Tạp chí toán học và tuổi trẻ số 298 tháng 4 năm 2002

Mffi*emeffi w&e*w$ee

e

H&eTffiH.[**r!ffi vd&s H*!t{;

Vn trildng ThUy
vi mOJ sO bi€n tip vi€n
ban Toin trong ngf,y ki ni€m
bio Toin hoc Tudi tre
PGS.

ffi

f,.'l*
*" *
';
*iq*
tr*[ f

€
ir

.l

i=i
!!h i+
t,,',,,8.::.:*

Ban
,i

bi0n

tipTri{n

xxn Ciio srd

,s',',
€ -J. r

&w


ei,ffi.,iffiffi ffi'.
"

€

*#,a#

,#v, i. j
s ,.ffiB.H
,j':."

,
c'

th:;

r^r=

f

i

TN-GI NR AS #-N-g 7N$NG
-I

Trong mAnh vudn cira m6t bAc nOng dAn

"s-

c6l

cdi

ao vi 4 cAy luu niOn A, B, C, D. B6c nhd 3 ban
Dinh, Khio, Nghi0n thidt kd ao mdi hinh vu6ng mh
vAn gifr lar 4 clry, trong d6 c6 it nhat 3 cdy trOn bd ao. (Tr€n bAn vd cAy
nim trOn canh hinh vu6ng vi chua t(nh dOn bd ao).
Ban Dinh thict kd ngay ao hinh vu6ng c6 1 canh di qua 2 cary A, D
nhu hinh 1. Bilc n6ng dAn chua thAt vita f .

; +

Ddnh cho bon doc
. Kh6o thiCt k0 : cdy A lh dinh, cbn 2 c0y B, C nlm tr0n 2 canh ao.
. NghiOn thidt kd: m5i cAy nam trcn m6t canh ao.

C

C6c ban h6y trinh bhy thidt kd cta KhAo, NghiOn cing v6i ciich vE vd
gtri vd Tda soan s6rn dd nhAn tang phdm.
!^/

Hinh

Gi6i tl6p : SAP XEP DEN TRANG TRI
Dd tien lip luAn ta ghi s6 vi tri cdc dbn

I

nhu hinh 2 vb,vi6t t6t ld d6 (D) ; vdng (Y), lam (L), tim (T). Khi quay hinh
di mot boi ctra 90" rhi vi tri c6c dinh (1 2 3 4) chuydn thinh c6c vi tri
(2 3 4 l). (3 4 I 2). (4 12 3) nhung vin Ih I c6ch sip xdp mhu dEn n€n ta chi
nOu ra cdc citch sip xdp mdu ddn b (I2 3 4) le dt.
X6t 3 trudng hop :

t) 6

trinfr vu6ng

(12

(D LT)
(DLT )

3 4) c6 dit 4 mdu thi

xiy

(D T L) (DL
(DTL ) (DT

O m6i trudng hop tron c6 2 cdchxdp

ra 6 trudng hgp

T)
L)

mlu nhu 6 hinh

6 trinn vu6ng (12 3 4) c6 dring 3 miu thi

D DL) (D DT)

L r) ( r- D)
LrLD) (lrl )
rDr ) (rDrL)

Hinh 2

xhy ra 12 trudng hqp

Bdn dd

:

u t.-]}

(oLDr)

rG-

(T D)
(LDL )
(T T L)

4 cdchxep miu nhu 6 hinh 4.
V0yc6 12x4=48c6chxOp.
:) 6 trinfr vuong (12 3 4) c6 dring 2 miu thi xiy ra 6 trudng hqp
m6i trudng ho. p tren c6 4
miu nhu & hinh 5. '

VAyc66x4=24cdchxlp.
Tdngc6ng c6:12+48+24=
84 cdch s6p xdp mhu ddn.
Hai ban c6lAp luAn dring duoc
nhAn

rG-

Hinh 3

sau

:

(s D ) (oLDL) (or or) ( t- L) ( r r)

c6ch xdp

ting phdm

li

4

(t"t

(

3.

O m6i trudng hqp trOn c6

6

,w

{

:

VAyc6 6x2=12cdchxdp.
Z)

rv\.

EllE-t

LT LT)

(F-t

Hinh 4

:

l)

Nguy€nVdn Hodng,
THP| Hoang H6a II,
Thanh H6a.
2) Luong Ti0n D[ng,
12A, THFrI NK Ngo Si Li0n,
Bdc Giang.
PHI PHI
11A,

.Ei

^q|(p €-t
Hinh 5


Nam tht?39

Todn hoc vd Tudi trd

Mathematics and Youth

Toa soqn

:

sd

2s8 (4-2002)
1878, phd Gidng V6, Hd Nfli

DT-Fax:04.5144272

Email :

THU TUONG CHINH PHU TA\G BANGKHEN
cHo raP cHl T0AN Flgc vA ruOl TRE
Ngdy 25.2.2002 Ph6 Thu tu6ng Nguy0n Tan Dfrng thay mit Thir tuong Chfnh phu da ki Qtryet
cho tap chi Toiin hoc vh Tudi tr6 dd cd nhidu
thlLnh tich trong cong tlrc, g6p phdn vio su nghi6p xay dqmg Chtr nghia xd h6i vh bio vC Td qudc.
Day lh mOt cOr mdc dr{nh ddu nhirng th}nh tich ctra c6n b6, cOng nh6n vi6n t4p chf trong nhfrng
nam gdn day da lien tuc a(Trorig c6c nam 1984 vi 1995 tap chi cfrng dd duoc nhAn 2 HuAn chuong Lao dong hang Nhi clra
Chu tich nudc tAng).
NhAn dip nhy Todn hoc vd Tudi tr6 chdn thdnh cim on su chi dao vi girip dd quy b6u cira B0 Gi6o
duc vd Dho tao, Nhd xudt bin Girlo duc, Hdi Toiin hoc Vi0t Nam vi su cOng t6c nhiOt tinh cira cdc
uy viOn HQi dong biOn tAp, c5c c6ng tdc vi6n cing quf d6c gi6 gdn xa.

dinliting Bang khen cfia Thri tuong-Chinh phii

THTT

TRONG SO NNY

O

Dinh cho Trung hoc co s0 - For Lower
Secondary Schools

Ng6 Etlc Minh - Tinh rru viOt cria dn
phu trong vi6c giii phrrong trinh chrla

o
o

dau

cin

NgO ViQt

Trung

Abel

-

Giai thrrtng to6n hoc

Tim hidu sAu th6m to6n hoc so cdp

-

Advanced Elementary Mathematics

Nguydn Minh'Hd,

-

@

To6n hoc vi ddi song - Math and Life
Nguy6n Duy Tidn - Nghich li chia giii
thr-t6ng cho hai ddu thri

@

D6'

@

Dd ra

@

Sii dung tich ngoii

Chudn bi thi

vio dai hoc - University

Entrance Preparation
Dd.ng Thanh Hdi - MQt sdphrrong ph6p
tim crlc tri cria him s6'

ki niy - Problems in this Issue
Til299, ..., T l0I 298, Lt, L2l298
GiAi bni ki trudc Solutions to
Previous Problems

Giii

cria hai vectd vD.o giii to6n

Nguy6n Vdn Th1ng - Tim cuc tri cria
hdm s6'nhi6'u biSn bing c6ch khio s6t
iAn lrtot tilng bidn

thi tuydn sinh l6p 10 kh6'i PTCTT
tnlbng DH Vinh 2001

@

cdc bdi cria s6 zg+

Di6n dirn day hoc todn - Math Teaching
Forum
DSng Hil.ng Thdng - V6' co s6 cria
phrrong ph6p nghiOm k6p

Ti6ng Anh qua c6c biri to6n - English
through Math Problems
NgO ViQt

Trung

-

Bei s6'52


rfruu Uu vrfrT
cua ru puuI
lo
'

IR0HB UIEC GtAr PllU0NG Tniillr

eDdnh eha cae

cHriA unu

TRUNG HtlG CO SO

(6/

lfly thla ad tam mdt dAu can thi d6n ddn
phuong trinh bAc cao vd khOng bidt crich giii.
Tuy nhiOn ndu bidt ci{ch dat an phu mdt c6ch
thich hqp thi c6 thd chuydn phuong trinh chrla

dAu can vd he phuong trinh hai dn nOn srl dung
du-o. c c6ch giii quen thuoc. Bhi vi0t niy dua ra
nhidu vi du minh hoa vdi c6c tinh hudng ttl don
gian ddn phfc tap nham hinh thnnh ki nang dit
An phu khi g4p phuong trinh chria dau can.
Bii to6n l. Gidi phaong trinlt

; -^{*a5=s

(t)

Didu kiOn : x + 5 > 0 <+x > -5.
Dat dn phU Jx+ S =y vdi y > 0.
TtI d6 Frf (1) chuydn thinh h0 phuong trinh

Trir theo timg vd (2) vd (3) du-o.c ?-yz+*-y
= 0 <) (x-y)(x+y+l) = 0. XAy ra 2 trudng hgp :
a) x - y = 0 hay. x = ! 20, thay vdo (2) c6

!1I * Jil
2'
b) x+y + | = 0 hayy = -x - I ) 0 thay vlo
(2) c6 xz + x - 4 = 0, giii ra du-o. c

.r=

+ 4y3 - 3y' - 3y = 0 + y(y+1)(4y'-3) = O
Vi y > 0 nOn PT trOn chi c6 nghiom : )r = 0,
4ya

t;

VJ

Va-

-.2

Thay vho (6) vdi didu kiOn 0 < x
xr =

1,

= 0 giAi raduocr, =

!.

dd

bii

Biri to6n 2. Gidi phuong trinh

Didu ki€n : -1
r"

1.

Vl - x" = y v6iy > 0.

Tir d6 Frf (4) chuydn thlnh h0 phuong trinh

Dit

dn phu

Day I) 2 hghi€m cua PT (4).
h-o.

p ta phAi dat 2 dn phu.

aa.-,..-::Vx+i735-x=5

m

lf35

(7)

- * =6.

(7) chuydn thinh h0 phuong trinh
(8)

(e)

Bidn ddi (9) thdnh (a+b)3 - 3ab(a+b) = f5.
(10)
h-o.p v6i (8) duoc ab = 6
Giai hC Irf (8), (10) du-o.c hai nghiem a1= 3,

bt = 2 vd. a2 = 2, b2 = 3. TiI d5 c6 x, = 27 yi
x2= 8lh 2 nghiOm cta PT (7).
Kdt

to6n 4. Gidi phuong trinh

1;1+'lva1=3
Giii.

(11)

Di6u kien x > -1.

Dat dn phq

{*1=o ,A Jx+l = b vdi

b > 0 thi PT (1 1) chuydn

la+b=3
Giii.

< 1 duoc

Biri to6n 3. Gidi philong trinh

Bii

-*(r. fi7)

Vi 2 nghiOm x1, x2thba mdn didu ki6n
n€n PT (1) c6 2 nghiOm rt, 12 nhu tr0n.

*z=
'2

T* d6

(3)

(6)

Til (5) c6 x > 0 nen 1 + y = x2(l + 2y12.Kdt

Giii. Dat 1l * = o ui

(2)

(5)

hqp vdi (6) duoc

C6 m6t sd trudng

Giii.

.t - x -5

NcO ouc MrNH
Hrii Phdng)

THCS NgO GiaTu,

t'tJl.v =x(t+2v)
t *2=t-y2

Vicc gi6i phuong trinh (PT) chrla ddu can
thuong gAy ra nhidu kh6 khin, phric tap : ndu
nAng lOn

clil

thlnh he phuong trinh

(t2)

(1 3)
la3 -b2 =-3
TiI (12) c6 b = 3 - a, thay thd vlo (13) duoc
o3 - o2 + 6a - 6 = 0 + (a -l)(a2 + 6) = 0
> a -1 = 0 = a= I,th6a mdn b =3 - a> 0.


TiI d6 c6 x -- 2 + a3 = 3 th6a mdn didu ki6n
dd bei.

Bii

GTAI THUdNG TOAN HQC

ABEL

to6n 5. Gidi phuong trinh

E-TI.n+1[2x

(14)

=1

NGO VIET TRLING

(Vi€nTodn hoc)

<* < nli - t.
A-ben N.H. (Niels Henrik Abel 1802- 1829) lit
nhd
to6n hoc Na Uy ndi tidng. 6ng c5 nhfrng
.,ElJ-z(t
v6i
a>
0
*,
D4t dn pr,, ,
=1li.o
nhfng d6ng g6p quan trong vho su phrit tridn
+,lr; =+li .b v6i b > o vh x = b4. Tt d6 PT todn hoc, dic bi6t trong dai sd. Nhidu kh6i niem
Giii.

Didu ki6n : 0

(14) chuydn thinh h0 phuong trinh

Itli<, + b'1=1
(_

(15)

laz +ba

(16)

=Jz-l

Dirng phuong phep th6,

tt

to6n hoc mang t6n 6ng nhu nh6m Abel, da tap
Abel, tich ph0n Abel, v.v. . C5 thd coi Abel vh
nhh to6n hoc Phr{p Ga-loa (Galois 1811-1832)

:

li

(15) vd (16) c6 PT:

I 2b .) ./ -Jr+1=o
----:-:+b'+bI

Jz llz

=@2+ r)2-

(u.#)'=o

=b2-b+l-+=0
\tz
Giei

-\=

lrf
,

4

Abel theo dd nghl cita khoa Toiin trudng DH

(dob>0)

Tdng hqp O-x16 (Oslo). Quy nhy c6 mOt

ndy duoc

n4li

-

\11 t

"-"-

=2b=l+
(th6a mdn b > 0 vi 0 < 4 - *li.*lzl
Tt

llz

d6 tfnh

duo. c

x = ba >

X6t di6u kion x
.

_.L

O.

< Jr-l

<> u4 <

rhri thay (r.Jo)-
Ji-t.

Suy ra PT

Cic vi dq tr6n cho thdy tinh uu viOt cira viOc
dat dn phu khi giii phuong trinh chrla dAu cf,n.
Viec dat dn phu nhu thd nho phu thu6c vho dang
cu thd cria phuong trinh. Cric ban hay thri sfc
bang c6ch giii cr{c phuong trinh dudi ddy v} tim
tbi, si{ng tao cdc bii to6n mdi.

nAI rAp
Bdi l. GiAi cric ohuone trinh sau
a) Vx - t + Vx -2 =112x -3

:

..:

b)Jr+Jx-Jl-x=1
Bdi 2. Giii vi biOn luAn phuong trinh.

Jr..+J*=o

nhtng cha d6 cira dai sd "hi€n dai". Ho sinh

ra vi lhm vi0c gdn nhu trong cirng mOt thdi ki,
cirng nghiOn crlu viOc giii cdc phuong trinh dai
sd bang cdn thfc, cing gap nhi6u bat hanh trong
cu6c sOng cfrng nhu trong to6n hoc vh cing mdt
rdt sdm.
Nhan ki ni6m 200 ndm ngdy sinh ctra Abel
nhd nu6c Na Uy vila thlnh l4p QuI giAi thuong

sO ti6n
ban ddu tuong duong vdi 22 tricu do la M!.
Tidn l6i cira n6 sE dir dd trao giii thubng hing
n6m cho cr{c nhd to6n hoc. MOt iry ban gdm c6c
nhd to6n hoc qudc td s6 duoc thinh lAp dd x6t
vi0c trao giii thuong.
Khi tuyOn b0 14p giii thuong Abel ngdy
231812001,6ng Thir tu6ng Na Uy Xt6-ten-boc
(Stoltenberg) n6i rang chring ta cdn quan tam
hon nfra dOn su ph6t tridn to6n hoc vI khoa hoc.
GiAi thuong nham muc dich dOng vi6n sinh viOn
toi{n vI nhfrng nhI to6n hoc, tao nOn su chri y
ctra c6ng ddng qudc td vd to6n hoc cfrng nhu
nhim n0ng cao uy tin cira Na Uy nhu li m6t ddt
nudc ctra tri thr1c.
D(ng ra giii thu&ng Abel dd duoc nhir vua
O-xca II (Oscar II) cira Thuy Didn vd Na Uy dti
nghi lap n6n tt ddu th6 ky 20. Nhmg kd hoach
niy d5 khOng duo. c thuc hi0n do liOn bang hai
nudc tan rd vio nam 1905. Nhh nudc Na Uy hy
vong rang giii thuong Abel v6 to6n hoc sE c6
vai trd tudng tu nhu giii thu&ng N6-ben (Nobel)
cho c6c m6n khoa hoc kh6c. Trong m6t cuOc
hop gdn dAy, Ban chdp hinh HOi Todn hqc quOc
tO dicoi vi0c lAp giii thuong Abel li d6 6n quan
trong nhdt cho su ph6t tridn to6n hoc trong
ntrG"u nam qru, .6 it d ta* thay ddi tinh hin[
ngay sau vii nam ddu, vd ld mOt tii sin quf br{_u
cho cdc thd hC to6n hoc tuong lai cirng v6i giii

thu&ng to6n hoc cao quy ld giii thuong Phin
(Fields) dLroc HOi Tor{n hoc qudc td trao 4 ndm

mOt ldn.


$rl orms
I

r[cn mou crlr rrar vort$

YA{l GHI TT}AN

NGUYEN MINH HA
Hd Noi)

(GV kh6'i PTCT DHSP
Od str dung duoc tfch ngoii ctra hai vecto nhu
m6t cOng cu trong vi0c giii c6c bii to6n hinh
hoc, trudc hdt cdn nim vftng dinh nghia vh c6c
tfnh chdt cira tich ngodi cira hai vecto (dd dang
trong THTT s6 294 thdng 1212001). MQt sd ki
thuAt sr? dung tfch ngodi cira hai vecto vdo giAi
toi{n cfrng d6 duoc gi6i thiOu trong THTT sd
295 (112002). Xin nhSc lai m6t s6 kidn thrlc :

Dinh nghTa. l)
i*6, 6*d, con

ini-littittin(i,i)

ini=O khi i=d

khi

hoic

i=0.

"

oc) =

I

,*M

n sT)

_ I S(ABC)ndu (AB, AC) c6hudngduong
-1
[-starcl nou (AB, AC) c6 hu6ng om
trong d6 S(A.BC) li di€n tich A,ABC, M thuOc BC.
Tinh chdlt.

i n(i -;)=i ni +i ni

t

(ri) "(t i,) = tt,l (i " ri)

r---_*

=

.l

di duo. c md r6ng cho da gi6c l6i.

Vi du L. Cho tam gidc ABC vd M ld
ndm trong tam gidc.

m\t didm

a) Chrtng minh rdng cdc sd do S(MBC)MA,
S(MCA)MB, S(MAB)MC ld do ddi ba canh cfia
mdt tam gidc ndo d6.
b)Tim M sao cho di€n tich tam gidc n6i trong
cdu (a) ld ldn nhdt.

M thu6c tam

121

= ls lt tc il.s

(M AB).slM

Bql

=

= S(MC A).S(M AB).S(M BC) <
S(MCA) + S( MAB) + S(MBC r)
.(
-1,
.,

khi M

khi giAi to6n hinh hoc.

gi6c ABC nOn c6c

tam gi6c MBC, MCA, MAB cing huong vd ci{c
4

ns(MAilMCl

o S(MCA) = S(MAB) =
e M ld trong tAm tam gi6,c ABC.

T6m lai : S(M) l6n nhat vd bang

6) SIMBCIMA + S1MCAIMB + StMAilMe =6
Bii vi6t niy trinh biy tidp mdt s6 bdi to6n
khdc nham lhm 16 hon hi6u luc cta tich ngoii

Y\

)lttr*ro1uE

Ding thric xiy ra

^;);.(; :;)i *(i "b); = o

a)

I

li

:

L

s(M) =

S(MBC)

Vdi MBC vd m6i didm M thi
s) stABCl = slMABl+ SIMBCI+ SIMCAI

Giii.

S(M), theo tinh chdt 3 c6

27

z)

d8ng thrlc tr6n

b) Kf hicu diOn tich ctra tam gi6c n6i trcn

= ls3te^ac)

I)ini=-ini
q (6

ba canh ctra mOt tam gir{c.

= l(slazca) .s(MAB).1tts nMCl =

l,*

2) slABCl=;(*

vecto S(MBC)lrtA, SutrcA)ttti, SQttAB)fr aoi
m6t kh0ng cirng phuong, theo tinh chdt 6 vd
dtnh nghia ph6p cQng vecto, ta thay :
S(MBC)MA, S(MC,4)MB, S(MAB)lilC le d0 dei

lI

trong

j,S'teAC>

am tam gi6c ABC.

Vi du 2. Cho bdh vecto;,ir,;,A.

Cht:rng

minh rdng :

(;"b)G "a)+(;";)(r" 6);(; "a)F,"1)= o
Giii: Theo tinh chAt 3 vd 4 c5
(; "b)(;"d)+(;";)@ "u)+F "a)p ";)=
:

= ; " ((;
=;

"4i)*;

"(F "

" ((; "4i,* (i "
=ind=0(dpcm)

r);)

.;((u";);)

i); + (b";)r)

Vi du 3. Cho ngfr gidc l6i AtA2AlAaA5 c6
di€n ticlt S. Dqt ai = S(AiAi+tAi+z) tl < i < S)
(Aa =At, Az =A).Tinh S theo a; (1

-<5).


GiAi. Ap dr,rng
kdt qui cua vi du

2

cho bdn

v6ctc,

ArH2 n A2Hy

=

(l)

=0

Lay di€im P bat ki. Tir ( 1)vd tinh chat 2 c6

(PH,-Fa,)"(*,*P[)=o
Ae;. a,$ nc6: > fa, " iAr-PH, n Pii =
= PA, "PA,-f4 "PH,
> zslPAt Ar) - zslrn,H r) =
(o,o;"n,,h)(A&"a{)*
= Fe,

:

A,4 , e,A1 ,

"F4-FA, "?H,

. (4o, " a,e-)(a'1' " a4f ) *
. @a"A,.4;X4t "Zuao) =o

r

Tuong tu c6

s[a,ere.].s[a,A,Ar]

=

o

(l

+

=

PHr-Fit n PH.
"
zslPl'Atl-zslrn..u,l

sle, 4e.,].s [e, eoa,f> o
]sIa, AzAo).s[a,arar]

:d
I

\

aiai+t

=o(a6:a1)

Vi

dU 5. Cho

i

Clt'o ttt gidc l6i AA2\A4 Goi H1,
H2, H j, Ha theo thrt u ld trtlc tdm cia c(c tarn
gidc A2AjAq, Aj\a,A1, AAAz: AA2A3. Chrtng
minh rdng : SIH fi2H jHa] = SIAAzAsA+J

Giai. Y\ H1, lI2 lh truc tam ciic tam gi6c
L A1A4
LA1A4

l.-

I

--.

= !(osnrc+rc^AD)
l = +Ae(eo-eE)
2\-2 \
1 ,_

=llacnBDl
2\t
Nhdn xit. Kdt qui tion lh su mo rQng cita
cong thrlc quen thu6c

SaBCD) =

:

!.ec.no.sin(AC, BD)
2

Vi du 6. Clto luc gidc l6i ABCDEF. Goi M,
N,

=) A,H,' ll A"H,

*(o.
" Bi)
2

Giei. SIABCDI= SIABCI+ SIACD]

Vi du 4.

)Attt,
lArH,

minlt

I

, o*,)
) )
\tsiss
l\tsiss ,/

:

rt gidc l6i ABCD. Chang

:

stABCD)=

l
AyA\A ,A1AaA1 n0n

cia (2), (3), (4) vi 6p dung

:

slAtAzAlAo ] = s [4, a zH rH q).

Ddthaysrl)a,.Vay:
) /-t

'=+l
'[rsi
(4)

"Pn;

cirng hu6ng v6i A1,42AqAa, suy ra

rdtrg

t>i<5

.(Lo,*,/[m\
r,, I -[ r

=

slAtA2Al] = s[n,arlr. ]
TU d6 LA.A2A1, LHtHzHl cirng hudng vi
S(44A3) = S(HtHzHt).
Tuong tU ta c5 S(A2A1A4) = S(HzHfi q) ,
S(AjA4At) = SGL.,H+HI). S(AaAtk) = S(H+Hilz)
viL tat cA c6c tam gi6c trOn ddu cing hudng vdi
AtA2A1A4, do d5 tfi gi6c HtHzHlH 4 cfing l6i,

/\

-l I r, l, . I
\ l
PAI

COng theo trmg vd

tinh chdt 5 c5

[s[a,a,a, ].sIa, 4eof, o
Tu(l).(2)suyra:
s ( A, k,\) s( a, aoa, ) - s(a,,qrlo ).s(a, a',q )
+ s(ererer).s(A, ArAo) = o -+
a rql - 6-a.t-a+)(S-a rat) + a5(S-a ;a) = 0
s2

(3)

PA;

= Pe. "PH._

I

=

=

)

ta, e, e., te1 AaA5 cirng hu6ng
I
Ta rhay , ] na, AzAq,LAtAsAlngugc hudng
Ite,e, er, LAtAi A4c[ng hudng

=

:

2slPA2A7)-zslrururl

sla,b*] s[A,A+] + s[a'er4].s[e,a'e.]

(z)

P threo thfi tu ld trumg didm cia AD , BE , C F

.

Chrtng minh rdng M, N, P cing thu6c mbt
dudng thdng khi vd chi khi S(ABCDEF) =
S(ACE) + S(BDF).


Giii.

ti+

KhOng mdt

.tdng_ __qlt
gih st
ABCDEF
c6 huong duong.

Theo tinh chdt
3vdVD5 c6:

Ufr nMF

A

2,

='

+ DE)"*(n * Dr)

=

i?,

=

!( L e,r n ec * ! on n ee *! eE n oF +
2\2
2
2
- \t__

---\

+1DB^DF)

t.

:I (-s(ACE ) + s(DABC) + s(ADEF) - s(BD4)
2'

= 1(rrou. DEF) - s(ACE)- s(BDr))

2'

YLy :.M, N,.P cDng thu6c m6t duong thing
<> MN nMP =0
e S(ABCDEF) = S(ACE) + S(BDF)
Dd rdn luy6n viOc s[t dung tich ngohi cia hai
vectd veo viOc giii to6n, mdi cdc ban lhm mdt
sd bli mp dudi day.
Bii tAp 1. Ctro tmn gi6,c ABC vd didm M ndo d6
khi{c ci{c dinh. Gie si AM cat BC tai A', BM cit
C A tai B' , C M cit AB tai C'. Chrmg minh r6ng :

M{ MY :=
MC
:AA' BB, CC,
Blri tep 2. Cho ttt gi6.c l6i ABCD c6 AD = BC.
Vd phfa ngoli n6 ta dung cdc tam gi6c ADE,
BCF cintai A, B sao cho dtrE = dEp . Cnue

minh rdng trung didm ctra cdc doan thing AB,
CD, EF thing hhng.

Bii tep 3. Cho t(r gi6cldi ABCD. AB cit CD tai
E ; AD cilt BC tai F. Chung minh rang trung didm
c[ra c6c doan thang A C , BD, EF thang hdng.
Bni tap 4. TrCn ba duirng thhng a, b, c theo thrl
tu c6 ci{c didm A, B, C chuydn dQng vdi vAn tdc
ddu. Bidt rang tai thdi didm ban dduA, B, C kh6ng
thing hhng. Chfng minh rang tdn tai khong qud
hai thoi didm md taid6 A, B, C thang hing.
Biri tAp 5. Cho tam gi6c ABC vI hai ciidm O,
O'. Gft sir AO cit BC rai A', BO cit CA tai B',
CO ci* AB tai C', AO' cat B'C'tai A", BO' cit
C'A' tai 8", CO' cat A'B' tai C". Chrmg minh
6

cA,

ABsao cho

,

Y=9N =4!
BC CA AB

a) Chung minh rang c6c dO diLi AM, BN, CP
thlnh ba canh cira m6t tam gi6c.
b) Tim vi tri cta M, N, P sao cho diOn tich
tam gir{c n6i trong cdu (a) nh6 nhAt.

Bni tAp 8. Cho tam gi6c ABC voi BC = a, CA
= b, AB = c. Ggi G, I, O theo thrl tu lh trong tam,
tAm dudng trdn nOi tiOp, tAm dudng trdn ngoai
tidp cfia tam gir{c. Chung minh rang
(a + b + c)(b- cXc - u)(u - b)
ctao

= )(sleecl+ S1DABC)+ SIADEF)+ stDBrl)
2'
=

rang c6c duong thhng A'A", B'8", C'C" ddng
quy hoac dOi m6t song song.
Bii tAp 6. Cho tam gi6c ABC. Cdc didm A',
B', C'theo thri tu thuOc c6.c canh BC, CA, AB.
Goi A", 8", C" ldn luot ld ci{c didm ddi xung
cira ci{c didm A, B, C theo thf tu qua c6c didm
A', B', C'. Chung minh rang :
s(4" B"C',) = 3S(ABC) + 4S(A', B',C)
Bii tAp 7. Cho tam gi6c ABC. C6,c didm M,
N, P theo thrl tu ch+y tron cdc dudng thing BC,

:

o15OIl

-

4sSlABCl
Blri tAp 9. Cho tam gi6c ABC. Cdc didm M,

N, P theo thf tu thuOc ci{c canh BC, CA, AB.
Ggi X Y, Z ldn luot ld trong t0m ci{c tam giiic
ANP, BPM, CMN. Chung minh rang :
IGBC)

. ?2 Nwzl

Bii tAp 10. Cho ttr gi6c ABCD nQi tidp. M li
m6t didm bdt ki. X, Y, Z, T , U , V theo thrl tu li
hinh chi6u cba M fiancdc dudng thhng AB, bD,
AC, DB, AD, BC. Goi E, F, G theo thri tu lh
trung didm cba W, Zf , W. Chung minh ring E,
F, G thing hlng.


\

..

,

'.t

\r\

,{

,\,

'[IM CUC'[RI CUi\ itAlA s0 l\ti-iliu 3ti NI

t',
I
\

U

,

,\

r

r$

illi Nl

TUI\IC
I}AI\IG CAC|I K|IAO $AT TAI\I !.UOI
I

NGUYEN VAN THONG
(CA/ THPT chuy€n Ld QLty Don, Dd Ndng)
Trong c6c cuOc thi chon hoc sinh gioi thudng
c6.bdi toiin tim gi6 tri lon nhat (GTLN) hay gi6
tri nh6 nhat (GTNN) cfia hdm s6'nli€u bidh phu
tlutdc ldn nhau. 6 ddp 6n thuong sri dung c6c
bat'ding thrlc dd udc-luong gid iri him so rdi
x6c din[ giri tri cia cdtc UlSn ttrl c6c ddu ding
thric xiy ra dOng thdi.
Dd tim cuc tri hlm sO nhi6u bidn c6 thd ding

philong phdp khdo sdt ldn luot tlng bi€h, ngfra
lh : tim GTLN (hoac GTNN) cfra hhm sd v6i
bidn thri nhdt vi cdc bi6n cdn lai coi ld tham s6,
r6i tim GTLN (GTNN) cira him sd v6i bidn thrl
hai vi trng vdi gi6 tri dd xdc dinh c[ra bidn thrl
nhat mi cr{c bidn cbn lai coi lh tham sd...
Dd thay 16 hi€u luc cira phuong phdp niy ta
thu 6p dung n6 vio gi6i mOt sd bhi to6n dang

nhy trong dd thi chon hoc sinh gi6i toin qudc

mOt sd

nim gdn dAy.

Bii to6n l.Xit

f(x,y) = 1l-x)(21)Ux -2y)
- {(x, y) I 0 Tim gid tri nhd nhdl crta hdm s6' f tan

mi6n D.

Ldi giii. Bidn ddi hdm sd dd cho thdnh
f(x, y) = 2(r-x)(2-y)t(2-y) - Z(r-x)l
Dit v - 1 - x vd'u = 2 - y, ta chuydn v6 tim
GTNN ctra hlm sd F(4, v) = -2uv2 + u2v tr€n
midn E= l(u,v) I 0< u<2,0
E

min I min(-2uv2
= oL(l
mdn didu kiAn abc + q
nhd't cia bidu thtc
D-

uaqua v.,thi
s'(v) = 0 khi v,,=
,, 14 md o < +a
o\,
4 I2
S'(v) ddi ddu tt duong sang Am, suy ra

ming(v) = min {S(0) ; S(1)} = min {0

;

0
- zul = Li - 7u do u2 - 2u = u(u - 2) <0.
V4y
mrn(uz -2u) = -1 khi
= 0' minF(u,v)
E

c2+l

-

1999)

Ldi giii. Bidn ddi gii thiet thhnh a + c =
a+c
(1)

=b(l-ac)>O=a<1 vhb=
Thay (1) vdo bidu tni p"a uislrifi'ouo.
2(a+c)2
2 +-+3
P_
-2 (2)
a2 +l c2 +l (l+a2X1+c2)

v6i 0

<, < !

(x + c)2

= .1 * (l+x2X1+c2)
x2 +l

ra coi c ld tham sd (c > 0)

C

Ta c6

Tr0n
nhdt Ih

f(x)

_l

X6t hdm g(v) = -2uvz + u2v v6i0 < v < 1 vh
coi u ld tham sd th6a min 0 < u < 2. Ta c6
S 'O) -- -4uv + ,r2 = u(-4v + u). Ta thdy

f

+Zcx _ l)
'(x) = -2c(x2

(l+

x2)2(l+c2)

thi /'(x) = o c6

[.':)
x. = -c

i. I
* .Jc'+

nghiOm duy

(3) vdi 0 < x.

( r.I

/'(x) ddi dau tir duong .urg arn nten
= f(x) < flx.,) =

dat cuc dai tai x,,

f + -S.

Tir dd theo (2) c5

:

J.2 +t

32c3
<--*
P=2f(x)-)a--:"
c'+l Jc2+l c"+l

=g(c)

X6t hdm sd g(c) vdi c > 0.

Ta c6 g'(c) =

2(1

-

8c2 )

(c2 + l)2 (3c +

u'

u=l,v=1.

oz+l b2+l

(Dd thi HS gi6iTHPT todn Edc biing A

Qua x,, thi

+u2fi1

+ c = b.Tim gid tri ldn

223

_

'

X6t hhm s6 /(x)

hdm sA'

tr€n midn D

Id min F\u,v)

Tt d6 min /(x, J) = 2min F(u, v) = -2 khi
DE
x=0,y= 1.
Bii to6n 2.Xet cdc s6'thuc duong a, b, c thda

Vdi r > 0 thi S'(c) = 0t4i .., =
thi

{
'v8
g'(c) ddi ddu tir duong sang Am nOn g(c,,) li

vd qua c,,


giii tri cuc dai, suy ra

p=
=

b

r = ,[*)=
Gia tri

"tJsl ]9.
3

[ darduoc khi c = +, r= J'
3Js2

Ji

ru

(o

li=r
(l)

4
|lxz>I ls

(3)

-1
(Z): yz>-

tri ldn nhd't cila

Ldi giii. Ttr didu

5

bidu thrtc

Til

Xiy

1* 1 vdi,
ho. p

y= -,2= "25'

rL thix> r, 4
l5z
JtS
1l)
felZZZ

t'52
-'1 *' I
4:
X6t him s6 s(z) vdi | <,

(5)

Tac6g'(;)=

=+
vts

il=4

2
J

(.-

vOithamsd

.-. , r -n
-'
m
' ] l- "
-'

l2r+my-r5=0

Bni 2. Tim GTNN, GTLN cira hhm sd
y = sin6x +
+ asirxcosx
"o16,
(De thi DH Throng nrui)

Vtina * Vti"E *

5

dat GTLN

+,J]COS

z
Hir
Noi 2000)
@A thi DH Bdch Khoa,

HD

sir dung

BDr

+63 ( a+b\3

] neu a+b>O
BAi 4. Khai tridn a-, ,r,,i" irr, = ll + ziD

Max(ao,

,rl.

Vti"Z

-le-ls^lc
J/COS- + ..tiCOS
\ z\ z !

thdnh dang ao

/1\

\t)

*1=4khix =2.--2
- 3'"-a
x z

8

2

hq

J

XZ

yz

2.2= l3

suvramaxP(x,y,z)=13.

HD: GiAi, bien luan

<

So siinh (5) vd (6) rrit ru I * ! , + ddng thdi

b)X6thdmsd

2

(b) ta c6

xiy ra khi v) chi khi x =

(6)

\5/

hOr=!+1

t

(Ki sau ddng ri€b)

thfic p-

.O khir. +.
+-*
4 z"
Vl5

I

aOry

Bni 3. Xr{c dinh dang cira MBC sao cho bidu

.

TiI d6 SQ) ld him giim vI flx) < s(z)
('t.i
sl

y22

@A thi DH Giao rhdng vdn rcii 2000)

= g(z)

:

vi

1*1.?,

10)

BAi 1. Tiry theo giri tri cia m, hdy tim GTNN
cta bidu thrlc
A = (x - 2y + l)2 + (2x + my + 5)2

theo(4)nen

.N€'u ?=r=3tn.o(t)rhi *r- 4 >,
5 JrS
t5z

theo(4) nOnflx) 3

(

frTET 5d Pnrfdnc PDKP ...
(Ti€pu'angll)

:

.Nduz

5

c6c k6t quA cfia (a)

> 0 vi tham

xz

ra 2 truong

zZ

t2

f+l

him sd/(x) =

nit ru

1*1=2 khiz= 2 .r=
y

Dau ding thric

t)

t

l.)

(e)

p(x,y,,t=(!*1'l*
2(!*ll .o *
\.x y) (y z)

kiQn (1) ve (2) suy ra

xlmaxlr, _ I
t5z)
,a , >

:

5Js2

thdi

(Dd thi HS gi6iTHPT todn quOc bdng B -2001)

a) X6t

(8)

2
. Net Z
=r=L thi /rrvr <
So sdnh (9) vn (10)

|
P(x, Y, r) = *? *1
xyz
(

ra

JS

Bii to6n 3.Xit cdc so'thilc duong x, y, z thda
ndn hd di6u ki€n sau :

Hdy tim gid

LAp luAn nhu c6u a) ta duo. c

suy

(3)

.Neuz>|trrl lrtg<2J5

theo (1), (3).

ls

TU di6u kien (1) vh
l' rl
y ) max 12,
t -l5z)

ar

'+47

+ otx +

art + ... + arrx". T\m

..., an).

tDi tli Hl' Ki thudt qudn slr 2000)
DS: I\4ax(rr,, . ur, .... tt,.\ = ar = Clt.28 = 126720.


/(x) = L*l,datduo.ckhix=
I
r .-rl
r e|
L 2 2)

TT

Do d6 Max

0 ptluuN6 pflflp
m0T60
U
,(\

rim sa$c TR! ctln HftM $0
DANG THANH HAI
(:@

W

PhbngkhOng -Khbngqudn SonTdy)

Bhi torin tim girl tri 16n nhdt (GTLN), gir4 tri
.h6 ;#;?Gri.{Ii) .[; *+ ta* io tuy mqi uidd
thric nho AO tnuong duoc dd cAp trong cdc ki thi
i;t;, ;irh "to caE truons Qai hec,"cao ding.
Trung hoc chuyen nghi€p. O bari vidt.niy xin
eiOi fiieu vdi cic ban-mQf vdi phuong phrip (PP)
ihons dunp. de eiii loai todn tren trong cdc dot
tni trlydn vio cdi trudng vai nam gdn day. Trudc
hei chLing ta nh[c lai :
56' rn vd M theo tht? tu duqc gqi ld Gf NN vd
Gf LN cita hdrn y = f(x) xdc dinh tr€n tdp D c
R ne'u m

f(x,,) =m:f(x1)=M.
tdi dua ra ba phuqng ph6p
Dudi dAv chring
-aann gia cric bat ding thrlc
thuong dirng dd
(BDTi : m < f(x) ( M, m6t cOng viOc rdt quan
trons dd t)m-GTLN, GTNN ctra hdm y = f(x)
r.en]-rlidn xdc dinh c[ra hhm s0 d5.

PPI: Phuong phdp udc luong gid tri hdm sd
Phuong phdp ndy chir ydu sir dung cdc bdt
ding thrii quen thu6c (Co-si, Bu-nhi-a-c0P;|ki,
...) [o4c cac gOf io bin dd udc lugng gir{ trf
cia hhm y = f(x); til d6 sir dung dinh nghra cira
GTNN, GTLN suy ra kdt qui cdn tim.
Vi du 1. Tim GTLN cira him sd
ll-r) = { + sin2, ,rrn [-1,

L2

2

(Di thi DH Kinh

l-1,1l
L 2 2)

thay rang v6i

him/(x)

moi,,

<

1.

L-;, i.]

-11

Tt d6flx) =

e )z

sin2 x

<

n+1.Ddu

4

nh6 nhdt.

(Di thi DHQG Ha Nai, kh6'i A -2000)
Ldi giii. Dd thi (C) ctra hdm s6 c6 tiOm cAn
drlng x = 1, ti€m can xi0n ! = x + 1' (ban doc tu
vE do thi)

Gii

OX=

sir

M(l + a,2 + o * !) la > 0) ld didm
a

th6a mdn didu ki6n bdi to6n. Phuong trinh tidp
tuydn v6i dd thi (C) tai M c6 dang
oz -l
(D) y=-G-l-a)+2+a+-.

I

ota
Goi E lh giao didm cira hai ti6m cAn, A, B li
cdc grao didh cira (D) v6i ci{c tiOm cAn d6. Khi

/

d5E(1,2), Al t,

\.

r\
Z+:l.A(t
a)

+2a,2+2a\.

SuyraP=EA+EB+AB=
a"-

so'+\-a. Ap

+zJza +

a

cho

C0-si

P>

a
cac

t-----=

z..lq'lz

+

.

zJi

Dau

I

e<

a =2,12a

la

lro'

la'

=

t

7

dqng

BDT

duong ta duoc
-s hay P >- 4*1, +

ding

thtic

xay

ra

!
2

I
u--:-=cla

\I L

T6m lai, chi c6 duy nhdt didm M th6a mdn
didu kicn bdi to6n

tvtltd6li u(.*L.z-f
"-

5'.'

.l)
,
llr)

vhMin P= 4{2+
PP2z Phuong phdp xdc dinh midn gid tri:
Cho hlm y = f(x) xdc dinh tron tap X c R' Dd

"GTl.lN

4

TE

l2 x=r
[Sln

I

va

L-)

L

1*
2

lx

ding thrlc xiy ra

nl *na,

. x
. [-+, +l co:
-<24
L

sin2x

f r

tam gi6c c6 chu vi

ducrng tiom cAn mot

(.>

t€'qudc ddn

re,

nhfrng didm trOn dd thi hdm sd c5 hodnh do lon
h* iruo cho tidp trySo iui didm dd tao vdi hai

2)

mra m,o khan, ta nghr tdi viec udc

luong gird tri cira

*-l'n*
Vidq 2.Chohdmso !=x+1
'
x-l

JstJz-tl

1"].

- 2000)
Ldi giii. BIi to6n dat ra ld tim GTLN cta m6t
rr*'p"n,ii- t+p, uie. iihAo sat him s6 d6 tren

rl

2

I

cira him y -- f(x) ta tim midn
tim GTLN,
y,
lh tim di6u ki6n ctra tham
tfc
gi6 tri cira hdm
y
iO y A0 phuong trinh = f(x) (dn ,r) c5 nghiem
tren tAp X.


Vi du 3. Tim GTNN, GTLN cria hhm sd
y = 3sinzx + 6sirurcosx
5cos2r

/_\
\

+

jzsinl 2x+!l +2

Ldi
hdm y.

55

- - :(l
2
y

-cos2x) + 3sin 2x

+ cos2x + sin2x

!

= 4sin2x

IA R.

-

+2
cosZx

+ 1.

TAp xiic dinh

Hai.dau tht ngang srlc choi m6r rrd choi (chi
thang ho4c thua) nghia lh khri nang chidn
thang cria ho lh nhu nhau. Ho thoa thlan vdi
nhau ring ai ld ngudi cidu tien th6ng 6 v6n thi
nh4n.todn b0 giii thudng. GiA su rang vi li do
d.o trd choi phAi dung tai o tinh hudng ngudi
1e9
thrl nhat thing 5 vi{n, cdn ngudi thrl trai t-tr6ig :
vi{n. Cdn phii phan chia giii nhu thd ndo l}
c.5.

:

<>1- 4,tq VAyMax1, = I + 4Ji,Miny= | - 4J,

hqp

PP3: Phuong phdp x6t chidu bi€n thi|n
dqng c6ng cu dao hi.m, thidt lAp bing bidn
. !t
thi€n cfia hdm y - f(x) ftAn rap x6c dinn cIa nO,
d6 suy ra GTLN, GTNN cria hdm sd dd.

GTNN cfia him

so

)=2sin8x +cosa2*

Lli

(Dd thi DHTdi chinh Kdtodn _2000)
giii. Ddt r = cos2x
< / < 1), him y tr6

(-l

':'l,\4 *ro,re [-1, 1)=D.
\2 )

t 11.-,r3-1
f'(t)= ql,t" -l(2))
+ i I ;f,r,l=o<+ r= !.

L

li

?

C6.nhi6u

f

kidn triii nguoc nhau. Dai da sd

cho rang cdn chia theo ti. t. S , :, c[ns c6 V kidn

cho rang cdn chia theo ri le Z : I 1v6i'iao tuan ta
ngudi thri nhat thdng nhidu hon 2 virn, ) lr'mor

phdn ba cua 6, nen ngudi rhrl nhai phdj nhAn
duoc l/3 gidi do 2 ldn th6ng cuoc, phin cdn lai
cria giii chia d6i, m6i ngudi mOt nua). Nhung
thgg
chia theo ti tC 7 : 1 II hop li hon ci (li
1a"
qur{ nhi !)

Gidi thich: Pa-xcan (pascal) vd

/ t
thhnhflr.l =21

3'

Lap bAng bidn rhiOn cua hlmflr) trOn

NGLYEN DUY TIEN
@HKWN_DHQG Hd N6i)

cta y

4sinZx - 4cosZx + (1-y) = 0 (l) cd nghi6m.
Frf (1) c6 nghiem khi vd chi khi
16 + t6 > (y- 1)2<+ y2 -Zy- 31 <0

Vi du 4. Tim GTLN,

CHO HAI DAU THU

-l1l+
-""- +
2'- cos2x)

Ta tim y dd phuong trinh (0n,r)

tt

N4KisK,gi
cHlA clAl THLIdNG

4)

(D€ thi nydn kh6i st Emn cdp phdn cl6i)
giii. Tru6c r0n ra lhm don giin hda

[-1, t].

phec-ma

(Fermat), doc lAp vdi nhau giAi thich vd ti le
trOn theo quan didm xric sudt. LAp lu6n cua
Fermat nhu sau : Ndu tidp tuc choi ihem 3 v,in
"gii tao" nfra thi ddi thi tfli trai muon lav duoc
tat ci giAi anh ta. phAi chien thdng ca 3 v6n ndy.
Vi vay xiic suAt thdng cuOc crja anh ta l) :

l11l
, i ,= g

vd do dd xiic suar th6ng cuOc c[ra

ddu thir thri nh{t ta

1. Oidu ndy chung 16 ri lc
8

: I ld hqp li nhat.
Clch giAi thich tr€n do pascal vd Fermat rim
ra vio n5m 1654. Ph6t ki€h nhy quan trong rdi
mrlc nhi6u ngudi cho rang l654lh'nam sinh-cta
7

Ttr bAng bi6n thiOn ta cd
c> *=I+kr
D2'
It c)cos2x=
Miny -.
D- 27
3

MS*y=3 <) cos}x=-l

(keZ)

Xin mdi cdc ban tiOp tuc rdn luy0n theo cdc
phuong phr{p tr6n qua ciic bii tap sau:

(Xem tidp trang B)
10

xi{c sudt)

Chit thtch
o Tac-ta-glia (Tarraglia) li ngudi dd nehi chia
theo ti lQ 2,: l. 1n4c dil 6ng ti ngudi kh6-tni gioi
(sau mOt d6m dd tim ra c6ng thric nghiem iria
phuong trinh bAc ba).
6=


fl

A\

ruyfx

u0 rru

/

4*

1.

SINH LoP to

xuot

THPT

cuuvBf{ T0Al{ -TIN

TRUONIG DAI HOC VII{H XAU
ruON roAN - voNG 1
(Thdi gian ldm bdi : 150 philt)

tting

I.

1) Tim m6t sd tu nhion c6 4 chfr sd
cila ro ao va tdt ch, c6c cht sd cira

CAu

.

Uarig ZOOf

2) Tim x sao

,
It-xJx
i

(r-Jx-

CAu

II.

cho
\
r][\/ t+xVn_Jil=s
+Vr ll ---------- -V.t l<-J

)

l[r+Jx

,l) r'

t-_==

+x+

IT

I -rlo -zJs -0

- 16n nhdt
2) Tim gi6 tri

thfc

A

-

vi

gir{ tri nh6 nhdt cira

v

sO

tt

1 ddn 8.

ciu v. a) cho day sd x1, x2, "', x,,,
mdncdcdiduki€n:
I
_ -z , - ,,k1 .,- I ?
*'=;'xn+t=*f;+*"vdin=l'i""'
ring

"'

th6a

:

|
l< I * 1 *...* x26s1+1 .2
x1*1 xr+l

b) Tim c6c sd tu nhiOn x vit Y th6a m6n
IOx2 + 29xy + 2ly2 = 2gg1
phuong trinh :
CAu

III. 6

vd phia ngodi tam gi6,c ABC,
ta dmg cdc hinh vu6ng ABB1A2, BCCPy,
CAA{2. Ggi M ld didm chinh gifia cita canh
Cnu

mat tdt ch cdc chfr

VI. a) Cho hai ducrng thing b vi c song
v6i nhau, A lh mOt didm khong thuOc hai

2

+3xy+2y2

3x2

ruON roAN - voNG 2
(Thdi gian ldm bdi : 150 phit)
CAu IV. C6 bao nhiOu sd tu nhicn c6 8 chfr
s0, trong d6 m6i sd ddu chia hdt cho 11 vi c6

Chung minh

1) Giei phucrng trinh

tl

bidu

mi
n6

2OO1

song
duong thing d5. Drrng tam gi6c ddu ABC sao cho
dinh B thuOc ducrng thing b, dinh C thuOc dudng

thing

c.

b) Cho tam gi6c ABC, dudng cao CH. Hai

AM vu6ng g6cv6i A/2.
2) Chrmg minh rdng ci{c dudng trung truc cta
c6c doan thhng A1A2, BP2, CP2 dOng quy tai

duong thing x vh y di qua C tuong rlng hqp vdi
hai dudng thhng CA, CB cdc g5c bang nhau sao
cho hinh chidu M cira A tren x, hinh chidu N cira
B tr€n y vb, H kh6ng thing hing' Chrlng minh
rang didm chinh gifra K ctra AB vd cdc didm M,

mot didm.

H, N nam tren mot dudng trdn.

flsng trudng hop tting qu6t khi ddu thir thrl
nhdt cdn phii thing thOm n viin nfa m6i nhin
duoc giii, cdn ddu thir thrl hai phii thang thOm
m vdnntramdi nhdn duoc giii thi x6c sudt th6ng

tao) cira Fermat dd cho m6t chimg minh don
giAn cira dinh li tuyOt diOu sau.
GiA sr] hai ddu thir choi trd choi vdi quy udc lh
ai thang trudc N v6n sE nh0n gi6i. Khi d6 ddu
thtr kh;i cu6c sE c6 cdc x6c sudt chidn thang
nhu nhau fiong2 lu6t choi sau dAY :
choi
- Hodc choi ldn luot (ngudi khai cuOc
ctl
ti6p
vd
ta,
anh
cta
v6n ddu, rdi dOn ddu tht
tuc choi tudn tu nhu th6),
- Ho[c choi theo cdch : ai th6ng vdn ndy sE
duoc tidp tuc choi v6n sau.
Truoc d6, ndrn 1976, Kinh-xton (Kingston) dd
chrmg minh dinh li ndy bang phuong ph6p tti hqp'

BC.
1) Chung minh rang

G=

o

cuOc cfia dau thir thrl nhat

lh

:

n+m-1

2-trm+t

Z 'l'*^-'
'i=n

(Sd vrin choi bang n + m
sd dong

khi

nang

-

ld z"n^-t

1

vI t6t ci

c6c bi6n

).

o Nim 1977 An-dec-son (Anderson) dd stl
dqng y tuong (tidp tuc choi thOm c6c v6n gii

t1


BdiT71298. GiAi phuong trinh;

3^= I +x+logie +2x)
LE ANH TUAN

DC RR HI IIRY

(G1, T H PT chuyd tr Vinh

cAc r6p rHCS

)

xi,=1,'r=f,

X

ruv6r rHANH
(Gl'SottTav, HdTc)1')

TnAN

BdiT2l298. Tim gi6 tri l6n nhdt cfra bidu thric
,2 + ,y +y2, trong d6 x, y li c6c se( thuc th6a

rr+1 'Xu
u+2

2002 x,,n, + 200L.r,, + 2000 r,,*,.x,,

vdi moi n = 0,1,2, ...
Hdy tim cOng thfc tdng qurit clia x,, theo

(GV THPT ChauThdnh A, BihTre)

DO THANH HAN
(G/ THPT chuy€n Bac Li€u)

ac.

TRANuduc soN

(G/ THPT

Bni T9/298. Goi AH vd r ldn luot ld duong

vi brin kinh duong trdn nOi tidp cria tam
gil,c ABC. Gqi pr li nria chu vi LABH vd r, li
br{n kinh duong trdn nOi tidp cfra n5. Goi p2 li
nfra chu vi MCH vi 12 li bdn kinh dudng trbn
cao

Bdi T31298. Chung minh rang ndu phuong
trinh axa + bx3 + r*2 - 2b* + 4a = 0 (a * 0)
c5 2 nghiQm x1, xr th6a mdn x, .xz = I rhi

nOi tidp cira n6. Chtnrg minh rang tam gi6c ABC
vuOng tai A khi vd chi khi

BC ThdiThuy,Thdi Binh)

BiiT4l298. Goi P li trung didm canh BC cia
tam gi6c ABC vd BE, CF ld hai dudng cao.
Duong thing qua A, vuOng g6c v6i PF, cdtt
ducrng thhng CF tai M. Duong thing qua A,
vuOng g6c v6i PE, cit dudng th&ng BE rai N.
Goi K vd G ldn luot li trung didm crta BM vi.
CN. Goi H li giao didm cria dudng thing KF vd
GE. Chrmg minh rang AH vu6ng g6c vdi E F.

(G/ THPT

+
r = Ptrt Pzti

2)

Pt+Pz

rnAN vAN VANG
chuy€n HingVuong. PhnTho)

Bni T10/298. Cho tf di€n gdn ddu ABCD (AB
= CD, AC = BD, AD = BC ). Goi R vi r ldn luot
ld br{n kinh ciic mit cdu ngoai tidp vi n6i tidp
cria trl di1n ABCD. Goi R,, vh r',, Idn luot li biin
kinh cr{c dudng trdn ngoai tidp vd nbi tidp cia
tam gii{c ABC. Chung minh rang
:

PHAM HOANG HA
(SV CLC K49

@<.'8
r^

khoaTodnTin, DHSP Hd N6i)

Bni T5i298. Cho hinh

vuOng ABCD. Ldy
didm E tr6n canh AB vd didm F trOn canh Cb
sao cho AE = CF. C{c dudng thhng BF vd CE
c6t dudng thhng AD ldn luor tai ALvd M. Ttm
qu! tich giao didm P cira cdc dudng thhng BM
vi CN khi E, F di chuydn rr€n ciic canh AB
vit CD.

(Trung tdm CNTT, PS/, Hd N6i)

2003

II
i=l

cAc r6p THPT
Tim sd du cira ph6p

T61298.

(r* *1

(Hd Noi)

?

u6 queNc vnrrn
(Hd Noi)

cAc
MQr

quA cdu khdi luong
m = 1,2 kg ndi bang

khdi luong vho m6t

thanh thing drmg

dang quay vdi vAn

cho zoo3.

- 3r2

Ding thrlc xhy rakhi nio

2 soi day kh6ng
chia

VUHAI

t2

tlR'

Bdi Lll298.

PHAMDINHTRUdNG

Bni

rr.

NGUYEN VAN THANH

:

lZx-yl<3vdlx_-3yl<1

5a'=2b'+

hilc

Bni TS/298. Ddy sd ( x,, )(rr = 0, 1.2. ...) duoc
xi4c dinh bdi :

Bni T1/298. Chung minh rang sd c6c chfr s6
vidt trong hQ thap ph6n ctra hai sd 2oO22oot vd,
2oo22oo1 +z2oot th bang nhau.

mdn2 di€u ki€n

P

tdc g5c ro. Hai dAy
buOc vAt bi cing vi

of vAr

ri


tao thenh mOt tam gi6c c0n ABC voi AB = AC =
I = I,5m, BC = 1,8m. Luc cang cira day AB lh
Ir = 35N. Cho g = 10m/s2. Hay xr{c dinh
a) Lr,tc cang

I,

b) Van t6c goc

ctra d0y AC.

c6c soi d6y ndi kh6ng gidn, khOng dAn di€n,
khdi lugng kh6ng dr{ng kd vh chidu ddi c6c
soi day bang do dii c6c canh tuong Ling ctra
tam giiic.
H0 thdng d{t

co.

^
*o*?[roH])G

1= m[t phing ngang, nhin v]

ci{ch diOn. Goi ABC = a, BC = l.HAy xdc dinh
gia tdc ctra c6c qui cdu ngay sau khi day BC bi

^

HAU

Bdi T21298. Ba qui cdu nh6 gidng hOt nhau
(m6i quA c6 khdi luong m, di0n tich 4) nam y€n
6 cdc dinh cfra tam gi6c ABC vuOng tai A nhd

drlt dOt ngOt.

NGUYEN XUAN QUANG
THPT chuv€rtVTnh Phrtc )

(C;V

wsffiffiII$ffiffiffi
FOR LOWER SECONDARY SCHOOLS

2001

l[i=l {t + ;2 ) by 2003.

Tll298. Prove that the numbers of digits of
the numbers 20022.mt and 2OO22o0t +
in decimal system are equal.

221\ot

written
T71298. Solve the equalion

the greatest value of the
expression .f + xy + y2 where x, y are real
numbers satisfying simultaneously the
conditions l2x - yl < 3 vh lx - 3yl < 1
T21298. Find

T31298. Prove that

3* =

I +.r + logi(l

T8/298. The sequence of numbers (r,, )
(n = 0, 1,2,

...) is defined by : xu =

if the equation
x

axl + bx1 + rr2

-

Zbx + 4a =

has two roots x1, x2 satisfying xt.x2
5a2

+ 2x)

O

(a +

O)

= 1 then

=2b2 + ac.

T41298. Let P be the midpoint of the side BC

of a triangle ABC and BE, CF be its two
altitudes. The line passing through A,
M.T\e
to PE,
cuts the line BE at N. Let K and G be
respectively the midpoints of BM and CN, let H
perpendicular to PF, cuts the line CF at
line passing through A, perpendicular

Xn+2

=

n+l'x

l,rr = !,
2'

n

2002 x,,*, + 2001 x,, + 2000 x,,*1.x,,

for every n = 0,1,2,

...

Calculate x,, in terms of n.

Tgl2g8. Let AH be the altitude and r be the
inradius of a triangle ABC. Let p1 be the
semiperimeter of triangle ABH and r'1 be its
inradius. Let prbe the semiperimeter of triangle
ACH and r, be its inradius. Prove that the
triangle ABC is right at ,4 if and only if

be the point of intersection of KF and GE. Prove
that AH is perpendicular to EF.

TSl298. Let ABCD be a square. Take a point
E on the side AB and a point F on the side CD
such that AE = CF. The lines BF and CE cut the
line AD respectively at N and M.Find the locus
of the point of intersection P of the lines BM
and CN when E moves on the side AB and F
moves on the side CD.
FOR UPPER SECONDARY SCHOOLS
T61298. Find the remainder of the division of
the product

Let ABCD be an

equifaced

tetrahedron (AB = CD, AC = BD, AD

= BC).Let

T101298.

R and r be respectively the radii of

the
sphere
circumscribed sphere and the inscribed

of the tetrahedron ABCD. Let

R,, and r,, be

respectively the circumradius and inradius of
triangle ABC. Prove that

R,, +
'tR'z

When does equality occur

4, ..[
- ""

1f

?

13

Li NgocTud'n,
78, HodngThi Mai,9C, THCS Nhr BA S!, Hoang H6a
; Quing Tri: Phan Hodng PhuongTrane,612,TIICS
Nguyen Trdi, Tx. DOng Hi, Pham HodngTh*dng,98,
THCS Tri0u DOng, Trieu Phong ; Quing Nam: Ngrlvdr
9A, THm NK Trdn Phri ; Thanh Hdat

glnr nnr xi Tnuoc

Dttong Tudn, 913, THCS Nguyen Du, Tx. Tam K) ;
Quing Ngai : Huj,nft COngThinh, gD, THCS Phd Van,
Drlc Phd, Phqm Minh Vaong, 9D1, THCS Nguydn

Bni Tf/293. Tim tdlt cd cdc sa'fi.t nhi€n n sao
2tt + 2" ld sa'chirh phtrtng
Ldi giii. Ctich t. Gii sir 28 + Ztt + 2n = a2 vdi

cho sd'28 +

NghiOm, Tx. Quing Ngdi ; Binh Dinhl. Nguy€n Y€n
Vu,8A8, THCS D6ng Da, Quy Nhon, Li Tan Khdnh.
8412, THCS If Hdng Phong, Quy Nhon; Tp. H6 Chi

Minh: Irrir Ddng Thinh, 642, TII Thuc nghiem SP.
Q.5 ; Cdn Thot Nguydn Minh Ludn, 8A, THCS
Nguy6n ViOt H6ng, Tp. Cdn Tho.

VIET HAI

aeNthi2"=o2-(28+}tt)

o

2" = o2

-

48'

e

2tt = (a

-

48)(o

+

48)

BdiTZl294. Tfnh gid tri bidu thric

a+l

la+48=2P

Tir d5 c6 {

la - 48 =2q

vdip,qeNvdp+q=n,p>q.

e 2q (z+-q - l) =2s .3
-q=5vhp-Q=2)p=7
)n=5+7=12.
Suy

ra 2P -

2q = 96

Thirlai c6 28 +Ztt +2t2 =802
Cdch 2.

.

Ndu n < 8. TiI 482 =28 + 2rr <28 + }rt +
2" <28 +zte +28 =2560 < 512 suy raa2 =28
+ 211 + 2" chi c6 rhd ldy gir4 tri 492 hoac 502,
nhung kh6ng cd n e N th6a mdn.
o Noun> 8 ditn = 8 + mth\a2 =28 +211 +
28+m - 2812* + 9) = 162 (2^ + 9) suy ra 2* + 9

b2 e 2* =b2 -32 vdi b e N. GiAi tidp
tuongtunhucdch l duo.c m=4. Dod5 n=12.

-

NhAn x6t. C6c ban sau c6 ldi giAi tdt :
Phri Tho: Vfi Minh Hodng,9A, THCS Giay, Phong
ChAu, Phir Ninh, ?'rrin Hfru Hi€'u,gA, TIICS Thi tran
SOng Thao ; V'inh Phtic : Pham Huy, HodngThi Minh
Hu€, lC, THCS Tam Duong, Bni Thi Thu Hi6n, 98,
THCS YOn Lac, Ngtry€n Quang Hd,7A4, THCS Vrnh
Tudng ; He NOi : Trdn Vi€t Tudn, 8H, THCS Le Quf
Don, Nghia TAn, Cdu GiAy ; Bdc Giang: Phan Tud'n
Anh,8A, THCS Drlc Giang, Yen D[ng ; Hi Nam: ki'r
Phan Binh,98, THCS Trdn Phri, Tx. Phfi Lf ; Nam
Dinh : Pham Duy Hidn, Nguydn Dtc Tdm, Hodng LA
Son, Vfr Khdc K!,9A7, THCS Trdn Dang Ninh, Tp.

Nam Dinh ; Hii Duong : Nguy,Sn Thi Quyan, 9Al,
THCS Chu Van An, Thanh Hd, Nguy1n Quang Anh,
8A,Ii Dinh Huy,9A, THCS Nguy,3n Trdi, Nam Srich ;
Hii Phdne z Bii Thi Thiy Linh, 9A, THCS Vinh

Phong, Vinh 86o, Pham Anh Minh, Vuong Anh
Chuyin, Nguyin Dt:rc Phuong, Trdn Xudn Dfrng,9A,

t4

trong dd a ld nghidm duong cria philong

tririt

+Jir-Ji

4rz

=O
Ldi giii. (cia HodngTudn Dfing,8A7, THCS
Phan Chu Trinh, BuOn Ma ThuOt, Ddc Ldc).

trinh

4x2 + J-2,
Phuong
nghiOm tr6i ddu nOn ta c5

-

nD

=

O c6 hai

4a2+Jia-Jr=o (1) (a>o)
a-l-2a+a2
thav oz = !_! vit a
TU ( t) de-J2J2
8

Goi bidu thrlc cdn tinh

li

S ta c6

L-

l-a

a+l
+a+l-a2
l-2a+o2 +8a+8

2.,12

a+3 |-a
L---l

2J', 2J'

F
/

Nhin x6t. C6 nhidu ban giii duoc biri ndy. Ciic ban
ldi giii tdt : Bdc Giang: Phan Tua'n Anh,
8A, THCS Drlc Giang, YOn D[ng ; Phri Tho :Vfi Minh
sau dAy c6

Hdne, 9A, THCS Giay, Phong ChAu, Phir Ninh; YInh
Phric : Pham Huy, HodngThi Minh Hu€ ,7C,THCS
Tam Drrong, Tam Duong ; Hii Phdng z Pham Anh
Minh,Trdn Xtdn Dfrng,9A, THPT NK TrZin Phri ; Hii
Duong : Li Dinh Huy,9A, THCS Nguy0n Trdi, Nam
Srdch ; Ciin Tho: Nguy€n Minh Ludn. 8A, THCS
Nguy€n Vi0t H6ng ; Hi T6y: Khudr l'dn Son, 9A'

THCS Thach That ; Thtra Thi6n ' Hue: Hd Vil Ngoc
Phtrong, g/4, THCS Nguy6n Tri Phuong, Hud ; Bac
Li6uz Nguy/n Vdn Tudn,9Al, THCS CL H0 Phdng'
Gi6 Rai.

rd NcuveN


cdr eAl xi rnuoc
Bdi T31294. G7i x ld s6' ldn nhdt trong ba sd
dtong x, y,-z.Tim gid tri nh6 nhdi cila bidu

tlu,rc

L

rhuc

+

v

t;

{1+r+

Ldi giii. Sir dung bat
sd trong c5n, ta c6 :

ding thrlc C6-si cho ciic

x

A= -*

Ldi giii. l. Khi Q nim ngodi dudng trdn
(Hinh 1). AQ cxt dudng trdn & B' vd ta c6 BB'lit

v

>

r + $rE . *t,_t_
y
\z
Xi

duong kinh. Ta

=

=#(;.0f, .'iEJ .(,- #);.
(w-:=)uE (r)
\

Sir

Bdi T41294. Cho tam giitc ABC ndi'ti€p
dudng trdn tdm O. Goi M ld didm ndo d6 tr€n
canh AC (M khdc A, C) Dudng fidng BM cdt

dudng trdn ldn nfta tai N. Dudng thdng qua A
vu1ng g6c vdi AB vd dudng thdng qua N vudrtg
g6c vdi NC cdt nhau tai didm Q. Chftng minh
rdng dadng thdng QM ludn di qua mdt didm c6'
difi khi M di chuydrt tr€n canh AC.

c6 ETB=

fr *t

=90". Ma

Nr + N2 = 90" n€n B' NC + N, = l$Q').
it hjc 1" tir si6c AB'NC nQi tidp nen
B'AC + B'NC = 180". Y4y B'AM = N2 (1)

zJz)\*

dung BDT C6-si cho 11 sd, ta co

*(;.4f..f) ,# \v,,
lll

lx 1v
-

11

z

-

(2)

2^,t2

mat khilc theo

cho n6n

gii

{ >l>1vlvi
nen

til ( I), (2) ra c6

zl,

--L>o>
zJz

x

v

-A
JZ-i

thiSt dd cho x = maxlx, y,

:

2,12

a,$*(r-+'l
.(w--L)
2.12 \
i \.
)
2V2

=

Hinh

I

Ldy N' ddi_",:rg ud{ qug_MQ c5 ngay
MN'Q = MNQ md MNQ+N2= 1800. Suy
(2)
ra MN'Q+ Nz = 180"

2^,12

=t+ Ji+1,E
Ding thric chi xAy ra khi x = J = z. Y4y gir{ tri

nh6nhat.tuata

t+ Ji*VIm,i x=y=z

NhAn x6t. Ciic ban sau dAy grii ldi giai c6 ddp sd
dring : Dic Lac : HodngTudh Dilng,8A7, THCS Phan
Chu Trinh, BuOn lr,'{a Thuot ; H} finh: Trdn Minh
Trong, 9C, THCS Trung Luong, Hdng Linh ; Quing
Trit, Ph.an Hodng Phildttg Trung. 612 THCS Nguyen
Trdi, Ding Hit, Pham HodngThudng, gB, THCS Trieu
DOng, Tri€u Phong ; CAn Tho: Nguy€n Minh Ludn,
8A, TIICS Nguy6n Viet Hdng, Tp. Cdn Tho ; Nam
Dinh: Vfr Khdc Ki vir Nguyin Dric Tam,9A7, THCS
Trdn Dang Ninh, Tp. Nam Dinh ; Bdc Giang : Phan
Tud'nAnh, 8A, THCS Drlc Giang, YOn D0ng ; Vinh
Phitc: N suydn Phi Cudng,98, TIICS_Ye: tu..
.

VU DINH HOA

Hinh 2

Tt(1)

vd (2) ta c6 MAQ+

Chung t6

tf

MN' Q=180"

.

gir{c AMN'Q nOi tiep

15


ctAr aAr xi

ra N' AQ = N' MQ = QMN (tinh chat ddi
xung). Mat khr{c jMN =& (gO. c6 canh
Suy

rnuoc

c6 dinh.

NhAn x6t.
1) Bii nny c6 thd dDng dinh

9"
MI 2
OC
QF +"

=

li

Men€lauyt dAo dd

tt

rsuy ra o. M. Q thanghins.

2) Cfrng c6 thd chrlng minh phAn -chfng vd dua vho
bhi toein Con budm suy ra Q, M, O thdng h?rng.
3) Giai tdt bei nay c6 c6c ban

:

Hii DuHn NOi : Ngb Minh Thdnh, 9A, THCS Nguy6n
TrudngT6 ; Nam Dinh: Ngryifl D[rc Tdm,9A7, Vfi
Khdc K!,9A7, THCS Trdn Dang Ninh ; Hii Phdng :
Pham Anh Minh,9A1, THPT Trdn Phi, Nguvdn Dtc
Phrong,9A, TI{P| Trdn Phri ; Thanh H6a : Hodng
Qudc Hodn,98, THCS Trdn Mai Ninh, Thanh H6a...

VU KIM THUY

Bni TSl294. Goi A y,d B ld citc giao didm cia
lrui dLrdng trdn tdm O bdn kinh n vd tdrn O' bdn
kinh R'. Ti{p tult€it chung cia hai dudng trdn
ndy ti€p xtic voi dudng trdn tdm O vd tdm O' ldn
luot tai T vci T'. Chrtng minh rdng didm B ld
trong tam cL)a tam gidc ATT'khi vd chi khi

/;

oo'="(R*R')
2'
Ldi giii. Gii sir l lI giao didm ctra AB vdTT' :
H ld giao didm cria OO' vit AB. Nhdn thay

hayTI2

N.BI
IT'
Mat khric Sorr + So,-r,t= !ro,
2
l_
I^
_
nen 5676. =

=N.BI (r)
(2)

Tuong tu cfrng c6T'12 =
TU (1), (2) suy ra IT =

EAN' = C . Suy ra
AB'NN'ld tr1 gir{c n6i tidp. V4y N'n[m tren
dudng trdn (ABC) vd NN' li day cung. Md
MQ L NN'nen duong thlng MQ di qua Am O
tuong fng vuong g5c) n€n

2. Khi Q nim trong dudng trdn (hinh 2) .
Cflilg minh tuong tu trcn, chi chf ]i rang

MAQ=MN'8 (do cilng bang CNB') vlt suy
ra AMQN' nOi tidp.

+=+
BI TI

MBI utATI, tird6:

=

,Soo'r,,

(3)
+

o'T').TI

,Soo'r,,

Yay IH.OO'= (R+ R').ff.

Til d5

t;
[f2 3 IA.IB
oo'= 9rn+R') e T=- €,
2tHt4lH'4
m2 -lnz
=i o

o
e

BH

IHz4lH23

B

li

3

-==L e st ='Lu

trong tam LATT'(dpcm).

NhAn x6t. l) Dd chfng minh dudng thing AB c6t
doanTT'tai trung didm cira doan thing d6 nhidu ban su
dung ddn khdi nicm phuong tich cira mot didm ddi vdi
mot dudng trdn. Ta nOn sir dung su ddng dang ctra hai
tam gi6c thi phi hqp vdi kidn thric toiin THCS.
2) Cdc ban sau day c6 lbi gini gon, s6ng srja, it ke
thom dudng phu :
Phri Tho : Vrt Minh Hodng,9A, THCS Giay, Phong
ChAu, Phir Ninh ; Hii Phdng : Bili Tud'n Anh, Trdn
Xudn Dfing, Pham Anh Minh,9A, THCS NK Trdn Phri ;

Hii

Dutrng : Li Dinh Huy. 9A, THCS Nguy€n Trai.
Nam Sdch ; Nam Dinh : Pham Duy Hidn,Vu Khdc Ky,
Hodng LA Son, Nguy€n Dirc Tdm, 9A7, THCS Trdn
Dang

Ninh

'H6 eueNc vmn

BdiT61294. Cho p ld s6'nguy|n t6'ldn hon 3.
Chrtng minh rdng

Cp*t
2O{\l

, -l -l

chia hei cho p1,

P'

trong d6 Cl, ld rti hop chdp k ctia il.

Ldi giii. (cira ban Hodng Ngoc Minh,1lA1,
TFIPT chuy6n HDng Vuong, Phri Tho)
Ta s6 chrlng minh kdt quA tdng qu6t sau : Ndu

p

li

sd nguy€n td lon hon 3

thi CP-l , - I
mp -t

chia

hdt cho pk*2 vdi moi m, t e N*.
Dd chrirrg minh ta sir dung kdt quA sau

Dinh li La-grdng : Cho s6' rtguy€rt td p. Xet
da thtlc P(x) = anx" + ar-tx"-l * ...a1x * a,,
vdi n e N*, a; eZ.
L6


orAr eAr
Gici sft cd n+l s6'nguy€rt dl, az, ..., d,,;1 sao
cho ai / a1 (mod p) n€fu i *i vd P(a;) =Q

(mod

p)

Vi = 1,2, ..., rt + 1. Khi d6 a;

=

Q

lmod p) Vi = 0. I,2, ..., rt.
Chrlmg minh bang quy n4p theo n. DC thay
dinh li dring vdi n = l. Gii sir dinh li dring vdi
moi da thrlc h€ sd nguy0n v6i bAc nh6 hon n. Ta
ki hiOu bAc cria G(x) ln deg G(x). Dat G(x) =
P(x) - a,,(x - ctt) ... (x - cn). Khi d6 deg G < n
vd G(a;) : 0 (mod p) (Vi = 1,2, ..., rr) do d5

G(x) :0 (mod p) > G(u,,*r) = 0 (mod p)
=
a,(cxr*1 - c[r)... (c[r+r - or) = 0 (mod p)
) o,,= 0 (mod p)
X6t da thfc

H(x)=P(x)- o,l'=a,r-tx" | +...+ a1xta.r.
Ta c6 deg H < nvh ll(cr;) = 0 (mod p) Yi = l,
2, ..., rt. Theo quy nap ao, ..., an-t: 0 (mod p)
Tro lai bbi to6n cta ta. X6t da thrlc

f(x) = (x-l)(x-2)...(x-pt+l) - l'-t - (p-1)! =
= ap-2f-2 + ... + af.
Ta c6 : f(i):0 (mod p) Vi = 1,2, ..., p -l
(dinh li Uyn-xon vd Phec-ma), do d6 (vi degf <
p-l) nOn theo dinh li La-gring vila chring minh
c6 ai= 0 (mod p) Vi = 1,2, ..., p - 2.

f?) = or-rf-'+

Lai c6

... + otp =

xirnuoc
NhAn x6t. DAy ld m6t bdi to6n kh6 nen c6 it ban giri
sau ddy c6 ldi giii r6t : Hd Hirtt
Cao Trinh, 11 To6n, THPT Hoirng Van Thu, Hda
Binh; Trdn Vd Huy, l1T, PTNK Tp. H0 Chi Ir{inh,

ldi gi6i den. C6c ban

Nguyin Thdnh Nam, 11T, THm Nguy6n Tr6i, Hii
Duong ; Nguy/n Hodi Phaong, l0T TIIPT chuyen Le

Qui D.n' Binh

DANG HUNG TIIANG

BdiT7l294. Day so"(a,,) (r, = 1,2, 3, ...)

azp2 +

atp ='arp

moi tt = 1 ,2, 3,

Ta c6

(*pr

: CP-l
mP -l

at

:0

voi rndi n = 2,3, ...
Ldi gifri. V6i m6i
-

t
(p

-

1)

tl

/

r\P-2

= ap-2\*pr )'

*

r

tt2

a2\^pr )

+ ...+

+

tt +

Vi (p-1)!
th&

aripz)

ud pk*2 nguy6n td

[cr-r
. -,),r'.'.
\ rnr-t
)

cing nhau, thhnh

(t

|

Zn(n +

-t)-

l \

D.,l;t+ |
I

2(n + l)2

'l;,

.2

(t
r )(r
lzJi 2.4Jj ) \2.+"6
t

I

2J'

1)

2(rt + l)z

I)

J'
n(n+l)(n+2)

tdng a1 * a2*

...

'l;,

Jr( r

t

)_

J,*? )-

1--

.2

Cdch khdc

=-r

-l+
)

z.e^,t4

- 2(n+l)2

I

NhAn x6t.

1

2rl2
Tit 2n2 ) n+l

r

s

s,,

I

)

rtOlXCI

2 [n1rr+1,1 \n+l)(n+21 )
a,,.

Ban Nguydn Ngoc Cudng, l0Al, THm Phan BOi
ChAu, Vinh, Ngh€ An da giii bdi to6n tdng quiit hon

sl
.

)

- 2,,rffi (rr n+2)

tl---lrl-

tl

chia hdt cho pk*2 (do

I

1

v6r a,, =

o,(*pk)*(*rr)o'

1

TtId6:ay*a2+,..+ ar<

-p+l)

.

L

+ ... + a,<

-_tt

n(n + 2)

-

2n'Jn+l

(mod p2)

c'-tr -1] = r(*or)*(*, * )'-'
mp-l

a2

D'l;i

I

.,1

-__

+

e N* ta c6

1

1

Ztf

+l

Ztr,ltt

=-_

|

v4y

ru

*

I
utt

,,2 (,,

...

\znz Jn+r

-l)...(mpk
(p - 1)!

a,, =

Chtlng minh rdng : a1

(modp3)

)

suy ra arp = O(modp3;

I

daoc xdc dinh bdi

-l/
= 0 (mod p'') do p > 3. Mat kh6c v\ ar:p ndn + +(
f(p)=

Dinh'

)a,<

A'

n*

1n+

ilJ;+ q'

trongd6

13p
suyra

,

p\+q)

2) C6c ban sau c6 ldi giii t6t : Hba Binh: Hd Hrtu
Cao Trinh, llT, Nguydn Ldm Tuy€n, l2T, THm NK
Hoing Van ThU

; Phti Tho: Vfi Minh Hodng, 94,

THCS GiAy, Phong ChAu, L€

Vi|t Ha, l0Al,

Hodng

11


cdr

eAr

lAl, THm HDng Vuong ; Vinh Ph(c :
Ngut,in ThanhHdi, l0Al, THm chuy€n. Nguy,in Phti
Crdng,98, THCS Y€n Lac : Ngh0 An: Nguyin Danh

Hdo, Nguydn Thi Nhung, L}Al, Hodng Thdi Hdo,
10A2, THIrI chuyOn Phan BQi ChAu ; Binh Dinh:
Nguyin Hodi Phaong, 10T, TI{PT chuyOn L€ Quj DOn;
Llro Cai: Hodng Vi€r, l0A, THPT Tx. Lio Cai ; Hd

xirnuoc

Ngoc Minh, I

Ndi: Ngd MinhThdnh,9.A. THCS Nguyen Trudng T6,
Ddng Da ; H6i Duong: Pham Thdnh Trung, 11T,
THPT Nguyen Trai, Tp. Hii Duong ; Hii Phbng :
Nguy€n Drtc Phrong, gA,

THm NK Trdn Phri ; Quing

Ninh: Ddo Anh Minh,llBl, THPT UOng Bi; Hi Nam:
Trdn Phan Binh,9B, THCS Trdn Phf, Pht Lf; Quing
Tri; Hodng Ti€h Trung, l0T, THPT chuyen Le Quj
chuyen Le Khidt

; Tp. H6 Chi Minh: Nguydn
- DHQG.

Ldm

Hrng, I lT, THPT NK DHKHTN

NCUYEN MINH DUC

Bni T81294. Tim cid tri cia bidu tht?c
a(x2 + y? ) + z2 trong c16 a ld hdng s6' thilc
duong vd x, y, z ld cdc bi|it s6'thda mdn di6u
ki€nxy+yz+zx=1.
Ldi giii. Vdi moi cr e (0, a) ta ddu c6
2-

tr))
It'i Y2)>'lZ'
+

tr
olr=
a(x2

+

(3)

vi

do

=o

+

+

z2 >

bo (x, y, z) rlng vdi
:

Tp. Hd Chi Minh: Nguydn AnhThiti, I lT-T, THPI
L€ Hdng Phong, NguyAn Khdc Dinh, l2T, Tran Vo
Huy, llT, THIrI NK DHQG Tp. HCM ; VInh Phfc :
H odng Lo n g, N guyd n T hanh. H di, l0 Al, T rd n Bd Bttch,
12A2, TIIPT chuyen ; Phri Tho : L€ Vi4t Hd, l0{l,
Hodng Ngoc Minh, I lAl, THPT chuy€n HDng Vuong ;
Hii Duong : Le Dinh Huy, 9A, THCS Nguy6n Trdi.
Nam S6ch, Nguy,ln Anh Ngoc, llAl, Pham Thdnh
Trung, I lT, TIIPT NK Nguy6n Trdi, Tp. Hii Duong ;
Binh Dinh: Nguy€n Hodi Phuong, 10T, THm Le Quf
DOn, Quy Nhon ; Hda Binh t Nguydn LdmTuyin,l2T,
THPT chuy0n Hoing Van Thu ; KhSnh Hdat Nguyin

Cho dudng trdn tdm O bail kinh
R vd s6' tu nhi€n ld n. Tim vi trl n didm A 1, A2,
..., A,, tr€n dudng trdn sao cho tdng cdc doan
ilfing A1A, + ArA., +...+ Ar-1A,, + A,,A,

tri ldn nhrilt

Ldi giii. Dat q^t=x; e [0, n] (i = 1,2, ...,
n), An+t : Al. Vi n 16 nOn ding thric AiOAi*t =
n khOng thti xiy ra vdi moi i = 1,2,..., n n€n ta
c5 bat ding thric.
(1)
\ xi

-r * JGs" ,taduoc

y2)

C6c ban saud6y c6 ldi giai dring

dat gid

f

I

BiiT9l294.

*2 +'y21 + z2 > z

o'*

NhAn x6t. Nhidu ban chi nOu
GTNN cira k.

9A7, THCS Trdn Dang Ninh.

Q)

COng theo trmg vd cira (1), (2) vh (3)

Chon cr sao cho

t__ :1 -.,h + 8d
['- 2 *,lt + 8a

Vinh ; Bdc Giang: Nguy€n Tud'n Huong, l2A2,THtr|
Vi€t Yen; Nam Dinh: Vn Khdc Kl, Nguydn DftcTdm,

(l)

t1u+yz+zx=lduoc

[" ,B),

{

NGUYEN VAN MAU

2r
avz + L ,- z.lLv,
' 2 \'lZ'

.

hoic

Hoa Cuong, 12T, THP| Le Quf DOn, Nha Trang ;
Ngh0 An : Trdn Ki€n Trung, 10A1, Todn - Tin, DH

:

or'*!>z^l?*,

2 \'lZ

-t

[,= ' tlnBo

l
-t*Jt*s,

(o4,OA^)

Mat khiic *, =

(mod 2n) Vi =

1,2,...,\t.
Vay min(a(

*' *y')*r')=-t*P

wi

I ,, = >, (oA, ,6i,) (mod 2nr
= ll
Vcrx

{cry

z
- .'lZ
Z

= ..12
-

xy+yz+zx=l
l8

= Z
1

f.= ' 154;
\r=-

*1

l'=

-t+Jt+gd
21lr+8a

\

*, =0 (mod 2n)

(2)

1
Tit (1), (2) suy ra

:

,

x;

l
Theo dinh lf hlLm sd sin

:

< (n-l)n (vi rr 16)


cAr aAr xi rnuoc

Z-:

3 2n11r tsinf
L

Z

A,A,*, =2R Z.

ll
3 2nprir9!:)n

co

t'i
u.l

lsis'r4 <

)''0<

(rr

- l)n<-7r.
2n 2'

)

n

song song vdi

Ding thrlc xiy ra

€
o

.{1 =

xz = ...= xn =@

-----:--r- -:--=r-

thidt diQn MPNQ ll
SB thi hidn nhiOn ta

,AP b
BQa
-=-,
NOu m4t phing
(MPNQ) kh6ng

2n

(Chf

Ldi giii. (dua theo Nguydn LdmTuydn,l2T,
TIIPT chuy6n Holng Vin Thu, Hda Binh)
a) Ndu mit phing

-l)n

cira

thi SB giao vdi

MP

(MPNQ) vd SB li D

n

ArOA, = AzOAt=... =

t,pi

_ @-l)x
tt

€ Ar, A2, ..., A,, ld cdc dinh c[ra n - gi6c ddu
nOi tidp duong trbn (O, R) vi A1A2, 4243, ...,
A,At ld cr{c dudng ch6o d}ri nhAt cira n - gi6c
deu oo.

taiDe(N0).Ap

NhAn x6t. Bhi to6n niy kh0ng kh6, tuy nhi6n chi c6
5 ban c6 kdt quA dring mi cfing kh6ng c6 ban n]ro cho
ldi giAi s6ng sta vd chinh x6c. NguyOn nhAn cira su vi€c
niy ld khOng ai biet tAn dung kh6i ru€m g6c dinh hudng
gifa hai v6cto dd l]rm to6n. Nhu chring ta dE thay 6 tr€n,

!

x; < (n-l)xdd duoc chrlng minh rdt

li M0-nO-la-uft

vho hai c4p

4
ryg!.8=*r
PA MB QC

ur'!=Y

MB NS

ban' giAi sai ddu kdt luAn rlng

e

A1A2...A,,ldda

gi6c ddu nQi tidp duhng trdn (O, R) !
3) C6c ban c6 kdt qu? dring li : Phri Tho: Hodng
Ngoc Minh,11A1, THPT chuyen Htng Vuong ; Ngh6
An : Nguydn Ddng Hdo,10Al, THPT chuy6n Phan B6i

Chiu;Hii

Duong : PhamThdnhTrung,

llT,

TIIPT

NK Nguy6n Tr6i ; Hda Binh: Ngaydrz limTuydn,l2T,

Hd Hilu CaoTrinh, 11T, THPT Hoing Van Thu.

NGUYENMINH HA

Bei T101294. Cho hinh ch6p tam gidc

d€u

SABC cd canh ddy AB= a vd canh b€n SA = b.
Goi M vd N ldn luot ld trung didm cdc canh AB
lri SC. Mdt mdt phdng hay ddi quay xung
quanh MN cdt cdc canh SA vd BC theo tht w d
P vd Q (P, Q khdng tring voi cdc d{nh).

ll =!a
'BQ
b) )hc dinh i ,d + sao clto thi€i di6n
a) Chrtng ntirtlt rdrtg

AS

MPNO c6 di€n tich nhd nhdlt.

(1)

(Dlnh li M6-n0-la-uft trong HHKG dp dung
vlo trl gi6c ghdnh SA.BC vd m[t phing (PMQN)).

gita hai v6cto.
nh6 nhdt

:

NS

ggn gang vd s6ng sira nhd kh6i ni€m g6c dinh hu6ng

(erer* $.At+...+4,Ar)

tam

gi6c vd ci{t tuydn :
(AABS, DPM); (ACBS, DNQ) ta duoc h€ thtlc

lli-
2) Da sd c6c

A

dung dinh

Tn d6 ta c6 kdt luan cdn thidt cho bii toi{n.

bat dingurt thric

SB,

ggi giao

=

-1

non tir (1) ta duoc

:

g=lg
PS

QC

L4i vi P vit Q nim trong c6c canh AS vd BC

-APBOAPPSAS

nfntaCO.

P.' QC = -=-=-.
BQ QC
-=----=

BC

VAvtac5APb
-BQa

(2)

b) Dat : s(MPNQ) = s,

l.<

1), ta duoc

Goi R

li

Y=9
AS BC

=1. (o <

, AF =XAS , BQ=?'E1

trung didm cira doan thing PQ,th\:
-l

-

2MR= AP + BQ

-

l.(AS + BC)=

2MN

(3)

(3) chung t6 MN di qua trung didm

PQ.

15,

huy

Tt

d6 suy ra : S(MNP) = S\MNQ)

=

2

(4)
S = 2S(MNP)
Bay gid ggi G lh trung didm cria MN, G
cfrng lh trong mm cria hinh ch6p SA.BC vb G

t9


cdr

eAr xi

thu6c dudng cao SH cua SABC. Dung GK I
AS tai K. Goi / lh trung didm cua SB, the thi
IMllSA, INIIBC vb do d6 GK L IM. Mat kh6c
mp(SAFO L BC n€n GK I 1N. TU d6 suy ra
GK L MN.
GK Lmp(lMN)

=

Do d5 khoing c6ch d(P, MN) > dn= KG vd
d(P, MN) - do= KG. Ttd6 vd (4) suY ra :S >
MN.GK vd minS - So = MN.GK.
Ta hay tinh

Od

f

),

theo a vd & trng v6i

S,,.

rang hai tam gi6c vu6ng'KSG vd HSA

d6ng dang vdi nhau n6n ta c6
Tir d5 ta duoc

:

.'K SG
SH SA

:

sK sH.sG
z( su\2
-I = 3b2 -a2
= _l _
sA SAz 4(sA) qu'

_=---

/5\

2

ivi

SH2

=

SA2

-

AHz = b'

- ll.3

KS a2+b2
AP _=l--=--_----:AK
^
5uvra
'
AS AS
AS 4bz
Khi P : K duoc x6c dinh b6i h0 thrlc (5) thi S
dat gid

tri nh6 nhat, hic d6
mlnJ

_ _a

NhAn x6t. 1) N6i chung c6c ban vAn dung kidn thfc
vd hinh hoc khOng gian (phdn quan h€ song song vi
vu6ng g6c c[ra dudng thing vi mat phdng) chrra thAt
nhudn nhuy€n, tri tu6rrg tuong khOng gian cdn bi han
chd. Didu d6 sE inh huong ddn kdt qui m6n hinh hoc
hoa hinh v?r v6 ki thuat khi c6c ban vho hoc & ciic
trudng dai hoc kI thuAt, xiy dung vi kidn trfc sau niy.

ci tich vecto (tich c6
nua dd giii bii to6n nhy, nhung ldi giii cdn dii,

2) Mot sd ban sir dung vecto,

hu6ng)
kh6ng gon. Ngodi ban LAm Tuydn, hai ban sau ddy c6

loi gidi tot

:

Hodng Ngoc Minh, 11A1, THPT chuyOn HDng
Vuong, Phri Tho ; Hd Hrtu Cao Trinh, 1l Todn, THPT
Hodng Van Thu, Hda Binh.

Chi c6 hai ban Hodng Ngoc Minh vit Li Cudng
Thinh,llAl, THm chuyen Hing Vuong, Phri Tho d€
cAp den vi tinh duoc gi6 tri nh6 nhAt S,,in cira dion t(ch
thiet dien MPNQ nhu dd ncu trcn.

NGUYEN DANG PHAT

Bdi Lll294. Cln doan mach kLC mdc na'i
fieb. Dat vdo hai ddu ctia n6 iieu di€n th€'xoay
chi€u u = UJTsinot , trong c16 t) kltotrg etdi
20

thay ddi. Khi a = o)t vd at = ah {
ot1 thi citc ddng dien tLtong ttng di trong moclt lct
i1 vd i2 c6 ctirrg cltdng dd hi€u dtlng (11 = I) vd
i 1 l€ch plru vdi i2 mil g6c 2a.

cbn a: c6

fid

Tim a = o3 ftheo at1 vd o) dd cudng dd
ddng di€n hi€u dung qua mach ld cttc dai
I)

- I-u-r)
2) Tinlr L vd C theo R, ct, cot, cot vd theo
R = 300Q, at = 200tr radls, a4 = 50n radls,
a= trll.
(1.1

Ldi giii. 1) Tt

11

=

raZl=22

12 suy

vi trr2 * col
= coll
' -+
azC)
orC= x(I ,rt- +].
ncn ta cd or,L-l = _-urtL.+ (l) suy

'

ra o1o2

=

'

,,C

(2). Khi

*

13

=

*r.C

trong mach c5

1,no*

c6ng hu6ng di6n, vi vAy ta c6 : rojl,

^= I

i

,i,4

(3).

ru

:

+ b2 )(3b2

8b

rnuoc

(2)

vi

= -!

(3) r)m duoc

=

:

,-, = Jor1r2
2) it tr6 pha hon a m6t g5c

xric dinh b6i

I

, - ,,C
-

(r),1,

igg, =


(4). Theo trcn i2 sdm pha so

R

'^-azL

1

vdi

(r)^ (

rz

m6t g6c Az, mirtg
L,vh
.

R

hon nfia, theo trOn, tBQr = tgQz <> 9, = 92. Nhu
v4y iz s6rn pha so v6i 11 m6t g6c qr + gz = Zer
= 2cr, suy ra gr - e2= d. TiI d6 vi (2), (4) rft

ta:L= Rtgu
0r - 0z

v1

C= 0l-oz

Thay sd ta duoc , L =

Rcolro2tgcr

?

(H) vi
TE

NhAn x6t. Cdc ban c6ldi giAi dring

, = '01
ZIT

1o;

:

Thanh H6a l LA Vdn Duong, 12,{1. THPT HAu
LOc I ; Ddng Nai: Nguydn Kim Huv, 12 Li I, THPT
chuyOn Luong Thd Vinh, BiOn Hdal Nghe An : Trdn
Dinh Trung, I lA Tin. PICT DH Vinh; Hcd Qu\t Huv,
Li Manh Crdng,Trdn QuangVn, llA7, THPT chuyen

I


,
cdl

I

Phan BQi ChAu, Vinh ; Hh
Hftu LId,1l Li, TFIPT NK

finh:

N guy,in Xudn

eAr

Dtc , La

Hi Tinh
MAI ANH

LZl294. Mbt ngudi nhin mOt vdt d phfa sau

ngudi d6 mdt khodng cdch d = 5m qua kfnh
mdt. Khi dd ngadi dy tring thdy 2 dnh cila vdt :
mdt d khodng cdch d1 = 5m, mQt d khodng cdch
dz = 517m. Quay mdt vd phia vdt rtgttdi dly nhin
thdl- qua kinh mdt dnh ctia n6 d khodng cdch
ds = 2,5m'

Hdy xdc dinh chiA't sudt cila thily tinh ldm

ri rnuoc
gudng" c5 dO tu : D' = 2O + ? v6i R lh bdn
R
kinh cira m6t thtl hai cira kinh. Gii thi6t rang
inh thu dugb do phin xa tt mdt tru6c cira kinh
nam 6 khoing cdch d1= 5m, khi ay mat ndy lir
mit ph[ng v\ d = 5m. Lric ndy doi vdi Anh thti

hai ta

"o

'

1- 1 = D, = zD * ?
dd2R

D = (tt-r\'[R
[l.1)

r) =

1

mat phing ncn

rhdh kinh cila kinh mdr.

Hu6ng d6n giii. Khi

quay mat vd phfa vAt
ngudi dy nhin thdy 6nh io cira n5. Ta c6 :

I

dd\

1

= D (1\,vdi D ld dd tu cira thAu kinh

Khi quay lung lai vAt, m6t

Anh cira vAt nhin thAy
do ph6n xa tren mat kinh hu6ng vd phia
ngudi, cbn Anh kia li do phAn xa trOn mat thrl
hai cria kinh. Vdi Anh thrl hai, cdc tia si{ng phii
di xuyOn qua thdu kfnh, phin xa trcn mat guong
vd sau d6 lai di qua thdu kinh ldn nta ; c6 nghra

duoc li

th Anh nhy duoc tao b&i h0 gh6p sdt "thAu kinh +

(d=5m; dz=
Ndu

gii

c[ra kinh

1
r

(rr

- l)

p1 uoi

(3) (mat rrudc l]r

R

= 0). TU(1), (2), (3), thay so

5

iI rnid.r = 2,5m,ta du-o^c rr = 1,5).

thidt ld sau khi phtn xa ffen mat tru6c

inh nam 6 khoiLne
"7cacn

to6n tuong tu ta lai tim duo.c n =

lm, thi tinh
0,75:

vO

li.

VAy rr = 1,5.
Nhan x6t. Ciic ban io toi giai cldy du vi dfng :
Vinh Phric l Ki6u Ngoc Thanh, 11B3, THm chuyOn
VInh Phtic ; Nghd An : Nguyin Thanh Tnng. 12A3,
TIIPT Phan Boi ChAu, Vinh.

MAI ANH

Ch8ng han, mOt qudc gia c6 tdc dO ting
tru&ng GDP binh quAn ld 9Va ndm

thi

72

,,=T=E(n5m)

?r(Wqr{?N1 GDP
Mot nhi kinh t€ dd dua ra cOng thfc dd tinh
s6 n6m ting gdp dOi tdng sAn phdm quOc nOi
(GDP) cira m6t qudc gia nhu sau

gdp doi GDP tit 200 USD/ngudi l6n 400
USD/ngudi. Du kidn t6c dO tang ffu&ng binh
qu1rr 1I T%olndmth\

:

,=?
o

trong d6

Nhu vAy, sau 8 ndm, GDP c[ra qu6c gia niy sE
tang gdp d6i.
Gii sir ndm 1992 mOt nudc c6 chiOn luoc tang

:

o gVo ld. tdc d0 tang trudng binh quAn GDP
trong giai cloan xem x6t,
o n llL sd nam dd tang gdp doi GDP.

GDP (Gross Domestic Product) lI tdng giri tri
thi trudng tdt ca cdc hlng h6a vI dich vu cudi
ctng dd du-o. c sin xudt ra trong nudc m6t nam.

7?.

n -- --1

:

10 (nam),

7

li

ddn ndm 2002 nudc d6 sE dat binh quAn
400 USD/ngudi.
COng thrlc tfnh to6n ren trong linh vuc kinh td

trlc

that don giin, mang tfnh chat udc luong, tr"ry
nhion trong thuc ti6n su tang trudng kinh tC di0n
ra kh6ng don giin chft ndo !
NGUYEN THUONG LANG

(6/

DH Kinh t€'Qu6'c ddn)

2t

I

VE CO SO
cila phuong phiip

I

G

ItlAI{

ltcttftvt rfP

DANG HLING THANG

(DHKWN -DHQG Hd N6i)

I

Theo quy dinh cira BO GD vh DT kd tt nam
hoc2000-2001 kh6ng duoc ding phuong phr{p
nghiom k6p dd giii quydt ci{c b}i to6n vd tidp
tuydn vi su tidp xfc cria hai dd thi vi n6 "thidu

co sd li thuydt". Thuc td cho thdy nhidu gir{o
viOn vi hoc sinh rdt hing tring khi kh6ng duoc
dDng phuong phdp nghiOm k6p, trong khi
phuong phr{p nIy thuc su c6 hiOu qu6.
Trudc hdt hdy nh6c lai khdi ni6m su tidp xfc
cfia dd thi hai hlm sd vd khr{i niOm nghiOm b6i

cfia da thric.

Dinh nghra.
1. DO thi hai him sd y =flx) vd y = g(x) du-o. c
goi ld tidp xdc v6i nhau tai didm c6 hodnh d6

^. lf7,,l=g(x,,)
-.
x=x^neu<
'(x")=
[f

8'(x")

(l)
.

2. Gie sit F(;r) li m6t da thrlc. Sd x,, duo. c goi
lI nghi€m b6i cria F(x) ndu F(,r) chia hdt cho
(* - ,")2 trlc li F(x) c6 dang
F(x)= (x-x.,)2.QQ)
b d6 Q(x) ld m6t da thric. Trong trudng hqp F(x)
lh m6t tam thfc bArc 2, nghiem bOi duoc goi ld
nghi€m kip.
Chri ), rang chring ta chi c6 khdi ni€m nghi€m
b1i ddi vdi da thftc.
Dinh lf sau dAy s6 chrlng t6 rang phuong ph6p
nghiOm k6p ld c5 co s6 toSn hoc khi chring ta
x6t su tidp xric cira hai dO thi c6c hdm phAn thrlc
hiru ti.

Dinh li. Clrc lrui plfirt rlulc ttfru ri

P(*)

-f(r) = QQ)

.

Khi dd d6 th! hai hdm s6'y = f(x) vd y = gk)
ti€p xric voi nhau tai didm c6 hodnh dQ x = x,,
khi vd chi khi phuong trinh
P(x)V(x) -Q0) U(x) = Q
c6 nghi€m bdi x = x,, vdi Q(x,,) * 0, V(x,,) * 0
(tftc ld x,, thuOc mi€'n xdc dinh cfia f(x) vd g(x)).
Vi da thrlc cfing li mOt phAn thtlc hfiu ti (khi
da thrlc d mAu thrlc li da thrlc hAng sO) nOn ta c6
22

cdc hO qui sau thudng dirng trong chuong trinh
phd thOng trung hoc.

qui.
HQl. Duong thing ! = kx + h Ld tidp tuydn
v6i dd thi him da thric y = P(x) khi vd chi khi
HO

phuong trinh P("r)

-

(kx

HQ2. Duong th&ng

!

+ lt) = g c6 nghi€m bOi.

= kx + lz lI tidp tuydn

vdi dd thi cira phAn thrlc hfru ti y. =

+-4

khi vd
QQ)
QG)&x + lr) = Q 96

chi khi phuong trinh P(x) nghiCm bQi x = x.vdi QQ) *0.

N5i ri0ng

:

HQ3. Duong thing
vdi dd thi hirn sd y

-

,

+ h lit tidp tuydn
^k,
ax + ox + c
khi vh chi
fitx+n
=

khi phuong trinh bAc hai
ax2 + bx + c - (mx + n)(kx+ lr) = 0
c6 nghi€m k6p (A = 0).
HQ4. Duong thing J = kx + h ld tidp tuydn
ax + b
v6i dd rhi hdm sd y khi vi chi khi

mx+n

phuong trinh bAc hai

ax+b-(mx+n)(kx+lr)=0
c6 nghiOm k6p (A = 0).

Dd chung minh dinh li ta cdn ddn bd dd sau
khi{ quen thu6c vh d0 ddng chrlng minh.
Bd di. Da thftc F(x) cd nghi€m bdi
vd chi khi F(x,) = F '(x,,) = 0.

x = x, khi

Qui vAy ndu .6. l) nghiCm b6i cira phuong

trinh F(x) = 0 thi do dinh nghra F(x) = (y - x)2
x 0(x). Suy ra F'(x) = (,r - x,12 9'1x1 +
+ 2(x - x")Q(x)
F(x") = F (x") = 0. D6o l4i
=
giA s[t F(x,) = F'(x.) = 0. Vi F(x,,) = 0
= F(r) =
(,r - x")G(x)
F '(x) = (,r - x,,) G'(x) + G(r)
=
=
G(x,,) = F'(x") = 0 3 G(x) = (x * x,) Q@).
Vay F(x) = (x - x)2 Q@) trlc ld F(x) c5 nghiOm
b6i x = x..


Chfmg minh dinh Ii. Gie su d6 thi ci{c him
sdflx) vd g(x) ti0p xric nhau tai didm c6 hohnh
dO

x = x". Khi d6 tir (1) c5

P(xo) _U(xo)
QQ,,) V(xo)
P' (x o)QQ,,)

-

BAI Sd 52
(2)

P(x,,)Q' Q,,)

Q'G,,)

=

_ U'(x,,)V(x,,) - U (xu)V'(x,,)
;r a;::;

,,
(r)

-

D4t q = Q@) + 0, v = V(x") * 0, p = P(x"),
tt = U(x), p'= P'(x), q'= Q'(x").
Tit qv + 0 vn (2) suy ra
P(xo)V(xo) = QQo)U(x,,) hay qu = pv (4)
Tt (3) c6 (p'q * pq')v2 - (u'v - ur)q' (5)
. Ndu pu * 0, nhAn theo tirng vd cira (4) vd (5)
duoc : (p'q - pq)v2qu = (u'v-uv)q2pv
<+ (p'v + pv)q2vu = (q'u + qu)vzpq
Theo (4) th\qu.qv - pv.qv + 0 n€n suy ra
p't' + pv' - cl'u + c1u' hay
P'(x.)V(x,,) + P(x")V'(x") =
(6)
= Q'(x.,)U(x,) + Q@)U'(x")
r Ndu pu = 0 thi ti (2) c6 p = u = O,thay vho
(5) duoc p'qv2 = r.ivqz > p'v = u'q. TU didu ndy

cing vdi P = Lt = 0 cf,ng suy ra (6).
T5m lai, tir (4), (6) vd bd ad ta kdt luAn rang
phuong trinh P(x)V(x) - Q@)U(x) = 0 c6
nghiOmbOi x=xo.
DAo lai ndu x - xn vdi Q@)V(x.) + 0 lI
nghiOm bQi cira phuong trinh P(x)V(x)
Q(x)U(x) = 0 thi tit (4) vh (6) bidn ddi ngu-oc lai
ta se suy ra (2) vi (3). Theo (l) d0 thi hai hdm s6
tidp xric vdi nhau tai didm c6 hodnh d0 x =.xn.
Chring tOi cho r6ng trong srich gi6o khoa
nen trinh bdy dinh li n6i rr€n (cd thd khong
chfng minh holc chfng minh v6i chfi nho) dd
ldm co s6 li thuydt cho phuong ph6p nghiOm
bOi. Tt d6 cho ph6p hoc sinh du-o. c str dung
phuong ph6p niy khi giAi c6c bli toi{n vd su tidp
xfc ctra dd thi hai him phAn thrlc hftu ti cflng
nhu bdi toiin vd tidp tuydn vdi dd rhi hlm phAn
thrlc hfru ti.

OUriu tiu
C6c ban doat gi6i c0a Cupc thi Vui hti 2001 vit
CuQc thi gidi Todn vd VQt Li tr6n THTT ndm hqc
2OOO-2OO{

hay ti6p tuc gfri dia chi m6i c0a minn vd

,ou.o,.n.

rrENG ANrr ouA cAc BAr ToAil

,r*

Problem. Let x1 3 x2 A ... 3 xn and y1 t Jt a
...Sln be two non-decreasing sequence ofreal
numbers. A permutation of yt,...,Jn is an
arrangement of the terms of this sequence in a
particular order. For an arbitrary permutation
!i1

, !i2..., linof y1,!2,...,y,r

prove that

x1 y,, +

xZ!,2+ ... * Jn !i,, S xtlt + xZJZ+ ... + xilyfi
and that equality holds

if

and only

if

x1 =

x, -

... = Xn Of )t - J2= ... = in.
Solution.

Put

S=

xtlil+

Suppose that

*rlitr.
y,, = li1. In the sum S we
xzliz+ ...

+

interchange !,,,und )iy . This transform

S

into

x2!ir+ ... * xjli,, + ... + xnyr.
We have S' - S = xn(!, - !i,,) + xj(yi, -yn) =
= (xr- x)(ln - )i,, ) > 0
S' =Jt1 !.i1*

because

xr2 xi and

y,,

) )in . Thus, S' > S. Next

find in S' the

summand containing Jn-t.
Suppose that y,,-1 - )t1 . In the sum S' we
interchange !ir_, and 1l;1 . This transform S'
into a new sum S". Similarly, it is easy to verify
that S" > S'. This process can be continued until
we obtain the sum xtlt * xzJz * ... I xn!,

which is larger than or equal to all previous
sums S, S', S",... So we have proved that
SIt is clear that equality holds if xy = x2 = ...
=XnOf

lt=JZ=...=!n.

Tir mdi:
decreasing = giam

real

= thuc (tinh tt)
permutation = ph6p hodn vi
term

particular

= sd h+ng

= d4c biQt, riCng (tinh

summand = hans ttl

tt)

interchange = ddi ;ho, thay rhe nhau (dOng tir)
(dgng tt)
similarly = mQt giich tuong tu,
cfrng nhu vay (ph5 tU)
= x6c minh, x6c nh1n (d.ng tt)
process = qu6 trinh

transform = bi€n ddi

verify

continue
previous

= ti€p tqc (dOng
= trudc (tinh tit)

tt)

NGO VIET TRUNG

23