Toán học và tuổi trẻ Số 202 (4/1994)

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Dưới đây là tạp chí Toán học và tuổi trẻ Số 202 (4/1994). Mời các bạn tham khảo bài giảng để nắm bắt những nội dung về mẹo nhỏ để giải bài toán về diện tích ở lớp 8; hình học hóa nội dung và cách giải một số bài toán đại số và một số nội dung khác.

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Nội dung Text: Toán học và tuổi trẻ Số 202 (4/1994)

J N\ffi Bo crAo DUC vA DAo rAo * Hgr roAN Hec vrEr NAM |l ---lG TAP ffi cui ne NcAy rs uANc rsANc 4poz1 1994 0 ffier mFo trH6 D€ @E#r efrl Eo6N 5 f,GH Ar,eb. o Dinh li lon Fermat at aA r\ va vi6c tim toi cf fn!. minh so cflp cua n0 t, 6 ffittH ffi$fl ffi{}a A'trt $$[ ffiUHffi ?A #,&flI[ $[AI '^P\t ,/ $$ mail ?ff.{H '$$Y f,l,tt $& I uHai effis$ki ia Thd,y uit. trb trudng.PTTH Quynh Luu 3,NghA An udi Tap chi Todn hgc uit. Tudi trd - s Ddm den #Ht$#ffiffi, ffffitn*r* mUsT ffi# fru'- iffif! uumr'eimrorug,*,*'u
TOAI{ HQC VA TUbI TRE MATHEMATICS AND YOUTH MUC LUC Trang o ctfrc bgn trung hgc co st Ddnh cho For Lower Secondary School Leuel Friends - Vu Hilu Binh : MQt mgo nh6 dd giAi Tdng biAn fiP : biri to6n v6 diQn tich d l6P 8 NGUYEN CANH TOAN - Phqm Brio ; Hlnh hqc hda nQi dung vd Phd tdng bidn tdP : cdch giAi mOt s6 biri to6n dai sd NGO DAT TLI o Gidi bdi l$ trrtic HOANG CHUNG Solutions of problems in preuious issue - c6c bai tt T1/198 d6n T10/198, L1/198, L2ll98 o Db ra hil ruit Hor ooNG BIEN TAP : Problems in this issue - Cilcbai tt Tll2O2 ddn T10/202,LU202,L21202 10 Nguy6n CAnh Todn, Hodng o 6ng kinh cdi ctfrch dqY vd hPc totin Chung, Ng6 Dat Tr1, LO Kh6c KaLidascope : Refonn of Marhs Teaching and' I*arning Nguydn Drtc Td,n; V6 mQt tidt day to6n 6 BAo, Nguy6n HUY Doan, - Nguy6n Vi6t Hei, Dinh Quang lOi ctrgn, l6p chuY6n 11 thi vdo dqi hoc HAo, Nguy6n XuAn HuY, Phan o Ddnh cho cdc bun chudn bi For College and, (Jniuersity Entrance Exam Preparert Huy KhAi, Vri Thanh Khi6t, Lo - Nguyfin Vd'n Mfuu: Dd thi tuydl sin-h vio Hei Khoi, Ng.ty6+ Ven Mau, Oiiiq" He nam 1993 (Kh6i A) L2 Tdng hgp NOi Hodngl6 Minh, NguY6n Kh6c o Bgn c6 biit Minh, Trdn Van Nhung, Do you know ? Nguy6n DEng Phdt, Phan - Nguydn Cd.nh Tod.n: Dinh lY Fermat Thanh Quang, Ta H6ng vI viec tlm tbi chrlng minh so cdp ctra nti 15 QuAng, D4ng tr{tng Thing, Vr1 - A"d" Ngq. S"".' Hai con s6 k}, la BIa 3 Drrong Thuy, Trdn T[enh - Nguy€n Dung; Chuong trinh rrit gon tinh Bia 4 Trai, L6 Bri Khdnh Trinh, Ng6 s6liQt Fibonacci ViQt Trung, Dang Quan Vi6n. o Gidi tri todn hpc Fun with Mathematics - Binh Phuong; GiAi dap bei Con s6 73 ki la - Nguydn Dinh Titng: Ddm ddn Trq. sd tba soqn : ' Bian @p uit, tri stl : \16 XfU T$UY 81 Trdn Hung Dqo, HA NQi DT: 260786 Minh DT: 356111 rrinn bdy: DOAN nbNc 231 Nguy6n Ven Cfi. TP Hd Chi
a x6t bii to6n diQn tich sau thu6c chrrong trinh ldp 8 : cD E Cho tam gi6c ABC. Qua di€m. D thuQc cqnh BC kA cdr dudng thd.ng song song udi cdc cqnh cfia tant gid.c tao thiuth hai tam gid.c nh6 c6 diQn tich 4cm2 (L sd uir.9cm2. Tinh d.iQn tich tam gid,c AtsC. V6i hoc sinh l6p 9, giai bbi todn tr6n kh6ng khd khen l6m : 0 BD DC TacdrBC+BC=1 D+t dt (ABC) = S. Cric tam giric EBD vb. (1) FDC E U ddng dang v6i AABC n6n : ,BDr2 4',DCtz 9 !trr /-:-\-=- \ac)-s'\.Bcl '-\--- -s (2) IF z -BD2DC3 E (U1. Fq = suY ra BC {-s 'BC = 15 (3) ,*23* rl-- EI Tt (1) vn (3) 1. Do d
T=rong hinh h'oc cdc tinh ctrdt dfnh luong cria suuv * sa,rmlr + s&vcP zr (Dd ,;hi tuydn sinh vio dai hqc. sd 45) thd hiQn bBng phuong trlnh. Trong ir6 tqa d6 Db ca.c ctia kh6ng gian h,ai chi6u mQt didm Gi6i : Cdc c6n thrlc trdn gdi cho ta ngti drrgc bidu di6n bing mQt cap sd sip th* tu ngay d6n c6ng thdc dO dai cria mdt dc4nth&ng' (r, y), dttdng th&ng li rn6t t$p hqp didm bidu di6n Troirg met ph&ng tga d0 Oxy ldy cic didm bdi mQt phrrong trinh d4ng ax * by * c = 0, dttbng M (Zcosx cosy, sin(r - / )) vn n*irum Hoc nmn ruor muNG vA cAcH GIAI "^r{i-r----d mor s6 mne r0AN sal sO PH.+.M EAo trbn lh rnQt tflp hop didm bi6u di6n bdi mQt N (2sinr siny, sin(r - Y)) ; !gl,! : phrrong trinh dane- {x - a}2 + (y - bF = R2... oM=d4coffi Trong kh6rig gian 3 chi6u, rn6t didm dr:qc bidu 0N: rj4sinTisin? *si#(r -y). di6n bing rn6t b$ 3 sS sip th13 tU (x, y, z). iludng th&ng l6L {(r, y, e)} mh Dgng hinh hlnh Y z-2, binh x'-r, !*lo ONPM (xem 2sin{r!) -- ---z' &bc hinh 2) -r/ z ,aii,,i ' r*4t phingl8r ift,.y, z)i rnh.A.r+By* Cz*D = {) Ta cci -/,/l troec rn?icllu ld {(r, y, e)} rnir (x - to)2 + (.y - OM+ON = an(e-y) - lo)z 4 (z - z,rlz = R:' OM+PM>-AP Do dti, v6i rnQt sd bii torin dai sfi, t'try theo Cfing d6 thdy cr{c dt ki€n ndu ta "li6n tddng" dd nh&n ra ciic P cd toa dd li r}{c didm hinh hqc cria nd, cd thd ta sE tim {2 (cosr cosy + 2cosa cosY 2si,zfrstn)/ +sirusiny); dr.tgc nhirng c6ch giAi ng6n gqn, dQc drio. Sau day le m6t sS vi dg. %in (rl)l; Hinh 2 do d 2 vd trrii crla (2) cd thd > rl4*"oF(r -.yi xem [i tich hai c+nh Vi dq 3 ; Ctw 3 sd x, y, z > 0 th&:. t'^zcn iw th;Zc : cria rnQt tam giric vb xyz(x*Y4z) = I {3) vl sg nhrr nhau n6n ta Tint gid tri nh.6 nhdt cila (x + y) {x + z). hey l6y mQt tam gi6c d6u ABC crj c6nh Giii : HQ thrlc (3) la tich cira 4 s6 gQi cho ta "Iigu tU0ng--ddn-qQUg-t]rlc H6-r6ng b&ng J. TrBn c6c cqnh 5 = ff1p - a)@ - b)(p--c.;. Ndu datx : P - d, AB, BC vh CA ldn lr-tqlt y _: p - b, z =p - c tlti x + Y +7 = P v& ldy cdc didm M, N vb Sz=xJzk+y+z)=i P sao choAM = r, EI{ Tr6n hinh 3 cdc s6 x, !, z ia d0 dai c'1c ti6p =e,CP=y.Tacdbdt tuy6n k6 trl dinh ctra tam giric ddn drldng trbn d&ng thtlc diQn tich ; ndi ti6p. Cqnh A-B *, *Y, AC = x * z. 1
a_ ' S"U dri hinh.:.. pfurong'Z v6 c6c d&ng thrlc trOn " LABC - '. lrr 1 tr"O ,, . .r' ', ..: =-rABACsinA I*z - .,2 1 l*": "i + x2 + u2 = u2 *y2 = I vd til lttL =.yz -2ABAC xu'+1,u:0n6-ur ' n€nAB.AC= thay uI y.h.o4c = (x*y)x x=utaed x(x*d> xy*uu=0. >2s*ur=z I "li F & ?rl cric vi du Hinh 3 tr6n ta nh&n thAy e*c bai tofn d6 VAy min[(r +y) (x * z)7 = 2 thitc chdt la Vi du 4. Tint gia tri nhd nhdt crtu nhfrng bhi todn hinh hec nhung F = r[zF -2, *, +rF+@ + jl, +l + +w Gi6i : Tuong tg nhu vi du 2, ta vidt lai duoc ph6t bidu b&ng "ng6n ngu dai s6", ndn khi giei phai cd nhin Ilinh 5 f: ra cdi "hdn hinh hoc" crla nei dd tt dd vin dung cdc kidn thrlc hinh hgc rni tim ra nhtng loi giAi ng6n ggn "tudng minh". Trong met phing toa d6 Oxy ldy cdc di6'rn Sau dAy li rn6t sd dd todn dd qic ban luy6n M(x, y = x), AQ, 1), 8(--f , *f,1, t+p: 1. Tim giri tri nh6 nhdt cria hdm c(*+ ,**r, sd tacd: y=Gz-r+1+frurffi;+1 F = MA+MB+MC Dd.PsdY=f; 2. Cho a. * h * c = 2 vd. ax *by * cz * 6 tim Cong d6 thdy gi6 tri nh6 nhdt cria tam gidcABC ddu vi F nh6 nhdt khi P = \fi6{ 4;ry + {{Sr a5zy + {{6V177 M tring vdi g6c O Dd.ps6:P:1A vAminF'=3khi 3. Cho 3 s6 r, y, z thba m6n hQ r = 0 (xem hinh 4). Ili.ruh A +*y *7,2 - o2 l*2 lf +w *22:bZ Yi a,l 5, Chrtng ninh rdrug ndu x2+ y2 = u3 lz2 +'zx + x2 = c2 *.uz =.7 (5r) udyy *yr-t = O (5a) thi x2 +u2 = y'+u'=7uitx.y*uu=0. tirrt, ,y *.yz * zx Gi*.i : Tryng rnet ph&ng toa dQ Ory phrrang Dd,p sd,1rS, S*r, trinh .r2 * y2 = I li phrrong trinh dudng trdn cci tAm O(0, O) bdn kinh bing l. Nhung ddng l* *yl nhdt nrat ph&ng tqa d6 Oxy vdimat phi.ng toa 4, GiA str f (r, y) = {T +x-z tn +iz d6 Ouu thl dudng trbn dd erlng li clrrdng trln cri phuong trinh ld uz + u2 = J. Tr6n dudng C:hdng minh r&ug v6i mqi a, b, c La cd trdn drilfly hai d$m M(x, y), Ntu, u) vd x6t hai f (a,c) < P (o, b) +f(b, c). vectd OM(x,g), QN(u,u). Tir (52) clia ta xu * yu 5. Chrlng minh r&ng vdi moi a, b, c, x, y, z .+€ = OM. ON = 0 n€n OM t ON. Thdnh thrl, ta cci : ndu M thuQc cung vu6ng I ihi N hoac nim at * by * cz *,17,7+ Or+ a\ti4 rra-4 , trong cung vu6ng 11, hoEc nirn trong cung 2 vu6ng /V, Nhrrng dir trong cung vu6ng nho ta -g(o+6+c).xk*y-rz) cring cr5 lrl = lul va n6u r, u ctng ddu thi w,ytrdiddu lul = lyl ho6cr,yctngdduthi Hudrug ddn : Ding tich vd hrrdng. y, u trrii ddu.
1t (---.Suvra: ht *t hk 111 hl' -!-r--L znzr' bhzr " "' J- {2o - t)hl 1111 11 Bei T1/X98. Chtng minh rd,ng c6 uO sd sd 111
Dinh, Phpm D*ng Hodn, Thanh Li6rg Nam HA ; Hodng (9A, Bd VIn Dirn, Hir NOi), l,e ngec Gidp (7A. D6ng S
- XA + XB >- AB V X e tryc hoi,nh (1) Le LOi, PTTI I t-am S
Orl = x3=,.. : f*1, x2 = x4= ... = ffn va 11 = xz (mod2) (6) n6u n ch&n. -..Hoq. nita, d5. thdy, ndu Mk * M**, vA 2) Nguqc l4i, d6 thdy, v6i cdc s6 nguyon 11, r!k), r!tt) , . . . , Ak) khong ddng tfidi rinfr'rihau x2t ....t rn th6a m6n (5) ndu n 16, hoec th6a (6) thl so ) sk*1. Suy ra, ndu r[k) ,xy),...,x!*) ndu n chin, ta sE cri rft l n.\[2 '2' L r Suy ra 2R(t + : V) c I{6u thi x$'"- 1)=xl"- t)=...=x{!"- r) ra 16 sirrA +sinB +sinC riz €TosA L u; (dtrng dinh li mAu thudn vdi cSch chgn /". + cosBt cosc ' o N6u z ch6n thi : hirm sd sin vi dang tht?c quen thuQc cosA r r! * xf"'- t) = x9,-1) -.. . = *#nr 1) (2) vd : r * cosB * cosC =I + gie sit A lA goc l6n nhdt xf"- r) = xt.-l) = . . . = *,9"-l) (B) J?) GiA str ,f" - ,) + x$"- 1) (*). Khi dd, v6i quy trong AABC , tac6 A > !o ; nann[ta AA.BC nhon, rldc ,1") = xi, Vi = 1, n, tt (2) c6 ncnf
do dci (4) dring, tray (3) drlng. Xdt hdm sd f(x) CIBCD, OCDA, ODAB, OABC vdV'1,V'2, V'3, x V'o ldn lirgt Id thd tich cric hinh chc5p cut cci sinr * 2co5 rndt aay ld BCD, cDA, DAB, ABC. Do cac tt x6c dinh tr6n do4n ln n1 p; zl "e diQn AtsCD , A, B 1, C r D I d6ng d4ng n6n : x cosr + 2sin, \,+vt_(*,)' _(y:n!:\' _= -1+sin7 3 v =[-] =i, A]i ,l c6f'(x) = ( < 0) hayf(r) nghlch lcosr +_2sin|)2 = (y::'\ = (, .#)'= (,.?)' bidn tr6n \AH ) W,;) ,lt Jt. hayv*v'r=v(1 +f,f = €t Vre\r;r)hay vz, fi, A v + svt * t * sinA * 2co5 ,t** vdi moi A € ;+4. oo d6 v't = Svr cosA + 2sin, V,;) 4f, _ J-- MQt c6ch tudng ttr, ta cri c6ng thrlc v'v2' K6t hop di6u nAy v6i (3), ta ra dpcm. cci (1) dring, suy : v?4 Nh{n x6t: 1. C6 68 ban tham gia giAi. trong s6 d6 c6 nrdrQng :V'. ?Vi+3 V + .$ = 1, 2, 3, 4). 63 ban giii dring. C6c bgn sau dAy c6 k3i giii t6t : LA NguyAn Chdt (ll'l - lamSoq Thanh I{6a), C/ru NpyAn Bnh (l2A- I r -.1 - l DHSP Vinh), Nguy€n Thn Caong (11A, Ly TrJ Trpng - Cdn i*i si4 * 11 , Tnay Nguv|n Minh Hdi (12 toin TrAn Phri - Hii Hung). suy ra i=t = i:l + ?"f i=l f,Z r=r 2. B}n Hodng Thi Tuy€t (l2T Lam Sdnq Thanh H6a) chilng minh "2 (ZR + r) a + b + c" rdi suy ra dpcm. B4n Ngd 3rn L w >3V+V 4*r, lA= Dtc buy (10T - PTNK - Hdi Hung) gqi OI, Ol, OK.ldn hlqt lir khoang c5ch tit o a6nfc, CA, AB r6i dlng bdt ding thtc 3VV61 Ts€-bu-sdp dd c6 ol.a + oIJ.b + oKc < !
hai hq.t theo phuong nd.m ngang uE hai chibu UAB = Ueu * UMB = Ry't + R/z = nguqc nhau udi uQn tdc tuong ilng ld, urvh, ur. I') (1) Chtng mink rdng hhi khod.ng cd.ch hai hqt lit. = RrIt + Rzgr + ; X - (ur+u) {ipfi thi udcto uQn t6c cia hai vd'Ir=Is-I'; hq.t d thiti di€m d.y se uudrug g6c udi nhau udi UAB = UeN * UNB = RjI3 * RrI, = g lit. gia tdc trgng trilimg. 86 qua stc cd.n cila khong ktti. = RzIz + R4Q3 - I') (2) Hudng d6n gi6i. Sau khi phAn tich chuy6'n Nhan (l) vii R.. * Ro vA nhAn (2) vdi ddng cria trlng hat (drroc n6m di theo phrrong Rt + R2, sau dd c6ng lai ta drJoc : nim ngang), di d6n nhQn xdt : d mgi thdi didm (Rl+R)+RJ+Ra)Uas: hai h4t d6u cirng nim tr6n mQt dtldng thing nim ngang vA khoAng crich gita chring d m6t =(Rr +Rr(R3 +R4X/r +/3) + @frt- RLR4)I' thdi didm , bdt ki sau khi n6m sE bing Theo giA thi6t Rfi,s - RtRt = 0, do dri X = ui * urt = (urt u2)t, til dci suy ra thdi (Er +Rr(.B: +Ra) Oidm r khi kioAng Lacn-x = (r, * u) ,[i,u1g "AB R.+ R2+ R3+ R4 /(vi.I, ff - +13=I) x Y to ) {;- ur*u,= ' t = -* I ( 1). Sau d
Bhi T7l202 : Cho tam thrlc bQc hai : f,(x) = atJ +' 616 * c vdi c5c h0 sei drrong vi a*b*c=7. Chrlng minh ring vdi moi sd nguyOn duong &thi; ;k._ .k f(x) '- tf('\x)l- Cric ldp PTCS rRAN vAN HANH BAi Tt/202 : Tim cric s6 nguy6n td dang Bei T8/202 : Ki hi€u N* la tAp c6c sd nguy6n drrong. Tim tdt cA cdc him 2tee4d q t7 @ eAr; bidu di6n drroc drr6i dang hiQu cdc ldp phrrong ctra hai sd tU nhi6n. /;.1[*+N* th6a m6n ddng thdi c6c di6u kiQn sau : ouc rAru NGLTYEN 1)f(n+r)>f(n)Vz€N* B.Ci T2l2O2: Chtlng minh rhng di6u kiQn cdn vd dn dd c6c s6 a, b, c ctng ddu ld : 2) fff(n)) = rr.+ 1994 V n c N* NGIJYBN MINH DUC ab*ac*bc>oua!+a+ ob ac !"!ro B}ri Tgi202: Quayhinh vu6ngABCD quanh NGUYE,N DE tAm O cira nrj mQt gdcg' , @ < g < 9U dtqc hinh vu6ng A'B'C'D'. Ta ggi giao didm cl0,a AB BAi T3/202 : Cho tam gSac ABC nQi tidp vdA'B'ldM, BC vdB'C'laA', CD vdC'D'ldP, drrdng trbn (O). Dgdng phdn gi6c trong cria DAvdD'A'l "q. g6e A c6t duang triin (O) t4i didm thrl hai D. MQt dudng trdn thay ddi lu6n lu6n di qua A, a) Tinh diQn tich ti gi6e MNPQ theo D c* c6c dtrdng th&ng AB, AC t4i cric didm a=ABvitg. b) V6i g6c A nirc thi chu vi phdn chung hai tudng itng M, N. Tim t4p hqp trung didm I hinh vu6ng bd nhdt. cta MN. DAO TRUONG GIANG onNc vr6N BAi T10/202 : Cho tri di6n d6uABCD. Tim quf tich nhtng didm M sao cho tdng binh C6c l6p PTTH phrrong cric khoAng c6ch tir dd d6n c6c mat cria trl diQn bang k2 cho tnrdc. Bai T4t2O2 : Cho ba s6 nguy€n NGUYBN MINH HA a,b, e, a ) O,ac *b2 = P = pl " p2.. . pm, trong d6 pt , . . . , pm lir c6c s6 nguyGn t6 kh6c Cdc dd VAt li nhau. Goi M(n) li sd cric e4p s6 r,gaybn (r, y) th6a min Bai Lt/202 : Anh sringm4t trdi chidu song song v6i mdt btlc tudngthing dtlng vd tao vdi a*+2bxy*ciF=n phrrong thing drlng gcic nhon a. Dtlng tnrdc Chrtng minh rang M(n) ld hfru han vi tudng, dirng m6t gudng phing, mQt em b6 M(n): M(fi .n) vdi mgi k > 0 chi6u rlnh s6ng vu6ng gde v6i m+t tudng. Khi dd m6t phing ctia guong t4o v6i m4t ddt mQt D€ dg uydn thi to6n qudc td n5m 1993 gdc bing bao nhi6u ? B,di T51202: GiAi hQ phuong trinh : NGUYEN DONG *l= ur g ca$?Yx2 BAi LZ|ZOZ: C
Fon f,ower Secondary Schools pm : '. : . I $::0F,':THIST t$Sil$H Tllz0lL.Find all prime numbers of the forrr ' : .:.. ..., .. :: .. i 219ef + 17 (n € AI) which can be writteni,as a , difference of cubes of two natural numbers T8l2O2. Solve the equation NGUYEN DUC TAN ' : * arccostinr: 1' arcsincosx TZ|2O2. Show that 3 numbers a, b, c are of de the same siga if and only if ob * bc * ca > 0 and NClIYF,N VAN M.\IJ 111 --= *;- +- > 0 mPAL"Iet f(x) = ax2 +bx * c withq b. c>0 ao ac cq ando+b*c:l.Provethat NGIJYEN DE TB|2O2. Let be given a triangle ABC and its f(x) > fft2h)l'r for every positive integer la, circumscribed circle (O). The inbisector of A TRAN VAN }IAN}I meets (O) again at a point D. A vafiable circle passing through A, D intersects AB aw)" AC at TBl202. Denote by N* the set of all positive M and N, respectively. integers. Find all functions /.'N**N* Find the locus cf the midpoints .I of J}fN. such that DANG VIE,N 1)f(n+1)>f(n)Vra€N* For Upper Secondary Schools 2) f(f(n)) -- tt + 1994 V n € N* T4l2O2. Let a, b, c be given integers with NGI]YEN MiNII DI]C a ) 0, ac -b2 = P = pt...p,r,, where Tgl202. Let A'B'C'D' be the image of the square ABCD by the rotation of angie P1., . . . , Pn are distinct prime numbers. Let M(rt) denote the number of pairs of integers p (0 < g < 9e\ about its center O. AB, BC, (x, y) for which o.x2 + 2bxy * cy2 = 77. CD and DAintersectA?', B'C', C'D' and D'A' at M, N, P and Q. respectively. Prove that M(n) is finite and fuI(n) = M(F Q for every integer k > o a) Calculate the area of MNPQ in c. : A,B, g IMO31. GI:O 3. b) Determine the value of ,p such ttrat the perimeter of the common part of two squares T5l2A2. Solve the system ls nilnlmum. {5 COS Jt X) Xt,Y= -'i- DAO'IRL,ONC Glz\NG {3 TlAn02. Let be given a regular I"=:COSfir- z9r tetrahedron ABCD. Find the locus of points M such that the sum ofsquares of distances from xs = {,g cos 7t x,1 M to each face of the tetrahedron ABCD is equal to a given value k2 . xl: 'VT cos lt xl NG(]YEN MINII I'IA g TO XUAN FIAI il #N{}KfiN T cho /l-T d5 dtJrlc dUa vao lnot s6 ti6t trong chrtdng trinh toarn ri 6 c5c lop chr-ry6n, lrip chon. ThUc ra, toirn b6 clrc ciiiu hi&r chia hdt n6i tren cri thd thitc hiOn trqn ven trong mot ti6r day : Ddu hi€u chia h6t cho 2k Sfj N= rrr"-1...r-+lrr--t vi .5k (k € AD viat dtroc tlU,fi dang N=r"r"-l-rt+1.10k+o*-.o, ma 10k : 2k.5k cAt cAcH DAY vA HQC TOAN chia htlt cho 2k vir 5k do vay N ; 2k ( hay ,5k ; ndr vir chi ndr \*nr{ ,rr*u- 2k tha.v -lk ; va chi nhfing s6 dn mrti chia hdt VH MOT TIET DAY TOAN 6 "r: cho ?k thay -sk ;. X6t cac tntdng hqp cv rhd l6p chon, l6p chuy6n k = I ta co ddu hieu chia h6t cho J, cho.5 k : 2 ta c6 ddu hifu chia hdt cho J, cho 2-5 Vie.c dpy- hec loen c6 khi bit ddu tri cdi cu rhe dd di k = 3 ta c6 ddu hiOu chia hdr cho 8. cho .11-s la ci-ic cldu ddn ciii tdng quit nhtlng cring c6 khi giAi quydt cii t6ng qudt hi€r.r chia hdt thrttrng sit dung. ta se dUOc cai cU thri mA c6 thd rtit ng[n duc,c thdi gian. Xin I-ly vgng n'ing thqc hi6n tidt d4y n2ry s6 dc'm den cho c6c n€u vi dp : em -hr2c sinh nhidu didu bd ich. Ddu hifu chia h6t cho 2k vi sk 1t e q li su tdng qudt cta c6c d6u hi&: chia hdt cho 2, cho -5. cho./, cho 25, cho 8, NGUl'I]N DIJC'I.,iN ll
tdnh cho ctfic bgn churin br. tht vtio ilat taoc \Al2\ DE THI TUYEN SINH VAO DAI HOC '-----N ^T- . ,r v r6nc HqP HA Nfi ruAnrt tges (KHOI A) .A - Phdn b6t buQc : 1) Chrlng minh ring AB t CD vit EF ld drrdng vudng gdc chung cliua AB vn CD. Tinh Cd,u L X6t hdm sd : y = (, - t)z (x - a)2 (L) E[theo x, o., b. 1) Kh6o srit s1r bidn thi6n vd vE dd th! cria him fng vdia = 0. s6 (1) fDlhir" r dd hai mat ph&ng tACDl vdlBCDl *,uHg gcic vdi nhau. Chrlng minh ring khi dti 2) Xdc dlnh o d6' dd thi cta hdm sd (1) cd td diQn ABCD cd thd tlch l6n nhdt. didm crre dai. 3) Chrlng minh ring vdi moi o, dd th! hdm Bai giai s6 (1) lu6n lu6n cci truc ddi xrlng song song vdi tryc tung, Cd.uI:1)Khio=0thly = qx2 - x)z Cdu II. 1) Xdc dlnh nz dd cdc bdt phrrong trinh sau c
2) m ?xe sdi crla ra dtrong n6n dd thi hdm sd Cdu III. 1) Di6u ki|n: cos2x * 0, cos\x * 0 {.*) cri didrn cgc dai khi vd chi khi phrrong trinh Ta cci : !' = A cri 3 nghi6m phdn bi6t. StgZx - 4tg\x = tg23x. tg2x !' = 2 {x - 1)'(x - a) (2x - a - 1) *3 (tgZx - tgsx) = tgSx tl + tgsx " tSZx) (L) J' = 0 cd 3 nghiQrn phAn bi6t L*o lo cosx * 0) vi r'6u n$tqc Xai thi ti phtrong trirrh ( 1) la*l *a*1 ta suy ratg 2x: tg 3x "+ f, * tg2?x: O (v6 li). la+1 Ta cri : I - *1 l" - tp2x - ts?x = tg3r *-7t*r tg3x (I) e=+J = 3)Dat x=X+o!-1 iT t$rt-C,r; 2 *-stgx tlf* .. '{* e2tgx [r'- (+)1' " = 1 - 3tg2x (stg2x- J) = 0 !=YthiY= Ddy lA hdm s6 chSn n6n X = 0 li. truc ddi xring ftgx : 0 -hay r a*l .= | Itsx= *l u I-g - 2 li, truc ddi xrlng. Cd.u II : 1) Di6u kiQn dd cric bdt phrrong V=kn(ke4 trlnh cri nghiOm ld : c= | I-t [r=+arctS\l S+kr(keZ) )'[L',:2*nt>o ca hai hC nghiQm ndy ddu th6qm6n di6u ki6n (x) 1u',= 7-2m>oom
10 Cd,u : AD N€N AF I CD IV (b) : 7) Do AC Cd.u IV(a) :S =j I tgxldx = BC=BDn6nBF tCD 0.1 YQy CD t (ABD + CD -L AIi vA CD t EF 1110 DoBC =AC+CE rABmiAts tCD+ = -J ltgxlax= - ! tgxd.x+ ! tgxd, ABl-(CDE)*ABLEF. 0,1 0,1 I Y\ tgx = lge . lnx, ta chi cdn tlnh nguy6n Y$y EF ln drrdng vu6ng gcic chung c&a AB hdm cria him s6 f(x) - lnx.la c6 : vd CD. {x ln x)' = lnx * 1 vdy lnx - (r ln x - x)' Suy ra : 110 s = tse t*t", *., l- txtn& = -Ol) ),, =9,9-8,t.lge. fr =4, rt) =4, 2) a) Dat rt'=4 ) D, ,) + rinh ,, ,6{' - Atr - AE me 62 AF! -- ot -Z i rb .y2 62 x2 er2 =7 + EF)) = az) -Z -l +-+++ Ta cti : = .lfN = MB +BC + CN -4 '+ + r, = r{ae=V--::F = (a * 1)et -t h{B = (1 - B'B . --) -+ ")---t (0
Tqp chi Tll&TT s6 9, 1993 th6ng brio ring "dinh li ldn Ferm,at dd drroc chdng minh". Ti6p do, trong sd t, 1994, lai th6ng b6o ti6p v6 vi6c de ph6t hi6n ra m6t F ,g ch6 hcing lrln trong chfng minh dri n6n dinh li niy v€rn tidp tuc ld m6t thrich thrie r a d m iJ dfii vdi todn hoc. O sd g, 1993, t6c giA th6ng brio vidt ; "... Sau chtlng moi. rrid- hi f) md th6t bai trong vi6c tim iai chrtng *inn uo qu€n
YOi n = 5, chrlng minh dd dga vdo dinh Iy nhau n6n mu6n cho ab(a2 +b1ld m6t chinh phttong thi ob vd a2 +b? d6u phAi ld chinh sau day : Ndu x5 * 15 = z5 thi mQt trong c6c ptruong. Nhtrng o, b cring nguy6n td cirng nhau s6 r, y, z phli chia h6t cho 5. Dinh lf niy dtrgc ,r6n o, & d6u phAi li chinh phttong ; vfy 6t ph6i md rQng nhu sau : n6,t n ld m6t sd nguyOn td cd nhtrng sd nguy6n drrong r, .s, f sao eho : vd 2n * I cflng nguy6n td thi tit * +f = zn a=fr,b=s2,a2+b2=t2 s€ suy ra ho6c r, ho6cy, hoQic z chia hdt cho n' Dodd:r1 +sl=t2 Sophie Germ.ain cbn md rQng hon nta dinh ly D6n dAy, ta chri Y rang vi o, b nguY6n t6 tr6n vd dirng dfnh ly cria minh dd chrlng minh cirng nhau vA ch6n, 16 kh6c nhau n6n r, s ctng ring ndu n lA m6t sd nguyOn td nh6 hdn 100 vQy. M4t khde, trong lQp luQn d tr6n, ta chtta thl phrrong trinh (1) kh6ng cd nghi6m dang sr? dgng giA thidt dngf Id luy thrla b{c 4 md (xo, lo, zo) trong d6 xo, ! o, zrd6u kh6ng chia chi m6i stt dsng giA thi6t zltattty thia b4c 2' hdt cho n. Nhrrng nhd to6n hqc h)ng danh da dd c6ng Vi vQytoin bQ lAp luSn d tr6n ddi vdi ,1,y1 srlc ra mh trong khoAng mOt th6 kj', cring chi va @?)2 cri thd lap tei irodn toin d6i vdi d4t ddn nhtng kdt quA khi6m t6n nhrr v63'' 14 , s4 vd l. Nntrng 14 +s4 < xt+yt vi vd thrl Nrii "Khi6m tdn" vi ti nhtng kdt quA dri cho ddn ch5 chtlng minh trgn vgn (v6i mqi n) cbn nhdt bing a2 +b2 tfe li bing p md ld xa vdi. Hai chfr "thdt bai" n6u ra d sd 9, 1993 p < p2 * q2 : 2r. z', = 4 +fi = x! +y[. ctra TH&TT n6n hidu nhu thd. Trong thQp kf 40 cira thd ki XIX )t ki6n srl drlng sd phrlc dd R6t cu6e, tt ch6 giA thidt ring (1) cd nghiQm phdn tich duqc/' * jt rathiLnh tich cta z thrla (4o,!o, zo) khi n = 4, ta suy ra dttgc srJ tdn s6 tuydn tinh duoc d6 xudt vi thAo luAn. Tr6n tai crla r, s, / cflng th6a mdn (1) ( mi6n lh nhin co sE dr5, Kumm,er xdy dtrng n6n lf thuydt cira vd thrl hai chl nhrr mQt chinh phuong chrl minh. Day ln c6i m6c md tt d6 cdc nhi to6n kh6ng phAi la trrly thrla bdc 4), chl khac ta hgc chuy6n nghiQp hdu nhrr tin ch6c ring 14 + . *2 + y[. Lai tidp t$c l4p lu6n dci, kh6ng thd ed chrlng minh so c6p eho dinh ly , "4 l6n Fermat Ndi "hdu nhd" cci nghia Id cung ta sE di d6n / +s4 < 11 +s4 v.v... vh crl thd chda khang dlnh hin. Bdi 16 Fe.rmat da ghi mdi. Di6u dci li v6 lf trong phpm vi c6c s6 ring 6ng ta tim drrgc chrlng minh nhrlng ti6c nguy6n duong khi ta xudt ph6t til cric s6 ring 16 srich kh6ng dt ch6 dd ghi. VA chang, r,, lo htu han. V6Y ta cci : trong llch stt todn hqc, de tr)ng cri nhrlng bdi Dinh li Il : Phuong'trinh (L) kh6ng cd torin nhu bdi to6n Varing md ldi giAi so cdp nghiQmkhin=4 lai ddn muQn hon c6c ldi giai cao cdp : nanr EIO qu6 : phrrong trinh (1) kh6ng c
LTS : K{ su QUAN NGQC SON nd.m nay 2. Sd 987654321 61 tudi, kh6ng may bi bQnh, h6ng cd hai mdt, Day ld "sd l6n ngugc" cira s6 123456789 de dd. nghi huu tit nam 1982. Tuy d.d. "nhibu nd.nt dudc gi6i thi6u d sd brio Ul9g2. Crlng gidng sdng trong b6ng t6i hod.n tod.n", nhung 6ng nhrr s6 niy, sd 987654321 cd dac tinh trrong "ud.n gid.nh nhiDu thiti gian suy nghi, tim. tbi, phd.t hiQn nhibu quy luQt li thn cfia cd.c con tu nhung phfc tap hon. D4c tinh ndy drroc sd' (theo thu 6ng giti Tba soqn ngiry 291911993). di6n dat nhtt sau : Cd.m dQng tru6c td,m guong lao dQng,lbng Khi nhAn s6 987654321 vdi c6c sd nhdn nh6 ydu khoa hgc, uit. c6 thd n6i lit. yAu diti, yAu tudi hon 100 ddng thdi th6a m6n 2 di6u ki6n sau : tri cila kl su,.tba soqn xin gidi thi€u biti bdo a. 56 dd khOng phf,i ld 3 vd b6i s6 cria 3. thi hai crta ki su udi ban doc. Trong mQt sd b5o trudc, s6 U1992 tOi de b. Tdng cac chri sd cta sd dci nh6 hon 9. Thi kct cd dlp gi6i thi6u con sd ki la 123456789. Ki quA s6ld m6t sd cri mudi ho?" 11udi m6t chr/ s6. ndy xin dugc gidi thi6u hai con sd ki la khric Ndu ld mtrdi ch(I s6 thi dci sO ld cric cht sd cring cd thu6c t{nh trrong tu. tit 0 ddn 9, kh6ng m6t chrt sd nAo tring nhau" t. Sd 12345679. Ch6c nhi6u bpn doc da bidt t6i con sti ndy vi m6t dac tinh cring kh6 ki Ia Ndu ln mudi mOt chir sd thi tdt y6u phAi cri cria nri. t6i thidu hai chrr s6 trirng nhau. Ta goi day In Khi nhAn con sd tr6n v6i cdc s6 nhdn bing kdt qud thqc. Ta thrrc hi6n thuAt to6n don giAn 9 vd b6i s6 ctra g nh6 hdn 82, thi kdt quA s6 ld nhtl sau : dem s6 ddu cria kdt quA thuc cOng mQt sd cri chin cht sd gidng nhau. vdi s6 cudi ctia chinh nd ta drloc mrJdi chil sd Thi du : 12345679 x 9 : 111 111 111 nri ta goi ld kdt qud. dd. dibu chinh. Va khi dd 12345675 x 63 = 777 777 777 mddi chrl sd niy lai trd thinh mudi chil sd Tuy nhi6n con s6 tr6n cdn cd m6t dac tinh kh6c nhau til 0 ddn 9. thltc ki la khric nila rnd it ai ngd t6i. Dac tinh Vi du : 987654327 X 8 = 7901234568 kl la ndy duoc phrit bi6'u nhrr sau : 987654321 x 13 = 12839506173 (kdt quA Khi nhAn s612345679 v6i cric sd nhdn nh6 thtlc) hon 82 ddng thdi kh6ng phAi ld sd 3 vi bdi sd 2839506173 * 1 = 2839506174 (kdt quA ctia 3, thi kdt quA s6 li mOt sd cd 8 ho[c 9 chu di6u chinh) s6 md kh6ng mOt chrf s6 nio tritng nhau". 987654321 x 61 : 60246913581 (kdt quA thrtc) O aay cring xin mr& ngodc ncii th6m, sd di 0246913581 * 6: 0246913587 (kdt quA c
cHudNg rRiNH RUr GgN rirun s6 uEr FtBoNAccl Trong b5o THTT s6 6 - 93, muc Tin hoc, tdc 976 d6 n6u chuong trinh sau dAy dd tinh cdc sd hang cria sd liQt Fibonacci. 1 Gidi ddp bdi M+ + MRC M- Con sd 73 kI le - MRC M+ ( I GiA stl a vd b ld hai phdn cria mQt sd N ndo L4p lai hai ddng cudi dd md ld bOi s6 cria s6 n phAi llm. N = ffi . Gid s{, N = 100a * b (1). Ta binh phrrong (1) l6n M6i dbng cta chuong trinh ldn lrrgt cho cic dtac 10.000a2 + 2o0ab + b2 Q). sd hang clia s6 liQt Fibonacci : 1, 1, 2,3, 5, 8,.... (2) cring la b6i s6 cria n. Nhdn (1) v6i 200a dtrgc : Kdt thric m6i ddng ta cd mQt sd h4ng tr€n khung s6. Sd h4ng ldn nhdt thu dudc tron m{iy 20.000a2 + 200ab (3) ; (3) cfing ld boi s6 tinh b6 tfi th6 so li ur, : 63245986. cl8,a n. Ldy (3) trrf cho (2) ta c6 : 10.000a2 - b2 @); (4) ciing ln bqi sd cia n. B4n Ng6 Kim Anh, l6p 12 42, PTTH Vi6t Tri da goi y chrrong trinh ndy cti thd rrit gqn Bi6t ring ta phAi tim o vi b sao cho a2 +b2 6) vd dtta ra mQt chrrong trinh khde ng6n hon. la bQi s6 cria z. Md (4) + (5) = 10-001a2 fi ring li bQi s6 cliua n. VAy n li m6t udc s6 cira 10,007. Ba rtdc s6 cta 10.001Id : n : I (ldi 1 giAi tdm thudng) ; n = 73 (ldi giAi da bi6t) vA lu- n = 137 Odi giei cdn tim). l*MRC Ta thtl nghiQm : Ldy N = 325 x 137 = I lai 2 dbng cudi 44.525. Trncli o - 445, b = 25 c6 a2 + b2 = "ap 198.025 + 625 = 198650 = 1.450 x 137. Nhrrng tidc ring chuong trinh niy khOng eiNH pHrJoNc phn hop v6i m6t s6 mriy th6 so. Vi du m6y SANYO OX 110. Tuy v6y y ki6n cria b4n Ngd Kim Anh de girip tdc gSh cAi tidn chrrong trinh tr6n thdnh chrrong trlnh sau dAy : offnn DEN M+ DOm thu trlng s6ng, gid reo MRC Ph6 phrrdng nhOn nhip, dbn treo sring ngdi M+ Vui chdn dao d6m ddn choi L4p lai hai ddng cu6i Hon ba trdm nggn, h6i ngtliri cd haY ? { ,tF K6t n6m trdn sd ddn niry r Chuong trinh'dE drrgc nit gqn ro rQt, kdt { Bdy dbn kdt mQt, thi hai ngqn thia thric m5i dbng ta vin drrqc mQt sd h4ng crla ]i Kdt chin cbn bdn nggn du i sd li6t. MOt chi ti6t nh6 ld : sd hang Idn nhdt Ngdn ngo em ddm tlt mil dEn sao thu dugc bay gid kh6ng phAi ld u, mi lA HOi ngrtdi tri sring, tii cao u:s : 39088169 Dtng chdn t{nh girip cri bao nhi6u dbn ? Tdc gSn hoan nghOnh vA ch6n thdnh crim NcuYEN oiNll ruNc on ban NgO Kim Anh dd e6p !. NGUYEN DONG ISSN: 0866 - 8035. Chi sd 12884 Gi6: 1200d M6 sd : 8BT04M4 In tai Xtr&ng Ch6bAn in Nhd xudt bAn GiSo duc. MQt nghln In xong vi grli lrru chidu thdng 4 11994 hai trdm tlbng