Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm TấN CHUYấN : N NH, TON NH V SONG NH TRONG CC BI TON V PHNG TRèNH HM Mó: TO19 PHN I M U: Lý chn chuyờn : Phng trỡnh hm (PTH) l mt ch thi hc sinh gii Quc gia, v õy l mt ch khú a phn hc sinh u cho rng cõu PTH l cõu c xp khú cao Trong hai nm gn õy thỡ PTH cng c xp vo cu trỳc thi HSG Tnh Tõy Ninh Nh vy, cỏc em hc sinh dự mun hay khụng thỡ thi vũng Tnh, cỏc em cng phi gp cõu hi v PTH Tuy ngun ti liu v PTH hin khụng him, nhng xõy dng thnh bi ging phự hp cho HS ca chỳng tụi ang ging dy thỡ cng cn phi h thng, thờm bt v ú l nhim v ngi vit ang c gng lm Tip ni bi vit ca chuyờn nm trc, tụi tip tc mng PTH nhng s trung vo cỏc loi PTH cú s dng tớnh n ỏnh, ton ỏnh v song ỏnh quỏ trỡnh gii Cỏc bi ging khỏc s c trỡnh by nhng quyn nghiờn cu tip theo i tng nghiờn cu: - Bi ging v n ỏnh, ton ỏnh v song ỏnh cỏc bi toỏn v phng trỡnh hm C s thc tin ca vic nghiờn cu chuyờn : - Do hc sinh gp nhiu khú khn quỏ trỡnh hc - Gúp phn hỡnh thnh h thng bi ging d hc cho hc sinh thi HSG - Mong mun to lp ti liu riờng cho trng v mng ny Cu trỳc chuyờn : Phn I Phn m u Phn II Phn ni dung A Phn lý thuyt B Cỏc vớ d: C Phn bi ngh C.1 Cỏc bi toỏn dựng tớnh n ỏnh ca hm C.2 Cỏc bi toỏn dựng tớnh ton ỏnh, song ỏnh ca hm Phn C Phn kt lun - M hng nghiờn cu Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm PHN II: PHN NI DUNG N NH, TON NH V SONG NH TRONG CC BI TON V PHNG TRèNH HM A PHN Lí THUYT nh x 1.1 nh ngha Mt ỏnh x f t X n Y l mt quy tc t tng ng mi phn t x ca X vi mt (v ch mt) phn t ca Y Phn t ny c gi l nh ca x qua ỏnh x f v kớ hiu l f(x) (i) Tp X c gi l xỏc nh ca f Tp hp Y c gi l giỏ tr ca f (ii) nh x f t X n Y c kớ hiu l f : X Y x a y = f (x) (iii) Khi X v Y l cỏc s thc, ỏnh x f c gi l mt hm s xỏc nh trờn X (iv) Cho a X , y Y Nu f ( a ) = y thỡ ta núi y l nh ca a v a l nghch nh ca y qua ỏnh x f { } ( ) (v) Tp hp Y = y Y x X , y = f ( x ) gi l nh ca f Núi cỏch khỏc, nh f X l hp tt c cỏc phn t ca Y m cú nghch nh n ỏnh, ton ỏnh, song ỏnh 2.1 nh ngha nh x f : X Y c gi l n ỏnh nu vi a X , b X m a b thỡ f ( a) f ( b) , tc l hai phn t phõn bit s cú hai nh phõn bit () () T nh ngha ta suy ỏnh x f l n ỏnh v ch vi a X , b X m f a = f b , ta phi cú a = b 2.2 nh ngha nh x f : X Y c gi l ton ỏnh nu vi mi phn t y Y u tn ti mt ( ) ( ) phn t x X cho y = f x Nh vy f l ton ỏnh nu v ch nu Y = f X 2.3 nh ngha nh x f : X Y c gi l song ỏnh nu nú va l n ỏnh va l ton ỏnh Nh vy ỏnh x f : X Y l song ỏnh nu v ch nu vi mi y Y , tn ti v nht mt phn t x X () y = f x Chỳ ý: Hm n ỏnh v liờn tc thỡ n iu Nu f ( v ( x )) = u ( x ) ú u ( x ) l ton ỏnh trờn thỡ f l ton ỏnh trờn Nu f ( f ( x )) = ax + b, x thỡ f l song ỏnh Nu f (u ( x )) f (v ( x )) = ax + b ( a 0) thỡ mi s thc luụn biu din c di dng f (u ) f ( v ) Núi cỏch khỏc x luụn cú u, v cho x = f ( u ) f ( v ) () () Nu f l ton ỏnh thỡ ta thng s dng a : f a = hoc a : f a = Mi a thc bc l u cú giỏ tr l i vi cỏc PTH cú cha f ( x f ( y)), f ( x ), f ( y) thỡ ta thng ly x = v thng thay x bi f ( x ), bi f ( y ) v thng phi tỡm cho c biu din ca f ( f ( x )) theo f ( x ) Ta thng bin i xut hin biu thc ax + b (a, b l cỏc hng s, a 0) v cho x thay i trờn suy biu thc ny cú giỏ tr l Tng quỏt hn, ta thng bin i xut hin a thc bc l, sau ú dng tớnh cht a thc bc l cú giỏ tr nh x ngc ca mt song ỏnh Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm 3.1 nh ngha nh x ngc ca f, c kớ hiu bi f , l ỏnh x t Y n X gỏn cho mi phn t y Y phn t nht x X cho y = f ( x ) Nh vy: f ( x ) = y f ( x ) = y 3.2 Chỳ ý a Nu f khụng phi l song ỏnh thỡ ta khụng th nh ngha c ỏnh x ngc ca f Do ú ch núi n ỏnh x ngc f l song ỏnh b Nu f l song ỏnh thỡ tn ti ỏnh x ngc f : Y X gỏn cho mi phn t y Y phn t ( ) () () nht x X cho y = f x Nh vy f x = y f x = y nh x hp Nu g : A B v f : B C v g A B thỡ ỏnh x hp f og : A C ( ) c xỏc nh bi: ( f og )( a ) = f ( g ( a ) ) Kớ hiu pn = p o p o o p 42 43 n Mt s kớ hiu : Tp cỏc s t nhiờn * : Tp cỏc s nguyờn dng Đ : Tp cỏc s hu t  : Tp cỏc s nguyờn  + : Tp cỏc s nguyờn dng : Tp cỏc s thc + : Tp cỏc s thc dng Đ + : Tp cỏc s hu t dng B CC V D: VD 1: (Dựng ụn cm song ỏnh) ỏnh Nu f : v f ( f ( x )) = ax + b, x ( a 0) thỡ f l song (nh lý ny dựng nhiu v sau) Gi s f ( x1 ) = f ( x ), ềo: ax1 + b = f ( f ( x1 )) = f ( f ( x )) = ax + b x1 = x Vy f l n ỏnh Vi mi y , luon ton taễ i x = f( yb yb f ( x) = f f ( ) = a +b= y a a Vy f l ton ỏnh, ú f l song ỏnh yb ) cho : a VD 2: (Dựng ụn cm song ỏnh) Xột tt c cỏc hm f , g, h : cho f l n ỏnh v h l ( ) ( ) song ỏnh tha iu kin f g ( x ) = h ( x ) , vi mi x Chng minh rng g x l mt hm song ỏnh ( ) l n ỏnh Tht vy vi x , x cho g ( x ) = g ( x ) suy f ( g ( x ) ) = f ( g ( x ) ) h ( x ) = h ( x ) x = x (do h l mt song ỏnh) Suy g l mt n ỏnh +) Ta chng minh g ( x )l ton ỏnh Tht vy vi mi x v h l mt song ỏnh nờn tn ti y cho f ( x ) = h ( y ) = f ( g ( y ) ) x = g ( y ) (do f l n ỏnh) Suy g l mt ton ỏnh +) Ta chng minh g x 1 2 2 Vy g( x ) l mt hm song ỏnh VD 3: (Dựng ụn cm song ỏnh) Xột tt c cỏc hm n ỏnh f : tha iu kin: ( ) f x + f ( x ) = x , vi mi x Chng minh rng hm s f ( x ) + x l mt song ỏnh Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm t g x = f x + x f x = g x x Khi ú t phng trỡnh ban u ta c: () () Do ú ta cú () () g ( x + f ( x )) ( x + f ( x )) = x g ( g ( x )) g ( x ) = x (1) g ( g ( x ) ) g ( x ) = x , x ( ) ( ( ) ( ) cho g x1 = g x2 +) Ta chng minh g l n ỏnh Tht vy vi x1 , x ) g g ( x1 ) g ( x1 ) = g g ( x ) g ( x ) x1 = x hay g l n ỏnh suy +) Ta chng minh g l ton ỏnh Tht vy vi mi x ta cú: f (x) f (x) v kt hp vi f l mt n ỏnh ta thu c: = f +f 2 f (x) f (x) f (x) = g ng thc ny chng t g l mt ton ỏnh x= +f Do ú g l mt song ỏnh hay f ( x ) + x l mt song ỏnh f ( x ) = f (x) {} VD 4: (Dựng ụn cm n ỏnh) Xột tt c cỏc hm f : + U tha ng thi hai iu kin sau: (i) f x + y = f x + f y , vi mi x, y + U ( ) ( ) () { (ii) S phn t ca hp x f ( x ) = 0, x + {} U {0}} l hu hn Chng minh rng f l mt hm n ỏnh ( ) ( ) {} f ( + 0) = f (0) + f (0) f (0) = Bng phng phỏp quy np ta d dng ch c: f nx = nf x , n * , x + U Thay x = y = vo phng trỡnh ban u ta c {} ( ) (1) ( ) ú theo iu kin (i) ta c: f ( x x ) + f ( x ) = f ( x ) f ( x x ) = (2) T (1) v (2) ta thu c: f ( n ( x x ) ) = nf ( x x ) = , vi mi n T ú kt hp vi iu kin Gi s x1 , x2 + U cho f x1 = f x2 Khụng mt tớnh tng quỏt ta cú th gi s x1 x Khi 2 1 * 2 (ii) ta suy x1 = x ( vẽ f ( 0) = vac oh u haễ n x ềef ( x ) = 0) Vy f l mt hm n ỏnh ( ) VD 5: (Dựng tớnh n ỏnh) Tỡm tt c cỏc hm f : tha f f ( x ) y = 3x + y + D thy f l mt n ỏnh ( ) C1: x thay y = 3x f f ( x ) + 3x = ( ) ( ) Do ú f f ( x ) + 3x = f f ( y ) + 3y , x , y f ( x ) + 3x = f ( y ) + 3y ( vỡ f l n ỏnh) f ( x ) + 3x = c Thay vo tỡm c c = f ( x ) = x y y C2: f f ( x ) y = 3x + y + = + 3x + = f f 3x y y y f ( x ) y = f 3x f ( x ) + x = f + , x , y 3 ( ) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm Suy f x = x + c thay vo tỡm c c = ( ) VD 6: (Dựng tớnh n ỏnh) Tỡm ( ) ( ) tt c cỏc hm f : s f x + y + f ( xy ) = f f ( x + y ) + xy , vi mi s thc x , y tha iu kin: (1) Trc ht ta chng minh f l mt hm n ỏnh Tht vy, xột hai s a,b bt kỡ Chn s cho s2 > 4a; s2 > 4b Khi ú phng trỡnh t2 st + a = cú hai nghim pbit l t1 , t2 t2 st + b = cú hai nghim pbit l t3 , t4 ( ( ) ( ) ( ) ) f s + f (a) = f f (s ) + a Trong (1) ln lt thay (x,y) bng ( t1 , t2 );( t3 , t4 ) ta c: f s + f ( b ) = f f ( s ) + b T ú nu f ( a) = f ( b) thỡ a = b suy f n ỏnh ( ) ( ) Thay y = (1) ta c f x + f ( ) = f f ( x ) f ( x ) = x + f ( 0) ( ) Vy f x = x + a , ú a l mt hng s VD 7: (Dựng tớnh n ỏnh) (IMO 1988) Tỡm tt c cỏc hm f : * * tha ng thc: ( ) f f ( m) + f ( n ) = m + n , vi mi m, n * ( ) Thay m = n vo ng thc trờn ta c f f ( n ) = 2n (1), v t ng thc ny ta cú: nu ( ) ( ) f ( n1 ) = f ( n2 ) f f ( n1 ) = f f ( n2 ) 2n1 = 2n2 n1 = n2 hay suy f l n ỏnh ền ( ( ) )) ( ( ( ) ( )) Ta cú 2n = n + n + = n + n f f n + f n + = f f n + f n , v f l n ỏnh nờn f ( n 1) + f ( n + 1) = f ( n) , n (2) () ( ) ( ) ( ) ( ) () suy ra: f ( n) = f (1) + ( n 1) a = an + b ; ú b = f (1) a (sp dc xung) Thay f ( n) = an + b vo phng trỡnh (1) ban u ta c a = 1, b = Vy f ( n) = n, n T ng thc (2) ta cú: f n f n = f n f n = = f f = a , * VD 8: (Balkan MO 2009) (i lin sau bi 7) Kớ hiu * l hp cỏc s nguyờn dng Tỡm tt c cỏc hm f : * ( ( m) + f ( n ) ) = m f f 2 * tha ng thc: + 2n , vi mi m, n * Nu m1 , m2 * cho ( ) ( ) f ( m1 ) = f ( m2 ) f f ( m1 ) + f ( n ) = f f ( m2 ) + f ( n ) (cng thờm, ri ly f cho v) m12 + 2n2 = m22 + 2n2 , suy m1 = m2 hay f l n ỏnh ( ) ( ) ( ) ( ) D thy vi mi n * , n ta cú: n + + n = n + n + (ngh cu to ) T ng thc kt hp vi phng trỡnh ó cho ta c: Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm f f ( n + ) + f ( n 1) = f f ( n ) + f ( n + 1) , ( ( ) ) ( ( ) ( ) ( ) ) f l n ỏnh nờn ta cú: f n + + f n = f n + f n + ( ) ( ) () = f ( n + 1) f ( n) + f ( n 1) = = f (3) f ( 2) + f (1) = a f ( n ) f ( n 1) = f ( n 1) f ( n ) + a f ( n 1) f ( n ) = f ( n ) f ( n 3) + a (1) T ng thc (1) ta cú: f n + f n + + f n = f ( n) f ( n 1) + f ( n 2) = 2 2 2 2 2 2 2 f ( n ) f ( n 1) = f ( ) f (1) + a ( n ) f ( n 1) f ( n ) = f ( ) f (1) + a ( n 3) f ( ) f (1) = f ( ) f (1) Cng tng v ca cỏc ng thc trờn ta c: ( ) f ( n ) f (1) = ( n 1) f ( ) f (1) + () () a ( n 1)( n ) T ng thc (2) ta suy f n cú dng: f n = bn2 + cn + d Mt khỏc phng trỡnh ban u cho m = n ta c: () ( (2) ) f f ( n ) = 3n2 (3) (4) T (3) v (4) ta thu c b = 1, c = d = Vy f n = n , vi mi n * Gii thớch: T (3) ta cú f ( n) bc nờn f(n) bc => gi f(n) = VD 9: (Dựng tớnh song ỏnh) VMO 2012 (cõu 7) Tỡm tt c cỏc hm s f xỏc nh trờn s thc , ly giỏ tr v tha ng thi cỏc iu kin sau: 1/ f l ton ỏnh t n ; 2/ f l hm s tng trờn ; 3/ f(f(x)) = f(x) + 12x vi mi s thc x Li gii (Theo chemthan Nguyn Ngc Trung) Nu f(x) = f(y) thỡ f(f(x)) = f((f(y)) nờn t phng trỡnh hm ta suy 12x = 12y, suy x = y Vy f l n ỏnh Theo bi, f l ton ỏnh t vo nờn t õy ta cú f l song ỏnh Gi f-1 l hm ngc ca f thỡ f-1 cng l hm tng Thay x = vo phng trỡnh hm, ta c f(f(0)) = f(0) Do f l song ỏnh nờn t õy suy f(0) = Ly f hai v ta suy f ( 0) = - t f n ( x ) = f ( f ( f ( x ))) , n ln, d thy f n l hm tng v f n ( 0) = - Xột dóy an vi a0 = f ( x ), a1 = x , an = f ( an1 ), vi n Thay x f ( an1 ) vo phng trỡnh hm, ta c an = an1 + 12.an Gii phng trỡnh sai phõn ny, ta tỡm c 4x f ( x) 3x + f ( x ) n ( 3) n + 7 * Xột vi x > 0, c nh Khi ú f n ( x ) > vi mi n (do f n l hm tng), 3x + f(x) > f n ( x ) = an+1 = Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm k k 4x f ( x) f ( x) 4x Cho n = 2k; n = 2k + , ta thu c: > , > 3x + f ( x ) 3x + f ( x ) Cho k + ta thu c x f ( x ) x , suy f ( x ) = x T ú f ( x ) = x vi mi x > * Vi x < 0, c nh Khi ú f n ( x ) < vi mi n, 3x + f(x) < Hon ton tng t ta cng suy f ( x ) = x vi mi x < Kt hp cỏc trng hp ta c f ( x ) = x , x Th li ta thy hm ny tha phng trỡnh hm ban u Vy f ( x ) = x , x l hm nht tha cỏc yờu cu bi toỏn VD 10: (IMO ShortList 2002) Tỡm hm s f : tha (ng dng ton ỏnh) ( ) ( ) f f ( x ) + y = x + f f ( y ) x , x , y () Phõn tớch: Nhỡn vo phng trỡnh hm ta thy nu tn ti s a : f a = thỡ thay ( ) ( ) ( ) ( ( x = a f f ( y ) a = f ( y ) a a , vỡ vy nu chng minh c f l ton ỏnh na thỡ bi toỏn c gii quyt Thay y = f x ta cú f f f ( x ) x )) = f ( ) x Do v trỏi l mt hm s bc nht nờn cú mim giỏ tr , ú f l ton ỏnh () T ú luụn tn ti a : f a = ( ) ( ) Thay x = a f f ( y ) a = f ( y ) a a () ( ) Do f l ton ỏnh nờn vi mi s thc x luụn tn ti y cho x = f y a Vỡ vy f x = x a Th li thy tha Chỳ ý : Cú th chng minh f ton ỏnh nh sau: y ta chn x cho x = f ( ) f (0) y = x= f f ( ) + = y = f ( ) f ( ) + = ton ỏnh ( ) f (0) y f (0) y thỡ f x = y nờn f l f f 2 ( ) (IMO 1999) Xỏc nh tt c cỏc hm f : VD 11: (ng dng tớnh ton ỏnh) món: f x f ( y ) = f f ( y ) + xf ( y ) + f ( x ) 1; x , y (1) ( ) ( ) ( ) ( tha ) Vit li gi thit di dng f x f ( y ) f ( x ) = f ( y ) x + f f ( y ) 1; x , y ( ) Ta thy nu y0 : f y0 thỡ v trỏi ca hm s l hm bc nht theo x nờn cú giỏ tr l ( ) T ú, vi mi s thc x luụn tn ti u, v cho x = f ( u ) f ( v ) f ( x ) = f f ( u ) f ( v ) ( ) () () xỏc nh f ta ch cn tớnh c f f ( u ) f ( v ) theo f u f v na l bi toỏn c gii quyt Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm Ta nhn thy f x 0khụng tha bi toỏn ( ) ( ) () T ú nu gi s f x l mt hm s tha bi toỏn thỡ tn ti s thc a cho f a ( ) ( ) Trong (1) cho x = a f x f ( a ) f ( x ) = f ( a ) x + f f ( a ) 1( ) V trỏi ca (2) l mt hm bc nht theo x nờn cú giỏ tr nờn v phi cng cú giỏ tr , ú vi mi s thc x luụn tn ti u, v : x = f u f v () () Vi s thc y bt kỡ, (1) thay x = f ( y ) f ( f ( y ) ) = f ( y ) + f ( ) f (0) 1 (3) f ( f ( y )) = f ( y ) + 2 Thay x = f ( u ) , y = v ta cú f ( f ( u ) f ( v ) ) = f ( f ( u ) ) + f ( u ) f ( v ) + f ( f ( u ) ) 2 ( ) 1 f ( u ) f ( v ) + f ( ) f ( x ) = x + f ( ) 2 f (0) 1 Do ú kt hp vi (2) suy f ( ) = f ( ) = f ( x ) = x 2 Th li thy tha Vy f ( x ) = x VD 12: (ng dng song ỏnh) Cho hm s f : * * l mt song ỏnh Chng minh rng tn = () ( ) () () ti bn s nguyờn dng a, b, c, d cho a < b < c < d v f a + f d = f b + f c Do f l mt song ỏnh t * n * nờn tn ti n cho f (1) < f ( n) { } M = n * : f (1) < f ( n) l khỏc rng ca * nờn tn ti phn t nh nht ca M, kớ hiu l b, {} b > v f ( b) > f (1) Gi c l phn t nh nht ca M \ b , kớ hiu l c, < b < c v f ( c ) > f (1) T f l song ỏnh nờn tn ti d * cho f ( d ) = f ( b) + f ( c ) f (1) T ng thc trờn suy f ( d ) f ( b), f ( d ) f ( c ) d > c > b > Do ú tn ti a, b, c, d * cho a = < b < c < d v f ( a) + f ( d ) = f ( b) + f ( c ) VD 13: (ng dng song ỏnh) (France 1995) Cho hm s f : * * () () l mt song ỏnh Chng () minh rng tn ti ba s nguyờn dng a, b, c cho a < b < c v f a + f c = f b Bi toỏn ny l kt qu trc tip t bi toỏn bờn trờn VD 14: (ng dng song ỏnh) Tỡm tt c cỏc hm f : tha ( ) f xf ( x ) + f ( y ) = y + f ( x ) , x , y Hng dn: Gi s f ( y1 ) = f ( y2 ) f xf ( x ) + f ( y1 ) = f xf ( x ) + f ( y2 ) ( ) ( y1 + f ( x ) = y2 + f ( x ) y1 = y2 f l n ỏnh (1) ( ) Cho x = f f ( y ) = y + f ( ) , y f l ton ỏnh (2) ) () T (1) v (2) suy f l song ỏnh t vo nờn tn ti nht s thc a : f a = ( ) Thay x = a f f ( y ) = y ( 3) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm T (2) v (3) f = Do ú f a = = f a = () ( ) Cho y = ta c f xf ( x ) = f ( ) () () ( x ) f ( f ( x ) f ( f ( x ))) = f ( f ( x )) f xf ( x ) = x f ( x ) = x , x f (x) = x Suy vi mi x thỡ f ( x ) = x Thay vo ta thy c hai hm s ny u tha nờn ta s chng minh nghim ca PTH l f x = x, x v f x = x, x ( ) ( ) () () Tht vy , gi s cú a , b : f a = a, f b = b ( ) ( Thay x = a, y = b f a2 b = a2 + b a2 + b ) ( 4a2 b = ( >< vi a 0,b ) Vy f ( x ) = x, x v f ( x ) = x , x Th li thy tha ) ( ) = f a2 b = a2 b ( vỡ f ( x ) = x , x ) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm C PHN BI TP NGH C.1 CC BI TON DNG TNH N NH CA HM (Dựng tớnh n ỏnh) Tỡm tt c cỏc hm s f : tha iu kin: ( ) f af ( y ) + bx = cx + dy + e, x , y ( a, b, c , d 0; b ac ) (1) Ta chng minh f n ỏnh trc Gi s f(x) = f(y), ú: cx + dy + e = f ( af ( y) + bx ) = f ( af ( x ) + bx ) = cx + dx + e x = y Trong (1) cho x = y = 0, ta c: e = f ( af ( 0)) C ho x = dy , ta c: c bdy f af ( y) = e = f ( af ( 0)) c bdy bdy M f n ỏnh nờn af ( y) = af ( 0) f ( y) = + f ( 0), y ( 2) c ac af ( 0) acf ( 0) be Trong (1) cho y = 0; x = , ềễ c f ( 0) = + e f ( 0) = b b b + ac bdx be Thay vo (2) c f ( x ) = + , x ac b + ac Th li thy ỳng, vy õy l hm s cn tỡm * T bi toỏn tng quỏt trờn, ta cú th gii c bi toỏn sau: Slovenie - 1999 Tỡm tt c cỏc hm s f : tha iu kin: f x f ( y ) = x y, x , y (1) ( ) Ta chng minh f n ỏnh trc Gi s f(x) = f(y), ú: x y = f x f ( y ) = f x f ( x ) = x x x = y ( ( ) C ho x = y , ta c: f ( y f ( y ) ) = 1, y Vy suy ra: f ( f ( ) ) = f ( y f ( y ) ) Trong (1) cho x = y = 0, ta c: f f ( ) = ) ( ) M f n ỏnh nờn f ( 0) = y f ( y) f ( y) = y + f ( 0) , y ( 2) Trong (1) cho y = 0; x = f ( 0), ềễ c f ( 0) = f ( 0) f ( 0) = Thay vo (2) c f ( x ) = x + , x Th li thy ỳng, vy õy l hm s cn tỡm ngh 30-4-2002 Tỡm tt c hm s f : tha iu kin: f 30 f ( y ) + x = 19 x + 75y + 2002, x , y (1) ( Ta chng minh f n ỏnh trc Gi s f(x) = f(y), ú: ( ) ( ) ) 19 x + 75y + 2002 = f 30 f ( y ) + x = f 30 f ( x ) + x = 19 x + 75 x + 2002, x x=y 10 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm Trong (1) cho x = y = 0, ta c: f 30 f ( ) = 2002 ( ) 75 75 y , ta c: f 30 f ( y ) y = 2002, y 19 19 75 Vy suy ra: f 30 f ( y ) y = f 30 f ( ) , y 19 75 10 M f n ỏnh nờn 30 f ( y) y = 30 f ( 0) f ( y) = y + f ( 0) , y ( 2) 19 19 f 30 f ( y ) + x = 19 x + 75y + 2002, x , y C ho x = ( ( ) ) 30 30 f ( 0), ềễ c f ( ) = 19 f ( 0) + 2002 4 287 572 f ( 0) = 2002 f ( 0) = 41 10 572 Thay vo (2) c f ( x ) = y + , x 19 41 Th li thy ỳng, vy õy l hm s cn tỡm Trong (1) cho y = 0; x = (n ỏnh) (T11/407 THTT thỏng - 2011) Tỡm tt c cỏc hm s f xỏc nh trờn , ly giỏ ( ) ( ) tr v tha phng trỡnh: f x + y + f ( y ) = f f ( x ) + y (*), x , y ( ) ( +) Cho y = ta c f f ( x ) = f x + f ( ) ) (1) ( ) ( ) f ( f ( y )) = f ( f ( y )) f ( y + f ( )) = f ( y + f ( )) +) Ta chng minh f l n ỏnh Tht vy nu y1 , y2 cho f y1 = f y2 T (1) v (2) ta cú (2) (3) () f ( f ( ) f ( y ) + y + f ( y )) = f ( f ( f ( ) f ( y ))) + 2y f ( y + f ( ) ) = f ( f ( f ( ) f ( y ) ) ) + y ( 4) Cho x = f f ( y) thay vo phng trỡnh (*) ó cho ta c T (4) ln thay y bi y1 , y2 ta c: ( f (y ) ( ( )) + f ( ) ) = f ( f ( f ( ) f ( y ) ) ) + y f y1 + f ( ) = f f f ( ) f ( y1 ) + y1 2 T hai ng thc ny kt hp vi (2) v (3) ta c y1 = y2 Vy f l mt n ỏnh ( ) () Do ú t (1) ta cú f x = x + f th li thy tha (n ỏnh) (BMO 2009) Tỡm tt c cỏc hm s f : tha ng thc: ( ) ( ) ( ) f xf ( y ) + f f ( x ) + f ( y ) = yf ( x ) + f x + f ( y ) , vi mi x , y +) Nu f ( x ) = vi mi x , th vo phng trỡnh ó cho ta thy tha ( ) ( ) +) Nu tn ti a cho f ( a ) Khi ú vi y1 , y2 cho f y1 = f y2 , t phng trỡnh trờn thay x bi a v y ln lt bi y1 , y2 ta c: 11 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm f af ( y1 ) + f f ( a ) + f ( y1 ) = y1 f ( a ) + f a + f ( y1 ) (1) ( v : ) ( ) ( f ( af ( y ) ) + f ( f ( a ) + f ( y ) ) = y f ( a ) + f ( a + f ( y ) ) (2) 2 () ) () T (1) v (2) ta c y1 f a = y2 f a y1 = y2 Vy f l mt n ỏnh ( ) ( ) Thay x = 0, y = vo phng trỡnh ta c: f ( ) + f f ( ) + f (1) = f ( ) + f f (1) , s dng f l () n ỏnh ta c f = () f ( xf ( ) ) + f ( f ( x ) + f ( ) ) = f ( x ) + f ( x + f ( ) ) Mt khỏc thay y = v phng trỡnh v s dng f = ta c: ( ền anh ) f f ( x ) = f ( x ) f ( x ) = x , x () ( ) Vy f x = 0, x hoc f x = x, x (n ỏnh) Tỡm tt c cỏc hm f : tha ng thi cỏc iu kin sau: ( ) (i) f f ( x ) + y = x + f ( y ) , vi mi x , y (ii) Vi mi x + tn ti y + ( ) () cho f y = x (27) ( ) Vi x1 , x cho f x1 = f x2 nờn t iu kin (i) ta c: ( ) ( ) f f ( x1 ) + y = f f ( x ) + y x1 + f ( y ) = x + f ( y ) x1 = x suy f l n ỏnh Thay x = y = vo phng trỡnh iu kin (i) ta c: ( ) f f ( ) = f ( ) f ( ) = (1) Thay y = vo phng trỡnh iu kin (i) v kt hp vi (1) ta c: ( ) ( ) ( ) f f ( x ) = x + f (0) f f ( x ) = x (2) Thay x bi f x phng trỡnh iu kin (i) v kờt hp vi (2) ta c: ( ( ) ) f f f ( x ) + y = f ( x ) + f (y) f (x + y) = f (x ) + f (y) () ( 2) ( ( )) Vi mi x > , tn ti y > cho x = f y = f f x ( ) (3) ền anh y = f ( x ) T ú suy vi mi x > thỡ f x > ( cm phn di) Ta chng minh f l hm ng bin Tht vy vi x1 , x , x1 > x v kt hp vi (3) ta cú: f ( x1 ) = f ( x1 x2 + x2 ) = f ( x1 x2 ) + f ( x2 ) > f ( x2 ) suy f l mt hm s ng bin ( ) Do hm s f ng bin v ng thc (2) ta thu c: f x = x , x (*) ềbieãn (Gii thớch (*) vỡ theo (2), nu f ( x ) > x f ( f ( x )) > f ( x ) x > f ( x ) ềbieãn ( ) Neãu f ( x ) < x f ( f ( x )) < f ( x ) x < f ( x ) (!) (!) ) Vy f x = x , x (n ỏnh) Chn i tuyn B Vng Tu 2013 Tỡm tt c cỏc hm s f : tha iu kin : f xy + f ( x ) + f x yf ( x ) = 2x (1) ( 12 ) ( ) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm ( ) ( ) ( ) Cho x = ta cú f ( f (0) ) + f ( yf (0) ) = f ( yf (0) ) = Nu f ( 0) thỡ t f ( yf (0) ) = f vụ lớ Cho x = 0, y = ta cú f f (0) + f f (0) = f f (0) = Do ú f ( 0) = ( ) Cho y = ta cú f f ( x ) + f ( x ) = x (2) * Tip theo ta cm f(x) n ỏnh T( 2) ta laãy f ( x1 ) = f ( x2 ), suy : x1 = f ( f ( x1 ) ) + f ( x1 ) = f ( f ( x2 ) ) + f ( x ) = x x1 = x => f l n ỏnh ( ) ( ) Cho y = ta cú f ( x + f ( x ) ) + f ( x + f ( x ) ) = 2x Suy f ( x f ( x ) ) = f ( x + f ( x ) ) x f ( x ) = x + f ( x ) f ( x ) = x * Cho y = ta cú f x + f ( x ) + f x f ( x ) = 2x (n ỏnh) (IMO Shortlist 2007) Tỡm tt c cỏc hm f : + + tha f x + f ( y) = f x + y + f ( y) ( ) ( ( ) ) ( ) Nu tn ti f < , ta thay y = , x = f ( ) ta c: f f ( ) = vụ lớ Nu tn ti : f ( ) = thỡ ta thay y = Khi ú f ( ) = vụ lớ Do ú f ( x ) > x , x t g( x ) = f ( x ) x , suy g : ( ) + + V f x + y + g( y) = f ( x + y) + f ( y) ( ) ( ) Do ú g ( x + y + g( y) ) = g( x + y) + y , tc l g ( t + g( y) ) = g(t) + y vi t > y > Hay f x + y + g( y) x + y + g( y) + g( y) = f ( x + y) ( x + y) + f ( y) y + y T õy ta cú g l hm n ỏnh ( ) ( ) ( ) ( ) { } Thy vy: Nu g y1 = g y2 g t + g( y1 ) = g t + g( y2 ) y1 = y2 vi t > max y1 , y2 Vi mi x , y > v t > x + y ta cú: g (t + g( x ) + g( y) ) = g (t + g( x ) ) + y = g(t) + x + y = g (t + g( x + y) ) Do ú t + g( x ) + g( y) = t + g( x + y) g( x ) + g( y) = g( x + y ) T õy, ta suy g( x ) = x hay f ( x ) = x , x (n ỏnh) Tỡm tt c cỏc hm f : ( 0; +) ( 0; +) tha f ( x + f ( y) ) = T gi thit suy f(x) > 0, x > Ta s cm f n ỏnh Gi s f ( x ) = f ( y) , ú: y x = f ( x + f ( y )) = f ( x + f ( x )) = xy + x +1 x2y + y = x2y + x x = y 13 y , x , y ( 0; +) (1) xy + Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm y ya Vi a > 0, xột: Do ú t (1) suy = a y = axy + a x = xy + ay ya x a f + f ( y) = a = f + f ( x ) , x > 2a; y > 2a ( 2) ay ax ya x a + f ( y) = + f ( x ), x , y > 2a ay ax 1 f ( x ) = f ( y ) , x , y > 2a x y f ( x ) = C , x , y > 2a (C : haậng soã) x Do ềo: f ( x ) = + C , x > 2a x M f n ỏnh nờn dn n: Vi mi x > 0, luụn tn ti s a > cho x > 2a, o ú theo (3) ta c f ( x ) = +C x + C , x > Thay vo (1) ta c: x y +C = , x , y ( 0; +) x + f ( y) xy + 1 y +C = , x , y ( 0; +) ( 4) xy + x + +C y T ( 4) cho x = y = ta ề ễ c: Vy f ( x ) = C = 1 +C = 2+C C = / Do f(x) > 0, x > nờn ta loi hm f ( x ) = cu bi l: f ( x ) = , x > x * Chỳ ý: S l sai nu lp lun: , x > Vy cú nht hm s tha cỏc yờu x y y = y = xy + x = xy + y Do ú t (1) suy ra: 10 (n ỏnh) Olympic 30-4-2006 Tỡm tt c cỏc hm f : tha f x + f ( y) + xf ( y) = x + xy + y, x, y ( ) Ta s cm f n ỏnh Gi s f ( y1 ) = f ( y2 ) , (1) cho x = 1, ú: f + f ( y) = + 2y, y ( ) (1) (2) Vẽ f ( y1 ) = f ( y2 ) nen f (1 + f ( y1 )) = f (1 + f ( y2 )) , bi vy, t (2) suy ra: + y1 = + y2 y1 = y2 Do ú f l n ỏnh Trong (1) ly x = ta c f ( f ( y)) = y, y ( ) Trong (1) ly y = ta c: f x + f (0) + xf (0) = x = f ( f ( x )) x + f ( 0) + xf ( 0) = f ( x ) ( f ền anh) t f(0) = a Theo trờn ta cú: f ( x ) = ( a + 1) x + a, x 14 (3) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm Thay (3) vo (1) ta c: ( a + 1)[ x + f ( y ) + xf ( y )] + a = x + xy + y, x , y ( a + 1)[ x + ( a + 1) y + a + x ( a + 1) y + ax ] + a = x + xy + y, x , y a = C hoễ n x = y = ta ề ễ c : ( a + 1)a + a = a = - Khi a = ta c f(x) = a - Khi a = -2 ta c f(x) = -x-2 Th li thy tha Vy cú hm s tha bi l: f ( x ) = x , x vaf ( x ) = x 2, x * Chỳ ý cỏc BT cú cỏch CM n ỏnh khỏc bit sau: 11 (n ỏnh) Tỡm tt c cỏc hm f : ( 0; +) ( 0; +) tha x [ f ( x ) + f ( y )] = ( x + y ) f ( yf ( x )), x , y > (1) Ta s cm f n ỏnh Gi s tn ti s dng x1; x cho x1 x vaf ( x1 ) = f ( x ) = a > x [ a + f ( y )] = ( x1 + y ) f ( ya) Trong (1) ln lt ly x = x1; x = x ta c: 12 x [ a + f ( y )] = ( x + y ) f ( ya) v Suy x1; x l nghim phng trỡnh tx ux v = ( t, u, v d ng) Dn n x1 x = < , n õy ta t gp mõu thun Vy f n ỏnh trờn ( 0; +) Trong (1) cho x = y ta c: x f ( x ) = xf ( xf ( x )) xf ( x ) = f ( xf ( x )), x > ( 2) Trong ( 2) cho x = ề ễ c f (1) = f ( f (1)) f (1) = 1 Trong (1) cho x = ề ễ c : + f ( y ) = ( y + 1) f ( y ) = yf ( y) f ( y) = , y > y Th li ta kt lun: Cú nht mt hm s tha yờu cu bi toỏn l f ( x ) = * Cú th cm n ỏnh nh sau: C nh y > 0, gi s x1; x > 0; f ( x1 ) = f ( x ) Ta co: , x > x x12 f ( yf ( x1 )) f ( yf ( x )) x2 = = = x1 + y f ( x1 ) + f ( y ) f ( x ) + f ( y ) x + y x = x2 x12 x2 = x1 + y x + y x1 x + y( x1 + x ) = ( volyvẽ x1 , x , y > 0) Va y cox1 = x 12 (n ỏnh) Tỡm tt c cỏc hm f : tha f (1 + x ) f ( y) = yf (1 + f ( x )), x, y (1) ( ) - T (1) ta thy rng nu f n ỏnh thỡ bi toỏn s rt d dng Tht vy, gi s f l n ỏnh T (1) ly y = c: f (1 + x ) f (1) = f (1 + f ( x )), x ( 2) ( ) - Do f n ỏnh nờn t (2) ta cú: (1 + x ) f (1) = + f ( x ), x suy f cú dng nh sau: f ( x ) = ax + b, x Thay vao (1) ềễ c : a[(1 + x )( ay + b)] + b = y[ a(1 + ax + b], x , y ab + b = - T (3) ln lt cho ( x; y) = ( 0; 0); ( x; y) = (1; 0), ềễ c: b=0 2ab + b = 15 ( 3) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm - Vy (3) tr thnh a[(1 + x )( ay)] = y[a(1 + ax )], x , y T õy ta cho x = y = c: 2a2 = a(1 + a) a2 = a a = hoaẻc a = - Th li thy f ( x ) = 0; f ( x ) = x tha cỏc yờu cu bi * Vn cũn li, nu f khụng n ỏnh, thỡ tn ti y1 y2 cho f ( y1 ) = f ( y2 ) K hi ềo: y1 f (1 + f ( x )) = f ((1 + x ) f ( y1 )) = f ((1 + x ) f ( y2 )) = y2 f (1 + f ( x )), x K eãt tn pti y1 racho : f (1f+( yf ()x))0=thẽ 0, t x Thay ềễ c : fn((1 + x ) f ( y )) = 0, x , y Nu y0 y2 suysao ( 4) ta cofva ( xo) =(10) ,tama u thua Vy t (4) phi cú: f(x) = Kt lun: Cỏc hm s tha bi l f ( x ) = 0; f ( x ) = x 13 (n ỏnh) Tỡm tt c cỏc hm f : tha f x + f ( y) = f ( x + xy) + yf (1 x ), x, y (1) ( ( 4) ) Kớ hiu P(u;v) ch vic thay x bi u v y bi v vo (1) + Trng hp f (1) P ( 0; x ) f ( f ( x )) = f ( 0) + f (1) x , x Suy f n ỏnh P ( 0; 0) f ( f ( 0)) = f ( 0) , m f l n ỏnh nờn f(0) = Vi x f ( x) f ( x) f ( x) ; x ) f (1 ) = = f ( 0) = (xem gii thớch bờn di) x x x Do ú: f ( x ) = x , x , kt hp vi f(0) = suy f(x) = x, x Th li thy tha + Trng hp f (1) = P ( 0; 0) f ( f ( 0)) = f ( 0) P( - T ú: P (1; f ( 0)) f (1 + f ( 0)) = f (1 + f ( 0)) + f ( 0) f ( 0) = P ( 0; x ) f ( f ( x )) = 0; P (1; f ( x 1)) f ( f ( x 1) + 1) = Thc hin P (1; x 1) f (1 + f ( x 1)) = f ( x ) Nh va y : f ( x ) = 0, x Th li thy tha Vy cỏc hm s tha yờu cu bi toỏn l: f ( x ) = x; f ( x ) = * Gii thớch: Vỡ cn trit tiờu: f ( x + f ( y)) vaf ( x + xy) hai veã, ta xet : x + f ( y) = x + xy x = 16 f ( y) y Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm C.2 CC BI TON DNG TNH TON NH, SONG NH CA HM (Cỏc bi di õy cú c tớnh l ngi ta hay chuyn f(x) qua trỏi, VP cũn li l bc nht ca x, dựng tớnh ton ỏnh) 14 (Ton ỏnh) Tỡm tt c cỏc hm f : tha: f x + f ( y) = x + f y + xf y , x , y (1) ( ) () () Gi s f l hm s tha bi, ú ta cú (1) Trong (1) ly x = ta c (2) f ( f ( y)) = f ( y), y Hay f ( f ( x )) = f ( x ), x ( ) () - Ta cú (1) tng ng vi: f x + f ( y) = x[1 + f ( y)] + f y , x, y * TH1: f ( y) ( hay f ( y) = 1, y ) D thy f ( y ) 1tha món(1) Vy f ( y) = 1, y tha bi * TH2: Ton taễ i y0 : f ( y0 ) K hi ềo: f ( x + f ( y0 )) = x[1 + f ( y0 )] + f ( y0 ), x ( 3) Cho x thay i trờn , v phi ca (3) l a thc bc nht theo x trờn (do + f ( y0 ) ), ú cú giỏ tr l Suy giỏ tr ca v trỏi ca (3) cng l Vy hm f cú giỏ tr l Do ú vi mi u , thỡ tn ti v cho f ( v) = u ( ) Vy vi mi u , ta cú: f ( u) = f ( f ( v)) = f ( v) = u Ngha l f ( x ) = x , x Th li thy hm s f ( x ) = x , x khụng tha (1) Vy cú nht hm s tha bi l f ( x ) = 1, x 15 (Ton ỏnh) Tỡm tt c cỏc hm f : tha: f x f ( y) = f x + x + f y , x , y (1) ( ) () () - Gi s f l hm s tha bi, ú ta t f(0) = a Ta cú: (1) f ( x f ( y)) f ( x ) = x + f ( y), x , y ( 2) - Trong (2) cho y = ta c: f ( x a) f ( x ) = x + a, x (3) - Cho x thay i trờn , v phi ca (3) l a thc bc nht theo x trờn , ú cú giỏ tr l Suy t , ton taễ i u, v cho t = f (u) f (v) ( tỡm hm f ta ch cn tỡm f (t) = f ( f (u) f ( v)) = ?) Thay x bi f(y) vo (1) ta c: f ( 0) = f ( f ( y)) + f ( y), y a f ( f ( x )) = f ( x ) + , x ( 4) Thay x bi f(x) vo (1) ta c: f ( f ( x ) f ( y )) = f ( f ( x )) + f ( x ) + f ( y ) = f ( x ) + a + f ( x ) + f ( y ) = ( f ( x ) f ( y )) + a, x , y Thay x bi f(x) - f(y) vo (1) ta c: f ( f ( x ) f ( y)) = f ( f ( x ) f ( y)) + f ( x ) = 2[ f ( x ) f ( y )] + 2a + f ( x ) = [ f ( x ) f ( y )] + 2a, x , y Ta cú: f ( t) = f ( f ( u) f ( v)) = ( f ( u) f ( v)) + 2a = t + 2a, t Hay f ( x ) = x + 2a, x Suy f ( f ( x )) = f ( x ) + 2a, x 17 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm a Kt hp vi (4) ta c 2a = a = Va y f ( x ) = x , x Sau th li ta cú kt lun: cú nht mt hm s tha bi l f ( x ) = x , x * Chỳ ý: Do t = f ( u) f ( v) nen f ( t) = f ( f ( u) f ( v)) = f ( f (u) f ( v) f ( v)) = f ( f ( u) f ( v)) + f ( u) f ( v) + f ( v) = f ( f ( u) f ( v)) + f ( u) Vy ta ny sinh nhu cu phi tỡm f ( f ( x ) f ( y )) nh ó trỡnh by trờn 16 (Ton ỏnh) x , y (17) Vi y1 , y2 ( ) tha: f f ( x ) + f ( y ) = f ( x ) + y + f ( y ) , f : Tỡm tt c cỏc hm ( ) ( f ( y1 ) = f ( y2 ) f f ( x ) + f ( y1 ) = f f ( x ) + f ( y2 ) cho f ( x ) + y1 + f ( y1 ) = f ( x ) + y2 + f ( y2 ) y1 = y2 Do ú f l mt n ỏnh ( ) ) suy Thay y = f x vo phng trỡnh ban u ta c: ( ( )) ( ) f ( x ) + f ( f ( x )) = f ( x ) f ( f ( x ) ) = f ( x ) , x f f (x) + f f (x) = f f (x) () (1) Thay x = f y vo phng trỡnh ban u ta c: ( ( ) ( ) ) f f f (y) + f (y) = f f (y) + y + f (y) ( ) f f ( y ) = f ( y ) + y + f ( y ) = y , y ( 2) Ly y0 , theo ( 2) thẽ x = f ( y0 ) : f ( x ) = y0 Suy f l mt ton ỏnh () f ( x ) = x , x Do ú vi mi x thỡ tn ti t cho x = f t T ng thc (1) ta cú: ( ) f f ( t ) = f ( t ) f ( x ) = x , x Vy (Ton ỏnh) (Indonesia TST 2010) Xỏc nh tt c cỏc s thc a cho cú mt hm s f : tha món: x + f ( y ) = af y + f ( x ) , vi mi x , y 17 ( ) (Bi ny khụng cn cm n ỏnh, cho dự nú l n ỏnh) D thy nu a = khụng tha Do ú a , thay y = vo ng thc trờn ta c: ( ) f f (x) = x + f (0) (1) a T ng thc (1) suy f l mt ton ỏnh nờn tn ti x cho f ( x ) = Khi ú t phng trỡnh () () ( ) () f ( y ) const ban u ta cú: x + f y = af y x = a f y , vi mi y T ng thc (2) thỡ s xy a = hoc () (2) +) Nu f y const thỡ khụng tha phng trỡnh ban u ( ) +) Nu a = thỡ ly f x = x , vi mi x tha bi toỏn Vy a = 18 (Ton ỏnh) ( ) ( Tỡm tt c cỏc hm f : tha: f xf ( x ) + f ( y) = y + f ( x ) 18 ) (1) Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm ( ) ( Cho x = , ta cú f f ( y) = y + f ( 0) ) (2) T õy, ta suy c f l n ỏnh v ton ỏnh Suy tn ti a cho f ( a) = ( ) Cho x = a ta cú f f ( y) = y kt hp vi (2) ta cú f ( 0) = hay a = ( ) ( ) Cho y = ta suy f xf ( x ) = f ( x ) ( ) ( ) ( Thay x bi f ( x ) ta cú f f ( x ) f ( f ( x )) = x hay f xf ( x ) = x = f ( x ) ( ) ) Suy f xf ( x ) + f ( y) = x + y ( Ta cú f (1) ) = f (1) = ( ) f (1) = 1, cho x = ta c f + f ( y) = + y ( ) ( Suy (1 + y) = f + f ( y) = + f ( y) Vỡ f (1) = nờn ta cú f ( y ) = y ( ) ( f ( y) y )( f ( y) + y + 2) = ) f (1) = 1, cho x = ta cú f + f ( y) = + y , t õy ta tỡm c f ( y ) = y 19 (Ton ỏnh) (Chn i tuyn PTNK 2013) Tỡm tt c cỏc hm s f : tho f ( x + y + f ( y )) = y + x f ( x ) x , y (1) ( ) Cho x = ta cú f y + f ( y) = 2y (2) nờn f ton ỏnh Do ú, tn ti a cho f ( a) = Trong (1), thay y = a, x = ta cú f ( a) = 2a a = suy f ( x ) = x = Trong (1), thay y = ta cú f ( x ) = x f ( x ) (3) Ta vit li (1) nh sau ( ) ( ) f x + y + f ( y ) = y + f x hay f ( x + y + f ( y) ) = 2y + f ( x ) (4) Trong (4), thay y = f (x) f ( x) ta cú: f ( x) f x + f ( x) f ( x) f + = x f ( x) f = f ( x) + f = x (5) 2 f (x) f ( x) f ( x) Do ú f ( x ) = f = + f = f ( x) 2 f ( y) T (4) ta thay y bi v s dng (5) ta cú f x y = f x f ( y) = f ( x ) + f y hay f x + y = f ( x ) + f ( y), x, y (6) Suy ( ) () ( ) T õy, ta cú f ( nx ) = nf ( x ), n S dng (3) v (6) ta cú ( ( ) (7) ) ( ) ( f ( x + 1) + ( x 1) = f ( x + 1) + f ( x 1) ) = ( x + 1) f ( x + 1) + ( x 1) f ( x 1) = ( x + 1) ( f ( x ) + f (1) ) + ( x 1) ( f ( x) f (1) ) 19 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm (8) = x + f ( x ) + xf (1) ( ) S dng (3), (6) v (7) ta cú ( ) ( ) ( ) f ( x + 1)3 + ( x 1)3 = f x + x = f x + f ( x ) = x f ( x ) + f ( x ) T (8) v (9) ta suy f ( x ) = f (1) x = ax Thay vo (1) ta tỡm c a = 1, a = Vy f ( x ) = x , f ( x ) = x l hai hm cn tỡm 20 (Song ỏnh) (9) Tỡm tt c cỏc hm f : tha: f ( f ( x ) ) + f ( f ( y) ) = 2y + f ( x y) (1) vi x , y f ( 0) nờn f ( x ) l hm song ỏnh Suy tn ti a f ( a) = v tn ti b f ( b ) = a Trong (1) thay x = y ta cú f ( f ( y) ) = y + Trong (1) cho y = b ta cú f ( f ( x ) ) = 2b + f ( x b ) = x + f ( 0) f ( 0) = x + m Thay vo (1) ta tỡm c m = Vy f ( x ) = x l hm cn tỡm Suy f ( x ) = x b + 21 (Song ỏnh) (Chn i tuyn Bỡnh Phc 2010 2011 ) Tỡm tt c cỏc hm s f : tha iu kin: f x + f ( y ) = y + x f ( x ) , (1) x , y ( ( ) f ( a) = ) Trong (1) cho x = ta cú f f ( y) = y (2) Suy ra, f ( x ) l song ỏnh Do ú, tn ti a cho ( ) Trong (1) thay x = a v s dng (2) ta c f a2 + f ( y ) = y = f ( f ( y ) ) Suy a2 + f ( y ) = f ( y ) a = hay f ( 0) = ( ) Trong (1) thay y = , ta cú f x = xf ( x ) (3) Trong (1) thay y xf ( x ) : ( ) ( ) f x + f ( xf ( x ) ) = x + f ( xf ( x ) ) = hay f x + y = f ( y ) + f ( x ) ( ) Do ú f x + y = f ( x ) + f ( y) 22 (Song ỏnh) (Phan Ngc Hin - C Mau) Tỡm tt c cỏc hm f:   tha : f ( x + f ( y)) = f ( x ) y, x , y  Cho x = ta cú f ( f ( y )) = y + f ( 0) , suy f l song ỏnh Do ú, tn ti a cho f ( a) = Thay y = a ta cú f ( x + f ( a)) = f ( x ) a a = f ( 0) = Thay y f ( y ) ta cú f ( x + f ( f ( y)) ) = f ( x ) f ( y) f ( x y ) = f ( x ) f ( y) (*) 20 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm Thay x f ( y ) ta cú f ( f ( y) + f ( y) ) = f ( f ( y) ) y = y y = f ( y ) + f ( y) = f ( y) = f ( y) Nờn (*) c vit li f ( x y ) = f ( x ) + f ( y) hay f ( x + z ) = f ( x ) + f ( z), x, z  T õy, suy f ( x ) = ax Thay vo phng trỡnh ban u ta tỡm c a2 = vụ nghim Vy khụng tn ti hm f tha yờu cu bi toỏn 23 ( (Song ỏnh) ) (IRAN TST 2011) Tỡm tt c cỏc song ỏnh f : cho: f x + f ( x ) + f ( y ) = f ( x ) + f ( y ) , vi mi x , y (42) Do f l mt ton ỏnh nờn vi mi x tn ti t cho f ( t ) = ( ) ( ) ( ) ( ) vo phng trỡnh ban u ta c: f x = f x + f 2t f 2t = x + f (x) + x Khi ú thay y=t (1) Thay x = y = 2t vo phng trỡnh hm ban u v kt hp vi (1) ta c: ( ) f 2t + f ( 2t ) + f ( 2t ) = f ( 4t ) f ( 4t ) = T (1), (2) v f l n ỏnh nờn ta cú: f ( 4t ) = f ( 2t ) 4t = 2t t = ( ) x + f (x) Vy f x = x , vi mi x * Chỳ ý: Cú th gii bng cỏch xem vai trũ ca x; y nh 21 (2) + x = f (x) = x Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm PHN III: TI LIU THAM KHO Nguyn Vn Mu, Phng trỡnh hm, NXB Giỏo dc, 2000 Nguyn Sinh Nguyờn, Chuyờn bi dng phng trỡnh hm, NXB Nng, 2006 Nguyn Trng Tun, Bi toỏn hm s qua cỏc kỡ thi Olympic, NXBGD, 2008 Lờ Honh Phũ, Chuyờn kho v a thc, Nh xut bn HQG Tp H Chớ Minh, 2003 Nguyn Vn Mu, a thc i s v phõn thc hu t, NXB Giỏo dc, 2002 Conhiagghin , Cỏc vụ ch Toỏn cỏc nc, Nh xut bn Hi phũng, 1993 Lờ ỡnh Thnh - Phng trỡnh sai phõn v mt s ng dng - NXB GD 2001 Nguyn Vn Nho - Tuyn chn cỏc bi toỏn t nhng cuc thi ụng Au - NXB GD 2003 Tuyn cỏc thi Olympic 30-4 lp 10 10 Tuyn cỏc thi Olympic 30-4 lp 11 11 Cỏc Toỏn hc v tui tr, t liu Internet 12 Trn Nam Dng, Chuyờn phng trỡnh hm a thc 13 Hõn - Ch a thc 14 Lờ Hi Chõu, Cỏc bi toỏn hc sinh gii toỏn PTTH ton quc (1962-1991), NXBGD 15 T sỏch THTT, Cỏc bi thi Olympic toỏn THPT vit Nam 1990 - 2006, NXBGD, 2007 16 Nguyn Vn Mu, Mt s Chuyờn Gii Tớch bi dng hs gii THPT, NXBGD, 2009 17 Nguyn Vn Mu, Mt s Chuyờn i S bi dng hs gii THPT, NXBGD, 2009 18 Din n mathlinks.ro Din n mathscope.org 22 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm PHN IV: MC LC PHN I M U: PHN II: PHN NI DUNG A PHN Lí THUYT B CC V D: C PHN BI TP NGH 10 C.1 CC BI TON DNG TNH N NH CA HM 10 C.2 CC BI TON DNG TNH TON NH, SONG NH CA HM 17 PHN III: TI LIU THAM KHO 22 PHN IV: MC LC 23 PHN V: THM NH CA GIO VIấN 24 23 Bi dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm PHN V: THM NH CA GIO VIấN , ngy thỏng nm 2015 Giỏo viờn thm nh 24 [...]... x b + 21 (Song ỏnh) (Chn i tuyn Bỡnh Phc 2010 2011 ) Tỡm tt c cỏc hm s f : tha món iu kin: f x 2 + f ( y ) = y + x f ( x ) , (1) x , y ( ( ) f ( a) = 0 ) Trong (1) cho x = 0 ta cú f f ( y) = y (2) Suy ra, f ( x ) l song ỏnh Do ú, tn ti a sao cho ( ) Trong (1) thay x = a v s dng (2) ta c f a2 + f ( y ) = y = f ( f ( y ) ) Suy ra a2 + f ( y ) = f ( y ) a = 0 hay f ( 0) = 0 ( ) Trong (1) thay... Thay vo (1) ta tỡm c a = 1, a = 2 Vy f ( x ) = x , f ( x ) = 2 x l hai hm cn tỡm 20 (Song ỏnh) (9) Tỡm tt c cỏc hm f : tha: f ( f ( x ) ) + f ( f ( y) ) = 2y + f ( x y) (1) vi x , y f ( 0) nờn f ( x ) l hm song ỏnh 2 Suy ra tn ti a f ( a) = 0 v tn ti b f ( b ) = a Trong (1) thay x = y ta cú f ( f ( y) ) = y + Trong (1) cho y = b ta cú f ( f ( x ) ) = 2b + f ( x b ) = x + f ( 0) 2 f ( 0) = x... y ) f ( ya) Trong (1) ln lt ly x = x1; x = x 2 ta c: 12 x 2 [ a + f ( y )] = ( x 2 + y ) f ( ya) v Suy ra x1; x 2 l nghim phng trỡnh tx 2 ux v = 0 ( t, u, v d ng) Dn n x1 x 2 = < 0 , n õy ta t gp mõu thun Vy f n ỏnh trờn ( 0; +) Trong (1) cho x = y ta c: 2 x 2 f ( x ) = 2 xf ( xf ( x )) xf ( x ) = f ( xf ( x )), x > 0 ( 2) Trong ( 2) cho x = 1 ề ễ c f (1) = f ( f (1)) f (1) = 1 1 Trong (1) cho... xf ( y) = x + xy + y, x, y ( ) Ta s cm f n ỏnh Gi s f ( y1 ) = f ( y2 ) , trong (1) cho x = 1, khi ú: f 1 + 2 f ( y) = 1 + 2y, y ( ) (1) (2) Vẽ f ( y1 ) = f ( y2 ) nen f (1 + 2 f ( y1 )) = f (1 + 2 f ( y2 )) , bi vy, t (2) suy ra: 1 + 2 y1 = 1 + 2 y2 y1 = y2 Do ú f l n ỏnh Trong (1) ly x = 0 ta c f ( f ( y)) = y, y ( ) Trong (1) ly y = 0 ta c: f x + f (0) + xf (0) = x = f ( f ( x )) x + f ( 0)... 0 ta cú f y + f ( y) = 2y (2) nờn f ton ỏnh Do ú, tn ti a sao cho f ( a) = 0 Trong (1), thay y = a, x = 0 ta cú f ( a) = 2a a = 0 suy ra f ( x ) = 0 x = 0 Trong (1), thay y = 0 ta cú f ( x 3 ) = x 2 f ( x ) (3) Ta vit li (1) nh sau ( ) ( ) f x 3 + y + f ( y ) = 2 y + f x 3 hay f ( x + y + f ( y) ) = 2y + f ( x ) (4) Trong (4), thay y = f (x) f ( x) ta cú: 2 f ( x) f x + 2 f ( x) f ( x) f... thay y = 0 , ta cú f x 2 = xf ( x ) (3) Trong (1) thay y xf ( x ) : ( ) ( ) f x 2 + f ( xf ( x ) ) = 0 x 2 + f ( xf ( x ) ) = 0 hay f x 2 + y = f ( y ) + f ( x 2 ) ( ) Do ú f x + y = f ( x ) + f ( y) 22 (Song ỏnh) (Phan Ngc Hin - C Mau) Tỡm tt c cỏc hm f:   tha : f ( x + f ( y)) = f ( x ) y, x , y  Cho x = 0 ta cú f ( f ( y )) = y + f ( 0) , suy ra f l song ỏnh Do ú, tn ti a sao cho f ( a)... 19 f ( 0) + 2002 4 4 287 572 f ( 0) = 2002 f ( 0) = 2 41 10 572 Thay vo (2) c f ( x ) = y + , x 19 41 Th li thy ỳng, vy õy l hm s cn tỡm Trong (1) cho y = 0; x = 4 (n ỏnh) (T11/407 THTT thỏng 5 - 2011) Tỡm tt c cỏc hm s f xỏc nh trờn tp , ly giỏ ( ) ( ) tr trong v tha món phng trỡnh: f x + y + f ( y ) = f f ( x ) + 2 y (*), x , y ( ) ( +) Cho y = 0 ta c f f ( x ) = f x + f ( 0 ) ) (1) ( ) (... dng Olympic 30/4 - Chuyờn Phng Trỡnh Hm C.2 CC BI TON DNG TNH TON NH, SONG NH CA HM (Cỏc bi di õy cú 1 c tớnh l ngi ta hay chuyn f(x) qua trỏi, VP cũn li l 1 bc nht ca x, dựng tớnh ton ỏnh) 14 (Ton ỏnh) Tỡm tt c cỏc hm f : tha: f x + f ( y) = x + f y + xf y , x , y (1) ( ) () () Gi s f l hm s tha món bi, khi ú ta cú (1) Trong (1) ly x = 0 ta c (2) f ( f ( y)) = f ( y), y Hay f ( f ( x ))... ( x ) + f ( y) hay f ( x + z ) = f ( x ) + f ( z), x, z  T õy, suy ra f ( x ) = ax Thay vo phng trỡnh ban u ta tỡm c a2 = 1 vụ nghim Vy khụng tn ti hm f tha yờu cu bi toỏn 23 ( (Song ỏnh) ) (IRAN TST 2011) Tỡm tt c cỏc song ỏnh f : sao cho: f x + f ( x ) + 2 f ( y ) = f ( 2 x ) + f ( 2 y ) , vi mi x , y (42) Do f l mt ton ỏnh nờn vi mi x tn ti t sao cho f ( t ) = ( ) ( ) ( ) ( ) vo phng... ỏnh) Tỡm tt c cỏc hm f : tha: f x f ( y) = 2 f x + x + f y , x , y (1) ( ) () () - Gi s f l hm s tha món bi, khi ú ta t f(0) = a Ta cú: (1) f ( x f ( y)) 2 f ( x ) = x + f ( y), x , y ( 2) - Trong (2) cho y = 0 ta c: f ( x a) 2 f ( x ) = x + a, x (3) - Cho x thay i trờn , v phi ca (3) l a thc bc nht theo x trờn , do ú cú tp giỏ tr l Suy ra t , ton taễ i u, v sao cho t = f (u) ... l song ỏnh yb ) cho : a VD 2: (Dựng ụn cm song ỏnh) Xột tt c cỏc hm f , g, h : cho f l n ỏnh v h l ( ) ( ) song ỏnh tha iu kin f g ( x ) = h ( x ) , vi mi x Chng minh rng g x l mt hm song. .. ) = y f ( x ) = y 3.2 Chỳ ý a Nu f khụng phi l song ỏnh thỡ ta khụng th nh ngha c ỏnh x ngc ca f Do ú ch núi n ỏnh x ngc f l song ỏnh b Nu f l song ỏnh thỡ tn ti ỏnh x ngc f : Y X gỏn cho mi... 2 2 Vy g( x ) l mt hm song ỏnh VD 3: (Dựng ụn cm song ỏnh) Xột tt c cỏc hm n ỏnh f : tha iu kin: ( ) f x + f ( x ) = x , vi mi x Chng minh rng hm s f ( x ) + x l mt song ỏnh Bi dng Olympic