Tuyén tap cdc dé thi mock test VMO — TST 1.1 Dé mock test sé 1 Dé thi Dé thi ngay nhdt Bai 1. Xét day s6 (up) xée dinh béi Uy = 2,Ung1 = 14+ (n+1)un vein =1,2,3,.-. 1. Chiing minh ring up = [en!| vai moi n > 1. 2. Chimg minh ring tén tai vo s6 s6 thuc ¢ > 0 sao cho limn{en!} = 1. Bai 2. Tap hop 2 gdm 2020 da thttc khac hing, hé sé nguyén duge goi 1a dep néu nhu vdi moi s6 nguyén m, déu tén tai mot da thtte P(x) nao dé trong 2 sao cho phuong trinh P(x) = m cé nghiém nguyén. 1. Ching minh ring tén tai tap hop 9 dep sao cho hé s6 bac cao nhat cia tat ca da thite trong d6 déu bang 2020. 2. Tén tai hay khong tap hop dep chi gdm cée da thite 6 bac lén hon 1? Bai 3. Véi k la s6 nguyén lén hon 1, trén bang c6 viét sin cdc sé 1,2,3,...,10*, 108 +1. 6 méi bude, cho phép xda di hai s6 a,b bat ly c6 sin trén bang rdi thay béi i (a,b), trong dé f 12 mét ham hai bién. Goi z 1a s6 nguyén dudng thu duge sau 10" bude. Chitng minh ring véi moi c4ch chon cp 386 6 mdi buéc : 1. Néu f(a, b) = ged(2020ab, a? + 506ab + b*) thi « = 1. 2. Néu f(a,b) = ged(a?b?-+2019, a?-+-b?+2021) thi x khong phai sé chinh phuong. 7 Scanned with CamScannerBai 4. Cho dudng tron (0, R) va day cung BC cé dinh khéc dutng kinh, diém A thay déi trén (O) sao cho tam giéc ABC khéng can. Goi BD, CE 1a c&c phan giéc trong va J 1a tam ndi tiép cia tam giée ABC. Gia sit DE cét BC tai F. Dung thang qua I vudng géc véi IF ct trung truc BC tai K. 1. Ching minh ring OK = 3R. 2. Ching minh ring khi A thay déi trén (O) thi dudng trung tuyén dinh I cha tam giéc IDE ludn di qua diém cé dinh. Dé thi ngdy hai Bai 5. Tim tat cd ham sé f : Rt > R* théa man déng thdi cdc diéu kién sau 1. f(a) tang ngat. 2. f(2x) > 2f (x), véi moi x > 0. 3. F(F(@)F(y) + 2) = (ef) + F(@) voi moi z,y > 0. Bai 6. Cho ttt gic ABCD khong phai hinh thang, noi tiép (O) va ngoai tiép (I). Goi M,N, P,Q lan lugt la tiép diém cita (I) v6i AB, BC,CD, DA. Hai duttng chéo AC, BD c&t nhau tai K. Goi H 1a hinh chiéu cia K trén QM. 1. Ching minh ring IH, AO c&t nhau tai mot diém thudc (0). 2. Goi giao diém cén lai cia IH véi (O) 1A Ay. Céc diém B,C), D; duge dinh nghia tuong ty. Ching minh ring 4,C,, B,D,, OI déng quy. Bai 7. Goi a 14 nghiém duong cia phuong trinh «+ : = 48. 1. Chtg minh rang véi moi sé nguyén duong n sao cho 9(2020(a?" + a-*")) > v(2019(a”" + a-*")). 2. Ching minh ring $ = a®” + 07%” 6 ft nh&t 5+ 2°° ude nguyén duong. Scanned with CamScannerL6i gidi chi tiét Bai 1. Xét day s6 (un) x4c dinh béi ty = 2, Ung = 1+ (n+ Lun voin = 1,2,3,... 1. Ching minh ring u, = len! véi moi n > 1. 2. Ching minh ring tén tai v6 s6 s6 thuc c > 0 sao cho limn{en!} = 1. Loi gidi. 1) Ta 06 Unt i, Un =1l+nu, > = = Wnt = 1+ Min > ED (tlt al nén suy ra Um ty 11 11 a utatgt ta ae “1. “1 hay Up = Er Tiép theo, vi day ms tang theo n va day héi tu vé e nén ta a k= 4 i ee phai o6 e >) 77,¥n. V6i moi m € Z* thi ko" 1 1 1< 1 Dern * oa arene Lid i f_4 1 gc saan sientt aaa < Ga nén } Tir d6 suy ra uy = [e- nl). 2) Tuong tu trén, ta cing c6 thé ching minh ring 1 fence tiy 2 n+1 “~n+1 ° (n+1)(n+2) nén 2n ate ents ee eee, wai <7" < + Gane Scanned with CamScanner10 CHUONG 1. TUYEN TAP CAC DB THI MOCK TEST VMO— TST Suy va lim n {e-n!} =1. Cust cing, ta chi en chon mot sf thue e> O ty yma {c} = {6 th 5 thé vdt c= m + e wi m €2*, Kh ds “lm nen) = tin fe-nl} ‘Vay nén tén tai vo 85 s6 thue ¢ > 0 théa man dé bai. o Bai 2. Tip hop 2 gdm 2020 da thitc khéc hing, he sb nguyén dude goi lb dep. néu nhut vi moi s6 nguyén m, déu tn tai mot da thite P(2) ndo d6 trong 80 cho phuang trink P(2) =m 6 nghigm nguyen. 1. Chimg mink ring tén tai tap hop 9 dep sao cho be s6 bac cao mht cia tht cd da thite trong d6 du bing 2020 2, Téa tal hay khong tap hep dep chi gdm efe da thie e6 bac lon bon 1? 118 gid 1) Ta on bo da the Bae mbt = {Ale =21m2 +k] ke 2,0 < ks 2019). ‘V6i s6 nguyén m bit ky thi gid stim chia 2020 durla & € {0,1,..., 2019} thi ta vide ‘m= 2020n-+k vim € Z. Khi dé x8 ring P(e) = me 20202 + k= 202004 & 6 nghigon nguyen z=, tha ma dB ba 2) Cou tra Wai la phi dink. ik st sing trong 9, céc da thie ddu 66 be 26 bic cao nbt ddu dvong. Nan xét. Do bie ea ede da thie du lén hom 1 nen khi 2 ~¥ 0, biéa thie P(e 1) ~ Pla) 6 thé Ién ty ¢ va P(e) dBag bién Didu nay ding do deg(P(e-+1)— P(2)) = deg P(2)~1 va he sb cao nhft vin dung. Xt 66 thuc AF didn 48 ci 2020 da the db tha min P(e-+1)— P(z) > 2020 va P(z) ding bién trén (I; +00) (hon 2021 6 nga Hen ip an hon pa PM) Bk... 4 2000 {ht 8 ring ng vot mi s5 k-+¥ nhw th, du phi ©6 mot da thie P(e) € 9 mi P(e) = b+ 4 cb aghiem nguyen, ate ‘Theo nguyen ly Dsichet th thn tl met da thie mo 46 trong 9 tha mi ne ; ‘Mo 46 trong 9 théa min vi ci hal sh, git hog Pla) = ki va Pls) = B+ j edu nghien neuen VO 0.5 < j'< 2000 Dat ede ngitind6 ln ht tha, Docke sO Ee the aah hom P(A), va vt P(e) dng bifn nen ta phil cd Mf 0-41. Suy to +d = PU)> PCO+1)> Pla) +2020 > 442000 > bj 11. DB MOCK TEST $6.1 n Diu v6 1y nay cho thy khong tén tai tap hep dep thda man d@ bat ‘Tratng hop 6 fe hon 2020 da thc ob hie sb bj cao mht dung tht r0 ring Dai tosn ‘vin ding (do ed da tht o6 he 6 eso nh am th tren mién 2 di Ion sé khong thé nha gi trj dung duge nia, a Bai 3. Vik ld sb nguyen Ion hon 1, trén bing 06 vibt sin ede 3b 1,2,3,--:, 10810 41, 6 mdi bue, cho phép xéa di ha 25 o,b bit Ky 06 sn trén bing r8i thay bai £(@,8), tong 6 f Ta mot him hai bin. Goi 2 Ik sb nguyén dong thn duge sou 10* buds. Ching minh ring v6 mai eéch chon ep 56 8 mdi bude 1. Neu (a,b) = gel 202008, a? + 506ab-+ 1) thi 2 = 1, 2. Néa f(o,8) = god(o?8? +2019, 0? + # + 2021) thi Kndng phil 6 chith phon Lei gids. 1) Ta xét ede trating hop sau: 4+ Néu a,b cing 1 th f(a,6) chin, + Néu a,b cing ch thi (0,6) chin. + Nu a,b Khe tin en 16181 (2,0) 18 Suy ra a6 lugng s6 18 hofe ght nguyen, hote gidm di 2 don vi; ma ban di o6 8 ob 18 nen ab x cubi cing phii lé. Gis st 2 > 1, goi py Ta mot we nguyen 88 cha x [Nbu p= 5 thi i 5] ged(20200b o? + 506ab +87) ne Slo? + 0b+ Bt ‘Ta e6 b8 dé quon thugc sau. Néu p= 3k-+2 la s6 nguyén 18a pla? + ab +0 tht a,b. (ching minh dB ding bing dinb iy Permat nbd, ca «° = 0 (aod p).) ‘Tir d6 uy ra cbc ob trite 46 dk sinh ra a,b cing chia hét cho , kéo theo th cl oe 6 ban din chia hit cho 6; v6 If. Tong tr voi 101. (Con bu p 5,201 th pla, ple? + 5060 + 6 se kéo theo p | 0,0. Tuomg, ttre fing suy ra v0 If. ‘Vay nti «Ih 66 18 va khong ob we nguyén t8 18 no, cing 18 ring 2 2) Vid? = 1 (mod 3) nfu Sa va o? =0 (mod 3) néu 3 | a nen tab # Néu 3], 3] btm 34 f(a). 1s Néu c6 ding moe trong hai sb a,b chia hét cho 3 thi 3| f(a, 8). + Néu 34a, 34 thi 3t f(0,8). Scanned with CamScanner2 CHUONG 1. TUYEN TAP CAC DB THI MOCK TEST VMO - TST Goi 7 hsb Iugng 6 chia dt cho 3. Sambi bube tinh chin 6 cia 7 Hh Bi, mi bon déu 7 = 202 1g nen a chia hét cho 3. Ngoki ra, = khong chia hét cho 9 vi 1a nguge ii, do tn tia, b so cho = = f(0,8)| 28? +2019 nén 2019 chin cho 98, didu niy v0 Ii Vay nn 2 khéng pha IA 6 chin phucng, a [Bai 4. Cho duting trdn (0,8) va day cung BC ob djnh khée dugg kinh, diémn [A thay d3itrén (0) sao cho tam giée ABC khong ean. Gol BD, CE th cbc pin fide trong vA J TA tim noi tigp cba tam gise ABC. Gia sit DE oft BC tal F. Duong thing qua J vudng géc voi IF eft trung truc BC tai K. 1. Ching minh ring OK = 3R. 2, Ching mink sing Ki A thay déi trea (O) thi dung trung tuyén dinh J cca tam gide IDE huon di qua diém o6 dinh, Lai gi. 4) Trude ét, ta phat bidu vA ching minh bé dé sav. BG bE. Cho tam gée ABC ni tifp dung tron (0) o6 dudng phan gide BD, CB, ‘eoi J Ia tam ducing tron bang tiép g6e A. Khi do OJ vuong gée voi DE. Ching mink, Goi U,V lin lug: Ih tem bang titp gic B,C cba tam giée ABC. Kbi 66 A,B,C Vin Ivor Ya chan dutng cao cia J,U,V trong tam giée JUV. Cée. diéin D, Elin hot la geo diém BU,CV vi AC,AB, gol 1 Ih gino diém BD voi CE thi Tia trye tam tam gise JUV. Ta 06 Poo) = DA-DC = DU- DI = Paxuovy tawong ty ta cing o6 Peyoy = Peyiuyy nén DE vuong,gée wi dutng nbi tam (O) va (IUV), Dé ding ching minh dung néi tim (0) va (UV) di qua J dya vio O 1a tam diting trin Buler cba tam giée JUV hay JO L DE. a 798 Ii bai tos, got MA teung diém eung nbd BC thi M 1 trang diém JT. Ly (O/ dbi sang vi O que M tht ti giée O'TOF Ii hinh binh hinb nén O'F || OT. Suy ra OLDE. Vi AL Wi phan gic ngoa cit tam gife ABC néa L(AI, DB) = -1. Trve giao din 1, v6ieéc nbn xét IM ALAIN 1 LI,10 LLD,I2. LB trong dé Zz Ma tha qua Z, song song vai OIC. Do a6, (Iz, 10',1K, IM) = -1 nea theo tinh cht cia cin did hoa, ta thfy O” 1a trung didm ein MO’ = O'K vwOK =3R. 6 MK nen OM = 11, DE MOCK TEST $6.1 3 2) Goi X,¥ lin gt It gino didm cin BD, CE véi (0). Gib st XY eft BC, DE lin lst tai Z,F. Ta théy X,¥ lin lugt TA trung diém efe cung nh AC, AB eta (0) hen XY.LAT, ma ATLAL nén Al. || ZF. Theo dinh Ij Pascal cho hai bo ba diém (AXC! i A a tuyén ota (0). Do db (fi3)-* 106 AF chinh It titp tuyén cia (0). ZPAL = ZBAL~ LFAB. Scanned with CamScanner‘Ta cing 06 x ‘A _ £B-26 ean ‘Vi thé nén ZPAL = ZALZ hay tit giée AFZL la hinh thang cn. Suy ra ZALF = AGF = LIZF (do YZ la trung trve Al). Ma AL || ZF nén suy ra DE | 12. ‘Ta d6 ouy ra nfu gol N Ta trang diém DB thi chim 1(NZ, DE) = -1. Gid sit INOBC = Q thi ta cing c6 1(ZQ, BO) = 1. Tit d6 suy ra 1Q chinh Ta duting bi eye cia Z di vei (O) nén né sé di qua cue clia BC, tite la giao diém hai tiép tuyén cba (0) 8 B,C. Ta 06 ditu phai ching minh, oO Nhan xét. Bai todn may 06 rét mbiéu efch tiép c€n. Riéng 3 céch cing mint DE || 1Z, ta 06 thé tan dyng diém J ta tam bang tiép 4 b6 a8 tren dé dua vé ching minh 1Z 1 OJ, vasit dung dinh ly Pascal va dinh ly Brocard cho tt gise toin phn, LALZ = £B~ BAL = £B~ G cau b, ta con 06 mot két qua rft tha vi va téng qust nhu eau Cho tam gide ABC 06 B,C cb dink va A di chuyén trén (O). Goi P 1a mot diém 6 dinh trén dudng trung trye cia BC. Chimg minh ring khi A di chuyén trén duing {rn (0) thi didng ding gée ola AP trong tam gise ABC luda di qua met di i Sm dl non ding the pi xiy en, t60 Ee cig” Pam Deas, JO ‘Trong gid thi, thay y = 1, taco MMe) +2) = 24(2) < f(22) nen tt tinh ding bién, ta dupe f(2) +2 < 2x hay f(z) <2. Mat We, i f(z) 22 ate ea ae) t = Sa iy the) 2 ane, fee) > ‘Thit ni ta thfy théa. Vay néa tht cd ce him a6 ein ima f(a)=2,¥e>0. 0 Ning x¢t, Bh toon cba mt cle ip hn Whe sau cng minh f(z) > 2.¥e > 1 sau * lim f(2) =0* va _lim_ f(z) = +00. © f lien tue ta O* va lien tue tai mot diém x > 0. © F toan dnb, ti dé tén tai f(a) ‘© Thay y =a, 06 f(2) 2f(z), véi moi x > 0. = iil) F(F@)F(W) + 2) = f(@F()) + Sz) voi moi x,y ¥ 0. 1L2i gidi. Theo gi thiét i) tht ham s6 f(2) kdéng bj chain trén vi véi moi sé nguyén duong n th} FO") 2 270") >... 2 274) va Jim, 2"/(1) = +00, Theo gia thi6t tht £(J(2) fy) +2) > sla ()) nen Sa) fu) +2 > 2f(v) Je — H(z)) <2. ‘Néu tn tai zo > 0 d8 zo — f(z) > 0 thi ti dénh gid tren, ta 06 = 10 <= Teas v6i moi y > Ojkéo theo f 1A ham s6 bj chan tren, mau thudn. Do dé, ta luén cé T@)zave>0 Tip theo, dit a = f(1), trong dé bai, thay z—y=1 thi f(a? +1) a. 6 a? +1 > 2a va f ting nén f(a? +1) > f(2a) > 2f(a). ea ea, pyr Sa) +02 2f(a) +02 f(a) >a Bi 6. Cho tt giée ABCD khéng phai hinh thang, noi tiép (0) va ngoal tibp (1). Goi M,N, P,Q ln lugt 1a tiép diém cia (I) v6i AB,BC,CD,DA, Hai ‘tong chéo AC, BD eit nhau tai K. Gol H I bh chiéu cia K tren QM. 1, Ching minh ring IH, AO eit shau tei mot diém thuge (0). 2. Goi giao diém odn Iai cia IH v6i (0) Wh Ay. Che diém Bi,Cy,Ds due inh nghia twang ty. Chiing mink ring iC}, ByDs,OF déng quy. Lai gid. Trude ht ta ching minh bé a8 sau: B6 Db. Cho tit gide ABCD noi tify trong dving trén (0) c6 hai dung chéo ‘vudng g6e vii nhau 6 K. Goi E, F,G, H Win hugt i hinh ehiéo ela K én ec canh AB,BC,CD, DA. Gi sit BG edt FH 8 L thi O, KL thing hing. Ghitag minh. ‘That viy, got X,Y, ZT lin lugt W trung diém cia AB, BO,CD, DA. Bling bién déi g6e, ta c6 ngay EFGH la tt giée noi tifp. Ngodl ra, vi AB Th déi song cia KOD nén diting cao KE cia KAB sé chia doi CD nén K,B,Z thing Ihang, Twomg ty thi KFT thing hing. Suy ra CKEF = LKBF = LKAT = LTKA= LKTZ nin tt gle BF27 nt tip, Dong tw fe bm dn Keys ra ta K.Y2.7 wa £,F,G,H ching thude mot dong trou. Ngoai ra, néu goi 7 Ta trang “Gide OK tht d& thy T thuge trang true e6c doan XE,YF,ZG,TH nen I ehh Ib tam cin dung tron néu tren Scanned with CamScannerBing bién déi g6c, ta 06 AEFC,AHGC noi tiép nén bing céch xét true ding phuong, ta 06 EF, AC,GH dong quy; dat diém 4618 Q. Tuong tu thi BH, FG, BD dong quy tai P. Ap dung dinh ly Brocard cho tit gide toan phin EFGH.PQ, ta 06 ILLPQ. Lai ob P,Q c6 cing phiang tich dém hai dong trdn (1), (0) nén PQ chinh Ja true ding phuong cia hai dung trén va PQLOT. Tu d6 suy ra 0,1, K,L thing hang. B6 d2 duge chimg minh a 1) Goi A',B",C’,D’ lin lugt la trung diém eta céc doan thing QM, MN,NP, PQ thi 16 ring céc diém ny theo thi ty cing thude eéc doan thing L4, 1B, IC, ID. Xét phép nghich dio © tim I, phutong tich r? v6i rd bin kin cia (J). DE thy Hing thong qua © thi cée diém 4, B,C, D+ A’, B',C',D, ma ABCD ni tigp nen A'BIC'D’ cing noi tiép. A tp =}, 1D wae Hon na, A'B' = SQN = C'D' va A'B' || QN | C'D’ nén tt ge A'B'C'D' cing 1 inh binh bho kt hop voi y6u 06 nb tp, sy ran6 1 hn eh nat, gi 9 ‘dudug tron ngoat tiép A'B'C'D. = me ‘Tiép theo, v1 QM th di song trong tam giée KINP nén trunng tuyén KC’ cia KNP s& vubng g6c v6i QM. Do d6, H,X,C’ thing hing. Do dé, ti giée A'D'CH noi tiép men H € 9. GoiT I gino didm cia tia HT vi (0) va Ay I giao dim com al cia ding thing HT ¥6i (0) thi A, Ay nim cing phis so wi BD. Vi 8: « (0) nén H 4 Ay. Lai c6 ©: A A’ nén LAA = ZHA'T = 90°, Didu may ching to AT chinh Ia dong kinh cia (0) ném AO di qua T. 2) Ta thy ring néu go G, F, Elin lug 8 hin chiéu a K len c6ccanh MIN, NP, PQ th tong tw tren, ta eb @ | GFE + By,C;, Dy. Suy ra.@: AiG; + (HIF) va BD, + (GIB) Ngoai ra © : OF 4 OF nén ta dua v8 ching minh OF, (HIF), (CIE) 06 met diém chung Khe I. Ditu niy tong dung véi vige ching minh O7 chinh IA trve ding phuong cia hai dvdng tron nay. Ap dng bé d 8 tren cho tt gide MN PQ voi chi F ring hai dvtng chéo cba n6 vudng gbe vi nhau tai X, ta suy ra giao diém J cba HF, GE #8 nkm trén IK, te lacing nlm trén OF. Ngoai ra, vi H,G,F, E mbt tidp nen JH JF = JG JE, ching 0 Jf cing phitong tich din hai dung tron (HF), (EGE). Tae day ta 66 db phi ching minh. o [hf xét, ‘Trong mo hinh tt gide Iudng tam 6 tren, dudng thing OF di qua diém Miquel cia tt giée toin phin MINPQ nén vige xét phép nghich dio tai tim dé sé thu duge nbibu két qué dep, thd vi. Scanned with CamScannera al ee TR te 1 IL i toag, Bai 7. Goi a la nghiem duong ofa phutong trinh + 7 1. Ching minh ring wéi moi s6 nguyén dutgng n sao cho Jn 2" so(2020(a" + 0-*")) > y(2019(a"" + a7"). 2. Chitng minh ring S = a*” + a7 06 it nhAt 5 - 2° wdc nguyén dutong, Loi gidé. Trude khi bit dau 183i gidi, xin nhéc lai mot s6 két qué quen thudc sau © y(n) va r(n) déu la ham nhan tinh. © 9(p*) = p* — p*-) véi p 1a 86 nguyén t6, k nguyén duong. © 7(p*) =k +1 voi pla s6 nguyen t6, k nguyen duong. d & Tinh cht nhan tinh téng quat (mn) = vlm)eln) ry trong dé d = ged(m,n). Suy ra néud>1thi wa Siva elma) > plm)oln). 1) Thuéc hét, ta 6 2020 = 2”- 5-101 va bling quy nap, ta thay céc sé hang cla day ‘tin = a" + a” da cho chi chia 4 dur 2 nén theo tinh cht nhan tinh téng quat 6 trén thi 2 9(2020un) > (4) - (5) - (101) - arn) = 16009(un). Ngoai ra, ¢(2019un) = 9(3)(673)p(2n) = 1344p(up) néu ged(673,u,) = 1. Néu ngugc lai thi ae y(2019u,) = 13449 (un) - wera) < 1600¢9(un). Do dé, trong moi trudng hgp thi ta déu c6 (2020u,) > y(2019un). 2) Dit ty = 0%" + 07%, khi d6 dé dang quy nap duge cong thite truy héi cia day nay 1A 41 = Tp — 3t_ Vi T = 48. Dé thy ©, chia hét cho 6 véi moi n tu nhién. Theo cdng thite trén thi n41 chia hét cho tq vi do ged (x? — 3, ¢ ) 4 22 —3 > aq, > 0 nén chic chin 86 2? — 3 sta udc cha tp. Do dé, ta c6 a = 3 cing nhu 6 mot uéc nguyén t6 méi khéng 1a U9(Zn41) = U3(@n) +1 nén vg(t99) = va(%9) +30 = 31 va 86 a9 C6 ft nh&t 30 uéc nguyén t6 18 phan biét, Khéc 3. Lai o6 24 v2 (go) = 4 va ta tinh duge | % | 23 nén 7 (90) 2 5-32-29 = 5. 995, Bai toén duge gidi quyét. Scanned with CamScanner