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 :[(x 1) 2y] (x + 2y 1) [(x 1) 2y] += 0 x 1 = 2y(x 1)2 + 4 +4y 2 + 4 To( Do chng minh ( ) lm cho biu thc trong ngoc ca nhn t lin hp >0 )-Vi x 1 = 2y ta thay vo phng trnh th hai ca h ban u ta c phng trnh: x+ x2 2x + 5 = 1 + 2 4x + 1 ngPhn vic cn li khng kh xin dnh cho bn c.o Bi ton 07 Gii h phng trnh: H(x + y + 1)(x + y + 1 + xy) = 12xy y 3x 2x2 1 + x 1 + y 2y 2 + xy = 1Ngc Ng Hong Ton i hc Y Dc Cn Th 13 14. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngPhn tch bi tonBn cht ca bi ton nm nh gi lng ca phng trnh th nht,khi quan st ta nhn thylng xy ca v phi yu hn x + y ca v tri nn ta ngh n vic dng AM GM a v tmcc chn ca xy.Tng t nh bi ton 1 ta a v dng nh gi 1 xy 1 (Li gii 1).1 11 DV li gii th 2,s kho lo nm vic bin i = phng trnhx + y + 1 x + y + 1 + xy 12 1u,vn cn li l dng bt ng thc AM GM chuyn v phng trnh x+y+14 1 4 12 sau dng o hm chng minh f (t) = 2 .T nhng nh tnh bant 12hu(x + y + 2) (t + 1)u ta i n li gii sau. T Li gii13x 2x2 1 0 x1Li gii 1 iu kin: 2Th1 + y 2y 2 0 1 y 1 2 Lc ny ta c:x + y + 1 1 + 1 + 1 = 1 > 0 n 22x + y + 1 + xy 1 + 1 + 1 + 1. 1 = 1 > 02222Tr Suy ra 12xy = (x + y + 1)(x + y + 1 + xy) > 0 xy > 0 M x > 0 nn suy ra y > 0 Do t phng trnh 2 ta d suy ra c xy 1n Mt khc, theo bt ng thc AM GM th t phng trnh 1 ta c:512xy = (x + y + 1)(x + y + 1 + xy) 33x.y.1.(2 x.y + 2 1.xy) = 12(xy) 6 Suy ra xy 1To Do xy = 1 x = y = 1 Th thy x = 1; y = 1 l nghim ca h. li x = 1 Vy . y = 1 ng 1 x1Li gii 2 iu kin: 2 1 y 1o2 +) Vi : (x + y + 1) (x + y + 1 + xy) = 0 h v nghim. H +) Vi : (x + y + 1) (x + y + 1 + xy) = 0 1 11 T phng trnh (1) cho ta : = x + y + 1 x + y + 1 + xy1211 1 4 Li c : x + y + 1 x + y + 1 + xyx + y + 1 (x + y + 2)2 14Ng Xt hm s : f (t) = , t [1; 3] , trong : t = x + y + 1 t (t + 1)214TY D 15. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng1 Lp bng bin thin cho ta : f (t) f (3) =12 x+1=y+1x=1 Vy (1) x+y+1=3y=1x=1 Th li cho ta nghim ca h : Dy=1Vy nghim ca h l (x, y) = (1; 1)Nhn xthu Li gii 1 t nhin hn hn,ch thun s dng bt ng thc nh gi,ci tinh t y l vic ta s dng bt ng thc AM GM ti T5 12xy = (x + y + 1)(x + y + 1 + xy) 3 3x.y.1.(2 x.y + 2 1.xy) = 12(xy) 6Th Ti sao ta khng p dng AM GM trc tip cho cc s m phi ghp cp li,ch v khi ta xt du ng thc th vic ghp cp gip ta bo ton du ng thc bi ton. S tinh ca li gii 2 nm v tr nh gi n(x + y + 2)2 4(x + y + 1 + xy) vi iu kin x, y 1. iu ny ng theo bt ng thc AM GM v ta cTr (x + y)2 (x + y)2 + 4(x + y) + 4 (x + y + 2)2x + y + 1 + xy x + y + 1 ++x+y+1= =44 4n Bi ton 08 Gii h phng trnh:(x + 6y + 3)xy + 3y = (8y + 3x + 9)y x2 + 8x 24y + 417 = (y + 3)y 1 + 3y + 17 To Phn tch bi tonng H cho gm cc phng trnh cn thc v a thc,vic ta nn lm l gii quyt cc cn thckh chu trn.V phng php thng dng nht l t n ph,nhng t n ph nh th no l on. l iu ta quan tm ?H Li giiTa vit li h phng trnh cho nh sau:(x + 6y + 3) xy + 3y = (8y + 3x + 9)y (1)Ngx2 + 8x 24y + 417 = (y + 3) y 1 + 3y + 17 (2)c Ng Hong Ton i hc Y Dc Cn Th 15 16. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Ta t a = x + 3; b = y vi a, b 0T ta vit li phng trnh (1) thnh : (a2 + 6b2 )ab = b2 (8b2 + 3a2 ) DVy ta c : b = 0 hay a3 + 6ab2 = 8b3 + 3a2 bVy ta c :hu b = 0 Suy ra y = 0 khng tho phng trnh (2). (a 2b)(a2 ab + 4b2 ) = 0 a = 2b TVi a = 2b x + 3 = 4yThay vo (2) ta c :4 (y + 4)(6 y) = (y + 3) y 1 + 3y + 17ThTheo bt ng thc AM GM ta c : (y + 4 + 6 y)4 (y + 4)(6 y) 4 = 202 nV ta c :(y + 3)y 1 + 3y + 17 3y + 17 3 + 17 = 20TrVy ng thc xy ra khi y = 1 thay vo ta c x = 1.Vy nghim ca h l (x, y) = (1; 1)n x3 y 3 + 5 (x + y)2 + 5x2 8 xy + 13x = 100 Bi ton 09 Gii h phng trnh:33 3 x2 + y 2 + xy 3x 4y + 4 = 0 Li gii ToPhn tch bi tony l dng ton hay v quen thuc vi vic kt hp hai cng c mnh l nh gi bt ngthc v o hm.Vi nhng dng ton nh th ny,cng vic ban u ta lm l tm cc chn ca cc ngbin x, y da vo vic nh gi Delta ca mt trong hai phng trnh ca h.Khi ta s c tiphai hng gii quyto 1. S dng o hm nh gi dng f (x) + g(y) = a vi min hoc max ca f (x) , g(y) c tngbng a. H 2. S dng bt ng thc nh gi dng A + B a vi A, B l hai biu thc no v a lhng s.Ng Li gii16 TY D 17. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng Ta vit li h cho nh sau x3 y 3 + 5 (x + y)2 + 5x2 8 xy + 13x = 100 (1) 33 3 x2 + y 2 + xy 3x 4y + 4 = 0(2)4D x0T (2) ta suy ra iu kin ca x; y l 73 y13Rt xy t (2) thay vo (1) ta c hu(3x3 + 18x2 + 45x) + (3y 2 3y 3 + 8y) = 108 (3)Xt hm s4T f (x) = 3x3 + 18x2 + 45x; x 0;34Suy ra f (x) 0 nnf (x) l hm ng bin vi x 0;3Th892 f (x) f 4 = (4) 9 3Xt hm s n 7g(y) = 3y 2 3y + 8y; y1; 3Da vo bng bin thin:Tr80 g(y) g 4 =(5)9 3 x = 4 n T (3), (4), (5) suy ra h c nghim 3 y = 4 3Th li thy nghim tha mn phng trnh th hai.4 4 H phng trnh c nghim (x; y) = ( ; )3 3ToBi ton tng t x4 + y 2 = 698 1. Gii h phng trnh 81 x2 + y 2 + xy 3x 4y + 4 = 0 ng x3 y 3 15(x y) (x + y)2 = x2 9y 2 15y + 942. Gii h phng trnh 4x2 + 4y 2 + 6x + 6y 2xy 9 = 0o x + y + 2(x2 + y 2 ) = 4 + 2xy H Bi ton 10 Gii h phng trnh : x 3x2 + 6xy + y3y 2 + 6xy = 6Li giiNgc Ng Hong Ton i hc Y Dc Cn Th 17 18. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng Phn tch bi ton Bi ton ny rt th v,nu ta khng nhn ra rng nu s dng cc bt ng thc qu mnh s dn ti lm kh bi ton v gn nh l a kt qu v con s 0.Bi ton c anh Nguyn TrungKin a ln trang www.k2pi.net trong mt thi gian khng nhn c li gii,sau y l li giica chng ti,mt li gii p dng ch bt ng thc c bn nhng kh tinh t.Mi cc bn cngDthng thc.Li giiTa vit li bi: hu x + y + 2(x2 + y 2 ) = 4 + 2xy T x3x2 + 6xy + y 3y 2 + 6xy = 6T phng trnh th hai,p dng bt ng thc AM GM ta c :Th x 3x2 + 6xy + y 3y 2 + 6xy 2 xy. 3x2 + 6xy. 3y 2 + 6xy 2. xy(9x2 y 2 + 18xy(x2 + y 2 ) + 36x2 y 2 ) 2. xy (9(xy)2 + 36(xy)2 + 36(xy)2 ) = 6xySuy ra xy 1 (1).n rng x. 9x. 3x + 6y y. 9y. 3y + 6xx 3x2+ 6xy + y 3y 2 + 6xy = + Tr 3 3Ta li c theo bt ng thc AM GM th : x. 9x. 3x + 6y y. 9y. 3y + 6x 12x2 + 6xy 12y 2 + 6xy + += 2(x2 + y 2 + xy)n336 6Vy ta suy ra:x2 + y 2 + xy 3M xy 1 nn x2 + y 2 2 .T phng trnh th nht ta c: 4 + 2xy x+ y + 4 2. 4 xy + 4.To Vy suy ra xy 4 xy 4 xy 1Hay xy 1 (2).T (1); (2) ta suy ra xy = 1.V t cc du bng bt ng thc ta c x = y = 1. ngVy tp nghim ca h phng trnh l S = (x; y) = (1; 1).o Bi ton 11 Tm tt c cc nghim dng ca h phng trnh : x + y + 2(2x2 5x 3) = y(1 y 5x) + 3 H 1 12 4 + 2x(12x + 1) + 2(y + 2)+ 4 + 2y(12y + 1) + 2(x + 2) = 16x16y145 Ng18TY D 19. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngNgun gcBi ton c a ln www.k2pi.net bi anh Con Ph Quen trong topic h sng to t cc thnhvin K2pi.netv li gii n hin nay cho bi ton ny l ca chng ti,mt li gii thn quen vinhng cng c bt ng thc a bi ton tr nn n gin hn nhiu.Thay v phn tch hng Dlm,chng ti mi cc bn c bi ton v t rt ra hng lm cho bn thn mnh.V nh th ccbn s hiu r hn cch gii v c nhng sng to nht nh.Chng ti ch i li gii t cc bn.Li giihu Ta vit li h phng trnh x + y + 2(2x2 5x 3) = y(1 y 5x) + 3 (1) T1 12 4 + 2x(12x + 1) + 2(y + 2) + 4 + 2y(12y + 1) + 2(x + 2)=(2)16x16y145 -Xt (1). t t =x + y;t > 0. Ta vit phng trnh (1) v dng:Th t + 4x2 10x 6 = y y 2 5xy +3Theo cch "chn qu" ta c rt y theo x, t nn ta c :y = t2 x Vy phng trnh trn bin i linthnh: t4 t2 + t 6 3 = 3(3 t2 )x (t 3)(t3 + 3t2 + 2t + 2 3 + 1) = 3( 3 t)( 3 + t).x Trn y cc bn c thc mc ti sao ti bit c cch phn tch nh th khng ? Tht ra,khi nhnvo phng trnh th hai,ta d nhn thy phng trnh trn l mt bt ng thc no ,nn iuta thit ngh lc ny c chng l x = y. Tht vy,bng kim tra n gin,ta c ngay mt nghim bi n3ton l x = y = .n y vi cch "chn qu" ta li c bin i nh trn l iu hon ton gii2thch c. Phng trnh trn cho ta : t = 3 (3) t3 + 3t2 + 2t + 2 3 + 1 = 3( 3 + t).x (4) ToChng ta hy tm gii quyt trng hp t =3 trc. Quay tr li phng trnh (2) ta c: 1 12 + =16x4 + 2x(12x + 1) + 2(y + 2) 16y 4 + 2y(12y + 1) + 2(x + 2) 145ngTheo bt ng thc Cauchy Schwarz ta c : 1 1 4 o + 16x4 + 2x(12x + 1) + 2(y + 2) 16y 4 + 2y(12y + 1) + 2(x + 2) 16(x 4 + y 4 ) + 24(x2 + y 2 ) + 4(x + y + 2)Ta bin i bt ng thc v :H16(x4 + y 4 ) + 24(x2 + y 2 ) + 4(x + y + 2) 290t t = x + y . Theo bt ng thc AM GM ta c cc nh gi :Ng x+y 2(x2 + y 2 )c Ng Hong Ton i hc Y Dc Cn Th19 20. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngx2 + y 2 2(x4 + y 4 )t u = 2 2. 4 x4 + y 4 . Vy ta c : 16(x4 + y 4 ) + 24(x2 + y 2 ) + 4(x + y + 2) 2u4 + 12u2 + 4(u + 2) 290DM 2u4 + 12u2 + 4(u + 2) 290 = (u 3)(2u3 + 6u2 + 30u + 94) = 0 do u = 3.Vy y ch l mt ng thc.3Vy du bng xy ra khi x = y = hu2-Ta gii quyt (4). t3 + 3t2 + 2t + 2 3 + 1 = 3( 3 + t).x TD nhn thy vi iu kin x, t > 0 ta c t3 + 3t2 + 2t + 2 3 + 1 + 3( 3 + t).x > 0 nn phngtrnh ny v nghim. 3 3Vy tm li h phng trnh c nghim duy nht (x; y) = ( ; ). 2 2ThNhn xt Cch to h t bt ng thc nh th ny gip ta c th c c nhng bi ton hay,cth ly mt bi bt ng thc trong thi th chuyn Khoa hc t nhin nm 2012 lm bi tonh nh sau:Tm nghim dng ca h sau nx + y = 2 112Tr2 + 9x4+2 + 9y 4= 2 + 6x 2 + 6y17Bi tp tng tnGii h phng trnh sau: x + y + z = 2013 1 +1+ 1= 1 + 1 + 13x + 2y 3y + 2z 3z + 2x x + 2y + 2z y + 2z + 2x z + 2x + 2y Tong oHNg20TY D 21. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngBI TP RN LUYN CHNG Ix + 2xy= x2 + y3x2 2x + 9. D1. Gii h phng trnh 2xyy += y2 + x3y 2 2y + 9(x + y)3 + 4xy = 32. Gii h phng trnhhu(x + y)4 2x2 4xy + 2y 2 + x 3y + 1 = 0.(2x + 3)4x 1 + (2y + 3)4y 1 = 2 (2x + 3)(2y + 3) T3. Gii h phng trnhx + y = 4xy. 1 + 2x2 +1 + 2y 2 = 2 1 + 2xy Th4. Gii h phng trnh x(1 2y) + x(1 2y) = 2 .3x + y 2 + 8xy = 16 2 x+y5. Gii h phng trnhn 23 2x + 2x = x + y y . 8y 3 3y42 Trx + y + 4 = 2xy6. Cho h phng trnh2x+y = m( x2 + y 2 + x + y + 5 + x + y). Tm m h c nghim (x, y) tho x, y 1. nx + x2 + y + 3 = 27. Gii h phng trnh2x + 4 + 3y + 8 = 13.3(x + y) = 2 |xy + 1|8. Gii h phng trnh To9(x3 + y 3 ) = |x3 y 3 + 1|x + y = 2439. Gii h phng trnh ( x + y) 1 +1=2ngx + 3yy + 3xx + 32 x y 2 = 34 10. Gii h phng trnh 4 x + 32 y + 6y = 24 oHNgc Ng Hong Ton i hc Y Dc Cn Th 21 22. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng 2 Chng II.BT NG THC V H PHNG TRNH 3 N S Bi ton 01 Gii h phng trnh: D + + = 33 (1) 1 11 xy zx + y + z = 1 (2)hux + y + z = 1 (2)xy + yz + zx = 7 + 2xyz (3) 27 TLi gii Thiu kin:x > 0, y > 0, z > 01Kt hp vi (2): x + y + z = 1 ta thy trong cc s x; y; z phi c t nht 1 s khng ln hn ,3 n 1khng mt tnh tng qut ta gi s z 3 1Do : z 0;Tr 3t:S = xy + yz + zx 2xyz = xy (1 2z) + z (x + y) = xy (1 2z) + z (1 z)nDo: 22 x+y 1zxy =2 2Vy: 21z1S(1 2z) + z (1 z) = 2z 3 + z 2 + 1 2 4ToXt hm s1f (z) = 2z 3 + z 2 + 14 ngTa c: 1 1 1 f (z) = 6z 2 + 2z = z (3z + 1) 0, z 0; 4 2 3oVy:1 7 1 f (z) f =, z 0;327 3 HDo : 7S 27Ng 2 H dng ny thng t cp trong thi i hc nn chng ti khng phn tch nhiu,cc c xemnh mt bi tham kho22 TY D 23. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngDu xy ra khi v ch khi: 1x = y, z = 3 DThay vo (2) ta c: 1 x=y=z= 3Th li ta thy tha mn h phng trnhhu 1 1 1Vy h phng trnh c nghim duy nht (x; y; z) =; ; 3 3 3 T Bi ton 02 Gii h phng trnh: 1 = x +1 Th xy z 1y= +1 yzx 1 x= +1 zxyn TrLi giiiu kin xyz = 0 .Nhn thy nu mt trong ba s x, y, z c mt s m,chng hn x < 0 thphng trnh th ba v nghim.Nu hai trong ba s x, y, z l s m,chng hn x, y < 0 th phngntrnh th hai v nghim.Vy ba s x, y, z cng du.1. Xt trng hp x, y, z > 0 th ta vit li h nh sau z = x2 y + xyTo x = y 2 z + yzy = z 2 x + zx ng Cng ba phng trnh li ta cx + y + z = (x2 y + y 2 z + z 2 x) + (xy + yz + zx) 6xyz ( )o Mt khc ta bin i h v dng z xy = x + z H x=y+x yz y =y+z zxNg zxy++= 2(x + y + z)xy yz zxc Ng Hong Ton i hc Y Dc Cn Th23 24. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngTh ta c x2 + y 2 + z 2 (x + y + z)22(x + y + z) = 6xyz x + y + z ( )( ) xyz 3xyzT ( ) v ( )( ) ta c x = y = z,t ta c nghim ca h l trong trng hp ny l D 2 2 2(x, y, z) =, , 2 2 22. Trng hp x, y, z < 0 ta t a = x; b = y; c = z ta chuyn v trng hp s dng vhu lm nh trng hp 1. 2 2 2 TVy nghim ca h l (x, y, z) = ,, 2 2 2 Bi ton 03a Gii h phng trnh: Th x + y + z = 0 x2 + y 2 + z 2 = 1 x + y 5 + z 5 = 5 6 536n a Ci bin thi i hc khi B nm 2012 TrLi giiVi x + y + z = 0 v x2 + y 2 + z 2 = 1 ta c n0 = (x + y + z)2 = x2 + y 2 + z 2 + 2x (y + z) + 2yz = 1 2x2 + 2yz 1Nn yz = x2 2 y2 + z2 1 x2 1 1 x2 6 6Mt khc yz = suy ra x2 do x()To 2 22 2 3 3Khi P = x5 + (y 2 + z 2 ) (y 3 + z 3 ) y 2 z 2 (y + z)1 2= x5 + (1 x2 ) [(y 2 + z 2 ) (y + z) yz (y + z)] + x2 2 x2 ng 1= x5 + (1 x2 ) x (1 x2 ) + x x2 2 + x2 1 x = 5 (2x3 x) 24 66 6Xt hm s f (x) = 2x3 x vi x suy ra f (x) = 6x2 1; f (x) = 0 x = 3 3 6o 66 6 6 6 6Ta c f =f = ,f=f =66 9 3 6 9 H 65 6Do f (x) suy ra P . 9 366 6Vy nghim ca h l x =;y = z = v cc hon v. 3 6Ng24TY D 25. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng BI TP RN LUYN CHNG II + 1 + = 33 1 11. Gii h phng trnhxy zDx + y + z = 1xy 3 = 92. Gii h phng trnhx + 3y = 6 hux 5 + y 5 + z 5 = 33. Gii h phng trnhTx 6 + y 6 + z 6 = 33(x2 + y 2 + z 2 ) = 14. Gii h phng trnh Thx2 y 2 + y 2 z 2 + z 2 x2 = xyz(x + y + z)336x2 y 60x2 + 25y = 05. Gii h phng trnh 36y 2 z 60y 2 + 25z = 0 n 236z z 60z 2 + 25x = 0Tr n Tong oHNgc Ng Hong Ton i hc Y Dc Cn Th 25 26. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngChng IV.TNG HP CC BI TON H GII BNG PHNG PHP BTNG THC Bi ton 01 Gii h phng trnh: D 11 1 +=1 + 2x2 1 + 2y 21 + 2xy x(1 2x) + y(1 2y) = 29hu thi hc sinh gii quc gia nm 2009 Li gii T 1 + 2xy 0 0 x 1 iu kin:x(1 2x) 0 2 0 y 1 y(1 2y) 0 2 Th 1111 12 2 Vi iu kin trn ta c :x2 ; y 2 =+ > 2 441 + 2xy 1 + 2x21 + 2y 232xy < 11Mt khc vi mi a, b [0; ] v ab < 1 ta lun c bt ng thc sau:n 211 2 + () 1+a2 1+b2 1 + ab TrTht vy bt ng thc () tng ng vi : 1 1 2 4 n2+ 2+ 01+a 1+b 1+a 2 . 1 + b2 1 + abTheo bt ng thc Cauchy Schwarz ta c : 22(1 + a2 )(1 + b2 ) 1 + ab 1 + a2 . 1 + b 2 1 + ab ToMt khc ta c :112(a b)2 (ab 1)ng+ =01 + a2 1 + b2 1 + ab (1 + ab)(1 + a2 )(1 + b2 ) op dng bt ng thc trn vi a = 2x; b =2y ta c 1 11 + 1 + 2x21 + 2xyH 1 + 2y 2ng thc xy ra khi x = y.Vi x = y thay vo h th hai ta c phng trnh:Ng 162x2 81x + 1 = 026 TY D 27. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng 81 + 591381 5913Gii phng trnh trn ta c: x = ;x = 32432481 + 5913 81 + 5913 81 5913 81 5913Vy nghim ca h l : (x; y) = (;); ( ;).324324 324324D Bi ton 02 Gii h phng trnh:(2x2 1)(2y 2 1) = 7 xy 2x2 + y 2 + xy 7x 6y + 14 = 0 hu Li giiD thy xy = 0 khng l nghim ca hVi xy = 0 ta vit h di dngT(2x 1 )(2y 1 ) = 7xy2 2 2 x + y + xy 7x 6y + 14 = 0Thiu kin phng trnh x2 + y 2 + xy 7x 6y + 14 = 0 c nghim theo n x l :7 = (y 7)2 4y 2 + 24y 56 0 y [1; ]3iu kin phng trnh x2 + y 2 + xy 7x 6y + 14 = 0 c nghim theo n y l :10 = (x 6)2 4x2 + 28x 56 0 x [2; ] n 31Xt hm s : f (t) = 2t l hm ng bin trn (0; +) nn:t7f (x).f (y) f (1)f (2) = .Tr2Kt hp vi phng trnh th nht ta c (x; y) = (2; 1) . Bi ton 03 Gii h phng trnh:n 44 2 2x + y ( x + y ) + y + x = 2y 4 x4 y 2 x2 x y 2x + y 6 8x + 6 = 0 Li giiToiu kin: x, y = 0Ta vit li h nh sau: 4 42 2 ng + y ( x + y ) + y + x = 2 (1) xy 4 x4y 2 x2x y 2 x + y 6 8x + 6 = 0 (2)o x yx2 y 2t t =+ suy ra t2 = 2 + 2 + 2 y xy x HMt khcx2 y 2 2 x4 y 4( + 2 ) = (t2 2)2 4 + 4 + 2 = t4 4t2 + 4y2 x yxNgTheo bt ng thc AM GM ta c :c Ng Hong Ton i hc Y Dc Cn Th 27 28. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngx2 y 2 + 2 hay t2 4 suy ra |t| 2y 2 x2Ta c v tri phng trnh (1) c bin i thnh g(t) = t4 5t2 + t + 4 , |t| 2. C g (t) = 4t3 10t + 1 = 2t(2t2 5) + 1Nhn xt:D 1. Nu t 2 suy ra 2t(2t2 5) 4(8 5) > 0 g (t) > 0 2. Nu t 2 suy ra 2t 4 ;2t2 5 3 ; 2t(2t2 5) 12 2t(2t2 5) 12 hay g (t) < 0Lp bng bin thin ta suy ra gi tr nh nht ca g(t) = 2 t c khi t = 2. hux yVy t phng trnh (1) ta c + = 2 t y suy ra x = y.y xThay x = y vo phng trnh th hai ta c :x6 +x2 8x+6 = 0 (x1)2 (x4 +2x3 +3x2 +4x+6) =T0 (x 1)2 [x2 (x + 1)2 + 2(x + 1)2 + 4] = 0 .Vy ta c nghim ca h l (x; y) = (1; 1) .Th Bi ton 04 Gii h phng trnh:x3 + 3xy 2 = x2 + y 2 + 2 (1)x4 + y 4 + 6x2 y 2 = 8 (2)nLi giiT phng trnh th hai ta c : Trx(x2 + 3y 2 ) = x2 + y 2 + 2 x > 0 x 4 = 8nNu y = 0 th h tr thnh :V nghim x 3 = x 2 + 2T suy ra y = 0 .Vit li h di dng: (x2 + y 2 )2 + (2xy)2 = 8 (3)To x2 + y 2 + 2 = x(x2 + y 2 ) + y(2xy) (4)T (4) ta c : ng (x2 + y 2 + 2)2 = [x(x2 + y 2 ) + y(2xy)]2 (x2 + y 2 )[(x2 + y 2 )2 + (2xy)2 ] = 8(x2 + y 2 ) ( )do (3)o (x2 + y 2 )2 4(x2 + y 2 ) + 4 0 H x2 + y 2 = 2Ngx2 + y 2 2xyDu ng thc trong ( ) xy ra khi = x2 = 1 x = 1. xy28 TY D 29. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngTh vo h ta c y = 1.Vy nghim ca h l (x; y) = (1; 1) . Bi ton 05 Gii h phng trnh: 1 + 1 x2 = x(1 + 2 1 y 2 ) (1)D 112 + = (2) 1+x 1+y 1 + xy Li gii hu iu kin:|x| , |y| 1 , xy > 0.T (1) suy ra 0 x 1.Do 0 y 1.TTheo bt ng thc Cauchy Schwarz ta c :2 11 1 1 +2+ (3) Th1+x1+y 1+x 1+yTa s chng minh 1 1 2 n +(4)1+x 1+y1 + xyTrTht vy: (4) (1 xy)( x y)2 0 niu trn ng vi mi x, y [0; 1].T suy ra : 1 1 2 +To 1+x 1+y1 + xy.ng thc xy ra khi v ch khi x = y.Th x = y vo (2) ta c : ng 1+ 1 x2 = x(1 + 2 1 x2 ) (5)o t x = sint, t [0; ] th 2 H (5) 1 + cost = sint(1 + 2cost)Ng tt tt t 2cos = 2sin cos [1 + 2(1 2sin2 ] (do t [0; ] cos > 0)22 22 2 2c Ng Hong Ton i hc Y Dc Cn Th 29 30. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHngt3t 2 3sin 4sin =2 22 D3t sin= sin24hu k4 t = 6 + 3 k4kZt= + T2 3 Vi t [0; ],ta c : 2 Th k4 1t = 6 + 3x= k4 2 t= +x=1 2 31 1nVy h cho c nghim (x; y) = ( ; ); (1; 1) .2 2 Bi ton 06 Gii h phngtrnh: Tr y 2 + (4x 1)2 = 3 4x(8x + 1) 40x2 + x = y 14x 1 nLi gii 1 iu kin:x 14 2t t = 4x (t ).H phng trnh tr thnh: 7 To y 2 + (t 1)2 = 3 t(2t + 1) (1) 5 t2 + t = y 7 t 1 (2) 2 42ngT (2) ta c y > 0 nn p dng bt ng thc AM GM ta c : 2t + 12t ++1 o3 3 2t + 121t(2t + 1) = 2t..1 =t+ 2 3 2HDo t phng trnh th nht ta c :11y 2 + (4x 1)2 t + y 2 t2 + 3t (3)22NgMt khc theo bt ng thc AM GM ta li c :30 TY D 31. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng7 7y2 + t 12 y t1 22Suy raD y 2 5t2 3t + 1 (4) huVT (3) v (4) ta c :15t2 3t + 1 t2 + 3t (2t 1)2 0T2 11 3Suy ra t = x = thay vo h ta tm c y = . 28 2Th 1 3Vy nghim ca h l (x; y) = ( ; ). 8 2 Bi ton 07 Gii h phng trnh: nx2 y 2 2x + y 2 = 0 (1)2x3 + 3x2 + 6y 12x + 13 = 0 (2)Tr Li gii T phng trnh (1) ta c :n2xy2 = 2 suy ra x 0. x +1Do x 0 nn p dng bt ng thc AM GM ta c :y 2 1 1 y 1 ( ) ToT (2) ta suy ra :2x3 3x2 + 12x 13 (2x 7)(x 1)2 y= = 1 (3)66 ngDo x 0 nn t (3) suy ra y 1 ( ).oT ( ) v ( ) suy ra y = 1.Thay y = 1 vo ta tm c x = 1.Vy nghim ca h l (x; y) = (1; 1) . H Bi ton 08 Gii h phng trnh: xy 2 xy +=y+1 x+1 xy + 1 5 3Ng + =4 x1y1 c Ng Hong Ton i hc Y Dc Cn Th 31 32. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Li gii iu kin :x, y > 1 xy > 1Ta chng minh D 1 1 2 +( )x+1 y+1 xy + 1 hu (x + 1)( xy + 1) + (y + 1)( xy + 1) 2(x + 1)(y + 1) T ( xy 1)( x y)2 0Lun ng vi mi xy > 1.ThTa c : x y2 xy +=y+1 x+1xy + 1 n x y 2 xy +1+ +1= +2Tr y+1 x+1xy + 1 1 1 2 xy + 1 (x + y + 1)(+)=2n x+1 y+1xy + 1Mt khc theo bt ng thc AM GM v ( ) ta cx + y + 1 2xy + 1 1 12 + Tox+1 y+1xy + 1 1 1 2 xy + 1 (x + y + 1)(+)2 x+1 y+1xy + 1ngng thc xy ra khi x = y.Thay vo h tm c x = y = 5. oVy nghim ca h l x = y = 5 . Bi ton 09 Gii h phng trnh:H x2 + y 2x2 + xy + y 2+=x+y2 3 x 2xy + 5x + 3 = 4xy 5x 3 Li giiNg32 TY D 33. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Ta c :11 113 x2 + xy + y 2 = (x + y)2 + (x2 + y 2 ) (x + y)2 + (x + y)2 = (x + y)222 244 DVy suy ra 3x2 + y 2x2 + xy + y 2 (x + y)2 (x + y)2 ++ 4x+y 2 3 43huT suy ra x = y 0Thay vo phng trnh th hai ca h ta c: T x 2x2 + 5x + 3 = 4x2 5x 3Th 2x2 + 5x + 3 x 2x2 + 5x + 3 6x2 = 0 n2x2 + 5x + 3 = 3x Tr 2x2 + 5x + 3 = 2x Do x 0 nn ta nhn2x2 + 5x + 3 = 2xSuy ra : n 1 22x + 5x + 3 = 0 x = 2x=3 Th li nhn nghim x = 3.ToVy nghim ca h l: (x; y) = (3; 3) . Bi ton 10 Gii h phng trnh:x + y + xy = z 22003 + 2z 22002 (1) ng 2004x4 + y 4 = 2z 2(2)(x + y)z1 = (z + 2004)xy (3)o Li gii H T phng trnh th hai ta c :Ng2004 2003 2z 2= x4 + y 4 2x2 y 2 xy z 2( )c Ng Hong Ton i hc Y Dc Cn Th33 34. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngTa li c (x + y)2 2(x2 + y 2 )D 2004 (x + y)4 2(x2 + y 2 )2 8(x4 + y 4 ) = 16z 2 hu 2002 x + y 2z 2( )TT ( ) v () ta c :Th 2003 2002x + y + xy z 2+ 2z 2 2002ng thc xy ra khi x = y = z 2 .H phng trnh tng ng vi : n x = y = z 22002 x=y=z=11Tr (2x)z1 = 1 1 2002 x = y = ; z = 2 22 n Bi ton 11 Gii h phng trnh: x2 + 2x 2 = y 2 4y 26x y + 11 +10 4x 2x2 = 0 Li gii To T phng trnh th hai p dng bt ng thc AM GM ta c :ng 4(10 4x 2x2 ) 4 + 10 4x 2x2y 6x + 11 = 10 4x 2x2 = 2 4 oThu gn phng trnh trn ta cHx2 10x + 2y 15 0 (1)Tip tc p dng bt ng thc AM GM cho phng trnh th nht ca h ta cNg y 2 4y 2 + 1 x2 + 2x 2 = y 2 4y 2 = 1(y 2 4y 2) 234 TY D 35. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngThu gn ta c 2x2 + 4x + y 2 + 4y 3 0 (2)DLy (1) cng (2) v theo v ta c :3x2 6x + y 2 + 6y + 12 0 3(x 1)2 + (y + 3)2 0 huT suy ra nghim ca h l (x; y) = (1; 3) .T Bi ton 12 Gii h phng trnh:2x + 1 2y + 1 = y x16x2 y 2 + 5 = 6 3 4x2 y + xTh Li giin 1iu kin: x, y . 2T phng trnh th nht ta c Tr 2x + 1 + x = 2y + 1 + y,hay n f (x) = f (y) vi f (t) =2t + 1 + t. 1D dng nhn thy rng f (t) l hm lin tc v tng nghim ngt trn , +nn f (x) = f (y) 2 s tng ng vi x = y.Thay vo phng trnh th hai ta cTo 16x4 + 5 = 6 4x3 + x. (1) 3 ngDo 16x4 + 5 > 0 nn t y ta suy ra 4x3 + x > 0, tc x > 0.By gi, s dng bt ng thc AM-GM, ta co33 3 4x (4x2 + 1) 2 4x + (4x2 + 1) + 24x3 +x=x(4x2 + 1) = .2 6 HKt hp vi (1), ta suy ra 16x4 + 5 4x2 + 4x + 3,hay tng ngNg2(2x2 + 2x + 1)(2x 1)2 0.c Ng Hong Ton i hc Y Dc Cn Th 35 36. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng1Nhng iu ny ch xy ra khi x = 2 . 1Mt khc, ta d thy x = 2 tha mn phng trnh (1). 1 1T y, ta i n kt lun h cho c nghim duy nht (x, y) =, . 2 2D Bi ton 13 Gii h phng trnh: x2 + 3 = 3 |xy| (2x 3y)2 + y 2 = 4 hu Li giiTTa vit li h phng trnh nh sau: x2 + 3 = 3 |xy| Th x2 + y 2 = 1 + 3xyMt khc ta c : n|xy| xyNnTr x2 + 3 x 2 + y 2 1Vy suy ra :y 2 4 |y| 2n. S dng bt ng thc AM GM ta c: x2 + 3 2 3 |x| 3 |xy|Nh vy ta c ng thc xy ra khi x = 3; y = 2 ToVy nghim ca h l: (x, y) = ( 3, 2) . Bi ton 14 Gii h phng trnh: x 2 = y + 2 ng y2 = z + 2 2 z = x + 2 o Li giiDng tm s:HBi ton trn c mt bn c nick luong_qt a ln din n toanphothong.vn v ti y nhnc hai li gii y tnh ngh thut t anh V Quc B Cn.Tht s khi c hai li gii ny,tcgi tht s chim nghim c nhiu iu,mi bi ton khi i ta iu nn c hng dn dt,khi tbt ln trang giy suy ngh,i hi ta phi c nhng cm nhn tht xc ng v c hng t duyNgtht gn.36 TY D 37. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng V li gii th nht,nt ngh thut nm nhng bin i lng gic kt hp vi nh gi bt ng thc a ta n kt qu.Cch lm gn v quen thuc,nhng mu cht nm vic ta quan st c min nghim bi ton l x, y, z 2 ta ngh n vic thm mt chn na x, y, z 2 vi vic lm trn ta a kho x, y, z [2; 2] v nh mt thi quen dn d,ta ngh n lng gic ho c v bi ton lng d x l hn.Vy vic t x = 2 cos l iuD d hiu. Li gii th hai l nt tinh ca ngi lm ton,vi vic nhn ra cc mi lin h ng thc (x2 1) = (x + 1)(x 1) x2 4 = (x 2)(x + 2) vi vic nhn ra tng mi lin h nh hu th v vic h ban u l i xng nn khi ta nhn vo th chc hn cc mi lin h ca (x 1)(y 1)(z 1) v (x + 2)(y + 2)(z + 2) s gip ta nh hng v gii bi ton trn n gin hn .TCch 1 Do x2 , y 2 , z 2 0 nn ta c y + 2, z + 2, x + 2 0 (da vo cc phng trnh ca h), suyra x, y, z 2.Th By gi, ta xt hai trng hp: 1. Nu 2 x, y, z 2 nt x = 2 cos vi [0, ]. .Ta c y = x2 2 = 4 cos2 2 = 2 cos 2.TrSuy ra z = y 2 2 = 4 cos2 2 2 = 2 cos 4. nT y a n2 cos = x = z 2 2 = 4 cos2 4 2 = 2 cos 8.T ta suy ra c: k2To 8 = + k2 = 7 k2 (k Z)8 = + k2 =9Vy ta suy ra cc nghim ca bi ton l : ng k2 = 7(x; y; z) = (2 cos ; 2 cos 2; 2 cos 4) vi k2 k Z=9o 2. Nu max{x, y, z} > 2: HKhng mt tnh tng qut, gi s x = max{x, y, z}..Khi , ta c x > 2, suy rax2 > 2x = x + x > x + 2 y + 2.NgMu thun vi nhng g c c t h. Vy trng hp ny khng th xy ra.c Ng Hong Ton i hc Y Dc Cn Th 37 38. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngCch 2 Ta xt ba trng hp: 1. Nu (x + 1)(y + 1)(z + 1) = 0: Khng mt tnh tng qut, gi s x + 1 = 0.D Khi ta c x = 1. T phng trnh th nht suy ra y = 1. T y v t phng trnh th hai, ta c x = y = z = 1. 2. Nu (x 2)(y 2)(z 2) = 0: hu Khng mt tnh tng qut, gi s x 2 = 0. Khi ta c x = 2. T phng trnh th nht suy ra y = 2.T T y v t phng trnh th hai, ta c x = y = z = 2. 3. Nu (x + 1)(y + 1)(z + 1) = 0 v (x 2)(y 2)(z 2) = 0:Th Ch rng h cho tng ng vi hai h phng trnh sau x2 1 = y + 1 n y2 1 = z + 1 2z 1 = x + 1 Trx 2 4 = y 2y2 4 = z 2 2 nz 4 = x 2 T h th nht, ta suy ra (x2 1)(y 2 1)(z 2 1) = (x + 1)(y + 1)(z + 1), v do (x + 1)(y + 1)(z + 1) = 0 nn suy ra (x 1)(y 1)(z 1) = 1. To T h th hai, ta thu c (x2 4)(y 2 4)(z 2 4) = (x 2)(y 2)(z 2), v do (x 2)(y 2)(z 2) = 0 nn ta c (x + 2)(y + 2)(z + 2) = 1.ng Nh vy, trong trng hp ny, ta thu c o (x + 2)(y + 2)(z + 2) (x 1)(y 1)(z 1) = 0, Hay tng ngH xy + yz + zx + x + y + z + 3 = 0 Nu x + y + z = 0: Do xy + yz + zx + x + y + z + 3 = 0 nn xy + yz + zx = 3. T y v t ng thcNg1 = (x 1)(y 1)(z 1) = xyz (xy + yz + zx) + (x + y + z) 1,38TY D 39. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng ta thu c xyz = 1. Vy ta c x + y + z = 0xy + yz + zx = 3xyz = 1D Suy ra x, y, z l ba nghim ca phng trnh X 3 3X + 1 = 0. hu D dng gii phng trnh ny, t tm c x, y, z. Th li vo h ban u loi nghim. Nu x + y + z = 1: Do xy + yz + zx + x + y + z + 3 = 0 nn xy + yz + zx = 2.T T y v t ng thc 1 = (x 1)(y 1)(z 1) = xyz (xy + yz + zx) + (x + y + z) 1, Th ta thu c xyz = 1. Vy ta cx + y + z = 1xy + yz + zx = 2nxyz = 1 Suy ra x, y, z l ba nghim ca phng trnh TrX 3 + X 2 2X 1 = 0. D dng gii phng trnh ny, t tm c x, y, z. Th li vo h ban u loi n nghim. Bi ton 15 Gii h phng trnh: x2 + 4 + x2 2xy + y 2 + 1 + y 2 6y + 10 = 5 2 3x2 z To log3 8xyz 3 = 10 log9 z 2 log3 y Li gii ngXt phng trnh: x2 + 4 + x2 2xy + y 2 + 1 + y 2 6y + 10 = 5Ta p dng bt ng thc Minkowski, ta c: o x2 + 4 + (y x)2 + 1 y 2 + 9 y 2 + 9 + (3 y)2 + 1 5 H x2 + 4 + x2 2xy + y 2 + 1 +y 2 6y + 10 5 Du bng xy ra khi v ch khi: x3=2 x = yx 2Ng y =3 y = 93y 4c Ng Hong Ton i hc Y Dc Cn Th 39 40. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng3 9Vy h c nghim l (x; y) = ( ; )2 4 Bi ton 16 Gii h phng trnh: D x2 yz y 2 xz z 2 xy+ 2 + 2 =02x2 + y 2 + z 2 2 2 2z + y 2 + x2 +x +z2y x2 y + y 2 z x y + 2 yz + 4 xz = 5zhu Li gii TTa s chng minhx2 yzy 2 xzz 2 xy+ +0 2x2 + y 2 + z 2 2y 2 + x2 + z 2 2z 2 + y 2 + x2 2(x2 yz) (y + z)2 rng: 2=1 2Th 2x + y 2 + z 22x + y 2 + z 2Thit lp cc biu thc cn li tng t. Ta quy v chng minh(y + z)2(x + z)2(x + y)2 3+ 2 + 22x2 + y 2 + z 2 2y + z 2 + x2 2z + x2 + y 2np dng bt ng thc Cauchy Schwarz ta c(y + z)2 y2 z2 Tr ( 2+)=3(x2 + y 2 ) + (y 2 + z 2 )x + y2 z2 + y2Vy bt ng thc c chng minh. ng thc xy ra khi v ch khi x = y = zThay vo h (2) ta c 2y 3 = 12 y = 3 6 n 3Vy h c nghim x = y = z = 6. Bi ton 16 Gii h phng trnh: To x 2 + y 2 = 1 125y 5 125y 3 + 6 15 = 0 Li giing Ta bin i phng trnh th hai v dng o 6 15y > 0 32y (1 y ) =H 2125y 6 x4 = 4.355p dng bt ng thc AM GM ta c:Ng3394.323 = 3(x2 + y 2 ) = y 2 + y 2 + y 3 + x2 + x2 5 y 6 x4 y 6 x4 55224 540TY D 41. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng 3 2 2 10 1010 10T ta c y = x .Do ta tm c nghim ca h l : (x; y) = ((;); (; )) 2 555 5 Bi ton 17 Gii h phng trnh:xyz = 32 D x2 + 4xy + 4y 2 + 2z 2 = 96( )x, y, z > 0Li giihuTheo bt ng thc AM GM ta c : T3x2 + 4xy + 4y 2 + 2z 2 = x2 + z 2 + z 2 + 4y 2 + 4xy 2xz + 4yz + 4xy 3 3 2.4.4x2 .y 2 .z 2 = 3 322 = 96T suy ra Th x2 + 4xy + 4y 2 + 2z 2 96 x = z ng thc xy ra khi v ch khi z = 2yn 2xz = 4yz = 4xy x = 2y Trz = 2yTh vo h ( ) ta c: x = 42y.y.2y = 32 y = 2 nz = 4Vy nghim ca h l (4, 2, 4) . Bi ton 18 Gii h phng trnh: x2 + y 2 = 2 (1) To z 2 + 2z(x + y) = 8 (2) z(x y) = 4 3 (3) ngLi giiT (1) v (2) ta c :oz2x2 + 2y 2 + z 2 + 2xz + 2xy = 12 x+= 6.2 Hp dng bt ng thc Cauchy Schwarz ta c : 222 zz zz(zy zx)2 (2y) x ++ (2x) y + (4x2 + 4y 2 )[ x + + y+ ] 22 22Ng Suy ra (zy zx)2 8.6 zy zx 4 3c Ng Hong Ton i hc Y Dc Cn Th41 42. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng zy zx 0 (4) Do (3) nn ta c x + z y+z 2 =2 (5) 2x 2y(5) 2x2 + 2y 2 = z(x + y) z(x + y) = 4 (6)Thay vo (2) ta c z 2 = 16 z = 4D Nu z = 4 x + y = 1x =1 + 3 x =1 3 hu Ta c : 2(loi) hoc 2 x2 + y 2 = 2y = 1 3 y = 1 + 32 2T Nu z = 4 x + y = 1x = 1+ 3x =1 3 Ta c : 2 hoc2(loi) x 2 + y 2 = 2y = 1 3 y = 1 + 3 Th 22 x = 1 3 1+ 3 x = 2 2 Vy h c hai nghim: y = 1 + 3hoc y = 1 3 n 2 2 z=4 z = 4 Tr Bi ton 19 Gii h phng trnh: x x2 + y y2 + z z2 (x y)(y z)(z x)(x + y + z) = 2 3 n 9 Li gii x x2 0 0 x 1 2iu kin: y y 0 0 y 1 To z z 2 0 0 z 1 x yGi s x = M axx, y, z ,c hai trng hp: x zng Nu x y z (x y)(y z)(z x)(x + y + z) 0 suy ra h v nghim o Nu x z y 2 3H Ta chng minh: P = (x y)(y z)(z x)(x + y + z) 9 Tht vy ta c : 4P = 4(x y)(y z)(z x)(x + y + z) 4(z y)(x z)(x + y + z)Ng = 2(z y)[( 3 + 1)(x z)][( 3 1)(x + y + z)]42 TY D 43. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng 31 2(z y) + ( 3 + 1)(x z) + ( 3 1)(x + y + z) 27 3 1 1 3 (2 3)38 32 3 =2 3x (3 3)y (2 3x) = P 27 2727 9 9Dx = 1Du ng thc tc xy ra khi y = 0z = 1 3 huDo :(2) th vo (1) ta thy nghim ny tho(1).Tng t cho cc trng hp cn li . 1T x = 1 x = 0 x = 1 3Vy h c ba nghim l y = 0; y= ; y=1 13 z = z=1 z=0 3 Bi ton 20 Gii h phng trnh:2x2 = y(x2 + 1) Th n3y 3 = z(y 4 + y 2 + 1) 44z = x(z 6 + z 4 + z 2 + 1)Tr Li giin Trng hp 1 Vi x = 0 th h c nghim x = y = z = 0 Trng hp 2 Vi x = 0 h c nghim th x > 0, y > 0, z > 0 Ga s (x, y, z) l nghim ca h ta c : To 2x2 = y(1 + x2 ) 2xy x y (v x2 + 1 2x) 3y 3 = z(y 4 + y 2 + 1) 3zy 2 y z (v y 4 + y 2 + 1 3y 2 )ng 4 4z = x(z 6 + z 4 + z 2 + 1) x4z 3 z x (v z 6 + z 4 + z 2 + 1) T h suy ra x = y = z oThay vo h ban u ta c hai nghim ca h l x = y = z = 0 ; x = y = z = 1 .H Bi ton 21 Gii h phng trnh: xy + 2(x4 + y 4 ) = 1 (1)n l s nguyn dng l,khc 1Ng xn+4 y n + xn y n+4 = 2 (2) 3n+2c Ng Hong Ton i hc Y Dc Cn Th43 44. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Li gii 2T (2) bin i ta c :(xy)n (x4 + y 4 ) = xy > 0 3n+2 0 < xy 1 4 42 2Khi t (1) vi lu :2(x + y ) 4x y ,suy ra :3 D 2 x4 + y 4 = (1 xy) 2Kt hp (1) vi (2),ta c : n 4 4n1 (1 xy)2(xy) (x + y ) = (xy) xy.2hup dng bt ng thc AM GM ta c 1 xy1 xy 2n1 xy + + 3 T (1 xy)1 2 2 2 22(xy)n1 xy. 2. = 2 3 3 3n+2 xy = 1 x = y = 3Thng thc xy ra khi v ch khi3 3x = y x = y = 3 3 3 3Vy h cho c cc nghim l x = y = x=y= .3 3n Bi ton 22 Gii h phng trnh:x + x + ....x = 2009 12 2009 Tr x8 + x8 + ... + x8 = x6 + x6 + ... + x6 12 2009122009 Li giinGi s(x1 , x2 , ..., x2009 ) l mt nghim ca h.Khng gim tnh tng qut gi s x2 x2 .... x2122009p dng bt ng thc Cauchy Schwarz ta c : 22009 x2 + x2 + .... + x212 2009 x1 + x2 + ....x2009 = 20092 x2 + x2 + .... + x2 2009 (1)12 2009Top dng bt ng thc Chebyshev cho cc b s (x2 , x2 , ..., x2 ) v (x6 , x6 , ..., x6 ) ta c12 2009 12 2009x2 + x2 + .... + x2 12 2009 x6 + x6 + .... + x612 2009 2009 x8 + x8 + .... + x812 2009 (2) ngT (1) v (2) suy rax8 + x8 + ... + x8 x6 + x6 + ... + x6 122009 122009ong thc xy ra ti x1 = x2 = ... = x2009 = 1.Vy nghim ca h l x1 = x2 = ... = x2009 = 1 . H Bi ton 23 Gii h phng trnh:xy(x + y) = x2 xy + y 2Ng 1 + 1 = 16x3 y 344TY D 45. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Li giiiu kin:x, y = 0 1 1t a = ; b = x y DT gi thit ta suy ra 1 1 1 1 1 1 + = 2 + 2 a + b = a2 ab + b2 ab a ba ab b 1 1 + 3 = a3 + b3 = (a + b)(a2 ab + b2 ) = (a + b)2hu x 3yTa c bt ng thc c bn sau: T(a + b)2 (a + b)2 a2 ab + b2 a+b0a+b4 4411 3 + 3 = (a + b)2 16 xyTh1ng thc xy ra khi a = b hay x = y =21Vy nghim ca h l x = y = .2n Bi ton 24 Gii h phng trnh: + + = 33 Tr111 x yzx + y + z = 1xy + yz + zx = 7 + 2xyzn ( )27 Li giiTa c 7( ) xy + yz + zx 2xyz =To 27Li cxyz (x + y z)(y + z x)(z + x y) = (1 2x)(1 2y)(1 2z)8(xy + yz + zx) 2 9xyz 4(xy + yz + zx) 1 2xyz ng 91 (x + y + z)2Ta phi chng minh xy + yz + zx =ng.33o1ng thc xy ra khi x = y = z = .3 1Vy nghim ca h l x = y = z = . H 3 Bi ton 25 Gii h phng trnh: x + a + y + a + z + a = 3 a2 + 1 a a>1 v a l hng sNg a x + a y + a z = 3 a2 1 ac Ng Hong Ton i hc Y Dc Cn Th 45 46. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ngLi gii t A = x + a + y + a + z + a, B = a x + a y + a z p dng bt ng thc Cauchy Schwarz ta c : A2 3(3a + x + y + z)D B 2 3(3a x y z)Cng hai v ca hai bt ng thc trn li ta c: A2 + B 2 18a ( ) huMt khc theo gi thit ta li c: a2 + 1 a2 1A2 + B 2 = 9 + = 18aTaa1Du ng thc trong h ( ) xy ra khi v ch khi x = y = z = .a1ThVy nghim ca h l x = y = z =a Bi ton 26 Gii h phng trnh: n 2014 1 + x1 + 1 + x2 + .... + 1 + x2013 = 2013 2013 1 x + 1 x + .... + 1 x2014Tr 122013 = 2013 2013 niu kin:1 xi 1(i = 1, 2, 3, ..., 2013).Ta c2 2 2014 2013=1 + x1 + 1 + x2 + 1 + x3 + .... + 1 + x2013 2013(2013 + x1 + x2 + ... + x2013 ) 2013Suy ra x1 + x2 + ... + x2013 1 (1)Li c To2 2 2014 2013=1 x1 + 1 x2 + 1 x3 + .... + 1 x2013 2013(2013 x1 x2 ... x2013 ) 2013Suy ra x1 + x2 + ... + x2013 1 (2)ngT (1) v (2) suy ra x1 + x2 + ... + x2013 = 1Do h phng trnh tr thnh o 1 + x = 1 + x = .... = 1 + x 122013 11 x1 = 1 x2 = .... = 1 x2013 x1 = x2 = ... = x2013 =2013x1 + x2 + ... + x2013 = 1H 1Vy nghim ca h l x1 = x2 = ... = x2013 = .2013Ng46 TY D 47. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Chng V.SNG TO H QUA CC BI TON BT NG THC 1 Tm nghim dng ca h sau: x + y + 2x2 + 2xy + 3y 2 = 4 D 3 x2 y = 32 5 13huBi ton trn c xut pht t bi ton sau y:Cho x, y l cc s thc dng tho mn x + y + 2x2 + 2xy + 3y 2 = 4.Tm gi tr ln nht ca ca biu thc x2 y. TBi ny c Michael Rozenberg a ln trang www.mathlink.ro v c cp n trong sch"Phn loi & phng php gii ton bt ng thc" ca cc tc gi Vasile Cirtoaje,V QucB Cn,Trn Quc Anh.Bi ton hay v c nhiu iu l th xung quanh bi ny,tuy nhin khun Thkh bi vit khng cho php chng ti bn n vn ny.Quay tr li h cho,khi quan st chchn ta c th ngh ngay n vic s dng bt ng thc gii quyt bi ton ny.Nhng y khngc g l hay khi sng to h kiu ny,ta tm mt bt ng thc c iu kin sau dng cc trca n lm mt phng trnh trong h th chc hn ta c th to ra hng trm bi nh th.Vic nquan trng l ta kho lo che i ci gi tr ln nht nhn khng p mt ny. 5Ta ngh n vic thm vo mt n s z = 1 vy bi ton nhn s p hn v y th v. 3Trrng ng thc xy ra ti x = 2y = 4z vy ta c th tm thm cc mi lin h cc bin x, y, z vinhau.T y ta c th to ra nhng h phng trnh kh t bi ton c bit trn nBi ton 01 Gii h phng trnh sau trn tp s dng :x + y + 2x2 + 2xy + 3y 2 = 4 (1)1.x2 y = 32z 3 (2) To3xz 2 + 6xz + 3z 3 + 15z 2 + 16z = 2x + 6 (3) Li giing T gi thit cho ta d dng nhn thy x + y < 4.Do phng trnh (1) c th vit thnh 2x2 + 2xy + 3y 2 = (4 x y)2 o T y bng mt s bin i ,ta cH (x + y)2 + 2(y + 2)2 = 40, hay 2Ngx 2 2++ (y + 2)2 = 202c Ng Hong Ton i hc Y Dc Cn Th47 48. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNHng Ti y s dng ln lt cc bt ng thc AM GM v Holder ,ta c 22 22 2xx 2x33 x2 y 2+ + 2++ (y + 2) 32+(2 + y) 2+2224 D T suy ra 23 x2 y 20 3 2 +, 4 hay 3hu 5x2 y 321 3 T (3) ta bin i v T[3(z + 1)2 5][x + z + 3] = 05 5 Vy ta c z =1 Suy ra x2 32z 3 . T suy ra nghim ca h l x = 2y = 4z = 413 3Th 2. Bng cch ta tip tc bin i phng trnh th (3) cho kh hn,ta c th a v vic xthm nh sau:f 3z 2 + 5z = f (2 z) n Vy ta s a v 3z 2 + 5z + 3z 2 + 5z = (2 z)3 + 2 z 3Tr 3z 2 + 5z + 3z 2 + 5z = 8 12z + 6z 2 z 3 + 2 z 3 z 3 3z 2 + 18z + 3z 2 + 5z 10 = 0 3 n Vy l ta c c mt h mi lx + y + 2x2 + 2xy + 3y 2 = 4x2 y = 32z 3 3z 3z 2 + 18z + 3 3z 2 + 5z 10 = 0ToGhi chVi nhng hng nh trn cc bn c th to ra nhiu h hay hn v kh hn na,lm cho bi toncng p hn. ngi t cc bi ton c v khng hay lm,th ta i t cc b vy.Trong sch "S dng phng php Cauchy Schwarz chng minh bt ng thc " cacc anh V Quc B Cn v Trn Quc Anh trang 194 c b nh lo Nu x 4y th x+y x + 5y H Vy ta c tng li to ra mt phng trnh c iu kin l x 4y phng trnh cn li cah s s dng bt ng thc trn.Nhng nu nh th th l qu,ta s bin i.Ng Thay x bi x2 xy + y 248 TY D 49. Chuyn : S DNG CC BT NG THC C BN GII H PHNG TRNH ng Thay y bi y 2 + 2x Vy ta c h phng trnh l: Bi ton 02 Gii h phng trnh sau :D x2 x(8 + y) 3y 2 + x2 3x + y 2 + 9 = 7 x2 xy + y 2 + y 2 + 2x = x2 x(10 + y) + 6y 2 huGhi ch:S tip tc cp nht34 CHC MNG NM MI 2013TThn Tr nTo ngo H3Rt mong nhn c s gp ca cc bn ln sau bi vit s c hon thin nhiu hn.Thnmn!Ng4 ABi vit c L TEX 2 bi Ngohoangtoan nick thienlonghoangde vi mc ch phi thng mi,mihnh vi sao chp nhm mc ch li nhun tuyt i khng c cho php.c Ng Hong Ton i hc Y Dc Cn Th 49