aaaaaaaaaaa

Li ni u Hc ton v lm ton l hai vn hon ton khc nhau. l hai mt

khng th tch ri ca ton hc, trong hc ton l c bn v lm ton l mt vn c bit quan trng. Hc ton s gip cho chng ta nm c nhng iu c bn nht v nhng vn dng ban u ca l thuyt c s. Lm ton ngha l o su suy ngh, pht trin mt bi ton mc t duy cao hn, nh s gip chng ta c mt ci nhn ton din v su sc hn v mt vn . V h qu tt yu ca vic o su suy ngh l nhng sng to ton hc nh nhng khi nim, nhng bi ton, nhng ng dng hay l thuyt mi. mi l mc ch su sc nht ca ton hc. Vi tinh thn , nhm nhng cu hc sinh trng THPT Chuyn Hong Vn Th Ha Bnh H cng nhau xy dng nn t Tp san Ton hc 2007 nhm mc ch ng vin phong tro hc ton trng Chuyn Hong Vn Th ni ring v cc bn hc sinh ca Tnh Ha Bnh ni chung. T bo c hon thnh vi s tm huyt, lng yu ton v hng ti mi trng c ca nhng hc sinh H tng hc tp di mi trng Hong thn yu. cng l mn qu m nhng cu hc sinh mun gi tng n cc thy c gio vi lng bit n su sc!

y l ln th hai Tp san ra mt, nhng vi quy m v ni dung phong ph

hn rt nhiu so vi ln ra mt trc . Ni dung ca Tp san l nhng bi vit vi ni dung tm ti, sng to, nhng kinh nghim, ng dng v nhng phng php hc ton. Hy vng rng d vi mt lng kin thc khng nhiu, nhng Tp san s mang li cho cc bn nhiu iu b ch v l th.

V kh nng ca Ban bin tp cn nhiu hn ch v thi gian c hn, nn

trong qu trnh bin tp, chc chn khng trnh khi nhng thiu st v nhiu im khng c nh mong mun, rt mong nhn c s thng cm v nhng ng gp xy dng ca cc bn c gi. V chng ti cng hy vng rng, vi truyn thng ho hng ca trng THPT Chuyn Hong Vn Th, cc bn th h sau s tip tc pht huy v khng ngng nng cao v th ca tui tr Ha Bnh trong mt bn b mi min t nc. Hy vng rng Tp san s c cc bn kha sau duy tr v hon thin hn na v mi mt. Ban bin tp xin c cm n tt c cc bn H tham gia v ng h nhit tnh t Tp san c ra mt ng nh d kin. Xin trn trng gii thiu cng bn c!

Chc cc bn thnh cng trong hc tp v thnh t trong cuc sng!

Ha Bnh thng 1 nm 2007 Ban bin tp

Tp san Ton hc 2007

Hi ng bin tp

Trng ban bin tp: Nguyn Lm Tuyn Ph ban bin tp: Bi L V Cng tc vin: Nguyn Thi Ngc, Lu Nh Ha, trn quang th

phm thI sn, nguyn duy hong

Mc lc

Phn 1. Sng to ton hc

Gii thiu phng php tnh mt s lp tch phn dng hm lng gic Cao Trung Chinh...

1 Tng qut ha bi ton - Th Thu H... 3 Xung quanh bi ton bt ng thc thi Ton Quc t 2005 Nguyn Anh Tun. 5 Th i tm bt ng thc trong tam gic Dng Th Hng Nguyn Nh Thng.. 9 Mt s tnh c Nguyn Lm Tuyn 12 S dng tnh cht hm n nh gii bi ton phng trnh hm Nguyn Thi Ngc. 15 Li gii cc bi thi Ton Quc t 2003 H Hu Cao Trnh ... 17 S phc vi hnh hc phng V Hu Phng.. 20 Phng trnh hm v s tr mt Bi L V 23 DHy s v s tr mt trn R+ H S Tng Lm. 26 Mt s bi ton s hc v dHy tng cc ly tha Trn Quc Hon. 28 Cn bng h s trong bt ng thc C-si Nguyn Lm Tuyn.. 30 Phng php s dng nh ngha tnh gii hn L Bo Khnh.. 35 im Lemoine trong tam gic L Vn nh... 38 Cu chuyn ng trn v elipse Lu Nh Ha.. 40 Mt s phng php xc nh gii hn ca dHy s Nguyn Lm Tuyn 41 Mt lp cc bi ton bt ng thc Nguyn Minh Phc. 46 Mt s khi nim v gc nh hng Trn Quang Th... 48 Tiu chun hi t tng qut Bi L V Nguyn Thi Ngc 52 ng dng nh l Stolz trong tm gii hn ca dHy s Ng Nht Sn. 55 ng dng ca mt bi ton tng qut Nguyn H Thut 57 Tp dt sng to ng Phng Hng. 59 Vn dng nh l sch gio khoa linh hot Trnh Anh Tun.. 61 M rng khi nim tm t c cho t din Hong An Giang 64 Phng php logic mnh Phm Phc Ln.. 66 Php chiu v ng dng ca php chiu Nguyn Lm Tuyn. 69 Mt s bi ton bt ng thc chn lc Lu Nh Ha 73 S dng ng thc chng minh bt ng thc V Vit Dng 75 Tip cn ton bng vt l Nguyn Lm Tuyn... 78 Bt ng thc Schur v ng dng Trng Quc Hng 81 Mt s bi tp v ton ri rc Bi Mnh Qun... 83 S dng hng im iu ha gii bi ton cc tr Trn Th Linh Phng.. 85

Phn II. Lch s v ng dng Ton hc

S pht trin ca s hc Phng Ngc Thng... 87 Ton hc v t ng ha Nguyn Lm Tuyn 90 Dng a thc pht hin li ng truyn Nguyn Lm Tuyn.. Cu trc t nhin Nguyn Thi Ngc..

93 95

Phn III. Ton hc v ngoi ng

Hc ton v ngoi ng Ng Thnh Long...... 97 Phng tch ca im vi ng trn Lu Nh Ha... 98 Php nghch o Lu Nh Ha... 99

Phn IV. Nhng bi ton hay v cc bi ton t sng to

Cc bi ton t sng to Nguyn Lm Tuyn.

103 Nhng bi ton hay Nhiu tc gi...

109

Phn i

Sng to Ton hc

Phn I - Sng to ton hc.

TAP S AN TO AN H OC - 200 7 1 TRNG THPT CHUYEN HOANG VAN THU - HOA BNH

Gii Thieu Phng PhapGii Thieu Phng PhapGii Thieu Phng PhapGii Thieu Phng Phap Tnh mot so lp tch phan dang ham lng giacTnh mot so lp tch phan dang ham lng giacTnh mot so lp tch phan dang ham lng giacTnh mot so lp tch phan dang ham lng giac

Thy cao trung chinh

GV. THPT Chuyn Hong Vn Th, Ho Bnh

gip hc sinh c thm nhng kin thc mang tnh h thng, ti xin gii thiu mt s lp tch phn dng hm s lng gic thng gp trong cc k thi tt nghip cng nh thi i hc. Hi vng qua bi vit ny, cc em c th rt ra nhiu iu b ch cho bn thn.

I. Dng dxxxf )cos,(sin . 1. Nu f(sinx, cosx) l hm hu t th t

t = tg2x.

2. Mt s hin tng c bit. - Nu f(-sinx, cosx) = - f(sinx, cosx) th t x = cost. - Nu f(sinx, - cosx) = - f(sinx, cosx) th t x = sint. - Nu f(-sinx, - cosx) = f(sinx, cosx) th t x = tgt.

Qua cc cch i bin nh trn, ta c th tnh cc tch phn mt cch n gin v nhanh chng. Sau y l mt s v d c th.

1. V d 1. Tnh I = xdx

sin.

Li gii.

t t = tg2x

22cos2

dxdtx

= ,

212

sint

tx

+= . Vy

I = xdx

sin = ln ln

2dt x

t c tg ct

= + = +

2.V d 2. Tnh I = 3 23

cos

sinx

xdx.

Li gii. t t = cosx xdxdt sin= . Ta c

I = -

3 2

21t

tdt =

4 23 3t t dt

=

7 13 7 33 33 33 cos 3 cos

7 7t t c x x c + = +

Cc bn hHy t gii hai v d sau:

3. V d3. Tnh I = dxxx

xx +

+42

53

sinsincoscos

.

4.V d 4.

Tnh I = + xxxx

dx22 coscossin2sin

Ch : y mi nguyn hm c hiu l trn mi khong ca tp xc nh. II. Dng xdxx

nm cossin . - Nu m hoc n l s nguyn dng l th tng ng ta t t = cosx hoc t = sinx - Nu m v n u l s nguyn dng chn th chng ta d dng s dng cng thc h bc v gc nhn i gii quyt bi ton. - Nu (m+n) l s nguyn chn th t t = tgx hoc t = cotgx.

Ty theo tng iu kin ca bi ton m ta c th chn la cch t cho ph hp. Sau y l mt s v d:



1.V d1. Tnh 4 5sin cosx xdx . Li gii. t t = sinx, ta c dt = cosxdx

Vy xdxx54 cossin =

= ( ) ( ) == dttttdttt 864224 21 = cttt ++ 975

91

72

51

= cxxx ++ 975 sin91

sin72

sin51

.

2.V d 2. Tnh 33

coscos

sinxx

xdx.

Li gii.

Ta c 33

coscos

sinxx

xdx=

43 3sin cosx xdx

t t = cosx (do m = 3, n = 43

), ta c

dt = - sinxdx. Vy 4

3 3sin cosx xdx

= - ( )4

2 31 .t t dt

= 2 43 3t t dt

= 5 13 33 3

5t t c

+

= 5 13 33 cos 3cos

5x x c

+

3. V d3. Tnh I = xdxx42 cossin .

Li gii. Ta s dng cng thc h bc:

1sinxcosx= sin 2

2x , 2

1 cos 2cos

2x

x+

=

v d dng gii quyt bi ton.

4.V d 4. Tnh I = 3 11 cossin xxdx

.

Li gii. D thy m =3

11 , n =

31

v

m + n = - 4 nn ta t t = tgx , ta c ngaydt = (1+tg2x)dx . Vy: I = 3 1211 cos xxtg

dx= 3 114cos xtgx

dx

= ( )

( )( )

22 112 3

112 3

11 .

1 .

tdt t t dt

t t

+= +

+

= 11 53 3t t dt

+

= 8 23 33 3

8 2t t c

+

= 8 23 33 3

8 2tg x tg x c

+

kt thc bi vit, ti xin a ra mt s bi tp cc em luyn tp thm v phng php trn. III. Bi tp.

Tnh cc tch phn sau:

a) I1 = dxxx

xx + cossin

cossin 2

b) I2 = + xxxdxsinsin

cos2

3

c) I3 = 1sincos

2sin23 xx

xdx

d) I4 = 3 23

cos

sinx

xdx

e) I5 = xxdx

2

4

sincos

./.

============================= Gio dc khng phi l s chun b cho cuc sng; Chnh gio dc l cuc sng.

Jonh Dewey



tong quat hoatong quat hoatong quat hoatong quat hoa Bai ToanBai ToanBai ToanBai Toan

Th Thu H Chuyn Ton K97 - 00

Sv. Khoa K ton Kim ton i hc kinh t Quc dn - H Ni

CCCCho cc bn - Nhng ngi H, ang v s tip tc gn b vi Ton hc trn con ng i tm v p lng ly ca n! Chc hn tt c chng ta u H tng kinh ngc v thn phc trc cc pht minh ca nhng nh ton hc v cng H tng hi, ti sao nhng kt qu p nh vy li khng phi do chnh chng ta sng to ra. Trong khi , trn thc t, nu chng ta c i mt vi nhiu trong s cc pht minh th chng ta c th tm ra li gii d dng trong tm kin thc ca mnh. Hay n gin hn, nhng bn yu ton H tng tham d gii bi trn tp ch Ton hc v Tui tr, H c bao gi cc bn mun tr thnh ngi ra ton hay cha? Hay bn cho rng l cng vic ca thy c, ca nhng ngi ang nghin cu ton hc? Cu tr li l khng phi! Chng ta u c th to cho mnh mt ci g trn nn tng nhng g chng ta H bit v H c, v ci chng ta cn ch l mt cht sng to. Ti mun cng cc bn th sc vi mt trong nhng phng php - phng php tng qut ha!

Khi cc bn gii xong mt bi ton, bn

hHy nn t ho mt cht v cch gii ca mnh v hHy t hi xem, liu cch gii c cn ph hp nu bn thay i chi tit bi. Theo ti, cch gii ti u phi l cch gii s dng t nht nhng d liu H c bi. Khi vi nhng gi thit khng cn thit, bn c th thay i n m cch gii vn gi nguyn. l mt cch tng qut ha. iu ny c v hi tri quy lut v cch lm l tng qut bi ton da trn cch

gii bi ton. Nhng ti ngh l rt t nhin v d lm. Chng ta hHy xem xt mt s v d: 1. V d 1. Tm hm f: [0;1] R, lin tc trong [0;1] tha mLn: 2f(x) 2x. f(x )

Ti xin a ra 2 cch gii khc nhau. a) Li gii 1. T f(x) 2x. f(x2) suy ra x.f(x) 2x2.f(x2) , x (0;1]

Thay x = 0 f(0) 0 t g(x) = x.f(x), x (0;1]

g(x) 2 g(x2)

)(21 2

1

xg g(x).

Bng quy np, ta chng minh c:

2n g(x) g(x n21

), n 1. V g(x) lin tc

trong [0; 1] nn+n

lim g(x n21

) = g(1).

1lim2nn+

. +n

lim g(x n21

) g(x)

0 f(x), x (0;1] (1)

Mt khc, vi x [21;1) ta c

f(x) 2x. f(x2) f(x2) ... f(xn2 )

f(x) +n

lim f(xn2 ) = f(0) 0 (2)

T (1) v (2) ta c f(x) = 0, x [ 12; 1)

Vi x(0;21). Bng quy np ta chng

minh c: f(x) 2nx 12 n

f(xn2 )


TAP S AN TO AN H OC 20 07 4 TRNG THPT CHUYEN HOANG VAN THU - HOA BNH

Vi n ln th f(x) 0 (3)

T (1) v (3) ta c f(x) = 0, x(0; 12)

Vy ( ) 0, (0;1]f x x= . V f(x) lin tc trong [0;1] nn f(x) = 0, x [ ]0;1 Nhn xt. T cch chng minh trn, ta thy: S 2 trong iu kin hon ton c th thay bng s a > 0 bt k, khi ta c bi ton: Bi ton 1.1. Tm tt c cc hm s f(x): [0;1] R, lin tc trong on [0;1] tha mLn iu kin: f(x)ax f(x2) , a > 0.

Hn na, ta c th thay i thnh bi ton

tng qut sau m li gii khng thay i.

Bi ton 1.2. Tm tt c cc hm s f: [0;1] R, lin tc trong [0; 1] tha mLn

iu kin f(x) ax 1 f(x ), trong >1.

b) Li gii 2. Do f(x) lin tc trong [0;1] nn f(x) c nguyn hm trong [0;1]. Gi F(x) l mt nguyn hm ca f(x) trong [0;1]. t g(x) = F(x) - F(x2) g/(x) = F/x) - 2x F(x2) = f(x) - 2x. f(x2) 0, x [0;1] g(x) l hm khng gim trn [0;1].

M g(0) = g(1) = 0 nn g(x) = 0), vi mi x [0;1]

F(x) =F(x2)= .... = F(xn2 )

F(x) = +n

lim F(xn2 ) = F(0), x(0;1)

f(x) = 0, x (0;1) Do f(x) lin tc trong [0;1] nn f(x) =

0, x [0;1]. Nh vy, theo cch gii th 2, ta c th

khi qut c bi ton nh sau:

Bi ton 2.1. Cho hm g: [0;1] [0;1] c o hm trong [0;1] tha mLn iu kin hm [ ]xxg )( n iu trn [0;1], g(0)=0 v g(1)=1. Tm tt c cc hm

s [ ]: 0;1f R , lin tc trong [0;1], tha mLn: f(x) g/(x). f(g(x)), x [0;1].

C bn s t hi ti sao li c th a ra

mt bi ton nh vy. Rt n gin: Bn hHy th tng qut ha bng cch thay x2 bng mt hm g(x) bt k, v p dng hon ton tng t cch trn bn s thy cn phi b sung gi thit c mt cch gii hon chnh. V nh ti H ni trn, cch tng qut ha bi ton y l xut pht t cch gii ch khng phi t bi. Tt nhin, vi gi thit qu c th nh trn s dn n thu hp hng tng qut ca bi ton, v c c mt bi thc s tng qut ti rt mong ch kh nng sng to ca cc bn. Sau y, mi cc bn cng theo di v d 2, cng vi 2 cch gii c v d trc, ti xin xut v d 3 kh th v:

2.V d 2. Gii phng trnh

f(x) -21f(

2x) = x2 trn tp tt c cc hm

lin tc trong on [-21;

31].

Li gii. Gi F(x) l mt nguyn hm ca

f(x) trong [-21;31]. t g(x) = F(x) - F(

2x),

ta c: g/(x) = f(x) -21f(

2x) = x2

g(x) = 31x3 + c.

V g(0) = 0 nn c = 0.

g(x) = 31x3. Vy F(x) = F(

2x) +

31x3

F(x) = 31x3+

31 (

2x)3 +... +

31 ( 12 n

x) +

+ F(n

x

2) =

218

x3(1 - n32

1) + F(

n

x

2), vi

mi x [-21;31]. Khi n ln, ta c:

F(x) = 218

x3 + F(0) x [-21;31].



f(x) = 78x3, x [-

21;31].

Th li thy ng. Nhn xt. Qua cch gii trn ta thy iu

u tin l gi thit x [-21;31] l khng cn

thit, ta c th m rng tp xc nh l

[ ]1;1 m kt qu khng thay i. Th hai, gi thit x2 cng c th khi qut thnh 1 a thc. Nh vy, ta c th khi qut nh sau:

Bi ton 2a. Cho g(x) l a thc bc n c tp xc nh l [-1;1]. Tm hm :[ 1;1]f R , lin tc trn R v tha mLn : f(x) -

31f(

2x) = g/(x).

Cc bn hHy th tm iu kin cho g(x) nu ta mun khi qut g(x) thnh mt hm lin tc bt k.

Tr li bi ton v d 1, vi cch gii trnh by bi ton v d 2, ta hon ton c th thay i gi thit f(x) - 2xf(x2) 0 bi f(x)- 2x f(x2) = g(x).

Cc bn hHy a ra mt bi c cc iu kin rng buc cho g(x) to thnh mt bi ton hon chnh.

Kt hp cc hng tng qut trn, ti xin xut mt bi ton tng qut hn:

3. V d 3.

Cho cc hm s g: [0;1] R,

[ ] [ ]: 0;1 0;1f trong g, h c o hm trn [0;1], h(0) = 0, h(1) =1 v g l a thc bc n .Tm hm f: [0;1] R, tha mLn:

f(x) - h/(x).f(h(x)) = g/(x). Mi cc bn hHy gii bi ton ny v tip

tc! Sau y l bi tp cc bn t luyn:

Bi tp. Cho f(x) c o hm trong (0;1), lin tc trong [0;1], ngoi ra f(0) = f(1) = 0. Chng minh rng tn ti mt s c (0;1) tha mLn iu kin:

f(c) = 1996.f/ (c). Chc cc bn thnh cng!

Xung Quanh Bai Toan Bat ang Thc

THI TOAN QUOC TE 2005

Nguyn anh tun Chuyn ton k97-00

Sv. Lp D2000VT, Hc vin Cng ngh Bu chnh Vin thng.

TTTTrong k thi Olympic Ton Quc t ln th 46 t chc ti Mexico c bi ton v bt ng thc (BT) nh sau:

Bi ton 1. Cho 3 s thc dng x, y, z tha mLn iu kin xyz 1. Chng minh rng:

5 2

5 2 2x x

x y z

+ + +

5 2

5 2 2y y

y z x

+ + +

+5 2

5 2 2z z

z x y

+ + 0 (1)

Li gii 1. BT (1) tng ng vi:

( ) ( )5 2 2 2 2 25 2 2

x y z x y z

x y z

+ + + +

+ + +

+ ( ) ( )5 2 2 2 2 2

5 2 2

y z x x y z

y z x

+ + + +

+ + +

+ ( ) ( )5 2 2 2 2 2

5 2 2

z x y x y z

z x y

+ + + +

+ + 0

5 2 21

x y z+ + + 5 2 2

1y z x+ +

+

+ 5 2 21

z x y+ + 2 2 2

3x y z+ +

(2)

Ta s chng minh:

5 2 21

x y z+ +

( )( )

2 2

22 2 2

3

2

y z

x y z

+

+ +.

Tht vy, theo gi thit xyz 1 ta c:

5 2 21

x y z+ + 4

2 2

1x y zyz

+ +



42 2

2 2

12x y z

y z+ +

+

(3)

p dng BT Bunhiacpxky ta c: 2

2 22 2

2y z y z

+ + +

24

2 22 22x y z

y z

+ + +

( )22 2 2x y z+ + 4

2 22 2

12x y z

y z+ +

+

( )

( )2 2

22 2 2

3

2

y z

x y z

+

+ +(4)

T (3) v (4) suy ra

5 2 21

x y z+ +

( )( )

2 2

22 2 2

3

2

y z

x y z

+

+ +.

Cng tng t:

5 2 21

y z x+ +

( )( )

2 2

22 2 2

3

2

z x

x y z

+

+ +

5 2 21

z x y+ +

( )( )

2 2

22 2 2

3

2

x y

x y z

+

+ +

Cng theo v cc BT trn ta thu c (2) pcm.

ng thc xy ra x = y = z = 1 Li gii 2. p dng BT Bunhiacpxky ta c:

( )5 2 2 2 21x y z y zx

+ + + +

( )22 2 2x y z+ +

5 2 21

x y z+ + ( )

2 2

22 2 2

1 y zx

x y z

+ +

+ +.

Thm hai BT tng t na:

5 2 21

y z x+ + ( )

2 2

22 2 2

1z x

y

x y z

+ +

+ + v

5 2 21

z x y+ + ( )

2 2

22 2 2

1x y

z

x y z

+ +

+ +

Ta suy ra:

5 2 21

x y z+ + + 5 2 2

1y z x+ +

+

+ 5 2 21

z x y+ +

( )

( )2 2 2

22 2 2

1 1 1 2 x y zx y z

x y z

+ + + + +

+ +

Mt khc t gi thit xyz 1

1 1 1x y z

+ + yz + zx + xy

2 2 2x y z+ + , do t BT trn suy ra (2) pcm. ng thc xy ra x = y = z = 1. Bng cch 2, ta chng minh c bi ton tng qut sau:

Bi ton 2. Cho n s thc dng 1 2, ,..., nx x x

( )3n tho mLn iu kin 1 2... nx x x 1. Chng minh rng:

2 11 1

2 11 2 3 ..

n n

n n n n

n

x x

x x x x

+

+

+ + + + +

+ 2 1

2 22 1

2 1 3 ..

n n

n n n n

n

x x

x x x x

+

+

+ + + + + +

+ 2 1

2 12 3 1..

n n

n n

n n n n

n n

x x

x x x x

+

+

+ + + + 0 (5)

Chng minh. Theo BT C-si v gi thit

1 2... nx x x 1 ta c:2 1

1nx +

21

2 3...

n

n

x

x x x

( ) 21

2 3

1...

n

n n n

n

n x

x x x

+ + + (6)



Mt khc, p dng BT Bunhiacpxky ra

c:

2

22

...

...

1

n nn nn

n

x xx x

n

+ + + + +

( ) 221

22

1...

...

n

n n

nn n

n

n xx x

x x

+ + +

+ +

( )21 2 ...n n nnx x x+ + + ( ) 21

2 32

11

...

...

n

n n n

nn n

n

n xx x x

x x

+ + + ++ +

( )

( )2

2

1 2

...

1...

n n

n

n n n

n

x xn

n x x x

+ +

+ + + (7)

T (6) v (7) suy ra:

2 11 2 3

1...

n n n n

nx x x x+ + + + +

( ) 212 3

2

11

...

...

n

n n n

nn n

n

n xx x x

x x

+ + + ++ +

( )

( )2

2

1 2

...

1...

n n

n

n n n

n

x xn

n x x x

+ +

+ + +.

Cng vi n -1 BT tng t khc, cng v vi v ta thu c:

2 11 2 3

1...

n n n n

nx x x x+ + + + +

+

+ 2 12 1 3

1...

n n n n

nx x x x+ + + + +

+ +

+ 2 11 2 1

1...

n n n n

n nx x x x+

+ + + +

1 2 ...n n n

n

n

x x x+ + +

1 22 11 2 3

...

...

n n n

n

n n n n

n

x x x

x x x x++ + +

+ + + + +

+ 1 22 12 1 3

...

...

n n n

n

n n n n

n

x x x

x x x x++ + +

+ + + + + +

+ 1 22 11 2 1

...

...

n n n

n

n n n n

n n

x x x

x x x x+

+ + +

+ + + + n

1 22 11 2 3

... 1...

n n n

n

n n n n

n

x x x

x x x x+ + + +

+ + + +

+

+ 1 22 12 1 3

... 1...

n n n

n

n n n n

n

x x x

x x x x+ + + +

+ + + +

+ +

+ 1 22 11 2 1

... 1...

n n n

n

n n n n

n n

x x x

x x x x+

+ + +

+ + + + 0

(5) ng thc xy ra 1x = 2x == nx =1 Mt dng tng qut khc ca Bi ton 1

nh sau: Bi ton 3. Cho s t nhin n 3 v 3 s thc dng x, y, z tho mLn iu kin

1xyz . Chng minh rng: 2

2 2

n

n

x x

x y z

+ + +

2

2 2

n

n

y yy z x

+ + +

2

2 2

n

n

z z

z x y

+ +

0 (8)

Hay l: 2 21

nx y z+ + + 2 2

1ny z x+ +

+

+ 2 21

nz x y+ + 2 2 2

3x y z+ +

Bng phng php tng t nh li gii 2 chng ta c th chng minh c Bi ton 3 ng vi n 8. Sau y ta chng minh trong trng hp 6n = : p dng BT Bunhiacpxky tng qut ta

c: ( )( )( )6 2 2 2 2 2 21 1x y z y z y z+ + + + + + ( )32 2 2x y z+ + 6 2 2

1x y z+ +

( )

( )22 2

32 2 2

1 y z

x y z

+ +

+ +.

Thm hai BT tng t na, suy ra

6 2 21

x y z+ + + 6 2 2

1y z x+ +

+

+ 6 2 21

z x y+ +

( ) ( ) ( )( )

2 2 22 2 2 2 2 2

32 2 2

1 1 1 (9)y z z x x yx y z

+ + + + + + + +

+ +

Ta s chng minh



( ) ( ) ( )2 2 22 2 2 2 2 21 1 1y z z x x y+ + + + + + + + ( )22 2 23 x y z+ + (10)

t u = 2x , v = 2y , t = 2z th ta c (10)

( )21 u v+ + + ( )21 v t+ + + + ( )21 t u+ + ( )23 u v t+ + 3 + 4 ( )y v t+ + + 2 ( )uv vt tu+ + + + 2 ( )2 2 2u v t+ + 2 ( )2 2 2u v t+ + + + 4 ( )uv vt tu+ + + ( )2u v t+ + ( )2u v t+ + - 4 ( )y v t+ + + + 2 ( )uv vt tu+ + - 3 0 (11)

T gi thit xyz 1, suy ra uv vt tu+ +

( )233 uvt = ( )433 xyz 3, do (11) ng v ta c (10). Vy t (9) v (10) ta c pcm.

Ti d on rng BT (8) ng vi mi n, mong cc bn cng quan tm ti vic chng minh bi ton ny. Sau y l mt bi ton mi m ti H pht hin ra trong qu trnh m rng bi ton trn.

Tm tt c cc s nguyn n sao cho bt ng thc (BT) sau ng vi mi x, y, z khc khng:

( ) ( )2 2 2 2n nx xy y y yz z+ + + + + + + ( )2 2 nz zx x+ + ( )2 2 23 nx y z+ + (12) Li gii. Vi n = 0 th BT (12) hin nhin ng vi mi x, y, z 0. Ta xt cc trng hp sau: i) Vi n < 0. t n = -m (m > 0) , khi BT (12) tr thnh:

( ) ( )2 2 2 21 1

m m

x xy y y yz z+ +

+ + + +

( ) ( )2 2 2 2 21 3

m m

z zx x x y z+

+ + + +

BT ny khng ng vi mi x, y, z 0. Tht vy, c inh x sao cho y0, 0z th

v tri + , trong khi v phi 23

mx ,

v l. ii) Vi n = 1. Khi (12) c dng:

( ) ( )2 2 2 2x xy y y yz z+ + + + + + + ( ) ( )2 2 2 2 23z zx x x y z+ + + + 2 2 2xy yz zx x y z+ + + + . BT ny ng vi mi x, y, z (12) ng. ii) Vi n = 2. BT (12) c dng:

( ) ( )2 22 2 2 2x xy y y yz z+ + + + + + + ( )22 2z zx x+ + ( )22 2 23 x y z+ +

( )3 3 3 3 3 32 x y y x y z z y z x x z + + + + + ( ) ( )4 4 4 2 2 2 2 2 23x y z x y y z z x + + + + +

( ) ( ) ( )4 4 41 1 1 02 2 2

x y y z z x + +

BT cui ng BT (12) ng vi 2n = .

iii) Vi 3n , ta s chng minh rng khi (12) khng ng. Tht vy, ta c (12)

2 2

2 2 2

n

x xy yx y z

+ +

+ + +

2 2

2 2 2

n

y yz zx y z

+ +

+ + +

+ 2 2

2 2 2 3n

z zx x

x y z + +

+ + (13)

Chn x = 1,1; y = 1; z = 0,1 th ta c: 2 2

2 2 2

n

x xy yx y z

+ +

+ + +

2 2

2 2 2

n

y yz zx y z

+ +

+ + +

+2 2 2 2

2 2 2 2 2 2

n n

z zx x x xy yx y z x y z

+ + + +>

+ + + + =

= 3,31 1,492, 22

n

n >

.

Bng quy np ta chng minh c 1,49 3n > vi mi 3n . T suy ra (13) khng ng vi mi x, y, z > 0 pcm.

T nhng phn tch trn cc trng hp trn ta i n kt lun: tt c cc s nguyn n phi tm l n = 0, 1, 2.



TH I TM MOT BAT aNG THC

TRON G TA M GIA C

DNG TH HNG Chuyn Ton K98 01

NGUYN NH THNG

Sv. Lp 3 CLC, K51 HSP H Ni 1

TTTTt c chng ta u bit n nhng nh

l, nhng kt qu l th hay nhng chng minh c o trong ton hc. V liu H c i ln bn H t hi v sao ngi ta li ngh ra nhng iu tuyt diu nh th? Tht kh c th tr cu hi ny mt cch tht chnh xc. Nhng nh th khng c ngha l chng ta s chu b tay! Mc ch ca bi vit ny l t chng ta ng v tr nhng nh kho c th i tm mt cht g , c th ch l mt tr chi cho ring mnh! Ti phi lu cc bn rng, chng ta s th lm nh khai khong, kho c, tm kim ch khng phi l nh pht minh bi c th, nhng g chng ta tm thy s khng c g l qu mi l!

Thng thng, c th tm kim, khai thc c, ta phi c mt khu m hay mt mnh t mu m.

Bn H bao gi n iu ny cha: Vi mi tam gic ABC v cc s thc x,

y, z R v vi mt im M bt k th:

0)( 2 ++

MCzMByMAx ? Khng qu c bit, nhng bn th bin

i li xem no! Khng my kh khn, bn c th nhn c bt ng thc (BT):

2 2 2( )( )x y z xMA yMB zMC+ + + + 2 2 2a yz b zx c xy+ + (*), vi a, b, c ln lt l 3 cnh BC, CA, AB ca tam gic ABC.

y chnh l khu m m chng ta s khai thc.

Mt. Ngay lp tc ta s gp mt h qu: x + y + z = 0 2 2 2a yz b zx a xy+ + , v vi x = b, y = c - a, z = a - b ta c mt kt qu quen bit: 2 ( )( )a c a a b + 2 ( )( )b a b b c +

2 ( )( ) 0c b c c a+ hai . Mt v d khc t tm thng hn l vi b s (x, y, z) = (1, 1, 1) th t (*) ta nhn c:

( )2 2 2 2 2 213MA MB MC a b c+ + + + Ta gp li mt kt qu quen thuc trong

tam gic. ng thc xy ra

++ MCzMByMAx = 0

0=++

MCMBMA M l trng tm tam gic. ba. Th vi b (x, y, z) = (a, b, c) ta c:

2 2 2( )( )a b c aMA bMB cMC+ + + + 2 2 2a bc b ca c ab+ + 2 2 2aMA bMB cMC abc+ +

Ta thy li kt qu c trng cho tm ng trn ni tip tam gic ABC. Nhng vn tng t c xt cho trc tm, tm ng trn ngoi tip, ng trn bng tip v. v ... Nh vy, vi mi b s (x, y, z) thay vo (*) ta s thu c mt BT. Cc bn hHy th chn vi b no nh! Bn. Nhng nh th th cng vic ca chng ta cha c g l th v c. Th ly

( )2( , , ) , ,1x y z a b= xem no! Ta c: ( )( )2 2 2 2 21a b a MA bMB MC+ + + +

2 2 2 2 2a b b a c a b+ + iu ny c v ng ng! Bn H ngh

ra cch no khc chng minh iu ng ng y hay cha?

Xin cc bn ng vi bc mnh v cng vic khai thc ca chng ta qu chm chp. Trn y chng ta H s dng cng ngh kh c k l khi thay b (x, y, z) bi nhng b s c nh. Vic lm ny khng phi l khng c li ch g nhng xem ra sn



phm ca chng ta cha c phong ph lm.

Sau y l mt vi ci tin nho nh nhng s mang li nhng kt qu bt ng. nm. Trong (*) thay

(x, y, z) = (a

MA,

bMB

, c

MC) vi

{ }; ;M A B C ta c: (

a

MA+

bMB

+c

MC)(aMA + bMB + cMC)

( ). . .

a b cabc

MB MC MC MA MA MB+ +

. . .MB MC MC MA MA MB

bc ca ab+ + 1. (1)

D thy khi M {A, B, C} th (1) tr thnh ng thc. Vy (1) ng vi mi M.

Suy din mt cht, chng ta s c ngay nhng kt qu quen thuc:

3MA MB MCa b c

+ +

3 32

a b cm m m

a b c+ +

2 3a b c

a b cm m m

+ +

133

nn n nMA MB MCa b c

+ +

y ma, mb, mc ln lt l di cc trung tuyn ng vi cc nh A, B, C ca tam gic ABC. Su. Chng ta cng th lm iu tng t

khi: (x, y, z) = (a

MA ,

bMB

, c

MC) ta c:

( )a b c aMA bMB cMCMA MB MC

+ + + +

( ). . .

a b cabc

MB MC MC MA MA MB

( )aMA bMB cMC + +

. . .MB MC MC MA MA MBbc ca ab

+ +

( )aMA bMB cMC .

l khi M khng nm trn cung ln BC cha A th: ( ) 0aMA bMB cMC + + > (BT Ptlm) nn ta thu c:

. . . 1MB MC MC MA MA MBbc ca ab

+ + (2)

Khi M BC khng cha A th

{ }; ;M A B C cng ng. Vy (2) ng vi im M bt k.

Ta H tm thy mt h hng ca BT Ptlm: Khi M BC th (2) v BT Ptlm l tng ng. Nhng tic rng, trong khi BT Ptlm th ai cng bit cn ngi anh em ny th chng my ai bit n! Tht ng thng! By. V nu chng ta thay

2 2 21 1 1( , , ) , ,x y z

MA MB MC

=

th t (*)

ta thu c:

( )2 2 21 1 1 . 1 1 1MA MB MC

+ + + +

2 2 2

. . .

a b cMB MC MC MA MA MB

+ +

( ) ( ) ( )2 2 2. . .MB MC MC MA MA MB+ + ( )2 2 2 2 2 213 a MA b MB c MC+ + (3)

(3) vn ng khi { }; ;M A B C . Bn th chng minh (3) khi M G xem

no, chc cng khng n gin lm! Tm. Trong (*) thay

1 1 1( , , ) , ,x y z

MA MB MC

=

ta c

( ) ( ). . .MB MC MC MA MA MB MA MB MC+ + + + 2 2 2a MA b MB c MC+ +



Gi s tam gic ABC t ti A, khi ta c: MB c MA, MC b MA.

Du bng xy ra M A. Do 2 2 2a MA b MB c MC+ +

( )2 2 2 ( )a b c MA bc b c + + bc(b+c) (4). ng thc xy ra M A.

Ta thy rng (4) ng c trong trng hp { }; ;M A B C v khi M A th c du ng

thc. Nh vy ta c bi ton: Cho tam gic ABC t A. Tm gi tr nh

nht ca biu thc.

( )( ). . .MB MC MC MA MAMB MA MB MC+ + + + Ta thy rng nhng bt ng thc thu c trn u ng vi mi M. Do khi thay M bi nhng v tr c bit ta li thu c kh nhiu kt qu, tnh cht th v.

Cc bn thy y, xut pht t (*), mi ln thay (x, y, z) bi mt b no , ta thu c mt kt qu mi. Cng vic ca nh tm kim l phi bit cht lc, gi li nhng g c gi tr t nhng th tng chng nh tm thng. C mt ln lm th, chng ta mi thy rng, c nhng g chng ta ang c hc hm nay khng phi l d. Ti hi vng sau bi vit ny, mi chng ta s rt ra mt iu g cho ring mnh c th hc mn Ton vui v hn, v nhng ai say m mun lm nhng nh kho c, hHy c bt tay vo cng vic ca mnh du bit rng chng ta c th chng thu lum c g to tt. Nhng, c mt iu ti tin chc l sau khi nhng ln nh th, bn s thy Ton hc cng ng yu hn.

Nu bn cm thy thch cng vic tm kim, my m nhng iu mi m (d ch cho ring mnh) ti c th gii thiu vi bn mt vi mnh t mu m, ph hp vi nhng g bn mong mun: A. Mnh t 1:

Mnh t ny i hi bn phi c mt vi s chun b v s phc. Trong s thc, ta c ng thc:

1))(())((

))(())((

))(())((

=

+

+

bcacbmam

abcbamcm

cabacmbm

c c ng thc trn hoc nhng ng thc tng t, bn c th da vo cng

thc ni suy Lagrng cho a thc v so snh h s. Chng hn, v d trn khai trin x2 ti a, b, c. Do cc s phc tnh ton nh s thc nn trong cc s phc cng c cc ng thc nh vy. Vn dng tnh cht nhHn (hoc chun) ca tng v tch cho mi s phc tng ng vi mt im. Chng hn t ng thc trn ta thu c: ( )( ) ( )( ) ( )( ) 1( )( ) ( )( ) ( )( )m b m c m c m a m a m ba b a c b c b a c a c b

+ +

1.

.

.

.

.

. ++CBCAMBMA

BABCMAMC

ACABMCMA

Theo cch , bn s thu c rt nhiu iu th v.

B. Mnh t 2:

Gi a, b, c l ba cnh ca tam gic

ABC tng ng. Cc im M, N nh trn hnh v.

Xut pht t kt qu a.NA + b.NB + +c.NC a.MA + b.MB + c.MC. HHy tm cch gii trn vn bi ton sau:

Cho tam gic ABC v x, y, z > 0 . HLy tm im M trong tam gic ABC sao cho:

S(M) = xMA + yMB + zMC nh nht. Chc cc bn cng H bit khi x = y = z

th ta c bi ton im Toricelli ca tam gic C. Mnh t 3.

Trc ht, cc bn hHy chng minh vi x + y > 0, y + z > 0, z + x > 0 th ta c :

ABCSzxyzxyzMCyMBxMA .4 ++++T hHy xy dng mt vi tnh cht mi!

Vn cn nhiu vng t mi ang ch in du chn cc bn. Chc thnh cng!



Mot S Mot S Mot S Mot S Tnh c

Nguyn Lm Tuyn Chuyn Ton K99 02

Sv. Lp iu khin T ng 1 - K47 H Bch Khoa H Ni

Trong cuc sng, nhiu iu th v i khi n vi chng ta mt cch nh nhng, man mc Lm cho ta thm yu i, yu cuc sng! Bi vit ny ti xin trnh by mt nim vui nho nh m ti tnh c c c khi ang th hn vi nhng bi ton hc ba. Thn tng ti cc bn, c bit l cc em lp chuyn Ton K01-04 nhng ngi rt tm huyt vi t bo ny. Hy vng rng qua bi vit ny phn no s gip ch cho cc bn trong qu trnh hc ton.

I/ Th thch.

Nm cn hc lp 10 Chuyn Ton, c hai bi ton khin ti rt trn tr. l hai bi ton thi hc sinh gii Quc gia, va quen li va l:

Bi ton HSG1. Cho a thc P(x) = x3 9x2 + 24x 27. Chng minh rng vi mi s t nhin n, tn ti s nguyn an sao cho P(an) chia ht cho 3n.

Bi ton HSG2. Cho a thc P(x) = x3 +153x2 -111x +38.

i) Chng minh rng vi mi s t nhin n, tn ti t nht 9 s nguyn a thuc on

2000[1;3 ] sao cho P(an) chia ht cho 32000.

ii) Hi trong on 2000[1;3 ] c tt c bao nhiu s nguyn a sao cho P(an) chia ht cho 32000. (Cc bn c th tham kho thm cc s tp ch Ton hc v Tui tr thng 01, 02, 09 nm 2001) Sau nhiu ngy suy ngh ti H pht hin ra mt cch chng minh, nhng kh di v ch cho ring Bi ton HSG1 (xin khng nu ra y).

II/ Tnh c.

Bng i mt thi gian ri tnh c lt li trang sch. tng cht ln, ti h bt vit nh H c lp trnh sn. Ti H c mt li gii mi cho Bi ton HSG1, nhng tt nhin l vi phong cch hon ton khc.

Li gii nh sau. Ta c P(x) = x3 9x2 + 24x 27

= (x - 3)3 3(x - 3) 9 P(3x+3) = 9(3x3 x 1).

Bi ton quy v vic chng minh: Vi mi n, tn ti bn N

* sao cho Q(bn) chia ht cho 3n. y Q(x) = 3x3 x 1.

Ta s chng minh iu ny bng quy np theo n. Vi n = 1 chn b1 = 2.

Gi s khng nh ng ti n. Ta c Q(bn+Q(bn)) =

= 3(bn+Q(bn))3 (bn+Q(bn)) 1

= ))()((9)13( 3 nnnnnn bQbbQbbb ++ + + )()(3 3 nn bQbQ = 33 3 ( )( ( )) ( )

n n n n nb Q b b Q b Q b + +

Chn bn+1 = bn+Q(bn) th Q(bn+1) chia ht cho 3n+1. Tm li ta c iu phi chng minh.

Vi vng em p dng cho Bi ton

HSG2 nhng khng thnh cng! Ti quyt nh quay tr li Bi ton HSG1 vi mc ch m rng n v H a ra c bi ton tng qut sau.

Bi ton A. Xt tp hp cc a thc c dng TTTT = { P(x) = ax3 + bx2 + cx + d / a 0, b 0(mod3), c 0(mod3), a + c



0(mod3)}. Khi vi mi s nguyn dng n, tn ti s nguyn an sao cho P(an) chia ht cho 3n.

Chng minh. Ta chng minh bng quy np. Vi n = 1, ta c P(0) = d, P(-1) =- a + b c + d - (a + c) + d 0(mod3), P(1) = a + b + c + d (a + c) + d(mod3). Lu rng trong 3 s hng lin tip ca mt cp s cng c cng sai khng chia ht cho 3, lun tn ti mt s chia ht cho 3. Vy vi n = 1, bi ton ng.

Gi s tn ti ak P(ak) chia ht cho 3k.

Ta c P(ak + hP(ak)) = a(ak + hP(ak))3 +

b(ak + hP(ak))2 + c(ak + hP(ak)) + d =

()

2 2 3 2

2

( ). 3 . . 3 . ( ) ( )( ) (2 ) 1

k k k k k

k k

P a a a h a a h P a h P a

bh P a ba c h

= + + +

+ + + +

Ta thy 2bak + c 0 (mod3) tn ti { }1;2h sao cho (2bak + c)h + 1 0(mod3).

T chn ak+1 = ak + hP(ak) th ta c P(ak+1) chia ht cho 3

k+1 (pcm). Tt nhin l cng vi xu hng , ti

tm cch m rng bi ton thm na, nhng qu thc l rt kh khn. Sau mt vi php th v d on ti a ra bi ton sau m theo ti, mt kha cnh no , n m rng cho Bi ton HSG1.

Bi ton B. Cho s nguyn t l p v a thc ( ) ( 1) 1pQ x p x x= . Chng minh rng vi mi s nguyn dng n, tn ti v hn s nguyn dng an m Q(an) chia ht cho pn.

Vi li gii cng ging nh cch chng minh Bi ton HSG1.

c bit trng hp n = p, ta c bi ton ring nhng dng nh li kh hn v vi bi ton mi ny, chng ta s khng d dng ngh ngay ti phng php quy np chng minh.

Bi ton C. Cho s nguyn t l p v a thc ( ) ( 1) 1pQ x p x x= . Chng minh

rng tn ti v hn s nguyn dng a m Q(a) chia ht cho pp.

T m, ti th tm mt li gii khc cho

bi ton mi ny. V cng tnh c ti a ra c mt li gii mi, v tt nhin l cng vi phong cch hon ton mi: s dng khi nim h thng d ca l thuyt ng d thc.

Bi ton C cng chnh l ni dung ca bi T8/336 trn Tp ch Ton hc v Tui tr thng 06/2005 do ti xut. Xut s ca bi T8/336 l nh vy v c l, cng l mt s tnh c.

Chng minh Bi ton C. Nhn xt: Gi tr ti pp im nguyn dng lin tip ca a thc Q(x) lp thnh mt h thng d y (modpp).

Tht vy, trong pp s nguyn dng lin tip, gi s c u > v sao cho

( ) ( )(mod )pQ u Q u p ( 1) 1pp u u ( 1) 1(mod )p pp u u p

( ) ( )( 1) 0(mod )p p pp u v u v p (*). Theo nh l Fermat nh, ta c

(mod )pu u p , (mod )pv v p . Do t (*) ta c ( )( 2) 0(mod )p u v p . Li c ( ); 2 1p p = , suy ra (mod )u v p .

Cng t (*) ta c

( ) ( ) ( )( )1 2 11 ... 1p p pu v p u u v v + + + 0(mod )pp . Mt khc (mod )u v p

( )( )1 2 11 ... 1p p pp u u v v + + + ( 1). 1 0(mod )pp p p . Suy ra (mod )pu v p . Ch l 0 < u v < pp u = v. Nhn

xt c chng minh. H qu l trong pp s nguyn dng lin

tip, tn ti duy nht mt s a Q(a) chia ht cho pp. V do hin nhin l trong tp hp v hn cc s nguyn dng, tn ti v s s a m Q(a) chia ht cho pp. Bi ton c chng minh.



Trong li gii Bi ton B ta ch ra c s tn ti ca an nhng H khng ch ra c c bao nhiu s nh vy. Ci th v cch gii Bi ton C khng ch l s mi l trong cch t duy m cn khc phc c im hn ch ca phng php trc . Kh bt ng vi li gii trn, ti cht nh n bi ton HSG2 m mnh cha gii c. em p dng phng php mi ny cho bi ton v ti H thnh cng!

Nhng ti li i t bi ton tng qut:

Bi ton D. Xt tp hp cc a thc c dng TTTT = { P(x) = ax3 + bx2 + cx + d / a 0, b 0(mod3), c 0(mod3), a + c 0(mod3)}. Khi gi tr ti 3n im nguyn dng lin tip ca a thc P(x) TTTT lp thnh mt h thng d y (mod3n).

Chng minh. Trong 3n s nguyn dng lin tip gi s c u > v m Q(u) Q(v) (mod3n) au3 + bu2 + cu + d av3 + bv2 + cv + d (mod3n) a(u3 - v3)+ b(u2 - v2) + + c(u v) 0(mod3n) (*).

Ta c b0(mod3), u3u(mod3), v3 v(mod3) nn t (*) a(u3 - v3)+ b(u2 - v2) + c(u v) 0(mod3) (a + c)(u - v) 0 (mod3) u v(mod3), do (a + c) 0(mod3).

Cng t (*) ta c (u v)[a(u2 + uv + v2) + b(u + v) + c] 0(mod3n).

M u v(mod3), c 0(mod3) a(u2 + + uv + v2) + b(u + v) + c 0(mod3) u v (mod3n). Vy u = v pcm.

H qu l: Trong 3n s nguyn dng lin tip tn ti duy nht mt s a Q(a) chia ht cho 3n.

y chnh l s tng qut cho Bi ton HSG2. C th, li gii ca bi ton HSG2 nh sau:

Li gii Bi ton HSG2. Ta c P(x) = x3 +153x2 -111x +38 TTTT. Gi s P(x) chia ht cho 32000 P(x) phi chia ht cho 3

x c dng 3k + 1 P(x) = P(3k + 1) = = 33(k3 + 52k2 22k +3).

* Nu k = 3m + 2 P(x) = 33(27m3 + 495m2 387m + 263)

khng chia ht cho 34 vi mi m. * Nu k = 3m + 1 P(x) = 34 (9m3 + 165m2 129m + 26)

khng chia ht cho 35 vi mi m. * Nu k = 3m P(x) = 34(9m3 + 156m2 + 22m + 1). Ta thy a thc Q(m) = (9m3 + 156m2 +

22m + 1) TTTT v 1 x 32000 0 m 31998 1. Vy P(x) chia ht cho 32000 x = 9m + 1 v Q(m) chia ht cho 31996.

Theo h qu ca Bi ton D suy ra: Trong 9.31996 s nguyn lin tip 0, 1, 2, ,

19983 1 tn ti ng 9 s nguyn a m Q(a) chia ht cho 31996 Trong on [1; 32000] tn ti ng 9 s nguyn a m P(a) chia ht cho 32000.

Ta cng d dng nhn ra l trong on

[1;3 ]n (n 1998) tn ti 3n 1998 s nguyn a m P(a) chia ht cho 32000. Bi ton HSG2 H c gii quyt trn vn!

III/ Li kt. Ch mt cht thay i bi theo

tng ca mnh cc bn c th to ra c nhng bi ton mi cng kh hc ba y ch! Vn t ra y l trong trng hp tng qut, a thc P(x) bc n th kt qu s ra sao? Bn thn ti cng cha c iu kin tm hiu thm, mong cc bn cng quan tm coi nh mt bi tp trc khi kt thc bi vit ny.

Nh vy y cc bn , t mt s tnh c

ti H gii c mt bi ton kh v tm ra c nhiu iu th v. Nhng c c s tnh c l c mt qu trnh n lc khng ngng v mt tri tim am m Ton hc mHnh lit. Cui cng xin chc cc bn thnh cng v tm ra c nhiu cng trnh cho ring mnh trong qu trnh hc tp v vn ln tt c cc lnh vc!



S D UN G T N H CHA T HA M N A N HS D UN G T N H CHA T HA M N A N HS D UN G T N H CHA T HA M N A N HS D UN G T N H CHA T HA M N A N H E GIAI BAI TOAN PHNG TRNH HAME GIAI BAI TOAN PHNG TRNH HAME GIAI BAI TOAN PHNG TRNH HAME GIAI BAI TOAN PHNG TRNH HAM

Nguyn thI ngc Chuyn ton k99-02

Sv. Lp T8 K48, Khoa in t Vin thng - H Bch Khoa H Ni

TTTTrong cc k thi Hc sinh gii, ta thng gp cc bi ton v gii phng trnh hm. y l dng ton kh quen thuc vi cc bn. Trong cun "Phng trnh hm" ca GS - TS.Nguyn Vn Mu, tc gi H cp tng i su v mt lp phng trnh hm. Trong phm vi bi vit ny, ti xin c nu ra mt phng php gii dng ton ni trn kh hiu qu. l phng php s dng tnh cht hm n nh. Trc ht, ti xin nu nh ngha v mt s nhn xt xoay quanh hm n nh: nh ngha hm n nh.

Hm s f : X Y x y = f(x) c gi l mt hm n nh nu x1, x2

thuc X m x1 x2 suy ra f(x1) f(x2) Nhn xt 1.

Cho f l mt hm s xc nh, lin tc trong khong (a,b), khi , nu cc s u, v thuc (a,b) sao cho u < v v f(u) < f(v) th vi bt k w thuc (u,v) lun c:

f(u) v hoc u < v, do vy f(u) < f(v) hoc f(u) > f(v) ngha l f(u) f(v). b) Nu f n nh th f n iu ngt: Vi a' < b' (a,b). Khi hoc f(a') < f(b') hoc f(a') > f(b') do vy ta s chng minh:

Hoc (i): Nu f(a') f(b') th f gim ngt. Xt (i), cho u < v; u, v (a,b) t w =

min {a';u }; z=max{b';v} khi a', b', u, v u thuc on [w, z] .

Theo h qu ca Nhn xt 1,v f n nh nn tng ngt trn [w,z]. V u v, u, v [w,z] nn f(u) < f(v) v u, v l hai im bt k (u < v) trn (a,b) nn f tng ngt trn (a,b).

Mun chng minh (ii) ch cn thay f bi f v lp lun tng t .



Nh vy l chng ta H c mt s nhn xt v h qu kh hay v hm n nh. Sau y xin c i vo mt s bi ton c th: Bi ton 1. Tm tt c cc hm s lin tc

:f R R tho mLn iu kin f(x.f(y)) =y. f(x) , x,y R.

Li gii. Cho x = y = 0 f(0) = 0. D thy f(x) 0 l mt nghim ca phng trnh hm. Xt f(x) 0. Cho x = y = 1 f(f(1)) =f(1) . Suy ra: f(x.f(f(1))) =f(1).f(x) =f(x.f(1))=f(x).

Vy f(1) =1. Gi s tn ti x1 x2 m f(x1) = f(x2) . Ta c f(x.f(x1)) = x1.f(x), xR v (x.f(x2)) = x2. f(x), xR. x1. f(x) = x2. f(x) , xR x1 = x2 (v f(x) 0 ), mu thun.

Vy f l n nh v do f lin tc nn theo Nhn xt 2 suy ra f n iu ngt .

C f(1) >f(0) vy f tng ngt (H qu ca Nhn xt 1)

C f(f(x.f(y))) =f(y.f(x)) = x.f(y), xR Nu f(x.f(y)) > x.f(y)

x.f(y) = f(f(x.f(y))) > f(x.f(y)) > x.f(y), v l Nu f(x.f(y))< x.f(y)

x.f(y) = f(f(x.f(y))) < f(x.f(y)) < x.f(y), v l Vy f(x.f(y)) = x.f(y) Thay x =1 f(1) =x,x Th li thy f(x) 0, f(1) x l hai

nghim ca phng trnh hm Bi ton 2. Tm tt c cc hm f(x) xc nh trn R c hu han nghim tho mLn:

f(x4+y) = x3.f(x)+f(f(y)), x,y R. (APMO- 2002)

Li gii. Cho x = 0 f(f(y))=f(y), yR f(x4+y) = x3.f(x) + f(y), x, y R. f(x4+y) = -x3.f(-x) + f(y), x, y R f(0) = 0

Cho y = 0 f(x4) = x3.f(x), xR Nu x0 0 sao cho f(x0) = 0

f(x 40 ) = 0. Nu x0 1 th tn ti dHy: x1 = x0 , xn=x

41n ,n = 2, 3, ....

y l dHy v s s hng khc nhau m f(x) nhn lm nghim. Tri gi thit.

Nu x0 = 1 tc l f(1) = f(-1) = f(0) th ta c f(2) = 2.f(1) = 0.

Vy 2 l nghim ca f, tri vi iu trn. Vy x = 0 l nghim duy nht ca hm

f(x). C f(x4+y) = x3.f(x) + f(y) = f(x4) + f(y) Nu x 0 f(x+y) = f(x) + f(y)

Nu x < 0 f(x+y) = f(-x-y) = -f(x-y) = -(f(-x) + f(-y)) = f(x) + f(y).

Vy f(x+y) = f(x) + f(y),x, y R. Gi s x1 x2 m f(x1) = f(x2)

f(x1+y) = f(x2+y),y R. f(x1- x2) = 0. Hay f nhn x1-x2 0 lm nghim (V l).

Vy f l n nh . Do f(f(y))=f(y) f(y) = y, yR. Th li thy f(y) y l nghim duy nht

ca phng trnh hm. Bi ton 3. Tm tt c cc hm f :N*N* tha mLn

f(m+f(n))= n+ f(m+ 2003),m,nN* Li gii. Gi s n1 n2 m f(n1) = f(n2) f(f(n1)+m) = n1 +f(m+2003), m,N* v f(f(n2)+m)= n2 +f(m+2003), m,N*, V l.

Vy f l n nh. Ta c: f(f(1)+f(n)) = n + f(f(1)+2003) =

n+1 + f(2003+2003) = f(f(n+1)+2003). T f(f(1)+f(n)) = f(f(n+1)+ 2003). Do f l n nh nn f(1) + f(n) = f(n+1) + 2003. Bng quy np ta suy ra: f(n) = an +b. Thay vo iu kin ca bi ta xc nh

c: a=1, b=2003. Vy f(n)=n+2003, n N*. Cui cng xin nu mt s bi ton m ta

c th s dng tnh cht hm n nh gii quyt . Chc cc bn thnh cng !

1. Tm tt c cc hm s f :QQ tho mLn :f(f(x)+y)=x+ f(y),x,yQ.

2. Tm tt c cc hm s f: RR tho mLn

f(y-f(x))=f(x2002-y)-2001.y.f(x), x,yR. (Chn hc sinh gii quc gia 2001-2002).



LI GIAI CAC BAI THI TOAN QUOC TE 2003

H Hu Cao Trnh Chuyn ton K99 02

Lp K6 CNKHTN Ton, HKHTN - HQG H Ni

TTTTi cng xin chia s mt s kinh nghim trong gii ton vi cc bn qua li gii thi ton quc t 2003. Hi vng rng qua y cc bn s rt ra cho mnh nhiu iu b ch. Bi s 1. Ly A l mt tp con 101 phn t ca tp S gm cc s t nhin t 1 n 1000000. Chng minh rng: tn ti cc s t1, t2,..., t100 thuc S sao cho cc tp

Ai = { }Axtx i + / , i = 1, 2,..., 100 i mt khng giao nhau. Li gii. Xt tp D = { }Ayxyx ,/ , ta thy D c nhiu nht 101.100 +1 phn t, trong chc chn cha phn t 0. Nhn xt: Ai, Aj c giao khc rng khi v ch khi ti - tj D (*), nn ta ch cn chn 100 phn t khng tha mHn (*). Ta chn bng quy np:

Vic chn 1 phn t l tm thng. Gi s chn c k phn t, k 99 khng vi phm (*). Nh vy, ta vn cn t nht 106 - 10101.k 1 s la chn na cho phn t th k+1. Tm li, ta c th chn c 100 s t1, t2,..., t100 khng vi phm (*). Theo nguyn l quy np, bi ton c chng minh. * Ch : Pht biu tng qut sau vn ng nh php chng minh tng t. Nu A l tp con k phn t ca tp S gm cc s t nhin t 1 n n v m l s nguyn

dng tha mLn: n > (m -1)( C k2 + 1), th tn

ti m s t1, t2,..., tm thuc S tha mLn cc tp Ai xc nh nh trn i mt khng giao nhau. Bi s 2. Tm tt c cc cp s nguyn

dng (a; b) tha mLn 12 32

2

+ baba

l mt

s nguyn dng.

Li gii. Gi s cp (a; b) tha mHn bi

ton. Do k = 12 32

2

+ baba

> 0 nn 2ab2 -

b3+1 > 0 v a 2b.

Mt khc, ta c a2 b2(2a - b) +1 > 0 nn hoc a > b hoc 2a = b (*)

Xt hai nghim (a1, a2) ca phng trnh a2 - 2kb2a + k(b3-1) = 0, vi gi s a1 a2 .Theo nh l Vi- et, ta c a1+a2 = 2kb

2 a1 kb2 > 0.

Hn na, t a1a2 = k(b3-1), ta c 0 a2 =

1

3 )1(a

bk 2

3 )1(kbbk

< b. Kt hp vi

(*) a2 = 0 hoc a2 = 22b .

- Nu a2 = 0 th b3 - 1 = 0 a1 = 2k, b =1.

- Nu a2 = 22b th k =

4

2bv a1 = 2

2b-

2b

.

Nh vy, ta H tm c (a;b) di dng (2t; 1) hoc (t; 2t) hoc (8t4- t; 2t), tN*. Th li u thy tha mHn. Ch : C th suy c (*) bng cch sau: Xt hm f(b) = 2ab2- b3+1, hm ny tng

trn

34

;0 a , gim trn

+;

34a

v ta c:

f(a) = a3 + 1 > a2, f(2a-1) = 4a2- 4a+2 > a2, f(2a+1) = - 4a2- 4a < 0.

Nu b a v )(2

bfa

nguyn dng th b =

2a. Tht vy, nu a b3

4ath f(b) f(a)

> a2 )(2

bfa

< 1, v l. Cn nu b > 3

4a

th :



i) Nu b > 2a + 1 th f(b) < f(2a+1) < 0, v l.

ii) Nu b2a-1 th f(b) f(2a-1) > a2, v l. Bi 3. Mi cp cnh i din ca lc gic li c tnh cht sau: khong cch

trung im ca chng gp 23ln tng

di ca chng. Chng minh rng tt c cc gc ca lc

gic bng nhau. Li gii. Cch 1. Ta s chng minh nhn xt sau: Nu

060QPR v L l trung im ca QR th

PL 23QR. Du bng xy ra khi v ch

khi PQR l tam gic u. Chng minh nhn xt. Ly S sao cho QRS

l tam gic u v P nm trong phn giao ca na mt phng b QR cha S vi phn hnh trn ngoi tip QRS.Vy P nm trong c (L, LS)

PL < LS = 23QR (pcm)

Tr li bi ton ca ta, gi lc gic li l

ABCDEF. Xt mt ng ni trung im N ca DE vi trung im M ca AB , AE BD = P. Khng mt tng qut gi s

060APB . V tng 3 gc to bi cc ng cho chnh lin tip bng 1800. Theo nhn xt trn, ta c :

MN = 23(AB+DE) PM+PNMN

Du bng buc xy ra theo gi thitABP l tam gic u. n lc , ta

c th gi s mt trong hai gc cn li to bi cc ng cho chnh 600 v chng minh tng t, ta c ngay ABCDEF c tt c cc gc bng nhau. Cch 2. S dng vect v nhn xt cch 1, cc bn hHy t chng minh (!). Bi 4. Cho ABCD l mt t gic ni tip. Gi P, Q, R l cc chn ng vung gc h t D xung BC, CA, AB. Chng minh: PQ=QR khi v ch khi phn gic ca

ADCABC , v ng cho AC ng quy. Li gii. Cch 1. Ta bit P, Q, R thng hng (ng thng Simsn). D dng c c :

~DCA DAR ~DAB DPQ ~DBC DRQ

BCBA

PQQR

BAPQDBBCQRDB

DPDR

DCDA

===

.

.

nn PQ

= QR BCBA

DCDA

= (1)

Theo tnh cht ca ng phn gic, (1) xy ra khi v ch khi phn gic

ADCABC , chia AC theo cng mt t s, tc l chng ng quy, (pcm).

Cch 2. Gi s phn gic ADCABC , ln

lt ct AC L v M. T BCBA

LCLA

= v

DCDA

MCMA

= LM khi v ch khi



=DCDA

BCBA

AB.CD = CB.AD (1)

t = ACB , = CAB , ta c: cc t gic PDQC v APQR ni tip nn PDQ hoc bng hoc bng 1800 - . QDR hoc bng hoc bng 1800 - . Theo nh l hm s sin ta c:

PQ = CDsin , QR = ADsin

PQ = QR DCDA

=

sinsin

Mt khc BABC

=

sinsin

nn ta c lun

PQ = QR ADCBCDAB .. = (2) T (1) v (2) suy ra pcm.

Bi 5. Cho n l s nguyn dng v x1 x2 ... xn l cc s thc.

a) Chng minh rng:

( )==

n

jiji

n

jiji xx

nxx

1,

222

1, 3)1(2

b) Chng minh rng du ng thc xy ra khi v ch khi x1,x2,...,xn l mt cp s cng. Li gii.

a) Khng mt tng qut, ta gi s

=

n

iix

1=0 (do 2 v ca bt ng thc (BT)

ch ph thuc (xi - xj)). Ta c :

( ) = 0. III. Dng hnh hc, lng gic ca s phc. Nhn. 1. Dng hnh hc.

Trong mt phng ta Oxy, nu im M c ta M(a;b) th ngi ta biu din n bi s phc z = a +ib, gi l nhHn ca im M.

Tm li: im M(a;b) c nhHn l z = a +

ib. Ngi ta cng ni vector OM

c nhHn l z = a +ib. 2. Dng lng gic.

Vn vi im M trn (khc gc ta ), ta xt gc nh hng ( , )Ox OM = . t r = 22 ba + ta c cos =

r

a, sin =

r

b, do

( )cos sinz r i = + (2) Biu din (2) c gi l biu din lng

gic ca s phc z. Ngi ta gi r l modul ca z, k hiu z

ng thi, gi l argument ca z, vit tt l argz, tt nhin argz nhn v s gi tr : argz = + k2pi , kZ, v ta thng dng [0; 2pi ]. Vi hai s phc

( )1 1 1 1cos sinz r i = + ( )2 2 2 2cos sinz r i = + ta c:



( ) ( )2 2 1 2 1 2 1 2cos sinz z r r i = + + + ( ) ( )1 1 1 2 1 2

2 2

cos sinz r iz r

= + cho tin, bt u t y ta k hiu

im bi ch in hoa v nhHn ca n l ch in thng. V d: im Z c nhHn l z = a + ib.

By gi, ta xt hai im Z1, Z2 c nhHn tng ng l

( )1 1 1 1cos sinz r i = + ( )2 2 2 2cos sinz r i = +

Lc ny th tng z3 = z1 + z2 s biu din v tr ca im Z3 m 3 1 2OZ OZ OZ= +

Hiu w = z1 - z2 s l nhHn ca 1 2Z Z

. Gc nh hng

( ) 11 2 1 22

, arg arg arg zOZ OZ z zz

= =

Vi 4 im Z1, Z2, U1, U2, ta c:

arg ( ) 1 21 2 1 21 2

, arg U UZ Z U UZ Z

=

IV. Tnh cht.

1) z + z = 2R(z)

2) 21zz = 1z 2z ; 21 zz + = 1z + 2z

3) z. z = z 2

4) 1 2 1 2Z Z U U

12

12

uu

zz

=12

12

uu

zz

V. Mt s bi ton.

Bi ton 1. Cho ba hnh vung ABCD, BEFC, EPQF nh hnh v. Chng minh

rng:ACD + AFD +AQD = 2pi.

Li gii. Dng h trc ta nh hnh v v nhn AB

lm vect n v ca trc honh.

Suy ra nhHn ca A, B, C, D, E, F, P, Q tng ng l: a = 0; b =1; c =1 + i; d = i; e =2; f = 2 + i; p =3; q =3 + i.

T : ACD AFD AQD + + =

= (AB,AC) + (AE,AF) + (AP,AQ) =

= argc + argf + argq =

= arg(c.f.q) =

arg(1 )(2 )(3 ) arg(10 )2

i i i i pi= + + + = = (pcm)

Bi ton 2. Cho t gic li ABCD. Dng ra pha ngoi cc tam gic cn ng dng ABP,BCD,CDR,DAS, m P,Q,R,S tng ng l cc nh cn. Chng minh: nu PQRS l hnh bnh hnh th ABCD cng l hnh bnh hnh.

Li gii.

t ( ),AB AP = , th th arg p ap b = hay l ( )cos sinp a AP i

p b AB = +

mABAP

=cos2

1nn

bpap

=21(1+tg ) suy

ra ( )( )1 12

p tg b a a= + + .

Tng t nh vy ta cng c:

( ) ( )1 12

q tag c b b= + + ,



( ) ( )1 12

r tg d c c= + +

( )( )1 12

s tg a d d= + +

Do PQRS l hnh bnh hnh nn s + q = p + r. T y d dng suy ra pcm.

Nhn xt. Bi ny khi ra cho chng ti, thy gio H cho iu kin nh hn l PQRS l hnh ch nht c th s dng nh l con nhm. Nhng r rng vi s phc, ta vn gii quyt bi ton mt cch bnh thng v cng ht sc t nhin. Tt nhin, s phc khng phi lc no cng l hng gii quyt tt cho cc bi ton, bi v trong tay cc bn cn rt nhiu v kh khc, nhng cc bn hHy c gng s dng n thng xuyn. kt thc, xin gii thiu vi cc bn mt bi thi hc sinh gii Quc gia m t ti H tm ra li gii bng s phc.

Bi ton 3. Xt cc tam gic ABC khng u c cc ng cao AD, BE, CF. Ly A/, B/, C/ sao cho AA/= k. AD; BB/= k.BE; CC/=k.CF.(k 0). Tm tt c cc gi tr ca k sao cho vi mi tam gic ABC khng u th ABC ~ A/B/C/.

(Thi HSGQG Bng A- 1995)

Li gii. Dng h trc ta nh hnh v, ta c ' .AA k AD= ' ( )a k d a a= + ' )

2k bc

a a b c aa

= + +

, v

12

bcd b c aa

= + +

. Dng h trc ta

nh hnh v, ta c: ' .AA k AD= ' ( )a k d a a= +

' )2k bc

a a b c aa

= + +

v

12

bcd b c aa

= + +

.Tng t

' )2k cab b c a b

b

= + +

,

' )2k ab

c c a b cc

= + +

Do ' '

' '

b ac a

=

aca

bcac

c

baca

k

aba

bcab

bacbak

+

++

+

++

2

2

=ac

ab

.( ) ( )

( ) ( )ac

cakbkab

bakck

21

21

+

+

, R rng,

~ ' ' 'ABC A B C tng ngvi mt trong

hai iu kin sau: ' '

' '

b ac a

=b ac a

(1) hoc l

' '

' '

b ac a

= b ac a

(2). Trong (1)

2 2 2 2c a c b b a b c+ = + ( )( ) ( ) 0a c b c b cb c b + + =

0ab bc ca+ + = .Nhngvi ( )1 1cos sina R i = + , ( )2 2cos sinb R i = + , ( )3 3cos sinc R i = + th ta c:

0ab bc ca+ + = R = 0, v l.

Vy: ~ ' ' 'ABC A B C (2)

( ) ( )12

kc a bb k

ab+

=

= ( ) ( )12

kb a ck

ac

+

23

k = .

Vy 23

k = l gi tr cn tm./.



Phng trnh ham

vavavava S Tru Mat

Bi l v Chuyn ton K01-04

Sv. Lp 04TT, Khoa Ton Tin, HKHTN - HQG TP H Ch Minh

Phng trnh hm l mt chuyn quan trng trong ton hc ph thng, c bit l trong cc k thi hc sinh gii. Cng nh nhiu vn khc trong Ton hc, phng trnh hm cng mang v p ring ca n. Bi vit ny xin c quan st v p di gc l mi lin h gia phng trnh hm v s tr mt. I / Tp hp tr mt. 1. nh ngha.

Cho hai tp hp A v B, AB. Ni A tr

mt trong B nu AxBx > ,0, sao cho a,

n0 sao cho: n > n0 ta c 12n

b a >

2nb > 2na + 1, n > n0

kZ 2nkb a> >

x R, >0, kZ, n N sao cho

2nk

x x + > >

S1 tr mt trong R (pcm). 2.2. V d 2. Cho tp hp

{ }2 / ,S a b a b N= . Chng minh rng S2 tr mt trong R.

Chng minh. Do lim( 1 )n

n n

+ =

= 1lim 01n n n

=

+ + nn > 0, tn ti

n0 sao cho nn +1 < , n > n0. Do 0lim( )

nn n

=

nn x R , > 0, n1 sao cho: nn 0 < , n > n1 . Xt dHy (Un) xc nh bi: Un = 10 nnn + x N. Ta c Un+1 - Un= 00 1 nnnn +++ <

v U0 = Um-1 + 2 > Um > x - 0, bB sao cho b > 2k

b(x + ) > b(x - ) + k, x[0; + ) aA b(x + ) > a > b(x - ) (do A c v hn phn t )

x+ > ba > x-

S3 tr mt trong [0; + ) 3. nh l. Cho hai hm s

( ), ( ) :f x g x X X , trong f(x) lin tc,



g(x) lin tc hoc n iu. V A l tp hp tr mt trong X. Khi , nu f(x) = g(x) x A th f(x) = g(x) x X. Chng minh. Ta c nhn xt: Vi tp A tr mt trong tp X th x X, tn ti dHy (xn) tha mHn xn A, n N v lim n

nx

+ = x.

Tht vy, nu xA th tm thng. Nu xA, theo nh ngha ta c:

AxXx > ,0, sao cho



kin bi ton. Vy k = 1 l gi tr duy nht cn tm. Bi ton 3. K hiu

S = 2 / ,

2 1p p Z k N

k

+

. Tm tt c cc

hm s f: R R lin tc v tha mLn cc iu kin:

i) x S th f(x)S, f2(0) S ii) f(xf(x) + f(y)) = y + f2(x), ) x, y S.

Li gii. Cho x = 0 f(f(y)) = f2(0) + y vi mi y S f n nh trong S.

Cho x = f(x) S, ta c:

( ) [ ] yxffyfxffxff +=+ 2))(()())(().( ( ) [ ] yaxyfaxxff ++=++ 2)()())(( , trong a = f2(0). Ti y, cho x = - a S, ta c f(f(y)) = y a= 0 f(0) = 0.

Cho y = 0 f(xf(x)) = f2(x) (1) Ti y, cho x = f(x), ta c

( ) [ ]2))(())(().( xffxffxff = f(xf(x)) = x2 (2)

T (1) v (2) suy ra f2(x) = x2 , x S Gi s tn ti y0 S ( 0 0y ) sao cho f(y0)

= - y0 th ti phng trnh hm ban u, cho y = y0 ta c:

f(xf(x) - y0) = y0 + f2(x)= x2+ y0

[ ] ( )20220 ))(( yxyxxff += (xf(x) - y0)

2 = (x2+y0)2

f(x) = - x, x S Vy f(x) = x, x S hoc f(x) = - x,

x S (*) Nhn xt.

1/. Theo v d 3, xt tp A={ }+ Zpp /2 ; B={ }Nkk + /12 , khi tp S/=

2 / ,2 1

p p Z k Nk

+ +

tr mt trong

{0;+ ). T y, d dng suy ra tp S tr mt trong R (**)

T (*) v (**) suy ra ( ) ,f x x x R= hoc ( ) ,f x x x R= . Th li thy tha mHn v

l 2 hm cn tm. 2/. Qua 3 bi ton trn ta thy mt s kt

qu v s tr mt H gip mt phn khng nh trong vic nh hng gii PTH.

Tip theo, ti thy rng s tr mt khng mnh bng s ph kn v ti H c gng tr li cu hi: Liu s ph kn c mi lin h no vi phng trnh hm? Bi ton 4. Tm tt c cc hm s

* *:f R R+ + tha mLn : f(x+y) = f(x2+y2) x, y > 0.

Li gii. t x + y = a2 , x2 + y2 = b ( )2 ( )f a f b= ), 2a > b a > 0

C nh a > 0, ta c: ( )f b const= , [ ;2 )b a a . K hiu n = [ ]aa nn 2;2 1

Do 1lim 2 lim 2n nn n

a a+ +

= = + nn tp

hp cc on n ph kn (0; + ) (v n 1+ n = { }an2 ) x > 0, tn ti n Z sao cho x n f(x) = const. Th li thy ng.

Vy f(x) = c, cR l hm cn tm. T tng ny, ta c th thit lp c

rt nhiu bi ton dng nh sau: Cho hai hm n bin *, :g h X X c

lin h vi nhau bi bt ng thc kp trong X. Tm tt c cc hm :f X X tha mHn:

( ()) ( ())f g f h= . V d nh bi ton sau:

Bi ton 5. Cho dLy hm s { }1)(xf n tha mLn ( ) : (1; ) (1; )nf x + + sao cho

1,22

>

+=

+ yxyxfyxfnn

n

n

n

Xc nh

2)(xf n .

(Bi ton ny xin b ngcho cc bn)

Trn y l mt s kt qu ti khai thc c v chc chn rng cn rt nhiu nhng kt qu p na, rt mong nhn c s trao i ca cc bn. kt thc bi vit, xin php c by t mt iu rt tm c m ti hc c t ngi thy ca mnh: Hc ton cng nh lm bt k mt vic g, chng ta hLy quyt tm tht cao, ti tin rng bn v ti s t c nhng kt qu tng xng vi s c gng ./.



day so va s tru mat day so va s tru mat day so va s tru mat day so va s tru mat TREN RTREN RTREN RTREN R

++++

H S Tng Lm Chuyn ton k01-04, ptnk

HQG TP H Ch Minh Sv. Lp CNKHTN K3, Khoa Ton Tin HKHTN - HQG TP H Ch Minh

I. nh ngha.

y ti xin c nu ra hai nh ngha ca s tr mt.

Xt cc tp M, X, R sao M X R. 1. nh ngha 1.

Tp XCM gi l tr mt trn tp X nu v

ch nu vi mi p, q X, p > q, tn ti m M sao cho p > m > q. 2. nh ngha 2.

Tp XCM gi l tr mt trn tp X nu v

ch nu vi mi x X, tn ti dLy sao cho

nlim xn = x.

Lu rng hai nh ngha trn l tng ng. Vic chng minh chng tng ng coi nh bi tp.

Mt s nh ngha khc: 3. nh ngha 3.

Ni rng tp A R b chn trn nu tn ti x R sao cho x z vi mi z A. Phn t z nh th c gi l cn trn ca tp A.

Gi s tp A b chn trn, z c gi l cn trn ng ca A nu z l cn trn b nht ca A. Cn trn ng ca A c k hiu l supA. 4. nh ngha 4.

Ni rng tp A R b chn di nu tn ti x R sao cho z x vi mi z A. Phn t z nh th c gi l cn di ca tp A.

Gi s tp A b chn di, z c gi l cn di ng ca A nu z l cn di ln nht ca A. Cn di ng ca A c k hiu l infA. 5. Tin v cn trn.

Mi tp A R , A b chn trn u c cn trn ng. II. Mt s tnh cht. 1. Tnh cht 1.

Tp Q cc s hu t l tr mt trn R. 2. Tnh cht 2.

Nu tp A tr mt trn R+ th tp

1,B r A

r

=

cng tr mt trn R+.

Vic chng minh hai tnh cht trn xem nh bi tp. III. Mt s bi ton m rng. 1. Bi ton 1.

Cho dLy s {an} dng, tng, khng b chn. Chng minh rng tp hp

A =

+Nnma

m

n

,/ tr mt trn R+.

Chng minh. Ta s dng b sau: B : Vi mi s > 0 v dLy {an} c tnh cht lim

nan = th ta cng c

nlim an

= . Xt p, q R+ tha mHn p > q. Suy ra

nlim (p - q)an = . Do , tn ti

s n0 N+ sao cho (p - q) 0na > 2

p 0na > [p 0na -1] > q 0na

Chn m0 = [p 0na ] 1, ta c p > 0

0

na

m > q.

Vy A tr mt trn R+. Nhn xt. p dng Bi ton 1, ta c Q+ tr

mt trn R+,

n

m

2 tr mt trn R+.

p dng tnh cht 2, ta c tp A

=

+Nnmm

an,, cng tr mt trn R+.

2. Bi ton 2. Cho { }na v { }nb l hai dLy s dng,

tng ngt v khng b chn. c bit, tp

hp { }nn aa +1 b chn. Chng minh rng tp hp

B =

+Nnmba

n

m,/ tr mt trn R+.



Chng minh. Xt p, q R+, p > q. t

{ }1n nM Sup a a+= , { }1,K Max a M= . V lim n

nb

= nn tn ti m0 sao cho

0( ) mp q b K > .

Ta c tn ti 0n sao cho

0 0 0n n npb a qb> > . Tht vy, do

{ }1,K Max a M= m lim nn

a

= nn suy ra

tn ti s t nhin N tha mHn

0 1N n Na b a

> > , hn na 0

( ) mp q b K > nn 0m N

pb a> . Chn 0n N= l xong.

Vy 0

0

n

m

ap q

b> > B tr mt trn R+.

Nhn xt. p dng Bi ton 2 ta c h qu sau:

Cho f: R+ R+ tha mHn: i) f kh vi trn R+. ii) lim ( )

xf x

=

iii) )1;0(\,)(' +< RxMxf Cho { }nb l dHy s dng, tng, khng b chn. Ta c:

B =

+Nnmbmfn

,/)( tr mt trn R+.

Do tp

+Nnmnm

,/2

tr mt trn R+.

3. Bi ton 3. Cho dLy s dng { }na tng

tha lim nn

a kn

= , k R+.

Cho dLy s dng { }nb tng, khng b chn. Chng minh rng tp hp C =

+Nnmba

n

m,/ tr mt trn R+.

Chng minh. Xt p, q R+, p > q.

Do lim nn

a kn

= nn chn > 0 sao cho

.

p q kp q

>+

. Ta c tn ti N sao cho:

.

na p qk kn p q

< +

, n N.

qpqp

+

.k + >n

an >qpqp

+

.k - , n N.

t

+

+

=

kqp

qq

2 <

=

+k

qpq

p2 .

p dng Bi ton 1, tn ti v s b (m,

n) sao cho v 0m sao cho 0

0

n

an>

qpqp

+

. k - . Nhn v

theo v, ta c: pba

qm

no



MOT SO BAI TOAN SO HOC veMOT SO BAI TOAN SO HOC veMOT SO BAI TOAN SO HOC veMOT SO BAI TOAN SO HOC ve

DAY TONG CAC LUY THADAY TONG CAC LUY THADAY TONG CAC LUY THADAY TONG CAC LUY THA Trn quc hon

Chuyn Ton K02 05, THPT Chuyn Nguyn Tr`i Hi Dng Sv. Lp K50CA - H Cng Ngh - HQG H Ni

TTTTrong bi bo ny chng ta cp n

mt s bi ton s hc lin quan n dHy tng cc ly tha. Cho 1 2, ,.., ma a a l cc s nguyn dng c nh cho trc. Xt dHy s sau 1 2 ...

n n n

n mu a a a= + + + trong n = 0, 1, 2, . Bi ton 1. Bit rng tp hp cc c nguyn t ca dLy nu l hu hn. Chng

minh rng 1 2 ... ma a a= = = .

Li gii. Trc ht t ( )1 2, ,..., md a a a= ta c th vit i ia db= , trong ( )1 2, ,..., 1mb b b = . Khi dHy

1 2 ...n n n

n mv b b b= + + + cng c hu hn cc c nguyn t l 1 2, ,..., kp p p

Chn x nguyn dng sao cho xip m> vi mi i = 1, 2, , k. Khi vi mi s t nguyn dng ln hn x, t

1

( ) ( )k

x

ii

n t t p=

= th vi mi i = 1, 2, , k ta c ( ) 0 1(mod )n t xs ib p vi mi s = 1, 2, , m. ( ( ) 0(mod )n t xs ib p khi v ch khi

sb chia ht cho ip ). Ch l cc ib nguyn t cng nhau v

x

ip m> nn d thy ( )n tu s khng chia ht cho xip vi mi i = 0, 1, , k. Do d c

1( )

1,

kx

n t ii

u p t x=

> , iu ny ch xy ra khi 1 2 ... 1mb b b= = = = hay

1 2 ... ma a a= = = . Bi ton c chng minh.

Xut s. Bi ton ny c t ra khi tc gi i tm li gii cho mt bi ton rt th v nhng hon ton khc trn mathlink contest (s c dp gii thiu vi bn c trong nhng dp khc). Sau mi n c chn lm thi gii ton online thng 10 nm 2006 trn trang web www.diendantoanhoc.net, c rt

nhiu bi ton tng qut cng nh ko theo t bi ton ny m cc thnh vin ca trang web ny H ra. Sau y ti xin gii thiu 4 trong s cc bi ton . Bn c c th tm li gii tng t bi ton trn (tuy c x l mt s k thut kh hn). Bi ton 1.1. Cho n > 1 l s nguyn dng. Chng minh rng vi bt k n s nguyn dng 1 2, ,..., na a a lun tn ti mt s

nguyn dng k sao cho s 1 2 ...k k k

na a a+ + + c mt c nguyn t khng l c nguyn t ca 1 2... nna a a .

Bi ton 1.2. Cho s nguyn dng { }na vi m l s nguyn dng cho trc v m s nguyn dng 1 2, ,..., mk k k . DLy s mi c xc nh nh sau:

1 1 2 2 ...n n n

n m mu k a k a k a= + + + . Chng minh rng dLy s { }nu c hu hn c nguyn t khi v ch khi 1 2 ... ma a a= = = . Bi ton 1.3. Cho m nguyn dng ln hn 1 v 1 2( ), ( ),..., ( )mP x P x P x l cc a thc h s nguyn khng m khng ng nht vi 0. Cc s nguyn dng 1 2, ,..., ma a a i mt phn bit. Xc nh hm s :f N N sao cho

1( ) ( )

mn

i ii

f n P n a=

= vi mi n N. Chng minh rng tp hp cc c nguyn t ca hm f l v hn. Bi ton 1.4. Hm :f N Z c gi l hm gn a thc nu f(m) f(0) chia ht cho m vi mi m. Khi vi mi 1 2, ,..., nf f f l cc hm gn a thc v cc s nguyn dng phn bit 1 2, ,..., na a a th tp cc s

nguyn dng c dng 1

( )n

ki i

if k a

=

c v s c nguyn t.



Bi ton 2. Bit rng vi mi n ln th nu

l s chnh phng. Chng minh rng m l s chnh phng. Li gii. Trc ht ta chng minh b sau: B : Cho a l s nguyn dng sao cho vi mi p nguyn t ln ta c a l s chnh phng theo modp. Khi ta c kt lun a l s chnh phng. Chng minh. Phn chng, gi s a khng l s chnh phng. Khi khng gim tng qut ta gi s a khng c c nguyn t chnh phng no khc 1 suy ra

1 2... ka p p p= vi 1 2 ... kp p p< < < l cc s nguyn t phn bit.

Theo bi ra tn ti 0N sao cho vi

mi s nguyn t p > 0N th 1a

p

=

hay

11

ki

i

pp

=

=

. Xt cc trng hp sau:

Trng hp 1: 1 2p > , t

( ) ( 1)( 1)41 1

1ik k p pi

i ii i

p p ps

p p p

= =

= = =

Nhn xt rng: Tn ti b s nguyn

dng sao cho 1( ; ) 1b p = v 1

1bp

=

(iu

ny l hin nhin v trong mt h thng d

thu gn 1mod p th c ng 1

2p

s chnh

phng 1mod p ). Khi ta ch vic chn s t nguyn dng sao cho

1

1(mod8)(mod )

1(mod )i

t

t b pt p

; 2,...i k=

Khi tn ti s nguyn t p > 0N sao cho 1(mod8 ... )kp t p p (Theo nguyn tc

Dirichlet). T y suy ra 1

1pp

=

v

1, 2,...,i

pi k

p

= =

nn ta s c s = -1

nhng vi ch rng p -1 chia ht cho 8 nn s = 1 theo cng thc trn. iu ny mu thun vi kt qu va c.

Trng hp 2: 1 2p = suy ra

( )2 1

1 82

1pp

p p

= =

t

1

2 2

k ki

i ii i

p pp ps

p pp p= =

= = =

( )2

2

( 1)( 1)18 41 (1)

ki

i

p pp

=

+=

R rng nu k = 1 hay a = 1 2p = ta c ngay mu thun (Chng hn, chn p nguyn t ln c dng 8n + 3). Ta ch cn xt khi k > 1. By gi lp lun tng t trng hp 1 s tn ti s nguyn t p sao cho p > 0N

v 1(mod8)p , 2

1pp

=

v

1, 3,...,i

pi k

p

= =

. T ta cng d thy

mu thun nh trn. B c chng minh. Tr li Bi ton 2, gi s tn ti N sao

cho mi n > N th na l s chnh phng. Xt tt c cc s nguyn t p > N + 1 v

{ }ip Max a> th 1 (mod )pu m p vy m l s chnh phng modp vi mi p nguyn t ln nn suy ra l s chnh phng (theo b trn). Bi ton c chng minh.

Xung quanh dHy s ny ti xin c xut 3 bi ton sau, rt mong s quan tm n li gii ca chng. Bi ton 3. Tn ti hay khng dLy s nguyn

1 2, ,...a a sao cho dLy s c v hn s hng khc khng v ng thi dLy s

1 2 ...n n n

n nu a a a= + + + c hu hn c nguyn t. (Ch rng nu dLy c hu hn s khc khng th n nh h qu ca bi ton 1 ta c tt c cc s hng khc khng ca dLy

na u bng nhau). Bi ton 4. Bit rng dLy nu c hu hn c s chnh phng. Chng minh rng

1 2 ... ma a a= = = Bi ton 5. Cho A l mt cp s cng dng ax + b vi a, b nguyn, a > 0. (a, b) = 1, x Bit rng dLy nu cha hu hn c nguyn t trong A. Chng minh rng

1 2 ... ma a a= = = ./.



CAN BANG HE SO TRONG BAT ANG THC CO-SI

Nguyn Lm Tuyn Chuyn Ton K99 02

Sv. Lp iu khin T ng 1 - K47, H Bch Khoa H Ni

S dng bt ng thc (BT) H bit m c bit l BT C-si l phng php thng c p dng gii cc bi ton v BT ni chung. Nhng bi ton cc tr, nht l trng hp c thm cc iu kin ph thng gy kh khn cho ngi gii trong vic c lng h s v xt iu kin du ng thc xy ra. Bi vit ny trnh by mt phng php nh gi thng qua BT C-si t , chuyn bi ton cc tr v vic gii mt phng trnh (PT) hoc h phng trnh (HPT) m vic gii quyt l d dng hoc c ng li r rng hn, l phng php cn bng h s. Cng t phng php ny, vi mt cht sng to, chng ta c th tng qut v to ra c nhng bi ton mi.

Trc ht xin nu li m khng chng minh hai BT quen thuc sau:

i) BT C-si tng qut:

1 2 1 2... ...n

n na a a n a a a+ + + ii) BT C-si suy rng:

1 1 2 2 ... n na a a + + +

( )( )1 2 1 21 ...1 2 1 2... ... n na a an na a a a + + ++ + + Trong hai BT trn th 1 2, ,..., na a a khng

m, 1 2, ,..., n dng v du ng thc xy

ra khi v ch khi 1 2 ... na a a= = = . Chng ta bt u t bi ton sau:

V d 1. Cho cc s thc dng ,x y tha mLn iu kin 3 3 1x y+ = (1). Tm gi tr ln nht (Max) ca biu thc

( ; )P x y x y= +

Phng php suy lun: S chnh lch v s m ca cc biu

thc 3 3x y+ v ( ; )P x y x y= + gi cho ta s dng BT C-si h bc ca 3 3x y+ . Nhng ta cn p dng cho bao nhiu s v l nhng s no? Cn c vo bc ca cc bin s x v y trong cc biu thc trn, ta thy cn phi p dng BT C-si ln lt cho

3x v 3y cng vi 5 hng s dng tng ng khc lm xut hin x v y . Mt khc do x, y dng v vai tr ca chng nh nhau nn ta d on ( ; )P x y t Max khi x y= . T (1) suy ra

3

12

x y= = v ta i n li gii

nh sau. Li gii. p dng BT C-si cho 6 s

dng: 1 s 3x v 5 s 12, ta c:

5 53 3 66

1 15. 6 . 6.22 2

x x x

+ =

Du = xy ra 3

12

x =

Tng t nh vy: 5 5

3 3 661 15. 6 . 6.22 2

y y y

+ =

Du = xy ra 3

12

y =

Cng theo v cc BT trn ta c:

( )53 3 6( ) 5 6.2x y x y+ + + (2) Du = xy ra

3

12

x y= = .



T (1) v (2) suy ra: 6 5( ; ) 2P x y x y= +

Du bng xy ra 3

12

x y= = , tha

mHn iu kin (1).

Vy { } 6 5( ; ) 2Max P x y = . V d 2. Cho cc s thc dng ,x y tha mLn iu kin 3 3 1x y+ (3). Tm gi tr ln nht (Max) ca biu thc

( ; ) 2P x y x y= + Phng php suy lun: v d 1, chng ta H nhanh chng d

on c Max ( ; )P x y t c khi x y= , t tnh c ,x y . Nhng trong bi ton ny, vai tr ca x v y l khng bnh ng. Tuy nhin ta hHy gi s ( ; )P x y t Max khi

x

y

=

=

no v d on , iu kin bin ca (3), tc l 3 3 1 + = (4). Ta vit:

( ) 553 3 3 3 265. 6 . 6.x x x + = ( ) 553 3 3 3 265. 6 . 6.y y y + =

Suy ra

( ) ( ) 5 53 3 3 3 2 25. 6. 6.x y x y + + + +

xut hin ( ; )P x y v phi, ta cn chn , sao c t l:

526. x :

526. y =1. x : 2. y

52

5

1 12 4

= =

(5)

Vy t (4) v(5) ta thu c HPT:

5

3 3

14

1

=

+ =

3 5

5

3 5

1

1 2 24

1 2 2

=

+

= +

Bng cch lm ngc li cc bc trn ta

s thu c { } ( )556( ; ) 1 2 2Max P x y = + Nhn xt. T cch phn tch trn ta thy c th thay i d kin ca bi ton sao cho HPT sau khi cn bng h s c th gii c. Chng hn nh cc bi ton di y: Bi ton 1. Cho cc s nguyn dng

, ,m p q sao cho { },m Max p q . HLy tm GTLN ca biu thc ( ; ) p qP x y ax y= + trong hai trng hp sau, bit rng a l hng s dng v x, y l cc bin s khng

m tha mLn iu kin 1m mx y+ :

i) 2

m qp +=

ii) 2

3m qp +=

Bi ton 2. Cho cc s thc dng a, b, c, d v cc s nguyn m, n tha mLn iu kin

0m n> > . Tm gi tr ln nht ca biu thc ( ; ; ) n n nP x y z ax by cz= + + trong

, ,x y z l cc bin s khng m tha mLn iu kin m m mx y z d+ + .

V d 3. Tm gi tr nh nht ca biu thc

( )2 2 2( ; ; )P x y z a x y z= + + . Trong a l s thc dng v x, y, z l cc bin s tha mLn iu kin 1xy yz zx+ + = (6) Phng php suy lun:

Do vai tr ca x v y l nh nhau nn ta d on ( ; ; )P x y z t Min khi

( 0)x y z = = > (7). p dng BT C-si cho hai s dng ta c

2 2 2 2x y xy xy+

( )22 2 2x z x z xz + 2 2

1 2x z xz

+

( )22 2 2y z y z yz + 2 2

1 2y z yz

+

T cc BT trn suy ra:



( ) ( )2 2 211 2 2x y z xy yz zx

+ + + + +

V phi ca BT trn l hng s, v vy ta cn tm c t l:

11 : 2 :1a

+ =

22 1 0a = 1 1 84

a

a

+ += ,

1 1 8 04

a

a

+= < loi.

Cng vi (6) v (7) ta c HPT:

1xy yz zxx y z+ + =

= =

( )2 22 1zx y z

+ =

= =

Gii HPT ny vi nh trn ta c:

( )( )

2

216

8 1 8 1 1

48 1 8 1 1

az

a a

ax y

a a

=

+ + +

= =

+ + +

Bng cch lm ngc li ta tnh c

{ } 4( ; ; )1 1 8

xy yz zxMin P x y za

+ += =

+ +Nhn xt. Bng cch lm tng t nh trn chng ta c th gii trn vn c bi ton tng qut hn sau: Bi ton 3. Cho cc hng thc dng a, b, c v cc bin s x, y, z tha mLn iu kin

1xy yz zx+ + . Tm gi tr nh nht ca biu thc 2 2 2( ; ; )P x y z ax by cz= + + .

V d 4. Xt cc s thc dng a, b, c tha mLn iu kin 21 2 8 12ab bc ca+ + . HLy tm gi tr nh nht ca biu

thc1 2 3( ; ; )P a b ca b c

= + + .

( thi chn TVN d thi IMO 2001) Phng php suy lun:

t 1 1 1

, ,a b cx y z

= = = . iu kin ca

bi ton t thnh 2 8 21 12x y z xyz+ + (9).

V ta cn tm Min ca biu thc ( ; ; ) 2 3P x y z x y z= + +

Gi s ( ; ; )P x y z t Min khi x zy z

=

=

p dng BT C-si suy rng ta c: 12 2 8 21xyz x y z + +

2 8 21x y z

+ +

( )1

82 2 8 21212 8 21 x y z

+ + + +

( )8 21 2 21 2 8 ,x y z A + + + (10) Trong biu thc ( ),A ch ph

thuc vo , . Cng theo BT C-si suy rng ta c:

( ), ,P x y z = x + 2y + 3z = 2 3x y z

+ +

( )1

2 2 332 3 x y z

+ + + +

= ( )( ) 12 3 2 3,B x y z + + (11) Trong biu thc ( ),B ch ph

thuc vo , . i chiu (10) v (11) ta thy cn chn

, sao cho c t l: ( ) ( ) ( ): 2 : 3 8 21 : 2 21 : 8 2 = + + +

8 218 2 32 21 28 2 3

+= +

+

=

+

2

2

2 8 24 6316 4 6 63

+ = +

+ = +

T PT th nht ( )22 63

8 3

=

. Thay

vo PT th hai ta c:



( ) ( )22 22 63 2 6316 4 6 63

8 3 8 3

+ = +

3 24 78 306 567 0 + = ( ) ( )22 9 2 48 63 0 + + =

92

= ( do 0 > ) 158

= . Khi ( ), ,P x y z t Min th tt c cc BT

trn u tr thnh ng thc, ngha l

2 8 21 12 39 52 4

18 25 3

x y z x

x z z y

y z z z

+ + = =

= = =

= = =

Ti y, im mu cht ca bi ton H c gii quyt v ta i n mt li gii tng i ngn gn cho bi ton nh sau:

Li gii. t 1 1 15 23 , ,4 3

x x y y z z= = = khi

iu kin (9) tr thnh

1 1 1 1 1 15 2 5 22.3 8. 21. 12.3 . .4 3 4 3

x y z x y z+ +

1 1 1 1 1 13 5 7 15x y z x y z + + . ( ) ( )1 1 1, , , ,P x y z P x y z= =

1 1 15 23 2. 34 3

x y z= + +

= ( )1 1 11 6 5 42 x y z+ + p dng BT C-si tng qut cho 15 s

dng ta c: 3 5 715

1 1 1 1 1 1 1 1 115 3 5 7 15x y z x y z x y z + + (12)

( ) ( )1 1 11, , 6 5 42P x y z x y z= + + 6 5 415

1 1 11

.15.2

x y z (13)

T (12) suy ra 6 5 41 1 1 1x y z , do t (13)

ta c ( ) 15, ,2

P x y z

ng thc xy ra 1 1 1 1x y z = = =

1 1 15 5 2 23 3, ,4 4 3 3

x x y y z z = = = = = =

1 4 3, ,

3 5 2a b c = = =

Vy Min ( ) 15, ,2

P a b c = .

Nhn xt. S d ta t cc bin mi 1 1 1, ,x y z l v ta H xc nh c b s (x,y,z)

( ), ,P x y z t Min. Mt khc vic xt du bng s tr nn d dng hn bu cc bin tham gia khi xy ra du ng thc l bng nhau v u bng 1.

Mt iu th v v ng ch y l cc BT (12), (13) tng i n gin, nhng qua php i bin H tr thnh BT khc phc tp hn rt nhiu. Chng ta hHy th vn dng iu ny to ra nhng bi ton mi rt th v, xut pht t b sau: B : Cho cc s thc , , , 0 v

, , , 0x y z t > . Khi ta c: i) Nu

( )x y z t xyzt + + + + + + th

( ) ( ) ( )x y z + + + + + + + + +( ) ( )3t + + + + + + (14)

ii) Nu

( ) ( ) ( )x y z + + + + + + + + + ( ) ( )3t + + + + + + th

( )x y z t xyzt + + + + + + (15) Chng minh. Trng hp 0 = = = = th b hin nhin ng. Ta xt khi

2 2 2 2 0 + + + > . i) p dng BT C-si suy rng ta c: ( ) xyzt x y z t + + + + + +

( )( ) 1x y z t + + + + + + 1x y z t + + + + + + + +

Nh vy: ( ) ( )x y + + + + + + + ( ) ( )z t + + + + +

( )3 + + +



( ) 1x y z t + + + + + + + + + + + ( )3 + + + ng thc xy ra 1x y z t = = = = .

ii) p dng BT C-si suy rng ta c: ( ) ( )3 x + + + + + +

( ) ( ) ( )y z t + + + + + + + + + ( )3 + + +

( ) 1x y z t + + + + + + + + + + + 1 x y z t + + + + + + + +

( ) 1x y z t xyzt + + + Nh vy:

x y z t + + + ( ) ( ) 1x y z t + + + + + + ( ) xyzt + + + ng thc xy ra 1x y z t = = = = . B c chng minh. S dng b trn bng cch thay vo

nhng gi tr c bit v bng nhng cch pht biu khc nhau, ta s c nhng kt qu khc nhau: - Vi 1, 0, 3, 5, 7t = = = = = , thay x, y, z, t ln lt bi 3x,

54

y , 23

z vo (14), sau

t 1 1 1

, ,a b cx y z

= = ta c Bi ton v

d 4. - Thay 1, 1, 1, 2, 3t = = = = = vo (14) v t

1 2 4, ,

2 3 3x y z

a b c= = = ta c bi ton:

Bi ton 4. Cho cc s thc dng a, b, c tha mLn iu kin 72ab + 9bc + 24ca + + 18abc 56. Chng minh rng: 3 10 16 15a b c

+ + . ng thc xy ra khi no?

- Thay 1 1 11, 1, , ,2 3 6

t = = = = = vo

(14) v t 1 2 4

, ,

2 3 3x y z

a b c= = = ta c bi

ton sau: Bi ton 5. Cho cc s thc dng a, b, c tha mLn iu kin ( )28 27 16a b c abc+ + . Chng minh rng:

5 10 22 64 9 9a b c

+ + . ng

thc xy ra khi no? - V khi xy ra ng thc hai Bi ton 4 v

5 u c 1 2 4

, ,

2 3 3a b c= = = nn khi kt hp

hai bi ton trn ta c: Bi ton 6. Cho cc s thc dng a, b, c tha mLn iu kin 72ab + 9bc + 24ca +

18abc 56 v ( )28 27 16a b c abc+ + . Chng minh rng:

17 19 166 214 9 9a b c

+ + .

ng thc xy ra khi no?

- Thay 1

, 1, 1, 2, 3tx

= = = = = vo

(14) v t 1 2 4

, ,

2 3 3x y z

a b c= = = ta c bi

ton sau: Bi ton 7. Cho cc s thc a, b, c dng

tha mLn iu kin 3 10 16 12 21

3 3a

a b c+ + + ,

chng minh rng 1 4 4 282

2 3 9a

a b c abc+ + + .

ng thc xy ra khi no? Bng cch thay i d kin bi ton theo

hng trn chng ta s c c rt nhiu bi ton mi. Cc bn hHy th tip tc suy ngh theo hng trn v theo hng tng qut cho trng hp nhiu bin hn na. kt thc bi vit ny, ngh cc bn gii mt s bi tp sau v hHy c gng m rng chng theo cch ca mnh. l mt vic lm thc s cn thit khi hc ton . Chc cc bn thnh cng!

(Xem tip trang 80)



PHNG PHAP S DUNG NH NGHAPHNG PHAP S DUNG NH NGHAPHNG PHAP S DUNG NH NGHAPHNG PHAP S DUNG NH NGHA

E TNH GII HANE TNH GII HANE TNH GII HANE TNH GII HAN

L bo khnh Lp 12A Ton, K01- 04 - THPT Chuyn Nguyn Hu, H Ty

Sv. Khoa Kinh t i ngoi, H Ngoi Thng H Ni Li Ban bin tp. Cc bn thn mn! Nh cc bn L bit, c rt nhiu phng php tnh gii hn ca mt dLy s. Mi phng php u c nhng im mnh c trng cho ring mnh. Tuy khng phi l mt phng php mi l, nhng s dng nh ngha tnh gii hn vn l mt phng php kinh in, n mang mt sc thi v v p ring. V cng ng nh tc gi bi bo ny nhn xt, y l mt phng php rt su sc v mt ton hc. vn dng thnh tho phng php ny th chng ta cn phi c mt ci nhn su sc v bn cht cng nh ngha ca l thuyt gii hn. Xin gii thiu cng bn c: I. nh ngha. Trc ht, chng ta hHy cng nhc li v nh ngha gii hn ca mt dHy s:

Cho dLy s thc ( )nx , a R . Ta ni lim nn

x a

= nu > 0, N sao cho n >

N th axn < .

II. Vi tnh cht tht c bn. - Gii hn ca mt dLy nu tn ti th duy nht. - lim n

nx a

= , a(p; q) th tn ti N sao cho

n > N, xn (p; q).

- Nu lim n

nx a

= , nx b> vi mi n > N th a

b. - Nu lim n

nx a

= , lim n

ny b

= v n nx y vi

mi n > N, th ab - Nu lim n

nx a

= th lim n

nx a

= . iu

ngc li khng ng.

- lim 0nn

x

= lim 0nn

x

= .

Trong cc bi ton v tm gii hn, ta c th dng mt s du hiu/phng php nh nguyn l n iu, nguyn l kp, ... Tuy nhin, trong mt s bi ton kh cn phi vn dng trc tip nh ngha chng minh nhng iu ny i hi chng ta phi hiu bit mt cc tng i su v gii hn. Ti xin bt u cc v d t c bn n phc tp. III. Mt s v d minh ha. 1. Bi ton1. Cho dHy s thc ( )nx khng m tha mHn lim 0nn

x

n= . Chng minh rng

{ }1,lim 0

ii n

n

Max x

n

=

= .

Li gii . T gi thit, suy ra vi mi > 0,

tn ti s t nhin m sao cho nx

n < ,

nm



n0 m tha { }

1,

0

ii nMax x

n

= <

Nu (m - 1) k 1 th ta c n

xk < kxk



Vi mi > 0, tn ti s t nhin m sao

cho LK

yn

mii

+ m.

Vi mi > 0, tn ti s t nhin n0 sao

cho nx K L

n0.

Chn N = Max{ }0;nm th vi mi n > N, ta c:

=

+

n

iini yx

11 =

=

+

N

iini yx

11 +

+=+

n

Niini yx

11

KLK +

+ L

LK +

= pcm

5. Bi ton 5. Cho dHy s thc dng ( )na tha mHn : 1) + > 0. T gi thit phn chng, ta thy c v s n nna > . Chn m > 1 sao

cho mma > 1 1(1 )m m ma a a + > > >

m

>

1+m

(1 + 1+m

) am-1 > 1+m

...

Tip tc qu trnh ny, ta c

2m

m

a >

+

2m

m

Ly tng li, ta c:

2

...m mm

a a

+ + >m

+....+

+

2m

m

>

4

Qu trnh ny c th tip din, suy ra ta c th chn c v s tng ri c ln

vt 4

+=

1na , Mu thunpcm.

6. Bi ton 6. Cho dLy s thc khng m ( )na tha mLn

+ 0, k hiu N(x) l

s s na x> . Chng minh rng 0lim ( )x

xN x

.

Li gii. Do + 0, n0

sao cho 210



iem LEMOINE Trong Tam Giac

L vn nh

Lp 3clc k51 ton, hsp h ni i

Chng ta hn ai cng bit n bi ton ni ting sau:

Bi ton 1:Tm im M trong mt phng

ABC sao cho tng MA+MB+MC t gi tr nh nht.

y l mt bi ton kh c t ra kh lu trc khi Toricelli - ngi u tin tm ra li gii. Th nhng khi nng cc i lng trong Bi ton 1 ln bc hai th vn li ht sc n gin . l ni dung bi ton m chng ta sau ny u bit rng, trng tm G

ca ABC l li gii duy nht ca n. Bi ton 1a:

Tm im M nm trn mt phng tam gic ABC sao cho i lng

2 2 2MA MB MC+ + t gi tr nh nht. Mt bi ton khc cng c t ra mt

cch rt t nhin t Bi ton 1. Cho im M nm trong tam gic ABC,

gi H, J, K ln lt l hnh chiu ca M trn cc cnh BC, CA, AB tng ng. Xc nh v tr ca im M sao cho tng S = MH + MJ + MK t gi tr nh nht.

Vic pht trin Bi ton 1 theo hng nng cc i lng ca tng S ln bc hai s dn ta ti khi nim sau, trong a, b, c l k hiu di ba cnh BC, CA v AB ca tam gic ABC. im Lemoine trong tam gic:

im L thuc mt phng cha tam gic ABC c gi l im Lemoine ca tam gic

nu 2 2 2. . . 0a LA b LB c LC+ + =

.

D thy rng dim Lemoine ca mt tam gic th nm trong tam gic . nh l sau cho ta mt tiu chun nhn bit im Lemoine. nh l 1:

Cho im L nm trong ABC . Gi H, J, K ln lt l hnh chiu ca L trn cc cnh BC, CA, AB tng ng. Khi L l im Lemoine ca ABC nu v ch nu L l trng tm ca tam gic HJK.

Chng minh ng thc ny da trn 2

ng thc vct quen thuc sau m vic chi tit ho khng c g l kh khn, cc bn hHy thit lp coi nh bi tp.

( ) ( ) ( ) 0S LBC LA S LCA LB S LAB LC + + =

,

v 0a b aLH LJ LKLH LJ LK

+ + =

, t

aaaaaaaaaaa

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