr ca cc im C, D sao cho tam gic MCD c din tch nh nht. Gii: Gi K l giao im ca CM v DB. MAC= MBK(g-c-g)MC= MK. DCK c ng cao DM l trung tuyn nn l tam gic cn, suy ra HDM MDB = . KMH CD . Do M thuc tia phn gic ca gc D nn: MH= MB= a. 1. .2MCDS CD MH = xyaaHKDMABC Do CD> AB= 2a v MH= a nn: 21.2 .2MCDS a a a = = 2.MCDS a CD Ax = Khi 0 045 , 45 AMC BMD = = . Vy2MCDS a = . Cc im C, D c xc nh tr Ax, By sao cho AC= BD= a V d 1.3:Cho tam gic ABC c gc B l gc t, im D di chuyn trn cnh BC. Xc 5nh v tr ca im D sao cho tng cc khong cch t B v t C n ng thng Ad c gi tr ln nht. Gii: Gi S l din tch ABC. Khi D di chuyn trn cnh BC ta c: ABD ACDS S S + = K, BE AD CF AD ta c : 1 1. . . .2 2AD BE ADCF S + =nn BE+ CF = 2SAD Do BE + CF ln nht AD nh nht. Nhm bin san lp 10 Tan 8 FEHABCD ng xin AD nh nht hnh chiu HD nh nht. Ta cHD > HB ( do 090 ABD > ) v HD = HB khi v ch khi DB. Nh vy khi D trng B th tng cc khong cch t B v t C n AD c gi tr ln nht. V d 1.4: Cho hnh bnh hnh ABCD. Qua A v ng thng d khng ct hnh bnh hnh. Gi B, C, D, ln lt l hnh chiu vung gc ca cc im B, C, D trn ng thng d. Xc nh v tr ca ng thng d tng BB + CC + DD c gi tr ln nht. Gii: Gi O lgiao im ca AC v BD. O l hnh chiu vung gc ca O trn d. ' , ' '/ / ' DD dBB d DD BB DDBB l hnh thang. dO'OB'C'D'CABD M' , ' '/ / ' OO dDD d OO DD v O l trung im BD ( ABCD l hnh bnh hnh) Do OO l ng trung bnh ca hnh thang DDBB ' '' ' ' 2. '2BB DDOO BB DD OO+ = + =' , ' '/ / ' OO dCC d OO CC v O l trung im AC.( ABCD l hnh bnh hnh) Do OO l ng trung bnh ca ACC '' ' 2. '2CCOO CC OO = =, ' A dOO d e nn OOs OA. Do BB + CC + DD = 4.OOs 4.OA ( khng i) Du = xy raO Ad vung gc AC ti A. Nhm bin san lp 10 Tan 9 Phng php 2: Quan h gia on thng v ng gp khc V d 2.1: Cho hnh ch nht ABCD v im E thuc cnh AD. Xc nh v tr cc im: F thuc cnh AB, G thuc cnh BC, H thuc cnh CD sao cho t gic EFGH c chu vi nh nht. Gii:Gi I, K, M theo th t l trung im ca EF, EG v GH. AEF vung ti A c AI l trung tuyn AI= 1.2EFKICABDFHEG Tng t MC= 1.2 GH . IK l ng trung bnh ca EFGIK=1.2FG. Tng t KM= 1.2EHDo : chu vi EFGH= EF + FG + GH +HE= 2(AI + IK + KM + MC) Ta li c: AI + IK + KM + MC > AC (so snh di on thng v ng gp khc) Suy ra: chu vi EFGH > 2AC ( khng i). Chu vi EFGH nh nht bng 2AC A, I, K , M, C thng hng. Nhn xt v phng php gii: bng cch v trung im cc cnh EF, GH, v trung im ca ng chp EG, ta tnh c chu vi ca t gic EFGH bng hai ln di ng gp khc AIKMC, di ng gp khc trn nh nht khi ng gp khc tr thnh on thng AC. V d 2.2: Cho tam gic ABC nhn. Dng mt tam gic c chu vi nh nht ni tipABC, tc l c ba nh nm trn ba cnh ca tam gic y. Gii:Cch 1: Xt tam gic MNP ni tip ABC mt cch ty ( M thuc AB, N thuc BC, P thuc AC). V E, F sao cho AB l ng trung trc ca NE, AC l ng trung trc ca NF. Chu vi MNP = NM + MP + PN = EM + MP + PF > EF M Nhm bin san lp 10 Tan 10 12AB CMNPEF Ta cn xt khi no th EF nh nht. Ta c 1 22 2 2 EAF A A BAC = + =EAF l tam gic cn c gc nh khng i nn cnh y nh nht khi v ch khi cnh bn nh nht. EF nh nht AE nh nht AN nh nht AN BC Nh vy chu vi tam gic MNP nh nht khi N l chn ng cao k t A, cn M v P l giao im ca EF vi AB, AC. Ta c nhn xt rng khi N l chn ng cao k t A th M v P cng l chn hai ng cao cn li ca tam gic. CM:Xt HMP: AB l ng phn gic ca gc EMH, AC l ng phn gic ngoi ca gc FPH. Ta c AB, AC gp nhau ti A nn HA l tia phn gic ca gc MHP. V AH HC nn HC l ng phn gic ngoi ti nh H. Theo trn AC l ng phn gic ngoi ti nh P, HC gp AC ti C nn MC l tia phn gic gc trong ti nh M.HABCMPEF MB v MC l cc tia phn gic ca cc gc k b nn MB MC. Tng t PC PB Vy chu vi tam gic MNP nh nht khi M, N, P l chn ba ng cao ca tam gic ABC. Do tam gic ABC nhn nn M, N, P thuc bin ca tam gic. Cch 2: Ly M, N, P ty trn AB, BC, CA v ni tm O ca ng trn ngoi tip tam gic ABC vi M, N, P. Gi R l bn knh ng trn ngoi tip tam gic, S l din tch tam gic. Khi : 1. .2OMBNS OB MN sNhm bin san lp 10 Tan 11 1. .2ONCPS OC NP s 1. .2OPAMS OA PM = 122131xOCABMPN Do OA = OB = OC = R nn 1. .( )2OMBN ONCP OPAMS S S S R MN NP PM = + + s + +Do chu vi 2SMNPR> VXy ra ng thc khi v ch khi OA MP, OB MN,OC NP. Ta s CM rng khi th AN, BP, CM l cc ng cao ca tam gic ABC. Tht vy, gi s OA MP, OB MN, OC NP. K tip tuyn Ax. Ta c 2C M =( cng bng gc BAx). Chng minh tng t 1C M = . Do 1 2M M = suy ra 2 3M M = . Nh vy MA l phn gic ngoi ca tam gic MNP. Tng t PA l ng phn gic ngoi tam gic MNP. Suy ra NA l ng phn gic ca gc MNP. Ta li c 1 2N N =nnNA BC . Chng minh tng t, BP AC CM AB . Tam gic MNP c chu vi nh nht khi v ch khi N, P, M l chn cc ng cao ca tam gic ABC. V d 2.3: Cho tam gic u ABC v trung im M ca AB. Trc tin An chn mt im N trn BC, tip Bnh chn mt im P trn AC. Mc tiu ca An l mun tng d = MN + NP + PM ln nht, cn Bnh mun tng d nh nht. Hi rng nu c hai u c cch chn tt nht th N v P l nhng im no? Gii V cc im D, E sao cho AC l ng trung trc ca MD, BC l ng trung trc ca ME. di ng gp khc DPNE bng d. Nhm bin san lp 10 Tan 12 EDMBPC AN D thy hoc PN + NE < PB + BE hoc PN + NE < PC + CE nn di ca ng gp khc DPNE khng vt qu di ca ng gp khc DPBE hoc di ca ng gp khc DPCE. Vy d ln nht th An phi chn N trng B hoc C. R rng tng d nh nht th Bnh phi chn P l giao im ca ND v AC. B'PDCAB trng NM Nhm bin san lp 10 Tan 13 Trong trng hp An chn N trng B th Bnh chn P l giao im ca BD v AC, khi d = 1d MB BP PM = + + . Cn trong trng hp An chn N trng C thBnh chn P l giao im ca CD v AC, chnh l C, khi d = 22 d MC CM MC = + = . hDC trng N trng PABM By gi ta so snh 1dv 2d . t MC = h th 2d = 2h (1). Qua B k ng thng vung gc vi AC, ct MP B. Ta c BP = BP nn : 1d = MB + Bp + PM = MB + BP + PM = MB +BM > BB = 2h (2) T(1) v (2) suy ra 1 2d d > . Do c hai ngi u chi ti u nn An chn N trng B c tng d ln nht, sau Bnh chn P l giao im ca BD v AC. V d 2.4:Cho hai im A v B nm trong gc nhn xOy. Xc nh im M trn tia Ox, im N trn tia Oy sao cho ng gp khc AMNB c di nh nht.Gii: Nhm bin san lp 10 Tan 14 yxNMB'A'OAB V cc im A, B sao cho Ox l ng trung trc ca AA, Oy l ng trung trc ca BB. di ng gp khc AMNB bng AM + Mn + NB = AM + MN + NB> AB. di ng gp khc nh nht trong trng hp M, N nm trn AB. Nhm bin san lp 10 Tan 15 Phng php 3: p dng bt ng thc trong ng trn tm cc tr V d 3.1:Cho hai im A v B nm trong na mt phng c b l ng thng d, hai im M,N thuc d v d di MN khng i. Xc nh v tr hai im M, N dng gp khc AMNB t gi tr nh nht. Gii: Dng hnh bnh hnh BNMB BB= MN = a (khng i); NB =MB, B c nh. Gi A l im i xng ca A qua ng thng d. Ta c AM =AM, A c nh. dB'A'ABM N Xt ba im A, M, B ta cAM + MB AB Do AM + MN + NB =AM+ MN +MB =( AM+ MB) + MN AB+ a (khng i) Du bng xy ra M[ '; '] A B e V d 3.2: Na ng trn (O;R) ng knh AB. M l im di ng trn na ng trn. Qua M k tip tuyn vi nang trn, gi D,C Ln lt l hnh chiu ca A; B trn tip tuyn y. Xc nh v tr ca im M din tch t gic ABCD c gi tr ln nht. Gii: Ta c: AD DC (gt), BC DC (gt) AD// BCABCD la hinh thang m D= 900 nn ABCD l hnh thang vung, OM DC nn OM // AD v O l trung im AB nn OM l ng trung bnh ca hnh thang ABCD 2AD BCOM+=Do . .2ABCDAD BCS DC OM DC+= =

Nhm bin san lp 10 Tan 16 DCEOA BM V AE BC. T gic ADCE l hnh ch nht( 090 ADC DCE AEC = = = ) DC = EAAEB= 900 E thuc ng trn ng knh AB, AE l dy cung ca ng trn (o) DC s 2R (trong ng trn n knh l dy ln nht) Do 2.2 2ABCDS R R R s =(khng i) Du bng xy ra AE l ng knh ca (O) OM AB M l trung im ca cung AB. V d 3.3: Cho ng trn (O;R) BC l dy cung c nh (BC= 2R). A l dim chuyn ng trn cung ln BC. Xc nh v tr ca A chu vi tam gic ABC ln nht Gii: ABCP= AB + AC + BC (BC khng i) Trn tia i ca tia AB ly D sao cho AD = AC Ta cAABC cn ti A 2 BAC ADC =MBAC khng i ADC khng i. Mt khc BDC khng i, BC c nh D thuc cung cha gc c s o 14 sd BCca (O) dng trn on thng BC. Nhm bin san lp 10 Tan 17 DOC BA ABCP ln nht (AB + DC) max BD max BD l ng knh ca cung cha gc ni trn. Khi BCD= 900. M ABC BDC ACB ACD + = + = 900 BDC ACD = ( AC = AD) Do ABC ACB = AB AC = A l trung im ca cung ln BC Nhm bin san lp 10 Tan 18 Phng php 4: p dng bt ng thc i s tm cc tr V d 4.1: Cho na ng trn (O;R) ng knh AB. M l im chuyn ng trn na ng trn. Xc nh v tr M MA +3MB t gi tr ln nht. Gii: 090 AMB = (gc ni tp chn na ng trn) Tam gic MAB c 090 M =nn theo nh l Pitago ta c: 2 2 2 24 MA MB AB R + = = OA BM p dng bt ng thc 2 2 2 2| | ( )( ) ax by a b x y + s + +Ta c: 2 2 23. | 3. | (1 3)( ) 4.4 4 MA MB MA MB MA MB R R + = + s + + = =3. 4 MA MB R + s hng s. Du = xy ra khi v ch khi3.MA MB = MAB Vl na tam gic u060 sdMA = V d 4.2: Cho tam gic ABC cn (AB = AC). Ly im D trn cnh BC ( D khc B,C ). Gi 1 2, r r ln lt l bn knh ng trn ni tip tam gic ABD v ACD. Xc nh v tr ca D tch 1 2r rt gi tr ln nht. Gii: Gi O l giao im 3 ng phn gic ca tam gic ABC, 1Ol tm ng trn ni tip tam gic ABD, 2Ol tm ng trn ni tip tam gic ACD. D thy 1O OB e , 2O OC e . V 1 2, r r > 0, p dng bt ng thc Cauchy, ta c : 2 21 21 2.2r rr r+s. Du ng thc xy ra khi v ch khi 1 2r r = . Nhm bin san lp 10 Tan 19 MHNKO2ODBCAO1 Khi 1 2O KB O NC = V Vsuy ra BK = CN. Suy tip ra BH = CM. T AH = AM. Vy 1 2AHO AMO = V V . Nn 1 2AO AO = , k 1 2, O I AD O J AD . D thy I trng J v 1 2. O I O J = T KD = DN. Vy D l trung im ca BC th tch 1 2r rt gi tr ln nht. Lc A, O, D thng hng. V d 4.3: Cho on thng BC c nh. A l im di ngsao cho tam gic ABC nhn. AA l ng cao v H l trc tm ca tam gic ABC. Xc dnh v tr A AA.AH t gi tr ln nht GiiXt AABH vAAAC c 0' ' 90 , ' ' BAH AA C A BH AAC = = = ( Hai gc nhn c cnh tng ng vung gc ) A'HABC Do AABH ~ AAAC HA/AC = AB/ AA AA. HA = AB. AC, Ta c :Nhm bin san lp 10 Tan 20 AB,AC= AB(BC - AB)=AB. BC AB2

= 2 2 2 22 2( ' . ) ( ' )4 4 4 2 4BC BC BC BC BCAB BC AB AB + = sVy AA. HA s 24BC(khng i) Du bng xy ra2BC= AB A l trung im BC A thuc trung trc ca BC VAABC nhn nn A nm ngoi ng trn ng knh BC. Nhm bin san lp 10 Tan 21 Phng php 5:ng dng din tch tm cc tr V d 5.1: Hy tm trong tam gic ABC mt im M sao cho tch cc khong cch t M n ba cnh c gi tr ln nht. Gii: Gi x, y, z ln lt l khong cch t m n ba cnh BC, AC, AB; , ,a b ch h h tng ng l ng cao xut php t cc nh A, B, C. Ta c:1ABC MBC MCA MABa b cx x xS S S Sh h h= + + = + + zxyHDEFBCAM Nh vy, cc s , ,a b cx x xh h h c tng khng i, do tch . .a b cx x xh h h ln nht (cng c ngha l x.y.z ln nht) khi v ch khi:13a b cx x xh h h= = =. Khi M l trng tm tam gic ABC. V d 5.2: Cho im M di chuyn trn on thng AB. V cc tam gic u AMC v BMD v mt phaca AB. Xc nh v tr ca M tng din tch hai tam gic u trn l nh nht. Gii:Cch 1: Gi K l giao im AC v BD. Cc tam gic AMC, BMD ng dng vi tam gic AKB. t AM = x, AB = a, 1 2, ,AMC BMD AKBS S S S S S = = =. Ta c:2 2 1 2( ) ; ( )S S x yS a S a= =nn 2 2 2 21 22 2 2( ) 12 2 2S S x y x y aS a a a+ + += > = =Xy ra du ng thc khi v ch khi x = y Do : 1 21min( )2S S S + = M l trung im ca AB Nhm bin san lp 10 Tan 22 12xyDCBAKM Cch 2:Ta c 2 21 23 3,4 4x yS S = = 22 2 21 23 3 ( ) 3( ) .4 4 2 8x yS S x y a++ = + > =21 23min( )8S S a x y + = = M l trung im ca AB. V d 5.3: Cho hnh vung ABCD c cnh bng 12 cm, E l trung im ca CD, im F thuc cnh BC sao cho CF = 4 cm. Cc im G v H theo th t di chuyn trn cc cnh AB v AD sao cho GH // EF. Xc nh v tr ca im G sao cho t gic EFGH c din tch ln nht. Tnh din tch ln nht . Gii:t EFGHS S = , BG = x v k hiu nh hnh v. AGH V ng dngCEF V12 2(12 )4 6 3AH AG AH x xAHCF CE = = =2(12 ) 12 2123 3x xDH + = =1 2 3 4 ABCDS S S S S S = =144-21 2. .(12 ) 4 12 (12 2 )2 3x x x +=21144 (144 24 ) 4 12 12 23x x x x + =21144 48 8 6 243x x x + =2 21 12 72 ( 3) 75 753 3x x x + + = + sNhm bin san lp 10 Tan 23 x3241EBADCFHG maxS = 75 khi v ch khi x = 3 Din tch ln nht ca t gic EFGH l 752cmvi BG = 3cm. V d 5.4: Cho hnh vung ABCD c AB = 6m, im E nm trn cnh AB sao cho AE = 2m. Xc nh v tr im F trn cnh BC sao cho hnh thang EFGH ( G thuc cnh CD, H thuc cnh AB v EH // GF // BD) c din tch ln nht. Tnh din tch ln nht . Gii: t BF = x, EFGHS S =.4x6-x6-xx422BAD CEHFG Ta c: ABCD AEH BEF CFG DGHS S S S S S = 22 2.36 4 4 (6 ) 4 S x x x = 272 4 4 36 12 4 x x x x = + 2 24 32 ( 2) 36 x x x = + + = +MaxS = 18 khi v ch khi x = 2 Vy BF = 2m. Khi 218EFGHS m =