Chuyen de he phuong trinh

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  • 1. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngCHUYN TON PH THNG S DNG CC BT NG THC C BN GII H PHNG TRNHD huT Thn Trn Tong oHNgc Ng Hong Ton i hc Y Dc Cn Th 1

2. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngLI NI U Trong th gii ton hc,cc mng kin thc lun c mi quan h hu c vi nhau. Nh ton hcRen Descartes i s ho hnh hc khi to ra th gii hnh gii tch . C th ni khi ta quan tmn mt vn no trong ton m li lng qun i cc lnh vc khc th tht l iu y ticDni,mun thnh cng phi bit chim nghim v hc hi nhiu iu,gia nhng khong khng gianbao la lun tn ti tnh yu p.Bt ng thc c xem nh l ti hp dn thu ht s quan tm ca cc ton th trn cc dinn,s huyn b ca hai cp du lun thch thc tr c v t duy ca ngi gii ton. Chng hunhng th,phn mn ny lun l cu nh cao trong cc thi hc sinh gii,Olympic v tuynsinh i hc. Cht nh n thi i hc khi A nm 2012 va qua l cu khng ch im ca huTht th sinh,nh th cho ta thy mc kh ca lp bi ton ny lun cao.Nu xem bt ng thc nh "ng hong" trong ton ph thng th h phng trnh nh mtc gi chn qu ht hn bao nhiu g si tnh,nhng chng th sn mi m tm v p ca nng,khg ca tri tim nng mong n lc no tm ra chn l (x; y) u ? Ta li cht nhn ra nuThcu bt ng thc trong i hc khng c hay khng qu kh th nng li vng bc kiu sa lmcho bao s t au u i chinh phc. c l l tnh yu p c cht nh nhng nhng lm phongba.S xa cch ca ni thnh tp np v chn thn qu bnh d y,c lc no li gp nhau ni ndng sng h hn ny,ni tnh cm nng m ca nhng i trai gi yu nhau.Tc gi xin lm con nh a l khch sang sng,ni nhp i b li vi nhau qua chuyn TrS DNG CC BT NG THC C BN GII H PHNG TRNHCho bi bit tham gia vo "Tuyn tp cc chuyn n thi i hc ca din n nwww.k2pi.net 1 " gp phn a cc th sinh ang c cht phn vn v cch hc v n tp tonnh th no l hiu qu thng tin vo cnh cng i hc pha xa kia c mt ti liu n tp btr c hai mng h phng trnh v bt ng thc,ng thi cng mun gi n c gi yu toncht gia v yu thng d nh b ny.iu c bit tc gi mun gi tng chuyn ny n ngi con gi m tc gi yu thngmang tn Trn Th Thu Dng Lp Dc B Kho 38 i Hc Y Dc Cn Th nh th hin tnh Tocm ca chnh bn thn mnh.Do khng phi theo nghip cm phn v kin thc ton cn hn hp nn chc hn sai st l iukhng th trnh khi c,rt mong nhn c s gp ca qu bn c qua a ch Ng Hong TonngLp YD1 kho 38 Trng i hc Y Dc Cn Th hoc email:Ngohoangtoan1994@gmail.com.Rt mong nhn c s quan tm ca cc bn ln vit chuyn sau c hon thin hn.Thn mn! oCn Th, ngy 01 thng 01 nm 2013H Ng Hong Ton1Tuyn tp d kin s tng hp v bin son cc bi vit di dng cc bi ging,chuyn thnhNgmt ebook hon chnh.D kin s hon thnh trong thng 3 nm 2013.Mi thc mc v ng k thamgia chuyn xin lin h trc tip ti forum:www.k2pi.net.2TY D 3. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngPhn 1:KIN THC CHUN B MT S BT NG THC THNG DNG Bt ng thc AM-GM:DCho a1 , a2 , ..., an l cc s thc khng m th ta c: a1 + a2 + ... + an n n a1 a2 ...anng thc xy ra khi v ch khi a1 = a2 = ... = an . huTuy nhin,khi gii ton ta hay quan tm nhiu n trng hp v .M ta thng c bitn di pht biu:T 1. Cho a, b 0 .Khi ta c:a + b 2 ab .ng thc xy ra khi v ch khi: a = b. Bt ng thc ny cn c vit di dng khc tng ng l:2Tha+b(a) ab2(b) (a + b)2 4ab (c) a2 + b2 2ab(a + b)2n2 2 (d) a + b 22. Cho a, b, c 0 Khi ta c: a + b + c 3 3 abc .ng thc xy ra khi v ch khi a = b = c. Tr Bt ng thc ny cn c mt s ng dng khc kh ph bin nh sau: Vi mi s thc a, b, c ta lun c:(a) a2 + b2 + c2 ab + bc + ca n (a + b + c)2(b) a2 + b2 + c2 3 2(c) (a + b + c) 3 (ab + bc + ca)(d) a2 b2 + b2 c2 + c2 a2 abc (a + b + c)(e) (ab + bc + ca)2 3abc (a + b + c) To Bt ng thc Cauchy-Schwarz:Vi hai b s thc ty a1 , a2 , ..., an vb1 , b2 , ..., bn ta c :(a2 + a2 + ... + a2 )(b2 + b2 + ... + b2 ) (a1 b1 + a2 b2 + ... + an bn )21 2 n 12na1a2 anng thc xy ra khi v ch khi = = ... = .ngb1b2 bnBt ng thc Cauchy-Schwarz dng EngelGi s a1 , a2 , ..., an l cc s thc bt k v b1 , b2 , ..., bn l cc s thc dng . Khi ta lun o a1 2 a2 2an 2(a1 + a2 + ... + an )2c :++ ... + b1b2 bnb1 + b2 + ... + ba1a2 anng thc xy ra khi v ch khi = = ... =Hb1b2 bnTuy nhin,khi gii ton ta hay quan tm nhiu n trng hp n = 2 v n = 3 .Khi ta gp mt s nh gi quen thuc sau:Cho a, b, c > 0 ta c:Ng(a + b + c)21. a2 + b2 + c2 3c Ng Hong Ton i hc Y Dc Cn Th3 4. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng 1 1 1 2. (a + b + c) + +9 a b c Bt ng thc Minkowski 111 + a1 , a2 , ..., an n p np n p v 1 < p + .Khi ap bp(ak + bk )p D Cho k +k b1 , b2 , ..., bn +k=1k=1 k=1 Nhng ta quan tm nhiu nht l cc bt ng thc quen thuc sau:hu 1.a2 + b 2 + c 2 + d 2 (a + c)2 + (b + d)2 2.a2 + b 2 + c 2 +m 2 + n2 + p2 (a + m)2 + (b + n)2 + (c + p)2 T 3. a1 2 + b 1 2 +a2 2 + b2 2 + ... +an 2 + b n 2 (a1 + a2 + ... + an )2 + (b1 + b2 + ... + bn )2 Th nTr n Tong oHNg4TY D 5. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Phn 2.CON NG I T BI TON N SUY NGM CA BN THN Chng I.BT NG THC V H PHNG TRNH 2 N S Bi ton 01 Gii h phng trnh: D 4 4 xyx +y 34= x+y 8(x + y) (1)13+ =4 xyhuPhn tch bi ton TCu hi t ra lc ny l khi ta nhn vo h ny,ti sao ta li ngh rng y l h gii bng phngphp Bt ng thc.Tht ra, iu ta quan tm n gi thit bi ton chnh phng trnh l x4 + y 4 xyth nht.S i xng hai bin x, y v c s xut hin ca i lng4 v .y l iu (x + y)x+y Thquen thuc trong cc bc nh gi bt ng thc.i lng (x + y)4 yu hn x4 + y 4 nn tanhn ra c V T a cn x + y th mnh hn xy nn ta ngh n vic nh gi V P a av V T a V P t a ra du ng thc.Vic cn li l gii phng trnh th hai khng qu kh.nLi Giiiu kin x, y > 0 Trp dng bt ng thc Cauchy Schwarz ta c: (x2 + y 2 )2 (x + y)4x4 + y 4 2 8 n 1Do v tri h (1) 8p dng bt ng thc AM GM cho v phi h (1) ta c xy3 xy31 = x+y 8 2 xy 88ToDu "=" xy ra khi v ch khi x = y 4Thay vo h (2) ta c = 4 x = 1.x ngVy nghim ca h l x = y = 1 .Nhn xtoVi vic bt u bc vo gii cc lp bi ton h bng bt ng thc,y c l l v d d tipcn vi cc bn lm quen phng php ny. tng trong sng cho h trn v phng php kt hp 1 Hchn V P V T l mt trong nhng bc khi u cho con ng chinh phc cc dng ton 8nh th ny.Vi vic nh gi nh th ta c th a bi ton kh hn cht na nh sau.Gii h phng trnh : 4x4 + y 4 + 6x2 y 2 = x3 x2 y 2Ng x4 y + y 4 x + 7x = 0c Ng Hong Ton i hc Y Dc Cn Th 5 6. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngBi tp tng t1.Gii h phng trnh sau: 2x + 4y = 32xy = 8 D thi hc sinh gii H Tnh nm 2008-2009.hu2.Gii h phng trnh sau: T 2(x + y)2 + 4xy 3 = 0 (x + y)4 2x2 4xy + 2y 2 + x 3y + 1 = 0Bi ton 02 Tm tt c cc cp s (x; y) khng m tha mn h:Th (2x + 4x2 + 1)( y 2 + 1 y) = 1 1 + 1 + 1 = 31 + 3x 1 + 2y 1 + 5x 1 + 4x nNgun gcTrBi ton c a ln trang www.k2pi.net bi anh Nguyn Trung Kin,bi ton chun mc v yl th ,nhng t duy cht ch trong cho ta nhiu iu suy ngm hng ti nhng bi ton vcch sng to h mi.Li gii n hin nay cho bi ton ny l t anh Con Ph Quen, mt li giip mang m cht ngh thut.n Phn tch bi ton1 11iu u tin khi ta nhn vo bi ton ny chnh l cc i lng + + l cc 1 + 3x 1 + 2y 1 + 5xi lng m v iu k th l 3x .4x .5x = 60x 64x vi 64x = 4x .Vy chc rng tn ti bt ng 3 To1 11 3thc c dng+ + h th hai l mt bt ng thc di dng b1+a 1+b 1+c 1 + 3 abc ton.Chnh iu ny nh hng phn no cho ta cch tip cn bi ton di cch nhn btng thc.ng Li gii oTrc tin ta cn rng :H y 2 + 1 y = 0;y2 + 1 y y2 + 1 + y = 1Tip n l mt bt thc quen thuc c dng trong bi ton ny nh mt b :Ng1 1 1 3+ + vi a, b, c 1 1+a 1+b 1+c 1 + 3 abc6TY D 7. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngVi nh gi th nht ta a phng trnh th nht trong h v phng trnh : 2x + 4x2 + 1 = y + y 2 + 1tDTi y xt hm s f (t) = t + t2 + 1, t 0. Ta c f (t) = 1 + > 0, t 0. t2 + 1T ta c :f (2x) = f (y) y = 2x.Vi kt qu ny cng vi nh gi th hai tc l b nu ra ta c :11 1 3 3 hux+ x+ x3 1+31+41+51 + 60 x 1 + 3 64xDu ng thc xy ra khi x = y = 0TNhn xt Li gii trn gip ta c li t duy p cho vic to ra cc bi ton hay,ci kh ca bi ton cnnm ch nh gi 60x 64x vi x 0.T bi ton trn ta cng c th to ra nhng bi ton n Thtng,v d nh bi ton sau.Gii h phng trnh sau:(2x + 4x2 + 1)( y 2 + 1 y) = 1(1 + 2x )(1 + 2y )(1 + 5x ) = (1 + 4x ) nBi tp tng t1.Tm nghim dng ca h : 3x4y2z+ +=1Trx+1 y+1 z+1 . 9 3 4 28 .x y z = 12.Gii h phng trnh nx + y + z = 1x4 + y 4 + z 4 = xyz Bi ton 03 Tm tt c cc cp s (x; y) dng tha mn h: 9 41 x2 + 1 To = 3 + 40x (1)22x + y x2 + 5xy + 6y = 4y 2 + 9x + 9(2) ng Ngun gcBi ton ny c bn Hi vi nick hoanghai1195 a ln din n www.K2pi.net trong thi giandi d nhn c nhiu s quan tm nhng cha c li gii no.Sau bn a li gii ca mnholn,mt li gii p v y tnh nh . H Phn tch bi toni lng cn lm ta c cm gic thy kh x l,cng vic ta cn lm l ph cc cn thc i, rng 41.2 = 82 = 92 + 12 vy theo bt ng thc Cauchy Schwarz ta c : 11Ng (92 + 12 ) x2 + |9x +| 2x + y (2x + y)c Ng Hong Ton i hc Y Dc Cn Th 7 8. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngM ta d on c x = y = 3 l nghim ca h nn n vic chn im ri ph cn thccn li nh sau:1 1.3 6 6|9x + | = |9x + | |9x +| 9x + 2x + y)9(2x + y) 2x + y + 92x + y + 9 DVic ph cn hon tt vn cn li l s dng gi thit cn li bi ton gii quyt vn trn.Li giihu41 2 116 + 80x 9 x += 3 + 40x 82 x2 + = (a)22x + y 2x + y9 TTheo bt ng thc Cauchy Schwarz ta c :1 1182 x2 + =(92 + 12 ) x2 + |9x +| Th2x + y( 2x + y)2 (2x + y)Theo bt ng thc AM GM ta li c : 1 1.3 6|9x +| 9x + 9x +(2x + y) 9(2x + y) 2x + y + 9 n phng trnh (a) c nghim thTr6 + 80x 18x2 + 9xy + 81x + 6 3x 2x2 xy + 6y 0 (3) 9 2x + y + 9Cng phng trnh (2) vi phng trnh (3) ta c: n x2 + 4xy 4y 2 + 12y 6x 9 0 (x 2y + 3)2 0 x + 3 2y = 0Vy du bng cc bt ng thc trn xy ra hay: x = y = 3.Th li ta thy tho mn h ban u. x=3ToVy nghim ca h :y=3Nhn xt Xt v tnh thc t bi ny rt kh i hi ngi gii phi thun thc k nng s dng c hai bt ngng thc AM GM v Cauchy Schwarz.iu ta quan tm l cch tc gi i t nhng nh gic bn i n bi ton ca mnh,khi cc bn c li gii trn c l cc bn thy c rng immu cht gii quyt bi ton nm cc nh gi thng qua vic chn im ri trong bt ng thcoAM GM v Cauchy Schwarz.Xin c ni thm cch chn c im ri nh th. Th nht H1 1 Vic c nh gi(92 + 12 ) x2 + |9x +| l do ta on c h( 2x + y)2 (2x + y) 91 c nghim x = y = 3 nn ta quan tm du ng thc xy ra l =. khi ta thay x1 Ng 2x + y x = y = 3 vo th iu ny ng,vy ta l gii c phn ng gi ny.8TY D 9. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Th hai1.3 6 Vi nh gi ny 9x + 9x + ta da vo im ri ca AM GM9(2x + y)2x + y + 9qua vic ph b cn thc.Thy biu thc trong cn c gi tr l 9 mun ph b cn thc nyta cn thm vo s 3 di mu th thm 3 trn t.Nh th khi p dng AM GM du ng Dthc vn bo ton.Bi tp tng tGii h phng trnh: x + 6xy y = 6 hu x3 + y 3 x + 6 2 2(x2 + y 2 ) = 3 x + xy + y 2 T Bi ton 04 Tm tt c cc cp s (x; y) dng tha mn h: 2x2 (4x + 1) + 2y 2 (2y + 1) = y + 32 x 2 + y 2 x + y = 1Th 2Phn tch bi tonnTa thy r vic nh gi bt ng thc qua s i xng cc bin x, y.Nhng iu quan trng l tann khai thc gi thit ny nh th no ? rng phng trnh th hai ca h c vit thnh 1 11 1(x )2 +(y + )2 = 1 vy nu t a = x ; b = y + th ta c ngay chn ca bin a, b [1; 1].Vic Tr 2 22 2cn li l bin i phng trnh th nht v cc i lng nh gi thch hp. nLi gii 11T phng trnh 2 ta c: (x )2 + (y + )2 = 1 22 1 111Vy nu ta t x = a; y + = b th x = a + ; y = b v a, b [1; 1] 2 222Lc ny thay vo phng trnh 1 ta c c:To8a3 + 14a2 + 8a + 4b3 4b2 = 30Hay ng (4a2 + 11a + 15)(a 1) + 2b2 (b 1) = 0(1)V a, b [1; 1] nn ta c 2 + 11a + 15)(a 1) 0 v b2 (b 1) 0(4a oa = 1a = 1Kt hp vi (1) ta suy rahocb = 0b = 1 x = 3 Ha = 1 * Nuth2b = 0 y = 1 2 a = 1x = 3*Nuth2Ng b = 1y = 1 2c Ng Hong Ton i hc Y Dc Cn Th9 10. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng3 1 3 1Vy nghim (x; y) ca h l ( ; ), ( ; ) .2 2 2 2Nhn xtTht trng hp khi bi ton trn c mt dng tng t kh kh nm trong thi th ln 1 ca dinn K2pi.net nh sau: DGii h phng trnh : (x + y) (25 4xy) = 105 + 4x2 + 17y 2 4 4x2 + 4y 2 + 4x 4y = 7hu 3a 13b + 1Li gii t x =;y =.Lc h tr thnh: 226b3 + 9b2 = 6a3 + 14a 20 (1) Ta2 + b 2 = 1Ta c (1) 3b2 (3 2b) = (a 1)(6a2 + 6a + 20) Th 3(1 a2 )(3 2b) = (a 1)(6a2 + 6a + 20) (a 1)(6a2 + 6a + 20 + 9 6b + 9a 6ab) = 01+) Vi a = 1 b = 0 x = 1; y =2 n+) Vi 6a2 + 29 + 15a 6b 6ab = 0 (2) ta c: V T (2) 6a2 + 29 15 6 3 = 6a2 + 5 > 0 nntrng hp ny phng trnh v nghim.1Vy phng trnh c nghim duy nht (x; y) = (1; )Tr2Bi ton tng tGii h phng trnh : nx 4 + y 4 = 2x3 2x2 + 2x = y 2.x + 2(y x 1) 19 + 1Bi ton 05 Gii h bt phng trnh:5 y2 + 1 2x + y 2 + y x + 1 = 3 To Li gii 1ng Phn tch : Nhn thy h ny c ba biu thc cha cn, ta suy ngh n vic t n ph b cn. Nhng bi ton t ra l t n ph nh th no? o R rng hai biu thc 2x + y 2; y x + 1 c mi lin h vi nhau nn ta ch cn t n ph cho mt trong hai biu thc ny v t n cn li l x 1.H Li gii di y la chn y x + 1 lm mt n, bn hon ton c th t u = 2x + y 2 Li gii x1 iu kin: yx+10Ng2x + y 2 010TY D 11. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng u= yx+10 x = v2 + 1t v = x10y = u2 + v 2Khi a v h bt phng trnh: v 2 + 1 + 2 (u2 + v 2 v) 19 + 15 (u2 + v 2 )2 + 1 D u + 2 (v 2 + 1) + u2 + v 2 2 = 3 (v 1)2 + 2 (u2 + v 2 ) 19 + 1 hu 5(u2 + v 2 )2 + 1 u+ u 2 + 3v 2 = 3 T bt phng trnh u ca h c chung nhn t (u2 + v 2 ) nn ta ngh n vic loi b(v 1)2 0, t bt phng trnh ny ta suy ra c: 191Th2 u2 + v 2 +5 (u 2 + v 2 )2 + 12 u2 + v 2 2 10 u2 + v 2 + u2 + v 2 + 12 0 u2 + v 2 2nMt khc s dng bt ng thc AM GM cho Tr u2 + 1 4 + u2 + 3v 2 3 (u2 + v 2 ) + 6 3.2 + 63=u+u2 + 3v 2 += =3 24 4 4Do vy cc du ng thc xy ra, tc n (v 1)2 = 0u=1yx+1=1 x=2 u=1 v=1x1=1 y=2 2= u 2 + 3v 2tha mn iu kin. ToVy h bt phng trnh c nghim duy nht (x; y) = (2; 2) . Li gii 2 2x + y 2 0 ngiu kin: 2y 2 + 2 0 x1 Ta c: o(x 2)2x + 2y 2 x 1 = + 2y 2y ()x+2 x1Do bt phng trnh (1) ng khi v ch khi bt phng trnh sau phi ngH 19 1+ 2 2y 5 y +1Ng (y 2)(10y 2 + y + 12) 0 y 2 ()c Ng Hong Ton i hc Y Dc Cn Th11 12. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng T phng trnh (2): x + 2y 1 + 2 (2x + y 2)(y x + 1) = 9 Ta c 2x + y 2 + 4(y x + 1) 9y V T x + 2y 1 += D 22 T suy ra 9y9y2()2 T ( ); ( ); () h bt phng trnh c nghim duy nht (x; y) = (2; 2)huNhn xt TCc nh gi trong bi ton u rt kho tuy nhin li gii 1 l tng minh v cho ta suy ngh phn.Nhng h bt phng trnh dng ny l mt v d in hnh cho li gii ton bt ng thctrong h.ThBi ton tng t 1. Gii h phng trnh: 2xy + 1 2y 2y n 20052xy y + 2006y = 1003Tr 2. Gii h phng trnh :x6 + y 8 + z 10 1x2007 + y 2009 + z 2011 1 n (y + 1)2 + y y 2 + 1 = x + 3 Bi ton 06 Gii h phng trnh: 2 x + x2 2x + 5 = 1 + 22x 4y + 2 To Phn tch bi ton ng Bi ton ny khng d nhn ra vic s dng bt ng thc nh th no,lp bi ton ny thnghay ph bin trong cc .Khi ta bin i qua mt s bc s a n vic dng bt ng thc chng minh cc phng trnh l c nghim hay v nghim.Qua bin i ta a v mt phng trnhol :(x 1)2 + 4 |x 1| (x 1) H2 y 2 + 1 > 2 |y| 2y.Vic pht hin ra dng nh gi dng A2 0 l cch a bi ton d chng minh hn bi nu tatm cch gii phng trnh (x 1) + (x 1)2 + 4 = 2y + 4y 2 + 4 l iu rt kh bi h trn cc cn thc v hai bin x, y.V th ta i n li gii sauNgiu kin : x 2y + 1 012TY D 13. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngT phng trnh (1) ca h ta c : (y 2 + 1) + 2y y 2 + 1 + y 2 = 2x 4y + 2 2 y+y2 + 1= 2x 4y + 2 (a) D T phng trnh th (2) ta li c : 2 2(x 1) +(x 1) + 4 = 4(2x 4y + 2) (b)huT (a) v (b) cho ta : T 22 2(x 1) + (x 1) + 4 = 4 y + y 2 + 1x + 2y 1 + (x 1)2 + 4 + 2 y 2 + 1 = 0 Th(3) (x 1) + (x 1)2 + 4 = 2y + 4y 2 + 4 (4)-Vi phng trnh (3) l :n(x 1)2 + 4 |x 1| (x 1)2 y 2 + 1 > 2 |y| 2y Tr x + 2y 1 +(x 1)2 + 4 + 2 y 2 + 1 > 0( )n- Vi phng trnh (4) ta c...