Tạp chí toán học và tuổi trẻ số 319 tháng 1 năm 2004

sdrchoffi& C ,* aggy , BS GlAo DUc vA stc T*o S t$t TSAN x$e vt*r N*r,t oANH cl"ro rnuN* xQc pxd TriSNc vA rRUNG lroc co sd , , f's; s ,1 ,F' ' #r l & f s $' dg ft qfs '=*#, Ch*c ful urxffi ffiffirw? ilfi#il w 30{}4 i 1' 6 , 4 TAP CHI VAN HOC VA TI]U TRT I,IUOI NA[{ HINH THANH 1/A PIIAT TRIEN Ngdy l611212003, tai tru so NXB Girio duc d6 dien ra LC ki ni€rn n iLIdi nam Tap chi Vcitr lrcc vil 'l'rtdi rre Den du bueii 10 c6 dai bidu cfia Ban Tu tu6ng V[n h6a Trung uong, 86 Van h6a Th6ng tin, B0 Gi[.]

sdrchoffi& C ,*

BS GlAo DUc vA stc T*o S t$t TSAN x$e vt*r N*r,t

oANH cl"ro rnuN* xQc pxd TriSNc vA rRUNG lroc co sd

g

ft.q fs '=*#,

TAP CHI VAN HOC VA TI]U TRT I,IUOI NA[{ HINH THANH 1/A PIIAT TRIEN

Ngdy l611212003, tai tru so NXB Girio duc d6

dien ra LC ki ni€rn n-iLIdi nam Tap chi Vcitr lrcc vil

'l'rtdi rre Den du bueii 10 c6 dai bidu cfia Ban Tu

tu6ng V[n h6a Trung uong, 86 Van h6a - Th6ng

tin, B0 Gi6o duc vi Dlo tao, ldnh dao NXB Gi6o

duc, cdc uy vi€n Hoi dong bien tAp 'fap chi, circ

nh) r,an, nhd gido, nhl nghien cLi'u thuoc ll6i Nhlr

vin ViOt Nam, Vi€n V[n hoc, trudng Dai hoc Su

ph4m, So Gi6o duc vir Diro tao m6t so tinh, th)nh

pho, cdc tntdng THPT, TIICS f{e NOi.

TS Nguydn Kim Phong, Ph5 Tcing biOn tAp

NXII Gi/ro duc, Tcing bien tAp T'ap chi VHTT dA

doc cliOn r,ern vd qLrir trinh hinh th)nh vh phiit tridn

cLra Tiip chi PGS TS Vu Duong Thuy, Ph6 Tcing

Giiinl cloc kiOm TCing bi€n tAp NXB Gi6o duc dd

phtit bi6u ch)ro mirng Nhidu dai bidu d[ phrit bidu

y kien d6ng g6p cho hoat dOng cua Tap chf Lanh

dao NXB Gi6o duc dl trao tlng phdm cho mot so

lry vi0n HDBT vd Cong tirc viOn c6 nhidu d6ng

g6p cho'I'ap chi

Clrring tOr xin trich d[ng 2 bdi phrit bidu

'Nhu cliu vd mOt td Tap chi phuc vu cho viOc

day - hoc mOn V[n trong nh] trudng xuat phdt til

thLrc td da1 hoc mOn Vln, vI didu d6 duoc khing

dinh khi NXB Gido duc cho ra mat Van hoc vd

'l- utii t t'e

Ddi vdi nhrdu em hoc sinh, Tap chi da tro thinh

noi nAng dd, uorn mdm cho nhi6u t)i ning van

hoc C6 nhidu grdo viOn day Vln, nhrdu cdn bd

nghiOn citu vln hoc trd tirng tAm su lfc cdn ld hoc

sinh phd th6ng, nhd 1/z7r lrcc t,dTtdi trd mh y6u

mon Van hon, dat kdt quA cao d mon Vdn vI sau

cirng da theo nghiep vin "

PGS TS VII DITONG THUY

(Ph6't'61tg GD, 7'dng bi€n ttip N\B Gido duc)

"Ngly lll1ll99l, BO truong BO Gi6o duc vd

Dho tao Trdn LIOng Qr"rAn ra quyet dinh so

ngBnCCB tiep nhAn bitoToirt ltoc't'c)7'udi tre

tit Vi0n Khoa hoc Vi€t Narn va giao cho NXBGD

cluAn li Ngay tit gid ph(rt [1' lanh dao NXBGI)mi\ trudc hit Id Gidm doc Trrin Triirn Phuong, dd

thay dair 1) mOt ccr hoi rtgan vittg vl khi clrttrc

quiin li mOt to- bdo Yc nrol nrun irue qtlitn tlollgnhat trong nhd trudng von c6 lich str hi\ng chuc

nam vh dd duoc Nhir nu6c tang thuong Iluin

chuong Lao d6ng, NXB se c6 tl'r€m didu kien d0

ph6t hly uy th;rii thuc luc, clil hoirn thdnh toi h,rn

nhiOm vu chinh tn cua m)nh, Chng ngiiy til gia)

ph(t ay, chring toi vAn thudng trao ddi y kicr voi

nhau mOt crich di dorn: Trul'dn th6ng tham rn1,

phr"tong D6ng rat ton trong su cAn doi, c5 !{AI

mon hoc chiem so gid nhidu nhat, 16 nlo motmOn c6 tiip chi r.ni nrOn kia lai khong; chftug ta tu

nhi6n duoc ban tang mot bdn hoa dep, tru6c rnatphAi ldrn cho n5 dep hon - ngay luong phrlt hinh

s6 Todn hctc t,r) Tudi trd dAu tiOn sau khi dua r,6NXBGD da tAng gfp 4 ldn so vdi khi cdn o ViOn

Khoa hoc Vi€t Nam - clen mOt hic n)o d6, moingudi sC cAm thay nhu thidu ving mOt crii gi d5

v) ddi h6i n6 xudt hiOn: Dd chinh \d td Van lrr,,c

viiTutii trd ! Bbi vay, b6n than su ra ddi cira l'arrhoc t,d T-udi trd da thd hr€n tinh chat gido duc,

mhu sic hoc dudng sAu dAm Chinh didu ay da

thm cho an phdm d6c d6o n).y luOn xdc dinh drtocdLing cdc toa clO cuu minh - VAN HCC, TUOITRE, HOC OUONC - tr0n con ciu'dng b6o chi

qua c6c ching dudng tii s6ch chuyOn d6, dac san rOi tap chi, ddn xdc drnh duoc ban sic riOng gr[ahhng loat t€n ciic NXB, tap^chi, tudn bio cirng

mang tU TRE,, nhu TRE, YEU T'RE, TAI HOATRE, VAN NGHE TRE ''

GS NGTJYEN KHAC PHI

(llgut,€n Ph(t GD,T(ing bi0n tr.ip NXB Gido rittc:)

didu kiOn sau : d' * 0, a2 + b2 + c2 * O vh, L, = b'2

- 4a'c'< 0, nghia li m6u thrlc trong (1) li tam

thtlc bAc hai kh6ng c6 nghiOm (lu6n duong holc

luOn Am) Ndu tim duoc midn gi6 trf G cira him

sd (1) thi cfrng bi6t duoc gi6 tri l6n nhA't

(GTLN) vi gir{ tri nh6 nhdt (GTNN) ndu c6 cria

hdm sd d6 Dd tim midn gi6 tri c[ra him sd (1)

c6 thd sri dung cr{c phuong phdp de nOu trong

bbi b6o Tim cuc tri mdt bidu th*c bdng nhidu

cdch ciat6.c glh Nguy6n Ngoc Khoa dlng trong

THTT s6 303 (912002) vi bli Sft dung phtong

pltdp tham bi€h dd tim cilc tri mdt bidu thfic cia

tdc glh Pham Thi Vict Th6i ding trong THTT sd

313 (7l2OO3) O day sit dung phuong phdp tirn

didu ki€n dd phuong trinh bdc hai cd nghi€m

Vdi m6i gi6 tri xo ctra (1) ta c6 gi6 tri

Ta cdn tim didu kiOn cira tham sd y dd m (2)

c5 nghi6m, tt d5 xiic dinh du-o.c midn gi6 tri G

o Vdi ) * 0 thi Pf (6) c6 nghiom

A

DK/(y) > 0 <+ yt sy syz

Dofl0) = b2 >0 = A/(0) < 0 = yt <0 <yz

VQy t4p gi6 tri cfia hlm s6 (5) ld G = lyr yz7 n6n GTNN cira y ld y, = Dt+

Dt-IrI (4) c6 nghi0m <> 41 = yzb'2-4ya'1yc'-c) =

y(y(b'z - 4a'c) + 4a'c) > O

a a'(a'x' +b'x+c')Theo trubng hqp 1 ta k6t luAn duo c :

N6u a'c > ac'thi G = (9, a -4(a'c-ac')f

, o 4(a'c-aa') n€n GTLN cira y th ya - ;-?

aa

b'(a'b-ab')

- Za'L' - 2(a'c-ac)

a'

PT(z) c6 dang yo'* + (yb'-B)x + yc' - C = O

Ya y + 0 PT nly c6 nghiCm e SO) ' L.y'

tr6n cfrng cho bidt vdi di6u ki6n nho cria ciic h0 s6 a, b, c, a', b', c' thi hhm sd (1) c6 GTNN hoacc6 GTLN

Tt khi vd Bo gi6o duc vh Ddo rao, rhu6c

Nhi xuAt bin Gi6o duc quin lf, tap chi To6nhoc vi Tudi tr6 dd kh6ng ngtng l6rn manh nhu

t[ng ki xudt b6n, c6 bia, tang ffang v]r thOm

nhlrng chuy0n muc mdi Bu6c sang tudi 40

tap chi tidp tuc c6 nhfrng phdt tridn mdi vd

nhAn du-o c nhtng dd xuAt, kidn nghi vd n6idung vh hinh thrlc cira ban viet vd ban doc gdn

xa, D6p rfug mong m6i cta ban doc vh duoc

su ddng f cria Cuc b6o chf 86 Van h6a Th6ng

tin tit I.2004 tap chi THTT sd tdng thOm 4

trang ruOt thlnh 28 trang m6i sd Tap chi s€

c6 didu kiQn dd dang nhidu hon c6c bii thu6cc6c chuy6n muc : Li.ch sft toSn hoc, To6n hoc

vh ddi sOng, Ban doc tim tbi, Thu ban doc,

Nhin ra thd gidi, Gi6i thi€u vd todn hochiQn dqi

ROn canh d6, tap chi m6 th6m,3 chuydn muc

mdi lh Phuong phdp gitii todn, Tit kho tdngtodn hoc vdTinh todn cdch ndo? Mu,cTt kho

tdng todn hoc sd, gdm cdc bhi vidt khai thSccdc vdn dd tuctng nhu d6 cfr dd tim ra cdi m6i,

ld c6c kh6m ph6 nho nh6 trcn con dudng ban

ddn lAu ddi toi{n hoc dd s6 Gidng nhu vi6nngoc 16ng l6nh nhrrng cbn nam trong khdi dd

xD xi th6 r6p Ban hdy lb ngudi tim ngoc trong

dd Tinh todn cdch ndo? \i, muc gi6i rhiOu c6c

ki ndng, phuong ph6p tinh to6n hqp li th6ng

minh, c6c kidn thrlc vI kinh nghiOm vd tfnhtoiin gdn dfng, udc lugng, nhfrng di6u rdr cdn

cho ciic ki su tuong lai vd ngudi lao d6ngtrong cu6c sdng Muc Phtrong phdp gidi todn

duo c t6ch riOng ra tD c6c chuyOn muc dd c6

N5 gdm c6c bli viOt mh nOi dung xoay quanh

vAn dd phan tich dd bei, tim ldi giAi, ph6n

dorin kdt qui vh trinh bdy cdch gihi O day

phuong phdp lit tiOu didm vh trong tdm cta

bhi vi0t Chuy6n muc nhy d6p ung yeu cdu ddimdi day vh hoc to6n & truong trung hoc

THTI chd doi nhAn duoc nhidu bdi vidt ciraban doc vd c6c rnuc m6i n6i trOn

c6m on c6c ban

THTT

KtsI(O TAI\IG

ToAw tsrorc

kin qudn

ngang

^cdi sim kh6ng c6 thd co ddn lai thhnh

m6t didm tren bd nrat cta slrn md khdng cat drit

sam (hinh xuydn, bidu thi bbi c6,i s5m, tu-o c coi

nhu rn6t bd mlt dorrg kin c,5 dirng m6t 16) Tir

thd ki 19, cdc nhd toi{n hoc bidt rang hinh cdu li

kh0ng gian duy nhdt hai chi6u d6ng kin c6 tinh

chat nhrr thd, nhinrg cii g\ xiy ra khi sd chidu

ldn hcrn ?

PoincarE n6u ra mOt

quy t6c tuong duong

vd hinh cdu ba

chidu Nguy€n tdc

nhy kh[ng dinh daithd rlng hinh cdubachi€u ld k'hdng gian

duy nhrit ba chidu

Iules Henri Poincard(18s4-1912)

N40t nhd toiin hoc Nga, Gri-g6-ri PO-ren-man

(Grigori Perelman), 6 Vi6n Steklov tai Saint

Peteisb,:urg (LB Nga), cla khing dinh ld chrmg

minh duoc ph6ng do6n Poincar6 Thuc raPerelman dd cOng b6 m6t loat bhi b6o tir thr{ng

11 narn 2002 NCu chrlng rninh cua Pereiman

trong hai nam kh6ng,cdn gAy tranh c5i nghi ngd

gi thi Perelman s6 nhAn giAi thu&ng 1 tri€u dOla

cta Vi€n Toi{n hoc Clay, o bang Massachusetts

Hoa K! Trong hai th6ng 3 vd 4 nam 2003,

Tomasz Mrowka nhd to6n hoc o MiT (Vien

COng ngh6 Massachusetts), du hOi nghi chuyOn

dd vd chrmg minh ctra Perelman, cho bidt m6i

khi c6 ai d6 nOu vdn dd b5c b6, Perelman ddu

Perelman dd suy nglfi til rd't ldu vd ra't sdu, khb

cd thd mdc sai ldm"

Nhfrng kdt qui md Perelman thu duo c cho

thay rang 6ng dd chrrng minh m6t ph6ng do6n

t(ing qu6t hon ph6ng doi4n Poincar6, vd vdn dd

hinh hoc nhfrng kh6ng gian ba chi6u dua ra

trong nhfrng nlm 1970 mh ph6ng dodn Poincar6

chi ld trudng hqp ri6ng HlLnh trinh c6 nhAn ctaGrigori Perelman c5 mOt sd didu giOng nhu

Andrew Wiles lh d5 khOng cho ciic ban ddngnghiOp bid.t, ihm vi6c mdt minh trong tdng s6t

n6c nhd dd giii dinh li Fermat Cbn Perelman

sdng cdm cung suOt 8 nlm gdn dAy & Nga,

kh6ng c0ng bd gi

Nhtug bdi brio ctta Perelman khing dinh d6

chung minh di6u mi ngudi ta goi lh "ph6ng

do6ir su hinh hoc h5a" (hoac ph6ng doi4n

Thurston), nOu dac tfnh ddy dir cria hinh hoc ci{c

khdng gian ba chi6u William P Thurston, giiio

su trudn.g dai hoc California 6 Davis, dd dua ra

nhiing phuong tiOn rlo c6c khoAng cr{ch gifla c6c

didm tren cdc mit 3 chidu, trong khi ddn tan

nhfrng nam 1970 chua c5 nhtng phuong ph6p

do niy (nhftng mat niy duoc xem x6t chi bang

ciich nhin tOpo).

dd duEe ch&ng minh

Phong dodn Poang-ca-r0 do nhd toiin hoc

Ph6p Giu,vn-lo Ang-ri Poang-ca-rO (Jules Henri

Poincar6) n6u ra ldn ddu tiOn nam 1904, lh mOt

vdn dd trung tam ctra tOp6 hoc TOpO hoc nghidn

criu ciic tinh chdt hinh hoc bdt bidn ctra mOt v2t

khi v0t nhy bi k6o, xoan hoic co iai mot cdch

liOn tuc Trong t6pO kh6ng c6 cdc khoing ci4ch

vd mot mat phing lh tuong duong""Oi Ud mat

gon s6ng, ching han, mOt hinh cdu nh6n tuong

duong vdi mOt hinh cdu m6p m6o, vdi didu ki6n

th bd mat d5 khong bi choc thring, kh6ng bi

dinh 6 didm niro d6 V6 rdng tao thd"nh bd mat

tr1,i ddt tao ra cdi mI c6,c nhit tOpO hoc goi lh

mot hinh cAu har chi6u N6 c5 mot rinh chat duy

nhAt : ndu quan xung quanh n6 mOt cr4ch bdt ki

mOt dAy chun kin tuong tuong vir din hdi thi

ngudi ta se cd thd thu gon ddn ddn dA1, 65 1u^n

luOn n5m trOn mat cdu r,6 mOt didm ctra hinh

cdu (di6u d5 tuong cluong vdi viOc coi hinh cA'u

nhu m6t bd mdt d6ng kin kh6ng c5 16).

Trong

trudnghcrp mOt

cdi sf,m

rhikh0ng

xiy ranhu thd:

m6t dAy

OT TII!TUYEll $II{II tOP 10 }IE T}IM C}ltIYE!{TRUOilG OtlItHTil tIA ilff ilAt{ 2OO3

mhy THI : TnAN Hoc

VONG l (Ddnh cho thi sinh thi vdo chuyin Todn, Li, Hda, Sinh)

(Thdi gian ldm bdi : 150 philt)

CAu I (2 didm).'Gi6i phuong trinh :

M, N ld hai didm tren nfta duong trdn (O) sao

CAu VI (2 didm) Cho phuong trinh :

x4 +2mx2 +4=0.

Tim gi5 tri cria tham sd m dd phuong trinh c6

4 nghiOm phAn bi€t x1, x2, 4, x4thba mdn

4444-^

X1 +X2-tX3tX4 = 3LCAu VII (2 didm) Giei he phuong trinhlz*' **y-y2 _ 5x+y+2=o

tam gir{c ABC ti6p xric vdi ciic canh BC, CA, ABtuong rlng tar cdc didm D, E, F Duong trbn tAm

clto hl thuOc cung AN vi tdng c6c khoAng cr{ch

tt A, B d6n duong thhng MN bang R.6 .

1 Tinh d0 dei doan MN theo R.

2 Goi giao didm cira hai day AN vd BM lit I,

giao didm crta cic duong thdng AM vi BN ld K

Chrlng minh rlng bdn didm M, N, 1, K cing

nAm ffCn rn6t dudrng trdn Tinh bi4n kinh cira duorrg trdn d6 theo R.

3 Tim gi6 tri l6n nhAt cua di0n tich tam gi6c

KAB theo R khi M, N tha1, ddi nhmg vdn th6amdn gi6 thidt cira bli to6n

CAu V (1 didm): x, !, z ld c6c s6 thuc thoamdn ditiu kiOn : x + y + z+ -xy + yz + zx = 6.

Chrlng minh rang ' 12 + y2 + z? >- 3

O' bing tidp trong g6c BAC cita tam gi6c ABC

tidp xric voi canh BC vd, phdn k6o ddLi cira c6c canh AB, AC tuong rlng tai c6c didm P, M, N

1 Chr,rng minh ring : BP = CD

2 Tr€n duong thhng MN taldy cdc didm 1 vI

K sao cho CKllAB, BIllAC Chung minh rang

cdc tir gi6c BICE vd BKCF ld cr{c hinh binh

hdnh.

3 Gqi ($ ln duong trdn di qua 3 didm I, K, P.

Chung minh ring (S) tidp xric vdi c6c dudngthhng BC, BI, CK

Cau X (l didm)

Sd thuc r thay Adi va th6a mdn didu kiOn

f + Q - x)2 >- 5 Tim 916 tri nh6 nhdt cta

bidu thrlc

p =./+ (3 - x1a + 6x213 - x72.

VONG 2 (Ddnh cho thi sinh thi vdo chuyAn Todn vd chuyAn Tin)

(Thdi gian ldm bdi : 150 phitr)

- Tu rhd ki 19, c6c nhd toiin hoc bidt rang c5 thd

g6n cho mdt loai khOng gian 2 chidu nho d5

"m6t da tap" (manifold), mQt cdu tnic hinh hoc

chat che vd 1u6n luOn ddng nhat Ddu nhtng

n[m 70 Thurston, ldn dAu ti6n dat duoc c6c kdt

qui vd da tap 3 chi6u vi nhAn giii thuong

Fields, giii thucrng cao quli nhat vd to6n hoc

COng trinh ctra Grigori Perelman cung cdp ydu

td cu6i cing ctra su mO ti ddy dir c6c bdt bidntopo ctra nhftng da tap 3 chi6u vi qua d6 gi6i

4

duoc van d6 ndi tidng cira Poincar6

Tomaiz Mro.;rka kdt luan : "Hodc Grigori

Ferelman dd tim daqc ldi gitii cfia bdi todn,

hodc dd cd nhfrng ilAit bO c6 y nghla dd tim ra

ldi gidi, nhfrng didu dd d€u giip ich cho fi't cd

chilng ta"

NGUYETS VAN THIEM

(Theo Le C ourri er I n t e r na tio na I 6-5 -2003

vd La Recherche thitng 7 -812003)

TRUryNG THPT CHUYEN TE QUY DON, EA NANG

@A thi dd dd.ng ffAnTHTT sd 3l8,thdtng 12 ndm 2003)Bni l a) Ta c6 :

P_ di+'E+zi +Ot.6+ z+Ji)

-HJi

Bni 3 a) Dat a * b = n (l), (a, b, n e

b + O) thi theo giA thidt , 1= n

Ydi n - 1 = -1 thi ru = 0, khOng th6a mdn (3)

Ydin - 1 = 1 thi il=2, tit(4) suy rab=2,1rtc

d6 a = 4 Thfr lai ta thdy (a, b) = (4,2) th6amdn

de bei.

b)Tac6:

(Ji+J1+21

b) Tt giA thi& i - 2y' =,ry, suy ra

(x + y)(x - 2y) = 0 Do x + y + 0 n6n x = 2y.

r- \! ?.'- y 1

VAvO=" r='r' .

x+y 2y+y 3

Bni 2 a) Dat t = 1G, phuong trinh (PT) dd

cho trd thdnh 2r2 - 5t - 3 = 0 PT ndy c6 hai

nshiem r, '2 = -! tt = 3.Do d6 PT dd cho c6 hai

Do (x1, yr) vg (xz, yz) ld hai nghiOm cta h€

phuong trinh, nOn tr hld hai nghi€m cfia (2)

TUdd !t+lz=_1,yryz= -:5

TU (1) cd x1 - xz= 3(yt - y2), suy ra

M = (xr * xz)2 + (y, - yr)'= 10(yr - yz)2.

Tt (3) c6 M = lO[(y;yr)z - 4ytyz) = 34.Bni 5 a) Ta c6 fie =il} 1.tng cnan fB)

ddB =6tr8(cnngch6n ffi).

Theo kdt qui a) 6iD =6iE ,suy ra :

(2)

(3)

Nvd

(2)(3)

DE DB

_-_ hav

DA DE

VO HONG TIEN(SdGD-DT Dd Ndns)

I

I

Nhu vAy van d6 d0 nhdm l6n khi sir dung quy

u6c ndy ld c6c con sd 0 R6c r6i nhdt li trudng

ho-p (iii) N6 c6 ngudn gdc thuc td ld con sd 30

chi sd hoc sinh cira m6t lop thi s6 0 nly l) hodn to),"n chinh xdc, trong khi s0 chi dan sd mOt d0

thi le 1300000 ngudi thi cdc con sd 0 chi Id

tuong ddi

Truong hqp (u) cfrng b6t ngudn tit mot thuc td.

Ndu n6i chidu dhi cr{i bin li 1,20m thi ta coi d0

chinh xdc cua ph6p do ld ddn cm, n6u ntii ld1,200 m thi d0 chinh x5c ld mm

Dd thuan ti0n ti nay vd sau ta s6 ding ki hiOu

= thay cho = nhung vdn hidu rlng ki hiOu d6 chi

so do dring ho[c gdn dring

Vf du o cdc cdch vidt sau ndu sd bOn tr6i ld sd

dfng thi c6 thd n6i (sau khi ldm trdn s6) :

7006 = 7000 chinh x6c ddn 1 chf sd c6 nghia

7006 = 7000 chinh x6c ddn 2 chfr sd c5 nghia

7006 = 7010 chinh xdc ddn 3 chft s6 c6 nghra

7436 = 7000 chinh x6c ddn 1 chfr sd c6 nghra 0,00609 = 0,006 chfnh x6c dOh 1 chfr sd c6 nghia0,00609 = 0,0061 chinh xiic ddn 2 cht sd c5 nghia6,009 = 6,01 chinh xi{c ddn 3 cht s6 cd nghra.Ban c6 thd xem SGK vd so s6nh vdi khr4i ni0m

chfl sd dr{ng tin

2.Dangti6u chudn (Standard Form)C6c nhi khoa hoc thudng phAi lhm vi€c vdi

c6c s6 rat l6n hoac rdt nh6 Vi du khoAng cdch

tit sao BOta ddn Tr6i ddt li :

dring 6 dang tiOu chudn Dang ti€u chudn cdn

goi ld ki hi€u khoa hoc (scientific notation)

Dang vidt khOng c6 lfry thira duoc goi li dangthdng thtdrtg (ordinary form)

3 C6c bni t4p 5p dunga) Vi dq 1 Tim gid tri ctia g6c x chinh xdc

deh 3 chfr s6'cd nglia n€i sinx = 0,45 vd

nudc, mOt didu dC nhAn thAy ld c6 nhidu bdi tap yOu cdu tinh to6n vh c6c ph6p tinh kh6 phrlc tap.

Trong nhidu blLi to6n co ciic yeu cdu khr{c nhau

vd do chinh xdc, vd c6ch xiic dinh vd thd hicnkdt quA Vi€c xiic dinh ket qui khi su dung mi4y

tinh cflng c5 yOu cdu nhu vAy Dd c6 thd tidp

cAn v6i cr{ch thi vi nhfrng y6u cdu cria dd thi

nhu thd, loat bhi ctra chring toi dd cAp ddnnhirng van dd vd y6u cdu vh ki nang tinh O day

cdc vi du kh6ng chri f nhidu ddn l6t 16o vd tuduy md chi ddi h6i su tuan theo nghiOm ngdt

cdc quy dinh vd kdt quA tinh to6n bOn canh suchon lua m6t ciich tinh hqp li vd nhanh nhAt, sai

s6 nh6 nhat., Ci4c ban chd coi thuong vdn dd

nIy n6u ban kh6ng mu6n c6ng lao tinh toi{nthdnh mAy kh6i khi dd k0t qui khr{c v6i y€u cdu cua dd bdi

Trudc hdt chring ta cdn nim vfrng hai khr{iniAm chfr s6' c6 nglia vit, dang ti€u chudn

1 Chf sd c6 nghia (Significant Figures)

Day ld quy u6c vh trong quy u6c ndy thrl tu

cht sd ld tinh tt b6n trdi sang phii

(i) Tdt ch cdc cht sd khiic khOng ld clu.i s6' c6 rrglia

(ii) Chfr sd 0 nam gitra cdc chfr sd khr{c khOng

ld chrt s6'cd nglfa

(iii) Trong m6t sd nguyen, ci{c chfr so 0 sau

chfr sd khdc kh6ng cuOi cing c6 thd lir cht sd c6

nghia hoac ld kh6ng c6 nghia (dd phan biOt cdnc5 ghi chf)

(iv) Trong mOt s0 thAp phAn, cr{c chfi s0 0

trudc cht s0 khric khOng ddu ti6n ld chfr sdklfirtg c6 nghTa.

(v) Trong mOt sd thip phAn, c6c chfr sd 0 sau chfr s6 kh6c khOng cudi cirng ld chfr s6'cb ngh1a.

6

Tinh toon coch noo ?

I

Mot so m6y tinh khong c6 phim lrin l-l ,i,t

dn phim ]invl truoc khi dn ffi oOi vdi mr{y

r -:-1

c6 DAL thi an lsin-r | 0.4s El dd c6 kdt qu6'

(Chf f 1) trong yOu cdu vidt chfr sO c6 nghia

hay vidrkdt qui dring ddn bao nhiOu chfi sd thQp ph-An v6n tuAn theo nguyOn t6c ldm trdn "5 ldy 4

b6"

Ch6'ng han, yOu cdu tim x ndu tanx = 2,4 vd'

0 < x < I vdi ket quA dring ddn 2 chfr s6 thAp

ggch chdo gidi han

gifra hai hinh trbn vd

dudng thdng vdi k€lt

qud chlnh xdc d€h 3cha so co nglua

Hudng ddn gidi

Tru6c tion, dd thudn

ti6n cho c6c bhi tinhto6n di6n tich chring ta

ldm quen vdi c6ng thrlctinh di0n tich hinh quat

Tro lai bIi tor{n, trong qu6 trinh tinh cdc phdp

tinh trung gian do y6u cdu cira dd bii ta cdnchon c6c kdt quA v6i 4 cht sd c6 nghra.

vdn Bi€lt rdng m1t don vi thi€n vdn ld 1,50 x

l08km Tinh khodng cdch trung binh theo km ti

Mdt trdng d€n Trdi cldt vd vi€1 k€i qud d dang

tidu chudn

Hudng ddn gidi Khoing ciich trung binh tit

Mat ffang ddn Tr6i ddt tinh theo km ld :

2,56 x 10-3 x1,50 x 108 = 3,84 x 10s (km)

Chf )i : Bdi nly kh6ng y€u cdu vd do chinh

x6c ciri chfr sd c6 nghia ncn k6t qui c5 thd vidt3,840 x 10s km Tuy nhi6n, dd bei cho 2 cht sdthap phan thi ta nOn dd kdt qui c6 2'chfi sd thap

phAn.

Chf y rang ndu dd bdi khOng chi r5 s6 cht sd

c6 nghia thi vdi 2bditodn ldn lucn cho n = 3,14

vi 7r = 3,142 ta ldy kdt quA tuong rlng sd ld 2 vd

3 chfr sd thAp phAn nhu sd n.

Bay gid mdi c6c ban hdy grhi cdc bhi tAp sau :

1 Bidu di6n cdc sd

(r) 2,149 chinh x6c ddn mOt cht sd thAp phAn.

(ii) 40S chinh x6c ddn hai chfi sO c6 nghia

(iii) 0,0054 o dang tiou chudn.

2.Ddi 486mm ra cm chinh x6c ddn hdng don

NGUYEN KHAC] MINH

Ngiy thi thrl nhdt (l2l3l2U$)

Bii 1: Cho hdm s6' f xdc dinh tr€n tdp hop s6'

thLrc R, ldy gid tri tr€n R vd thda man di6u ki€n

f(cotgx)=sin2x+cos2x

vdi moi x thudc khodng (0; tr).

Hdy tim gid tri nhd nhdr vd gid'rri ldn nhdlt

cua lfim so' g(x) = [(sinzx).f(cos'x) tr€n R.

Biri 2: Cho tam gidc nhon ABC ndi tiy'p dudng

rrdn tdm O Tr€n dudng thdng AC ldy cdc didm

M, N sao cho MN = AC Gqi D ld hinlt chi€lu

vudng g6c cila M tr€n dadng thdng BC, E td hintt

chi€|u vudrtg g6c crta N t€n dudng thdng AB.

ll Chrtng minh rdng trtrc tAm H cila tam giitc

ABC ndm tr€n dadng trdn tam O' ngoai ti€p

rum gidc BED

2l Cht?ng minh rdng trung didm ctia doan

rhdng AN d6'i xt?ng vdi B qua trung didm cfia

doan fidng CO'

Bii 3: Xit s6'rtguy€n n > l Ngadi m mu6'n tb

tdt cd cdc s6'trt nhi€n bdi hai mdu xanh, dd sao

cho cdc cliitt k'i€n sau duoc ddng thdi thda mdn;

il Mdi s6'duoc fi bdi mAt mdu, vd mdi mdu

cldu daoc cling'ctd t6 vb s6'sd;

rI DAP AN tsii 1: Tac5:

f(cotgx) = sin2x + cos2x Vx e (0; n)

- 2cotsx cots2x - 1

<?i(COIpX) = -: ; -: + cotg'x+1 -*; cotg'x+l

sina2x +32sin22x-32

D4r u = (ll4)sin22x DO thdy, khi x chay

qua R thi u chay qua.[0; 1i4] Vi vAy, ttt (1) taduoc:

min g(x) = min h(u) r,i max g(x) = max h(u),

Tir d6, v6i luu ! rang v6i m6i t e R d6u t6n

tai x e (0; n) sao cho cotgx = t, ta duoc:

(u2 -2u + 2)2

Dc ding chung minh duoc h'(u) > 0 Vu e

t0; ll4l Suy ra hdm h(u) ddng bidn tr0n

VteR.

trong d6 h(u) =

Ta c6: h'(u) =

2

BAI{G B

(Cuc Khdo thi vd Kidm dinh CLGD - Bo GD&DT)

I DE THI

iil Tdng cila n sd dai mdt khdc nhau cing

mdu ld s6'c6 cilng mdu d6.

Hdi cd thd thuc hi€n dtoc phip t6 mdu n6itr€n hay kh6ng, ndu:

ll n=2002?

2l tt = 2003 ?

Ngiy thi thir hai (L3B1}A03)

Bii 4: Hdi c6 tdn tai hay kh6ng cdc s6'rtgtty,An

x, y, u, v, t thda mdn didu ki€n sau:

Bii 6: Cho s6'thuc a /0, vd cho ddy sa'iltac{xn}, n= 1,2,3, ,xdc dinhbdi:

\=0 vd xn+1(xn+a) =a+l vdi

moi n= 1,2,3, ,

I I Hdy tim sd hang tdng qudr crta day {xn} .2l Chrtng minh rdng ddy {xn} cd gidi han hfruhan khi n -) + o Hdy tim gi6i han d6.

l0 ll4] Vi vay, trcn [0; ll4]ta c5: min h(u1 =

Chf f : Ddi v6i bdi ldm cfia thi sinh, y€u

cdu trinh bby chi tidt vioc chrlng minh h'(u) > 0

Vu e [0; U4]

Bhi 2: 1/ Gqi K ld giao didm cira c6c dudng

thang MD vh NE Dd thAy, dudng trdn dudng

kinh BK ngoai tidp A BED

Ta c5:

AH // MK (vi cirng vuOng g6c vdi BC)

CH // NK (vi cing vudng g5c vdi BC)

Suy ra: IHAC = Z KMN vh Z ACH

= Z MNK

'fil ciic ding thrlc tron va gii thidt AC = MN

suy ra A AHC = A MKN Do d5 d(K, AC) =

d(H, AC) Me K vi H ndm cing phia doi vdi

duong thing AC, nOn KH ll AC Suy ra KH I

BH Do d6 H nam trcn dudng trdn dudng kinh

BK ngoai tidp A BED

2/ Ttr chrrng minh cta phdn trcn hidn nhien c6

O' th trung didm cria doan thing BK Goi I 1)

trung di<lm ctra doan thing AN Dc thdy, hinh

chidu vu6ng g6c cira I trOn BA vi tr6n BC tuong

rmg ld trung didm ctra EA vdL DC Do d6, hinh

chidu vuOng g5c cira vecto O;i trdn BA vi trcn

BC tuong ung bang U2vecto BI vn l?vecto

Bd Suy ru OT = BO Do d6, tt gi6c BO'IO

ld hinh hinh hhnh Vi thd, c6c didm B vb I ddi

xrmg vdi nhau qua trung didm cria doan thing

oo'.

Bhi 3: ll Xdt n = 20A2 Ta sE chrlng minhrang cAu tri ldi cho cAu h6i cira bli toiin trong

trudng hop ndy ld"khbng"

ThAt vay, gi6 sri, nguoc iai, ta c5 thd to tdt cac6c sO tu nhiOn bang hai mdu xanh, d6 sao cho

m6i so duoc tO b&i m6t mdu, m5i mdu duoc

clilng dii to vo sd sd, vi tdng cira 2002 sd d6i

mOt khdc nhau cing mlLu lh so c6 cing mhu d6.

Khi d6, do c6 vO s6 sd duoc t6 bdi miu xanh vi

c6 vO sd sd duoc tO b6i mhu d6 nOn:

+ TOn tai sd a, mi a, duoc tO b&i mdu xanh

vd sd b, = &r * 1 duoc to b6i mdu d6;

+ T6n tai sd b, > b, mi b, duoc t6 b0i mlLu d6 vd sd az bz+ 1 duoc t0 b&i mhu xanh;

+ Tdn tai sd a., > a, mh a., duoc t0 b&i mduxanh vd s6 b.l = a? + I duoc t0 b(yi mdu d6;

+ Tdn tai s0 b, > b., mir b, duoc tO b&i mdud6 vi sd zr=bq+ 1 duoc to b0i mdu xanh;

+ Tdn tai sd aron, ) szooo mI aroo, duoc tO bdi

miu xanh vh sO bzoor = a200r +1 duoc tO b6i

G) 2AO2 sd a,, 22, , ftzoor, ar,,,,, duoc t6

bdi mdu xanh; vh 2002 s6 b, , b, , , b,uu, , brou,

duo c tO boi mdu d6.

(ii) b2k-'r =&zv-r+ 1 vd bzr=az*- 1 vdimoi

k=1,2, ,1001.

Tn didu kiqn (i) suy ra sd a = at + a2 + +

szoor * arn,,, dudc tO b&i mhu xanh, vi sd b = br

+ b2 + * bzoo, *bz,o, duoc t0 b6i mhu d6.

Tir didu kign (ii) d0 dlng suy ra a = b Do d6

a vd b phAi duoc to cilng mdu, mAu thu6n vdididu vta nhan duoc & tron Tir d6 suy ra di6u

"chn chrlng minh

2l Y6i n = 2003, x6t cdch t6 miu sau: T0 tat

ci cdc sd ch6n b0i mhu xanh vh t6 tdt cAl c6c sO

16 b&i mhu d6 Dc thay, ciich to mdu vila nou

th6a mdn tdt ce cdc yOu cdu cria bhi to6n VAy,

trong trudng h-o p ndy cAu trA ldi cho cAu h6i cira

bii ra lb "cd"

Bhi 4: o D6 thdy, vdi a lh mOt sd nguyOn tiy

y, ta c6:

[l fmod8) ndu a=+l(mod4)

-l

a'= 10 (mod8) neu a=0 (mod4)

I

[a(mod8)neua=2(mod4)

Suy ra, v6i a, b ld hai s6 nguyOn tiry 1f, ta c6:

{Z;t;S (mod8) ndu a=+116s64;

I

a2 +b2: it;O;4 (mod8) ndu a=0 (mod4)

I

[S;+;O (modS) ndu a=2(mod4)

o GiA sir tdn tai c6c s6 nguydn x, y, u, v, t

th6a mdn cdc h€ thrlc ctra dd bai Khi d5, do x,

x + 1, x + 2, x + 3 lAp thdnh m6t hO thing du

ddy dri modulo 4 nln, theo trOn, ph6i tdn tai sd

nguyOn m sao cho

m e {2;1;5}n {1;0; 4} n {5:4; U = A

Didu vO lf vira nhAn duoc cho thay gii srl n€u

trOn liL sai, nghra ld khdng t6n tai cdc s6'nguy€n

x, y, u, v, t thda mdn cdc h€ thrlc cila dd bdi

trong tl6 Q(x) ld mor da thrlc vdi h6 sd thuo

cfia bien x Din tdi:

trong d6 c li mOt hing sd thuc tiry y

Ph6p thft truc tidp cho thdy cic da thrlc P(x)

vira tim duoc 6 trOn th6a mdn hO thfc cira dd

bli, vi do d6 chring lit tdt ch cdc da thfc cdn

Ar--ct

D4t p,=0, pr=cr+ l, er = l, qr=o;tu ' Pr ' = * cia sir x1 cd dang

co x,= T v3 xr= gz

xx = * Khi d6 tir (9) suy rir X1+ 1 cd dang

Qt'

Pk+t Xk+1 = ;;, trongd6 pk+r =(a+ l)qp vd

Qk + t = crQi + pr< Vi vdy, theo nguyen li quiIl&p, Xn c6 dang xn = * Yn = 1,2,3

Qntrong d6 {pn} vd {qn} x6c dinh b&i: n' =_Ol

Qkm ileb rang 24)

ehuan bi

fhi vao

Dryx age

@BBNW@ffi

Od dap fng nhu cdu d6ng dio cira cdc ban

chudn bi thi t6t nghiCp THPT vlL Dai hoc, Cao

ding Tap chi THTT sE ldn luot dang m6t sO d6 crta-citc gir4o viOn gidu kinh nghiOm nham gifp

c6c ban tu rbn luyOn cilc phuong phr{p giAi cdc

loai bdi thudng gap trong dd thi, vd di6u khOng

k6m phdn quan trong li c6ch trinh biy ldi giAi

bei thi trong thdi gian cho ph6p Phdn hu6ngd6n giii sE dlng trong sd b6o tidp theo.

1) Kh6o si{t vd v6 d6 thi hdm so Tr0n (l!| ldy 6 didm phan biot A; (i = 1' "'' 6)

Zj Corl ld giao didm ctra hai dudng tigm c1n sao

-cho : A1A2 ll A4As ; AzAt ll AsAa' Chttng

cia (C) Uay -viet phuong trinh hai duong thing minh ring : A3Aa ll A1A6'

di qua 1 sao cho

"^h.irrg "O fre sd g6c nguyon vi 2) Cho trl di0n ABCD c6 biin kinh mlt cdu n6i

hinh chf nhAt tidp tI r Chrlng minh rang : Veaco rTf

CAu II (2 didm)

1) Bang dinh nghia, hdy tfnh d4o hdm cira cau v' (2 didm) r .2 t

hdmsd:f(x)=lxl3 +e* taididmx=0 1)Timx>0saocfro: 1-I J-dr =1.

2) BiQn luAn theo m, mi6n x6c dinh cira him

1) Cr{c g5c cira tam gi6c ABC th6a mdn didu

Chrmg minh r[ng tam giSc ABC d6u.

2 Grei h0 phuong trinh :

2) C6 bao nhi6u sd tu nhi6n c6 dfng 2004

cht s6 mr tdng c6c chfr tu oun**G

vu HA(HdTdy)

' ::1 :.:; rilili'r,fii.,::, ,,,-'": ,,i ,

THTT So 320 (212004)

I-Ing dqng c[ra m6t hQ thrlc truy hdi

o Ptruotrg phdp viet cdt tuydn cria dudng cong.

H,rOrg a3n giii dd thi chgn hgcsinh gioi

todn TFIPT qudC gia bAng A nam 2003

: utl' 1ritd$6n sm top,jo H 1 $p ua

Noi 2003,'j",.Ti d$n'rgi ldd thi tuydn sinh l9p t0 PfcTT DHKIITN HA NQi 2003

e Dd tU 0n thi DH sd 2 Huong dan giii dd tu

0n thi DH sd l.

Todn hoc vi co cdu C6c loai ddng tidn

Mdi c6c ban dat mua THTT tai c6c co s&

Trong c6c bli to6n vd phuong trinh (PT), bdt

phuong trinh (BPT), hq FrI , hc Bm m gap khri

nhidu bii toi{n chrla tham so v6i y6u cdu : Tim

didu ki6n (DK) ddi v6i tham so dd PT, he PI,

BPT, hC BIrI dA cho th6a mdn mOt di6u ki6n

rdng bu6c nio d6.

Trong sd c6c cdch gidi bhi to6n ftAn vi€,c tim

didu ki€n cdn cho tham sd th6a m6n DK rdng

bu6c n6i tr0n li rat cdn thidt Th6ng thudng ciic

bidu thrlc giii tich c5 trong IrI, he-PI, ePf, hc

BPT dn ddu m6t tinh ch6t ndo d5 Ta cdn phdt

]ti€n vd khai thdc tinh chdt dy dd tim ra mdi

quan h€ ddc bi€t hodc m1t rdng bu6c ddi vdi

tham s6', ti db tim ra didu ki€n cdn, d0y li chia

kh6a dd giAi quy6t bii torin

GiAi loai toiln niy thudng qua bdn budc sau :

c NhAn x6t vd tinh chdt cira nghiOm cta PT,

he F'r, iiPr, ho BF,r

o Tir nh4n x6t tr0n vh DK rhng buOc suy ra

mot didu kion cdn cria tham sd.

c Vdi gir{ tri tim duoc cria tham s0 cdn chrlng

t6 rang F'f, hq FrI, BPT, hc BFrI th6a man DK

rhng buoc

o Ket IuAn.

ThOng qua ciic bhi to6n cu thd chring t6i se

lim s6ng t6 viOc khai thric ciic tinh chat dn d0u

chrla trong c6c bidu thrlc"dd giii mOt sd bdi todn

hay thu6c dang nly

Bhi to6n l Cho PT ; 4x +2 = m.2' ,sinxx (l)

Tim gid rri crta ilmm s6'm dd PT (1) c6

nghi€m duy nhdlt.

Ldi giii o Nhdn xdt.W (1) <+

NhAn thAy Frf (2) c6 mOt tfnh chdt dac bi6t :

N€'u xo ld nghi€m cila PT thi I - xo cfing ld

o Kdt ludn; PT dd cho c6 nghi€m duy nhdt

khi vd chi khi m = 2't5 .

Bii todn 2 Cho h€ phuong trinh

HUi}IH DUY THUY

(6/ THPT Tdng Bat Ud, noai Nhon,

Binh Dinlt\

nghidm cfia W d6 That vAy ndu thay xo b&i

1 - xo thi sinn(1 - ,16) = sin(n - fiJu) = sinrx,,

nOn PT (2) v6n c6 dang nhu trudc

o Do d6 didu ki€n cdn dd m (1) c6 nghi€m

i+y2+r2 = 2(x +y) Til d5 suy ra x = y = z = O,

hay h0 c6 nghiOm duy nhdt (x, y, z) = (0, 0, 0).

K€'t ltfin: H0 da cho c6 nghiOm duy nhdt

cira BPT vdi DK xo+ l a > 0 Khi d5

duy nhdt khi vd chi khi o = Ji-l .

Bii to6n 3 Cho he Pf

I

lr' *1 y_l)2 + z2 = m+2x (2)1'

{1

[(x+y+rzsin' z) l(l-m)ln(l-ry)+1] = Q (3)

Tim gid tri cila tham s6'm dd hQ c6 nghi€mduy nhdr.

Ldi giii Nh4n x6t rang ru (2) cira h0 c6 thd

vidt lai dudi dang :

o Ta chrlng minh m = 1nghi0m duy nhdt Thgc vQy

t

Xic dinh gid tri cila tham s6' m dd h€ cdnghi€m duy nhdlt.

Ldi giii Nhan x6t ratng I

lxem ti€p trang 26)

thi he dd cho c6

: vdrm = 1 PT (3)

o' .(*n

Ki nay

I

I

Bni T1/319 (Lop 6) Tim nghiOm nguyOnduong cria phuong trinh.sau :

(GV THCS Ngb Mdy, Phn Ctu, Binh Dinh)

Bdi T3/319 Tim udc s6 nguy6n td p cfia sd

237 - r = r37 438 953 4ir,bidt rangp < 300.

r.E eueNc NAu

(Cao hot khoaTodnTin DHKHTN Tp HCM)

Bhi T4l319 Giii phuong trinh sau vdi n lh sdnguy6n duong dd cho :

^;

116x4" +l)\y4" +l)(24il +l) = 32*2ttr2ttr2n

x.'NGUYENANH THUA].{

nhon v6i AB = c, BC = a, CA = b Tim gi5 tri

nh6 nhdt cira bidu thric

(a+ b)(b+ c)(c + a)abc

rnAN nONc rrfN (6/ THCS Ho p Hda,Tam Duong,Vinh Philc)

Bni T7l319 Hai dudng trbn tAm O binkinh R

vd tdm O'bin kinh R'c6t nhau tai A vi B Ttdidm C trOn tia ddi ctra tia AB k6 cdc ti6p tuydn

CD vd CE v1t duong trdn tAm O (D, E ld c6c

tidp didm vd didm E nam trong dudng trdn tAm

O) AD vd AE c6t dudng trbn tAm O' ldn nfra 14

ldn lucn tai M vd, N Chung minh rang duongthhng DE cit MN tai trung didm cira MN

NGUYEN DUC TAN (Tp H6 Chi Minh)

cAc LoP TRUNG Hec psd rnoxc

Bdi T8/319 Mot td gidy kd o vuOng gdm rii

hing ru clt (nt + n > 3) T6 mdu k 6 vuOng sao

cho mOt 0 bdt ki ndu kh6ng duoc t0 mhu thi c6

it nhdt mot didm chung v6imOt O duoc t0 mdu.

Tim gi6 rri nh6 nhdt cira [.

DINH VAN KHAM

(W THPT LtongVanTuy, Ninh Binh)

Bhi T9/319 Chtrng minh rang

Bni T10/319 Tim ta't cit cdc him sd/: N* -+

N* th6a mdn didu kion

t ^ ^ '3

zlf w2 *,,2))" = yz 1m).f (ri+ f2 (n).f (m)

v6i moi mkhdc n.

TO MINHHOANG

(SV K5, HE CNKIITN, DHKIITN Hd NOi)

Bni T11/319 Goi AD, BE, CF ld cric dudngcao c[ra mOt tarn gi6c nhon ABC C6,c cap doan thhng AD vd EF , BE vit FD , C F vd DE cat nhau

tai M, N, P theo thf tu Ki hi6u S th'di6n tichtam gi5Lc Chung minh rang :

1 .suxp.

S,qnc - 9ou, - 8cos A.cos.B.cos C.5466.

TRAN NGOC DIEP (K51A, khoaTodn, DHSP Hd Ndi)

Bii T12l319 X6t cric hinh ch6p rti gir4c ddu

SABCD v6i g5c phang & dinh m = cr (0 < cr <

60') G6c nhi dic.n canh bOn baog q.Xdc dinh cr

Ad Uidu th.frc P = cos3rp-gcosrp dat gir{ trt lon nhdt.

sd cONc DTJNG

(W THPT Tr(in Hung Dao, BinhThudn)

cAc oi vAr li

Bni LU319 MOt ld xo c6 d6 ctrng t=80(N/m)

DO dei tu nhi6n / = 20(cm), m6t ddu cd dinh tai

-r ddu kia m6c vho mQt vat khdi lugng