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Bồi dưỡng kiến thức học sinh giỏi phương trình hàm: Phần 1

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Phần 1 tài liệu Bồi dưỡng học sinh giỏi phương trình hàm trình bày các nội dung: Phương trình hàm số với phép biến đổi đối số, giải phương trình hàm trên tập số thực, phương trình đa thức. Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Bồi dưỡng kiến thức học sinh giỏi phương trình hàm: Phần 1

  1. 515 NGUYEN TAI CHUNG B452D BOI OaSNG HOC SINK GIOI • * Phan loai toan Phvfdng t r i n h h a m & bat phifdng t r i n h h a m * PhUdng phap giai Phifdng t r i n h h a m & bat phiidng t r i n h ham Cac de t h i Quoc gia, K h u viJc, Quoc te 4 fm 1 ^ 1 1 h H 1 1 H -1 -2 CD DVL.013493 oca CEG NHA XUAT BAN DAI H O C Q U O C GIA H A NOI H a WQI
  2. •si's NGUYEN TAI CHUNG BOI DOSNG HOC SINH GIOI^ PHIfdNG TRINH HAM * Phan loai toan Phvfdng t r i n h ham & bat phvfdng t r i n h ham * Phvfdng phap giai Phifdng t r i n h ham & bat phvfdng t r i n h ham * Cac de t h i Quoc gia, K h u vvfc, Quoc te • •• f • N H A XUAT BAN OAI HOC Q U 6 C GIA H A N 6 I
  3. HHii XUfiT Bi^N Dfll HOC Q U O C G I R H A NOI 16 Hang Chuoi - Hai Ba TrUng - Ha Npi Dien thoai : Bien t a p - Che ban: (04) 39714896; Hanh chinh: (04) 39714899; Tong bien t a p : (04) 39714897 Fax: (04) 39714899 Ldi noidau Chiu track nhi^m xuat ban 1 Phildng t i i i i h liain la mot chuycn dc quaii troiig thuoc clnrdiig t i i i i h chuyen toan trong rac trirdng T H P T chuyen. Cac bai toan lien quan den phuong Gidm doc - TSng bUn tap : TS. P H / l M T H j T R A M t r i n h ham cuug la nhiing bai tap kho, thudng gSp trong cac k i t h i hoc sinh gioi toan ciia cac cap tinh-thanh, quoc gia, khu vuc, quoc te va Olympic hpc sinh, sinh vien. Bien tap BUI B I C H - Qudc DAT Toan phitfing t r i u h ham phong phu, da dang, phiic help nen kho phan loai Chi' ban va he thong hoa thanh cac chuyen de rieng biet. Tuy vay, chiing t o i c6 gS.ng C O N G TY K H A N G V I E T •'I toi da sap xep hdp l i de giiip ban doc tiep can tilfng budc, tijfng miic dp kien Trinh bay bia C O N G TY K H A N G V I E T tinic va luyen tap giai toan. Phan 16n ciic bai toan troug sach uay duuc tuycn clipn t i l cac k i t h i : Olympic toan ciia cac nirdc, k h u virc va quoc te, Olympic toan Sinh Vien Tongphdt hdnh vd doi tac lien ket xuat ban: toan qu6c, Olympic 30/04, t h i hoc sinh gioi quoc gia, t h i chon doi tuygn quoc gia d u t h i toan quoc te, t h i hpc sinh gioi ciia cac t i n h thanh, t h i vao he K i s U tai nang cua D H B K Ha Npi va t h i vao he C i i nhan t ai nang cua C O N G TY T N H H MTV D H K H T N - D H Q G Ha Npi, hoac tren Tap chi Todn hoc va Tuoi tre. M o t so D!CH VU V A N H O A K H A N G VI|T bai toan khac la do tac gia t u bien soan. Nhan day x i n cam On tac gia ciia cac bai toan ma chung t o i da trich chon. Nhfmg Icii giai dUa ra dua tren tieu Dia chF: 71 Dinh Tien Hoang - P D a Kao - Q.1 : TP H C M chl t u nhieii, de higu. Tuy nhien, do chua h i n la Idi giai hay nhat va ngan Dien thoai: 08. 39115694 - 39105797 - 39111969 - 39111968 Fax: 08. 3911 0880 gpn nhat, hy vpng cung vdi cuon sach, ban doc se c6 dupc nhiing 16i giai hay hdn. Email: khangvietbool
  4. Tot nhat, doc gia nen tit minh giai cac bai tap c6 trong sach nay. Tuy nhien dg giup ban doc thay va lam chii nhung ky xao tinh vi khac, cac bai tap deu da duoc giai s&n (tham chi la v6i nhieu each giai) v6i nhiing mufc do chi tiet khac nhau. Npi dung sach da c6 ging tuan theo y chu dao xuyen suot; Biet diTdc Idi giai cua bai toan chi la yeu cau dau tien, quan trQng hdn la lam the nao di giai dvtdc no, each ta xii ly no, nhiing suy luaii iiao to ra "c6 ly", cac ket luan, nhan xet va liTu y ma bai toan dua ra... Chifdng 1 Hy vpng quyen sach nay la tai li^u tham khao c6 I'ch cho hpc sinh cac Idp chuyen toan b?Lc trung hoc pho thong, giao vien toan, sinh vien todn cua cac trudng DHSP, DHKHTN ciing nhir la tai lieu phuc vu cho cac ky thi hpc Phtfdng trinh ham vdi phep sinh gioi toan THPT, thi Olympic toan sinh vien gifla cac trirdng Dai hoc. Cac ban hoc sinh, sinh vien, giao vien va nhfing ngu6i quan tarn khac c6 bien doi doi so the se tim thay thieu sot d cuon sach nay trong qua trinh su dung. Do vay, s\l gcp y tren tinh than khoa hoc va xay diJng tii phia cac ban la dilu chling 1.1 Mot so kien thufc ve ham so va anh xa * toi luon mong ddi. Hy vong rkng tren h\ldc dirdng t i m toi, sang tao toan hoc, ban doc sc tim dudc nhfmg y tit6ng tot hdn, mdi hdn, nhhn bfi sung cho cac Muc dich cua quyen sach nay la cung cap cac k i thuat giai toan ve phudng Idi giai dUdc trinh bay trong quyen sach nay. trinh ham va thong qua viec giai toan nhSm khac sau mot so kien thtic rat quan trong ve ham so, ren luyen t u duy, chuan bi cho cac ki thi hoc sinh gioi, van dung nhOng y tucing, suy luan trong quyen sach nay vao cac phan mon Tac gia khac ciia toan hoc ciing nhxt cac mon hoc khac. Vdi muc dich tranh nang ne Ths : Nguygn Tai Chung f;ho da so ban doc ngay t\t luc dau tien khi doc sach, trong bai nay chi trinh bay rat it mot so kien thutc chung nhat ve ham so va anh xa, con nhiing kien thiic thong dung khac dtt^c diing vao giai toan se dildc trinh bay vao ngay phan dau cua cac bai tildng ling cua tinig clnfdng, hoac ngay tru6c ho$c sau cac bai toan cu the, ho$c trong cac nhgn xet, chu y, luu y. 1.1.1 Anh xa Dinh nghla 1. Cho hai tg,p hap A va B. Neu c6 mot quy tdc f nao do sao cho vdi moi a & A tUdng ling vdi dung mot phan titb & B thi ta noi f Id mot anh xa tic A den'B, ki hieu Id f : A B. Phan tri b ggi Id anh cua a va viet ldb = f{a). Chu y 1. Neu cho anh xa f : A -* B thi ta thudng quan tam den hai tap hap sau day: Tap hap f{A) = {f{a)\a e A} {goi la anh cua tg,p A, hay goi Id tap gia tri cua anh xg. f) vd tap hap = {a€ A\ = 6} {gpi Id nghjch anh cua b). * 3
  5. 1.1.2 Dcfn a n h , t o a n a n h , song a n h Day so chinh Id hdm so rd tap xdc dinh Id X = N* {hoac X = N), chdng han day so {xn}t=i ^ ''^^ x„ = n - Vn"^ +5, Vn = 1, 2 , . . . , /d hdm so: D i n h n g h i a 2. Anh xa f : A —* B diMc (]oi Id ddn dnh neu v&i. m.oi ai,a2 E A sao cho a\ ta c6 f{a\) ^ /(a2)- f : N* ^ R ; , , " ^ /(") = n - \/r),2 + 5 C h u y 2. Aiih xa f : A B Id dm dnh niu / ( a i ) = f{a2) keo theo a\ 0 2 - {trong day so tren thi f{n) dd dupe ki hieu bdi x „ ) . D i n h n g h i a 3. Anh xa f : A ^ B diMc goi Id todn dnh neu vdi moi b £ B luon ton tai a G A sao cho f(a) = b. 1.1.4 M o t so b a i t o a n a p d u n g C h u y 3. Anh xa f : A ^ B Id todn dnh khi vd chi khi f{A) = B. ^ B a i toan 1.1. Cho hdm so f nhu sau: / :R D i n h n g h i a 4. Anh xa f : A B goi Id song dnh neu f iriCa Id dan dnh, ,.2 + 5 Vila Id todn dnh. a) Hay tinh gid tri cua hdm so tai 2 vd tai -\/5. C h u y 4. Anh xa f : A ^ B Id song dnh khi vd chi khi vdi moi h € B luon b) Hdy chi ra tap xdc dinh vd tim tap gid tri cua hdm so dd cho. ton duy nhdt a G A sao cho f{a) = b. c) Hdm so f CO phai Id dan dnh khong? c6 phdi Id todn dnh khong? co phai D i n h n g h i a 5. Gid siC f : A —* B la mot song dnh. Khi do dnh xa. cho tuang Id song dnh khong? vi sao? I'ng moi phdn tu y £ B vdi tao dnh x = f~^{y) cua no dilcic goi la dnh xa d) Tim / - ' ( 8 ) . ngMc cua f vd ki hieu Id f~^. G i a i . T a c6 f{x) = x^ + 5, Vx € R. «) /(2) = 2'^ + 5 = 9, /(-\/5) = ( - y 5 ) 2 + 5 = 1 0 . |ir 1.1.3 H a m s6 b) Tap xac dinh: R. Tap gia tri ['5; +CXD). c) Ham so / khong phai la drtn anh Y\\=^ - \g / ( I ) = / ( - I ) = 6. Ham D i n h n g h i a 6. Cho X cR vdY cR. Khi do dnh xa f : X ^ Y dUdc goi la mot ham so tic tap X din tap Y. s6 / khong phai la toan anh vi khong ton tai x de cho + 5 = - 2 . Ham so / khSng phai la song anh (do / khong phai la ddn anh). C h u y 5. Cho hdm so f : X ~* Y. Khi do: d) / 1(8) = {x e K|/(x) = 8} = {xeR|x=^ + 5 = 8} = { - ^ 3 , v ^ } . • Tap X goi la tap xdc dinh cua hdm so f. B a i t o a n 1.2. Cho day hiCu han XQ = \,xi = - 5 , X 2 = 7. Hdy xdc dinh tap xdc dinh, tap gid tri, gid tri Idn nhdt, gid tri nho nhdt cua ddy so do. • Neu xo € X thl f{xo) goi Id gid tri cua hdm so f tai XQ. G i a i . Tap xac dinh la {0,1,2}. Tap gia tri la {1, - 5 , 7}. G i a tri Idn nhat la • Tap hop f{X) goi Id tQ,p gid tri cua hdm so f. 7, gia tri nho nhat la - 5 . • 2/0 id mot gid tri cua so f khi vd chi khi phuang trlnh f{x) = yo c6 B a i t o a n 1.3. Hdy chi ra mot hdm so Id nghi^m. Hay noi each khdc la: phuang trinh f{x) = yo c6 nghiem khi vd chi khi yo thuoc tap gid tri cua hdm f. a) Dan dnh; b) Todn dnh; c) Song dnh; d) Khong phdi Id dan dnh; • f Id todn dnh o phuang trinh {an x) y = f{x) {vdi x e X,y e Y) c6 nghiem. c) Khong phdi Id todn dnh; f) Khong phdi la song dnh. ' • f la song dnh phuang trinh {an x) y = f{x) {vdi x e X,y e Y) eo G i a i . Ham so / : R ^ E cho hd\g thiic / ( x ) ^ 2x - \a la ddn anh, nghiem day nhdt. vira la toan anh, vira la song anh. Ham so / : R [2; + 0 0 ) cho bdi cong thiic / ( x ) = x^ + 3 vita khong phai la ddn anh, vita khong phai la toan anh, vita khong phai la song anh. 4
  6. Bai toan 1.6 (Kosovo National Mathematical Olympiad 2011, Grade 9). Bai toan 1.4. Chiing minh rang neu ham so / : K —• iR thoa man diiu kien f (/(x)) = ax + b, Vx G K (a 7^ 0) thi f la song anh. '^^'^ Cho ham so / : K\} R thoa, man f{x) = i + - . Ham so f c6 phdi la. Giai. Gia suf f{xi) = f{x2). Khi do ddn anh khong ? vi sao ? axi+b= f{f{xi)) = f{f{x2))=aX2 + b=>xi=X2. Giai. V i 2 e R\} , ^ e R\, 2 ^ vh f{2) = / (^^^ nen / khong phai Vg-y / la ddn anh. Vcii moi y e M, luon ton t^i x = f sao cho la ddn anh. . ^ > ' B a i toan 1.7 (Kosovo National Mathematical Olympiad 2011, Grade 11). y-b = a.- Cho ham so / : R -+ K thoa man /(sinx) = sin(2011x). Tim /(cosi). Vay / la toan anh, do do / la song anh. Giai. Ta c6: /(cosx) = / ( s i n ( | - x)j = sin (2011(^ - x ) ) /, , Lifti y. Ket qua bai toan 1.4 nay dudc sii dung nhieu. = sin (IOOSTT + ( ^ - 201 I x ) ) ,i ' ; Bai toan 1.5. Cho ham 50 : [0; Ij —• [a; b\ sau: = - sin ( ( ^ - 2011x)) = - cos(2011x). tfiit) = {I - t)a + tb, V«e [0; 1]. a) Chiing minh rdng ^ : [0; 1] —• [a; b] la song anh. Bai toan 1.8. Cho cdc so thuc a, b, c, d sao cho ad - be ^ Q vd c ^ 0. Xet b Tim ham ngMc ip~^ cua ip. ham so: Giai. a) Ta can chiing minh / la ddn anh va la toan anh de suy ra / la song anh. ..^ ax + b X cx + d • Chiing minh tp : [0; 1] [a; b] la ddn anh. Gia sii tiJ2 e [0,1] sao cho v>(«i) = v?( [a;b] la toan anh. , , ,v axi + b ax2 + b ' ' ' Do do : [0; 1] —• [a; b] la song anh. cxi +d CX2 + d ; * 6) Do : [0; 1] [a; b] la song anh nen ton tai ham nguqc: ^acxiX2 + adxi + bcx2 +bd = acxiX2 + adx2 +bcxi + bd 1, ,^ I
  7. ... „ , , ,„ .,„„ ,f tJ::j;*.;;;r;ir::.S/'^! Ma nd - be # . 0 nen suv m a;, = xo Dk\v t-, cn, , - b) Tap xac dinli cua ham so la [-1;8] va '"
  8. B a i t o a n 1.13 (HSG Gia Lai nam hpc 2009-2010). Tim cdc ham so f thoa G i a i . Gia sil / la ham so thoa mail cac yen cau de bai. Dat t = - =^ x = - m.dn diiu kien: f {x) + xf {1 - x) = x^y-x eM.. X t' Thay vao (1) dUdc f{t) = ? + J l + ^, V< 7^ 0. T h i i lai diing. Vay c6 duy G i a i . Gia siir ham so / thoa man f (x) + xf {1 ~ x) — x^, Vx e M. (1) Trong (1), thay x bdi 1 - x, t a diTdc nhat mOt ham sp thoa man cac yeu cau de bai la / ( l - x ) + (l-x)/(x) = (l-x)2,Vx6R. ' (2) T . ( l ) v . ( 2 ) t a c 6 h e { / ( ! ) + ^fi^--)-^' ;^^^^^^^_^^^^_^. Taco: B a i t o a n 1.11. Tim ham so f : thoa man dieu kien D = 1 X = 1 - x - l - x ^ 7^0 - 1 1-x f , Va: ^ 0. (1) V x^ X = x^ - x ( l - 2x -t- x^) = -.7;^ + 3x^ - X Dm = (1-x) 2 1, G i a i . Gia siir ham so / thoa man cac yen cau de bai. Dat < = a: - - . K h i do 1 = 1 - 2x + x^ - x^ -h X * = x^ - 2x - I - 1 . X 1 - X (1-X)2 t' = x^ Bdi vay \ - z ( x ~ - \ = ^ x ' - \ t' + M. x^ \ i -x3+3x2-x ,Vx € (3) De y rftng t = x - -
  9. D a p so. Ham so can t i m la / ( x ) = l , V x € K. G i a i . Troiig (1) thay x hd\ t a dUdc: / ( - ) + 2f{x) = 3~. Dat f(-) = g{x), ( f{x) + 2yix) = 3x ^ • ' " B a i t o a n 1.17. Tim tat cd cdc ham so / : E —• R thoa man , ta dUdc he I + ^(^) = I . *^ ... ,;u xf{x)+f{~x) = x + a,\/x&R. ' •!. 3x 2 D = 1 2 = - 3 . Df = = 3x - - , D , = = 6x. D a p so. Ham so can t h n la / ( x ) = — ^ 2 — - ^ , V x e R. ^ 1 2 1 .r 2 1 1 N h a n xet 1. Cdc bdi todn 1.15, 1.16, 1.17 con c6 the diMc t.rmh bay gidng Vay / ' ( x ) = X , g(x) = 2.x . thoa man (1) va thoa man / ( - ) = g{x). nhu bdi todn 5.38, bdi todn 5.39 trong bdi dung day so gidi mot so dang phuang Do do /(./:) = - - X, V.r ^ 0. Bdi vay ' . ^; ' Q trinh ham. B a i t o a n 1.18 (Kosovo TST-2011). Tim cdc hdm / : R\} R thoa 2 2 2 /(X) = / ( - X ) ^ - - X = - - + X < ^ X - - 4 : > x 2 = 2 ^ r ^ = ^ - X L X = -v/2. mar*. / - — = x, Vx ^ { - 1 , 1 } . (1) Do do tat CH cat- ngliieni cua ])hitc(ng trinh / ( x ) = / ( - x ) la: V2, - \ / 2 . G i a i . Dat = t k i n do x = va = - . l h a y vao 1): B a i t o a n 1.15 (Slovak 2003-2004). Tzm cdc ham so / : R ^ M
  10. Thay X = — - vao (3) ta dUOc /(85) = 2.85 - 3 = 167. Theo (1) ta C O / ( I - x) = 2x - x"" - x 2 / ( a : ) , the vao (2) dUdc LvTu y. Phep thay x bdi 1 - a: vho (1) dUdc tim ra nhu sau; T a can thay x (1 - xf [2x - x " - x-'Six)] + J{x) = 2(1 - x ) - (1 - x)* y bdi g{x) vao (1), v6i g{x) se tim sau, sao cho van xuat hi§n /(x^ - 3x + 5), ^ [1 - (1 - x ) V ] fix) = 2(1 - x) - (1 - x)' - (1 - x)2(2x - x") / ( i ^ + 1 + 3). Muon v$,y ta xet phaong trinh ^ [l - X + x2] [l + x - x2] fix) = (1 - x) [2 - (1 - x ) 3 - (1 - x)(2x - x"); + X + 3 = - 3x + 5 4x = 2 2x 1 44> X = 1- X. [X^ - X + ] J [x2 - X - 1] fix) = (1 - X)(l + x3)(x2 - X - 1) >-'.^ • t ; Bdi v$,y ta se chpn g{x) = 1 - x. Phep doi' bien nay chSc chin se thanh cong. 4 ^ [X^ - X - 1] /(X) = (1 - x2)(x2 - X - 1). B a i t o a n 1.20 (Kyrgyzstan National Olympiad 2009). Gia sv! rdng ham so Nhu vay vdi x ^ a vh x ^ b, trong do a, b \k hai nghiem cua phudng trinh f -.R-^R thoa man dieu kien ' •" x 2 _ x - 1 = 0, ta C O fix) = 1 - x^. Tiep theo ta xac dinh / ( a ) , /(6). Theo djnh li Viet ta c6 a + 6 = 1 va a6 = - 1 . Thay x = a va x = 6 vao (1) ta dUdc ^'/{x"^ + X + 3) + 2f{x^ - 3x + 5) - 6x^ - 10x+ 17,Vx € R. (1) a V ( a ) + fib) = 2a - a^ ( a + l ) / ( a ) + /(6) = - a 2 Hay tim /(2009). { b'^fih) + /(a) =>(a + 1) [-b~ ^2b-h^ - 1 - (6+ l ) / ( 6 ) ] + fib) Th+imb) + fio-) = -h^--^ = -a^ - 1 \ Hufdng d i n . Xem bai toan 1.20. =^ [(a + l ) ( t + 1) - Ij /(6) = a^' + 1 - (a + l)ib'^ + 1) . B a i t o a n 1.21. Gia. sii rdng / : R —> R /d ham so thoa man dieu kien =J«(aV - l ) / ( 6 ) = a + 2 - ( a + l ) ( 6 + 2 ) = a + 2 - ( a 6 + a + fe+a + 2) f {x' - 5 x + 1) + 5/ (x2 + X - 5) = x2 - 9, Vx e R. =)^0./(6) = a + 2 - (-1 + 1 + « + 2) = 0 => Q.fib) = 0. Hay tim /(2011). Bdi vay ta c6 fib) = c, / ( a ) ^ 2b - b^ - b^c, vdi c tuy y. Tom lai: { 1 - x^ neu X ^ avk X ^ b G i a i . G i a sii / la ham so thoa man cac yen can de bai. Trong (1) thay x bdi (3) 2 - X ta dUdc: / (x^ + x - 5) + 5/ (x^ - 5x + l ) = - 4 i - 5, Vx 6 R. (2) Nhan hai ve ciia (2) vdi - 5 roi cong vdi (1) ta dildc ceR tuy y ngu x = 6 26 - 6^ - neu x - a -24/ ( x 2 - 5 x + l) = -4x2 + 20x+16,VxeR trong do a, b la hai nghiem ciia jihudng trinh x^ - x - 1 = 0. T h i i lai thay 'f* ^ 6 / (x^ - 5x + 1) = x^ - 5x - 4, Vx 6 R thoa man. Vay / ( x ) dUdc xac dinh bdi (3) la tat ca cac ham so can tim. 6/ (x2 - 5x + 1) = ( i ^ - 5x + l ) - 5,Vx e R. ^ (3) 1.2.3 M o t s6 k i thuat thU'dng d i i n g Khong CO nlnrng dinh li ciing nhit cac thuat toan chung de giai phitdng trinh ^. , . , •> 5 ± v/8065 Xet phitdng trinh: x^ - 5x + 1 = 2011 /(2011) = — ? l ^ 3 • 0 childng 1, ta se di sau vao viec giai phifdng trinh ham dita tren cac dac triTng ciia ham so sd cap. Trong do hay dung kl thuat tim cac nghiem rieng B a i t o a n 1.22 ( H S G quoc gia-2000). Tim cac ham so / : R -» R thoa man: ddn gian nhu ham hang, luun bac nhat, ham da tliiic,... Di;ta vao cac nghiem rieng, chiing ta se c6 hilu bigt hdn ve ham can tim va c6 t h i c6 dildc cac ^^f{^) + / ( I - T ) = 2x - x^ Vx € R. (1) phitdng hitdng giai phUdng trinh ham da cho. • T i l (hildng 2 trc) di ta hay diing cac ki thuat sau: • G i a i . Trong (1) thay x bcii 1 - x ta dUdc - Tinh cac gia tri dac biet ciia ham so fix), chang han: /(O), / ( ± 1 ) . (1 - x ) V ( l - x) + / ( x ) = 2(1 - x) - (1 - x ) ^ V I e R. (2) /(2),... Doi khi /(O) ho$c / ( I ) khong tinh dUdc, ta c6 the dat chung 14
  11. bkng cac chft (tham so). K i thuat nay dac biet hay sx't dung khi gap B?ii todn 1.25. Tim tat ca cac ham so / : R —» R thoa man phudng trinh hain vdi cap bien tu do, se dvtdc de cap nhieu d chUdng 2. f{a^)+fih^)+f{c^)=fi3abc),^a,h,ceR. (*) - Nghien ciJu cac tinh chat dac biet cua ham so can tim nhu ddn anh, Giai. T i i (*) cho a = 6 = c = 0 ta dudc 3/(0) = /(O) ^ /(O) = 0. TiJt (*) cho toan anh, song anh, chan, le, tuan hoan, ddn dieu, dau,... 6 = c = 0 ta dvtdc / (a^) = 0, Va G R. Dat x = a^, khi do / ( x ) = 0, Vx G R. - Khai tha:c tfnh doi xiing trong phudng trinh ham da cho. Thuf l?i.i thay ham so / ( x ) = 0, Vx G R thoa man cac yeu cau de bai. - Lcri giai ciia bai toan giai phUdng triuh ham thudng ditdc bat dau bang Bai toan 1.26 (Slovenia National Olympiad 2010). Tim tat cd cac ham menh de "Gia sii ton tai ham so / thoa man cac yeu cau de bai". Khi f : [0, +oo) —> [0, +oo) thoa man tim dttdc biiu thiic ciia ham so nghiem, ta phai thii lai roi mdi dUdc ket iy + l ) / ( x + y) = f ( x / ( y ) ) , Vx, y G [0; +oo). (1) lu$,n. Giai. Trong (1) cho x = 0, ta dvtdc 1.2.4 PhiTdng phap the gia tri dac biet (?/ + !)/(?;) =7(0). Vy G [0;+cx)) Day la phUdng phap thacJng dimg nhat de giai phUdng trinh ham va se dudc (2) de cap nhieu hdn 6 bai 2.1: Giai phUdng trinh ham bang phep the (d trang 93). Trong muc nay ta chi d§ cap den mot so bai toan de nhkm giup cho ban Thi't lai : thay (2) vao (1), ta dudc doc lam cjuen v6i vigc giai phUdng trinh ham. Khi the gia tri dac biet cAu lim y la cac gia tri do phai nSm trong tap xac dinh, viec the gia tri phai c6 tinh ke thila. Bai toan 1.23. Tim tat cd cac ham / : R R thoa man f(^^ + y) + fix - y) = fix) + 6xy ^/fii) + x\ y e R. (1) Giai. Gia sut ham so / thoa man cac yeu cau ciia de bai. Trong (1) cho y = 0 «iL±ilfL = :k±iL,Vx,.6(0;+cc) ta dUdc fix) = x^, V i G R. Thii lai thay thoa man. X + y + 1 ax + y+l Bai toan 1.24. Tim tat cd cac ham / : R -> R thoa man ^ [ X + y + 1 = ax + y + L Vx, y G [0; +oo) fix + y)+ fix -y) = fix) + 2f{y) + x\, y 6 R. (1)
  12. B a i t o a n 1.28 (D? nglii t h i Olympic 30/04/2009). Tim tat cd cdc ham s6 B a i t o a n 1.31 (Turkey National Olympiad First Round 2008). Tim tdt c& / : IR ^ R thoa man /(O) = 1 va cdc ham f : (0, +oo) (0, +oo) thoa man [fix + y)r' = [nx)r' + + 2xy, Vx, y e K. (1) 10(x + y) f{x)f{y) - fixy) - 90 = -, V x , y € (0;+oo). (1) xy G i a i . Gia si'r ham / thoa man cac ycu cau dc bai. Tit (1) lay .r = 0 t a dudc G i a i . T i r (1) cho y = 1, t a ditrtc [ny)r' = l + y'..yyeR / ( x ) / ( l ) - fix) - 90 = l ^ i ^ t i ) , Vx € (0; +oo) ^/(.T) = '"Vl+^r^ e R. (2) ^[f{\) - l ] / ( x ) = 10 f 1 + + 90, Vx G (0; +oo). (2) \j T h u lai: Vdi ham so / xac diiili bc3i (2) t a co /(O) = = 1 va /(I) = 11 [/(a:)]'""-' + y' + 2xiy = 1 + + + 2xy = {x + y f + \ Trong (1) c'ho x = y = 1, t a dUdc: [/(l)]^ - / ( I ) - HO = 0 « .Ai) = - 1 0 . = [/(•'• + .'y)]^""^ V.r. .V e R (thoa man (1)). . K h i / ( I ) = 11, thay vao (2), t a dtfdc /(x) = 1 0 + ^, Vx e (0; +oo). T h i i lai Vay CO tluy uhat mot ham so thoa man cac yeu c5u de bai la thay thoa man. • K h i / ( I ) = - 1 0 , thay vao (2), t a durtc ' _ /(.r) = ' " V l +.7:2, V.r e R. 10 1 \0 - ^ , V x e (0;+a3). (3) B a i t o a n 1.29. Tun tdt cd cdc hdni so / : R R thoa man /{•'•f{z) + y) = zf{x) + y, yx,y.zeR. (1) Thay (3) vao (1), t a drtdc 90 10 / \\0 10 1 \ 90 G i a i . Ttr (1) clio x = = 0 ta dime f { y ) = y, Vy e R hay f { x ) = x, Vx G R. 10 A 1^ + 1 + -90 T h u lai thay thoa man. Vay c6 duy nhat mot ham so th()a man cac yen cau . 11 * 11 11 \ V ) do hai la /(".;:) = x, V.r G R. .l^_+J^i,V.x,yG(0;.foo). (4) B a i t o a n 1.30 (Switzerland Final Round 2010). Tim tdt cd cdc ham so xy / :R R fhda man: f (/(x)) + / (/(y)) = 2y + f(x - y ) , Vx, y e R. (1) Ta thay (4) khong dung, chang han k h i x = y = - . Vay ham so xac d i n h bdi G i a i . Dat a = /(O). Trong (1) cho y = x, ta diWc / ( / ( x ) ) = x + - , Vx e R. (3) khong thoa man (1). Vay (1) trc) thanh: Kelt luan : ham .so thoa man yeu cau de bai l a /(x) = 10 + - , Vx e (0; +oo). \+ ^ + y + ^ =2(/ + / ( x - ? / ) . V x , i / e R B a i t o a n 1.32 (Romania Team Selection Test 2011). Tim tdt cd cdc ham A'o / ; M R thoa man *^' 0. (2) # /(x) = x , V x € R . Fix, x) 2 f i x ) = /(2x) + /(3x), Vx > 0. (3) 18 19
  13. P(0; x) ^ 2/(0) = fix) + f{2x), Vx > 0. (4) • H a m s6 / ditdc goi la ham tuan hoan nhan t i n h chu k y a (a ^ { 0 , 1 , - 1 } ) P(0; 2x) => 2/(0) = f{2x) + /(4,T), VX > 0. (5) tren M neu M c Df (Dy la tap xac dinh ciia ham /) va TiJt (5) t a CO /(4:;:) = 2/(O) - /(2.r), thay vao (2) ta d\xac . '• f Vx G A / a ± i x G A/ \) = / ( x ) , V x G Af. 2/(2x) = /(3x) + 2/(0) - /(2x) ^ /(3x) = 3/(2x) - 2/(0). (6) • Trong muc nay t a se giai quyet van de t i m tat ca cac ham so tuan hoan, Thay (6) vao (3) t h u ditdc phan t u a n hoan (cong t i n h , nhan t i n h ) chu k i cho tru6c, mot so dang md rong, k i thuat dita dang mci rong ve dang cd ban va mot s6 bai toan 2/(x) = /(2x) + 3/(2x) - 2/(0) ^ 2/(2./;) = /(:/:) + /(O). (7) lien quan den ham t u a n hoan. Nhiing van de tren se ditdc giai quyet Thay (7) vao (4) dan t d i : thong qua cac bai toan cu the va cac nhan xet, chii y, l u u y. y. 4/(0) = 2/(x) + fix) + /(O) ^ fix) = /(O), Vx > 0. (8) B a i toan 1.33. Tim tat cd cac ham so / : R R thoa man v-.i vv/^^ Trong (1) thay ij bdi 2y, t a ditrtc / ( X + 1) = / ( X ) , V X G R . (1) 2 / ( x ) = / ( x + 2t/) + /(x + 4 y ) , V x G R , V j / > 0 . (9) G i a i . Gia sii / la ham so thoa man de bai, khi do t a c6 ( I ) . Bang phitdng phap quy nap, t a chitng m i n h ditdc: /(x + n) = /(x), Vx G R, Vn G Z. T i t (9) va (1) suy ra: /(x + y) = fix + 4y), Vx e R, Vy > 0. (10) Dodo: / ( X ) = : / ( [ X ] + {X}) = / ( { X } ) , V X G R , (2) Trong (10) thay x bcii -4y, t a ditdc / ( - 3 y ) = /(O), Vy > 0. T\X day thay y d day [x] la p h l n nguyen cua x va { x } la phan thap phan ciia x. Tiep theo, bdi |, t a d U d r / ( - y ) = / ( O ) , Vy > 0. Vay/( 0) tren M neu M c Df iDj Ik tap xac dinh ciia /(x + 1) = ^ ({x + 1}) = y ( { x } ) = / ( x ) , Vx G R. ham /) va i / Vx e A/ => X ± 0 , 6 M Vay t a k i t luan: Tat ca cac ham so can t i m la /(x) = g ( { x } ) , Vx G R, trong \x + a) = / ( x ) , V x G M . do g la ham so t u y y, xac dinh tren ni'ta khoang [0; 1). : • Cho / la mot ham tuan hoan tren M . K h i do T ( T > 0) dUdc goi la chu B a i t o a n 1.34. Cho a 7^ 0. Tim t&t ca cac ham so / : R —» K thoa man l\i c d sd ciia ham / ngu / tuan hoan vdi chu k i T ma khong tuan hoan . / ( x + a) = / ( x ) , V x G R . • (1) vdi bat cu: chu k i nao be hdn T. • Ham so / dUdc goi la ham phan tuan hoan cong t i n h chu ky a (a > 0) G i a i . Ta CO ' ^ .'SJ , , w o ' r • - : . t i e n M ngu M c Dj ( D / la tap xac dinh cua ham /) va (1) /(ax + a) = / ( a x ) , V x G R / Vx e A/ => X ± a G M ff(x) = 5(x + l ) , V x G R (vdi 5(x) = / ( a x ) ^ / ( x ) = 3 (^)). t) • 0-'. I /(x + a) = - / ( x ) , V x G A / . 20 21
  14. Ap dung kct qua bai toan 1.33 ta d\Wc g{x) = /i ({./:}), V.r 6 R, trong do h la Hxidng d a n . Dua (1) ve phUdng trinh ham cd ban F ( i + a) = Fix). Ta c6 ham so t u y y, xac dinh tren nuta khoang [0; 1). Bdi vay hix + a) = hix) hix) = ^ [hix) + hix + a)], ma hix + a)+ hix) khong "dong dang" vdi fix + a) - / ( x ) nen ta phai "nang C h i i y 6. Hai bai toan 1.33, 1.34 ?/ nghla rat I6n, no giup ta tlw. ducfr tat bac" nhir sau: cd cdc ham s6 tudn hodn vd phdn tuan hodn c6 chu ki cho trudc. . , , 1 ,/ N 1 . X , / N (x + a)hix) xMx) B a i t o a n 1.35. Tim tat cd cdc ham s6 / : R —> IR thoa man diiu kien hix) = -.ahix) = -ix + a-x) hix) = '—^ ^—^ a a a a f{x + a)-fix) = h{x),yxeR, (1) _ ix + a)hix + a) _ xhix) trong do h Id ham phdn tudn hodn cong tinh chu ki a tren E . a a % Do do ta bien dfii (1) thanh ' '>^i ( i ) ' ^ Hi^dng d i n . Dua (1) ve phildng t r i n h ham cd ban F{x + a) = F{x). Ta c6 a a ' ''?k h{x) = -h{x + a)^h{x)=^-[h{x)-h{x + a)\. a a Vay t a bicn d6i (1) thanh
  15. L t f u y. Ngoai viec dita tren bai toan 1.37, he thurc Vay h{x) = ^ ^ / i ( x + 7 ) - ^/i(x), Vx G R. Do do ' J"'"^ cosx=— cos(x + ZTT) - — cosa; ZTT ZTT ( 1 ) ^ fix + 7 ) - /(x) = /i(x), Vx G R 4=^ /(x + 7 ) - fix) = ^ h i x + 7 ) - ^/i(x), Vx G con dmc tun nh\t sau: Gia sU cos x = a{x + 27r) cos(.r + 2TT) ~ ax cos .r. K h i do cos a; =^ «(a; + 27r) cos a; - axcosx => 1 = 2na =J- a = ^ /(x + 7 ) - ^ / i ( x + 7 ) = /(x) - 'i^hix) , Vx G Bai toan 1.39. Ttm tdt cd car. ham s6 f-.R-^R thua •man di&u kitn ^ six + 7) = gix), Vx G R ( v d i gix) = /(x) - ^ / i ( x ) ) . (2) f{x + a)~ f{x) = P{x), VX-GM, (1) TiT ( 2 ) , sii: dung k i t qua bai toan 1 . 3 4 d trang 2 1 t a ditdc • . Irony do. P{x) Id da thiic bdc n. ,, flW = fc({y}).VxGR, Hifdng d t n . Ta bigu dign P ( x ) = Q{x + a) - Q{x), v6i la da thiic bac n + 1 , sau do dung phUdng phap dong nhat he so de t i n i cac he so cua Q(.T). troiig do k la ham so tfiy .y, xac dinh tren niVa khoang [0; 1). Vay t a t ca cac Lvfu y- C u the cho dang nay la bai toan 1 . 6 9 d trang 42, ban doc hay xem ham so th(m m a n yeu cau de bai deu c6 dang de ntnn ditcJc each bieu dien P{x) = Q{x + a) - Q ( . r ) . 2015 \ Bai toan 1.40. Tim f.dt cd cac ham so / : E'—» R thoa man dieu kien + 1 + k Vx G /(:r + l ) - / ( x ) = 2 - ^ V x e R . (1) trong d6 k la ham s6 t u y y, xac dinh tren niila khoang [0; 1 ) . Hu'dng dan. Ta o6 C h u y 8. Sau day ta sc xet mot s6 bdi todn c6 cdch gidi rdt ddc hiet, ban doc /(.r + l ) - / ( x ) = 2 - ^ V x e R cdn ndm b&t vi no xudt Men rkt nhtiu trong cuSn sdch nay, do la sU dung ^ / ( x + 1 ) - fix) = 2 ' - ^ - 2'-/(x + l ) + 2 ' - ( ^ + i ) = /(x) + 2^-^, V x e R a = h 4^ a = - id + b) : a = - 6
  16. D g t h a y n e u h a m / t h o a m a n (3) t h i t h o a m a n (1). N e u h a m / t h o a m a n (1) Thuf l a i : V d i m p i x < 0, k h i d o • m"'^ t h i d o (2) n e n / c6 d ^ n g ( 3 ) . V a y (1) ^ (3). T i r d o h a m so c a n t i m c6 d a n g /(5x) = /i(log5(-5x)) = / i ( l + l o g 5 ( - x ) ) = /i(log5(-x)) = fix). I ' f{x) = ~[g{x)~9{x + a)]yxeR V a y k h i X < 0 t h i / ( x ) = h ( l o g 5 ( - x ) ) (/?. l a h a m t u a n h o a n c h u k y 1 t r e n R ) . ( t r o n g d o g la h a m so t u a n h o a n c p n g t i n h c h u k y 2a t r e n K, t u y y ) . K e t l u a n : H a m so / can t i m l a ( g (logs a:) khi x > 0 B a i t o a i i 1.43. Tim cdc ham so / : K —> IR thoa man /(a:)+/{4-i) =0,VxG]R. (1) c (c l a h a n g so t i i y y ) khi x = 0 ' ' % G i a i . T a c6 * t'r i /i(log5(-x)) k h i X < 0. ( t r o n g d o g, h l a cac h a m so t u a n h o a n c h u k y 1 t r e n R, t h y y ) . (l)^/(x) = i[/(x)-/(4-x)l,Vx€R B a i t o a n 1.45. Ttm tdt cd cdc ham so / : R -+ R thoa man (1) ' f{x)=^-[h{x)-h{4-x)],yx€R /(-2013x) = / ( x ) , V x 6 R . G i a i . T h e o (1), t a co / (2013^x) = / ( - 2 0 1 3 ( - 2 0 1 3 x ) ) = / ( - 2 0 1 3 x ) , Vx G R! ^ { h l a h a m so xac d i n h t r e n E , t u y y. v a i (1), t a d,t0c / ( X ) ^ / ( 2 0 1 3 ^ x ) , V x G R. V a y (1) t u d n g d u d n g vPi H a m so can t i m c6 d a n g f{x) = ^ [h{x) - h{4 - x)]yx £ R, t r o n g do h l a / /(x) = i [ / ( x ) + / ( - 2 0 1 3 x ) ] , V x G R (2) h a m so xac d i n h t r e n R, t u y y. \ = / ( 2 0 1 3 2 x ) , V x G R. B a i l o a n 1.44. Tim tat cd cdc ham so / : R —> R thoa man X e t h a m so: / ( x ) = ^ [(/(x) + ( ; ( - 2 0 1 3 x ) ] , V x G R. (3) , /(5x) = / ( x ) , V x € R . (1) t r o n g (to g la h a m so tu
  17. B a i todn 1.46. Tim tat ca cdc ham so / : ]R\} —>• R thoa man diiu kien trong do hi, /12 la nhfmg ham so xac dinh tren mita khoang [0; 1), t i i y y. M a theo tren t a c6 /(x) = logax + g (logsx), Vx > 0 nen /(3x) - /(x) = 0 o g 3 |x| - [log3 |x|),Vi ^ 0. (1) { log2X + hi ( { l o g 2 ( l o g 2 x ) } ) khi x > 1 G i a i . Dat h{x) = ^ l o g g |a;| - [logg |x|] = ^ { l o g g |x|}. K h i do /i(3x) = /{!og3|3x|} = ^ { l + l o g 3 | x | } = v/{log3|:c|} = Hx), Vx ^ 0. c (c la hang so t i i y y) khi x = 1 log2X + /i2 ( { l o g 2 ( - l o g 2 x ) } ) khi 0 < X < 1 ^3) tiiy y k h i x < 0. (l)^/(3x)-/(x) =/i(x),Vx^O Thuf lai: Gia sii ham / xac dinh bcii (3). Vdi x e R t i i y y. ^ /(3x) - fix) = (log3 |3x| - log3 |x|) /i(x), Vx ^ 0 . Neu X > 0 t h i 4^ > 1 va 2 ^ > 1 nen • , ^ /(3x) - h{x) log3 |3x| = /(x) - h{x) log3 |x|, Vx ^ 0. / ( 4 ^ ) = log24^ + hi ({log2 (log24^)}) = 2x + hi ( { l o g 2 x } ) . (2) f{2^) + x = log22^ + hi ({log2 (log22^)}) +x = 2x + hi ( { l o g 2 x } ) . Dat g{x) = / ( x ) - /i(x) log3 |x|, thay v a c (2) t a ditdc Suyra/(4-) = /(2^)+x. g(3x) = p ( x ) , V x 7 ^ 0 . • Neu X = 0 t h i 4^ = 2 ^ = 1, do do / ( 4 ^ ) = c, / ( 2 ^ ) = c, hien nhien thoa (1). (3) • Neu X < 0 t h i 0 < 4"^ < 1, 0 < 2 ^ < 1 nen T i r (3) ti(>n lianli tudiig ti.r i i l m bai toan 1.44 ci trang 26 t a d u ^ c / ( 4 ' ) = l0g24- + h2 ((log2 ( - l 0 g 2 4 - ) } ) = 2X + h2 ( { I 0 g 2 ( - X ) } ) . f{2n + X = log22^ + h; ({log2 ( - l o g 2 2 ^ ) } ) + X = 2 x + /I2 ( { l o g 2 ( " X ) } ) . _ / '"(loggx) khi X > 0 ^^•^^ ~ I 7z(log3(-x)) k h i X < 0. Suy r a / ( 4 - ) = / ( 2 - ) + x . Nlut vay ham so xac dinh bdi (3) thoa man (1). Do do tat ca cac ham so (trong do m , n l a c a c h a m so tuan hoan c h u k y 1 tren M). Vay thoa man yen can de bai den c6 dang (3). / ( X ) - ^ ( X ) + /l(x) l0g3 |x| B a i t o a n 1.48. Tim cdc ham so f tuan hoan chu ki 3 tren R vd thoa man _ f m (logs x) + ^ { l o g 3 x } ] o g 3 X khi x > 0 /(x) + /(x + l ) + / ( x + 2 ) = 0 , V x e R . (1) I " ( l 0 g 3 ( - : c ) ) + v / { l 0 g 3 ( - x ) } l 0 g 3 ( - x ) k h i X < 0. B a i t o a n 1.47. Tim tat ca. cdc ham so / : IR —» R thoa man , G i a i . Theo gia thiet ta co / tuan hoan chu k i 3 tren R va thoa (1), tiic la /(4^) = /(2^) + x, Vx 6 R . I /ja;) = - / ( x + l ) - / ( x + 2 ) , V x e (1) + 3) = / ( x ) , V x 6 G i a i . Ta c6 ''' / /(^) = \T) - fix + 1) - fix + 2)1 Vx € R ! ^ ^2) ( l ) ^ / ( 4 ^ ) - 2 x = /(2^)-x. Vx€R \) = / ( x ) , V x e R . ' ' ^ ,9(2x) = g{x): Vx 6 R (v6i g(x) = / (2^) - x ) . (2) Ta se chutng m i n h (1) t i t d n g dUdng v d i T i t (2), sir d y n g bai toan 1.33 d trang 21 v a tien h a n h tUdng t u nhir bai toan 1.44 d trang 26 t a dUdc fix) = i \2gix) - c,(x + 1) - yix + 2)], Vx G R (3) { { y la hara t u S n h o a n CQUg t i n h c h u k i 3 t r e n R, t u y y. hi ({iog2x}) khi X > 0 Gia si't fix) CO dang (3). K h i do vdi mpi x G R ta c6 c (c la hang so tiiv y) khi x = 0 /i2({log2(-x)}) ' khix
  18. Tif (3), thay x bdi x + r ta diroc: gix) = ff (x + (r - s)), Vx G R . (4) f f{:r + 2) = l^[2g{x + 2)-g{x)-g{x+l)l Tir (3), thay x boi x + s ta dudc: ^ ( x ) = g (x + (s - r)), V i G R . (5) T i l (4) va (5) suy ra g la ham so tu^n hoan vdi chu ki T = |r - s| > 0. V i the fix + 3) = i [2q{x) - q{x + 1) - fl(.T + 2)] = f{x). vdi moi x G R ta c6 agix) - x = gix-y-giy)) =gix + T-y-giy)) Do (to j{x) + f{x + \) -1-/(x + 2) = 0, / ( x + 3) = / ( x ) . Vay / ( . r ) thoa man (1). = agix+ T)-ix + T) = agix)-x-T. (6) Giasi'r f(x) tlioa man (1). K h i do do (2) nen f{x) c6 dang (3). Vay (1)
  19. G i a i . De thay (1) titdiig clildng v(3i: f{x) = ^ [/(i) - / ( - i ) ] , V i € K. (2) B a i toan 1.55. Tim tat cd cdc ham so / : R —• M thoa man X€t ham s6: /(x) = J bCx) - »(-a:)],Vx € R , (3) f{-x) + af{x) = byxeR. : ^ (1) trong do g la ham so tily y tren R . K h i do de thay ham / thoa man (1). (vdi a, b Id hai so thuc cho trUdc). ^ Ngitcjc l^i neu ham / thoa man (1) t h i / c6 d9.ng (3) do (2). V$,y ham so can G i a i . Trong (1) thay x bcii -x t a ditdc: f{x) + a / ( - x ) = 6, Vx G R. (2) t i m CO dang f{x) = ^ [g{x) - g{-x)],Vx € R , vdi g la ham so t u y y tren R . Dat /( ^x) = , ; ( x ) . Tfr (1) va (2) ta c6 he: { }gf++4(^j I 6. , (3) B a i t o a n 1.53. Tim tat cd cdc ham so f : thoa man Taco: a 1 b 1 a b = b{a-l). f{x) = fl- ,Vxe D = = a2 - 1, = = h{a-\), Dg = 1 b (l)i 1 a b a \x TrtfcJng h d p 1 . « 7^ 1 va a 7^ - 1 . K h i do £> 7^ 0. He (3) c6 nghiem duy n h i t G i a i . Dg thay (1) tUdng dudng vdi f{x) = ^ fix) + / ,Vx€]R*. (2) b{a- 1) b Xet ham so: /(x) = ^ g{x) + g D a2- 1 a+ 1 ,VxeR*, (3) b{a - 1) b D «2- 1 a+ 1 trong do g la ham so t u y y tren R * . K h i do dg thay ham / thoa man (1). Ngudc lai neu ham / thoa man (1) t h i / thoa man (2), do do / c6 dang (3). Vay ham so can t i m c6 dang: Thi'r lai t h i y fix) = ^' , Vx e R thoa man de bai. a+ 1 Trtrdng h d p 2. a = 1. K h i do (1) t r d thanh: /(x) + / ( - x ) = 6, Vx e R. (4) 9{x) + g ,Vx€R*, Dat fix) - ^ = g{x),Wx e R. Thay vao (4) t a dUdc trong do g la ham so tuy y tren R*. ' * gix) + ^+g{-x) + ^ = byx£m B a i t o a n 1.54. Tim tat cd cdc ham so R thoa man
  20. Thijf 19.1 thay thoa man. Tom l?ii, ta c6 ket qua: f}. • Neu a 7^ 1 va a 7^ - 1 t h i C O duy nhat mot ham so thoa man de bai la B a i t o a n - 1 . 5 7 (De nghi t h i Olympic 30/04/2011). Tim tat cd cdc ham so ^ . IK biet rhng f la ham so ch&n vd a + l f{xy) - fix)f{y) = 3 [fix + y)- 2xy - 1], Vx, y € R. (1) • Neu a = 1 t h i t i t ca cac ham so thoa man dg bai la X X G i a i . T i t (1) thay x b6i - va y bdi - ta ditdc f{x) = ^ + 1 [hix) - h{-x)],yx e M (/i la ham s6 t u y y xac dinh tren R)j • Neu a = - 1 va 6 7^ 0 t h i khong c6 ham so nao thoa man de bai. (2) f Ngu a = - 1 va 6 = 0 t h i tat ca cac ham so thoa man d l bai- la / ( i ) = - \k{x) 4- fc(-2;)],Vi e E (A; la ham so t u y y xac dinh tren R ) . T i t (1) thay x bdi | va y bdi t a dUdc B a i t o a n 1.56. Tim tat cd cac ham 50 / : R —» R thoa man diiu kien , Vx,2/e (3) f{x) r, + f{-x)= , 2010COSX ^ ^ , V x e R . /(-T)-/(f)^H)- VX-' + 1 Do / la ham chSn nen (3) viet lai G i a i . Gia su / la ham so thoa man cac yeu cau de bai. K h i do ta co 2010 cosX / ( - ) = 3' /(0) + y - l , Vx,y € (4)^ f{x) + f{-x) = — , Vx 6 R Vx^ + 1 ^, , ^, , 1005cosx 1005cos(-x) ^ Lay (2) t r f t (4) theo ve t a ditdc: /(x) = x^ + / ( O ) , Vx e R. (5) V^X^ V'(-X)2 + 1 TiJt (1) cho y = 0 ta C O 1005 cosx / ( O ) - / ( x ) / ( 0 ) = 3 [/(x) - 1], V x e R 1005 c o s ( - x ) n-x)- ,VxeR. (1) ^ [/W - 1 ] [ / ( o ) + 3] - 0, Vx e R. 1-,-^ \x , , N I J / ( O ) 7^ - 3 t h i fix) = 1, Vx e R, ham so nay khong thoa (1). Vay D^t fix) - —p== = gix),^x e R. T i t (1) suy ra ^(x) la ham s5 thoa man / ( O ) = - 3 va t i t (5) ta c6 /(x) = x^-3, Vx e R. ThiV lai thay thoa man. Do do /?Tl C O duy n h i t mot ham s6 thoa man cac yeu cau de bai la /(x) = x ^ - 3 , Vx e R. d i l u kien: g{x) = - g ( - x ) , V x e M. (2) Tfr (2), si't dung bai toan 1.52 cl tr
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