Tuyen tap Phuong trinh He phuong trinh MS

7/31/2019 Tuyen tap Phuong trinh He phuong trinh MS

1/383


2/383


3/383

Din n MATHSCOPE

PHNG TRNH

H PHNG TRNHCh bin: Nguyn Anh Huy

26 - 7 - 2012


4/383

Mc lcLi ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Cc thnh vin tham gia chuyn . . . . . . . . . . . . . . . . . . . . . . . . 8

1 I CNG V PHNG TRNH HU T 10

Phng trnh bc ba . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Phng trnh bc bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Phng trnh dng phn thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Xy dng phng trnh hu t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Mt s phng trnh bc cao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2 PHNG TRNH, H PHNG TRNH C THAM S 32

Phng php s dng o hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Phng php dng nh l Lagrange - Rolle . . . . . . . . . . . . . . . . . . . . . . 42

Phng php dng iu kin cn v . . . . . . . . . . . . . . . . . . . . . . . . . 46Phng php ng dng hnh hc gii tch v hnh hc phng . . . . . . . . . . . . . 55

Hnh hc khng gian v vic kho st h phng trnh ba n . . . . . . . . . . . . . 76

Mt s bi phng trnh, h phng trnh c tham s trong cc k thi Olympic . . . 81

3 CC PHNG PHP GII PHNG TRNH 93

Phng php t n ph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Mt s cch t n ph c bn . . . . . . . . . . . . . . . . . . . . . . . . . . 93

t n ph a v phng trnh tch . . . . . . . . . . . . . . . . . . . . . . . 94t n ph a v phng trnh ng cp . . . . . . . . . . . . . . . . . . . . 101Phng php t n ph khng hon ton . . . . . . . . . . . . . . . . . . . . 103Phng php s dng h s bt nh . . . . . . . . . . . . . . . . . . . . . . . 108t n ph a v h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . 109Phng php lng gic ha . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Phng php bin i ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Phng php dng lng lin hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Phng php dng n iu hm s . . . . . . . . . . . . . . . . . . . . . . . . . . 138

Phng php dng bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Mt s bi ton chn lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

3
http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


5/383

4

4 PHNG TRNH M-LOGARIT 158L thuyt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

Phng php t n ph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158



Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

5 H PHNG TRNH 177Cc loi h c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

H phng trnh hon v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

Phng php t n ph trong gii h phng trnh . . . . . . . . . . . . . . . . . . 206



Phng php h s bt nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231K thut t n ph tng - hiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

Phng php dng bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

Tng hp cc bi h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258H phng trnh hu t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258H phng trnh v t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

6 SNG TO PHNG TRNH - H PHNG TRNH 297

Xy dng mt s phng trnh c gii bng cch a v h phng trnh . . . . 297S dng cng thc lng gic sng tc cc phng trnh a thc bc cao . . . . 307

S dng cc hm lng gic hyperbolic . . . . . . . . . . . . . . . . . . . . . . . . . 310

Sng tc mt s phng trnh ng cp i vi hai biu thc . . . . . . . . . . . . . 312

Xy dng phng trnh t cc ng thc . . . . . . . . . . . . . . . . . . . . . . . . 318

Xy dng phng trnh t cc h i xng loi II . . . . . . . . . . . . . . . . . . . 321Xy dng phng trnh v t da vo tnh n iu ca hm s. . . . . . . . . 324Xy dng phng trnh v t da vo cc phng trnh lng gic. . . . . . . . 328

S dng cn bc n ca s phc sng to v gii h phng trnh. . . . . . . 331S dng bt ng thc lng gic trong tam gic . . . . . . . . . . . . . . . . 338S dng hm ngc sng tc mt s phng trnh, h phng trnh. . . . . 345

Sng tc h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

Kinh nghim gii mt s bi h phng trnh . . . . . . . . . . . . . . . . . . . . . 353

7 Ph lc 1: GII TON BNG PHNG TRNH - H PHNG TRNH 362

8 Ph lc 2: PHNG TRNH V CC NH TON HC NI TING 366

Lch s pht trin ca phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 366C my cch gii phng trnh bc hai? . . . . . . . . . . . . . . . . . . . . . 366Cuc thch chn ng th gii ton hc . . . . . . . . . . . . . . . . . . . . 368Nhng vinh quang sau khi qua i . . . . . . . . . . . . . . . . . . . . . . . 372
http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


6/383

5

Tu s mt s nh ton hc ni ting . . . . . . . . . . . . . . . . . . . . . . . . . 376Mt cuc i trn bia m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376Ch v l sch qu hp! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376Hai gng mt tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377Sng hay cht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378

9 Ti liu tham kho 381
http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


7/383

Li ni uPhng trnh l mt trong nhng phn mn quan trng nht ca i s v c nhng ng

dng rt ln trong cc ngnh khoa hc. Sm c bit n t thi xa xa do nhu cu tnhton ca con ngi v ngy cng pht trin theo thi gian, n nay, ch xt ring trong Tonhc, lnh vc phng trnh c nhng ci tin ng k, c v hnh thc (phng trnh hu t,phng trnh v t, phng trnh m - logarit) v i tng (phng trnh hm, phng trnhsai phn, phng trnh o hm ring, ...)

Cn Vit Nam, phng trnh, t nm lp 8, l mt dng ton quen thuc v cyu thch bi nhiu bn hc sinh. Ln n bc THPT, vi s h tr ca cc cng c gii tchv hnh hc, nhng bi ton phng trnh - h phng trnh ngy cng c trau chut, trthnh nt p ca Ton hc v mt phn khng th thiu trong cc k thi Hc sinh gii, thii hc.

c rt nhiu bi vit v phng trnh - h phng trnh, nhng cha th cp mt

cch ton din v nhng phng php gii v sng to phng trnh. Nhn thy nhu cu cmt ti liu y v hnh thc v ni dung cho c h chuyn v khng chuyn, Din nMathScope tin hnh bin son quyn sch Chuyn phng trnh - h phng trnh mchng ti hn hnh gii thiu n cc thy c gio v cc bn hc sinh.

Quyn sch ny gm 6 chng, vi cc ni dung nh sau: Chng I: i cng v phng hu t cung cp mt s cch gii tng qut phngtrnh bc ba v bn, ngoi ra cn cp n phng trnh phn thc v nhng cch xy dngphng trnh hu t. Chng II: Phng trnh, h phng trnh c tham s cp n cc phng phpgii v bin lun bi ton c tham s ,cng nh mt s bi ton thng gp trong cc k thiHc sinh gii. Chng III: Cc phng php gii phng trnh ch yu tng hp nhng phngphp quen thuc nh bt ng thc, lng lin hp, hm s n iu, ...vi nhiu bi tonm rng nhm gip bn c c cch nhn tng quan v phng trnh.Chng ny khng cp n Phng trnh lng gic, v vn ny c trong chuyn Lng gic ca Din n.

Chng IV: Phng trnh m logarit a ra mt s dng bi tp ng dng ca hms logarit, vi nhiu phng php bin i a dng nh t n ph, dng ng thc, hm niu, ... Chng V: H phng trnh l phn trng tm ca chuyn . Ni dung ca chng


8/383

7

bao gm mt s phng php gii h phng trnh v tng hp cc bi h phng trnh haytrong nhng k thi hc sinh gii trong nc cng nh quc t. Chng VI: Sng to phng trnh - h phng trnh a ra nhng cch xy dng mt bihay v kh t nhng phng trnh n gin bng cc cng c mi nh s phc, hm hyperbolic,hm n iu, . . .

Ngoi ra cn c hai phn Ph lc cung cp thng tin ng dng phng trnh, h phngtrnh trong gii ton v v lch s pht trin ca phng trnh.

Chng ti xin ng li cm n ti nhng thnh vin ca Din n chung tay xy dngchuyn . c bit xin chn thnh cm n thy Chu Ngc Hng, thy Nguyn Trng Sn,anh Hong Minh Qun, anh L Phc L, anh Phan c Minh v h tr v ng gp nhng kin qu gi cho chuyn , bn Nguyn Trng Thnh v gip ban bin tp kim tra ccbi vit c mt tuyn tp hon chnh.

Nim hi vng duy nht ca nhng ngi lm chuyn l bn c s tm thy nhiu iub ch v tnh yu ton hc thng qua quyn sch ny. Chng ti xin n nhn v hoan nghnhmi kin xy dng ca bn c chuyn c hon thin hn. Mi gp xin vui lngchuyn n [email protected]

Thnh ph H Ch Minh, ngy 11 thng 7 nm 2012

Thay mt nhm bin sonNguyn Anh Huy


9/383

Cc thnh vin tham gia chuyn

hon thnh c cc ni dung trn, chnh l nh s c gng n lc ca cc thnh vin cadin n tham gia xy dng chuyn :

Ch bin: Nguyn Anh Huy (10CT THPT chuyn L Hng Phong - TP HCM)

Ph trch chuyn : Nguyn Anh Huy (10CT THPT chuyn L Hng Phong - TP HCM),

Nguyn An Vnh Phc (TN Ph thng Nng khiu- TP HCM)

i cng v phng trnh hu t: Hunh Phc Trng (THPT Nguyn Thng Hin TP HCM), Phm Tin Kha (10CT THPT chuyn L Hng Phong - TP HCM)

Phng trnh, h phng trnh c tham s: thy Nguyn Trng Sn (THPT Yn M A Ninh Bnh), V Trng Hi (12A6 THPT Thi Phin - Hi Phng), nh V Bo Chu(THPT chuyn L Qu n - Vng Tu), Hong B Minh ( 12A6 THPT chuyn Trni Ngha - TP HCM), Nguyn Hong Nam (THPT Phc Thin - ng Nai), Ong Th

Phng (11 Ton THPT chuyn Lng Th Vinh - ng Nai) Phng php t n ph: thy Mai Ngc Thi (THPT Hng Vng - Bnh Phc), thy

Nguyn Anh Tun (THPT L Qung Ch -H Tnh), Trn Tr Quc (11TL8 THPTNguyn Hu - Ph Yn), H c Khnh (10CT THPT chuyn Qung Bnh), on ThHo (10A7 THPT Long Khnh - ng Nai)

Phng php dng lng lin hp: Ninh Vn T (THPT chuyn Trn i Ngha -TPHCM) , inh V Bo Chu (THPT - chuyn L Qu n, Vng Tu), on ThHa (THPT Long Khnh - ng Nai)

Phng php dng bt ng thc: Nguyn An Vnh Phc (TN Ph thng Nng khiu-TP HCM), Phan Minh Nht, L Hong c (10CT THPT chuyn L Hng Phong - TPHCM), ng Hong Phi Long (10A10 THPT Kim Lin H Ni), Nguyn Vn Bnh(11A5 THPT Trn Quc Tun - Qung Ngi),

Phng php dng n iu: Nguyn Anh Huy (10CT THPT chuyn L Hng Phong- TP HCM), Hong Kim Qun (THPT Hng Thi H Ni), ng Hong Phi Long(10A10 THPT Kim Lin H Ni)

Phng trnh m logarit: V Anh Khoa, Nguyn Thanh Hoi (i hc KHTN- TPHCM), Nguyn Ngc Duy (11 Ton THPT chuyn Lng Th Vinh - ng Nai)

Cc loi h c bn: Nguyn Anh Huy (10CT THPT chuyn L Hng Phong - TP HCM)


10/383

9

H phng trnh hon v: thy Nguyn Trng Sn (THPT Yn M A Ninh Bnh),Nguyn Anh Huy (10CT THPT chuyn L Hng Phong TP HCM), Nguyn nh Hong(10A10 THPT Kim Lin - H Ni)

Phng php bin i ng thc: Nguyn nh Hong (10A10 THPT Kim Lin - HNi), Trn Vn Lm (THPT L Hng Phong - Thi Nguyn), Nguyn c Hunh (11

Ton THPT Nguyn Th Minh Khai - TP HCM) Phng php h s bt nh: L Phc L (i hc FPT TP HCM), Nguyn Anh Huy,

Phan Minh Nht (10CT THPT chuyn L Hng Phong TP HCM)

Phng php t n ph tng - hiu: Nguyn Anh Huy (10CT THPT chuyn L HngPhong TP HCM)

Tng hp cc bi h phng trnh: Nguyn Anh Huy (10CT THPT chuyn L Hng PhongTP HCM), Nguyn Thnh Thi (THPT chuyn Nguyn Quang Diu ng Thp), Trn

Minh c (T1K21 THPT chuyn H Tnh H Tnh), V Hu Thng (11 Ton THPTNguyn Th Minh Khai TP HCM)

Sng to phng trnh: thy Nguyn Ti Chung (THPT chuyn Hng Vng Gia Lai),thy Nguyn Tt Thu (THPT L Hng Phong - ng Nai), Nguyn L Thu Linh (10CTTHPT chuyn L Hng Phong TP HCM)

Gii ton bng cch lp phng trnh: Nguyn An Vnh Phc (TN Ph thng Nng khiu-TP HCM)

Lch s pht trin ca phng trnh: Nguyn An Vnh Phc (TN Ph thng Nng khiu-TP HCM), Nguyn Hong Nam (THPT Phc Thin - ng Nai)


11/383

C I: I CNG V PHNG TRNHHU T

PHNG TRNH BC BA

Mt s phng php gii phng trnh bc ba

Phng php phn tch nhn t:Nu phng trnh bc ba ax3 + bx2 + cx + d = 0 c nghim x = r th c nhn t(x

r) do

c th phn tch

ax3 + bx2 + cx + d = (x r)[ax2 + (b + ar)x + c + br + ar2]

T ta a v gii mt phng trnh bc hai, c nghim l

b ra b2 4ac 2abr 3a2r22a

Phng php Cardano:Xt phng trnh bc ba x3 + ax2 + bx + c = 0 (1).Bng cch t x = y a

3, phng trnh (1) lun bin i c v dng chnh tc:

y3 +py + q = 0(2)

Trong : p = b a2

3, q = c +

2a3 9ab27

Ta ch xt p, q= 0 v p = 0 hay q = 0 th a v trng hp n gin.t y = u + v thay vo (2), ta c:

(u + v)3 +p(u + v) + q = 0

u3 + v3 + (3uv + p)(u + v) + q = 0 (3)

Chn u, v sao cho 3uv +p = 0 (4).Nh vy, tm u v v, t (3) v (4) ta c h phng trnh:u

3 + v3 = qu3v3 = p

3

27

Theo nh l Viete, u3 v v3 l hai nghim ca phng trnh:

X

2

+ qX p3

27 = 0(5)

t =q2

4+

p3

27

10


12/383

11

Khi > 0, (5) c nghim:u3 = q

2+

, v3 = q

2

Nh vy, phng trnh (2) s c nghim thc duy nht:

y = 3

q2

+

+ 3

q2

Khi = 0, (5) c nghim kp: u = v = 3q2

Khi , phng trnh (2) c hai nghim thc, trong mt nghim kp.

y1 = 23

q

2, y2 = y3 =

3

q

2

Khi < 0, (5) c nghim phc.Gi u30 l mt nghim phc ca (5), v30 l gi tr tng ng sao cho u0v0 =

p

3.

Khi , phng trnh (2) c ba nghim phn bit.y1 = u0 + v0

y2 = 12

(u0 + v0) + i

3

2(u0 v0)

y3 = 12

(u0 + v0) i

3

2(u0 v0)

Phng php lng gic ho - hm hyperbolic:Mt phng trnh bc ba, nu c 3 nghim thc, khi biu din di dng cn thc s lin quann s phc. V vy ta thng dng phng php lng gic ho tm mt cch biu din

khc n gin hn, da trn hai hm s cos v arccosC th, t phng trnh t3 + pt + q = 0 () ta t t = u cos v tm u c th a () vdng

4cos3 3cos cos3 = 0

Mun vy, ta chn u = 2p

3v chia 2 v ca () cho u

3

4 c

4cos3 3cos 3q2p

.

3p

= 0 cos3 = 3q2p

.

3p

Vy 3 nghim thc l

ti = 2

p3

. cos

1

3arccos

3q

2p.

3p

2i

3

vi i = 0, 1, 2.

Lu rng nu phng trnh c 3 nghim thc th p < 0 (iu ngc li khng ng) nn cngthc trn khng c s phc.Khi phng trnh ch c 1 nghim thc v p = 0 ta cng c th biu din nghim bng cng

thc hm arcosh v arsinh:

t =2|q|

q.

p3

cosh

1

3.arcosh

3|q|2p

.

3p

nu p < 0 v 4p3 + 27q2 > 0.


13/383

12

t = 2

p

3. sinh

1

3.arsinh

3q

2p.

3

p

nu p > 0

Mi phng php trn u c th gii quyt phng trnh bc ba tng qut. Nhng mc chca chng ta trong mi bi ton lun l tm li gii ngn nht, p nht. Hy cng xem quamt s v d:

Bi tp v dBi 1: Gii phng trnh x3 + x2 + x = 1

3

Gii

Phng trnh khng c nghim hu t nn khng th phn tch nhn t. Trc khi ngh ticng thc Cardano, ta th quy ng phng trnh:

3x3 + 3x2 + 3x + 1 = 0

i lng 3x2+3x+1 gi ta n mt hng ng thc rt quen thuc x3+3x2+3x+1 = (x+1)3.Do phng trnh tng ng:

(x + 1)3 = 2x3hay

x + 1 = 32xT suy ra nghim duy nht x =

11 + 3

2

.

Nhn xt: V d trn l mt phng trnh bc ba c nghim v t, v c gii nh kho lobin i ng thc. Nhng nhng bi n gin nh th ny khng c nhiu. Sau y ta s i

su vo cng thc Cardano:

Bi 2: Gii phung trnh x3 3x2 + 4x + 11 = 0

Gii

t x = y + 1 . Th vo phng trnh u bi, ta c phng trnh:y3 + 1.y + 13 = 0

Tnh = 132

+

4

27 .1

3

=

4567

27

0p dng cng thc Cardano suy ra:

y =3

13 +4567272

+3

13 4567272

Suy ra x =3

13 +4567272

+3

13 4567272

+ 1.

Nhn xt: V d trn l mt ng dng c bn ca cng thc Cardano. Tuy nhin cng

thc ny khng h d nh v ch c dng trong cc k thi Hc sinh gii. V th, c l chngta s c gng tm mt con ng hp thc ha cc li gii trn. l phng php lnggic ho. u tin xt phng trnh dng x3 + px + q = 0 vi p < 0 v c 1 nghim thc:


14/383

13

Bi 3: Gii phng trnh x3 + 3x2 + 2x 1 = 0Gii

u tin t x = y 1 ta a v phng trnh y3 y 1 = 0 (1). n y ta dng lng gicnh sau:

Nu |y| 0 khng kh chng minh phng trnh c nghim duy nht:

Bi 5: Gii phng trnh x3 + 6x + 4 = 0

Gii

tng: Ta s dng php t x = k

t 1t

a v phng trnh trng phng.

php t ny khng cn iu kin ca x, v n tng ng k(t2 1) xt = 0. Phng trnh

trn lun c nghim theo t.Nh vy t phng trnh u ta c

k3

t3 1t3

3k3

t 1

t

+ 6k

t 1

t

+ 4 = 0


16/383

15

Cn chn k tho 3k3 = 6k k = 2Vy ta c li gii bi ton nh sau: Li gii:

t x =

2

t 1

t

ta c phng trnh

22t3 1t3+ 4 = 0 t6 1 + 2t3 = 0 t1,2 = 31 3

2

Lu rng t1.t2 = 1 theo nh l Viete nn ta ch nhn c mt gi tr ca x l

x = t1 + t2 =

2

3

1 + 3

2+ 3

1 3

2

. 2

Bi 6: Gii phng trnh 4x3 3x = m vi |m| > 1

GiiNhn xt rng khi |x| 1 th |V T| 1 < |m| (sai) nn |x| 1. V vy ta c th tx =

1

2

t +

1

t

.

Ta c phng trnh tng ng:1

2

t3 +

1

t3

= m

T :t =

3

m

m2 1 x = 12

3

m +

m2 1 + 3m

m2 1 .Ta chng minh y l nghim duy nht.

Gi s phng trnh c nghim x0 th x0 [1, 1] v |x0| > 1. Khi :4x3 3x = 4x30 3x0

hay(x x0)(4x2 + 4xx0 + 4x20 3) = 0

Xt phng trnh:4x2 + 4xx0 + 4x

2

0 3 = 0

c = 12 12x20 < 0 nn phng trnh bc hai ny v nghim.

Vy phng trnh u bi c nghim duy nht l

x =1

2

3

m +

m2 1 + 3

m m2 1

.

Bi tp t luynBi 1: Gii cc phng trnh sau:a) x3 + 2x2 + 3x + 1 = 0b) 2x3 + 5x2 + 4x + 2 = 0c) x3 5x2 + 4x + 1 = 0


17/383

16

d) 8x3 + 24x2 + 6x 10 36 = 0Bi 2: Gii v bin lun phng trnh:

4x3 + 3x = m vi m RBi 3: Gii v bin lun phng trnh:

x3 + ax2 + bx + c = 0

PHNG TRNH BC BN

[1] Phng trnh dng ax4 + bx3 + cx2 + bkx + ak2 = 0 (1)

Ta c

(1) a(x4 + 2x2.k + k2) + bx(x2 + k) + (c 2ak)x2 = 0 a(x2 + k)2 + bx(x2 + k) + (c 2ak)x2 = 0

n y c hai hng gii quyt:Cch 1: a phng trnh v dng A2 = B2:Thm bt, bin i v tri thnh dng hng ng thc dng bnh phng ca mt tng, chuyncc hng t cha x2 sang bn phi.Cch 2: t y = x2 + k y kPhng trnh (1) tr thnh ay2 + bxy + (c 2ak)x2 = 0

Tnh x theo y hoc y theo x a v phng trnh bc hai theo n x.

V d: Gii phng trnh: x4 8x3 + 21x2 24x + 9 = 0 (1.1)

Cch 1:(1.1) (x4 + 9 + 6x2) 8(x2 + 3) + 16x2 = 16x2 21x2 + 6x2 (x2 4x + 3)2 = x2

x2 4x + 3 = xx2 4x + 3 = x

x2 5x + 3 = 0x2 3x + 3 = 0

x =5 13

2

x = 5 + 132

Cch 2:

(1.1) (x4 + 6x2 + 9) 8x(x2 + 3) + 15x2 = 0 (x2 + 3)2 8x(x2 + 3) + 15x2 = 0t y = x2 + 3. (1.1) tr thnh: y2 8xy + 15x2 = 0 (y 3x)(y 5x) = 0

y = 3x

y = 5x

Vi y = 3x: Ta c x2 + 3 = 3x: Phng trnh v nghim

Vi y = 5x: Ta c x2 + 3 = 5x x2 5x + 3 = 0

x =5 13

2

x =

5 +

13

2

Vy phng trnh (1.1) c tp nghim: S =

5 +

13

2;

5 132

Nhn xt: Mi phng php gii c li th ring. Vi cch gii 1, ta s tnh c trc tip m


18/383

17

khng phi thng qua n ph. Vi cch gii 2, ta s c nhng tnh ton n gin hn v t bnhm ln.

Bi tp t luynGii cc phng trnh sau:1) x4 13x3 + 46x2 39x + 9 = 02) 2x4 + 3x3 27x2 + 6x + 8 = 03) x4 3x3 6x2 + 3x + 1 = 04) 6x4 + 7x3 36x2 7x + 6 = 05) x4 3x3 9x2 27x + 81 = 0

[2] Phng trnh dng (x + a)(x + b)(x + c)(x + d) = ex2 (2) vi ad = bc = m

Cch 1: a v dng A2 = B2(2) (x +px + m)(x2 + nx + m) = ex2(ad = bc = m, p = a + d, n = b + c)

x2 +p + n

2x + m n p

2x

x2 +

p + n

2x + m +

n p2

x

= ex2

x2 +p + n

2x + m

2=

n p2

2+ e

x2

Cch 2: Xt xem x = 0 c phi l nghim ca phng trnh khng.Trng hp x = 0:

(2) x + mx

+px + mx

+ n = et u = x +

m

x. iu kin: |u| 2|m|

(2) tr thnh (u + p)(u + n) = e. n y gii phng trnh bc hai theo u tm x.

V d: Gii phng trnh: (x + 4)(x + 6)(x 2)(x 12) = 25x2 (2.1)

Cch 1:(2.1) (x2 + 10x + 24)(x2 14x + 24) = 25x2

(x2 2x + 24 + 12x)(x2 2x + 24 12x) = 25x2

(x2 2x + 24)2 = 169x2

x2 2x + 24 = 13xx2 2x + 24 = 13x

x2 15x + 24 = 0x2 + 11x + 24 = 0

x = 8x = 3

x =

15

129

2Cch 2:

(2.1) (x2 + 10x + 24)(x2 14x + 24) = 25x2Nhn thy x = 0 khng phi l nghim ca phng trnh.


19/383

18

x = 0 : (2.1)

x +24

x+ 10

x +

24

x 14

= 25

t y = x +24

x |y| 46. (2.1) tr thnh:

(y + 10)(y 14) = 25 (y + 11)(y 15) = 0

y = 11y = 15

Viy

= 11

: Ta c phng trnh:x +

24

x= 11 x2 + 11x + 24 = 0

x = 3x = 8

Vi y = 15: Ta c phng trnh:

x +24

x= 15 x2 15x + 24 = 0 x = 15

129

2

Phng trnh (2.1) c tp nghim S =

3; 8; 15

129

2;

15 +

129

2

Nhn xt: Trong cch gii 2, c th ta khng cn xt x = 0 ri chia m c th t n phy = x2 + m thu c phng trnh bc hai n x, tham s y hoc ngc li.

Bi tp t luyn

Gii cc phng trnh sau:1) 4(x + 5)(x + 6)(x + 10)(x + 12) = 3x2

2) (x + 1)(x + 2)(x + 3)(x + 6) = 168x2

3) (x + 3)(x + 2)(x + 4)(x + 6) = 14x2

4) (x + 6)(x + 8)(x + 9)(x + 12) = 2x2

5) 18(x + 1)(x + 2)(x + 5)(2x + 5) =19

4x2

[3] Phng trnh dng (x + a)(x + b)(x + c)(x + d) = m (3) vi a + b = c + d = p

Ta c (3) (x2 + px + ab)(x2 +px + cd) = mCch 1:

(3)

x2 + px +

ab + cd

2

+ab cd

2 x2 +px +

ab + cd

2

ab cd

2 = m

x2 + px +ab + cd

2

2= m +

ab cd

2

2Bi ton quy v gii hai phng trnh bc hai theo x.Cch 2:

t y = x2 +px iu kin: y p2

4. (3) tr thnh:(y + ab)(y + cd) = m

Gii phng trnh bc 2 n y tm x.

V d: Gii phng trnh: x(x + 1)(x + 2)(x + 3) = 8 (3.1)

Cch 1:


20/383

19

Ta c(3.1) (x2 + 3x)(x2 + 3x + 2) = 8

(x2 + 3x + 1 1)(x2 + 3x + 1 + 1) = 8

(x2 + 3x + 1)2 = 9

x2 + 3x + 1 = 3

x2 + 3x + 1 = 3

x2 + 3x 2 = 0x2 + 3x + 4 = 0

x = 3 172

Cch 2:

(3.1) (x2 + 3x)(x2 + 3x + 2) = 8t y = x2 + 3x y 9

4(3.1) tr thnh:

y(y + 2) = 8

y2 + 2y

8 = 0

y = 2

y = 4(loi) y = 2

Vi y = 2: Ta c phng trnh:

x2 + 3x 2 = 0 x = 3

17

2

Phng trnh (3.1) c tp nghim: S =

3 + 17

2;3 17

2

Bi tp t luynGii cc phng trnh sau:1. (x + 2)(x + 3)(x 7)(x 8) = 1442. (x + 5)(x + 6)(x + 8)(x + 9) = 40

3.

x +1

4

x +

3

5

x +

1

20

x +

4

5

=

39879

400004. (6x + 5)2(3x + 2)(x + 1) = 355. (4x + 3)2(x + 1)(2x + 1) = 810

Nhn xt: Nh dng (2), ngoi cch t n ph trn, ta c th t mt trong cc dngn ph sau: t y = x2 +px + ab t y = x2 +px + cd t y =

x +

p

2

2 t y = x2 +px +

ab + cd

2

[4] Phng trnh dng (x + a)4 + (x + b)4 = c (c > 0) (4)

t x = y a + b2

. (4) tr thnh:y +

a b2

4+

y a b

2

4= c


21/383

20

S dng khai trin nh thc bc 4, ta thu c phng trnh:

2y4 + 3(a b)2y2 + 2

a b2

4= c

Gii phng trnh trng phng n y tm x.

V d: Gii phng trnh: (x + 2)4 + (x + 4)4 = 82 (4.1)

t y = x + 3. Phng trnh (4.1) tr thnh:(y + 1)4 + (y 1)4 = 82 (y4 + 4y3 + 6y2 + 4y + 1) + (y4 4y3 + 6y2 4y + 1) = 82 2y

4

+ 12y2

80 = 0 (y2

4)(y2

+ 10) = 0 y2 = 4 y = 2Vi y = 2, ta c x = 1Vi y = 2, ta c x = 5Vy phng trnh c tp nghim: S = {1; 5}

Bi tp t luynGii cc phng trnh sau:

1. (x + 2)4 + (x + 8)4 = 2722. (x +

2)4 + (x + 1)4 = 33 + 12

2

3. (x + 10)4 + (x 4)4 = 285624. (x + 1)4 + (x 3)4 = 90

[5] Phng trnh dng x4 = ax2 + bx + c (5)

a (5) v dng A2

= B2

:

(5) (x2 + m)2 = (2m + a)x2 + bx + c + m2

Trong , m l mt s cn tm.Tm m f(x) = (2m + a)x2 + bx + c + m2 c = 0. Khi , f(x) c dng bnh phng camt biu thc.

Nu 2m + a < 0 : (5) (x2 + m)2 + g2(x) = 0 (vi f(x) = g2(x))

x2 + m = 0

g(x) = 0

Nu2m + a > 0 : (5) (x2 + m)2 = g2(x) (vi f(x) = g2(x)) x2 + m = g(x)

x2 + m = g(x)


22/383

21

V d: Gii phng trnh: x4 + x2 6x + 1 = 0 (5.1)

Ta c:(5.1) x4 + 4x2 + 4 = 3x2 + 6x + 3 (x2 + 2)2 = 3(x + 1)2

x2 + 2 =

3(x + 1)

x2 + 2 =

3(x + 1)

x2 3x + 2 3 = 0x2 +

3 + 2 +

3 = 0

x =

3

4

3 52

x =

3 +

4

3 52


3

4

3 52

;

3 +

4

3 52

Bi tp t luynGii cc phng trnh sau:1. x4 19x2 10x + 8 = 02. x4 = 4x + 13. x4 = 8x + 74. 2x4 + 3x2 10x + 3 = 05. (x2 16)2 = 16x + 1

6. 3x4

2x2

16x 5 = 0Nhn xt: Phng trnh dng x4 = ax + b c gii theo cch tng t.Phng trnh = 0 l phng trnh bc ba vi cch gii c trnh by trc. Phngtrnh ny c th cho 3 nghim m, cn la chn m sao cho vic tnh ton l thun li nht. Tuynhin, d dng nghim m no th cng cho cng mt kt qu.

[6] Phng trnh dng af2(x) + bf(x)g(x) + cg2(x) = 0 (6)

Cch 1: Xt g(x) = 0, gii tm nghim v th li vo (6).

Trng hp g(x) = 0: a

f(x)

g(x)

2+ b.

f(x)

g(x)+ c = 0

t y =f(x)

g(x), gii phng trnh bc hai ay2 + by + c = 0 ri tm x.

Cch 2: t u = f(x), v = g(x), phng trnh tr thnh

au2 + buv + cv2 = 0 (6)Xem (6) l phng trnh bc hai theo n u, tham s v. T tnh u theo v.

V d: Gii phng trnh: 20(x 2)2 5(x + 1)2 + 48(x 2)(x + 1) = 0 (6.1)


23/383

22

t u = x 2, v = x + 1. Phng trnh (6.1) tr thnh:

20u2 + 48uv 5v2 = 0 (10u v)(2u + 5v) = 0

10u = v

2u = 5v

Vi 10u = v, ta c: 10(x

2) = x + 1

x =

7

3 Vi 2u = 5v, ta c: 2(x 2) = 5(x + 1) x = 1

7


7

3; 1

7

Nhn xt: Nu chn y =

f(x)

g(x)Vi f(x) v g(x) l hai hm s bt k (g(x) = 0), ta s to

c mt phng trnh. Khng ch l phng trnh hu t, m cn l phng trnh v t.

Bi tp t luynGii cc phng trnh sau:1. (x 5)4 12(x 2)4 + 4(x2 7x + 10)2 = 02. (x 2)4 + 3(x + 3)4 4(x2 + x 6)2 = 03. 4(x3 1) + 2(x2 + x + 1)2 4(x 1)2 = 04. 2(x2 x + 1)2 + 5(x + 1)2 + 14(x3 + 1) = 05. (x 10)4 15(x + 5)4 + 4(x2 5x 50)2 = 0

[7] Phng trnh bc bn tng qut ax4 + bx3 + cx2 + dx + e = 0 (7)

Phn tch cc hng t bc 4, 3, 2 thnh bnh phng ng, cc hng t cn li chuyn sang vphi:

(7) 4a2x4 + 4bax3 + 4cax2 + 4dax + 4ae = 0 (2ax2 + bx)2 = (b2 4ac)x2 4adx 4ae

Thm vo hai v mt biu thc 2(2ax2 + bx)y + y2 (y l hng s) v tri thnh bnh phngng, cn v phi l tam thc bc hai theo x:

f(x) = (b2 4ac 4ay)x2 + 2(by 2ad)x 4ae + y2

Tnh y sao cho v phi l mt bnh phng ng. Nh vy, ca v phi bng 0. Nh vy taphi gii phng trnh = 0. T ta c dng phng trnh A2 = B2 quen thuc.

V d: Gii phng trnh x4 16x3 + 66x2 16x 55 = 0 (7.1)

(7.1) x4

16x3

+ 64x2

= 2x2

+ 16x + 55 (x2 8x)2 + 2y(x2 8x) + y2 = (2y 2)x2 + (16 16y)x + 55 + y2

Gii phng trnh = 0 (8 8y)2 (55 + y2)(2y 2) = 0 tm c y = 1, y = 3, y = 29.Trong cc gi tr ny, ta thy gi tr y = 3 l thun li nht cho vic tnh ton.


24/383

23

Nh vy, chn y = 3, ta c phng trnh:

(x2 8x + 3)2 = 4(x 4)2

x2 8x + 3 = 2(x 4)x2 8x + 3 = 2(x 4)

x2 10x + 11 = 0x2 6x 5 = 0

x = 3 14x = 5 14

Phng trnh (7.1) c tp nghim S = 3 + 14;3 14; 5 + 14;5 14 Nhn xt: V d trn cho ta thy phng trnh = 0 c nhiu nghim. C th chn y = 1nhng t ta c phng trnh (x2 8x + 1)2 = 56 th khng thun li lm cho vic tnh ton,tuy nhin, kt qu vn nh nhau.Mt cch gii khc l t phng trnh x4 + ax3 + bx2 + cx + d = 0 t x = t a

4, ta s thu c

phng trnh khuyt bc ba theo t, ngha l bi ton quy v gii phng trnh t4 = at2 + bt + c.

Bi tp t luyn

1. x4 14x3 + 54x2 38x 11 = 02. x4 16x3 + 57x2 52x 35 = 03. x4 6x3 + 9x2 + 2x 7 = 04. x4 10x3 + 29x2 20x 8 = 05. 2x4 32x3 + 127x2 + 38x 243 = 0

PHNG TRNH DNG PHN THC[1] Phng trnh cha n mu c bn

t iu kin xc nh cho biu thc mu. Quy ng ri gii phng trnh.

V d: Gii phng trnh: 12 x +

x

2x 1 = 2 (1.1)

iu kin: x= 2; x

=

1

2.

(1.1) 2x 1 + x(2 x)(2 x)(2x 1) = 2 2x 1 + 2x x

2 = 2(4x 2 2x2 + x)

3x2 6x + 3 = 0 x = 1(tha iu kin)Vy phng trnh (1.1) c tp nghim S = {1}

[2] Phng trnh dng x2 + a2x2

(x + a)2= b (2)

Ta c:

(2) x ax(x + a)

2 + 2x. axx + a

= b

x2

x + a

2+ 2a.

x2

x + a+ a2 = b + a2


25/383

24

t y =x2

x + a. Gii phng trnh bc hai theo y tm x.

V d: Gii phng trnh: x2 + 9x2

(x + 3)2= 7 (2.1)

iu kin: x=

3.

(2.1)

x 3xx + 3

2+ 6.

x2

x + 3= 7

x2

x + 3

2+ 6.

x2

x + 3= 7

t y =x2

x + 3. Ta c phng trnh

y2 + 6y 7 = 0 y = 1

y = 7

Nu y = 1: Ta c phng trnh x2 = x + 3 x = 1

13

2 Nu y = 7: Ta c phng trnh x2 + 7x + 21 = 0 (v nghim)Vy phng trnh (2.1) c tp nghim: S =

1 +

13

2;

1 132

Nhn xt: Da vo cch gii trn, ta c th khng cn phi t n ph m thm bt hngs to dng phng trnh quen thuc A2 = B2

Bi tp t luyn

Gii cc phng trnh sau:

1. x2 +4x2

(x + 2)2= 12

2. x2 +25x2

(x + 5)

2= 11

3. x2 +9x2

(x 3)2 = 14

4.25

x2 49

(x 7)2 = 1

5.9

4(x + 4)2+ 1 =

8

(2x + 5)2

[3] Phng trnh dng x2 + nx + a

x2 + mx + a+

x2 + qx + a

x2 +px + a= b (3)

iu kin:x2 + mx + a = 0x2 +px + a = 0

Xt xem x = 0 c phi l nghim phng trnh khng.


26/383

25

Trng hp x = 0:

(2) x +

a

x+ n

x +a

x+ m

+x +

a

x+ q

x +a

x+p

= b

t y = x +a

x. iu kin: |y| 2

|a| ta c phng trnh

y + ny + m

+ y + qy +p

= b

Gii phng trnh n y sau tm x.

V d: Gii phng trnh: x2 3x + 5

x2 4x + 5 x2 5x + 5x2 6x + 5 =

1

4(3.1)

iu kin: x = 1, x = 5.x = 0 khng phi l nghim ca phng trnh.

Xt x = 0 :

(3.1) x +

5

x 3

x +5

x 4

x +

5

x 5

x +5

x 6

= 14

t y = x +5

x |y| 25, y = 6. Phng trnh (3.1) tr thnh:

y 3y 4

y 5y 6

=

1

4 2

y2 10y + 24=

1

4 y2

10y + 16 = 0

y = 2 (loi)y = 8T ta c phng trnh

x +5

x= 8 x2 8x + 5 = 0 x = 4

11


4 +

11;4 11 Nhn xt: Cc dng phng trnh sau c gii mt cch tng t: Dng 1: mx

ax2 + bx + d+

nx

ax2 + cx + d= p

Dng 2: ax2 + mx + cax2 + nx + c

+ pxax2 + qx + c

= b

Bi tp t luyn

Gii cc phng trnh sau:1)

4x

4x2 8x + 7 +3x

4x2 10x + 7 = 1

2)2x

2x2 5x + 3 +13x

2x2 + x + 3= 6

3)3x

x2 3x + 1+

7x

x2 + x + 1=

4

4)x2 10x + 15x2 6x + 15 =

4x

x2 12x + 55)

x2 + 5x + 3

x2 7x + 3 +x2 + 4x + 3

x2 + 5x + 3= 7


27/383

26

Tng ktQua cc dng phng trnh trn, ta thy phng trnh hu t thng c gii bng mt trongcc phng php:[1.] a v phng trnh tch[2.] t n ph hon ton[3.] t n ph a v h phng trnh[4.] a v ly tha ng bc (thng l dng A2 = B2)[5.] Chia t v mu cho cng mt s[6.] Thm bt to thnh bnh phng ngTuy nhin, c mt s dng phng trnh c nhng phng php gii c trng. Nhng phngtrnh ny s c trnh by c th hn nhng phn khc.

Bi tp tng hp phn phng trnh hu t

Phn 1:1) x3 3x2 + 18x 36 = 02) 8x2 6x = 1

23) x3 4x2 4x + 8 = 04) x3 21x2 + 35x 7 = 05) x3 6x2 + 8 = 0Phn 2:1) 6x5 11x4 11x + 6 = 02) (x2

6x)2

2(x

3)2 = 81

3) x4 + (x 1)(3x2 + 2x 2) = 04) x4 + (x + 1)(5x2 6x 6) = 05) x5 + x2 + 2x + 2 = 06) (x2 16)2 = 16x + 17) (x + 2)2 + (x + 3)3 + (x + 4)4 = 2

8) x3 +1

x3= 13

x +

1

x

9)

x 1

x 2

+

x 1x

2

2

=40

9

10) x(3 x)x 1

x +

3 xx 1

= 15

11)1

x+

1

x + 2+

1

x + 5+

1

x + 7=

1

x + 1+

1

x + 3+

1

x + 4+

1

x + 6


28/383

27

XY DNG PHNG TRNH HU TBn cnh vic xy dng phng trnh t h phng trnh, vic xy dng phng trnh t nhngng thc i s c iu kin l mt trong nhng phng php gip ta to ra nhng dngphng trnh hay v l. Di y l mt s ng thc n gin.

4.1 T ng thc (a + b + c)3 = a3 + b3 + c3 + 3(a + b)(b + c)(c + a) (1) :

V d: Gii phng trnh: (x 2)3 + (2x 4)3 + (7 3x)3 = 1 (1.1)

Nhn xt: Nu t a = x 2, b = 2x 4, c = 7 3x. Khi ta c phng trnh: a3 + b3 + c3 =(a + b + c)3. T ng thc (1), d dng suy ra (a + b)(b + c)(c + a) = 0. T , ta c li gii:(1) (x 2)3 + (2x 4)3 + (7 3x)3 = [(x 2) + (2x 4) + (7 3x)]3

(3x 6)(3 x)(5 2x) = 0 x = 2x = 3

x =5

2


2;3;5

2

Vi bi ton trn, cch t nhin nht c l l khai trin ri thu v phng trnh bc ba. Tuynhin, vic khai trin c th khng cn hiu qu vi bi ton sau:V d: Gii phng trnh: (x2 4x + 1)3 + (8x x2 + 4)3 + (x 5)3 = 125x3 (1.2)

4.2 T mnh 1a +1b

+ 1c

= 1a + b + c

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

V d: Gii phng trnh: 1x 8 +

1

2x + 7+

1

5x + 8=

1

8x + 7(2.1)

iu kin: x = 8, x = 72

, x = 85

, x = 78

T bi ton (2), ta c:

(2.1) (x 8 + 2x + 7)(x 8 + 5x + 8)(2x + 7 + 5x + 8) = 0 x =

1

3x = 0

x = 157

Phng trnh c tp nghim: S =

1

3; 0; 15

7

4.3 T ng thc a3 + b3 + c3 3abc = (a + b + c)(a2 + b2 + c2 ab bc ca) (3) :

V d: Gii phng trnh 54x3 9x + 2 = 0 (3.1)Ta tm cch vit v tri ca phng trnh di dng x3 + a3 + b3 3abx. Nh vy th a, b l

nghim ca h phng trnh a3 + b3 =

2

54

a3.b3 =1

183


29/383

28

a3, b3 l nghim ca phng trnh

t2

2

54t +

1

182= 0 t = 1

54

2 a = b = 1

3

2

Khi phng trnh cho tng ng vi

(x + a + b) x2 + a2 + b2 a x bx ab = 0

x +2

3

2

x2 2

3

2x +

1

18

= 0

x =2

3

2

x =1

3

2

Vy (3.1) c tp nghim S =

23

2;

1

3

2

4.4 T bi ton Nu xyz = 1 v x + y + z = 1x + 1y + 1z th (x 1)(y 1)(z 1) = 0 (4) :

V d: Gii phng trnh: 110x2 11x + 3 =

1

2x 1 +1

5x 3 + 10x2 18x + 7 (4.1)

iu kin: x = 35

, x = 12

Nhn xt: Nu t a = 2x 1, b = 5x 3, c = 110x2 11x + 3

Th ta c: abc = 1 v a + b + c =1

a+

1

b+

1

cT (3.1) (2x 1 1)(5x 3 1)

1

10x2 11x + 3 1

= 0

x = 1

x =4

510x2 11x + 2 = 0

x = 1

x =4

5

x =11 41

20


1;

4

5;

11 4120

Qua cc v d trn, ta c th hnh dung c bn vic s dng ng thc xy dng phngtrnh. Hi vng da vo vn hiu bit v kh nng sng to ca mnh, bn c c th to ranhng phng trnh p mt v c o hn na. Sau y l mt s bi tp t luyn t ccng thc khc.

Bi tp t luyn

1. (x 2)6 + (x2 5x + 4)3 = (2x2 + 3)3 + (5 9x)32. (x2

2x + 3)5 + (8x

x2 + 7)5 = (9x + 5)5 + (5

3x)5

3. (x35x+4)2(127x5x2)+(x3+2x2 +7)(7x2 +12x9) = (x3+7x2+7x5)2(2x2+5x+3)4. (x 5)4(4x x2) + (2x + 7)4(x2 3x + 12) = (x2 2x + 7)4(12 + x)5.

(x2 5x + 3)3(4 9x)(x2 12x + 7) +

(x2 + 4x 1)3(9x 4)(x2 3x + 3) +

(7x 4)3(12x 7 x2)(x2 + 3x 3) = 7


30/383

29

6.1

x2 + 4x + 3+

1

x2 + 8x + 15+

1

x2 + 12x + 35+

1

x2 + 16x + 63= 5

7. (x2 8x + 5)7 + (7x 8)7 = (x2 x 3)7

MT S PHNG TRNH BC CAONh ton hc Abel chng minh rng khng c cng thc nghim tng qut cho phngtrnh bc cao (> 4). y cng khng phi l dng ton quen thuc ph thng. V th bivit ny ch cp n mt s phng trnh bc cao c bit, c th gii bng bin i s cp.

Bi 1: Gii phng trnh x5 x4 x3 11x2 + 25x 14 = 0

GiiPhng trnh cho tng ng

(x5 2x4) + (x4 2x3) + (x3 2x2) + (9x2 + 18x) + (7x 14) = 0(x 2)(x4 + x3 + x2 9x + 7) = 0 x = 2 x4 + x3 + x2 9x + 7 = 0 ()

Xt (*) ta c(

)

(x4 + x3 + x2

9x + 6) + 1 = 0

(x4 x3 + 2x3 2x2 + 3x2 3x 6x + 6) + 1 = 0 (x 1)2(x3 + 3x + 6) + 1 = 0 (v nghim)

Vy phng trnh c tp nghim S = {2} 2

Bi 2: Gii phng trnh x6 7x2 + 6 = 0 ()

Gii

R rng ta khng th on nghim ca phng trnh ny v bc cao v h s xu. Mt cch tnhin ta t

6 = a. Lu rng ta hi vng c th a (*) v phng trnh bc hai theo a, do

ta phn tch 7 = a2 + 1. Cng vic cn li l gii phng trnh ny:t

6 = a, khi

() x6 x2(a2 + 1) + a = 0 a2x2 a + x2 x6 = 0 ()

() c: = 1 + 4x2

(x6

x2

) = (2x4

1)2

nn c nghim:a1 = x2

a2 =1 x4

x2


31/383

30

Vy ()

x2 =

6

1 x4 =

6x2

x = 4

6

x =

5

2

3

2

Vy phng trnh c tp nghim S = 4

6;

5

2 3

2 . 2Bi 3: Gii phng trnh x6 15x2 + 68 = 0 ()

Gii

Do x = 0 khng tho (*) nn x = 0. Vit li (*) di dng

x3 +

68

x3=

15

x x3 + 2

17

x3=

17 2x

t a =

17 > 0 ta c phng trnh

x3 +2a

x3=

a2 2x

x2a2 2a x6 2x2 = 0 ()

Coi (**) l phng trnh n a ta tm c nghim a = x2 (loi do a > 0)a =

2 + x4

x2

Vy ta c

17 =2 + x4

x2 x4

17x2 + 2 = 0

x2 =

17 3

2 x =

17 32

Kt lun: (*) c tp nghim S =

17 32

2

Bi 4: Chng minh phng trnh x5 5x4 + 30x3 50x2 + 55x 21 = 0 c nghim duynht

x = 1 +5

2 5

4 +5

8 5

16

Gii

t f(x) = x5 5x4 + 30x3 50x2 + 55x 21.Ta c f(x) = 5x4 20x3 + 90x2 100x + 55 = 5(x2 2x + 3)2 + 10(2x 1)2 > 0 x, do phng trnh f(x) = 0 c khng qu 1 nghim.Ta s chng minh nghim l x = 1 + 5

2 54 + 58 516:

tx = 1 +

5

2 5

4 +5

8 5

16

5

2x =5

2 +5

4 5

8 +5

16 2 x + 5

2x = 2

5

2 1 x + 1 = 5

2(2 x) (x + 1)5 = 2(2 x)5

Khai trin biu thc trn, sau rt gn, ta c phng trnh:

x5 5x4 + 30x3 50x2 + 55x 21 = 0


32/383

31

Vy ta c iu phi chng minh. 2

Bi 5: Chng minh phng trnh sau c 7 nghim thc:

g(x) = x9 9x7 + 3x6 + 27x5 18x4 27x3 + 27x2 1 = 0 ()

Gii

t f(x) = x3 3x + 1 th g(x) = f(f(x)). Ta s tm nghim ca g(x) = f(x).f(f(x)): Nghim ca f(x) l: f(x) = 0 3x2 3x = 0 x = 1. tm nghim ca f(f(x)) ta tm nghim ca f(x) = 1 v f(x) = 1: f(x) = 1 x3 3x + 2 = 0 x {2; 1} f(x) = 1 x3 3x = 0 x {0; 3}Nh vy tp nghim ca phng trnh g(x) = 0 l {2; 3; 1;0;1; 3}.

Suy ra g(x) c ti a 7 nghim. Li c:

g(x) khi x g(2) = 3 > 0 g(x) c 1 nghim trong (; 2)g(3) = 1 < 0 g(x) c 1 nghim trong (2; 3)g(1) = 19 > 0 g(x) c 1 nghim trong (3; 1)g(0) = 1 < 0 g(x) c 1 nghim trong (1;0)g(1) = 3 > 0 g(x) c 1 nghim trong (0; 1)g(3) = 1 < 0 g(x) c 1 nghim trong (1; 3)g(x) + khi x + g(x) c 1 nghim trong (3; +)

Nh vy g(x) = f(f(x)) c 7 nghim thc. 2


33/383

C II: PHNG TRNH, H PHNGTRNH C THAM S

PHNG PHP S DNG O HM

L thuyti vi bi ton tm iu kin ca tham s phng trnh f(x) = g(m) c nghim min D tada vo tnh cht: phng trnh c nghim khi v ch khi hai th ca hai hm s y = f(x)v y = g(m) ct nhau. Do bi ton ny ta tin hnh theo cc bc sau:Bc 1: Lp bng bin thin ca hm s y = f(x) .Bc 2: Da vo bng bin thin ta xc nh m ng thng y = g(m) ct th hm sy = f(x).Ch : Nu hm s y = f(x) lin tc trn D v m = Min

Df(x) , M = Max

Df(x) th phng

trnh: k = f(x) c nghim m k MSau y l mt s bi tp v d:

Bi tp v d

Bi 1:Tm m phng trnh

x2 + x + 1 x2 x + 1 = m c nghim.

Gii

Xt hm s f(x) =

x2 + x + 1 x2 x + 1 vi x R cf(x) =

2x + 1

2

x2 + x + 1 2x 1

2

x2 x + 1Ta s tm nghim ca f(x):

f(x) = 0 2x + 12

x2 + x + 1 2x 1

2

x2 x + 1 = 0

(2x + 1)

x2 x + 1 = (2x 1)

x2 + x + 1

x +1

2

2 x 1

2

2+

3

4

=(x 1

2)2

x +1

2

2+

3

4

x +1

2

2=

x 1

2

2 x = 0

Th li ta thy x = 0 khng l nghim ca f(x). Suy ra f(x) khng i du trn R, mf(0) = 1 > 0 f(x) > 0x R . Vy hm s f(x) ng bin trn R.Mt khc: lim

x+f(x) = lim

x+

2xx2 + x + 1 +

x2 x + 1 = 1 v limx f(x) = 1.

32


34/383

33

Da vo bng bin thin ta thy phng trnh cho c nghim khi v ch khi m (1;1) 2

Bi 2: Tm m phng trnh 4

x2 + 1 x = m c nghim

Gii

KX: x 0Xt hm s f(x) = 4x2 + 1 x vi x 0 ta c f(x) lin tc trn [0;+)Li c: f(x) =

x

2 4

(x2 + 1)3 1

2

x 0.

Suy ra hm s f(x) nghch bin trn [0;+)Mt khc: lim

x+f(x) = 0

Do phng trnh cho c nghim khi v ch khi m (0;1] 2 . Nhn xt: i khi ta phi tm cch c lp m a phng trnh v dng trn.

Bi 3: Tm m phng trnh sau c nghim:4x4 13x + m + x 1 = 0 (*)

Gii

Ta c:

() 4

x4 13x + m + x1 = 0

1 x 0x4 13x + m = (1 x)4

1 x

4x3 6x2 9x = 1 m

Xt hm s f(x) = 4x3 6x2 9x vi x 1

Ta c: f(x) = 12x2 12x 9, f(x) = 0 x = 32

x = 12

.

Da vo bng bin thin suy ra phng trnh c nghim khi v ch khi m 32

. 2

Bi 4: Tm m phng trnh sau c nghim: x

x +

x + 12 = m(

5 x + 4 x)

Gii

KX: x [0;4].Khi phng trnh tng ng vi

(x

x +

x + 12)(

5 x 4 x) = m

Xt hm s f(x) = (x

x +

x + 12)(

5 x 4 x) lin tc trn on [0; 4].Ta c: f(x) =

3

2

x +

1

2

x + 12

1

2

4 x 1

2

5 x

> 0 x [0; 4]Vy f(x) l hm ng bin trn [0;4].Suy ra phng trnh c nghim khi v ch khi 2

3(

5

2) m 12 2 .

Bi 5: Tm m h sau c nghim: ()

2x2

1

2

45x3x2 mxx + 16 = 0


35/383

34

Gii

Ta c:

()

x2 4 + 5x3x2 mxx + 16 = 0

x [1; 4]3x2 mxx + 16 = 0

m = 3x2 + 16

x

x

Xt f(x) =3x2 + 16

x

xvi x [1; 4]. Ta c:

f(x) =6x2

x 3

2

x(3x2 + 16)

x3=

3

x(x2 16)2x3

0x [1; 4]

Nh vy m = f(x) nghch bin trn [1; 4], do f(4) m f(1) 8 m 19Vy h c nghim khi v ch khi m [8; 19] 2

Nhn xt: Khi gp h phng trnh trong mt phng trnh ca h khng cha tham sth ta s i gii quyt phng trnh ny trc. T phng trnh ny ta s tm c tp nghim(i vi h mt n) hoc s rt c n ny qua n kia. Khi nghim ca h ph thuc vonghim ca phng trnh th hai vi kt qu ta tm c trn.

Bi 6: Tm m h sau c nghim:

72x+x+1 72+

x+1 + 2007x 2007

x2 (m + 2)x + 2m + 3 = 0

GiiTa c:

72x+x+1 72+

x+1 + 2007x 2007 72+

x+1(72x+2 1) 2007(1 x) ()

Nu x > 1 72+x+1(72x2 1) > 0 > 2007(1 x).Suy ra (*) v nghim.

Nu x 1 72+x+1(72x2 1) 0 2007(1 x). Suy ra (*) ng .

Suy ra h c nghim khi v ch khi phng trnh x2 (m + 2)x + 2m + 3 = 0 c nghim vi

x [1;1], hay phng trnh m =x2

2x + 3

x 2 c nghim vi x [1;1].Xt hm s f(x) =

x2 2x + 3x 2 vi x [1;1], c

f(x) =x2 4x + 1

(x 2)2 = 0 x = 2

3

Da vo bng bin thin suy ra h c nghim khi v ch khi m 2 3.

Bi 7: Tm m h phng trnh sau c nghim: () x y + m = 0y +

xy = 2

Gii


36/383

35

KX: xy 0. T h ta cng c y = 0.

()

x y + m = 0

xy = 2 y

x y + m = 0

x =y2 4y + 4

y

y 2

m = y y2 4y + 4

y=

4y 4y

(y 2)

Xt hm s f(y) =4y

4

y (y 2) ta c f(y) =

4

y2 > 0 y = 0, suy ra hm s f(y) ng bintrn cc khong (; 0) v (0; 2].Mt khc, lim

yf(y) = 4, lim

y0+f(y) = ; lim

y0f(y) = +

Suy ra h c nghim khi v ch khi m (; 2] (4;+) 2

Bi 8: Tm m phng trnh c ng hai nghim phn bit:

x4 4x3 + 16x + m + 4

x4 4x3 + 16x + m = 6 ()

Gii

Ta c:() 4

x4 4x3 + 16x + m = 2 m = x4 + 4x3 16x + 16

Xt hm s f(x) = x4 + 4x3 16x + 16 vi x R. Ta c:

f(x) = 4x3 + 12x2 16; f(x) = 0

x = 1x = 2

Da vo bng bin thin suy ra phng trnh c hai nghim phn bit khi v ch khi m < 27.

Bi 9: Tm m phng trnh m

x2 + 2 = x + m () c ba nghim phn bit.Gii

T (*) ta c: () m = xx2 + 2 1

Xt hm s: f(x) =x

x2 + 2

1

vi x R.

Ta c: f(x) = 2 x2 + 2x2 + 2(

x2 + 2 1)2

; f(x) = 0 x = 2 .

Da vo bng bin thin phng trnh c nghim khi v ch khi 2 < m < 2 2

Bi 10: Tm m phng trnh: mx2 + 1 = cos x () c ng mt nghim x

0;

2

Gii

Ta thy (*) c nghim th m 0. Khi

mx2 + 1 = cos x m = cos x 1x2

2m =sin2

x

2x2

4


37/383

36

Xt hm s f(t) =sin2t

t2vi t

0;

4

Ta c

f(t) =2t2 sin t cos t 2tsin2t

t4=

2sin t(t cos t sin t)t3

=sin2t(t tan t)

t3< 0 t

0;

4

Suy ra hm s f(t) nghch bin trn (0;

4).

Da vo bng bin thin suy ra phng trnh c ng mt nghim trn (0; 4 ) khi v ch khi82

< 2m < 1 12

< m < 422

Bi 11: Tm m h phng trnh c ba cp nghim phn bit:

()

3(x + 1)2 + y m = 0x +

xy = 1

Giiiu kin xy 0Ta c

()

3(x + 1)2 + y = m

xy = 1 x

3(x + 1)2 + y = m

y =x2 2x + 1

x

y 1

m = 3(x + 1)2 + x2 2x + 1

x

m 3 = 3x2

+ 6x +

x2

2x + 1

x

Xt hm s: f(x) = 3x2 + 6x +x2 2x + 1

x(x 1) ta c

f(x) =6x3 + 7x2 1

x2= 0

x = 1x =

12

x =1

3

Da vo bng bin thin ta thy h phng trnh c ba nghim phn bit khim [4; 15

4] [20

3;12] 2

Nhn xt: Khi t n ph ta phi tm min xc nh ca n ph v gii quyt bi ton nph trn min xc nh va tm. C th:* Khi t t = u(x)(x D), ta tm c t D1 v phng trnh f(x, m) = 0 (1) tr thnhg(t, m) = 0 (2). Khi (1) c nghim x D (2) c nghim t D1.* tm min xc nh ca t ta c th s dng cc phng trnh tm min gi tr (v min xcnh ca t chnh l min gi tr ca hm u(x)).* Nu bi ton yu cu xc nh s nghim th ta phi tm s tng ng gia x v t, tc l mi

gi tr t D1 th phng trnh t = u(x) c bao nhiu nghim x D.

Bi 12: Tm m phng trnh m(

x 2 + 2 4x2 4) x + 2 = 2 4x2 4 c nghim


38/383

37

Gii

KX; x 2Ta thy x = 2 khng l nghim ca phng trnh nn ta chia hai v phng trnh cho 4

x2 4:

m

4

x 2x + 2

+ 2

4

x + 2

x

2

= 2 ()

t t = 4x + 2

x 2(t > 1). Khi (*) tr thnh: m1

t+ 2

t = 2 m = t2 + 2t

2t + 1.

Xt hm s f(t) =t2 + 2t

2t + 1(t > 1) ta c f(t) =

2t2 + 2t + 2

(2t + 1)2> 0 t > 1.

Vy hm s f(t) ng bin trn (1;+).Vy phng trnh c nghim khi v ch khi m > 1 2 Nhn xt: Trong cc bi ton trn sau khi t n ph ta thng gp kh khn khi xc nhmin xc nh ca t . tm c iu kin ca n ph t, chng ta c th dng cng c hm

s, bt ng thc, lng gic ha. . .

Bi 13: Tm m phng trnh log23 x +

log23 x + 1 2m 1 = 0 c nghim trn

1; 33

Gii

t t =

log23 x + 1. Do x

1; 33

1 t 2Phng trnh cho tr thnh: t2 + t = 2m + 2Xt hm s f(t) = t2 + t vi 1 t 2, ta thy f(t) l hm ng bin trn [1; 2]

Vy phng trnh c nghim khi v ch khi 2 2m + 2 6 0 m 2 2Bi 14: Xc nh m h sau c 2 nghim phn bit

log3(x + 1) log3(x 1) > log34log2(x

2 2x + 5) mlog(x22x+5)2 = 5

Gii

iu kin : x > 1. T bt phng trnh th nht ca h ta c: log3

x + 1

x 1 > log32 x (1; 3).t t = log2(x2 2x + 5)(t (2;3))v phng trnh th hai ca h tr thnht +

m

t= 5 t2 5t = m

T cch t t ta c: Vi mi gi tr t (2; 3) th cho ta ng mt gi tr x (1; 3). Suy ra h c2 nghim phn bit khi v ch khi phng trnh t2 5t = m c 2 nghim phn bit t (2; 3).Xt hm s f(t) = t2 5t vi t (2; 3). Ta c f(t) = 2t 5; f(t) = 0 t = 5

2. Da vo bng

bin thin ta c, h phng trnh c 2 nghim phn bit khi v ch khi m (6; 254

) 2

Bi 15:( thi H khi B - 2004) Tm m phng trnh c nghim:

m(

1 + x2

1 x2 + 2) = 2

1 x4 +

1 + x2

1 x2 ()


39/383

38

Gii

iu kin: 1 x 1Trc tin, ta nhn thy rng: (1 x2) (1 + x2) = 1 x4 v 1 x2 + 1 + x2 = 2 nn ta c phpt n ph nh sau:t t =

1 + x2 1 x2

Phng trnh cho tr thnh:

m (t + 2) = t2 + t + 2 t2 + t + 2

t + 2= m (1)

Do

1 + x2

1 x2 t 0Mt khc: t2 = 2 21 x4 2 t 2Ta xt hm s: f(t) =

t2 + t + 2t + 2

, t 0; 2, ta c:f(t) = t2 4t(t + 2)2

0

Vy hm f(t) nghch bin trn on

0;

2

. M hm s lin tc trn

0;

2

nn phng trnh

cho c nghim x khi phng trnh (1) c nghim t 0; 2.iu ny tng ng vi:min f(t) m max f(t) t

0;

2

f

2 m f(0)

Vy cc gi tr m cn tm l

2 1 m 1 2.

Bi 16: Tm m phng trnh sau c nghim:

xx + x + 12 = m 5 x + 4 xGii

Cng ging nh nhng bi ton trc, bi ny ta ngh ngay l phi a bi ton v dngf(x) = m ri s dng tng giao gia 2 th v suy ra iu kin m.Ta gii bi ton nh sau:iu kin: 0 x 4

Phng trnh cho tng ng vi:x

x +

x + 125

x +

4

x

= m

t f(x) =xx + x + 12

5 x + 4 x x [0; 4]Tuy nhin, nu n y ta kho st hm s ny th c v kh phc tp v di dng. V vy tas gii quyt theo mt hng khc. rng, ta c tnh cht sau:

Vi hm s y =f(x)

g (x). Nu y = f(x) ng bin v y = g (x) nghch bin th y =

f(x)

g (x)ng

bin.Ta vn dng tnh cht trn nh sau:Xt hm s g (x) = x

x +

x + 12 ta c:

g(x) = 32x + 1

2

x + 12> 0 nn hm g(x) l ng bin.

Xt hm s h(x) =

5 x + 4 x ta c:h(x) = 1

2

5 x 1

2

4 x < 0 nn hm h(x)l nghch bin


40/383

39

Vy, theo tnh cht trn ta c hm y =g (x)

h (x)ng bin x [0; 4].

Do phng trnh f(x) = m c nghim khi v ch khif(0) m f(4) 2 15 12 m 12 2.Bi 17: Gii v bin lun phng trnh sau theo m:

x2 2x + m2 = |x 1| m (1)

Gii

Xt x 1 ta c:

(1)

x2 2x + m2 = x 1 m

x 1 m 0x2 2x + m2 = (x 1 m)2

x 1 + m

2mx = 2m + 1

Nu m = 0: h trn v nghim Nu m = 0 x = 2m + 1

2m

Ta c x 1 + m 2m + 12m

1 + m 2m2 + 1

2m 0 m

2

2 0 < m 0

Kt hp iu kin trn ta c 0 < m 0, x nn chia hai v ca phng trnh cho x2 x + 1 ta c:x2 + x + 1

x2 x + 12

x2 + x + 1

x2 x + 1 = m


41/383

40

t t =x2 + x + 1

x2 x + 1 ,1

3 t 3

Phng trnh trn tr thnh: t2 t = my l mt phng trnh bc hai n gin nn vic kho st xin dnh cho bn c.iu lu y l iu kin ca t. Thc cht y ta tm gi tr ln nht v nh nht

ca biu thcx2 + x + 1

x

2

x+ 1

(c th dng phng php min gi tr).

Bi 19: Tm m h sau c nghim thc:x +

1

x+ y +

1

y= 5

x3 +1

x3+ y3 +

1

y3= 15m 10

Gii

Cch 1:

Nhn vo h ny, ta thy ngay c hng i l phi t n ph v gia cc i lng x + 1xv x3 +

1

x3dng nh c mi lin h vi nhau. Vi tng , ta c php t nh sau:

t

a = x +

1

x

b = y +1

y

, (|a| 2, |b| 2).

Ta c

x3 +1

x3=

x +

1

x

3 3

x +1

x

= a3 3a

y3

+

1

y3 = y + 1y3

3y + 1y = b3 3bKhi , h cho tr thnh:

a + b = 5

a3 + b3 3 (a + b) = 15m 10

a + b = 5

ab = 8 m

D thy a, b l nghim ca phng trnh X2 5X + 8 m = 0 X2 5x + 8 = m (1) Xthm s f(X) = X2 5X+ 8, |X| 2, ta c:f (X) = 2X 5 f (X) = 0 X = 5

2

T , k bng bin thin v ch rng h cho c nghim thc khi v ch khi phng trnh(1) c nghim |X| 2. Ta tm c: 7

2 m 2 m 22.

Cch 2:Ta nhn thy rng phng trnh th nht khng c cha tham s nn ta s xut pht tphng trnh ny. Khai trin phng trnh ny ra, ta c:

x3 y3 + 4 (x y) = 9y + 8 3x2 + 6y2 x3 + 3x2 + 4x = y3 + 6y2 + 13y + 8

(x + 1)3

+ (x + 1) = (y + 2)3

+ (y + 2)

Xt hm s f(t) = t3 + t, d thy l hm s ny ng bin nn

f(x + 1) = f(y + 2) x + 1 = y + 2 x = y + 1


42/383

41

T y, thay x = y + 1 vo phng trnh th hai ta c:

15 + 2y y2 = 2m +

4 y2

(5 y) (y + 3)

4 y2 = 2m

n y tng r, ta ch cn chuyn v tng giao gia hai th.

Bi 20: Tm m h sau c nghim thc: x3 + (y + 2) x2 + 2xy = 2m 3x2 + 3x + y = m

Gii

tng: h ny ta quan st thy bi ton cn cha r ng li no v c hai phngtrnh trong h u cha n tham s m. V vy i n hng gii quyt tt ta nn bt uphn tch hai v tri trong hai phng trnh trong h. C th ta c:

x3 + (y + 2) x2 + 2xy = x3 + yx2 + 2x2 + 2xy = x2 (x + y) + 2x (x + y) = (x + y)

x2 + 2x

Mt khc:

x2 + 3x + y = x2 + 2x + x + y

R rng bc phn tch ny ta tm ra li gii cho bi ton ny chnh l t n ph. Li gii:

t: a = x2 + 2x 1

b = x + y

ta c h phng trnh

a + b = m

ab = 2m 3

a2 3 = (a + 2) m (1)b = m a

T phng trnh (1) trong h ta c:a2 3a + 2

= m (2)

H cho c nghim khi v ch khi phng trnh (2) c nghim a 1.Xt hm s: f(x) =

x2 3

x + 2

vi x

1

n y ta ch cn lp bng bn thin. Cng vic tip theo xin dnh cho bn c.

Bi tp t luyn

Bi 1: Tm m phng trnh tan2x + cot2x + m(cot x + tan x) = 3 c nghimBi 2: Tm m phng trnh

x +

x + 9 = 9x x2 + m c nghimBi 3: Tm m phng trnh

3 + x +

x + 6 18 + 3x x2 = m c nghimBi 4: Tm m phng trnh x3 4mx2 + 8 = 0 c 3 nghim phn bit.

Bi 5: Tm m phng trnh x3 + 3x2 + (3 2m) x + m + 1 = 0 c ng mt nghim lnhn 1.Bi 6: Tm m phng trnh sau c ng 2 nghim thc phn bit:4x2 2mx + 1 = 38x3 + 2x


43/383

42

PHNG PHP DNG NH L LAGRANGE-ROLLE

L thuyt

nh l Rolle: : Nu f(x) l hm lin tc trn on [a; b], c o hm trn khong (a; b)

v f(a) = f(b) th tn ti c (a; b) sao cho f(c) = 0.T ta c 3 h qu: H qu 1: Nu hm s f(x) c o hm trn (a; b) v f(x) c n nghim (n l s nguyndng ln hn 1) trn (a; b) th f(x) c t nht n - 1 nghim trn (a; b). H qu 2: Nu hm s f(x) c o hm trn (a; b) v f(x) v nghim trn (a; b) th f(x)cnhiu nht 1 nghim trn (a; b). H qu 3: Nu f(x)c o hm trn (a; b) v f(x) c nhiu nht n nghim (n l s nguyndng) trn (a; b) th f(x) c nhiu nht n + 1 nghim trn (a; b).Cc h qu trn vn ng nu cc nghim l nghim bi (khi f(x) l a thc) v cho ta

tng v vic chng minh tn ti nghim cng nh xc nh s nghim ca phng trnh, vnu nh bng mt cch no ta tm c tt c cc nghim ca phng trnh th ngha lkhi phng trnh c gii.

nh l Lagrange: : Nu f(x)l hm lin tc trn on [a; b], c o hm trn khong

(a; b)th tn ti c (a; b) sao cho f(c) = f(b) f(a)b a .

Sau y l mt s ng dng ca hai nh l trn:

Bi tp v d

Dng nh l Lagrange - Rolle bin lun phng trnh:

Bi 1: Chng minh rng phng trnh a cosx + b cos2x + c cos3x = 0 lun c nghim vimi a,b,c.

Gii

Xt f(x) = a sin x +b sin2x

2 +c sin3x

3ta c f(x) = a cosx + b cos2x + c cos3xM f(0) = f() = 0 x0 (0; ) : f(x0) = 0, suy ra iu phi chng minh. 2

Bi 2: Cho a + b + c = 0. Chng minh rng phng trnh sau c t nht 4 nghim thuc[0; ]: a sin x + 9b sin3x + 25c sin5x = 0

Gii

chng minh f(x) c t nht n nghim ta chng minh F(x) c t nht n + 1 nghim vi F(x)

l mt nguyn hm ca f(x) trn (a; b) (c th phi p dng nhiu ln)Xt hm s f(x) = a sin x b sin3x c sin5x, ta c:

f(x) = a cos x 3b cos3x 5c cos5x; f(x) = a sin x + 9b sin3x + 25c sin5x


44/383

43

Ta c f(0) = f(

4) = f(

3

4) = f() = 0

Suy ra x1 (0; 4

), x2 ( 4

;3

4), x3 ( 3

4; ) : f(0) = f(x1) =

f(x2) = f(x3) = 0

x4 (x1; x2), x5 (x2; x3)|f(x4) = f(x5) = 0m f(0) = f() = 0, suy ra iu phi chng minh. 2

Bi 3: Chng minh phng trnh x5 5x + 1 = 0 c ng ba nghim thc.Gii

t f(x) = x5 5x + 1 th f(2) = 21 < 0, f(0) = 1 > 0, f(1) = 3 < 0, f(2) = 23 > 0nn t y suy ra phng trnh cho c ba nghim thcGi s c nghim th t ca phng trnh. p dng nh l Rolle ta c:f(x) = 0 5x4 5 = 0. Phng trnh ny c hai nghim thc nn suy ra mu thun.Vy ta c iu phi chng minh 2.

Bi 4: Cho a thc P(x) v Q(x) = aP(x) + bP(x) trong a, b l cc s thc, a =0.

Chng minh rng nu Q(x) v nghim th P(x) v nghim.

Gii

Ta c deg P(x) = deg Q(x)V Q(x) v nghim nn deg Q(x) chn. Gi sP(x) c nghim, v deg P(x) chn nn P(x) ct nht 2 nghim. Khi P(x) c nghim kp x = x0 ta c x0 cng l mt nghim ca P(x) suy ra Q(x) c

nghim. Khi P(x) c hai nghim phn bit x1 < x2: Nu b = 0 th hin nhin Q(x) c nghim. Nu b =0 : Xt f(x) = eab xP(x) th f(x)c hai nghim phn bit x1 < x2 v

f(x) =a

bea

bxP(x) + e

a

bxP(x) =

1

bea

bx[aP(x) + bP(x)] =

1

bea

bxQ(x)

V f(x) c hai nghim suy ra f(x) c t nht 1 nghim hay Q(x) c nghim.Tt c trng hp u mu thun vi gi thit Q(x) v nghim. Vy khi Q(x) v nghim th

P(x) v nghim2

Bi 5: Gi s phng trnh sau c n nghim phn bit:a0x

n + a1xn1 + ... + an1x + an = 0, (a0 = 0)

Chng minh (n 1)a21 > 2na0a2Gii

t a0xn + a1xn1 + ... + an1x + an = f(x)Nhn xt f kh vi v hn trn R nn suy ra

f(x) c n 1 nghim phn bit.f(x) c n 2 nghim phn bit.. . . . .f[n2](x) =

n!

2a0x

2 + (n 1)!a1x + (n 2)!a2 c 2 nghim phn bit.


45/383

44

Nhn thy f[n2](x) c > 0 nn ((n 1)!a1)2 2n!a0(n 2)!a2 > 0Suy ra iu phi chng minh. 2

Bi 6: (VMO 2002) Xt phng trnhni=1

1

i2x 1 =1

2, vi n l s nguyn dng. Chng

minh rng vi mi s nguyn dng n, phng trnh nu trn c mt nghim duy nht ln

hn 1; k hiu nghim l xn

Gii

Xt fn(x) =ni=1

1

i2x 1 1

2, ta c: fn(x) lin tc v nghch bin trn (1;+)

M limx1+

fn(x) = +, limx+

fn(x) = 12

fn(x) = 0 c mt nghim duy nht ln hn 1. 2

Dng nh l Lagrange -Rolle gii phng trnh:

Bi 7: Gii phng trnh 3x + 5x = 2.4x ()Gii

Nhn xt: x = 0; x = 1 l nghim ca phng trnh (*).Gi x0 l nghim khc ca phng trnh cho. Ta c:

3x0 + 5x0 = 2.4x0 5x0 4x0 = 4x0 3x0 ()

Xt hm s f(t) = (t + 1)x0 tx0, ta c () f(4) = f(3)V f(t) lin tc trn [3; 4] v c o hm trong khong (3; 4), do theo nh l Rolle tn tic (3; 4) sao cho

f(c) = 0 x0[(c + 1)x01 cx01]=0

x0 = 0

x0 = 1(loi)

Vy phng trnh (*) c tp nghim S = {0; 1}. 2

Bi 8: Gii phng trnh 5x 3x = 2x ()Gii

Nhn xt: x = 0; x = 1 l nghim ca phng trnh (2). Gi x0 l nghim khc ca phngtrnh cho, ta c:

5x0 5x0 = 3x0 3x0 (2a)Xt hm s: f(t) = tx0 tx0, khi : (2a) f(5) = f(3)V f(t) lin tc trn [3;5] v c o hm trn (3; 5), do theo nh l Lagrange lun tn tic (3; 5) sao cho

f(c) = 0 x0(cx01 1)=0 x0 = 0

x0 = 1

(loi)

Vy phng trnh (*) c tp nghim S = {0; 1} 2

Bi 9: Gii phng trnh: 3x + 2.4x = 19x + 3 ()


46/383

45

Gii

Ta c() 3x + 2.4x 19x 3 = 0

Xt hm s: y = f(x) = 3x + 2.4x 19x 3 ta c: f(x) = 3x ln 3 + 2.4x ln 4 19 ta cf(x) = 3x(ln 3)2 + 2.4x(ln 4)2 > 0,

x

R hay f(x) v nghim, suy ra f(x) c nhiu nht 1

nghim, suy ra f(x) c nhiu nht 2 nghim.M f(0) = f(2) = 0 do (*) c ng hai nghim x = 0, x = 2 2

Bi 10: Gii phng trnh: (1 + cos x)(2 + 4cosx) = 3.4cos x ()

Gii

t t = cos x, (t [1; 1])

Ta c: () (1 + t)(2 + 4t) = 3.4t (1 + t)(2 + 4t) 3.4t = 0Xt hm s: f(t) = (1 + t)(2 + 4t) 3.4t ta c:f(t) = 2 + 4t + (t 2)4t ln 4, f(t) = 2.4t ln 4 + (t 2)4tln24Li c: f(t) = 0 t = 2 + 2

ln 4, suy ra f(t) c nghim duy nht.

Suy ra f(t) c nhiu nht hai nghim, ngha l f(t) c nhiu nht ba nghim.

Mt khc d thyf(0) = f(1

2) = f(1) = 0, do f(t) c ba nghim t = 0,

1

2, 1.

Kt lun: Nghim ca phng trnh (*) l: x =

2+ k2, x =

3+ k2, x = k2, k Z 2

Bi 11: Gii phng trnh 3cosx 2cosx = 2cosx 2cos x ()Gii

Xt hm f(t) = tcos t cos , f(t) = cos (tcos1 1)Ta nhn thy f(3) = f(2) v f(x) kh vi trn [2;3] nn p dng nh l Lagrange ta c:

c [2; 3] : f(c) = f(3) f(2)1

cos (ccos1 1) = 0

T ta suy ra nghim ca phng trnh (*) l x =

2+ k,x = k2(k

Z) 2


47/383

46

PHNG PHP DNG IU KIN CN V

L thuyt

Bi ton:

Cho h phng trnh (hoc h bt phng trnh) cha tham s c dng:

(I)

f(x, m) = 0

x Dxm Dm

hoc (II)

f(x, m) 0

x Dxm Dm

Trong x l bin s, m l tham s, Dx, Dm l min xc nh ca x v m.Yu cu t ra: ta phi tm gi tr ca tham s m h (I) hc (II) tha mn mt tnh chtno .

Phng php gii:

Bc 1 (iu kin cn): Gi s h tha mn tnh cht P no m u bi i hi. Khi, da vo c th ca tnh cht P v dng ca phng trnh ta s tm c mt rng buc no i vi tham s m v rng buc y chnh l iu kin cn c tnh cht P. iu c nghal: nu vi m0 khng tha mn rng buc trn th chc chn ng vi m0, h khng c tnh cht P.

Bc 2 (iu kin ): Ta tm xem trong cc gi tr ca m va tm c, gi tr no lmcho h tha mn tnh cht P. bc ny ni chung ta cng ch cn gii nhng h c th khngcn tham s. Sau khi kim tra, ta s loi i nhng gi tr khng ph hp v nhng gi tr cnli chnh l p s ca bi ton.

Nh vy, tng ca phng php ny kh r rng v n gin. Trong rt nhiu bi ton vbin lun th phng php ny li th hin u th r rt. Tuy nhin, thnh cng ca phngphp cn nm ch ta phi lm th no pht hin iiu kin cn mt cch hp l v chniu kin mt cch ng n.

Bi tp v d

S dng tnh i xng ca cc biu thc c mt trong bi ton

Bi 1: Tm m phng trnh sau c nghim duy nht

4

x + 4

1 x + x + 1 x = m (1)

Gii

iu kin cn:Gi s (1) c nghim duy nht x = D thy nu (1) c nghim x = th (1) cng c nghim x = 1 . V nghim l duy nht


48/383

47

nn = 1 = 12

Thay =1

2vo (1) ta tm c m =

2 + 4

8.

iu kin :Gi sm =

2 + 4

8, khi (1) c dng sau:

4

x +4

1 x + x + 1 x =

2 +

4

8 (2)

Theo bt ng thc AM-GM ta c:

x +

1 x

2 v 4

x + 4

1 x 4

8

Do (2) x = 1 x x = 12

.

Vy (1) c nghim duy nht th iu kin cn v l m =

2 + 4

8 2

Bi 2: Tm a v b phng trnh sau c nghim duy nht

3(ax + b)2 + 3(ax b)2 + 3a2x2 b2 = 3b (1)Gii

iu kin cn:Gi s (1) c nghim duy nht x = x0, khi d thy x = x0 cng l nghim ca (1). Do t gi thit ta suy ra x0 = 0. Thay x0 = 0 vo (1) ta c :

3

b2 =3

b

b = 0

b = 1

iu kin : Khi b = 0, (1) c dng:

3

a2x2 +3

a2x2 +3

a2x2 = 0 a2x2 = 0Do (1) c nghim duy nht khi v ch khi a = 0 Khi b = 1, (1) c dng:

3

(ax + 1)2 +

3

(ax 1)2 +

3

a2x2 1 = 1 ()t u = 3ax + 1; v = 3ax 1, ta thy:

()

u3 v3 = 2u2 + uv + v2 = 1

u v = 2u2 + uv + v2 = 1

u = 1

v = 1

ax + 1 = 1

ax 1 = 1 ax = 0

Vy (*) c nghim duy nht khi v ch khi a = 0Tm li, phng trnh (1) c nghim duy nht th iu kin cn v l

a = 0; b = 0b = 1

2

Bi 3: Tm m h sau c nghim duy nht:

7 + x +

11 x 4 = m

4 310 3m7 + y +

11 y 4 = m

4 310 3m


49/383

48

Gii

iu kin: 7 x, y 11; 7427 m

10

3Tr theo v hai phng trnh ta c:

x + 7 11 x =

y + 7

11 y

Xt hm s: f(t) = t + 7 11 t; 7 t 11 ta c:f (t) =

1

2

t + 7+

1

2

11 t > 0 Vy hm s ng bin, suy ra: f(x) = f(y) x = y.Thay vo mt trong hai phng trnh ca h ta c:

7 + x +

11 x 4 = m

4 310 3m ()

iu kin cn:Ta thy l nu x0 l mt nghim ca phng trnh th 4 x0 cng l nghim ca phng trnh.Nn h cho c nghim duy nht khi v ch khi

x0 = 4 x0 x0 = 2

Thay vo phng trnh (*) ta c:4 310 m = m 2 ()

Gii phng trnh (**) ta tm c m = 3. iu kin :

Vi m = 3, ta thu c h phng trnh:7 + x + 11 x = 67 + y + 11 y = 6

V x = y nn ta ch vic gii phng trnh

7 + x +

11 x = 6 x = 2Vy m = 3 l gi tr cn tm h cho c nghim duy nht. 2

Bi 4: Tm a,b h sau c nghim duy nht:

xyz+ z = a

xyz2 + z = b

x2 + y2 + z2 = 4

Gii

iu kin cn:Gi s(x0; y0; z0) l nghim ca h phng trnh cho th (x0; y0; z0) cng l nghim. Do

tnh duy nht nn x0

=

x0; y

0=

y0

x0

= y0

= 0 Thay tr li vo h , ta c:

z0 = a

z0 = bz20 = 4

T y ta suy ra a = b = 2 hoc a = b = 2 iu kin :


50/383

49

Xt hai trng hp sau: Nu a = b = 2: Khi h c dng:

xyz+ z = 2 (1)

xyz2 + z = 2 (2)

x

2

+ y

2

+ z

2

= 4 (3)

Ly (1) (2) ta c xyz(1 z) = 0, t (1) li c z= 0 do xy(1 z) = 0* Nu x = 0 z = 2 y = 0* Nu y = 0 z = 2 x = 0

* Nu z = 1 x2 + y2 = 3xy = 1 .

H trn c nghim (x1; y1) = (0;0).V vy ngoi nghim (0, 0, 2), h cn c nghim khc(x1; y1; 1) do h khng c nghim duy nht. Trng hp ny khng tha mn.

Nu a = b = 2:Khi h c dng:

xyz = 2xyz22 + z = 2x2 + y2 + z2 = 4

Tin hnh lm nh trng hp trn ta i n:* Nu x = 0 z = 2 y = 0* Nu y = 0 z = 2 x = 0* Nu z = 1

x2 + y2 = 3xy = 3Ta thy t h phng trnh trn, ta suy ra x2 + y2 < 2|xy| nn h v nghim.Vy trong trng hp ny h c duy nht nghim (x; y; z) = (0, 0, 2)Vy iu kin cn v h phng trnh cho c nghim duy nht l a = b = 2 2

S dng im thun li

Bi 5: Tm a phng trnh sau nghim ng vi mi x:

log2

a2x2 5ax2 + 6 a = log2+x2 3 a 1 ()Gii

iu kin cn:Gi s (*) ng vi mi x. Vi x = 0 ta c log2

6 a = log2 (3

a 1)

Li c: 1 a 6

a 1 < 3

a 1 + 6 a = 3 a {2; 5}

iu kin :


51/383

50

Nu a = 2 th () log2(2 12x2) = log2+x2 2 ()R rng (**) khng ng vi mi x, v log2(2 12x2) c ngha th phi c 12x2 < 2 Nu a = 5 th () log2 1 = log2+x2 1 (lun ng)Vy a = 5 l iu kin cn v (*) ng vi mi x 2

Bi 6: Tm a h phng trnh n (x; y) c nghim vi mi b:2bx + (a + 1)by2 = a2(a 1)x3 + y2 = 1Gii

iu kin cn:Gi s h c nghim vi mi b, thay b = 0 ta c

a2 = 1(a 1)x3 + y2 = 1Do iu kin cn l a = 1 iu kin :

Nu a = 1: ta c h

2bx + 2by2 = 1y2 = 1Khi b >

1

2h v nghim. Vy trng hp ny loi.

Nu a = 1: ta c h2bx = 12x3 + y2 = 1

H trn lun c nghim (x; y) = (0; 1)Vy a = 1 l iu kin cn v h phng trnh c nghim vi mi b 2

Bi 7: Tm a h phng trnh n (x; y) c nghim vi mi b:

(x2 + 1)a + (b2 + 1)y = 2

a + bxy + x2

y = 1

Gii

iu kin cn:Gi s h c nghim vi mi b, thay b = 0 ta c

()

(x2 + 1)a = 1

a + x2y = 1

a = 0; x2y = 1

x2 + 1 = a + x2y = 1 a {0; 1}

iu kin :

Nu a = 0 : ta c

(b2 + 1)y = 1 (1)bxy + x2y = 1 (2)


52/383

51

Nu b = 0 b2 + 1 = 1 nn t (1) c y = 0, nhng khng tho (2). Vy trng hp ny loi.

Nu a = 1: ta c

x2 + (b2 + 1)y = 1bxy + x2y = 0H trn lun c nghim x = y = 0.Vy a = 1 l iu kin cn v h cho c nghim vi mi b 2

Bi 8: Tm iu kin ca a,b,c,d,e,f hai phng trnh n (x; y) sau l tng ng:ax2 + bxy + cy2 + dx + ey + f = 0 (1)x2 + y2 = 1 (2)Gii

iu kin cn:

Ta thy (x; y) = (0; 1) , (1;0) , 12 ; 12 , 12; 12 l nghim ca (2). Do (1)cng phi c cc nghim trn.

Nh vy

c + e + f = c e + f = a + d + f = a d + f = 0a + b + c +

2d +

2e + 2f

2=

a + b + c 2d 2e + 2f2

= 0

Gii h trn ta tm c iu kin cn ca bi ton l ()b = d = e = 0a = c = f = 0

iu kin :

D thy vi (*) th (2) trng vi (1).Vy (*) l iu kin cn v (1) (2) 2

Bi 9: Cho phng trnh x3 + ax + b = 0 (1)Tm a, b phng trnh trn c ba nghim phn bit x1 < x2 < x3 cch u nhau.

Gii

iu kin cn:

Gi s phng trnh (1) c 3 nghim khc nhau x1, x2, x3 tha gi thit x1 + x3 = 2x2Theo nh l Viete vi phng trnh bc 3 ta c: x1 + x2 + x3 = 0 3x2 = 0 x2 = 0Thay x2 = 0 vo (1) ta c b = 0 iu kin :Gi sb = 0 , khi (1) tr thnh:

x3 + ax = 0 x(x2 + a) = 0 (2)

Ta thy (2) c 3 nghim phn bit nu a < 0. Khi cc nghim ca (2) l

x1 = ax2 = 0

x3 =a


53/383

52

Cc nghim trn cch u nhau nn iu kin cn v (1) c nghim tha mn bi lb = 0, a < 0 2

Bi 10: Cho phng trnhx3 3x2 + (2m 2) x + m 3 = 0

Tm m phng trnh c ba nghim x1, x2, x3 sao cho x1 0 khi x1 < x < x2Suy ra f(1) > 0 hay m 5 > 0 m < 5. iu kin :

Gi sm < 5.Do limx

f(x) = nn tn ti < 1 m f() < 0Li c: f(1) = m 5 > 0 v f(x) lin tc nn ta c < x1 < 1 sao cho f(x1) = 0Ta c: f(0) = m 3 < 0 ( do m < 5)Vy tn ti 1 < x2 < 0 sao cho f(x2) = 0Mt khc, do lim

x+f(x) = + nn phi c > 0 sao cho f() > 0.

T , tn ti x3 m 0 < x3 < sao cho f(x3) = 0.Nh vy, phng trnh f(x) = 0 khi m < 5 c 3 nghim x1, x2, x3 tha mn x1 < 1 < x2 0 (vi x1 < < x2 < < x3)


55/383

54

Phng trnh f(x) = 3x2 + 2ax + b = 0 (1) c hai nghim dng.Nh vy

= a2 3b > 0P =

b

3> 0

S =2a

3> 0

a2 > 3b

b > 0

a < 0

()

b > 0

a < 3b

(*) l iu kin cn ca bi ton. iu kin :Gi s a, b tho mn (*) th r rng phng trnh 3x2 + 2ax + b = 0 c 2 nghim dng0 < <

Suy ra hm s f(x) = x3 + ax2 + bx + c c cc i ti x = v cc tiu ti x = .Do > 0 nn tm c x1 (c; ) sao cho f() < f(x1) < f() phng trnh f(x) = f(x1)c ba nghim dng. t c = f(x1) th phng trnh x3 + ax2 + bx + c = 0 c 3 nghimdng.

Vy (*) l iu kin cn v tn ti c sao cho phng trnh x3 + ax2 + bx + c = 0 c 3nghim dng.T , ta suy ra: b 0 hoc a 3b l iu kn cn v sao cho phng trnhx3 + ax2 + bx + c = 0 c khng qu 2 nghim dng. 2

Bi tp t luyn

Bi 1: Tm m hai phng trnh sau l tng ng x2 + (m2 5m + 6) x = 0 v x2 +2 (m

3) x + m2

7m + 12 = 0

Bi 2: Tm m phng trnh sau c nghim duy nht:|x + m|2 + |x + 1| = |m + 1|Bi 3: Tm m phng trnh sau c nghim duy nht:

x + 3 +

6 x

(3 + x) (6 x) = m

Bi 4: Tm a h sau c ng mt nghim:

x2 + y2 1 a x + y 1 = 1x + y = xy + 1

Bi 5: Tm a h sau c ng mt nghim:

x2 + 3 + |y| = a

y2 + 5 + |x| =

x2 + 5 +

3

a

Bi 6: Cho h phng trnh (x + y)4 + 13 = 6x2y2 + m

xy

x2 + y2

= m

Tm m h c nghim duy nht.

Bi 7: Tm s thc m sao cho h:

x3 my3 = 1

2(m + 1)2

x3 + mx2y + xy2 = 1c nghim (x; y) tho x + y = 0


56/383


57/383

56

Bi 2: Bin lun s nghim ca h phng trnh

(x p)2 + (y q)2 = rax + by = c (r>0) Phn tch: t u = x p; v = y p a v bi ton 1.

Bi 3: Bin lun s nghim ca h phng trnh px2 + qy2 = r

ax + by = c(p, q, r > 0)

Phn tch: H cho gm phng trnh ng elip (E) v phng trnh ng thng d.Nh vy ta cn kho st s giao im ca d v (E). S dng php co - dn bin (E) thnhng trn (C) v bin d thnh d, ta a v kho st s im chung ca d v (C). Ta bitrng s im chung khng ln hn 2.

Phng php: t u = x; v =

q

py h cho tr thnh

()u

2 + v2 =r

p

au +

b

p

qv

= c

T y lm tip nh bi 1. Nu pht hin ng thng i qua mt im nm trong (E) th hlun c 2 nghim phn bit.

Bi 4: Bin lun s nghim ca h phng trnh

p(x k)2 + q(y h)2 = r

ax + by = c

(p, q, r > 0)

Phng php: t u = x k; v = y h a v bi ton 3.

Bi 5: Bin lun s nghim ca h phng trnh vi p, q,r > 0; k = 1:px2 + kxy + qy2 = rax + by = c Phng php: Bin i phng trnh u v dng mu2 + nv2 = r a v bi ton 4.

Bi 6: Kho st tnh cht nghim ca h

x2 + y2 = rax + by = c (r > 0) Phng php: Coi mi nghim ca h l mt im vi to l cp s . Bin i hphng trnh v iu kin ca h thnh ngha hnh hc. iu kin bi ton thng lin quann mt s tnh cht nh tp hp im thuc phn chung ca cc na mt phng hoc mintrn, min elip hoc khong cch gia 2 im, tch v hng ca 2 vect c lp ra t ccim , s o gc to bi 2 vect hoc 2 ng thng.

Bi 7: Kho st tnh cht nghim ca h phng trnh

px2 + qy2 = rax + by = c (p, q, r > 0)


58/383

57

Phng php: t u = px; v = qy h cho tng ngu2 + v2 = rau + bv = c

Trong a =

a

p ; b =b

q. Bi ton a v bi 6.

a phng trnh v t mt n sang h phng trnh hai n

Bi 1: Kho st nghim ca phng trnh

a bx2 = kx + m vi a, b > 0

Phng php: t y = a

b x2 khi ta c h

()

x2 + y2 =

a

bkx by = my 0

Bi ton a v kho st nghim ca h ().

Bi 2: Kho st nghim ca phng trnh a b(x + c)2 = kx + m vi a, b > 0

Phng php: t z = x + c; y =

a2 + bz2 v a v bi 1.

Bi 3: Kho st nghim ca phng trnh a sin t + b cos t = c vi t [; ]

Phng php: t x = sin t; y = cos t ta c h

()

ax + by = c

x2 + y2 = 1

Bi ton a v kho st nghim ca h () vi x, y tho iu kin xc nh.

Bi 4: Kho st s nghim ca phng trnh p

a + bx + q

c + dx = m vi bd < 0; q = 1

Phng php: Coi b > 0 d < 0. t y =

a

b+ x; z =

cd

x ta c h

()

y2 + z2 =a

b

c

dpby + qdz = my 0

z 0


59/383

58

Bi ton a v kho st s nghim ca h () gm cung trn AnB nm trong min gc P xcnh bi cc min x 0, y 0 v ng thng d.

Bi tp v d

Cc bi tp v h tuyn tnh v phn tuyn tnh

Bi 1: Bin lun s nghim ca h

(x 1)2 + (y 2)2 = 5x + my = m + 1

Gii

t u = x 1; v = y 2 ta c h cho tng ng:u2 + v2 = 5u + mv = m

(C)

O

(0,-1)

H trn gm phng trnh ng trn (C) : x2 + y2 = 5 tm l gc ta v bn knh l 5v phng trnh ng thng d : x + my + m = 0 ph thuc m. Hn na, ng thng d luni qua im c nh A(0; 1) nm trong (C) nn d lun ct (C) ti 2 im phn bit vi mim.2

Bi 2: Tm m h c nghim x > 0; y > 1:

()x2 + y2 = 42x y = mGii


60/383

59

d3 d2 d1

= 1

x

1

n

A( 3 , 1)

M(m

2; 0)

M2 M1

B(0; 2)

O

iu kin ca bi ton xc nh min gc P vi x > 0; y > 1. () gm phng trnh ng trn(C) : x2 + y2 = 4 v phng trnh ng thng d : 2x y = m ph thuc m. Phn chung ca(C) v P l cung AnB c cc u mt A(

3;1) v B(0;2). Yu cu bi ton l tm m d v

cung AnB c t nht 1 im chung.

Phng trnh ng thng d1, d2 i qua A, B v cng phng vi d l 2x y = 23 1 v2x y = 2. Cc ng thng ct Ox ti M1(

3 1

2; 0) v M2(1;0). Cc im nm

trong (C) nn giao im th hai ca d1, d2 vi (C) thuc na mt phng P1 = {(x; y) : y < 0}.V vy d1, d2 ch c mt im chung vi AnB l A v B.

Gi Q l di mt phng bin d1, d2, khi Q cha AnB. ng thng d ct on thngM1M2 ti M(

m

2; 0) khc M1, M2.

T suy ra 2 < m < 23 1 l tp hp cc gi tr m tho gi thit. 2

Bi 3: Tm m h c 2 nghim phn bit tho x1x2 + y1y2 > 0:

()x2 + y2 = 4x + y = m

Gii

d

H

O

B

Gi s h c hai nghim phn bit l (x1, y1) v (x2, y2), khi ta c: Gi A(x1; y1), B(x2; y2),


61/383

60

th do OA.

OB = OA.OB. cos

OA,

OB

= x1x2 + y1y2

Nn suy ra:cos

OA,OB

> 0 AOB < 2

Hn na do (x1, y1) v (x2, y2) u l nghim ca h nn cc im A v B u nm trn ngtrn (C) : x2 + y2 = 4 v ng thng d : x + y = m, nh vy AB l mt dy ca ng trn(C), ng thi AB cng chnh l ng thng d.Nh vy h c nghim tha iu kin x1x2 + y1y2 > 0 th AB phi ct ng trn (C) tihai im phn bit, hn na phi tha mn AOB

2 22 > |m| > 2 2

Bi 4: Tm m h c 2 nghim phn bit tho (x1 x2)2 + 4(y1 y2)2 = 3:

(

)

x2 + 4y2 = 16

x my = m

Gii

H () gm phng trnh elip (E) : x2

16+

y2

4= 1 v ng thng d : x my = m ph thuc m.

-1

O

Ta c ng thng d lun i qua im c nh A(0; -1) hn na A nm trong (E). Vy d lunct (E) ti 2 im phn bit, do h () c 2 nghim phn bit vi mi m.Gi B(x1; y1), D(x2; y2) v t u = x; v = 2y (1). Cng thc (1) xc nh mt php dn vi h


62/383

61

s 2 bng trc dn Ox. Php dn bin cc im B, D thnh B(u1; v1), D(u2; v2) v h ()tr thnh:

()u2 + v2 = 162u mv = 2m

To cc im B, D l nghim ca (). H gm phng trnh ng trn (C) v phng

trnh ng thng d

l nh ca d qua php dn .

Q'

P'

-2

O

Mt khcBD2 = (u1u2)

2 + (v1 v2)2 = (x1 x2)2 + 4(y1 y2)2 = 3Ngha l d ct (C) ti 2 im B, D sao cho BD =

3 hn na (C) c tm l gc ta nn

d2O/d +BD2

4 = R2 = 16 Suy ra dO/d

=

61

2 , do d = |2m

|4 + m2 nn tm c m = 1223 2Bi 5: Tm m h c 2 nghim phn bit v (x1 x2)2 + (y1 y2)2 t gi tr nh nht:

()x2 + y2 = 42x + my = m + 2

Gii

(C)

d

BM

O


63/383

62

H () gm phng trnh ng trn (C) : x2 + y2 = 4 v ng thng d ph thuc vom. Gi A, B l giao im ca d v (C) th ta cn tm m AB2 t gi tr nh nht. DoM(1; 1) d m v M nm trong (C) nn d lun ct (C) ti 2 im phn bit A, B.Ta c: AB Min d OM ().Gi u = (m; 2) l vect ch phng ca d th

() u .OM = 0 2 m = 0 m = 2Do vi m = 2 th h c hai nghim phn bit tha (x1x2)2 +(y1y2)2 t gi tr nh nht.

Bi 6. Bin lun s nghim ca h phng trnh sau:|x| + 2 |y| = 4(x 2a) (y a) = 0Gii

Xt h phng trnh: |x| + 2 |y| = 4 (1)(x 2a) (ya) = 0 (2)Ta thy rng cc im tha mn (1) l bn cnh ca hnh thoi ABCD trong :

A((4;0); B (0;2); C(4;0); D (0; 2)ng thi, cc im tha mn (2) nm trn hai ng thng d1 : x = 2a v d2 : y = a. Snghim ca h hai phng trnh (1) v (2) chnh l s giao im ca bn cnh hnh thoi vi

hai ng thng ni trn.Trc tin ta tm xem khi no 3 ng thng x = 2a; y = a v |x| + 2 |y| = 4 ng quy. Gi(x0; y0) l im chung ca ba ng thng ny, khi ta c h sau:

x0 = 2a

y0 = a

|x0| + 2 |y| = 4

x0 = 2a

y0 = a

4 |a| = 4

|a

|= 1

x0 = 2; y0 = 1

x0 = 2; y0 = 1

-4

-2

4

2

y = 1

x = 2

D

B

CO

y = a

x = 2a

-4

-2

4

2

D

B

CO


64/383

63

Da vo th trn ta c kt lun sau:

Nu |a| > 2: H (1), (2) v nghim.

Nu |a| = 2: H (1), (2) c hai nghim.

Nu |a| < 2 v |a| = 1: H (1), (2) c 4 nghim.

Nu |a| = 1: H (1), (2) C 3 nghim.

a phng trnh v t v h phng trnh

Bi 7: Bin lun s nghim ca phng trnh

m9 x2 x + 5m = 0 (1)

Gii

t y =

5 x2 ta c h

()

x2 + y2 = 9

x my = 5my 0

H () gm phng trnh ng thng d ph thuc m v na ng trn (C) xc nh bi hbt phng trnh x2 + y2 = 9y 0Ta c A(0; 5) d m do A nm ngoi (C)

(C)

d

-3 35m

M2 M1OM


65/383

64

Cc im M1(3; 0), M2(3;0) l u mt ca ng knh ca (C). Cc tia AM1, AM2 l binca min gc P. ng thng d ct M1M2 ti M(5m; 0) nn vi m tho 3 5m 3 3

5

m 3

5th d v na trn ca (C) c duy nht 1 giao im. Trng hp ny phng trnh c 1

nghim.Vi cc gi tr cn li ca m th d v na trn ca (C) khng c giao im nn phng trnh

v nghim. 2

Bi 8: Bin lun theo m nghim ca phng trnh sau:

4 x2 = mx + 2 m ()

Gii

Ta bit rng s nghim ca phng trnh () l s giao im ca hai ng y = mx + 2 m

v y = 4 x2.Hn na v y =

4 x2 x2 + y2 = 4, y 0 nn th ca y = 4 x2 l na ng trn

tm l gc ta bn knh bng hai v nm pha trn trc honh.Cn y = mx + 2 m l mt h ng thng lun i qua im c nh A(1; 2). Ta nhn thyc hai tip tuyn vi ng (C) : y =

4 x2, mt l ng thng y = 3 song song vi trc

honh, v tip tuyn AD.

2

-2 21O

B C

E

Gi B(2, 0) v C(2, 0) l hai u mt ca ng knh BOC, gi s m1, m2, m3, m4 tngng l h s gc ca cc ng thng AC, AD, AB, AE th ta c cc iu sau:

m1 = tan ACO = 2.

m2 =

tanDCO =

tanEAD =

tan2OAE =

4

3

.

m3 = tan ABO = 23

.

m4 = 0.


66/383

65

T suy ra:

1. Phng trnh () c hai nghim

0 < m

2

3

2 m

4

3

2. Phng trnh () c mt nghim

m >

2

3m < 2m = 0

m = 43

3. Phng trnh () v nghim 4

3 < m < 0 2

Bi 9: Bin lun s nghim ca phng trnh sau theo a:a

9 x2 + x

x a

3

= 0

Gii

t y =

9 x2 x2 + y2 = 9Phng trnh (4) tng ng:

x2 + y2 = 9, y 0

(ay + x)

x a3 = 0D thy phng trnh th nht biu din phn pha trn trc honh ca ng trn tm l gcta bn knh l 3 cn phng trnh th hai biu din hai ng thng x = a3 v y = 1

a.x

(nu a = 0).S nghim ca h chnh l s giao im ca hai ng thng vi na ng trn. Ta ch cnxt khi a > 0 (v khi a < 0 ta c kt qu tng t v khi a = 0 th (4) x = 0 v lc hc 1 nghim).

Thy rng ng thng y = 1a

x nn n i qua O v v vy lun ct na ng trn ti mt

im. Ta ch cn quan tm n v tr tng i ca x = a

3 vi ng trn v ng thngy =

1

ax.

Do y 0 nn nu ng trn v hai ng thng ng quy th phi c x = a3 > 0 vy = 1

ax < 0. iu ny l v l. Do s khng c trng hp ba ng ny ng quy. Nh

th nhn vo th ta s thu c kt lun:


67/383

66

x = a 3

a > 3

y = -1

ax

3

O

x = a 3

0 < a < 3

y = -1

ax

3

O

Do ta c kt lun sau:

Phng trnh (4) c mt nghim

a = 0

|a| > 3

Phng trnh (4) c hai nghim 0 < |a| < 3 2

Bi 10: Bin lun s nghim ca phng trnh:

9 2x x2 = x + m (()

Gii

Quan st mt cht ta thy bi ny c cu trc hi khc cc bi trn nhng tht ra chng nhnhau:

()

10 (x + 1)2 = x + mt z = x + 1; y =

10 z2 ta c h:

y2 + z2 = 10

y

z = m

1

y 0

Bi ton a v bin lun s im chung ca ng thng y = z+ m 1 v na ng trny2 + z2 = 10y 0n y tng t nh Bi 1, xin dnh cho bn c gii quyt tip phn cn li ca bi ton.

Bi 11: Bin lun s nghim ca phng trnh

2sin t (m + 3) cos t = m 1, t

3;

2

3


68/383

67

Gii

t x = sin t; y = cos t v a iu kin ca bi ton thnh:

x2 + y2 = 1 (i)

2x (m + 3)y = m 1 (ii)3

Documents

Tuyen tap Phuong trinh He phuong trinh MS