Phuong Trinh Bac Cao

Nguyn Tt Thu

Chng II: Phng trnh bc cao

Bi 1. Phng trnh bc ba

I. Tm tt l thuyt1. Phng trnh c dng: 3 2ax bx cx d 0 (1), trong a, b, c, d lcc s thc cho trc a 0 .2. Cch gii: By gi ta i xt cch gii phng trnh (1).V a 0 nn ta c th chia hai v ca phng trnh (1) cho a. Do vy ta chcn i gii phng trnh dng : 3 2x ax bx c 0 (2) .t ax y

3 , khi (2) tr thnh : 3y py q 0 (3)

Trong :2 3a 2a 9ab 27cp b; q3 27

.t 2 327q 4p . xt s nghim ca (3), ta kho st s tng giaoca hm s 3f (y) y p.y q vi trc Ox.Ch hm bc ba ct Ox ti Mt im hm lun n iu hoc CD CTf .f 0 Hai im CD CTf .f 0 Ba im CD CTf .f 0 Xt hm s 3f (y) y py q , ta c: 2f '(y) 3y p .* Nu p 0 f '(y) 0 f (y) l hm ng bin f (y) 0 c mtnghim .

* Nu pp 0 f '(y) 0 y3

v CD CT p pf .f f .f3 3 27

.

T y ta c cc kt qu sau:* Nu 0 (3) c nghim duy nht. tm nghim ny ta lm nh sau :t y u v , khi (3) tr thnh: 3 3u v (3uv p)(u v) q 0 Ta chn u,v sao cho: p3uv p 0 uv

3 , lc 3 3u v q ta c h:

Nguyn Tt Thu

33 3

3 3

3 3

pu v

u , v27u v q

l nghim phng trnh:

32 pX qX 0

27 (4)

(4) c hai nghim: 1q

27X2

; 1

q27X

2

3 33 31 2

q q27 27y u v X X

2 2

(*)Cng thc (*) gi l cng thc Cardano.

* Nu 0 , khi (3) c hai nghim, mt nghim kp ( 3 qy2

hoc

3 qy2

) v mt nghim n. Tc l: 33

qy2

y 4q

hoc

3

3

qy2

y 4q

(**).

* Nu 0 , khi (3) c ba nghim phn bit v ba nghim ny nmtrong khong p p( 2 ;2 )

3 3 . tm ba nghim ny ta t py 2 cos t

3 ,

vi t ( 0; ) ta a (3) v dng: cos3t m (5), trong 3 3qm2p p

.Gii (5) ta c ba nghim 1 2 3t , t , t 0; , t y suy ra ba nghim caphng trnh (3) l :

1 1 2 2 3 3p p py 2 cos t ; y 2 cos t ; y 2 cos t3 3 3

(***).Ch : Trong mt s trng hp gii phng trnh bc ba ta i tm mtnghim ri thc hin php chia a thc v chuyn phng trnh cho vphng trnh tch ca mt nh thc bc nht v m t tam thc bc hai.

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II. Cc v d

V d 1: Gii phng trnh : 34x 3x 9 0 .Gii: Ta thy phng trnh c mt nghim 3x

2 (dng MTBT) nn ta

bin i phng trnh : 2 3(2x 3)(2x 3x 3) 0 x2

.

V d 2: Gii phng trnh : 3x 3x 1 0 .Gii: Ta c: 2 3 327q 4p 27 4( 3) 135 0 nn phngtrnh c duy nht nghim:

3 3 3 3q q 1 5 1 527 27

x2 2 2 2

.

V d 3: Gii phng trnh : 3x 3x 1 0 (1).Gii:Ta c: 2 3 327q 4p 27.1 4.3 81 0 nn phng trnh c banghim thuc khong ( 2;2) . t x 2cos t vi t (0; ) (2) tr thnh: 3 3 18cos t 6cos t 1 4cos t 3cos t

2

1 2cos3t cos t k

2 3 9 3 .

V t (0; ) nn ta c: 1 2 37 5t ; t ; t9 9 9 .

Vy phng trnh c ba nghim: 5 7x 2cos ; x 2cos ; x 2cos9 9 9 .

V d 4: Tm m phng trnh sau c ba nghim phn bit3 2x (2m 1)x (3m 2)x m 2 0 (1).

Gii: V tng cc h s ca phng trnh bng 0 nn phng trnh c nghimx 1 nn :

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22

x 1(1) (x 1)(x 2mx m 2) 0

x 2mx m 2 0

Phng trnh (1) c ba nghim phn bit 2f (x) x 2mx m 2 chai nghim phn bit khc 1

2' m m 2 0

m 1f (1) m 1 0

.

Vy m 1 l gi tr cn tm.

Ch : S nghim ca PT : 2(x )(ax bx c) 0 ph thuc vo snghim ca tam thc: 2f (x) ax bx c (a 0) . C th* Nu f (x) c hai nghim phn bit , tc l: 0

f ( ) 0 th phng

trnh c ba nghim phn bit.* Nu f (x) c hai nghim phn bit, trong mt nghim bng , tc l:

0f ( ) 0 th phng trnh c hai nghim: x ,

cx

a .

* Nu f (x) c nghim kp khc , tc l:0

b2a

th phng trnh c

hai nghim x v bx2a

.

* Nu f (x) c nghim kp x , tc l:0

b2a

th phng trnh c

mt nghim x .* Nu f (x) v nghim th phng trnh c ng mt nghim x .

V d 5: Tm m th hm s sau ct trc Ox ti hai im phn bit:3 2y 2mx (1 2m)x (3 8m)x 4m 2

Gii:Ta c phng trnh honh giao im:

3 22mx (1 2m)x (3 8m)x 4m 2 0 (2)

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2(x 1) 2mx (1 4m)x 4m 2 0

2x 1

f (x) 2mx (1 4m)x 4m 2 0

Yu cu bi ton (2) c hai nghim phn bit.TH 1: f (x) c hai nghim phn bit, trong c mt nghim

bng 1. iu ny c 2m 0

148m 24m 1 0 m2

f (1) 2m 1 0

.

TH 2: f (x) c mt nghim khc 1. Khi xy ra hai kh nng

Kh nng 1:m 0

3 60 m12

1 4m 14m

.

Kh nng 2:m 0

m 0f (1) 0

.

Vy cc gi tr ca m cn tm l: 1 3 6m ; m 0; m2 12

.

V d 6: Chng minh rng phng trnh : 3 2x ax bx c 0 a 0c 3 nghim 33 227c 2a 9ab 2 a 3b (1).Gii: Gi s phng trnh c ba nghim. Ta chng minh (1).* Nu ba nghim ca phng trnh trng nhau th p q 0 (1) ng.* Nu ba nghim phng trnh ch c hai nghim trng nhau hoc banghim l phn bit. Khi ta c: 2 3 27q 4p 0 ,

( trong :2 3a 2a 9ab 27cp b; q3 27

)3 3 3 2 3

22

4p p 2a 9ab 27c (a 3b)p | q | 2 | | 227 27 27 27

.

Nguyn Tt Thu

33 227c 2a 9ab 2 a 3b pcm.T cch chng minh trn ta suy ra c nu c (1) th phng trnh c banghim .

V d 7: Cho phng trnh 3 2x ax bx c 0 a 0 (1) c banghim phn bit. Chng minh phng trnh sau ch c hai nghim thcphn bit. 3 2 2 24(x ax bx c)(3x a) (3x 2ax b) (2).Gii:Gi 3 2f (x) x ax bx c . Khi (2) c vit di dng:

2 22f (x).f ''(x) f '(x) g(x) 2f (x)f "(x) f '(x) 0 .Ta c: (3)g '(x) 2f (x)f (x) ( (3)f l o hm cp ba ca hm f).Gi 1 2 3x x x l ba nghim phn bit ca f(x), ta c:

1 2 3g '(x) 12(x x )(x x )(x x ) g '(x) c ba nghim 1 2 3x , x , xBng bin thin.

x 1x 2x 3x g '(x) 0 + 0 0 +g(x)

2g(x )

1g(x ) 3g(x )V 2i i if (x ) 0 g(x ) f '(x ) 0 i 1, 2,3 , nn t bng bin thinsuy ra phng trnh g(x) 0 ch c hai nghim phn bit pcm.

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Bi 2. Phng trnh bc cao

Phng trnh bc cao y ta xt l phng trnh c bc cao hn 3. Phngphp chung gii phng trnh bc cao l ta tm cch chuyn v phngtrnh c bc thp hn, thng chng ta chuyn v ph ng trnh bc hai. lm iu ny ta thng s dng cc phng php sau:

1. Phng php a v dng tch: Tc l ta bin i phng trnh :

F(x) 0 f (x) 0f (x).g(x) 0g(x) 0

. a v mt phng trnh tch ta thng dng cc cch sa u:Cch 1: S dng cc hng ng thc a v dng

2 2 3 3a b 0, a b 0,... Cch 2: Nhm nghim ri chia a thc: Nu x a l mt nghim caphng trnh f (x) 0 th ta lun c s phn tch: f (x) (x a)g(x) . don nghim ta da vo cc ch sau:Ch : * Nu a thc n n 1n n 1 1 0f (x) a x a x ... a x a c nghimnguyn th nghim phi l c ca 0a .

* Nu a thc n n 1n n 1 1 0f (x) a x a x ... a x a c tng cc h sbng 0 th a thc c mt nghim x 1* Nu a thc n n 1n n 1 1 0f (x) a x a x ... a x a c tng cc h s

chn bng tng cc h s l th a thc c mt nghim x 1 .Cch 3: S dng phng php h s bt nh. Ta thng p dng chophng trnh trnh bc bn.

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V d 1: Gii phng trnh : 4 2x 4x 12x 9 0 (1) .Gii:Ta c phng trnh 4 2x (2x 3) 0 (1.1)

22 2

2x 2x 3 0(x 2x 3)(x 2x 3) 0x 2x 3 0

x 1, x 3 . Vy phng trnh c hai nghim: x 3; x 1 .

Nhn xt: Mu cht ca cch gii trn l chng ta nhn ra hng ng thcv bin i v phng trnh (1.1). Trong nhiu phng trnh vic lm xuthin hng ng thc khng cn d dng nh vy na, lm iu ny ihi chng ta phi c nhng nhy cm nht nh v phi thm bt nhnghng t thch hp.

V d 2: Gii phng trnh : 4 2 x 13x 18x 5 0 .Gii: Phng trnh 4 2 2(x 4x 4) (9x 18x 9) 0

2 2 2(x 2) (3x 3) 0 2 2(x 3x 5)(x 3x 1) 0 2

2x 3x 5 0

x 3x 1 0

3 29x

23 5

x2

.

Vy PT cho c 4 nghim: 3 29x2

; 3 5x2 .

Ch :1) Chc hn cc bn s thc mc lm sao m ta bit cch tch nh trn ?!.Tht ra th chng ta lm nh sau:Phng trnh 2 2 2 2(x m) (13 2m)x 18x 5 m 0 .Ta chn m sao cho biu thc trong du phn tch c hng ng thc,

c iu ny ta phi c:2 2 3 2

' 9 (5 m )(13 2m) 0 2m 13m 10m 16 0 , phngtrnh ny c mt nghim m 2 , do ta c th phn tch nh trn.Vi phng trnh bc bn tng qut 4 3 2x ax bx cx d 0 (I) ta cng

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c th bin i theo cch trn nh sau:4 3 2x ax bx cx d 0 4 3 2ax 2 x bx cx d

2

Ta cng thm hai v ca phng trnh mt lng:2

2 2a( 2 )x a x4

2

2 2 2 2a a(x x ) ( 2 b)x (a c)x d2 4

(1.I).By gi ta ch cn chn sao cho VT ca (1.I) phn tch thnh hng ngthc, tc l :

22 2a0 (a c) 4( 2 b)( d) 0

4

3 2 2 28 4b 2(ac 4d) c a d 4bd 0 (2.I)y l phng trnh bc ba nn bao gi cng c t nht mt nghim. Khi ta s a phng trnh (1.I) v phng trnh tch ca hai tam thc bc hai, ty ta gii hai tam thc ny ta c nghim phng trnh (I).2) V mt l thuyt th ta c th gii c mi phng trnh bc bn theocch trn. Tuy nhin trn thc t th nhiu lc vic gii khng c d dngvy, v mu cht quan trng nht ca cch gii trn l tm . Mc d (2.I) c cch gii nhng khng phi gi tr lc no cng p, nn s khkhn cho cc php bin i ca chng ta.

V d 3: Gii phng trnh : 4 3 22x 10x 11x x 1 0 (4).Gii: Ta c phng trnh:

3 2 2 28 4b 2(ac 4d) c a d 4bd 0 3 2 58 22 0

4 , phng trnh ny c nghim: 1

4 .

Do vy 2 2 2 25 1 1 3 9 1 3(4) (x x ) x x ( x )2 4 4 4 16 2 4

2

2 22

2x 4x 1 01(x 2x )(x 3x 1) 02 x 3x 1 0

,

2 2x

2 v 3 13x

2 .

Nguyn Tt Thu

Ch : a v phng trnh tch ngoi cch to ra hng ng thc trn, tacn c cch khc l s dng phng trnh h s bt nh. Chng hn xt vd trn. Ta phn tch:

4 3 2 2 22x 10x 11x x 1 (x ax b)(2x mx n) Khai trin ri ng nht cc h s ta c c h phng trnh :

2a m 102b n am 11an bm 1nb 1

. T phng trnh cui ta chn: n 1;b 1 , thay vo ba

phng trnh u ta c:2a m 10am 10

a m 1

ta thy h ny v nghim, do ta

chn n 1;b 1 , thay vo ta gii c m 4 v a 3Vy: 4 3 2 2 22x 10x 11x x 1 (x 3x 1)(2x 4x 1) .

V d 4: Tm m phng trnh sau c bn nghim phn bit.4 3 2 2x x (3m 4)x 5x 2m m 3 0 (5).

Gii: Khi gp bi ton ny c l cc bn s suy ngh khng bit nn x ltheo hng no? V phng trnh ny khng c nghim c bit, nu sdng phng trnh phn tch bnh phng th vic gii phng trnh (2.I) erng s khng i n kt qu ! Vy phng php h s bt nh th sao? Ch n h s t do ca phng trnh ta thy: 22m m 3 (m 1)(2m 3) ,iu ny dn ti ta ngh n phn tch VT ca phng t rnh v dng:

2 2(x ax m 1)(x bx 2m 3) (mc c l lm gim s n cn tmxung cn 2 n). ng nht h s ta c h phng trnh :

a b 1a 1

2m 3 m 1 ab 3m 4b 2

a(2m 3) b(m 1) 5

.

Vy 2 2(5) (x x m 1)(x 2x 2m 3) 0

Nguyn Tt Thu

2

2x x m 1 0 (a)x 2x 2m 3 0 (b)

(5) c bn nghim phn bit (a) v (b) u c hai nghim phn bit vchng khng c nghim chung.* (a) v (b) cng c hai nghim phn bit

a

b

4m 3 0 3m

' 2m 2 0 4

* Gi s (a) v (b) c nghim chung l 0x , khi 0x l nghim ca h:2 00 0

22 0 00 0

3x m 4 0x x m 1 0x x m 1 0x 2x 2m 3 0

0

2

m 4x

3m 4m 13 0

,

h ny v nghim (a) v (b) khng c nghim chung.Vy 3m

4 l nhng gi tr cn tm.

Nhn xt : Vic nhn thy 22m m 3 (m 1)(2m 3) l mu cht hnch kh khn trong vic phn tch ra tha s. y l mt tnh cht ca athc rt hay c s dng trong vic phn tch mt a thc thnh cc nhnt. C th : Nu tam thc bc hai (tng t cho a thc)

2f (x) ax bx c c hai nghim 1 2x , x th ta lun c s phn tch1 2f (x) a(x x )(x x ) . Vi phng trnh trn ta khng s dng c tnh

cht ny v v tri l mt a thc bc 4 khng c nghim c bit. Tuy nhinnu chng ta nhy bn th ta thy VT ca phng trnh li l mt tam thcbc hai i vi n l tham s m. Tc l ta c:

2 2 4 3 2(5) 2m (3x 1)m x x 4x 5x 3 0 (5)Tam thc ny c :

2 2 4 3 2 2 2(3x 1) 8(x x 4x 5x 3) (x 4x 5) (5) c hai nghim 2 2 21 3x 1 x 4x 5 x 2x 3m 4 2

v 22m x x 1 . Do vy ta c:

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22 x 2x 3(5') 2(m x x 1)(m ) 0

2

2 2(x x m 1)(x 2x 2m 3) 0 . y l phng trnh m ta vabin i trn.

V d 5: Gii phng trnh : 3 2 3x 3x 3x 16. x 9 0 .

Gii: t 3y x , ta c : 9 6 3y 3y 3y 16y 9 0 2 6 4 3 2(y 1)(y y 1)(y 2y 2y 4y 2y 9) 0 2 3 2 4 2 2(y 1)(y y 1) (y 1) 2y 3y (y 1) 7 0 2

y 1(y 1)(y y 1) 0 1 5y

2

.

Vy nghim ca phng trnh cho l:x 1; x 2 5; x 2 5 .

V d 6: Gii phng trnh : 6 4 3 25x 16x 33x 40x 8 0 .

Gii:Ta c phng trnh 2 4 3(5x 10x 4)(x 2x 5x 2) 0

2 2 2(5x 10x 4)(x 3x 1)(x x 2) 0 2 2(5x 10x 4)(x 3x 1) 0

2 1,2

23,4

3 5x

x 3x 1 0 55 55x 10x 4 0

x2

.

Vy phng trnh cho c bn nghim: 3 5 5 5x ; x2 5 .

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V d 7: Tm m phng trnh 4 2 2x x 2mx m 0 c bn nghimphn bit.

Gii:PT: 4 2 2 2x (x m) 0 (x x m)(x x m) 0

2

2x x m 0 (a)x x m 0 (b)

.

Phng trnh cho c bn nghim phn bit (a) v (b) u c hainghim phn bit v chng khng c nghim chung.

(a) v (b) c hai nghim phn bit 1 4m 0 1 1m1 4m 0 4 4 .

Gi s (a) v (b) c nghim chung l 0x20 0 020 0

x x m 0 x 0m 0x x m 0

.

Vym 0

1 1m

4 4

l nhng gi tr cn tm.

V d 8: Tm m phng trnh : 4 2 2x 2x 2mx m 2m 1 0 .1) C nghim ln nht ? 2) C nghim nh nht ?

Gii:Gi s 0x l nghim ca phng trnh cho, khi phng trnh:

2 4 20 0 0m 2(x 1)m x 2x 1 0 lun c nghim m

2 2 2 2 20 0 0 0 0 0' (x 1) (x 1) 0 (x x 2)(x x ) 0

00 x 1 .*

20x 0 m 2m 1 0 m 1

*2

0x 1 m 4m 4 0 m 2 Vy nghim nh nht ca phng trnh l x 0 c c khi m 1 nghim ln nht ca phng trnh l x 1 c c khi m 2 .

Nguyn Tt Thu

2. Phng php t n ph:t n ph l phng php kh hu hiu i vi cc bi ton i s , tronggii phng trnh bc cao cng vy , ngi ta thng t n ph chuynphng trnh bc cao v phng trnh c bc thp hn. Mt s dng sau yta thng dng t n ph.Dng 1: Phng trnh trng phng:

4 2ax bx c 0 (a 0) (1).Vi dng ny ta t 2t x , t 0 ta chuyn v phng trnh :

2at bt c 0 (2).Ch : S nghim ca phng trnh (1) ph thuc vo s nghim khng mca (2).

Dng 2: Phng trnh i xng (hay phng trnh hi quy):4 3 2 2ax b x cx k b x k a 0 (k 0) .

Vi dng ny ta chia hai v phng trnh cho x 2 (x 0) ta c phng trnh :2

22

k ka(x ) b(x ) c 0

xx .

t kt xx

vi t 2 k ta c :2

2 2 22

k kx (x ) 2k t 2k

xx

thay vo ta c phng trnh: 2a(t 2k) bt c 0 .Dng 3: Phng trnh : (x a)(x b)(x c)(x d) e , trong a b c d Phng trnh 2 2[x (a b)x ab].[x (c d)x cd] e .t 2t x (a b)x ta c: (t ab)(t cd) e .Dng 4: Phng trnh 2(x a)(x b)(x c)(x d) ex , trong ab cd .Vi dng ny ta chia hai v phng tr nh cho 2x (x 0) ,phng trnh 2 2 2[x (a b)x ab].[x (c d)x cd] ex

ab cd[x a b].[x c d] ex x

. t ab cdt x xx x

.Ta c phng trnh : (t a b)(t c d) e .Dng 5: Phng trnh 4 4(x a) (x b) c .

Nguyn Tt Thu

t a bx t2 ta a v phng trnh trng phng.

Ch : phng trnh c no 4a bc 2( )2

Ngoi nhng dng trn trong mt s phng trnh bc cao ta c th s dngphng php lng gic ha nh vo cng thc biu din cos nx qua cos x .

V d 9: Gii cc phng trnh:4 3 21) 2x 5x 6x 5x 2 0 4 42) (x 1) (x 3) 2

3)x(x 1)(x 2)(x 3) 24 4) 2(x 2)(x 3)(x 4)(x 6) 6x 0 Gii:1) Ta thy 0x khng l nghim phng trnh nn chia hai v phngtrnh cho 2x ta c: 2 2

1 12(x ) 5(x ) 6 0xx

.

t 1t x , (| t | 2)x

2 2 221 1

x (x ) 2 t 2xx

Ta c : 2 2t 2

2(t 2) 5t 6 0 2t 5t 2 0 1t (l)

2

*

21t 2 x 2 x 2x 1 0 x 1x

l nghim duy nht caphng trnh.2) t x t 2 ta c:

4 4 4 2(t 1) (t 1) 2 t 6t 0 t 0 x 2 Vy phng trnh c nghim duy nht x 2 .

Ch : Vi bi 2 ta c th gii bng cch khc nh sau: Trc ht ta c

BT:44 4a b a b

2 2 vi a b 0 .

p dng BT ny vi : a x 1, b x 3 VT VP . ng thcxy ra khi x 2 .3) Ta c phng trnh 2 2(x 3x)(x 3x 2) 24 .t 2t x 3x ta c :

Nguyn Tt Thu

2t(t 2) 24 t 2t 24 0 t 6, t 4 .*

2t 6 x 3x 6 0 Phng trnh v nghim.*

2t 4 x 3x 4 0 x 1, x 4 .Vy phng trnh c hai nghim x 1; x 4 .4) Phng trnh 2 2 2(x 4x 12)(x x 12) 6x 0 V x 0 khng l nghim ca phng trnh nn chia hai v phng trnhcho 2x ta c: 12 12(x 4)(x 1) 6 0

x x

t 12t xx

, ta c:2 t 1(t 4)(t 1) 6 0 t 3t 2 0

t 2 .

*2 x 412t 1 x 1 x x 12 0

x 3x .

*2t 2 x 2x 12 0 x 1 13

Vy phng trnh c bn nghim: x 3; x 4; x 1 13 .

V d 10: Tm m phng trnh : 2(x 1)(x 3)(x 5) m (1) c nghim.

Gii:Phng trnh (x 1)(x 5)(x 1)(x 3) m

2 2(x 4x 5)(x 4x 3) m t 2 2t x 4x (x 2) 4 4 ,ta c phng trnh :

2(t 5)(t 3) m t 2t 15 m (2) .Phng trnh (1) c nghim (2) c nghim t 4 .Vi 2 2t 4 t 2t 15 (t 1) 16 16 (2) c nghimt 4 m 16 .

V d 11: Tm m phng trnh : 4 3x 4x 8x m (1) c bn nghimphn bit.

Nguyn Tt Thu

Gii:Phng trnh 2 2 2(x 2x) 4(x 2x) m t 2 2t x 2x (x 1) 1 1 . Phng trnh tr thnh: 2t 2t m (2).Phng trnh (1) c bn nghim phng trnh (2) c hai nghim phnbit t 1 .Xt hm s : 2f (t) t 2t vi t 1 , ta c bng bin thin:

t -1 1 f(t)

3 -1

Da vo bng bin thin 1 m 3 .

V d 12: Gii phng trnh :2 2 2 33(x x 1) 2(x 1) 5(x 1) .

Gii: V x 1 khng l nghim ca phng trnh nn chia hai v cho3x 1 ta c:

2

2x x 1 x 13 2 5

x 1 x x 1 .

t2

2x x 1 2t 3t 5 3t 5t 2 0x 1 t

1t 2, t

3 .

*2 3 13t 2 x 3x 1 0 x

2

*21t 3x 2x 4 0

3 phng trnh v nghim.

V d 13: Gii phng trnh :6 5 4 3 2x 3x 6x 21x 6x 3x 1 0 .

Gii:y l phng trnh bc 6 v ta thy cc h s i xng do ta c thp dng cch gii m ta gii i vi phng trnh bc bn c h si xng.

Nguyn Tt Thu

Ta thy x 0 khng l nghim ca phng trnh . Chia hai v ca phngtrnh cho 3x , ta c: 3 23 2

1 1 1x 3(x ) 6(x ) 21 0

xx x .

t 1t x , | t | 2x

. Ta c : 2 2 3 22 31 1

x t 2; x t(t 3)x x

.Nn phng trnh tr thnh : 2 2t(t 3) 3(t 2) 6t 21 0

3 2 2 t 3t 3t 9t 27 0 (t 3) (t 3) 0t 3 .

*21 3 5t 3 x 3 x 3x 1 0 x

x 2 .

*2 3 5t 3 x 3x 1 0 x

2 .

Vy phng trnh c bn nghim 3 5 3 5x ; x2 2

.

V d 14: Gii phng trnh : 2(x 1)(x 2)(x 3) (x 4)(x 5) 360 .Gii:Phng trnh 2 2 2(x 6x 5)(x 6x 8)(x 6x 9) 360 t 2t x 6x , ta c phng trnh : (y 5)(y 8)(y 9) 360

2 2 x 0y(y 22y 157) 0 y 0 x 6x 0x 6

Vy phng trnh c hai nghim: x 0; x 6 .

V d 15: Gii phng trnh :3 2 2(64x 112x 56x 7) 4x 4 (1).

Gii:Mt trong nhng cng c gii quyt phng trnh bc cao l t n phbng cc hm lng gic. C c s thun li l da vo cng thc

biu din cos nx qua cos x . C th:

n[ ]2

n 2kn,k

k 0cos nx a .cos x

.

Nguyn Tt Thu

Trong k n 1 2k kn,k n kn

a ( 1) 2 Cn k

.T cng thc ta c:

7 5 3cos 7x 64cos x 112cos x 56cos x 7 cos x , iu ny gi cho chng ta t 2x cos t . Tuy nhin t c nh vy th phichng minh c phng trnh ch c nghim trong [0;1] ?.Tht vy: T phng trnh (1) 4x 4 x 1 .Ta xt x 0 . t x y y 0 , ta c phng trnh :

3 2 2(64y 112y 56y 7) 4 4y (2).Nu 20 y 1 VP(2) 8 7 VT(2) (2) v nghim.Vi 3 2 2y 1 (64y 112y 56y 7) 4 4y (2) v nghim.Vy ta chng minh c (1) nu c nghim x th x [0;1] .t 2x cos t; t [0; ]

2 , khi (1) tr thnh:

6 4 2 2 2(64cos t 112cos t 56cos t 7) 4(1 cos t) 7 4 3 2 2 2(64cos t 112cos t 56cos t 7 cos t) 4sin t cos t (Do

cos t 0 khng l nghim ca phng trnh nn ta nhn hai v phngtrnh vi 2cos t )

2 2cos 7t sin 2t cos14t cos( 4t) k

t14t 4t k2 18 914t 4t k2

t k10 5

.

V t [0; )2 nn ta c: 1 2 3 4 5 65 7 3t , t , t , t , t , t18 6 18 181 10 10

.Vy phng trnh cho c 6 nghim: 2i ix cos t (i 1, 2,...,6) vi

it (i 1, 2,..,6) l cc nghim va tm trn.

V d 16: Gii phng trnh : 3 3 3(x 5x 5) 5x 24x 30 0 .Gii:Ta c: 3 35x 24x 20 5(x 5x 5) x 5 nn phng trnh

Nguyn Tt Thu

3 3 3(x 5x 5) 5(x 5x 5) 5 x . t 3u x 5x 5 Ta c h:

32 2

3u 5u 5 x (u x)(u ux x 6) 0 u xx 5x 5 u

3 2x 4x 5 0 (x 1)(x x 5) 0 x 1 l nghim duy nht

ca phng trnh cho.

V d 17: Cho phng trnh : 5 4 3 22x x 10x 2x 8x 2 0 1) Chng minh rng phng trnh (1) c ng 5 nghim.2) Gi ix (i 1,5) l 5 nghim ca (1). Tnh tng :

5i

5 4i 1 ii

x 1S2x x 2

.Gii:1) Xt 5 4 3 2f (x) 2x x 10x 2x 8x 2 l hm lin tc

3 1 5f ( 2) 5 0;f ( ) 2 0;f (0) 1 0;f ( ) 0;2 2 8

1 175f (1) 0;f (3) 02 2

.T suy ra phng trnh (1) c 5 nghim thuc cc khong:

3 3 1 1( 2; ), ( ;0), (0; ), ( ;1), (1;3)2 2 2 2

.2) V ix l nghim ca (1) nn ta c:

5 4 3 2i i ii i2x x 2 2(5x x 4x )

5i

3 2i 1 i ii

x 11S2 5x x 4x

.Mt khc: x 1 1 2 5

x(x 1)(5x 4) 4x 9(x 1) 36(5x 4)

5 5 5

i ii 1 i 1 i 1 i

1 1 1 1 1 1S 48 x 9 x 1 72x

5

.

Ta c: 1 2 3 4 5f (x) (x x )(x x )(x x )(x x )(x x ) 5

ii 1

f '(x) 1f (x) x x

v 4 3 2f '(x) 10x 4x 30x 4x 8 .

Nguyn Tt Thu

5 5

i ii 1 i 1

f '(0) 1 1 f '(0) 4f (0) x x f (0)

; 5ii 1

1 f '(1) 12x 1 f (1)

;5

i 1 i

4f '( )1 1290054 4 4789

x f ( )5 5

77406S 43101 .

V d 18: Chng minh rng nu a thc bc n f (x) c n nghim phn bitth a thc g(x) f (x) f '(x) cng c n nghim phn bit.Gii: Xt hm s xh(x) f (x).e h(x) c n nghim phn bit

1 2 nx x ... x (y chnh l nghim ca f(x)).Ta c: xh '(x) (f (x) f '(x))e .Theo nh l Lagrang ta suy ra h' x c n 1 nghim

1 1 2 2 n 1 n 1 nx y x y ... x y x .M nghim ca h' x chnh l nghim ca g(x) f (x) f '(x) . g(x) l a thc bc n c n -1 nghim nn g(x) s c n nghim.Ch : S dng tnh cht trn ta c th sng tc ra nhng bi ton khc.Chng hn1) p dng tnh cht trn i vi a thc g(x), ta c:

a thc: g(x) g '(x) f (x) 2f '(x) f "(x) c n nghim phn bit.2) p dng tnh cht trn i vi a thc n 1F(x) x f ( )

x

a thc: n n 1 n 21 1F(x) F '(x) x f ( ) nx x f '( )x x

c n nghim

t 1tx

th ta c phng trnh : 2f (t) nt t f '(t) 0 c n nghim.

Bi tpBi 1: Gii cc phng trnh sau1) 3 22x 3x 1 0 2) 3 2x x 1 0 3) 3x 3x 1 0 4) 3 2x 6x 2x 1 0 Bi 2: Gii cc phng trnh sau:

Nguyn Tt Thu

4 31) x 4 x 8 x 5 2) (x 1)(x 2)(x 3)(x 4) 3 4 43) (x 4) (x 6) 82 2 2 24)(x x 1) 3x 3x 1 0

45)x 4x 1 2 2 2

2x 2 x 2 x 46) 2 2 5 0x 1 x 1 x 1

4 37) x 2x 4x 4 0 3 2 28) (32x 48x 18x 1) x 1

9) 2 4 2x (x 1)(x 2) 1 0 10)2 2 2 2128x (4x 1)(8x 1) 1 2x 0

1x 0

2

.

Bi 3: Tm m phng trnh :1) 3 2x (2m 1)x (m 3)x 3m 3 0 c 3 nghim phn bit.2) 3 2 2x 2mx m x m 1 0 c hai nghim phn bit.Bi 4: Cho phng trnh 3 2x ax bx c 0 c ba nghim phn bit.Chng minh rng phng trnh sau cng c ba nghim phn bit:

3 2x (a 6)x (b 4a 3)x a c 2b 0 .Bi 5: Chng minh rng nu a,b,c,d tha mn 2 38a d 4abc b 0 th tac th chuyn phng trnh 4 3 2ax bx cx dx e 0 v phng trnhtrng phng.Bi 6: Tm m phng trnh sau c 4 nghim phn bit.1)(x 1)(x 3)(x 5)(x 7) m 4 2 22) x m x 2mx 1 0 .Bi 7:Gii v bin lun cc phng trnh sau1) 4 3 2x 4x (m 4)x 2mx 2m 0 2) 4 3x 2x x m .Bi 8: Cho phng trnh bc n: n n 1n n 1 1 0a x a x ... a x a 0 trong n n 1 1 0a 0,a ,..., a ,a l s thc tha : n n 1 1 0a a a... a 0

n 1 n 2

Chng minh phng trnh cho c t nht mt nghim thuc 0;1 .Bi 9: Chng minh phng trnh c 4 nghim phn bit ix ;i 1,4 4 3 2f x 2x 6x 10x 5x 3 0 v hy tnh tng

24 i

2i 1 i

2x 1Sx 1

.

Nguyn Tt Thu

Bi 3 . nh l Vit i vi phng trnh bc cao

1. nh l Viet cho phng trnh bc baCho phng trnh : 3 2ax bx cx d 0 (a 0) (1).a) nh l Viet thun: Nu (1) c ba nghim 1 2 3x , x , x th ta c:

1 2 3

1 2 2 3 3 1

1 2 3

bx x x

a

cx x x x x x

a

dx x x

a

.

b) nh l o ca nh l Viet: Nu ba s x, y, z tha mn :x y z axy yz zx bxyz c

th x, y, z l ba nghim ca phng trnh :

3 2t at bt c 0 .2. nh l Viet cho phng trnh bc caoNu a thc n n 10 1 n 1 nf (x) a x a x ... a x a ( 0a 0) c nnghim 1 2 nx , x ,..., x th ta c:

11 2 n

0

21 2 1 3 n 1 n

0

n n1 2 n

0

ax x ... x

a

ax x x x ... x x

a

...

ax x ...x 1

a

.

Ta k hiu :n 1

1 1 2 n ii 1 0

aS x x ... x xa

;

22 i j

1 i j n 0aS x xa

; 1 2 k1 2 k

k kk i i i

1 i i ... i 0

aS x x ...x ( 1)a

Nguyn Tt Thu

kS l tng cc tch chp k ca n s ix v c gi l cc a thc i xngs cp ca cc nghim.

V d 1: Gi 1 2 3x , x , x l nghim ca PT : 3x 5x 1 0 . Tnh:2 2 21 2 3A x x x ; 2 2 2 2 2 21 2 2 3 3 1B x x x x x x ; 4 4 41 2 3C x x x

2 2 2 2 2 21 2 1 2 2 3 2 3 3 1 3 1D x x (x x ) x x (x x ) x x (x x ) .

Gii:Ta d dng chng minh c phng trnh cho c ba nghim phn bit.

p dng nh l Vit ta c:1 2 3

1 2 2 3 3 1

1 2 3

x x x 0x x x x x x 5x x x 1

.

Ta c: 21 2 3 1 2 2 3 3 1A (x x x ) 2(x x x x x x ) 0 2( 5) 10 2

1 2 2 3 3 1 1 2 3 1 2 3B (x x x x x x ) 2x x x (x x x ) 25 .2 2 2 2 2 2 2 2 2 2 21 2 3 1 2 2 3 3 1C (x x x ) 2(x x x x x x ) 10 2.25 50

V: 2 2 2 21 2 1 2 1 2 1 2 3 1 2x x (x x ) 2x x x x x 2x x 2 2 2 2 2 2

1 2 3 1 2 3 1 2 2 3 3 1D x x x (x x x ) 2(x x x x x x ) 50

Ch : 1) Cc biu thc A, B, C, D trn gi l cc a thc i xng babin. Mt tnh cht quan trng ca cc a thc i xngba bin l chng lun biu din c qua ba a thc i xng ba bin scp.C th nu ta t p x y z; q xy yz zx; r xyz th ta c mt sbiu din sau:

2 2 2 2x y z p 2q 3 3 3 3x y z p 3pq 3r

2 2 2 2(xy) (yz) (zx) q 2pr 4 4 4 4 2 2x y z p 2p q 2q 4pr

xy(x y) yz(y z) zx(z x) pq 3r 2 2 2 2 2 2 2 2xy(x y ) yz(y z ) zx(z x ) p q 2q pr

Nguyn Tt Thu

3 3 3 3 2(xy) (yz) (zx) q 3pqr 3r .2) Gi 1 2 3x , x , x l nghim PT : 3 2ax bx cx d 0 .t n n nn 1 2 3S x x x , khi ta c h thc:

n n 1 n 2 n 3aS bS cS dS 0 .

V d 2: Tm iu kin phng trnh : 3 2x ax bx c 0 c banghim, trong tng hai nghim gp k 1 nghim cn li.

Gii: * Gi s phng trnh c ba nghim 1 2 3x , x , x trong tng hainghim gp k nghim nghim cn li.

Theo nh l Vi t, ta c:1 2 3

1 2 2 3 3 1

1 2 3

x x x a

x x x x x x bx x x c

.

Ta c: 1 2 3 2 3 1 3 1 2(x x kx )(x x kx )(x x kx ) 0 1 2 3a (k 1)x a (k 1)x a (k 1)x 0

3 21 2 3a a (k 1)(x x x ) 2 3

1 2 2 3 3 1 1 2 3a(k 1) (x x x x x x ) (k 1) x x x 0 3 3 2 3a a (k 1) ab(k 1) c(k 1) 0

2 3 3ab(k 1) a k c(k 1) (1).Vi a,b,c tha mn (1), ta tm iu kin phng trnh c ba nghim lc. T (1) 32

k abc a

k 1(k 1) thay vo phng tr

nh ta c:

3 2 32

k abx ax bx a 0

k 1(k 1)

22

a ka ka(x ) x x b 0k 1 k 1 (k 1)

phng trnh c ba nghim 2 2 22 2

k a 4ka 4b 0(k 1) (k 1)

Nguyn Tt Thu

2 2 2(k 4k)a 4(k 1) b 0 (2).Vy iu kin a,b,c l:

2 3 3

2 2 2ab(k 1) a k c(k 1)(k 4k)a 4(k 1) b 0

.

Ch : 1) Vi k 2 th ba nghim ca phng trnh trn lp thnh cp scng.

Vy phng trnh 3 2x ax bx c 0 c ba nghim lp thnh cp scng nu v ch nu:

3

29ab 2a 27c

a 3b 0

.

2) n y chc cc bn s t t ra cu hi l a,b,c phi tha mn iu king phng trnh c ba nghim lp thnh cp s nhn ? Cu tr li dnhcho cc bn .

V d 3: Gii h phng trnh : 2 2 2

3 3 3

x y z 3

x y z 21

x y z 57

.

Gii: V h cho gm ba phng trnh l nhng a thc i xng ba binnn ta biu din ba phng trnh qua ba a thc i xng c bn .t a x y z;b xy yz zx;c xyz , h tr thnh:

2

3

a 3 a 3 x y z 3a 2b 21 b 6 xy yz zx 6

c 8 xyz 8a 3ab 3c 57

x, y, z l ba nghim ca phng trnh : 3 2t 3t 6t 8 0

2(t 1)(t 2t 8) 0 t 1; t 2; t 4 .

Vy nghim ca h cho l:x 1y 2z 4

v cc hon v.

Nguyn Tt Thu

V d 4: Gii h phng trnh sau :1 1 1 8

x y z3x y z

1 1 1 118x y z

x y z 91 1 1 728

x x y y z z27x x y y z z

(THTT).

Gii: t 1 1 1a x ; b y ; c zx y z

.

Khi : 2 21 1x ( x ) 2 a 2;x x

2 31x x a(a 3) a 3a

x x

H cho tr thnh:

2 2 2

3 3 3

8a b c

3118

a b c 69

728a b c 3(a b c)

27

8a b c

3ab bc ca 0abc 0

( y ta s dng cc ng thc:

2 2 2 2a b c (a b c) 2(ab bc ca) v 3 3 3 2 2 2a b c (a b c)(a b c ab bc ca) 3abc )

a, b,c l nghim ca phng trnh : 3 2t 0

8t t 0 83 t

3

.

Xt phng trnh : 1X 0 X 1X

Phng trnh 1 8X 3X 8 X 3 0 X 93X

.T ta c nghim ca h cho l: (x; y;z) (1;1;9) v cc hon v.

Nguyn Tt Thu

V d 4: Cho phng trnh 3 2 3x ax bx c 0 c ba nghim v b ac ,c 0 . Chng minh rng phng trnh c ng mt nghim ln hn c.

Gii: Trc ht ta thy rng nu 1 1x c x c 0 . Do yu cu biton tr thnh chng minh trong ba s 1 2 3x c; x c; x c c ng mt sdng. iu ny dn n ta i xt tch: 1 2 3(x c)(x c)(x c) .Gi 1 2 3x , x , x l ba nghim ca phng trnh cho.

Theo nh l Vi t, ta c:1 2 3

1 2 2 3 3 13

1 2 3

x x x a

x x x x x x b

x x x c

.

Ta xt: 21 2 3 1 2 3 1 2 3(x c)(x c)(x c) x x x c (x x x ) 3 2

1 2 2 3 3 1c(x x x x x x ) c c a cb c(ac b) 0 (do gi thit)Nu c ba nghim 31 2 3 1 2 3x , x , x c x x x c v l (do 31 2 3x x x c ),vy trong ba s 1 2 3x c; x c; x c , tn ti ng mt s dng trongba nghim ca phng trnh c ng mt nghim ln hn c.

V d 5: Gi s phng trnh 3 2x ax bx c 0 (1) c ba nghim phnbit khc 0. Chng minh rng phng trnh : 3 2 2x bx acx c 0 (2)cng c ba nghim phn bit.

Gii: Gi 1 2 3x , x , x 0 l ba nghim ca phng trnh (1).

Theo nh l Viet, ta c:1 2 3

1 2 2 3 3 1

1 2 3

x x x a

x x x x x x bx x x c

.

Gi s (2) c ba nghim 1 2 3y , y , y , ta c:

1 2 3 1 2 2 3 3 1y y y b x x x x x x Dn n ta d on ba nghim ca (2) l 1 2 2 3 3 1x x ; x x ; x x .t 1 1 2 2 2 3 3 3 1y x x ; y x x ; y x x , ta c:

Nguyn Tt Thu

1 2 3

1 2 2 3 3 1 1 2 3 1 2 32 2

1 2 3 1 2 3

y y y by y y y y y x x x (x x x ) acy y y (x x x ) c

1 2 3y , y , y l ba

nghim ca phng trnh : 3 2 2x bx acx c 0 .V l ba nghim phn bit nn 1 2 3y , y , y cng l ba nghim phn bit

Vy phng trnh 3 2 2x bx acx c 0 c ba nghim phn bit.

Ch : Khi gp bi ton cho phng trnh : 3 2x ax bx c 0 (1) c banghim v yu cu chng minh phng trnh 3 2x mx nx p 0 (2)cng c ba nghim ta thng lm nh sau:Gi 1 2 3x , x , x l ba nghim ca (1), ta chng minh 1 2 3f (x , x , x );

1 2 3g(x , x , x ) ; 1 2 3h(x , x , x ) l ba nghim ca (2) bng cch chng minh:1 2 3 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3 1 2 3

1 2 3 1 2 3

1 2 3 1 2 3 1 2 3

f (x , x , x ) g(x , x , x ) h(x , x , x ) mf (x , x , x )g(x , x , x ) h(x , x , x )f (x , x , x ) g(x , x , x )h(x , x , x ) nf (x , x , x )g(x , x , x )h(x , x , x ) p

V d 6: Cho a, b 0 dng v phng trnh 3 2x x ax b 0 c banghim. Chng minh rng:

1) 10a b27

2) 7a 2b27

.Gii:Gi thit bi ton cho phng trnh c ba nghim v yu cu chng ta chngminh BT gia cc h s nn ta ngh n chuyncc BT thnh cc BT ca ba nghim . Gi 1 2 3x , x , x l ba nghimca phng trnh . V 1 2 3a,b 0 x , x , x 0

Ta c:1 2 3

1 2 2 3 3 1

1 2 3

x x x 1x x x x x x a

x x x b

Bi ton tr thnh cho ba s 1 2 3x , x , x 0 v c tng bng 1. Chng minh:

Nguyn Tt Thu

1) 1 2 2 3 3 1 1 2 310

x x x x x x x x x27

.

p dng BT Csi ta c:3

1 2 31 2 3

x x x 1x x x

3 27

21 2 2 3 3 1 1 2 3

1 1x x x x x x (x x x )

3 3

1 2 2 3 3 1 1 2 31 1 10

x x x x x x x x x3 27 27

.

2) 1 2 2 3 3 1 1 2 37

x x x x x x 2x x x27

.

1 2 3 1 2 37(1 2x )x x x (x x )27

1 2 3 1 17P (1 2x )x x x (1 x )27

Gi s 1 1 2 3 1 11

x min{x , x , x } x 1 2x 03

2

2 31 1 1

x xP (1 2x ) x (1 x )

2

21

1 1 11 x(1 2x ) x (1 x )

2

3 2 21 1 1 12x x 1 (3x 1) (6x 1)7 7P

4 27 108 27 (pcm).

ng thc xy ra 1 2 31

x x x3

.Ch : Khi gp cc BT v h s ca phng trnh bc ba (cng nh bccao) ta c th s dng nh l Viet chuyn BT cn chng minh v BTcc nghim ca phng trnh . Hn na ta thy cn ng sng tc ranhng bi ton dng ny l xut pht t mt BT i xng ba bin, s dngnh l Viet ta chuyn BT v BT gia cc h s ca phng trnh bcba. Chng hn t bi ton:Cho x, y, z 0 tha mn xy yz zx xyz 4 . Chng minh

x y z xy yz zx .Ta chuyn thnh bi ton nh sauCho phng trnh : 3 2x ax bx b 4 0 c ba nghim khng m.Chng minh a b . V d sau y cng l mt sn phm ca cc h lm trn.

Nguyn Tt Thu

V d 7: Cho phng trnh 3 2x ax bx c 0 c ba nghim . Chngminh: 3 2 312ab 27c 6a 10 (a 2b) (HSG QG 2001 ).Gii:Gi 1 2 3x , x , x l ba nghim ca phng trnh cho. Theo Viet, ta c:

1 2 32 2 2 2

1 2 2 3 3 1 1 2 3

1 2 3

x x x a

x x x x x x b a 2b x x xx x x c

BT cn chng minh tr thnh: 2 2 36a(a 2b) 27c 10 (a 2b)

2 2 2 2 2 2 31 2 3 1 2 3 1 2 3 1 2 36(x x x )(x x x ) 27x x x 10 (x x x ) (1).

* Nu 2 2 21 2 3x x x 0 (1) ng.* Nu 2 2 21 2 3x x x 0 . V (1) l BT thun nht nn ta ch cn chngminh (1) khi 2 2 21 2 3x x x 9 . V (1) tr thnh:

1 2 3 1 2 32(x x x ) x x x 10 (2)

Gi s 21 2 3 1| x | | x | x x 3 2 2 22 3 1

2 3x x 9 x

x x 32 2 .

Gi P l VT(2) 1 2 3 2 3P x (2 x x ) 2(x x ) 2 2 2 2

1 2 3 2 3P x (x x ) (2 x x ) 4 2

2 3 2 3(9 2x x ) (2 x x ) 4 t 2 3t x x 3 t 3 v

2 2 2P (9 2t)(8 4t t ) (t 2) (2t 7) 100 2P 10 P 10 . ng thc xy ra 1 2

3

x x 2x 1

.

Hay (a;b;c) l hon v ca b (2t;2t; t), t 0 .

V d 8 Cho phng trnh 3 2ax x bx 1 0 vi a 0;a b , c banghim dng. Tm gi tr nh nht ca

2

25a 3ab 2P

a (b a) (HSG QG 1999).

Nguyn Tt Thu

Gii:Gi m, n, p l ba nghim thc dng ca phng trnh cho.p dng nh l Viet ta c:

1m n p

a

bmn np pm a,b 0

a

1mnp

a

.

Ta c:2

3(m n p) 3(mn np pm)m n p 3 mnp

2

3

1 3b 1ba 3aa

11 1 a3 3 3a a

.

Cch 1:

Xt hm s:2

25a 3ab 2f (b)

a (b a) v

i 1b (0; ]3a

. Ta c:2

2 22a 2f '(b) 0 f (b)

a (b a) gi

m trn 1b (0; ]3a

2

21 3(5a 1)f (b) f 3.g(a)3a a(1 3a )

Xt hm g(a) vi 1a (0; ]

3 3 , ta c:

4 2

2 2 215a 14a 1 1g '(a) 0 a (0; ]

3 3a (3a 1)

1g(a) g 4 3 f (b) 12 33 3

.ng thc c1

a ;b 33 3

Khi m n p 3 . Vy minP 12 3 .

Cch 2: Ta c22 2 2

22

b 1 1 25 3 2 5a1 1 3(5a 1)a a aP b 1a a a(1 3a )1 1a 3a

Nguyn Tt Thu

3 2

236 3a 15a 12 3a 3P 12 3

a(1 3a )

2

2(3 3a 1)(12a 3 3a 3)

a(1 3a ) .

Do 21 1 140 a 12a 3 3a 3 12. 1 3 027 93 3

v

3 3a 1 0 P 12 3 0 P 12 3 . ng thc xy ra1

a3 3

b 3

V d 9: Cho phng trnh n n 11 n 1 nx a x ... a x a 0 c nnghim dng. Chng minh cc bt ng thc sau:

1)n

n 1n 1n

aa

n

2)21 n 1

n

a an

a

.

Gii:Gi 1 2 nx , x ,..., x 0 l nghim ca phng trnh. Theo nh l Viet ta c:

n 1n 1 1 2 n 1 2 3 n

nn 1 2 n

1 1 2 n

0 ( 1) a x x ...x ... x x ...x0 ( 1) a x x ...x0 a x x ... x

.

p dng BT Csi , ta c: n 1n 1 nn 1 1 2 n 1 2 3 n 1 2 n( 1) a x x ...x ... x x ...x n x x ...x

n

n 1 n(n 1)n(n 1) n 1n 11 2 n n

a( 1) x x ...x 1 an

n

n 1n 1n

aa

n

.

2) 1 2 n 1 2 n 1 2 3 n1 n 1n 1 2 n

(x x ... x )(x x ...x ... x x ...xa aa x x ...x

n 1nn 1 2 n 1 2 n 21 2 n

n x x ...x .n x x ...xn

x x ...x

.

Nguyn Tt Thu

V d 10: Cho phng trnh n n 11 n 1 nx a x ... a x a 0 c nnghim, trong 1 1a , 2 n 1a 2

. Chng minh rng cc nghim caphng trnh nm trong on 1 1a ;a 2 .

Gii: Gi 1 2 nx , x ,.., x l n nghim ca phng trnh cho. Khi yu cubi ton cn chng minh: 1 i 1 1 i 1a x a 2 a 1 x 1 a 1

2 2i 1(x 1) (a 1) . iu ny dn ti ta c cch gii nh sau:

Theo nh l Viet ta c:n

i 1i 1

x a

;n

i j 2i, j 1i j

x x a

2n n2 2

1 i i 2i 1 i 1

a x x 2a

n 2 2

i 1 2i 1

x a 2a

n n n2 2 2

i i i 1 2 1i 1 i 1 i 1

(x 1) x 2 x n a 2a 2a n

2 21 1 2 1a 2a 1 [2a (n 1)] (a 1) .

(Do 2n 1

a2 ) 1 i 1a x a 2 pcm.

V d 11: Cho phng trnh sau c n nghim ( n 3 ). Hy tm cc nghim ?

2n n 1 n 2 n 3

3 n 1 nn n

x nx x a x ... a x a 02

.

Gii: Ta thy phng trnh ch c ba h s ca n n 1 n 2x , x , x l nhnggi tr c th cn nhng h s khc chng ta cha xc nh c. Do gii phng trnh ny ta phi da vo mi quan h gia cc h s ca

n n 1 n 2x , x , x , iu ny gi cho chng ta s dng nh l Viet. Tht vy:Gi 1 2 nx , x ,.., x l n nghim ca phng trnh cho.

Theo nh l Viet ta c:n

ii 1

x n

;2n

i ji, j 1i j

n nx x

2

.

Nguyn Tt Thu

2n n n2

i i i ji 1 i 1 i, j 1

i jx x 2 x x n

.

n n n2 2i i i

i 1 i 1 i 1(x 1) x 2 x n n 2n n 0

ix 1 i 1, n . Vy phng trnh c n nghim trng nhau: x 1 .

Nhn xt: Bi ton trn l mt trng hp c bit ca bi ton trong v d10. Vi cch lm tng t hoc t s nh gi gia cc tng cc tch chp k

kS ta c th sng tc ra nhng bi ton tng t.

V d 12: Cho a thc f (x) c bc 2007 v c 2007 nghim dng. Chngminh rng phng trnh sau cng c 2007 nghim dng.

2 2(1 2007x)f (x) (x 2007x 1)f '(x) x f "(x) 0 (1).Gii: Gi 1 2 2007x , x ,..., x 0 l 2007 nghim ca f(x).Trc ht ta chng minh b sauB : Phng trnh f (x) f '(x) 0 cng c 2007 nghim dng .Chng minh:Xt hm s xg(x) f (x).e g(x) c 2007 nghimdng (chnh l nghim ca f(x))(Nu a thc f(x) c nghim bi th cng chnh l nghim ca f(x), do ta ch cn i xt cc nghim khng la nghim bi nn ta ch xt f(x) c2007 nghim dng phn bit)Theo nh l Roll g '(x) c 2006 nghim 1 2 2007y , y ,..., y tha mn :

1 1 2 2 2006 2006 2007x y x y ... x y x iy 0 i 1,2006 .

M xg '(x) e (f (x) f '(x)) nghim ca g '(x) chnh l nghim caphng trnh : f (x) f '(x) 0 (2)(2) l phng trnh bc 2007 li c 2006 nghim nn (2) s c 2007 nghim ,gi nghim l 2007y . Ta chng minh c 2007y 0 . Tht vy:t 2007 2006 20051 2 2006 2007f (x) x a x a x ... a x a .V f(x) c cc nghim dng nn ia 0 i 1,2007 .

Nguyn Tt Thu

2006 20051 2005 2006f '(x) 2007x 2006a x ... 2a x a

20072007 2006f (x) f '(x) x ... a a .

Theo Viet 1 2 2007 2007 2006 2007y y ...y a a 0 y 0 B c chng minh.Tr li bi ton:t h(x) f (x) f '(x) h(x) c 2007 nghim dng phn bit vh '(x) f '(x) f "(x) . 2007 1F(x) x h( )

x cng c 2007 nghim dng.

p dng b trn cho F(x) , ta suy ra phng trnh F(x) F '(x) 0 c2007 nghim dng. Hay phng trnh :

2007 2006 20051 1 1F(x) F '(x) x h 2007x h( ) x h '( ) 0x x x

(3) c 2007

nghim dng . t 1tx

. Khi (3) tr thnh:

2007 2006 20051 2007 1h(t) h '(t) 0

t t t

2(1 2007t)h(t) t h '(t) 0 (4)

Thay h(x) bi f(x) vo (4), th phng trnh sau c 2007 nghim dng: 2(1 2007t)[f (t) f '(t)] t f '(t) f "(t) 0

2 2(1 2007t)f (t) (t 2007t 1)f '(t) t f "(t) 0 Ta c pcm.

Nguyn Tt Thu

Bi tp:Bi 10: Tm m phng trnh : 3 2x 3x (5m 1)x 3m 0 c banghim lp thnh cp s cng.Bi 11: Cho phng trnh 3 2x px qx p 0 c ba nghim thc khngnh hn 1 v p,q 0 . Tm gi tr ln nht ca biu thc q 3F

p .

Bi 12: Cho phng trnh 3 2x ax bx c 0 (1) c ba nghim thcphn bit. Chng minh phng trnh sau cng c 3 nghim thc phn bit :

3 2 2 2 2 2 4x (b 2ac)x (a c 2bc )x c 0 (2).Bi 13: Gi s c ABC nhn tan A, tan B, tan C l ba nghim caphng trnh: 3 2x px qx p 0 (q 1) . Chng minh: p 3 3 vq 9 .Bi 14: Cc s a,b,c phi tha mn iu kin g phng trnh sau c banghim lp thnh cp s nhn: 3 2x ax bx c 0 .Bi 15: Cho phng trnh 3 2x ax bx c 0 c ba nghim m, n, p . Vinhng gi tr no ca a,b,c 3 3 3m ,n ,p l ba nghim ca phng trnh :

3 3 2 3 3x a x b x c 0 .Bi 16: Gi , , l 3 nghim ca phng trnh : 3x x 1 0 . Tnh gitr ca biu thc: 1 1 1P

1 1 1 .

Bi 17: Gii h :

2 2 2

4 4 4x y z 9

x y z 33xyz 4

Bi 18: Chng minh phng trnh: 3 29x 36x 45x 17 0 c ba nghiml di ba cnh ca mt tam gic t.Bi 19: Cho phng trnh : n n 1n 1 1 0x a x ... a x a 0 (1) c nnghim, bit ia 1 i 0,1, 2,..., n 1 . Chng minh n 3 .

Nguyn Tt Thu

Hng dn gii v p s chng IIBi 1: 1) 1x ; x 1

2

2) 3 31 25 3 69 25 3 69x3 2 2

3) 2 4 8x 2cos ; x 2cos ; x 2cos

9 9 9

4) 3 3563 56313 1327 27

x2 2

.

Bi 2:1) PT 2 2 2(x 2x) 4(x 2x) 5 0 . S: x 1 ; x 1 6 2) t 2t x 5x . S: 5 13x

2 .

3) t x t 5 . S: x 3; x 7 .4) t 2t x x 1 . S: 1 5x 1; x 2; x

2 .

5) PT 4 2 2x 2x 1 2(x 2x 1) 0 22 2(x 1) 2x 2 0

2 4 2 2x 1; x

2 .

6) t x 2 x 2a ;bx 1 x 1 .

Ta c: 2 22a 5ab 2b 0 (2a b)(a 2b) 0 S: 9 73x2

.7) PT

2 2 2(x x) (x 2) 0 2 2(x 2)(x 2x 2) 0 x 2 8) T phng trnh 0 x 1 . t 2x cos t , t [0; ]

2 .

Phng trnh tr thnh: 2 2cos 6t sin t .S: 2 2 23 5x cos ; x cos ; x cos ;

14 14 14 2x cos

10 ; 2 3x cos ; x 0

10 .

9) PT 2 2 2 2x (x 1)(x 1)(x 2) 1 0

Nguyn Tt Thu

4 2 4 2(x x )(x x 2) 1 0 4 2 2 1 5(x x 1) 0 x

2 .

10) T iu kin 1 x 02

, gi ta t: 2x cos t, t ( ; )2 . Khi ta

c: 2 2 2 232cos t(cos t 1)(2cos t 1) 1 cos t 0 8 6 4 2128cos t 256cos t 160cos t 32cos t 1 cos t

1 2 3 44 6 2 8

cos8t cos t t ; t ; t ; t7 7 3 9

Vy phng trnh c bn nghim: i1

x cos t (i 1, 4)2

Bi 3:

1) PT 2(x 1)(x 2mx 3m 3) 0 2x 1

x 2mx 3m 3 0

S:

4m

53 21 3 21

m ( ; ) ( ; )2 2

.

2) PT 2 2 3xm (2x 1)m x 1 0 ,phng trnh c 2(2x 1)

2x x 1(m x 1)(m ) 0x

2(x m 1) x (1 m)x 1 0 2

x m 1

x (1 m)x 1 0 (*)

TH1: (*) c hai nghim phn bit trong mt nghim x m 1 2

2m 2m 3 0 3

m2(1 m) (1 m)(1 m) 1 0

TH2: (*) c nghim kp x m 1

2m 2m 3 0 m 1m 1 m 3m 1

2

S: 3m 1;m ;m 3

2 .

Nguyn Tt Thu

Bi 4: t 3 2f (x) x ax bx c , ta chng minh c phng trnhf (x) 2f '(x) f "(x) 0 c ba nghim phn bit (p dng v d 21),t tac pcm.Bi 5: t x y m thay vo phng trnh ta c:

4 3 2 2ay (4am b)y (6m a 3mb c)y 3 2(4am 3bm 2mc d)y 4 3 2am bm cm dm e 0 (1).

(1) l phng trnh trng phng khi h sau c nghim m

3 24am b 0

4am 3bm 2mc d 0

3 2

bm

4a

b b b4a 3b 2c d 04a 4a 4a

3 2

bm

4ab 4abc 8a d 0

pcm.

Bi 6:1)Phng trnh 2 2(x 8x 7)(x 8x 15) m .t 2 2t x 8x 7 (x 4) 9 9 PT tr thnh: 2t 8t m 16 m 9 .2) PT 4 2 2 2x (mx 1) 0 (x mx 1)(x mx 1) 0 S: | m | 2Bi 7:1) Phng trnh 2 2 2(x 2x) m(x 1) m 0 .t 2t (x 1) 0 x 1 t . Phng trnh tr thnh:

2 2(t 1) mt m 0 t (m 2)t m 1 0 (*)(*) c m(m 8) Nu 0 m 8 (*) v nghim phng trnh v nghimNu m 0 Phng trnh c hai nghim x 0; x 2 Nu m 8 Phng trnh v nghim

Nguyn Tt Thu

Nu m 0 (*) c hai nghim tri du phng trnh cho c hainghim

m 2 m(m 8)x 1

2 .

Nu m 8 (*) c hai nghim m phn bit PT cho v nghim.2) PT 2 2 2(x x) (x x) m .t 2 21 1 1 1 1t x x (x ) x t

2 4 4 2 4 .

PT tr thnh: 2f (t) t t m 0 (*).T bng bin thin ca f(t) ta suy ra

* Nu 1m (*)4

v nghim Pt cho v nghim

* Nu 1m4

PT cho c hai nghim 1 3x2

* Nu 1 5m (*)4 16

c hai nghim 1t4

Phng trnh cho c

bn nghim: 1 3 2 1 4mx2 4

.

Nu 5m (*)16

c ng mt nghim 1t4

Phng trnh cho c

hai nghim : 1 3 2 1 4mx2 4

.

Bi 8: Xt hm s n 1 n 2n n 1 1 0a a af (x) x x ... x a

n 1 n 2 ,

ta thy f(x) tha mn iu kin nh l Lagrang trn [0;1] v f (0) f (1) 0 n n 1

n n 1 1 0f '(x) a x a x ... a x a c mt nghim (0;1) pcm.Bi 9: Ta c f ( 2) 33; f ( 1) 4; f (0) 3; f (1) 6; f (5) 278

f (x) c bn nghim. Ta c:2 4 3 2 2 2i i i i i i i iix 2x 1 2x 6x 9x 3x 4 (2x 1)(x 3x 4)

2i

2 i i i ii

2x 1 1 1 1 1(x 1)(x 4) 5 x 4 x 1(x 1)

Nguyn Tt Thu

i i

1 1 15 1 x 4 x

i 1, 2,3, 4 24 4 4i

2 i ii 1 i 1 i 1i

2x 1 1 1 1 15 1 x 5 4 x(x 1)

1 f '( 1) f '(4) 1 1 9 375 f ( 1) f (4) 5 4 149 569 .

Bi 10: Gi s phng trnh c ba nghim 1 2 3x , x , x lp thnh cp s cng

th ta c: 1 2 3 21 3 2

x x x 3x 1

x x 2x

, thay vo phng trnh ta

c: 1m2

.

Vi 1m2

, phng trnh tr th nh: 2(x 1)(2x 4x 3) 0 x 1 .Vy khng tn ti m tha mn iu kin bi ton.Bi 11: Gi 1 2 31 x x x l ba nghim ca phng trnh cho. Khi theo Viet, ta c: 1 2 3 1 2 3

1 2 2 3 3 1

x x x x x x p (1)x x x x x x q (2)

..

T (1) tn ti tam gic ABC 1 2x tan A; x tan B; 3x tan C ,vi A B C

4 2 .

F co t A cot B cot C 3cot A cot Bcot C 2sin A cos(B C) cos A

cot A 3cot Acos(B C) cos A cos(B C) cos A

3 31 A 3 A 1 3F 3tan tan 3t t f (t)A 2 2 2 2t 22 tan

2

.

(Trong At tan2

v 1t [ 2 1; ]3

, do A ;2 8 6

).

Ta c: 221 9f '(t) 3 t 0 f (t) f ( 2 1) 8 4 2

22t

Vy max F 8 4 2 , t c khi 3A ;B C4 8 .

Nguyn Tt Thu

Bi 12: Gi 1 2 3x , x , x l ba nghim ca (1). Theo Viet, ta c:1 2 3

1 2 2 3 3 1

1 2 3

x x x a

x x x x x x bx x x c

.

t 2 221 1 2 2 2 3 3 3 1y x x ; y x x ; y x x . Ta chng minh c1 2 3y , y , y l ba nghim ca (2).

Bi 13: Theo nh l Viet:p (t anA tan B tan C) tan AtanB tan Cq tan A tan B tan B tan C tan C tan A

V ABC nhn nn ta c: tan A tan B tan C 3 3 p 3 3 23q 3 tan A tan B tan C 9 .

Bi 14: Gi s phng trnh c ba nghim 1 2 3x , x , x lp thnh cp s nhn:

1 2 3 322

1 3 2

x x x cx c

x x x

.

Thay vo phng trnh ta c:3 2 3 2 3 3 33a c b c 0 a c b c 0 c(ca b ) 0 (*). Nu 2c 0 (1) x(x ax b) 0 d thy phng

trnh ny c ba nghim lp thnh cp s nhn a b 0 . Nu

3

3b

c 0 (*) ca

( a 0 ), khi (1) tr thnh:

3 3 4 2 3 3 2 2 2 2 2a x a x a bx b 0 (ax b) a x a (a b)x b 0 2

2 22

bx

a

bx (a b)x 0 (**)

a

(**) c hai nghim

22 2 3 3

24b(a b) 0 (a ab 2b)(a ab 2b) 0a

.Khi ta thy (1) c ba nghim v ba nghim ny lp thnh cp s nhn.

Nguyn Tt Thu

Vy iu kin ca a,b,c cn tm l: 3 3

2 2 2 2

a b c 0

a c b 0

a (a b) 4b 0

.

Bi 15: Theo Viet: m n p a;mn np pm b;mnp c V 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3m n p a ;m n n p p m b ;m n p c Mt khc:

3 3 3 3m n p (m n p) 3(m n p)(mn np pm) 3mnp 3 3a ( a) 3ab 3c c ab . Khi :

3 2 3 2 2x ax bx c 0 x ax bx ab 0 (x a)(x b) 0

2x a

x b

phng trnh c ba nghim th b 0 .

Vy iu kin cn tm l:c abb 0 .

Bi 16: Ta c: 1 1 1P 3 21 1 1

t x t 1 ,phng trnh tr thnh: 3 2t 3t 1 0 .Gi 1 2 3t , t , t l ba nghim ca phng trnh ny :

1 2 2 3 3 1

1 2 3 1 2 3

t t t t t t1 1 1P 3 2 2 0 P 3t t t t t t

.

Bi 17: Ta c:2 2 2 2 2 2 2 2 2 2 4 4 42(x y y z z x ) (x y z ) (x y z ) 2 2 2 2 2 2x y y z z x 24

Do :

2 2 2

2 2 2 2 2 2 2 2 2

2 2 2

x y z 9

x y y z z x 24 x ; y ;z

x y z 16

l nghim ca phng

trnh : 3 2t 9t 24t 16 0 2(t 1)(t 4) 0 t 1; t 4 .Th li ta thy nghim ca h l: (1; 2;2),( 1; 2; 2), ( 1;2;2) v cc hon v ca ba cp nghim .

Nguyn Tt Thu

Bi 18: Gi f(x) l a thc v tri ca phng trnhTrc ht ta chng minh phng trnh c ba nghim phn bit dng. Tht

vy: 9 19f (0) 17; f (1) 1; f ( ) ; f (2) 1 f (x)5 125

c ba nghim

dng phn bit 1 2 39 9

x (0;1), x (1; ), x ( ;2)5 5

.Theo Viet ta c: 1 2 3 1 2 3x x x 4 x x 4 x

1 2 3 3 1 2 3x x x 4 2x 0 x , x , x l di ba cnh ca mt tamgic, ta gi tam gic l ABC.

Mt khc: 2 2 2 2 21 2 3 1 2 1 2 3 33

34x x x (x x ) 2x x x 16 8x

9x

Xt hm s 34g(t) 16 8t9t

vi 9t ;25

, c2

2 234 72t 34 9g '(t) 8 0 t ;2

59t 9t

9 202g(t) g( ) 05 405

2 2 21 2 3x x x 0 tam gic ABC l tamgic t.Bi 19: Gi 1 2 nx , x ,..., x l n nghim ca (1). Theo Viet, ta c:

n

i n 1 i j n 2i 1 i j

x a 1; x x a 1

2

n n2i i i j

i 1 i 1 i jx x 2 x x 3

(Do

n2i

i 1x 0

).

V ix 0 i 1, n i

1 (i 1, n)

x l phng trnh :

n n 10 1 n 1a x a x ... a x 1 0

.

Tng t, theo Viet :2

n n

2 i i ji 1 i 1 i ji

1 1 12 3x x xx

p dng bt ng thc Csi, ta c:

Nguyn Tt Thu

n n2 2i 2

i 1 i 1 i

19 x n n 3x

(pcm).