BO G|AO DUC VA OAO TAO * HOI TOAN HOC VIET NAM , itffii ',W #+ Hinh I Hinh 2 qidi itiP ' TiT,,{ EUIO ['IG TtsONq Mf, EUJNG Tr€n hinh 2 c6 Z duong di (mhu lam) tt ngohi vdo bui cAy, chring chi khd,c nhau 6 doan dudng di til A ddn B' f en ttint 3 chi c6 1 duong di (mhu lam) ttr didm 1 ddn didm 2 Dd tim du o c ducrng di ng6n nhat cdn thuc hi0n cdc nguy€n dc' chung sau dAY 1 KhOng di lap lai dudng dd di qua 2 Ndu g4p ng6 cut thi b6 doan dulng til ng6 cut ddn ngd ba, ngd tu, , ddu ti6n J Ngu c6 duirn[.]
Trang 1BO G|AO DUC VA OAO TAO * HOI TOAN HOC VIET NAM
Trang 2nguy€n dc' chung sau dAY :
1 KhOng di lap lai dudng dd di qua
2 Ndu g4p ng6 cut thi b6 doan dulng til ng6 cut ddn ngd ba, ngd tu, , ddu ti6n
J Ngu c6 duirng kh6p kin md c6 2 dudng khdc.ctng
gdp duong khdp kin o I aidm th'i b6 du]ng.khep \i' {6' Iai, ndu JO z ai rng kh6c gap dudng kh6p k(n o 2 diem
ptran Uigt A, B thi"ta b6 toan duong AB dei hon cua
dudng kh6p kin
+ fni tim duo c tdt ca cac dudng di ddn dfch cAn so
s6nh dd chon ra duong di ng6n nhAt.
N6u c6 so dd m0 cung ta td mdu cdc doan dttdng bd
di theo c6c nguydn thc 2 vd 3 thi so dd m0 cung cbn- lai sE
don giin hon-ntri6u, tir d6 dc tim ra ducrng di tdi dich'
0l/0/t/q 0/ C[i4 Ql/i/t/ lt'li
RE/t/ Sirtt CO
TrCn mQt bln cd kich thu6c 4 x 200i 0 vu6ng' mQt quAn.m6
"6^;hd ;i';6i budc ti 6 vuong nhy den 6 vu6ng kh6c,theo
;rrdrn; ch6o cira hinh chfr nhat kich thudc 2 x 3 O (hodc 3 x 2
6) nhu hinh 1.
Ddnh cho bon dgc
1) Hdv chi ra ciich qudn md bu6c li€n tie'pJil vi-trf xuat phi{t
duoc ch'on di qua tat i6 cdc 6 cua bhn cd, moi o dung mot lan'
2) Cau trA ldi trOn c6 dting vdi vi trf xuAt ph6t khdc khOng ?
Tang phdm dinh cho lbri giAi ddy dir, giii thich 16 rhng'
N6u khOng c6 so dd m€ cung, trlc li phAi di trong
m0 cung thuc, cdn phAi d6nh dd:u c6c ng6 rE (chdng hanJau * ie di duo.", ddu - li loai b6) khi thuc hi6n-cdcnsuven t6c tren Truong hqp niy tdt c6 ich khi sri dung
o1t'tdc' tay trdi, tric li khi di tay tr6i ludn cham tulng m0
"urg, quy-t6c nhy girip ta di
lAn luot het cdc ngh dudng'
Nf,iOu ban tham gia gi6i bii niy n]rung kh6ng giAiHinh 3 tn?ctr troac chi dua ra-l duong di 6 hinh 2'
C6c ban sau c6 ldi gi6i ddy dt ducv c nhAn tang phdm :
l.Trdn Binh Minh,l1A1, THPI chuy€n Nguyen TAt Thhnh' Y€n Bdi ;
2)'Nguydn Huy Hing,gA, THCS Lap Tryh,VInh
Phtic-z\ t't"stiyl" uiin o*i,x6m2K8 thitran Thanh Chuong' NghC An ;
+jf iaiTrung Ki€n, L0 To6n, THPT chuycn Th6i Nguy0n'
pHI pHI
Trang 3Todn hoc vd Tudi trd
Ndm thu39s6 2e5 (1-2002) Tda soqn : 1878, phd Gidng V6, He Nfli
DT : 04.5142648 - 04.5142650 FAX : 04.5142648
Email : toantt@hotmail.com
O Dirnh cho Trung hoc co sd - For Lower @
Secondary Schools
Hod.ng Vdn Dd.c - odi uidn dd chfng minh
bdt ding thrlc c6 didu ki€n
O Ti6ng Anh qua c6c bii todn - English
through Math Problems
NgO ViQt Trung - Bai s6'a9
Nguy|n Minh Hd - Mgt sdbai to6n rlng
dung cria tich ngoii cria hai vecto
Nhin ra thd gidi - Around the World
D0' thi Oiympic To6n mt6c Anh n[m
2001
Chudn bi thi vio dai hqc - University
Entrance Preparation
Phq.m Vd,n Thu6.n - Phrrong phAp dai
s6''h6a dd chrlng minh bdtt&ng-thiic
trong tam giAc.
DO'thi tuydn sinh m6n To6n trudng Dai
hoc Thriy sin nim 2001
Dd ra ki niry - Problems in this Issue
@
@
wcuvBN cANH ToAN, HoANb csUttc, NcO DAT rrI lE xtrAc eAo, NduyEN HUy DoAN, r.rcuvErv vrBrHAt, otrytr QUANc HAo, NouvEn xuAN nw, pHeN nw EHAr, w rHar.rr rnlfr,lE sAr tcH0r, xcwfo
vAu-rraAu, noANc r-Er,,rrun NcuyEF{ KHAa rt{rNH, rRAN vAN NHLTNG, r.rcuy6x oAxc puAr, pHAN THANH
queuc, ia Hdnc quAnc, oAwc stmc nr$lc, vU ow,rc rHW, rnAN rHAnH rnai, r-E eA KHANH Trudng hqn hitn tdp : NGUYEN vE.r ufu inu 4 tai *qn, it ry6Nc xvir.Th,t, hien : vtt xrrt,t rnOv
T ri su : vu eNn TtrU rnfNn ruvdr rnexc i rtin my, NcuvEN rHI oellH
Dai di€n phia Nam : rxlN Cni sdu,2S t NeuydnVdn C&,TP Hd Chi Minh W : 08.8323044.
Di6n tlin day hoc to6nForum
LA Thdng Nhdt - Lai
tidp xric cria hai dd thi
CAu lac b0 - Math ClubSai ldm & tIAu? Where's the Mistakes?
Bia 2:Todn hoc mu6n miu -Multifarious
Mathematics gudng di cOa quAn md tr6n bAn cdBia 3 : Giii tri todn hoc - Math Recreation
Bia 4 :.tl6i nghi C6ng tZrc vi6n vd Mrlng tho
GS Nguy6n C6nh Todn 75 tudi
Trang 4X6t bhi to6n chimg minh bdt ding thric
2x2 + xy+ 5 < 9 vdi didu ki6n 3x + Y = 4.
Bii niy c6 thd dua vd vi6c tim gi6 tn lon nhdt
cira mdt bidu thrlc hai bidn P(x, y) = 2x' + xy +
5 v6ri didu ki6n 3x + y = 4, chi kh6c li bhi to6n
ban ddu cho bidt th6m m6t h[ng sd lon hon
ho[c bang gi6tri l6n nhAt cira bidu thrlc P(x, y)
Dd gi6i bii niy ta thay thd y = 3 - x vio bidu
thfc F(x, y) vh duo c tam thft bac hai cira m6t
#; i;; '='
-t * 4x + 5 = -(n - 2.):2 + 9 Dhng
thfc x6y rakhi x = 2.
Trong trudng hop thay bidn d6n ddn hlm sd
T(x) c6 bac lEn tiim Z -horc
hbm s6 c6 nhidu
bidn thi vicc tim gi6 tri l6n nh{t, nh6 nhdt rdt
kh6 khAn v6i kidn thrlc toi{n b6c trung hoc co
s& Ching han :
Bni 1 Chung minh rdng ndu a + b = 4 thi
o4+b4>32
Nhdn x6t rdng m6t bidu thrlc nhidu bidn
thudng dat gi6 trl nh6 nhat (hoac lon nhdt) khi
m6t # bien" c6 gi6 trf bang 0 hoac tat cA cdc
bidn c6 gir{ tri bang nhau Di€u n}y goi ! cho ta
cdch ddi bidn nhu sau.
Ldi giii Do a + b = 4 n6n c6 thd dAt
MOH BTNN
o
, !, ,, r A a
BAT OANC TFIUC CO DIIU IflTN
HOANG VAN OAC
(6/ T HCS T dn V i€t, Binh Giang,
Hdi Duong)
= (1+a)2 + (l+b)z + (l-a-b)z + (1+aX1+b) +
(l+h)(l-a-b) + (l-a-b)( 1 +a)
Yqv t + y2 + 12 + ry + yz + zx> 6Bei 3 Arc a+b+c+d = 1 Chfng minh r6ng :
Thay vio vdtrdi BDT duo.c :
Trang 5Bei 4 Cho a * b = c + d Chung minh ring
,2+*+cd>3ah
Ldi giii Do a+b = c*d ndn ta dat c = a + x vd
d = b-x v6i x tiry y.Tac6
ding thfc nhidu bidn kEm theo didu ki6n cira
c6c bidn li kh6o 16o ddi bidn dd c6 thd dua vi6c
x6t m6t bidu thrlc phrlc tap vd m6t bidu thrlc
quen thu6c, don gi6n hon vi phi hqp vdi kidn
thrlc bAc trung hoc co s6 C6c ban hdy tht stt
dung phucrng ph6p ddi bidn dd giAi c6c bli tap
du6i dAy.
Bni 1 Cho c6c s6 duong x, y thba mdn x+y=1
Tim gid tri nh6 nhdt cta bidu thrlc
c- I - 3
"- x\y2- 4*yBni 2 Cho x + y + z = 3 Tim gi6 tri nh6 nhdt
rang:
27a?+tob3>945
13
24 31 42
T can be divided
into 4 pairs, and
only one square ofeach pair can be occupied by a knight As aconsequence, no more than 4 knights can bearranged in each of these rectangles such that
none of them lies on a square controlled by
another Therefore, the total number of knightswhich cap be arranged in such a way on the chessboard it at most 4xB = 32.
control = kidm so6t, khdng chd(dQng tn)
obvious = hidn nhi6n (tfnh tt)
suffice = chi cdn (d6ng tt)
purpose = mUc dfch, f dinh
divide = chia c6t, phen ra (d6ng tt) rectangular - vu6ng g6c (tinh tt)
section = phdn, bQ phan
wide = rQng
situate = du-d c dat (d6ng tt)
occuPy = chidm, cu ngu (dQng tn)consequence = hg quA, hAu qui
rectangle = hinh chf nhflt
NGO VIET TRUNG
nAr sd ngProblem What is the greatest number ofknights which can be arranged on an 8x8
chessboard in such a way that none of them lies
on a square controlled by another ?
Solution Since a knight on a white squarecontrols only black squares, it is obvious that 32knights can be arranged in such a way that none
lies on a square controlled b5, another To do
this, we only need to put a knight on each whitesquare Now, it suffices to show that anarrangement using more than 32 knights is notpossible For this purpose, let us divide the chessboard into 8 rectarigular sections, each two
lquares wide and four squares high:
Trang 6mor sti mfu roAru tlnc DUNG cua
,\t
TICH NGOAI CUA HAI VECTO
Trong THTI s6 294 (1212001) d6 gi6i thi6u
vd tich ngohi cfra hai vecto vi mdt sd tfch ch6t
cira n6 Nttng tinh chdt d6 s6 duo.c sir dung dd
giii nhidu bii to6n hinh hoc dudi diy Ctui y
iang tich ngoii ctra hai vecto cirng phuong biog
sd 0.
Bii niy xin gi6i thi6u v6i ban doc nhfrng ki
thuAt quan trqrE nhdt dd giii -or Uii to6n^ bing
tfch ngoii Nhtng ki thudt niy duo c thd hiOn
qua c6c vf du sau.
Vi du lz Cho tt gidc l6i ABCD c6 AD cdt BC
tai E Goi I, J ldn luqt ld trung didm cfia AC,
BD Chfing minh rdng: S(EIJ) =)S1eaCOl
huong (mQt ket qu6 kh6ng d6 chung minh) Ndu
khdng s& dung tich ngohi thi ph6p chtrng minh
r6t d6 phu thu6c vio hinh v6 do d6 phii x6t qu6
nhidu truirng hgrp.
Vi dg 2; Cho tam gidc ABC Cdc didm A', B',
C' theo thil tu thudc c'dc' duong thdng BC, CA,
AB A",8", C" theo th* tt ld trung didm cfia cdc
doan AA', BB', CC' Chilng minh rdng:
TU d6 suy ra: S(A"B"C") = | S1e'n'C')
Vi dU 3 Cho tam gidc ABC Qua didm M hdt
Tt d6 suy ra: S(EIJ) = | slencol.
NhQn xit: Ld\ giii sir dgng tich ngoii kh6ng
1
nhfrng cho ta bidt S(EIJ) = a S(ABCD) m) cdn
cho ta bidt tam gi6c EIJ vi tir gi6c ABCD ctng
4
Trang 7slMBCl , stMCAl , slMABl _^
:-:-L.-MAr MBr MCt
Gidi: Goi d ld, vlc to chi phuong don vi cira
A (Ban doc tu vE hinh) Ta c6:
Vi dq 4z Cho luc gidc' l6i ABCDEF Goi M,
N, P, Q, R, S theo tht fir ld trung didm cila cdc
canh AB, DE, CD, FA, EF, BC Ch*ng minh
rdng: MN, PQ, RS ddng quy khi vd chi khi
S(AEC) = S(BFD)
Gidi: Ban doc tu vE hinh
Lay didm O bdt ki, ta thdy:
"oE + oE nOfr * OE "oE
+ Oe nOF +od nO1-,+Ob nOe +Oo nOA
+ oE n-oi * oE noe + oF
"oE + oF "oe
= (OA n OE + OE
" OC + Od n
OA) (oB noF +oF no6*oD noi)
= Z(SIOAE] + SIOEC) + SIOCAI) - 2 (SlOBFl
Vi dq 5: Cho tam gidc ABC vd didm M khdc'
A, B, C Cdc duong thdng qua M,ldn laot vubngg6c'voi MA, MB, MC, cdt cdc dudng thdng BC,
CA, AB tai A1,81,C1 Chfing minh rdng A,,8,,
C, rhdng hting
Gidi: Ban doc tu v6 hinh Ta c6:
E=srMAtBl AtC SIMAPI =yg!!g
C6 thd xiry-ra m6t trong hai trudng hgp g6c
(MAr, MA) = 90o (mod 360o) hodc
<Uer,lrtel = -90o (mod 360o) Tuy nhi€n,trong ci hai trulng hgp ta ddu c6 :
'i"(MAt , MB) _"""(ua , un) ()\
sin(rvr,a1 , Mq-
"or(rrZe MC) \'t
Tt (1) vi (2) suy ra:
MB.cos(MA,MB)A,B
(3)ArC MC.cos(MA,MC)
Tuong tu nhu vay:
uc.""t(ue, t ta)
MA.cos(rr,*)
ua.cos(ue ,tute) ua.""t(tutc,tuta)
NhAn theo trrng vd c6c dang thric (3) (4) (5),
theo dinh lf M€n6laulit thi ba didm A,, B,, C,thing hIng
Vi dq 6z Cho tam gidc' ABC -Gid sft M, N , Ptheo thfi til thubc cdc dqdng thdng BC, CA, AB,
Trang 8chftng minh rdng: AM, BN, CP ddng quy hodc
ddi mdt song song khi vd chi khi:
sin(AM,AC) sin(BN,BA) sin(CP,CB)
Giii: Tlteo dinh li Xdva ta thdy: AM, BN, CP
ddng quy trolc d6i mOt song song
Nhdn xit: Kdt quA tr€n rdt c6 loi trong viOc
chfng minh su ddng quy cira ba ducrng thing
N6 du-o.c goi li dinh lf X6va dang sin vh duo.c
chrlng minh m6t c6ch chit ch6 nhu tr6n
Vi dB 7: Cho tam gidc ABC.Voi mdi dtdm M
c'hilng minh rdng:
sD,Bqffi +sMCAlffi +sMAslMd = d.
Gidi: Dat i = SIMBCI MA + SIMCAI ffi
Ch,i i ring, trong ba v6c to MA, MB, MC
ta lu6n chon duoc hai vdcto kh6ng cing
BAi 2 Dudng trdn (.9) nAm rong duong trbn (I)
vi tidp xric (Q tai A Tt m6t didm P k}rdc A tr6n(7) v6 hai dAy cung PQ vdPR cria (I) sao cho hai
dAy ndy tidp xric (,9 theo thf tu tai Xvd L Churrg
minh r[ng iU Oenbang hai ldn g6cMy.Bei 3 Quan cd tctr6min6 li m6t hinh gdm 4
hinh wdng don vi duo.c gh6p v6i nhau ba cdc
canh chung
" !) Hai quAn cd dud.c xem li m6t ndu chring li
Anh cia nhau qua m6t ph6p quay trong iratphang Chung minh rdng c6 7 loai quAn cd
t€tr6min6
WWW WW
WW-WTru
IWWWW
2) MQnh dd sau dring hay sai, hdy chtrng
minh: C6 thd xdp 7 quan- cd ietromin6'd6i m6i
kh6c loai vio m6r hinh cht nhAt kich thudc 4 x
7 sao cho kh6ng c6 hai hinh vu6ng nio chdngl€n nhau.
Bei 4 Ddy sd (a,) dury.c x6c dinh b&i :
an=ft+lJ;l v6i n nguydn duong va { Jn } u
sd nguy6n ge" J; nhdt Xdc dinh sd nguyCn
duong k bd nhdt sao cho ciic sd hqng : a1r, a1r,a1i
ak+2000 lAp thinh m6i diy 2001 sd tu nhi6n
li6n tidp
Bei 5 a, b, c vI R li ba canh vi b6n kinhdqolg trdn ngoai tidp cira mdt tam gi6c Chfngminh ring tam gi6c li vu6ng khi vi chi khi :
Ngrrdi gid thiau : tnAN ANH DONG(w r H Pr
""J!{,!ff,s r h€' v i nh'
Trang 9PHTJONG PTIAP DAI Sd HOA
PHAM vAN rHUAN(S/ K324, DH Ngoai ngfr, DHQG Hd Noi)Ldi eiii cho vi6c chrirrg minh ciic BDT trong
am giic c5 thd tr& non Eqn dep nhd viec bid[
di0n sina, cosa vi tga ' theo t = E1 "2 c6 rong
SGK lcrp 11.
Ndutaquyudc ,=q,,4,y=W! "2' "2 ,r=l 2"(r,
y, z ddu duong) thdng nhdt cho tat ci c6c bii
to6n minh hoa du6i dAy thi nhidu hO thrlc luong
gi6c trb thinh hd thrlc dai sd vi b)i todn BDT
luong gidc c5 thd chuydn vd BDT dai sd vi c6c
BDT dai s6 c6 thd g-o i f cho ta hu6ng chung
minh, ngodi ra vi6c trinh biy bi6n ddi dai sd dd
phrlc tap hon bidn ddi luong gi6c
C6c kdt qui sau day li rdt quen thu6c (ban
doc tu chttng minh) vi duo c s& dung nhidu
trong bii vidt niy
Yor A, B, C ld cdc g6c cira tam gi6c ABC th\ :
( cosA/[ cosB/( cosC )
Lli giii V6i c6ch quy u6c trdn vi do tam
gi6cABC nhonn6n x,!,2 ddunh6hon l,tac6:
Ding thrlc & (3) xAy ro (1 I = ! = z e tam
gi6c ABC ddu.
Nhdn xdt R5 rlng ldi gi6i trdn c6 uu didm
hon cdch ldm thuong thay li khai tridn yd trdicira(3) d6n d6n viOc phAi tim giri tri nh6 nhAt
cia cdc bidu thrlc thlnh phdn khr{ phric t4p.
Vi dU 2 Chtlng minh rdng trong moi tam giitt'ABC ta ludn cd
111
(4)e-+-+->3(x+y+z) (5)
xyz (t I l)2 3(x+v+z\
\x y z) xyz
*( t-1.j' *[,r-r)' *( !-1)', o
\x y) [y z) (z x)
Ding thfc xhy rakhi x = ! = z >AABC ddu.
Vi du 3.Voi A, B, C ld c'dc g6t' t'ila milt tam
gidc ABC, chftng minh:
( t l l \f A B c\ l_+_+_ ll cots_+cots_+cote_ |
\sinA sinB sinC/\ "2 "2 "2)
Trang 10111 * _) "A .B ,C
phii bidn ddi tucrng duong Ching han :
Vi dU 4 Chilng minh rdng voi moi tam gidc
_f _ l\
a y' 1 1 I \ry yz zx)
t-) x2 +y2 +22 -(xy+yz+zx), (t r)' (t r)2 (r r)2 (x y) (y z) (z x)
Phri Chidn vi bii T81286 tr6n THTT 412001.Cudi cing mdi c6c ban llm th6m c6c bdi tap
sau, trong d6 dd-vidt ggn ta ki hi0u A, = +, n,=!-c,=9 ' 2' 2
Bei l Cho tam gi6c ABC Chung minh ring:
Trang 111) cotgA+cot gp+cotgC + cotgA lcotgBlcotgCl
.z( ' * I * ', ')
(sin A sinB sinC/
2) cotgA, + cotg8l + cotgCl >
> Z(cotgA + cotgB + cotgC) + rE
3) cotgA + cotgB + cotgC + 3.6 <
\sinA sinB ,r-J(*sa +ccEq +oa$r)
> z(antgqcotfpt + cotgft cotgC, + mtgC,cotg,ar )
Bdi 2 Cho tam gi6c ABC nhon Chrlng minh:
nl,r+ ]-)(r*'')[,*' )
=
\, cosA][ cosA/( cosC )
> (cotgA, + cotgB, + cotgCl )2
\ (tgz lrtgz B, + tgz Brtgz c, + tg2 crtg2 Ar)
"
" (teA+ tg8 + tBC)' >
> cotg2 A1 + cotg2Bl + cotgzCl
Bni 3 Cho tam gi6c ABC, tim gi6 tri nh6 nhdt
cira :
1)P= cotg,41+ cotgB, + cotgCl
2\ Q= sinA sinB -*- sinC
i/,#, *tgB, +tgC,
Bni 4 Cho tam gi6c ABC nhon, tim gi6 tri
nh6 nhAt c[ra bidu thrtc :
_+_+_
sinA sinB sinC
tEA+tgB +cotgfi + mtgfi cotgct + cotgCrcotg.Al
ClAt gAl XI TRU$C gidp trang2o)Mdc th€m I di€n trd R = I (Q ldn luot vdo ;
.2 didm AB, thdy G1 chi 0,05 (A)
2 didm BD, thdy Gz, Gs chi 0,05 (A)
.2 didm CD , thdy G j chi 0,05 (A)
.2 didm AC, thdy G1, Gz chi 0,05 (A)
.2 didm BC, thdy G2 chi 0,05 (A)Hdy tlnh cdc' di€n trd r, g1, Sz, Bs.
Ldi giii Cudng dO dbng diOn mach chinh :
I = E- Khi m6c R = 1(o) vdo 2 didm AB,r+R,
Ki hi€u 1 ln so chi cira Gr, ta c6 Ir= L l
Bz(&- 8e) +r(gr -8:)=0=9, =9, Q)
Tt (3) vn (6) suy ra : Bt= Bz (8) Tt (3), (4),(f vn (8) tim duoc : r = 0 ; Br = Sz= 8: = 4 (O)
NhAn x6t Ta thay dt ki€n mic R vtro AC (hoac BD)
THPT Hdng Linh ; Vinh Phric : KimThanhThiy, Chu Anh Dfing, llA3,TrdnTudn Dfing, Nguyin Ndng An, 12A3, THPT chuyOn Vinh Phric.
MAI ANH
Trang 12TRUSITG DAI IroC TITUY S,i:u ruAm zoor
D Giei phuong triirh :
CAU I 1) Ban doc tU gi6i
* Tt hinh vE ta thdy duong A c6t dd thi y =
/(lxl) tai bdn didm phin biQt k$i vi chi khi :
kue<k.k**
10
trong d6 : kro ld, hO sd g6c cria dudng thing
MA v6i A(0, -2) n€n kro =
fi: t
kMN ld, h0 sd g6c cira dudng thing MN du-o.c
x6c dinh khi A tidp xric vdi nhdnh tr6i cira dd
thi hem y = f(lxl) tai didm N (xo; )o) v6i
x^ " < 0 , [-*3 l-34 +3xo +3= -2= ky,y kmwGo -Z)
e (lJx+t -tl-t) -o e lJx+t-,1=,
Giii ra d6n ddn kdt luAn : Nghi€m cua (l) ln
x=-lhodc x=3.
^2.
CAu I 1) Kh6o s6t him sd: y = (x + l)2(x-2)
2) Cho dudng th&ng A di qua didm M (2; 0) vn
c6 hC sd g6c li k Hdy x6c dinh tdt ciL clc gi6 tri
cria k dd duong thing A cit dd thi cua him s6
sau tai bon didm phdn biet :
UU6NC DAN GIAI
r/tloez 'l2x + log J2x)logrxz +
Cf,u lV :
Cho trl diQn SPQR v6i SP L SQ, SO -L SR,
Sft I SP Goi A, B, C theo thrl tu li trung didm
Trang 132) Phuong trinh c6 nghi€m , = -1+Zkn
Tucmg t{ : SB = CA, SC = AB
3 Ndu a <2hodc a> 4: (1) v0 nghiem
Cdch 2 Khi a > 0 phuong trinh (1) c6 didu
ki€n0 < 2x <a o O.L =l dAtZ =dcoSg,
Hudng ddn gidi :
MAITHANG(@ DH Thiy Srin Nha n'ang)
,"r,1;rta-"t)
Trang 14cAc l6p rHCS
Bdi Tll295 Tim biy sd nguydn td sao cho
tich cira chring bang tdng cdc liy thta bdc s6u
nhat vi sd l6m nhdt trong n sd duong xt, x2, .,
xn (n > 2) Chung minh rdng :
(x, + x" -r- ' \2
4 q \^l ' ^l ' "'nn/ a')g
x, + 2x2 * .! n*n*
BUI THE HUNG
(S/ K34B khoaTodn - DHSP Thdi Nguy€n)
BiLiT4l295 Cho luc gidc l6i A1A24A44A6
c6 cic canh ddi di6n song song v6i nhau Goi
Br, Bz, A, tdn luot'li gia6 aid;r cira rirng c6p
duong ch6o A,Aa vd A,Ar, ArA,vit A3A*, A.rAuvd,
A,Ao Goi C1,Cz, C3 ldn luot li trung didm cira
cdc doan thing ArAu, AtA4, AA,s Chrlng minh
rang c6c duorng thing 8,C,, BrC.r, BrC, d6ng
quy
'EN,?flLSI-j)"
Bni T5/295 Cho tam gi6c ABC (AB < AC s
BC) Tim qu! tfch nhfrng didm M nam fong
tam gi6c sao cho tdng cdc khoing crich tt didm
M d€n ba canh cira tam gi6c lu6n bing m6t
TNANTUANANH
(tr khoaTodn -Tin20A0, DHKHTN - DHQG Hd N6i )
BdiT7l295 H6i c6 tdt cabao nhi6u da thrlc
P, (r) bac n chdn th6a mdn c6c didu ki6n :
NGITYEN VI6T LONG
(GV THPT Lam Son,Thanh H6a)
Bei T8/295 Tim tAt ch cdc him sd/ : D -+ Dtrong d5 D = ll, +o) th6a mdn didu ki0nf(*.f(y)) = y.f(x) v6i moi x, y thuoc D
VU THI HUE PHT'ONG
(SV K33B khoaTodn, DHSP Thdi Nguyln)
Bifi T91295 Goi S, R, r ldn lugt ld di€n tich,b6n kinh dubng trbn ngoai tidp vi b5n kinhduong trbn nQi tidp ctra tam gi6c ABC DAt a =
BC, b = AC, c = AB Chrtng minh rdng
-25AB
- R22< (b +('-a)sln-+(a
-h+t')sln-+(a + h- c )sin9< S
2r
M6i ding thrlc xiy ra khi nio ?
Lt]U XUAN TiNH
(@ THPT Lam Son,Thanh H6a)
Bei T10/295 Cho trl dien ABCD ndi tidp m6tmdt cdu tam O bdn kinh R Hdy x6c dinh vi tri dilm M trOn m6t cdu sao cho tdng ci{c binhphucrng c6c khoang c6ch til didm M t6i c6c dinhcira trl di6n dat gi6 tri lon nh{t, nh6 nhdr
DO BA CHU (GV THPT Ddng Htrng.Jld,Thdi Binh)
cAc oi v4r r,Y
Bii Lllzgs MOt elecffon dang chuydn d6ng
v6i v1n tdc vo = '105 m/s thi ba! vio virng'c3
di6n trudng ddu vi tt
trudng ddu ; vecto v6n t6c 1 E
C tE ua [rE (rrintr
@B
v6)
X6c dinh dO lon vAn t6c
cira dlectron & thli didm
vo