cac-chuyen-de-bd-hsg-ntb-nhatrang-2013.pdf

7/23/2019 CAC-CHUYEN-DE-BD-HSG-NTB-NHATRANG-2013.pdf

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Mc lc

Li ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Trn Nam DngNguyn l cc hn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Trnh o Chin, L Tin Dng

Mt s dng tng qut ca phng trnh hm Pexider v p dng . . . . . . . . . . . . . . . . . . . . 16

L SngXy dng mt lp phng trnh hm nh cc hng ng thc lng gic . . . . . . . . . . . . . 24

L Th Anh oanTnh n nh nghim ca mt s phng trnh hm Cauchy . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Trn Vit TngMt s lp phng trnh hm a n sinh bi phi ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

L Sng, Nguyn inh HuyT cng thc Euler n bi ton s phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Nguyn Th TnhMt s ng dng ca phng trnh Pell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Hunh B LcPhp th lng gic l cng c gii ton trong cc bi thi chn hc sinh gii . . . . . . . . . . 79

Nguyn Trung HngS dng vnh cc s nguyn gii mt s bi ton s hc . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Phm Th Thy HngNi suy theo yu t hnh hc ca th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

L Sng, V c Thch SnBt bin nh l mt phng php chng minh v ng dng trong gii ton . . . . . . . . . . . 108

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L Th Thanh HngMt s dng ton lin quan n dy s c quy lut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Trng Vn imVn dng tnh n iu trong cc bi ton tm gii hn dy s v gii phng trnh, bt

phng trnh, h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134Hunh Tn Chung dng mt s nh l c bn ca gii tch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

L Vn ThnMt s phng php gii h phng trnh .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

Hunh Kim Linh, T Hng KhanhMt s bi ton v a thc trong cc k thi hc sinh gii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

Nguyn Vn NgcMt s bi ton v chia ht i vi cc a thc i xng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

Hunh Duy ThyNt p hm s tim n trong bi ton bt ng thc, bi ton tm gi tr ln nht vgi tr nh nht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

Nguyn Ti ChungThm mt phng php mi chng minh bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

T NguynMt s vn v php nghch o trong mt phng v ng dng . . . . . . . . . . . . . . . . . . . . . 213

Trn Vn TrungS dng mt s tnh cht ca nh x gii bi ton phng trnh hm s............235

Nguyn Hu Tm - Hong T QuynT gic lng tip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

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Li ni u

Ha nhp vi tui tr c nc hot ng si ni k nim ngy thnh lp on thanh nin

Cng sn H Ch Minh v thi ua lp thnh tch cho mng ngy sinh ca Bc H knh yu,

tin ti k nim 37 nm ngy gii phng Nha Trang v thc hin cc chng trnh i mi gio

dc ph thng, S Gio Dc v o to Khnh Ha phi hp vi Hi Ton hc H Ni ng

t chc Hi tho khoa hc Cc chuyn Ton hc bi dng hc sinh gii THPT khu vc

Duyn hi Nam Trung b v Ty nguyn.

y l hi tho ln th hai theo tinh thn cam kt ca cc tnh duyn hi Nam Trung b

v Ty Nguyn v vic hp tc pht trin kinh t - vn ha v x hi. S Gio dc v o

to Ph Yn tin hnh t chc Hi tho ln th nht vo ngy 18-19/4/2011 ti thnh ph

Tuy Ha v lin kt bi dng hc sinh gii v bi dng hc sinh gii mn ton trng Trung

hc ph thng Chuyn cc tnh duyn hi Nam Trung B v Ty Nguyn. Ti Hi tho ln th

nht thng nht giao cho S Gio dc v o to Khnh Ha t chc Hi tho ln th hai.

y l nt sinh hot truyn thng mi v sinh hot chuyn mn, v giao lu hp tc trong gio

dc, o to v cc sinh hot hc thut khc. V thc t, gi y, ti vng duyn hi Nam

Trung b v Ty Nguyn ny xut hin ngy cng nhiu nt thnh tch ni bt, c hc

sinh t gii ton Olympic quc t. Nm nay, nhiu i tuyn t gii cao trong k thi hc sinhgii quc gia. Cc tnh k Lc, Ph Yn mnh dn c i tuyn tham d k thi Olympic

H Ni m rng bng ting Anh v t gii cao.

Khu vc Duyn hi Nam Trung b v Ty nguyn gi y thc s khi sc, to tin

vn ln tm cao mi, ch ng hi nhp, snh vai ngang bng vi cc khu vc khc trong

c nc.

Hi tho khoa hc ln ny c tin hnh t 14-15/4/2012 ti thnh ph Nha Trang, Khnh

Ha hn hnh c n tip nhiu nh khoa hc, nh gio lo thnh, cc nh qun l, cc chuyn

gia gio dc v cc nh ton hc bo co ti cc phin ton th v cc cn b ch o chuyn

mn t cc s Gio dc v o to, cc thy gio, c gio b mn Ton cc tnh, thnh khu

vc Duyn hi Nam Trung b v Ty nguyn ang trc tip bi dng hc sinh gii mn Ton

bo co ti cc phin chuyn ca hi tho.

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Ban t chc nhn c trn 30 bo co ton vn gi ti hi tho. Song do khun kh

rt hn hp v thi gian, khu ch bn v thi lng ca cun k yu, chng ti ch c th a

vo k yu c 22 bi, nhng bi cn li s c ch bn gi qu i biu khi thc hin

chng trnh bo co chuyn chnh thc ca hi tho.

Ni dung ca k yu ln ny rt phong ph, bao gm hu ht cc chuyn phc v vic

bi dng hc sinh gii ton t i s, gii tch, hnh hc, s hc n cc dng ton lin quan

khc. Bn c c th tm thy y nhiu dng ton t cc k olympic trong nc v quc t,

mt s dng ton v hm s, l thuyt ni suy, cc tr, ...

Ban t chc xin chn thnh cm n s hp tc v gip ht sc qu bu ca qu thy

gio, c gio v c bit l ton th t ton ca trng THPT chuyn L Qu n Nha Trang,

Khnh Ha c c cun k yu vi ni dung thit thc v rt phong ph ny.V thi gian chun b rt gp gp, nn cc khu hiu nh v ch bn cun k yu cha

c y , chi tit, chc chn cn cha nhiu khim khuyt. Rt mong c s cm thng

chia s ca qu i biu. Nhng kin ng gp lin quan n cun k yu ny xin gi v

a ch: Trng THPT Chuyn L Qu n, s 67 Yersin, Nha Trang, Khnh Ha. Email:

[email protected].

Xin trn trng cm n.

Nha Trang ngy 25.03.2012

Nguyn Vn Mu

Ch tch Hi Ton hc H Ning trng ban t chc hi tho

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NGUYN L CC HNTrn Nam Dng, Trng i hc KHTN Tp HCM

Bi vit ny c pht trin t bi vit Cc phng php v k thut chng minh mchng ti trnh by ti Hi ngh Cc chuyn Olympic Ton chn lc ti Ba V, H Ni,thng 5-2010 v ging dy cho i tuyn Olympic Vit Nam d IMO 2010. Trong bi ny, chngti tp trung chi tit hn vo cc ng dng ca Nguyn l cc hn trong gii ton.Mt tp hp hu hn cc s thc lun c phn t ln nht v phn t nh nht. Mt tp conbt k ca N lun c phn t nh nht. Nguyn l n gin ny trong nhiu trng hp rt cch cho vic chng minh. Hy xt trng hp bin! l khu quyt ca nguyn l ny.

1 Mt s v d m uTa xem xt mt s v d s dng nguyn l cc hn

V d 1. C 3 trng hc, mi trng cn hc sinh. Mi mt hc sinh quen vi t nht n + 1hc sinh t hai trng khc. Chng minh rng ngi ta c th chn ra t mi trng mt bnsao cho ba hc sinh c chn i mt quen nhau.

Li gii. Gi A l hc sinh c nhiu bn nht mt trng khc. Gi s bn nhiu nht nyl k. Gi sA trng th nht v tp nhng bn quen A l M = {B1, B2, . . . , Bk} trngth 2. Cng theo gi thit, c t nht 1 hc sinh C trng th 3 quen vi A. V C quenkhng qu k hc sinh trng th nht nn theo gi thit C quen vi t nht n + 1 k hcsinh ca trng th hai, t N =

{D

1, D

2,...,D

m}l nhng ngi quen C trng th hai th

m n + 1 k. V M, N u thuc tp hp gm n hc sinh v |M| + |N| k + n + 1k = n + 1nn ta c M N = . Chn B no thuc M N th ta c A,B,Ci mt quen nhau.V d 2. Chng minh rng khng tn ti sn l, n > 1 sao cho 15n + 1 chia ht cho n

Li gii. Gi s tn ti mt s nguyn l n > 1 sao cho 15n + 1 chia ht cho n. Gi p l c snguyn t nh nht ca n, khi p l. Gi sk l s nguyn dng nh nht sao cho 15k 1chia ht cho p (s k c gi l bc ca 15 theo modulo p).V 152n1 = (15n1)(15n+ 1) chia ht cho p. Mt khc, theo nh l nh Fermat th 15p11chia ht cho p. Theo nh ngha ca k, suy ra k l c s ca cc s p 1 v 2n. Suy rak|(p1, 2n). Do p l c s nguyn t nh nht ca n nn (n, p1) = 1. Suy ra (p1, 2n) = 2.Vy k|2. T k = 1 hoc k = 2. C hai trng hp ny u dn ti p = 7. Nhng iu nymu thun v 15n + 1 lun ng d2mod 7

Trong hai v d trn, r rng vic xt cc trng hp bin em n cho chng ta nhngthng tin b sung quan trng. Trong v d th nht, vic chn A l hc sinh c s ngi quennhiu nht mt trng khc cho ta thng tin s ngi quen ca C trong trng th hai tnht l n + 1 k. Trong v d th hai, do p l c s nguyn t nh nht nn p 1 nguyn tcng nhau vi n l bi s ca p.Bi tp

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1. Cho n im xanh v n im trn mt phng, trong khng c 3 im no thng hng.Chng minh rng ta c th ni 2n im ny bng n on thng c u mt khc mu sao chochng i mt khng giao nhau.2. Trn ng thng c 2n + 1 on thng. Mi mt on thng giao vi t nht n on thngkhc. Chng minh rng tn ti mt on thng giao vi tt c cc on thng cn li.3. Trong mt phng cho n > 1 im. Hai ngi chi ln lt ni mt cp im cha c nibng mt vc-t vi mt trong hai chiu. Nu sau nc i ca ngi no tng cc vc t v bng 0 th ngi th hai thng; nu cho n khi khng cn v c vc t no na mtng vn cha c lc no bng 0 th ngi th nht thng. Hi ai l ngi thng cuc nu ching?4. Gi sn l s nguyn dng sao cho 2n + 1 chia ht cho n.a) Chng minh rng nu n > 1 th n chia ht cho 3;b) Chng minh rng nu n > 3 th n chia ht cho 9;c) Chng minh rng nu n > 9 th n chia ht cho 27 hoc 19;d) Chng minh rng nu n chia ht cho s nguyn t p = 3 th p 19;e)* Chng minh rng nu n chia ht cho s nguyn t p, trong p = 3 v p = 19 th p 163.

2 Phng php phn v d nh nhtTrong vic chng minh mt s tnh cht bng phng php phn chng, ta c th c thm

mt s thng tin b sung quan trng nu s dng phn v d nh nht. tng l chngminh mt tnh cht A cho mt cu hnh P, ta xt mt c trng f(P) ca P l mt hm cgi tr nguyn dng. By gi gi s tn ti mt cu hnh P khng c tnh cht A, khi stn ti mt cu hnh P0 khng c tnh cht A vi f(P0) nh nht. Ta s tm cch suy ra iumu thun. Lc ny, ngoi vic chng ta c cu hnh P0 khng c tnh cht A, ta cn c micu hnh P vi f(P) < f(P0) u c tnh cht A.

V d 3. Cho ng gic li ABCDE trn mt phng to c to cc nh u nguyn.a) Chng minh rng tn ti t nht 1 im nm trong hoc nm trn cnh ca ng gic (khcvi A, B, C, D, E) c to nguyn.b) Chng minh rng tn ti t nht 1 im nm trong ng gic c to nguyn.c) Cc ng cho ca ng gic li ct nhau to ra mt ng gic li nh A1B1C1D1E1 bn trong.Chng minh rng tn ti t nht 1 im nm trong hoc trn bin ng gic li A1B1C1D1E1.

Cu a) c th gii quyt d dng nh nguyn l Dirichlet: V c 5 im nn tn ti t nht2 im X, Y m cp to (x, y) ca chng c cng tnh chn l (ta ch c 4 trng hp (chn,chn), (chn, l), (l, chn) v (l, l)). Trung im Z ca XY chnh l im cn tm.

Sang cu b) l lun trn y cha , v nu XY khng phi l ng cho m l cnh th Z cth s nm trn bin. Ta x l tnh hung ny nh sau. rng nu XY l mt cnh, chnghn l cnh AB th ZBCDE cng l mt ng gic li c cc nh c to u nguyn v tac th lp li l lun nu trn i vi ng gic ZBCDE, ... Ta c th dng n bin chngminh qu trnh ny khng th ko di mi, v n mt lc no s c 1 ng gic c imnguyn nm trong.Tuy nhin, ta c th trnh by li l lun ny mt cch gn gng nh sau: Gi s tn ti mtng gic nguyn m bn trong khng cha mt im nguyn no (phn v d). Trong tt c

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cc ng gic nh vy, chn ng gic ABCDE c din tch nh nht (phn v d nh nht). Nuc nhiu ng gic nh vy th ta chn mt trong s chng. Theo l lun trnh by cua), tn ti hai nh X, Y c cp to cng tnh chn l. Trung im Z ca XY s c to nguyn. V bn trong ng gic ABCDE khng c im nguyn no nn XY phi l mt cnhno . Khng mt tnh tng qut, gi s l AB. Khi ng gic ZBCDE c to cc nh

u nguyn v c din tch nh hn din tch ng gic ABCDE. Do tnh nh nht ca ABCDE(phn v d nh nht pht huy tc dng!) nn bn trong ng gic ZBCDE c 1 im nguynT. iu ny mu thun v T cng nm trong ng gic ABCDE.Phn v d nh nht cng l cch rt tt trnh by mt chng minh quy np ( y thngl quy np mnh), trnh nhng l lun di dng v thiu cht ch.

V d 4. Chng minh rng nu a, b l cc s nguyn dng nguyn t cng nhau th tn ticc s nguynx, y sao cho ax + by = 1.

Li gii. Gi s khng nh bi khng ng, tc l tn ti hai s nguyn dng a, b nguynt cng nhau sao cho khng tn ti x, y nguyn sao cho ax + by = 1. Gi a0, b0 l mt cp snh vy vi a0 + b0 nh nht (phn v d nh nht).

V (a0, b0) = 1 v (a0, b0) = (1, 1) (do 1.0 + 1.1 = 1) nn a0 = b0. Khng mt tnh tng qut,c th gi sa0 > b0. D thy (a0 b0, b0) = (a0, b0) = 1. Do a0b0 + b0 = a0 < a0 + b0 nndo tnh nh nht ca phn v d, ta suy ra (a0 b0, b0) khng l phn v d, tc l tn ti x, ysao cho (a0 b0)x + b0y = 1. Nhng t y th a0x + b0(y x) = 1. Mu thun i vi iu gis. Vy iu gi s l sai v bi ton c chng minh.Bi tp5. Gii phn c) ca v d 3.6. Trn mt phng nh du mt s im. Bit rng 4 im bt k trong chng l nh camt t gic li. Chng minh rng tt c cc im c nh du l nh ca mt a gic li.

3 Nguyn l cc hn v bt ng thcNguyn l cc hn thng c p dng mt cch hiu qu trong cc bt ng thc c tnh

t hp, dng chng minh tn ti k s tn s tha mn mt iu kin ny .

V d 5. (Moscow MO 1984) Trn vng trn ngi ta xp t nht 4 s thc khng m c tngbng 1. Chng minh rng tng tt c cc tch cc cp s k nhau khng ln hn 14 .

Li gii. Ta cn chng minh rng vi mi n 4 s thc khng m a1,...,an, c tng bng 1,ta c bt ng thc

a1a2 + a2a3 + ... + an1an + ana1 14

.

Vi n chn (n = 2m) iu ny c th chng minh d dng: t a1 + a3 + ... + a2m1 = a; khi, r rng,

a1a2 + a2a3 + ... + an1an + ana1 (a1 + a3 + ... + a2m1) (a2 + a4 + ... + a2m) = a(1 a) 14

.

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Gi sn l v ak l s nh nht trong cc s cho. ( thun tin, ta gi s1 < k < n1 - iuny khng lm mt tnh tng qut khi n 4.) t bi = ai, vi i = 1,...,k 1, bk = ak + ak+1v bi = ai+1 vi i = k + 1,...,n 1. p dng bt ng thc ca chng ta cho cc s b1,...,bn1,ta c:

a1a2 + ... + ak2ak1 + (ak1 + ak+2)bk + ak+2ak+3 + ... + an1an + ana1 1

4 .

Cui cng, ta s dng bt ng thc

ak1ak + akak+1 + ak+1ak+2 ak1ak + ak1ak+1 + ak+1ak+2 (ak1 + ak+2)bk, suy ra iu phi chng minh.nh gi trn y l tt nht; du bng xy ra khi 2 trong n s bng 12 , cn cc s cn li bng0.

V d 6. Cho n 4 v cc s thc phn bita1, a2, . . . , an tho mn iu kin

ni=1

ai = 0,ni=1

a2i = 1.

Chng minh rng tn ti 4 s a,b,c,d thuc{a1, a2, . . . , an} sao cho

a + b + c + nabc ni=1

a3i a + b + d + nabd.

Li gii. Nu a b c l ba s nh nht trong cc ai th vi mi i = 1, 2, . . . , n ta c btng thc

(ai a)(ai b)(ai c) 0Suy ra

a3i (a + b + c)a2i (ab + bc + ca)ai + abc vi mi i = 1, 2, . . . , n .

Cng tt c cc bt ng thc ny, vi ch ni=1

ai = 0,ni=1

a2i = 1 ta c

ni=1

a3i a + b + c + nabc.

By gi nu chn d l s ln nht trong cc ai

th ta c

(ai a)(ai b)(ai d) 0vi mi i = 1, 2, . . . , n . V cng thc hin tng t nh trn, ta suy ra bt ng thc v phica bt ng thc kp cn chng minh.

V d 7. Tng bnh phng ca mt 100 s thc dng ln hn 10000. Tng ca chng nhhn 300. Chng minh rng tn ti 3 s trong chng c tng ln hn 100.

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Li gii. Gi s 100 s l C1 C2 ... C100 > 0. Nu nhC1 100, th C1 + C2 + C3 >100. Do ta c th gi s rng C1 < 100. Khi 100 C1 > 0, 100 C2 > 0, C1 C2 0, C1 C3 0, v vy

100(C1 + C2 + C3) 100(C1 + C2 + C3) (100 C1)(C1 C3) (100 C2)(C2 C3)= C2

1+ C2

2+ C

3(300

C1

C2

)

> C21 + C22 + C3(C3 + C4 + . . . + C100)

C21 + C22 + C23 + . . . + C2100) > 10000.Suy ra, C1 + C2 + C3 > 100.Bi tp7. Trong mi ca bng 2 n ta vit cc s thc dng sao cho tng cc s ca mi ct bng1. Chng minh rng ta c th xo i mi ct mt s sao cho mi hng, tng ca cc s cnli khng vt qu n+14 .8. 40 tn trm chia 4000 euro. Mt nhm gm 5 tn trm c gi l ngho nu tng s tinm chng c chia khng qu 500 euro. Hi s nh nht cc nhm trm ngho trn tng s

tt c cc nhm 5 tn trm bng bao nhiu?

4 Nguyn l cc hn v phng trnh DiophantTrong phn ny, ta trnh by chi tit ba v d p dng nguyn l cc hn trong phng

trnh Fermat, phng trnh Pell v phng trnh dng Markov.

V d 8. Chng minh rng phng trnhx4 + y4 = z2 (1) khng c nghim nguyn dng.

Li gii. Gi s ngc li, phng trnh (1) c nghim nguyn dng, v (x,y,z) l nghimca (1) vi z nh nht.(1) D thy x2, y2, z i mt nguyn t cng nhau(2) T nghim ca phng trnh Pythagore, ta c tn ti p, q sao cho

x2 = 2pq

y2 = p2 q2z = p2 + q2

(3) T y, ta li c mt b ba Pythagore khc, v y2 + q2 = p2.(4) Nh vy, tn ti a, b sao cho

q = 2ab

y = a2 b2p = a2 + b2

a, b nguyn t cng nhau(5) Kt hp cc phng trnh ny, ta c:

x2 = 2pq = 2(a2 + b2)(2ab) = 4(ab)(a2 + b2)

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(6) V ab v a2 + b2 nguyn t cng nhau, ta suy ra chng l cc s chnh phng.(7) Nh vy a2 + b2 = P2 v a = u2, b = v2. Suy ra P2 = u4 + v4.(8) Nhng by gi ta thu c iu mu thun vi tnh nh nht ca z v:

P2 = a2 + b2 = p < p2 + q2 = z < z2.

(9) Nh vy iu gi s ban u l sai, suy ra iu phi chng minh.Phng php trnh by trn cn c gi l phng php xung thang. y c l l

phng php m Fermat ngh ti khi vit trn l cun sch ca Diophant nhng dng chm sau ny c gi l nh l ln Fermat v lm in u bao nhiu th h nhng nhton hc.

V d 9. Tm tt c cc cp a thcP(x), Q(x) tha mn phng trnh

P2(x) = (x2 1)Q2(x) + 1(1)Li gii. Khng mt tnh tng qut, ta ch cn tm nghim trong tp cc a thc c h s

khi u dng.Nu (x + x2 1)n = Pn(x)+x2 1Qn(x)(2) th (x x2 1)n = Pn(x)x2 1Qn(x)(3)Nhn (2) v (3) v theo v, ta c

1 = (x + x2 1)n(x

x2 1)n = (Pn(x) +

x2 1Qn(x))(Pn(x) x2 1Qn(x))

= P2n(x) (x2 1)Q2n(x)Suy ra cp a thc Pn(x), Qn(x) xc nh bi (2) v (3) l nghim ca (1). Ta chng minh yl tt c cc nghim ca (1). Tht vy, gi s ngc li, tn ti cp a thc P(x), Q(x) khngc dng Pn(x), Qn(x) tha mn (1). Ta xt cp a thc (P, Q) nh vy vi degQ nh nht. t

(P(x) +x2 1Q(x))(x x2 1) = P(x) + x2 1Q(x)(4)Th r rng(P(x)

x2 1Q(x))(x +

x2 1) = P(x)

x2 1Q(x)

Suy ra (P, Q) cng l nghim ca (1).Khai trin (4), ta thu c P(x) = xP(x) (x2 1)Q(x), Q(x) = xQ(x) P(x). Ch lt (1) ta suy ra (P(x) xQ(x))(P(x) + xQ(x)) = Q2(x) + 1. V P(x) v Q(x) u c hs khi u > 0 v degP = degQ + 1 nn ta c deg(P(x) + xQ(x)) = degQ + 1. T y, dodeg(Q2(x) + 1) 2deg(Q) nn ta suy ra deg(Q(x)) deg(Q) 1 < degQ.Nh vy, theo cch chn cp (P, Q) th tn ti n sao cho (P, Q) = (Pn, Qn).Nhng khi t (4) suy ra

P(x) + x2 1Q(x) = (P(x) +

x2 1Q(x))(x +

x2 1)= (x +

x2 1)n(x +

x2 1)

= (x + x2 1)n+1

Suy ra (P, Q) = (Pn+1, Qn + 1), mu thun.Vy iu gi s l sai v ta c iu phi chng minh.

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V d 10. Tm tt c cc gi trk sao cho phng trnh(x + y + z)2 = kxyz c nghim nguyndng.

Li gii. Gi sk l mt gi tr cn tm. Gi x0, y0, z0 l nghim nguyn dng ca phngtrnh

(x + y + z)2 = kxyz(1)

c x0 + y0 + z0 nh nht. Khng mt tnh tng qut, c th gi sx0 y0 z0.Vit li (1) di dng x2 (kyz 2y 2z)x + (y + z)2 = 0,ta suy ra x0 l nghim ca phng trnh bc hai

x2 (ky0z0 2y0 2z0)x + (y0 + z0)2 = 0(2)

Theo nh l Viet x1 = ky0z0 2y0 2z0 x0 = (y0+z0)2

x0cng l nghim ca (2). T

(x1, y0, z0) l nghim ca (1). Cng t cc cng thc trn, ta suy ra x1 nguyn dng. Tc l(x1, y0, z0) l nghim nguyn dng ca (1). T tnh nh nht ca x0 + y0 + z0 ta x1 x0. Ty ta c

ky0z0 2y0 2z0 x0 x0 v (y0 + x0)2x0

x0

T bt ng thc th hai ta suy ra y0 + z0 x0. T , p dng vo bt ng thc th nht,ta c ky0z0 4x0.Cui cng, chia hai v ca ng thc x20 + y

20 + z

20 + 2x0y0 + 2y0z0 + 2z0x0 = kx0y0z0 cho x0y0z0,

ta cx0

y0z0+

y0x0z0

+z0

x0y0+

2

z0+

2

x0+

2

y0= k.

T suy ra k4 + 1 + 1 + 2 + 2 + 2 k, tc l k 323 . Suy ra k 10.Ch nu x0 = 1 th y0 = z0 = 1 suy ra k = 9. Nu k = 9 th x0 2 v nh gi trn trthnh

k4 + 1 +

12 + 2 + 1 + 2 k suy ra k

263 , suy ra k 8Vy gi tr k = 10 b loi.

Vi k = 1 phng trnh c nghim, chng hn (9, 9, 9)Vi k = 2 phng trnh c nghim, chng hn (4, 4, 8)Vi k = 3 phng trnh c nghim, chng hn (3, 3, 3)Vi k = 4 phng trnh c nghim, chng hn (2, 2, 4)Vi k = 5 phng trnh c nghim, chng hn (1, 4, 5)Vi k = 6 phng trnh c nghim, chng hn (1, 2,3)Vi k = 8 phng trnh c nghim, chng hn (1, 1, 2)Vi k = 9 phng trnh c nghim, chng hn (1, 1, 1)Ngoi ra, ta c th chng minh c rng trng hp k = 7 phng trnh khng c nghim

nguyn dng (xin c dnh cho bn c).Vy cc gi tr k cn tm l k = 1, 2, 3, 4, 5, 6, 8, 9.

V d 11. (CRUX, Problem 1420) Nu a,b,c l cc s nguyn dng sao cho

0 < a2 + b2 abc cChng minh rng a2 + b2 abc l s chnh phng.

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Li gii. Gi s ngc li rng tn ti cc s nguyn dng a,b,c sao cho 0 < a2 + b2abc cv k = a2 + b2 abc (1) khng phi l s chnh phng.By gi ta c nh k v c v xt tp hp tt c cc cp s nguyn dng (a, b) tha mn phngtrnh (1), tc l ta xt

S(c, k) =

{(a, b)

(N)2 : a2 + b2

abc = k

}Gi s (a, b) l cp s thuc S(c, k) c a + b nh nht. Khng mt tnh tng qut c th gisa b. Ta xt phng trnh

x2 bcx + b2 k = 0Ta bit rng x = a l mt nghim ca phng trnh. Gi a1 l nghim cn li ca phng trnhny th a1 = bc a = (b

2k)a .

Ta c th chng minh c rng (bn c t chng minh!) a1 nguyn dng. Suy ra (a1, b)cng thuc S(c, k).Tip theo ta c a1 = (b2 k)/a < a2/a = a, suy ra a1 + b < a + b. iu ny mu thun vicch chn (a, b).

Bi tp9. Chng minh rng phng trnh x3 + 3y3 = 9z3 khng c nghim nguyn dng.10. Chng minh rng phng trnh x2 + y2 + z2 = 2xyz khng c nghim nguyn dng.11. (IMO 88) Nu a,b,q= (a2 + b2)/(ab + 1) l cc s nguyn dng th q l s chnh phng.12. (PTNK 03). Tm tt c cc s nguyn dng k sao cho phng trnh x2 (k2 4)y2 = 24c nghim nguyn dng.13. (Mathlinks) Cho A l tp hp hu hn cc s nguyn dng. Chng minh rng tn ti tphp hu hn cc s nguyn dng B sao cho A B,

xB

x =xB

x2

14.* (AMM 1995) Cho x, y l cc s nguyn dng sao cho xy + x v xy + y l cc s chnhphng. Chng minh rng c ng mt trong hai s x, y l s chnh phng.15. (IMO 2007) Cho a, b l cc s nguyn dng sao cho 4ab

1 chia ht (4a2

1)2. Chng

minh rng a = b.16. (VMO 2012) Xt cc s t nhin l a, b m a l c s ca b2 + 2 v b l c s ca a2 + 2.Chng minh rng a v b l cc s hng ca dy s t nhin (vn) xc nh bi

v1 = v2 = 1 v vn = 4vn1 vn2 vi mi n 2.

5 Nguyn l cc hn trong t hpTrn y chng ta xem xt cc v d p dng ca nguyn l cc hn trong mnh t

mu m nht dnh cho nguyn l cc hn. Nguyn l cc hn c th c ng dng chngminh mt qu trnh l dng (trong cc bi ton lin quan n bin i trng thi) trong biton v th, hay trong cc tnh hung t hp a dng khc. Cc i tng thng c emra xt cc hn thng l: on thng ngn nht, tam gic c din tch ln nht, gc lnnht, nh c bc ln nht, chu trnh c di ngn nht ...Di y ta xem xt mt s v d:

V d 12. (nh l Sylvester) Cho tp hp S gm hu hn cc im trn mt phng tha mntnh cht sau: Mt ng thng i qua 2 im thuc S u i qua t nht mt im th bathuc S. Khi tt c cc im ca S nm trn mt ng thng.

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Kt lun ca nh l nghe c v hin nhin nhng chng minh n th khng h n gin.Chng minh di y ca Kelly c chng ti tham kho t WikipediaGi s phn chng l tn ti mt tp hp S gm hu hn im khng thng hng nhng ming thng qua hai im trong S u cha t nht ba im. Mt ng thng gi l ngni nu n i qua t nht hai im trong S. Gi s (P,l) l cp im v ng ni c khongcch dng nh nht trong mi cp im-ng ni.Theo gi thit, l i qua t nht ba im trong S, nn nu h ng cao t P xung l th tn

Hnh 1:

ti t nht hai im nm cng mt pha ca ng cao (mt im c th nm ngay chnng cao). Trong hai im ny, gi im gn chn ng cao hn l B, v im kia l C.Xt ng thng m ni P v C. Khong cch tB ti m nh hn khong cch tP ti l, muthun vi gi thit v P v l. Mt cch thy iu ny l tam gic vung vi cnh huyn BCng dng v nm bn trong tam gic vung vi cnh huyn P C.Do , khng th tn ti khong cch dng nh nht gia cc cp im-ng ni. Ni cchkhc, mi im u nm trn ng mt ng thng nu mi ng ni u cha t nht baim.

V d 13. V d 13. (Trn u ton hc Nga 2010) Mt quc gia c 210 thnh ph. Ban ugia cc thnh ph cha c ng. Ngi ta mun xy dng mt s con ng mt chiu nigia cc thnh ph sao cho: Nu c ng i t A n B v t B n C th khng c ng it A n C.Hi c th xy dng c nhiu nht bao nhiu ng?

Li gii. Gi A l thnh ph c nhiu ng i nht (gm c ng i xut pht t A vng i n A). Ta chia cc thnh ph cn li thnh 3 loi. Loi I - C ng i xut pht tA. Loi II - C ng i n A. Loi III: Khng c ng i n A hoc xut pht t A. tm = |I|, n = |II|, p = |III|. Ta c m + n +p = 209.D thy gia cc thnh ph loi I khng c ng i. Tng t, gia cc thnh ph loi 2 khngc ng i.

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S cc ng i lin quan n cc thnh ph loi 3 khng vt qu p(m + n). (Do bc caA = m + n l ln nht).Tng s ng i bao gm:+ Cc ng i lin quan n A: m + n+ Cc ng i lin quan n III : p(m + n)

+ Cc ng i gia I v II: mnSuy ra tng s ng i nh hn mn + (p + 1)m + (p + 1)n (m + n + p + 1)2/3 = 2102/3.Du bng xy ra vi th 3 phe, mi phe c 70 thnh ph, thnh ph phe 1 c ng i

n thnh ph phe 2, thnh ph phe 2 c ng i n thnh ph phe 3, thnh ph phe 3 cng i n thnh ph phe 1.

V d 14. Trong quc hi M, mi mt ngh s c khng qu 3 k th. Chng minh rng c thchia quc hi thnh 2 vin sao cho trong mi vin, mi mt ngh s c khng qu mt k th.

y l mt v d m ti rt thch. C nhiu cch gii khc nhau nhng y chng ta strnh by mt cch gii s dng nguyn l cc hn. tng tuy n gin nhng c rt nhiu

ng dng (trong nhiu bi ton phc tp hn).Ta chia quc hi ra thnh 2 vin A, B mt cch bt k. Vi mi vin A, B, ta gi s(A), s(B) ltng ca tng s cc k th ca mi thnh vin tnh trong vin . V s cch chia l hu hnnn phi tn ti cch chia (A0, B0) sao cho s(A0) + s(B0) nh nht. Ta chng minh cch chiany tha mn yu cu bi ton.Gi s rng cch chia ny vn cha tho mn yu cu, tc l vn c mt ngh s no cnhiu hn 1 k th trong vin ca mnh. Khng mt tnh tng qut, gi s ngh s x thuc A0c t nht 2 k th trong A0. Khi ta thc hin php bin i sau: chuyn x tA0 sang B0 c cch chia mi l A = A0 {x} v B = B0 {x}. V x c t nht 2 k th trong A0 vA khng cn cha x nn ta cs(A) s(A0) 4 (trong tng mt i t nht 2 ca s(x) v 2 ca cc k th ca x trong A0)V x c khng qu 3 k th v c t nht 2 k th trong A0 nn x c nhiu nht 1 k th trongB0 (hay B), cho nn

s(B) s(B0) + 2T s(A) + s(B) s(A0) + s(B0) 2. Mu thun vi tnh nh nht ca s(A0) + s(B0). Vyiu gi s l sai, tc l cch chia (A0, B0) tha mn yu cu bi ton (pcm).Bi tp17. Cho 2n im trn mt phng, trong khng c 3 im no thng hng. Chng minh rngnhng im ny c th phn thnh n cp sao cho cc on thng ni chng khng ct nhau.18. Trong mt phng cho 100 im, trong khng c ba im no thng hng. Bit rng baim bt k trong chng to thnh mt tam gic c din tch khng ln hn 1. Chng minhrng ta c th ph tt c cc im cho bng mt tam gic c din tch 4.

19. Trn mt phng cho 2n + 3 im, trong khng c ba im no thng hng v khng c4 im no nm trn mt ng trn. Chng minh rng ta c th chn ra t cc im ny 3im, sao cho trong cc im cn li c n im nm trong ng trn v n im nm ngoing trn.20. Trong mt phng cho n im v ta nh du tt c cc im l trung im ca cc onthng c u mt l cc im cho. Chng minh rng c t nht 2n 3 im phn bit cnh du.21. Ti mt quc gia c 100 thnh ph, trong c mt s cp thnh ph c ng bay. Bit

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rng t mt thnh ph bt k c th bay n mt thnh ph bt k khc (c th ni chuyn).Chng minh rng c th i thm tt c cc thnh ph ca quc gia ny s dng khng qu a)198 chuyn bay b) 196 chuyn bay.22*. Trong mt nhm 12 ngi t 9 ngi bt k lun tm c 5 ngi i mt quen nhau.Chng minh rng tm c 6 ngi i mt quen nhau trong nhm .

Ti liu tham kho[1] Nguyn Vn Mu, Cc chuyn Olympic Ton chn lc,Ba V , 5-2010 .

[2] on Qunh ch bin, Ti liu gio khoa chuyn ton - i s 10, NXB GD, 2010.

[3] http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-n-4.html

[4] vi.wikipedia.org/wiki/nh l Sylvester-Gallai

[5] www.mathscope.org

[6] www.problems.ru

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MT S DNG TNG QUT CA PHNGTRNH HM PEXIDER V P DNG

Trnh o Chin, Trng Cao ng S Phm Gia LaiL Tin Dng, Trng THPT Pleiku, Gia Lai

Phng trnh hm Pexider (PTHP) l phng trnh hm tng qut trc tip ca Phng trnhhm Cauchy quen thuc. Bi vit ny cp n mt s dng tng qut ca PTHP v vi pdng ca n trong chng trnh Ton ph thng.

1 Phng trnh hm PexiderPTHP c bn gm bn dng di y (li gii c th xem trong [1] hoc [2])

Bi ton 1.1. Tm tt c cc hm sf, g, h xc nh v lin tc trnR tha mn iu kinf(x + y) = g (x) + h (y) , x, y R. (1)

Gii. Nghim ca phng trnh (1) l

f(t) = ct + a + b, g (t) = ct + a, h (t) = ct + b; a,b,c R.Bi ton 1.2. Tm tt c cc hm sf, g, h xc nh v lin tc trnR tha mn iu kin

f(x + y) = g (x) h (y) , x, y R. (2)Gii. Nghim ca phng trnh (2) l

f(t) = abec

t, g (t) = aec

t, h (t) = bec

; a,b,c Rhoc

f 0, g 0, h CR,trong CR l tp hp cc hm s lin tc trn R,hoc

f 0, h 0, g CR.Bi ton 1.3. Tm tt c cc hm sf, g, h xc nh v lin tc trnR+ tha mn iu kin

f(xy) = g (x) + h (y) , x, y R+. (3)Gii. Nghim ca phng trnh (3) l

f(t) = m.lnt + a + b, g (t) = m.lnt + a, h (t) = m.lnt + b; m,a,b,c R.Bi ton 1.4. Tm tt c cc hm sf, g, h xc nh v lin tc trnR+ tha mn iu kin

f(xy) = g (x) h (y) , x, y R+. (4)Gii. Nghim ca phng trnh (4) l

f(t) = abtc, g (t) = atc, h (t) = btc; a,b,c R.

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2 Mt s dng tng qut ca Phng trnh hm PexiderTy theo mc kin thc, PTHP c nhiu dng tng qut khc nhau. Di y l mt s

dng tng qut ca phng trnh (1) gn gi vi chng trnh ca h ph thng chuyn Ton.Bi ton 2.1. Tm tt c cc hm s f, fi (i = 1, 2,...,n) xc nh v lin tc trn R tha

mn iu kinf

ni=1

xi

=

ni=1

fi (xi), x, xi R. (5)

Gii. y l dng quy np mt cch t nhin ca Bi ton 1.1. Nghim ca phng trnh(5) l

f(t) = at +

ni=1

ai, fi (t) = at + ai; a, ai R.

Tng t Bi ton 2.1, ta cng c th gii c phng trnh hm dng

fni=1

aixi

=

ni=1

aifi (xi), x, xi R, ai R.

Bi ton sau y l mt dng tng qut kh c bn, m phng php quy np khng thp dng trong li gii. Mt s phn chng minh c s dng mt s kin thc c bn, khngqu kh, ca i s tuyn tnh v Phng trnh vi phn, thuc chng trnh c s ca Toncao cp.Bi ton 2.2. Tm tt c cc hm s f, fi, gi (i = 1, 2,...,n) xc nh v tn ti o hm(theo mi bin s c lp x, y) trnR tha mn iu kin

f(x + y) =

nk=1

fk (x) gk (y), x, y R, n 2. (6)

Gii. Ta gii bi ton ny trong trng hp n = 2. Trng hp n 3 c gii tng t.Xt phng trnh hm

f(x + y) = f1 (x) g1 (y) + f2 (x) g2 (y) , x, y R, (7)trong cc hm f, f1, f2, g1, g2 xc nh v tn ti o hm (theo mi bin s c lp x, y)trn R.

Khng mt tnh tng qut, ta lun c th gi thit rng cc h hm {f1 (x) , f2 (x)} v{g1 (x) , g2 (x)} l c lp tuyn tnh. Ta c

fx (x + y) = f

1 (x) g1 (y) + f

2 (x) g2 (y) ,

fy (x + y) = f1 (x) g

1 (y) + f2 (x) g

2 (y) .

V fx (x + y) = fy (x + y), nn

f1 (x) g1 (y) + f

2 (x) g2 (y) = f1 (x) g

1 (y) + f2 (x) g

2 (y) . (8)

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Ngoi ra, v {g1 (x) , g2 (x)} l c lp tuyn tnh, nn tn ti cc hng s y1 , y2 sao cho g1 (y1) g1 (y2)g2 (y1) g2 (y2)

= 0.Thay y1, y2 vo (8), ta c

f1 (x) g1 (y1) + f

2 (x) g2 (y1) = f1 (x) g

1 (y1) + f2 (x) g

2 (y1) ,

f1 (x) g1 (y2) + f

2 (x) g2 (y2) = f1 (x) g

1 (y2) + f2 (x) g

2 (y2) .

V nh thc nu trn khc 0, nn h phng trnh ny c nghim duy nht f1 (x), f

2 (x).Do , ta c th biu din f1 (x) v f

2 (x) qua f1 (x) v f2 (x) di dng

f1 (x) = a11f1 (x) + a12f2 (x) , f

2 (x) = a21f1 (x) + a22f2 (x) . (9)

Mt khc, thay y = 0 vo (7), ta c

f(x) = c1f1 (x) + c2f2 (x) . (10)

Suy raf (x) = c11f1 (x) + c12f2 (x) , f

(x) = c21f1 (x) + c22f2 (x) . (11)

- Nu

c11 c12c21 c22

= 0, th t (10) v (11), ta thu c phng trnh vi phn tuyn tnhthun nht

a1f (x) + a2f

(x) = 0,

trong a1 v a2 khng ng thi bng 0. Gii phng trnh vi phn ny, ta tm c f(x).Tt c cc hm f(x) ny u tha mn (7) nn l nghim ca phng trnh.

- Nu c11 c12

c21 c22

= 0, th t (10) v (11), ta c th biu din f1 (x) v f2 (x) bi mt t hptuyn tnh ca f (x) v f (x). Thay biu din ny vo (5), ta thu c phng trnh dng

f(x) + a1f (x) + a2f

(x) = 0.

Gii phng trnh vi phn ny, ta tm c f(x) .Tt c cc hm f(x) ny u tha mn (7) nn l nghim ca phng trnh. Bi ton

c gii quyt.Di y l mt s trng hp c bit m phng trnh (7) tr thnh mt s phng trnh

hm c bn. Nhng phng trnh ny kh ni ting v c li gii hon ton s cp (c th

xem trong [1] hoc [2]).- Vi f1 (x) = f(x), g1 (y) 1, f2 (x) 1, g2 (y) = f(y), phng trnh (7) tr thnh Phng

trnh hm Cauchyf(x + y) = f(x) + f(y) , x, y R.

- Vi f1 (x) = g (x), g1 (y) 1, f2 (x) 1, g2 (y) = h (y), phng trnh (7) tr thnh Phngtrnh hm Pexider

f(x + y) = g (x) + h (y) , x, y R.

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- Vi f2 (x) 1, phng trnh (7) tr thnh Phng trnh hm Vinczef(x + y) = f1 (x) g1 (y) + g2 (y) , x, y R,

- Vi f1 (x) = f(x), g1 (y) = g (y), f2 (x) = g (x), g2 (y) = f(y), phng trnh (7) trthnh phng trnh hm dng lng gic (v mt nghim ca phng trnh ny l f(t) = sint,g (t) = cost)

f(x + y) = f(x) g (y) + g (x) f(y) , x, y R.

3 p dngPTHP tng qut c nhiu p dng trong vic nghin cu mt s vn lin quan ca Ton

ph thng. Sau y l mt p dng lin quan n cc php chuyn i bo ton yu t gc camt tam gic.Bi ton 3.1. Tm tt c cc hm s f, g, h xc nh v lin tc trnR tha mn iu kinsau: NuA,B,C

R, A+B +C = , thA1+B1+C1 = , trong A1 = f(A), B1 = f(B),

C1 = f(C).Gii. Gi s cc hm s f, g, h xc nh v lin tc trn R tha mn iu kin trn. Ta c

A1 + B1 + C1 = f(A) + f(B) + f(C) = f( B C) = f(B) f(C) . (12)

t F (x) = f( x), G (x) = 2

g (x), H(x) = 2

h (x). Khi , phng trnh (12) cdng

F(B + C) = G (B) + H(C) . (13)

Phng trnh (13) chnh l Phng trnh Pexider bit. Nghim lin tc tng qut ca phng

trnh ny l

F(x) = ax + c1 + c2, G (x) = ax + c1, H(x) = ax + c2,

trong a, c1, c2 R.Do

f(x) = a ( x) + c1 + c2, g (x) = 2

ax c1, h (x) = 2

ax c2. (14)

t a = k, c1 + c2 + a = , 2

c1 = , 2

c2 = . Th th k + + + = 1. Khi, bi (14), ta thu c

f(x) = kx + , g (x) = kx + , h (x) = kx + ,

trong k + + + = 1.R rng cc hm s f, g, h nu trn tha mn iu kin ca bi ton.

Bi ton 3.2. Tm tt c cc hm s f, g, h xc nh v lin tc trnR tha mn iu kinsau: Nu A,B,C 0, A + B + C = , th A1, B1, C1 0, A1 + B1 + C1 = , trong

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A1 = f(A), B1 = f(B), C1 = f(C) .Gii. Tng t cch gii trn, ta tm c

f(x) = kx + , g (x) = kx + , h (x) = kx + ,

trong

0,

0, k +

0, k +

0, +

1, k + +

1.

Kt qu ca Bi ton 3.2 c nhiu p dng trong cc php chuyn i bo ton yu t gctrong tam gic, chng hn cc H qu sau y m phn chng minh dnh cho bn cH qu 3.1. NuA, B, C l ba gc ca mt tam gic, thA1, B1, C1 xc nh nh sau

A1 =B + C

2, B1 =

C+ A

2, C1 =

A + B

2

cng l ba gc ca mt tam gic.H qu 3.2. NuA, B, C l ba gc ca mt tam gic tha mn max {A,B,C} <

2, tc l

tam gicABC nhn, thA1, B1, C1 xc nh nh sau

A1 = 2A, B1 = 2B, C1 = 2Ccng l ba gc ca mt tam gic.H qu 3.3. NuA, B, C l ba gc ca mt tam gic, thA2, B2, C2 xc nh nh sau

A2 =A

2, B2 =

B

2, C2 =

+ C

2

cng l ba gc ca mt tam gic, trong C2 l gc t.H qu 3.4. NuA, B, C l ba gc ca mt tam gic, trong C l gc t, thA2, B2, C2xc nh nh sau

A2 = 2A, B2 = 2B, C2 = 2 C

cng l ba gc ca mt tam gic.H qu 3.5. Nu tam gicABC c ba gc nhn (hoc vung tiC), thA3, B3, C3 xc nhnh sau

A3 =

2 A, B3 =

2 B, C3 = C,

cng l ba gc ca mt tam gic t (hoc vung ti C3).H qu 3.6. Nu tam gicABC c gcC t (hoc vung), thA3, B3, C3 xc nh nh sau

A3 =

2 A, B3 =

2 B, C3 = C,

cng l ba gc ca mt tam gic nhn (hoc vung tiC3).

By gi, m rng mt cch t nhin cc bi ton trn, ta c cc bi ton sauBi ton 3.3. Tm tt c cc hm s f, g, h xc nh v lin tc trnR tha mn iu kinsau: NuAi R,

ni=1

Ai = (n 2) , thni=1

Ai = (n 2) , trong Ai = f(Ai).Gii. Tng t cch gii Bi ton 3.1, trong phng trnh hm cm sinh chnh l Phngtrnh hm Perxider tng qut. Cc hm s tm c l

fi (x) = k0x + ki (n 2) (i = 1, ..., n, n 3) ,

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trong n

j=0

kj = 1.

Tng t, m rng Bi ton 3.2, ta thu cBi ton 3.4. Tm tt c cc hm s fi (i = 1,...,n,n 3) xc nh v lin tc trnR thamn iu kin sau: Nu0

Ai

2,ni=1 Ai = (n 2) , th0 Ai 2,

ni=1Ai = (n 2) ,trong Ai = f(Ai).Gii. Tng t cch gii Bi ton 3.2, trong phng trnh hm cm sinh chnh l Phngtrnh hm Perxider tng qut. Cc hm s tm c l

fi (x) = k0x + ki (n 2) (i = 1, ..., n, n 3) ,trong 0 ki (n 2) 2, 0 2k0 + ki (n 2) 2.

Thu hp gi thit ca Bi ton 3.4, ta thu cBi ton 3.5. Tm tt c cc hm s fi (i = 1,...,n,n 3) xc nh v lin tc trnR thamn iu kin sau: Nu 0

Ai

,

n

i=1Ai = (n 2) , th0 A

i

,

n

i=1 Ai = (n 2) ,trong Ai = f(Ai).Gii. Tng t cch gii Bi ton 3.4, trong phng trnh hm cm sinh chnh l Phngtrnh hm Perxider tng qut. Cc hm s tm c l

fi (x) = k0x + ki (n 2) (i = 1, ..., n, n 3) ,trong 0 ki (n 2) 1, 0 k0 + ki (n 2) 1.

T nhng kt qu trn ta thy rng, vi ba gc ca mt tam gic cho trc, c th tora c ba gc ca mt tam gic mi v do c th suy ra c nhiu h thc lng giclin quan n cc gc ca tam gic . Hn na, bng cch phi hp nhng phng php khcnhau, ta cn c th to ra c nhiu ng thc v bt ng thc lng gic khc, v cng

phong ph. Sau y l mt vi v d.Gi s rng, ta chng minh c cc h thc sau y v xem chng l nhng h thc"gc" ban u

sin A + sin B + sin C 3

3

2, (15)

cos A cos B cos C 18

(16)

0 < sin A sin B sin C 3

3

8, (17)

sin2A + sin 2B + sin 2C = 4 sin A sin B sin C. (18)

p dng H qu 3.1 vo (15), ta c

sin

A

2

+ sin

B

2

+ sin

C

2

3

3

2.

Nh vy, ta to c bt ng thc sau

Bt ng thc 1. cosA2

+ cosB

2+ cos

C

2 3

3

2.

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cos

A

2

cos

B

2

cos

C

2

1

8.

Nh vy, ta to c bt ng thc sauBt ng thc 2. sin A2

sinB

2sin

C

2 1

8.


sin2

A

2

+ sin 2

B

2

+ sin 2

C

2

= 4 sin

A2

sin B

2sin

C2

.

hay

sin( A) + sin ( B) + sin ( C) = 4 sin A2

sin B

2sin

C2

.

Nh vy, ta to c ng thc sau

ng thc 1. sin A + sin B + sin C = 4cosA2

cosB2

cosC2

.

By gi, sng tc thm nhng h thc a dng hn, ta tip tc khai thc nhng kt qutrn, chng hn t Bt ng thc 2 ta c

8sinA

2sin

B

2sin

C

2 1 32sin A

2sin

B

2sin

C

2.cos

A

2cos

B

2cos

C

2 4cosA

2cos

B

2cos

C

2

4

2sinA

2cos

A

2

2sin

B

2cos

B

2

2sin

C

2cos

C

2

4cosA

2cos

B

2cos

C

2

4sin A sin B sin C

4cos

A

2cos

B

2cos

C

2. (19)

Nh vy, ta to c bt ng thc sau

Bt ng thc 3. sin A sin B sin C cosA2

cosB

2cos

C

2.

Bi (18) v ng thc 1, t (19), ta c bt ng thc sauBt ng thc 4. sin2A + sin 2B + sin 2C sin A + sin B + sin C.

Ta tip tc khai thc Bt ng thc 4. Nhn xt rng, nu tam gic ABC l tam gic nhnth, p dng H qu 3.2 vo Bt ng thc 4, ta c

sin2( 2A) + sin 2 ( 2B) + sin 2 ( 2C)

sin(

2A) + sin (

2B) + sin (

2C)

sin4A sin4B sin4C sin2A + sin 2B + sin 2C.Nh vy, ta tip tc to c bt ng thc sauBt ng thc 5. sin2A + sin 2B + sin 2C+ sin 4A + sin 4B + sin 4C 0.

By gi, p dng H qu 3.3 vo Bt ng thc 4, ta c

sin

2.

A

2

+ sin

2.

B

2

+ sin

2.

+ C

2

sin A

2+ sin

B

2+ sin

+ C

2.

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Ta to c bt ng thc sau

Bt ng thc 6. sin A + sin B sin C sin A2

+ sinB

2+ cos

C

2.

By gi, gi s tam gic ABC c gc C t. p dng H qu 3.4 vo Bt ng thc 1, ta c

cos2A2 + cos2B2 + cos2C 2 332 .

Ta to c bt ng thc sau

Bt ng thc 7. cos A + cos B + sin C 3

3

2

C >

2

.

Tip theo, gi s tam gic ABC nhn (hoc vung ti C). p dng H qu 3.5 vo (17), tac

0 < sin

2 A

sin

2 B

sin( C) 3

3

8

Ta c bt ng thc sau

Bt ng thc 8. 0 < cos A cos B sin C 338

C

2

.

By gi, gi s tam gic ABC c gc C t (hoc vung). p dng H qu 3.6 vo (15), tac

0 < sin

2 A

+ sin

2 B

+ sin ( C) 3

3

2

Ta c bt ng thc sau

Bt ng thc 9. 0 < cos A + cos B + sin C 3

3

2

C

2

.

TI LIU THAM KHO

[1] J. Aczl (1966), Lectures on Functional equations and their applications, Chapter 3, pp.141-145, Chapter 4, pp. 197-199.

[2] Nguyn Vn Mu, Mt s lp phng trnh hm a n hm dng c bn, K yu Hitho khoa hc "Cc chuyn chuyn Ton bi dng hc sinh gii Trung hc ph thng", HNi, 2011.

[3] D.S. Mitrinovic, J.E. Pecaric and V. Volenec (1989), Recent advances in geometric inequal-ities, Mathematics and its applications (East European series), Published by Kluwer AcademicPublishers, the Netherlands, Chapter V, pp. 64-69.

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XY DNG MT LP PHNG TRNH HM NHCC HNG NG THC LNG GIC

L Sng, Trng THPT Chuyn L Qu n - Khnh HaTrong cc k thi i hc cu hi v phng trnh, bt phng trnh thng c ch ,th trongcc cu hi ca thi chn hc sinh gii quc gia hay quc t cc bi ton v phng trnhhm cng chim phn trng tm. Trong bi vit ny chng ti th lin h kin thc v lnggic hc trong chng trnh ph thng a n mt s bi ton c nghim l hm slng gic

1 Cc hm s lng gic trong chng trnh ton v vitnh cht

sin(x y) = sin x cos y sin y cos x, x, y R(1)sin(x + y)sin(x y) = sin2 x cos2 y sin2 y cos2 x, x, y R(2)cos(x y) = cos x cos y sin x sin y, x, y R(3)

T (2) a n cng thc ca phng trnh hm n l hm sin

g(x + y)g(x y) = g2(x) g2(y) vi mi x, y R.T (3) ta cng t c cng thc ca hm cosin (phng trnh hm dAlembert )

f(x + y) + f(x y) = 2f(x)f(y) vi mi x, y R.

Ngoi ra t mt s cng thc lng gic m ta cng on c nghimf(2x) = 2f2 (x) 1, f(3x) = 4f3 (x) 3f(x),x R.

Quy c: fn(x) = [f(x)]n.

Bn Phng trnh hm c bn : Trong cc bi ton sau phn nhiu trc khi i nkt qu thng phi qua trung gian l cc phng trnh hm c bn sauCc phng trnh Cauchy

A(x + y) = A(x) + A(y) (I)

E(x + y) = E(x).E(y) (II)

L(xy) = L(x) + L(y) vi x > 0 (III)F(xy) = F(x).F(y) vi x > 0 (IV)

Ta c ln lt cc nghim l A(x)=ax ,vi a=f(1) c gii bi A.L.cauchy 1821

E(x) = exp(ax) hay E(x) = 0L(x) = alnx hay L(x) = 0F(x) = xc hay F(x) = 0

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2 Phng trnh hm dAlembert Hm cosinBi ton 1. Tm cc hmf(x) xc nh v lin tc trn v tha mn cc iu kin

f(x + y) + f(x y) = 2f(x) f(y) , x, y Rf(0) = 1, x0 R : |f(x0)| < 1Li gii. V f(0) = 1 v f(x) lin tc trn R nn > 0 sao cho f(x) > 0, x (, )Khi theo (2) vi n0 N ln th

f(x02n0

) > 0 f( x02n0

) < 1 (do phn chng )

Vy tn ti x1 = 0, x1 = x02n0 sao cho

0 < f(x1) < 1.f(x) > 0, x ( |x1| , |x1|) , f(x1) = cos , 0 < < 2

T (1) suy ra f(2x1) = 2f2(x1) 1 = 2cos2 1 = cos 2Gi sf(kx1) = cos k, k = 1, 2,...,n N+. Khi

f((n + 1) x1) = f(nx1 + x1)

= 2f(nx1) f(x1) f((n 1) x1)= 2 cos n cos cos(n 1) = cos (n + 1) .

T suy ra f(mx1) = cos m, m N+ v f(x) l hm chn trn R v nh vy

f(mx1 = cos m, m Z(3)Do tnh tr mt trong R , f(x) v cos x l cc hm lin tc trn R nn f(x) = cos ax,a RTh li ta thy f(x) = cos ax (a = 0) tha mn cc iu kin ca bi ton .Nhn xt 1. Thay trong gi thit |f(x0)| < 1 bi ton 1 bi |f(x0)| > 1 ta c nghim cabi ton l f(x) = cosh(x), y l hm cosin hyprebol m ta khng kho st chng trnhhc ph thngNhn xt 2. Khi f l hm kh vi, ly o hm theo y hai ln , ta c f(0) = 0, f(x) =k.f(x), k hng Nu k = 0 th f(x) = ax + b;Nu k > 0 th f(x) = c sin bx + d cos bx, c,d hngTf(0) = 1, f(0) = 0 suy ra d = 1, bc = 0, b = 0 th f hng; c = 0 th f(x) = cos x

Nu k < 0 th f(x) = c sinh bx + d cosh bx vi b2

= k, iu kin f(0) = 1, f

(0) = 0.Vy nghim l f(x) = cos bx, b s thc.

nh l 1. (nh l nghim ca Phng trnh dAlembert)Cho f : R R hm lin tc v tha mn iu kin

(C) f(x + y) + f(x y) = 2f(x)f(y), vi mix, ythf(x) = 0, f(x) = 1, f(x) = cos ax, hayf(x) = coshbx,a,b R.

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Bi ton 2. (IMO1972). Tmf : R R lin tc v tha mn cc iu kin

f(x + y) + f(x y) = 2f(x)f(y), vi mix, y

Chng minh rng nuf(x) = 0, |f(x)| 1, x R th|g(y)| 1, y R

Li gii. Do f b chn

f(x) = 0 |f(x)| |g(y)| = 2 |f(x + y) + f(x y)| |f(x + y)| + |f(x y)| 2M, x R

Bi ton 3. Tm tt c cc hm lin tc D R tha

f(x + y) =f(x) + f(y)

1 f(x)f(y)

Li gii. Ta bit rng hm f(x) = tan x tha bi.V th nu t A(x) = arctanf(x) th A(x + y) = A(x) + A(y) 2k .Suy ra A(x) = kx mod 2 , t dn n f(x) = tan kcx.

Bi ton 4. Tm cc hmf(x) xc nh v lin tc trnR v tha mn cc iu kin

f(x) + f(y) = f

x + y

1 xy

, x, y : x + y > 0

Li gii. t x = cotgu, y = cotgv, 0 < u, v <

Thx+y1xy = cotg (u + v)

Hay A (u + v) = A (u) + A (v) , 0 < u, v < trong A (u) = f(cotgu) .

f(x) = karcctgx, k , x R (i)

Th li ta thy hm f(x) xc nh theo (i) tha mn cc iu kin ca bi tonKt lun f(x) = karcctgx, k R, x RBi ton 5. Tm cc cp hmf(x) vg(x) xc nh v lin tc trnR v tha mn iu kin

f(x y) = f(x)g(y) + f(y)g(x)

Li gii. Nghim ca bi ton lf(x) = cg(x) = 1 c2 hay

f(x) = cos kxg(x) = sin kx , k R

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Bi ton 6. (Putnam1991)Cho hai hm f : R R, g : R R, f(x), g(x) khc hng, kh vi v tha mn iu kin

f(x + y) = f(x)f(y) g(x)g(y)g(x + y) = f(x)g(y) + g(x)f(y), x, y R

v f(0) = 0 .Chng minh rng f2(x) + g2(x) = 1

Li gii. Ta ch cn chng minh rng H(x) = f2(x) + g2(x) l hngTht vy ly o hm theo y ri thay y = 0ta c f(x) = g(0)g(x) vg(x) = g(0)f(x)Do 2f(x)f(x) + 2g(x)g(x) = 0 suy ra f2(x) + g2(x) = CNgoi ra f2(x + y) + g2(x + y) = (f2(x) + g2(x))(f2(y) + g2(y)) ,nn C2 = C.Nhng C = 0,nn C = 1Nhn xt.T gi thit ca bi ton ta thy hai hm f(x) = cos x, g(x) = sin x l nghim cabi ton, nn ta t E(x) = f(x) + ig(x) ,t gi thit bi ton ta c E(x + y) = E(x)E(y)V vy ta c 1 cch gii khc nh sau

Do E hm kh vi nn E(0) = ib, E(x + y) = E(x)E(y) Ly o hm y ,ri cho y = 0 ta cE(x) = Ceibx,TE(0 + 0) = E(0)E(0), ta rt ra c C = 1.Cui cng f2(x) + g2(x) = |E(x)| =

eibx

2= 1

Bi ton 7. a.Tm cc cp hmf : R R, g : R R lin tc v tha mn iu kinf(x y) = f(x)g(y) f(y)g(x)g(x y) = g(x)g(y) + f(x)f(y), x, y R

p s. Nghim ca bi ton l f(x) = sin bx,g(x) = cos bx, b s thc hayf(x) = g(x) = 0

b. f(x + y) = f(x)g(y) + f(y)g(x)

g(x + y) = g(x)g(y) f(x)f(y), x, y Rp s . Nghim l f(x) = ax/2 sin bx,g(x) = ax/2 cos bx, a > 0, b s thc hay f(x) =

g(x) = 0.

Bi ton 8. Tm cc cp hmf(x) vg(x) xc nh v lin tc trnR v tha mn[f(x) + g(x)]2 = 1 + f(2x), f(0) = 0f(x + y) + f(x y) = 2f(x)g(y), x, y R

p s . Nghim ca bi ton l f(x) = sin bx, g(x) = cos bx, bs thc hay f(x) = g(x) = 0

3 Phng trnh hm c cha hm s lng gic3.1 Trc ht xt hm thc f xc nh vi mi x,y thuc R tha

f(x + y) + f(x y) = 2cosxcosy, (3.1)

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Kt qu 1. f(x) l nghim ca 4.1 khi v ch khi f(x) = cos xKt qu 2. Phng trnh f(x + y) + f(x y) = 2sin x sin y , khng c nghimKt qu 3. Phng trnh f(x + y) + g(x y) = 2 sin x sin yc nghim f(x) = c cos x, g(x) = cos x cDo cng thc bin i 2sin x sin y = cos(x y) cos(x + y) suy ra f(u) + cos u = cos v g(v)Kt qu 4. Phng trnh f(x + y) f(x y) = 2 sin x sin y c nghimf(x) = c cos x, c l hngKt qu 5. Phng trnh f(x + y) + f(x y) = 2 cos x sin y khng c nghimKt qu 6. Phng trnh f(x + y) + g(x y) = 2 cos x sin y,c nghim f(x) = c + sin x, g(x) = sin x cKt qu 7. Phng trnh f(x + y) + f(x y) = 2sin x cos y ,c nghim f(x) = sin x3.2 Xt hai hm thc f ,g xc nh trn R tha

f(x + y) + g(x y) = 2 sin x cos y, (3.2)Kt qu 8. Nghim ca (3.2) l f(x) = c + sinx, g(x) = sinx cCho y = 0, ri thay y =

y vo (3.2)

ta c f(x) + g(x) = 2sin x v f(x + y) + g(x y) f(x y) + g(x + y) = 03.3 Cho hm thc f(x) tha f(x + y) + f(x y) = 2f(x)cos y (3.3) .Khi hm g(x) = f(x) d cos x tha g(x + y) = g(x)cos y + g(y)cos x (3.4)Kt qu 9. Nghim tng qut ca (3.4) l g(x) = b sin x, do f(x) = b sin x + d cos xTht vy, cho x = 0, ri hon i x, y trong (3.3) ta c f(y) + f(y) = 2d cos y v 2f(x +y) + f(x y) + f(y x) = 2f(x)cos y + 2f(y)cos xTc l 2f(x + y) + 2d cos(x y) = 2f(x)cos y + 2f(y)cos x. Suy ra (3.4) c nghim (3.4) , dng tnh kt hp ca hm g(x)

g(x + y + z) = (g(x)cos y + g(y))cos x + g(z)cos(x + y)

= g(x) cos(y + z) + (g(y)cos z+ g(z)cos y)cos x

Suy ra g(x)sin y sin z = g(z)sin y sin x, mi x,y,zC nh y, z vi sin z= 0, ta c g(x) = b sin x3.4. Cho hai hm f : R R, g : R R tha

f(x y) f(x + y) = 2g(x)sin y, vi mi s thc x, y(3.5)khi v ch khi f(x) = a cos x d sin x + c, g(x) = a sin x + d cos xLi gii. Cho x = 0, ri hon i x, y trong (3.5) ta c f(y) f(y) = 2d sin y v

f(x y) f(y x) = 2g(x)sin y 2g(y)sin x

Tc l 2d sin(x y) = 2g(x)sin y 2g(y)sin x.Suy ra (g(x) d cos x)sin y = (g(y) d cos y)sin x, tc l g(x) = d cos x + a sin x (c nh y).Thay g(x) vo (3.5) ta c kt qu3.5.Lp lun tng t ta cng c vi phng trnh hm

sin(x + y) = f(x)sin y + f(y)sin x (3.6)

sin(x + y) = g (x)sin y + g (y)sin x (3.7)

f(x + y) = g (x)sin y + g (y)sin x (3.8)

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Kt qu 10. Nghim (3.6 ) l f(x) = cos x ,nghim ca (3.7 ) l f(x) = cos x + d sin x, g(x) =cos x b sin x , nghim ca(4.8) l f(x) = a cos x = g(x)3.6. Dng tnh kt hp ca hmXt f(x + y)f(x y) = sin2(x) sin2(y), (3.9)Kt qu 11. (4.9) c nghim l f(x) = k sin x, k2 = 1

Bi ton 9. Nghim ca f(x + y)f(x y) = f2(x) sin2(y), (3.10)L f(x) = a sin x hay f(x) = b cos x + d sin x,a,b,d hng tha a2 = 1 = b2 + d2

Li gii. Cho x = y trong (4.10).Xt 2 trng hp f(0) = 0 hay khc khngTrng hp khc khng ta a v dng f(u)f(v) = bf(u + v) + sin u sin v.Sau xt bf(u + v + z) = f(u)(f(v)f(z) sin v sin z) sin u(sin v cos z+ cos v sin z)

Tnh cht 1. Cho hm f(x)xc nh v lin tc trn R v tha mn iu kin

(S)f(x + y + 2d) + f(x y + 2d) = 2f(x)f(y), vi mi x,y,d hng khc khngth f l hm s l

Tnh cht 2.

(i) Nu f(0) = 1 v f(d) = 1 th hm f c chu k l d(ii) Nu f(0) = 1 v f(d) = 1 th hm f c chu k l 2d(iii) Nu f(0) = 1 th f(d) = 0 hm f c chu k l d

Bi ton 10. Tm hmf(x) tha mn (S*)

Li gii. Trng hp (i) v (ii) th f tha phng trnh dAlembert ca hm (C)Trng hp (iii) th g(x) = f(x + 2d) l nghim ca phng trnh (C)g(x + y) + g(x y) = 2g(x)g(y), g(x) c chu k 4dKt qu. Nghim ca bi ton (S*) l f(x) = cos 2nxd hayf(x) = cos (2n+1)xd .

4 Phng trnh hm sin (S)f(x + y)f(x y) = f2(x) f2(y),vi mi x, y

Tnh cht 3. Hm f khc khng, tha (S) l hm lBi ton 11. Cho f : R R lin tc tha (S)Thf(x) = cx , f(x) = c sin bxhayf(x) = sinh bx

Li gii. Do f lin tc, kh vi. Ly o hm ln th nht theo y, ln th hai theo xSuy ra f(x) = kf(x) . Nh vy ta c kt qu

Bi ton 12. (Corovei) Cho hai hmf : R R, g : R R khc khng thaf(x)g(y x) = f

y

2

g

y

2

f

x y

2

g

x y2

, x, y R()

Khi g l nghim ca (S) v

g(x) = A(x), f(x) = c+dA(x)(1), g(x) = b(E(x)E(x)), f(x) = c(E(x)E(x))+dE(x)(2)trong A, E tha phng trnh c bn, b vc l cc hng s.

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Li gii. Thay x bi x + y, v y bi 2x vo (*) f(x + y)g(x y) = f(x)g(x) f(y)g(y)(3),sau i ch x v y vi nhau ta c f(x + y)g(y x) = f(x + y)g(x y),suy ra g(x) hm l ( do f khc khng)Ly y = x ri thay y bi y trong (1) , tr (*) cho (3)

f(x

y)g(x + y)

f(x + y)g(x

y) = g(y)(f(

y) + f(y))

Xt 2 trng hp f(0), f(0) = 0, th f v g trong (1) l nghim; f(0) khc khng, f v g trong(2) l nghimKt qu Trng hp hm lin tc, nghim khc khng l

f(x) = bx + c, g(x) = ax;

f(x) = c sin ax + d cos ax,g(x) = b sin ax;

f(x) = b sinh ax + d cosh ax, g(x) = b sinh ax.

5 M rng phng trnh hm dAlembert dng lng

gic ca W. H. Wilson1919f(x + y) + f(x y) = 2f(x)g(y), (1)vi mi x, yf(x + y) + g(x y) = h(x)k(y), (2) vi mi x, y

Nhn xt khi f(x) = 0 th g(x) l mt hm ty Nn ta xt f(x) = 0 ,nn c a sao cho f(a) = 0Trong (1) thay x = a, y = y ta c g(x) l hm s chn. Nh phng php tch f thnh 2hm chn f1 v hm l f2 Wilson thu gn c f1(x) = kg(x),vi k = f1(0) v g tha mnhm dAlembert

nh l 2. (nh l Wilson.) Nghim tng qut ca phng trnh(1) lf(x) = 0, g(x) ty , hayf(x) = k cos bx + Csin bx vg(x) = cos bx, hayf(x) = k cosh bx + Csinh bx vg(x) = cosh bx, hayf(x) = k + Cx vg(x) = 1

Nhn xt. Trong phm vi ca bi ny chng ti ch nhm xt phng trnh (1) c nghimdng lng gic, dng (2) c xt tng qut trong bi khc

Bi ton 13. Nu hm f thaf(x y) = f(x)f(y) + g(x)g(y), x, y R.th cng tha phngtrnh dAlembertf(x + y) + f(x y) = 2f(x)cos y, x, y R.Li gii. Do bi ton 3.5

Bi ton 14. ( Bnh nh 2009 )Tm tt c cc hm s f xc nh trn R thaf(x + y) + f(x y) = 2f(x)cos y, x, y R

.

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6 M rng cc phng trnh hm c bn dng

f[G(x, y)] = F(f(x), f(y))

Trong bi vit ny, chng ti ch xt trng hp c bit G(x,y)=x+y, tc l dng phngtrnh hm f(x + y) = F(f(x), f(y)) , x , y RTnh cht 4. F c tnh kt hp tc l F[F[u, v], w] = F[u, F[v, w]] = f(x + y + w)Trng hpF(u, v) = Auv + Bu + Bv + C, F l a thc i xng, do tnh kt hp ta c AC = B2 BKhi A = 0 th B = 1 bi ton c dng f(x + y) = f(x) + f(y) + Ct A(x) = f(x) + C, bi ton c a v A(x + y) = A(x) + A(y) (phng trnh Cauchy I)Do f(x) = A(x) C. Khi A=0 , bi ton c dng

f(x + y) = Af(x)f(y) + Bf(x) + Bf(y) +B2 B

A=

[Af(x) + B] [Af(y) + B] BA

t E(x) = Af(x) + B, bi ton c a v E(x + y) = E(x).E(y) (phng trnh Cauchy II)Do f(x) = E(x)BA

Mt s bi ton lin quanCc nghim tng qut c a ra y trong lp hm lng gic c xc nh trong cckhong hm s lin tc v n iu trong khong xc nh

6.1 f1(x + y) =f1(x) + f1(y)

1 f1(x)f1(y) ,

6.2 f2(x + y) =f2(x)f2(y) 1f2(x) + f2(y)

,

6.3 f3(x + y) =f3(x) + f3(y) 2f3(x)f3(y)

1 f3(x)f3(y) ,

6.4 f4(x + y) =f4(x) + f4(y) 1

2f4(x) + 2f4(y) 2f4(x)f4(y) 1 ,

6.5 f5(x + y) =f5(x) + f5(y) 2f5(x)f5(y)cos a

1 f5(x)f5(y) ,

6.6 f6(x + y) = f6(x)f6(y)

1 f2(x)

1 f2(y)

C cc nghim ln lt l

f1 (x) = tan kx,f2 (x) = cot kx,f3 (x)

=1

1 + cot kx, f4 (x) =

1

1 + tan kx, f5

(x) =sin kx

sin(kx + a), f6 (x) = cos kx.

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7 Lng gic ha bi ton phng trnh hmBi ton 15. (Putnam2000) Tm hm f(x) xc nh v lin tc trong [1, 1] v tha mn

f 2x2 1 = 2xf(x) , x [1.1]Li gii. Ta c x = 1, x = 12 tha mn 2x2 1 = x suy ra f(1) = f(12) = 0Nn f cos 23 + 2n

= 0.Tf(cos 2a) = 0 ta suy ra f(cos a) = 0

t x = cos a, Ta c f cos2k 23 + 2n

= 0, n Z, k NHn na, tp cc s 2k(23 + 2n) tr mt trong R v f(cos x) lin tc suy ra f(cos r) = 0,mir

Bi ton 16. Tm hmf(x),chn, lin tc trong ln cn im O xc nh

f(x) 0 khi 0 x /2,f(x) 0 khi/2 x ,

v tha (*) f(2x) = 1

2f2 2 - x ,x R, thf(x) = cos x, mi x.Bi ton 17. Tm cc hmf(x) xc nh v lin tc trn [1, 1] v tha mn cc iu kin

f(x) + f(y) = f

x

1 y2 + y

1 x2

, x, y [1, 1] (18)

Li gii. t x = sin u, y = sin v, u, v 2 ,

2

Th x

1 y2 + y1 x2 = sin (u + v)

Khi c th vit (18) di dng

g (u + v) = g (u) + g (v) , u , v

2,

2

vi g(u) = f(sin u)

Suy ra f(x) = aarc sin x,

x

[

1, 1] , a

R (i)

Th li ta thy hm f(x) xc nh theo ( i ) tha mn cc iu kin ca bi ton .

Bi ton 18. Tm cc hm sf(x) xc nh v lin tc trn [1, 1] v tha mn iu kin

f

xy

1 y2

1 x2

= f(x) + f(y) , x, y [1, 1]

Li gii. t x = cos u, y = cos v, u, v [0, ] .Khi sin u 0, sin v 0

xy

1 y2

1 x2 = cos (u + v) , u, v [0, ]Phng trnh hm cho c th vit di dng

f(cos u) + f(cos v) = f(cos (u + v)) , u, v [0, ]t f(cos u) = g (u) ta c

g (u + v) = g (u) + g (v) , u, v [0, ]Do vy, g (u) = au,a = const, f(x) = aarccos xTh li, ta thy hm s ny tha mn bi ton .

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Bi ton 19. Tm cc hmf(x) xc nh v lin tc trnR v tha mn cc iu kin

f(x) + f(y) = f

x + y1 - xy

, x,y R : |xy| < 1(17)

Li gii. t x = tgu,y = tgv,2 < u, v 0 sao cho

d (f(xy), f(x)f(y)) < , x, y Gv do , tn ti mt ng cu M : G H sao cho

d (f(x), M(x)) < ,

x

G.

1 Tnh n nh ca phng trnh hm (Cauchy) cngtnh

Trc ht ta nhc li phng trnh hm (Cauchy) cng tnh (A)

f(x + y) = f(x) + f(y) (A)

Gi s hm f : X Y tha mn (A), vi X v Y l hai khng gian Banach. Khi f cgi l hm cng tnh.

nh l 1. Gi s , hmf : X Y tha mn, vi mi > 0, ta cf(x + y) f(x) f(y) , x, y X. (1)

Khi tn ti gii hn sauA(x) = lim

n2nf(2nx) (2)

vi mix X v tn ti duy nht hm cng tnh A : X Y tha mnf(x) A(x) , x X. (3)

Chng minh. Thay x = y vo (1) ta c

1

2

f(2x) f(x)

1

2

. (4)

S dng phng php quy np, ta c

2nf(2nx) f(x) (1 2n). (5)

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Tht vy, trong (4) ta thay x bi 2x, ta c

12

f(22x) f(2x) 12

.

Khi [ 1

2f(22x) 2f(x)] [f(2x) 2f(x)] = 1

2f(22x) f(2x) 1

2

hay

122

f(22x) f(x) 12

f(2x) f(x) 122

,

nn 1

22f(22x) f(x)

1

2+

1

22

,

do

1

2n

f(2nx)

f(x)

1

2

+1

22

+

+

1

2n = 1

1

2n .

By gi ta s chng minh dy { 12n f(2nx)} l dy Cauchy vi mi x X. Chn m > n, khi

12n

f(2nx) 12m

f(2mx) = 12n

12mn

f(2mn.2nx) f(2nx)

12n

1 1

2mn

=

1

2n 1

2m

.

Do dy { 12n f(2nx)} l dy Cauchy v do Y khng gian Banach nn tn ti A : X Y saocho A(x) := lim

n2nf(2nx) vi mi x X, hay

A(x) 12n

f(2nx) 12n

.

Tip theo ta cn chng minh A l hm cng tnh. Thay x, y bi 2nx v 2ny trong (1), ta c

12n

f(2n(x + y)) 12n

f(2nx) 12n

f(2ny) 12n

vi n Z+, x , y X. Cho n , ta cA(x + y) A(x) A(y) .

Vi mi x X, ta cf(x) A(x) = [f(x) 1

2nf(2nx)] + [

1

2nf(2nx) A(x)]

f(x) 12n

f(2nx) + 12n

f(2nx) A(x)]

1 12n

+

1

2n= .

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Cui cng, ta cn chng minh A duy nht. Gi s tn ti mt hm cng tnh A1 : X Y thamn (3). Khi , vi mi x X,

A(x) A1(x) = 1n[A(nx) f(nx)] + [A1(nx) f(nx)]

2

n theo (3)

Vy A1 = A.

nh l 2. Vi mi dy s thc bt kan tha mn

|an+m an am| < 1, n, m Z+, (6)

th tn ti gii hn hu hnA := lim

n

an

nv

an nA < 1, n Z+.

Chng minh. p dng nh l 1 cho Y = R. C nh x X v t an := ( 1 )f(nx), n Z+.Khi , theo (1) dy (an) tha mn (6). t

A := limk

akk

=

1

limk

f(kx)

k

,

Theo nh l 1, ta c

|1

f(nx) nA(x)| < 1,

vi mi n Z+ v mi x X hay

|f(x) nAx

n

| < .

V theo (1), vi mi x, y X ta c

|A(x + y) A(x) A(y)| = limn

f(nx + ny)

n f(nx)

n f(ny)

n

lim

nn

= 0,

Do |f(x) A(x)| < , x X.

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2 Tnh n nh ca phng trnh hm (Cauchy) nhntnh

Trong phn ny nghin cu phng trnh

f(xy) = f(x)f(y) (M)

Gi s hm f : X Y tha mn (M), vi X v Y l hai khng gian Banach. Khi f cgi l hm nhn tnh.

nh l 3. Gi s > 0, S l mt na nhm vf : S C sao cho

|f(xy) f(x)f(y)| , x, y S. (1)

Khi

|f(x)

| 1 +

1 + 4

2=: ,

x

S. (2)

hocf l hm nhn tnh vi mix, y S.Chng minh. Trong (2), ta c 1+

1+4

2 =: hay 2 = v > 1. Gi s (2) khng xy ra,

tc l tn ti a S sao cho |f(a)| > , hay |f(a)| = + , vi > 0 no . Trong (1), chnx = y = a, ta c

|f(a2) f(a)2| (3)Khi

|f(a2)| = |f(a)2 (f(a)2 f(a2))| |f(a)

2

| |f(a)2

f(a2

)| |f(a)|2 theo (3)= ( + )2 = ( + ) + (2 1) + 2 (do 2 = )> + 2 (do > 1)

Bng php chng minh quy np, ta c

|f(a2n)| > + (n + 1), n = 1, 2, . . . .

Vi mi x,y,z S,|f(xyz) f(xy)f(z)| , v |f(xyz) f(x)f(yz)|

Ta c

|f(xy)f(z) f(x)f(yz)| |f(xyz) f(xy)f(z)| + |f(xyz) f(x)f(yz)| 2

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v|f(xy)f(z) f(x)f(y)f(z)| |f(xy)f(z) f(x)f(yz)|

+ |f(x)f(yz) f(x)f(y)f(z)| 2+ |f(x)|

Suy ra |f(xy) f(x)f(y)|.|f(z)| 2+ |f(x)|.Chn z = a2

n

, ta c

|f(xy) f(x)f(y)| 2+ |f(x)||f(a2n)| .

vi mi x, y S v mi n = 1, 2 . . . . Cho n , ta c f(xy) = f(x)f(y), x, y S. Vy fl mt hm nhn tnh.

3 Cc v d p dngV d 1. Nghim ca phng trnh Jensen.Bi ton 1. Tm hm f : R R tha mn phng trnh sau

f

x + y

2

=

f(x) + f(y)

2x, y R (1)

Thay y = 0 vo (1), ta c

fx

2

=

f(x) + f(0)

2x R (2)

Khi p dng (1) v (2), ta c

f(x) + f(y)

2= f

x + y

2

=

f(x + y) + f(0)

2

hayf(x) + f(y) = f(x + y) + f(0), x, y R.

t A(x) = f(x) f(0). Ta c A(x) + A(y) = A(x + y), x, y R. Vy A l mt hm cngtnh trn R nn f(x) = A(x) + , trong = f(0).Ch . Nu bi ton c thm gi thit: hm f lin tc th nghim tm c s l f(x) = ax+,vi a, l cc hng s ty .

Tip theo ta xt tnh n nh nghim ca phng trnh (1).

Mnh 1. Gi s hmf tha mnfx + y2

f(x) + f(y)

2

(3)vi l s dng ty cho trc v vi mix, y R. Khi tn ti duy nht mt hm cngtnhA : R R sao cho

|f(x) A(x) f(0)| 4, x R.

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Chng minh. Thay y = 0 vo (3), ta cfx2

f(x) + f(0)

2

x R.Do fx + y

2

f(x + y) + f(0)2

x, y R.Ta c f(x) + f(y)

2 f(x + y) + f(0)

2

f(x) + f(y)2

f(x + y)2

+

f(x + y)2

f(x + y) + f(0)2

2

hay |f(x + y) + f(0) f(x) f(y)| 4. (4)t g(x) = f(x) f(0). Thay vo (4), ta c

|g(x + y) g(x) g(y)| 4Theo tnh n nh ca hm cng tnh, tn ti duy nht hm cng tnh A sao cho

|g(x) A(x)| 4.Ta c

|f(x) A(x) f(0)| = |g(x) A(x)| 4.V d 2. Nghim ca phng trnh Cauchy hai n hm.

Bi ton 2. Tm cp hm f, g : R R tha mn phng trnh sauf(x + y) = g(x) + g(y) x, y R (5)


f(x) = g(x) + g(0) x R,

hay f(x) = g(x) + , vi = g(0). Do g(x) = f(x) vi mi x R.Thay vo phng trnh (5), ta cf(x + y) = f(x) + f(y) 2 (6)

t f(x) = A(x) + 2. Phng trnh (6) tr thnh

A(x + y) + 2 = A(x) + 2 + A(y) + 2 2

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vi mi x, y R. tG(x) = g(x) g(0), (12)

vi mi x, y R. Th vo (11) ta c

|G(x + y)

G(x)

G(y)

| 2,

x, y

R.

Theo nh l v tnh n nh ca hm cng tnh, tn ti duy nht mt hm cng tnh A : R Rsao cho

|G(x) A(x)| 2, x R. (13)T (12) v (13) ta c

|g(x) A(x) g(0)| 2, x R. (14)T (8), (9) v (14) ta c

|f(x) A(x) f(0)| = |f(x) g(x) g(0) + g(x) A(x) g(0) + 2g(0) f(0)|

|f(x)

g(x)

g(0)

|+

|g(x)

A(x)

g(0)

|+

|f(0)

2g(0)

| + 2 + = 4.

V d 3. Nghim ca phng trnh Pexider.

Bi ton 3. Tm tt c cc hm f , g , h : R R tha mn phng trnh sauf(x + y) = g(x) + h(y) x, y R (15)


f(x) = g(x) + h(0) x R,hay f(x) = g(x) + , vi = h(0). Do g(x) = f(x) vi mi x R.

Thay x = 0 vo (15), ta c f(y) = h(x) + , vi = g(0), hay h(x) = f(x) vi mix R.

Phng trnh (15)tr thnh

f(x + y) = f(x) + f(y) , x, y R. (16)t f(x) = A(x) + + thay vo phng trnh (16), ta c

A(x + y) + + = A(x) + + + A(y) + +

hayA(x + y) = A(x) + A(y) x, y R.

Vy A l mt hm cng tnh trn R nn

f(x) = A(x) + + g(x) = A(x) + h(x) = A(x) +

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T phng trnh (27), ta c f(x) 0, x R. Gi s tn ti x0 R sao cho f(x0) = 0.Khi

f(x0 + y

2) =

f(x0)f(y) = 0 y R,

hay f(x) = 0 vi mi x R.Xt f(x) > 0, x R. Khi ly logarit hai v ca phng trnh (27), ta c

ln f(x + y

2) =

ln f(x) + ln f(y)

2, x, y R

t g(x) = ln f(x) ta c

g(x + y

2) =

g(x) + g(y)

2, x, y R

hay g l mt nghim ca phng trnh Jensen, tc l g(x) = ax + b. Suy ra nghim ca phngtrnh (27) l f(x) = eax+b vi a, b R.

Tip theo ta xt tnh n nh nghim ca phng trnh (27).

Mnh 4. Gi s hmf : R R+ tha mn

|f(x + y2

)

f(x)f(y)| (28)

vi mix, y R v|f(x) f(x)| (29)

vi, l cc s dng ty cho trc. Gi s tn tif(a)1, khi tn ti mt hmE : RR+ sao cho

|E(x + y)

E(x)

E(y)

| ,

x, y

R (30)

v|f(x) 1

2(E(x) E(x))| , x R (31)

vi, l cc hng s no .

Chng minh. t m = supxR

f(x)f(a). T iu kin (29) th m l hu hn. Khi , ta c

f(x)f(a)

f(x)f(a) + |

f(x)f(a)

f(x)f(a)|

m + |f(x a2

) f(x)f(a)| + |f(x + a

2) f(x)f(a)|

+ |f(x a2

) f(x + a2

)| m + 2 +

t h : R R+ thah(x) =

f(x)f(x), x R.

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Khi h l mt hm chn v

|h(x) f(x)| =

f(x)|

f(x)

f(x)| 2 m2

f(a), x R, |h(x)

f(a)| m. (32)

t E : RR+ tha mn

E(x) = h(x) +

f(a) x Rp dng (32) ta c

|E(x + y) E(x)E(y)| = |h(x + y) +

f(a) h(x)h(y) (h(x) + h(y))

f(a) f(a)| |h(x + y)| + |h(x)h(y)| + |(h(x) + h(y))

f(a)| + |f(a)|

|h(x + y) f(x + y)| + |f(x + y)| + |h(x)h(y)f(a)f1(a)|+ |h(x)

f(a)| + |h(y)

f(a)| +

f(a) + |f(a)|

2

m2

f(a)+

mf(a) +m2

f(a)+ 2m + f(a) + |f(a)| =

v

|f(x) 12

(E(x) E(x))| =

|f(x) h(x) + h(x) 12

(h(x) + h(x))

f(a)|

|f(x) h(x)| +

f(a) 2 m2

f(a)+

f(a) = .

Ti liu tham kho[1] Nguyn Vn Mu, Phng trnh hm, Nh xut bn Gio dc, 1997.

[2] Pl. Kannappan, Functional Equations and Inequalities with Applications, Springer, 2009,295-323.

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MT S LP PHNG TRNH HM A N SINHBI PHI NG THC

Trn Vit Tng, Trng THPT Trn Ph - Nng

Trong ton hc ph thng cc bi ton v phng trnh hm l cc loi ton thng mi v rtkh, thng xuyn xut hin trong cc thi hc sinh gii quc gia, Olympic Ton khu vcv Quc t, Olympic sinh vin gia cc trng i hc v cao ng. Lin quan n cc dngton ny l cc bi ton v cc c trng khc nhau ca hm s v cc tnh cht lin quan vichng. h thng cc phng trnh hm, cn thit phi h thng cc kin thc c bn v nng caov cc dng phng trnh hm cng nh cc ng dng ca chng.i vi cc bi ton v phng trnh hm vi nhiu n hm trong cc lp hm c th: lin tc,kh vi, tun hon, li lm,... cn nm c mt s k thut v bin i hm s, kho st cctnh cht c bn ca hm thc v cc php bin hnh trn trc thc.

1 Phng trnh hm sinh bi phi ng thc a2 + b2 g(a + b)h(a b)

Bi ton 1. Tm cc hm s f , g , h xc nh v lin tc trn R tha mn iu kin

f(x2 + y2) = g(x + y).h(x y), x, y R. (1)Gii. Xt trng hp g(0) = 0.

Cho y = x , phng trnh cho tr thnh

f(2x2) = 0, x R

Suy raf(x) = 0, x 0.

Thay f(x) vo phng trnh cho ta c

g(x + y).h(x y) = 0, x, y Rg(u).h(v) = 0, u, v R.

Do g(x) = 0

h(x) lin tc ty trn Rhoc

h(x) = 0

g(x) lin tc ty trn R

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Vy nghim trong trng hp ny l :

f(x) = 0 vi x 0f(x) = q(x) vi q(x) lin tc ty trong (; 0] v q(0) = 0

g(x)

0

h(x) l hm lin tc ty

hoc


g(x) l hm s lin tc ty v g(0) = 0h(x) 0

Xt trng hp h(0) = 0

Cho y = x, phng trnh cho tr thnhf(2x2) = 0, x R

Suy raf(x) = 0, x 0.

Th f(x) vo cho ta c

g(x + y).h(x y) = 0, x, y Rg(u).h(v) = 0, u, v R.

Do g(x) = 0

h(x) lin tc ty trn R; h(0) = 0hoc

h(x) = 0

g(x) lin tc ty trn R.

Vy nghim ca phng trnh trong trng hp ny l :


g(x) l hm lin tc ty

h(x) 0hoc


g(x) 0h(x) l hm lin tc ty vi h(0) = 0

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Xt trng hp g(0) = 0 v h(0) = 0 .Ta c f(0) = 0.Cho x = y, phng trnh cho tr thnh

f(2x2) = g(2x).h(0), g(2x) =f(2x2)

h(0) , x R, g(x) =f(

x2

2

)

h(0) , x R.

Cho x = y, phng trnh cho tr thnh

f(2x2) = g(0).h(2x), h(2x) =f(2x2)

g(0), x R, h(x) =

f(x2

2)

g(0), x R.

Thay g(x) v h(x) vo phng trnh ta c

f(x

2

+ y

2

) = f[

(x + y)2

2 ].f[

(x

y)2

2 ].

1

g(0)h(0), x, y R.

t

u =(x + y)2

2

v =(x y)2

2

.Khi ta c

f(u + v) = f(u).f(v).1

g(0)h(0), u, v 0. (2)

t f(u) = g(0)h(0)F(u), u 0.Phng trnh (3.2) tr thnh

g(0)h(0)F(u + v) = g(0)h(0)F(u).g(0)h(0)F(v).1

g(0)h(0), u, v 0

F(u + v) = F(u)F(v), u, v 0.

Ta c

F(u) = au u 0; a > 0f(u) = b.au vi b = g(0)h(0); a > 0

f(x) = b.a

x

, x 0; a > 0, b = 0.

Suy ra g(x) =f(

x2

2)

h(0), x R = g(0).ax22 x R = m.ax22 vi m = g(0).

h(x) =f(

x2

2)

g(0), x R = h(0).ax22 , x R = n.ax22 vi n = h(0).

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Vy nghim ca phng trnh trong trng hp ny l :

f(x) = b.ax vi x 0; a > 0; b = 0f(x) = q(x) vi q(x) lin tc ty trong (; 0] v q(0) = 0

g(x) = m.ax2

2 ,

xR

h(x) = n.ax2

2 , x RTm li nghim ca bi ton l


g(x) 0h(x) l hm lin tc ty

hoc


g(x) l hm s lin tc ty v g(0) = 0h(x) 0

hoc

f(x) = 0 vi x 0f(x) = q(x) vi q(x) lin tc ty trong (

; 0] v q(0) = 0

g(x) l hm lin tc ty h(x) 0

hoc


g(x) 0h(x) l hm lin tc ty vi h(0) = 0

hoc

f(x) = b.ax vi x 0; a > 0; b = 0f(x) = q(x) vi q(x) lin tc ty trong (; 0] v q(0) = 0

g(x) = m.a

x2

2 , x R

h(x) = n.a

x2

2 , x R

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Bi ton 2. Tm cc hm s f , g , h xc nh v lin tc trn R tha mn iu kin

f(x2 + y2) = g(x + y) + h(x y), x, y R. (3)

Gii. Nghim ca bi ton l

f(x) = ax + b + c vi x 0f(x) = q(x) vi q(x) lin tc ty trong (; 0] v q(0) = 0

g(x) = ax2

2+ b, x R

h(x) = ax2

2+ c, x R

vi b = g(0); c = h(0)

Bi ton 3. Tm cc hm s f, g, h xc nh v lin tc trn R tha mn iu kinf(x2 + y2) = g(x2) h(y2), x, y R. (4)


f(x) = ax + b vi x 0f(x) = q(x) vi q(x) lin tc ty trong (; 0] v q(0) = 0

g(x) = ax + c, x Rh(x) = ax d, x R

2 Phng trnh hm sinh bi ng thc a2 b2 = (a +b)(a b)

Bi ton 4. Tm cc hm s f , g , h lin tc v xc nh trn R tha mn iu kin

f(x2 y2) = (x + y)g(x y), x, y R. (5)

Gii. Cho y = 0, phng trnh cho tr thnh

f(x2) = x.g(x), x R.Nu x = 0 th

f(x) = 0. (6)

Nu x = 0 , ta cg(x) =

f(x2)

x.

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Do , phng trnh cho tr thnh

f(x2 y2) = (x + y)f[(x y)2]

x y , x = yf(x2

y2)

x + y =

f[(x

y)2]

x y , x = yf(x2 y2)

x2 y2 =f[(x y)2]

(x y)2 , x = y.

t h(x) =f(x)

xvi x = 0. Khi , ta c

h(x2 y2) = h[(x y)2].Cho x = y + 1 , ta c

h(2y + 1) = h(1), y Rh(x) = a, x = 0.

Do f(x) = ax vi x = 0. (7)

Kt hp (10) v (7) ta cf(x) = ax, x R.

Suy ra

g(x) =ax2

x

= ax,

x

= 0.

Vy nghim ca bi ton l

f(x) = ax

g(x) = ax

Bi ton 5. Tm tt c cc hm s f , g , h xc nh v lin tc trn R tha mn iu kin

f(x2 y2) = g(x y) + h(x + y), x, y R. (8)


f(x) = a, x Rg(x) = b,

x

R

h(x) = c, x Rtrong a,b,c R; a = b + c.


f(x2 y2) = g2(x) h2(y), x, y R. (9)

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f(x) = mx + a b, x R; a,b,m 0g(x) =

mx2 + a

h(x) = mx2 + b

hoc

f(x) = mx + a b, x R; a,b,m 0g(x) =

mx2 + a

h(x) = mx2 + b

hoc

f(x) = mx + a b, x R; a,b,m 0g(x) = mx2 + ah(x) =

mx2 + b

hoc

f(x) = mx + a

b,

x

R; a,b,m

0

g(x) = mx2 + ah(x) = mx2 + b

3 Mt s bi ton phng trnh a n hm khcBi ton 7. Tm cc hm s f , g , h lin tc v xc nh trn R tha mn iu kin

f(x) g(y) = xh(y) yh(x), x, y R. (10)Gii. Cho x = y, phng trnh cho tr thnh

f(x) g(x) = 0 f(x) = g(x), x R.Cho y = 0, phng trnh cho tr thnh

f(x) g(0) = x.h(0), x Rf(x) = x.h(0) + g(0), x Rf(x) = ax + b vi a = h(0); b = g(0).

Thay f, g vo phng trnh cho ta c

(ax + b) (ay + b) = xh(y) yh(x), x, y Rax ay = xh(y) yh(x), x, y R

a

y a

x=

h(y)

y h(x)

x, x, y R

a

x h(x)

a=

a

y h(y)

y, x, y R.

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Suy ra

a

x h(x)

x= C vi C l hng s

h(x) = Cx + ah(x) = cx + a vi c = C.

Th li phng trnh ta thy f , g , h tha mn.

Vy nghim ca phng trnh l

f(x) = g(x) = ax + b

h(x) = cx + a

Bi ton 8. Tm tt c cc hm s f, g xc nh v lin tc trn R tha mn iu kin

f(x) f(y) = (x + y)g(x y), x, y R. (11)

Gii. Nghim ca bi ton lf(x) = ax2 + b

g(x) = axvi mi a, b R.

Bi ton 9. Tm tt c cc hm s f, g xc nh v lin tc trn R tha mn iu kin

f(x) + f(y) + 2xy = (x + y)g(x + y), x, y R. (12)

Gii. Nghim ca bi ton l f(x) = x2 + ax v g(x) =

x + a vi x = 0c vi x = 0

.

Bi ton 10. Tm tt c cc hm s f, g xc nh v lin tc trn R tha mn iu kinf(x).g(y) = x2 y2, x, y R. (13)

Gii. Nu tn ti x0 sao cho f(x0) = 0. Khi ta c

0 = f(x0).g(y) = x20 y2, y R (v l).

Suy raf(x) 0, x R.

Tng t ta cng cg(x) 0, x R.

Cho x = y, phng trnh (15) tr thnh

f(x).g(x) = 0 f(x) = 0 hoc g(x) = 0 (loi do f(x) 0; g(x) 0).

Vy khng tn ti cc hm f, g tha mn bi ton.

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f(x + y) + g(x y) = h(xy), x, y R. (14)


f(x) =mx2

4

+ b,

x

R

g(x) = mx2

4+ a, x R

h(x) = mx + a + b, x R.

Bi ton 12. Tm tt c cc hm s dng f , g , h xc nh v lin tc trn R tha mn iukin

f(x + y).g(x y) = h(xy), x, y R. (15)Gii. Do f , g , h l cc hm s dng nn phng trnh (??) tng ng

ln f(x + y) + ln f(x y) = ln h(xy), x, y R.

t

ln f(x) = F(x)

ln g(x) = G(x)

ln h(x) = H(x)

. Khi ta c F(x + y) + G(x y) = H(xy), x, y R.

Ta c

F(x) =mx2

4+ b, x R

G(x) = mx2

4+ a, x R

H(x) = mx + a + b, x R.

. Suy ra

f(x) = emx2

4 +b, x Rg(x) = e

mx2

4+a, x R

h(x) = emx+a+b, x R.Th li ta thy cc hm f , g , h tha mn iu kin bi ton.

Vy nghim ca bi ton l

f(x) = emx2

4+b, x R

g(x) = emx2

4+a, x R

h(x) = emx+a+b, x R.

Bi ton 13. Tm tt c cc hm s f, g xc nh v lin tc trn R tha mn iu kinf(x) + g(x) + f(y) g(y) = sin x cos y, x, y R. (16)


f(x) =1

2(sin x cos x), x R

g(x) =1

2(sin x + cos x) + a, x R.

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Ti liu tham kho[1] Nguyn Vn Mu, 1997, Phng trnh hm, NXB Gio Dc

[2] Nguyn Vn Mu, 2006, Cc bi ton ni suy v p dng, NXB Gio Dc.

[3] Nguyn Vn Mu, 2009, Phng trnh hm vi nhm hu hn cc bin i phn tuyn tnh,K yu HNKH "Cc phng php v chuyn ton s cp" ti Bc Giang, 27-29/11/2009.

[4] Christopher G. Small, 2000, Functinal equations and how to solve them, Springer.

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T CNG THC EULER N CC BI TON SPHC

L Sng, Nguyn inh Huy, Trng THPT chuyn L Qu n - Khnh Ha

p dng s phc trong hnh hc phng c nhiu ti liu ca tc gi on Qunh,NguynHu in nh l sch gio khoa dng tham kho.Trong bi vit ny chng ti cp n mtphng php rt hiu qu trong gii h phng trnh, chng minh ng thc, hay dng tnhtng c gi l phng php s phc. Trc xt thm khai trin chui Taylor xy dngcng thc Moivre m trong sch gio khoa c c t cng thc nhn s phc dng lng gicv chng minh quy np I Khai trin Taylor v cng thc Moivre

1 Khai trin Taylor v cng thc MoivreCho z l s phc

ez = 1 +z

1!+

z2

2!+

z3

3!+ .... +

zn

n!+ ...

Trng hp c bit, vi mt gc x, ta c

eix = 1 + ix

1! x

2

2! ix

3

3!+

x4

4!+ ....

Phn thc v phn o ca eix ln lt l

Re(eix) = 1 x2

2!+

x4

4! x

6

6+ , Im(eix) = x

1! x

3

3!+

x5

5! x

7

7!+

Hai chui trn l khai trin Taylor ca cos x v sin x.Ta c c cng thc Euler nh sau eix = cos x + i sin x v ta c mt cng thc tuyt p: ei =1 Ngoi ra do einx = (e(ix)n) nn ta suy ra cng thc Moivre: cos nx+i sin nx = (cos x+i sin x)nV d m uTm khai trin Taylor ca hm f(x) = excos cos(x sin ) ti im 0, vi l tham s.Gii. t g(x) = ex cos sin(x sin ) v vit :

f(x) + ig (x) = ex cos (cos (x sin ) + i sin(x sin ))= ex cos .eix sin = ex(cos +i sin )

Dng cng thc Moivre v khai trin Taylor:

1 +x

1!(cos + sin ) +

x2

1!(cos 2 + i sin2) + + x

n

n!(cos n + i sin n) +

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Khai trin v rt gn ta thu c

f(x) = 1 +cos

1!x +

cos2

2!x2 + + cos n

n!xn +

Sau y l mt s bi ton p dng phng php s phc

2 Chng minh ng thcBi ton 1. Chng minh 1

cos60+ 1

sin 240+ 1

sin 480= 1

sin 120

Li gii. t z = cos 60 + i sin60. Ta c z15 = iM cos60 = z

2+12z , sin12

0 = z41

2iz2 , sin240 = z

812iz4 , sin48

0 = z16112iz8

ng thc cn chng minh tr thnh chng minh 2zz2+1 2iz2

z41 +2iz4

z81 +2iz8

z161 = 0. Quy ng mus, thu gn Ta c z16 1 iz(z14 + 1) = 0z16 1 iz(z14 + 1) = 0 tc l iz 1 i2 iz = 0iu ny hin nhin ng.Bi ton 2. Cho a,b,c l cc s thc tha cos a + cos b + cos c = sin a + sin b + sin c = 0Chng minh rnga) cos2a + cos 2b + cos 2c = sin 2a + sin 2b + sin 2c = 0b) 3(cos(a + b + c) = cos 3a + cos 3b + cos 3c v 3(sin(a + b + c) = sin 3a + sin 3b + sin 3c

Li gii. t x = cos a + i sin a, y = cos b + i sin b, z = cos c + i sin cT gi thit ta c x + y + z = x1 + y1 + z1 = 0 suy ra xy + yz + zx = 0a) x2 + y2 + z2 = (x + y + z)2 2(xy + yz+ zx) = 0. Suy ra kt qub) x3 + y3 + z3 = 3xyz. Suy ra kt qu

Bi ton 3. Chng minh ng thc:1 + i tan t

1 i tan tn

=1 + i tan nt

1 i tan nt, n 1

Li gii. Nu ta nhn t v mu v bn tri vi cos t, v v bn phi bi cos nt, th ta c ucng thc:

eit

eit

n= e

int

eint

Bi ton 4. Chng minh ng thc

n0 +n

k +n

2k + ... =2n

k

k

i=1 cosnj

k

cosnj

k

Li gii. Let C1, C2, . . . , C k l k nghim ca cn n v, tc l, Cj = cos 2jk + i sin2jk

, j =1, 2,..,kNh vy

kj=1

(1 + Cj)n =

ns=0

ns

kj=1

Cj

= k

nk

j=0

n

jk

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V 1 + Cj = 2cosjk cosjk + i sin

jk

Nn p dng cng thc Moivre

k

j=1(1 + Cj)

n =k

j=12ncosn

j

k cos

nj

k+ i sin

nj

k So snh phn thc 2 v ta c kt qu

Bi ton 5. Chng minh ng thc

1

n2

+

n4

n6

+ ... = 2

n2

cos

n

4+ i sin

n

4

, n 1

Li gii. Dng cng thc Moivre, ta c:

(1 + i)n =

2cos

4

+ i sin

4n

= 2n2 cos

n

4

+ i sinn

4 Khai trin (1 + i)n v cho hai v bng nhau bi phn thc,ta c c iu phi chng minhban u.

3 Tnh tch v tngBi ton 6. Chng minh tngcos 27 + cos

47 + cos

67 = 12

Li gii. Nu z = cos 7 + i sin7 th z

7 = 1 iu cn chng minh tng ng vi

12

z+ 1

z

+ 1

2

z2 + 1

z2

+ 12

z3 + 1

z3

+ 12

= 0

Nhn cho 2z3 , sp xp cc s hng z6 + z5 + z4 + z3 + z2 + z+ 1 = z71z1 = 0

Bi ton 7. Tnh tngT = 3

cos 29 +3

cos 49 +

3

cos 89

Li gii. Xt phng trnh z9 1 = 0 c 9 nghim zk, k = 0, 1, . . . , 8 c tng l 0Phng trnh t4 + t3 3t2 2t + 1 = (t + 1) (t3 3t + 1) = 0 c 4 nghim:

2cos2

9 , 2cos4

9 , 2cos6

9 , 2cos8

9

t t = z+ 1z , phng trnh (t3 3t + 1) = 0 c 3 nghim 2cos 29 , 2cos 49 , 2cos 89

Dng cc biu thc i xng nghim phng trnh bc 3, ta c tng l 3

3 3

962

Bi ton 8. Tnh tch P = cos cos400cos800

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Li gii. . t z = cos 200 + i sin200

Ta c 2cos200 = z+ 1z

, 2cos400 = z2 + 1z2

, 2cos800 = z4 + 1z4

, z9 = 1Suy ra (z z8) (z2 z7) (z3 z5) = (z z2 + z3 z4 + z5 z6 + z7 z8) = 1Bi ton 9. Tnh tchsin

2nsin 2

2nsin 3

2n sin (n1)

2n

Li gii. Xt a thc P(X) = X2n 1C cc nghim xk = cos kn + i sin

kn , k = 0, 1, , 2n 1, x0 = 1, xn = 1, xk = x2nkkhi 1

k n 1Khi P(X) = (X2 1)

n1k=1

(X xk)(X xk) = (X2 1)n1k=1

(X2 2cos kn + 1), do xk + xk =2cos kn , xkxk = 1Chia 2 v cho x2 1 , ta c

x2n2 + x2n4 + x4 + x2 + 1 =n1

k=1x2 2x cos k

n+ 1

.

Ly x = 1, ta c n = 2n1n1k=1

1 cos kn

=2n1n1k=1

2sin2 k2n , suy ran1k=1

sin k2n =n

2n1

Bi ton 10. Tnh tng:n1

cos x +

n2

cos2x + ... +

nn

cos nx

Li gii. Gi tng cn tm l S1 v cho

S2 = n

1

sin x + n

2

sin2x + ... + n

n

sin nx.

Dng cng thc Euler, ta c th vit

1 + S1 + iS2 =

n0

+

n1

eix + ... +

n2

ei2x

Nh tnh nhn ca ly tha, ta c:

n

k=0

nk eixk = 1 + eixn = 2cosx2ne ix2

n

Tng trong cu hi l mt phn thc ca khai trin lun b hn 1, iu ny dn n kt qutng l

2ncosnx

2cos

nx

2 1

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Bi ton 11. (USAMO 1999) Tnh (cos )(cos 2)(cos 3) . . . (cos 999)vi = 21999

Li gii. Xt bi ton tng qut, vi n l s nguyn l, hy tnh:

S = (cos )(cos 2)(cos 3) . . . (cos n) vi =2

2n + 1

Chng ta c th cho = ei...? v S = 2nn

k=1

k + k

.

Khi k + k = 2n+1k, k = 1, 2,..,n, chng cha:

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