dat an phu giai phuong trinh chua can - tran tri quoc

Trn Tr Quc THPT NGUYN HU PH YN

CHUYN T N PH GII PHNG TRNH V H PHNG TRNH V T

Nh cc bn bit trong chng trnh Ton THPT th phng trnh v h phng trnh v tlun l mt ch kinh in, bi th nn n lun xut hin trong cc k thi ln nh thi i hc vcc k thi hc sinh gii ln nh. Trong phng php dng n ph gii ton lun l mt cng cmnh v hu ch. Hm nay bi vit ny s trnh by mt s phng php t n ph gii quytcc bi ton.

Ni dung: t biu thc cha cn bng biu thc mi m ta gi l n ph, chuyn v phngtrnh theo n mi. Gii phng trnh n ph ri thay vo biu thc tm nghim ban u.

Phng php: Gm c cc bc sau:Bc 1: Chn cch t n ph, tm iu kin xc nh ca n ph. lm tt bc ny phi c squan st, nhn xt mi quan h ca cc biu thc c mt trong phng trnh ri a ra biu thcthch hp t n ph.Bc 2: Chuyn phng trnh ban u v phng trnh theo n ph, thng l nhng phng trnh bit cch gii, tm c nghim cn ch n iu kin ca n ph.Bc 3: Gii phng trnh vi n ph va tm c v kt lun nghim.

Thnh vin tham gia chuyn :1-Trn Tr Quc 11TL8 THPT Nguyn Hu, Ph Yn2-H c Khnh 10CT THPT Chuyn Qung Bnh.3-on Th Ha 10A7 THPT Long Khnh, ng Nai4-Thy Mai Ngc Thi THPT Hng Vng, Bnh Phc.5-Thy Nguyn Anh Tun THPT L Qung Ch, H Tnh.

u tin ta cng gii cc v d c bn sau:C l nhiu bn quen vi bi tp dng loi ny nn mnh ch mun nhc li 1 tI-t n ph a v phng trnh theo n ph:Dng 1

Pt c dng ax2 + bx+ c =px2 + qx+ r trong

a

p=b

q

Cch gii : t t =px2 + qx+ r, t 0

Ti s a ra vi v d cc bn n li v y l phn kh d

Gii cc phng trnh sau1/(H Ngoi Thng-2000) (x+ 5)(2 x) = 3x2 + 3x2/(H Ngoi ng 1998) (x+ 4)(x+ 1) 3x2 + 5x+ 2 = 63/(H Cn Th 1999)

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

4/ 4x2 + 10x+ 9 = 5

2x2 + 5x+ 35/ 18x2 18x+ 5 = 39x2 9x+ 26/ 3x2 + 21x+ 18 + 2

x2 + 7x+ 7 = 2

Dng tip theo cng rt quen thucDng 2PT c dng P (x) +Q(x) + (

P (x)Q(x)) 2P (x).Q(x) + = 0 ( l s thc)

Cch gii t t =P (x)Q(x) t2 = P (x) +Q(x) 2P (x).Q(x)

Page 1


Bi 1: Gii phng trnh 1 +2

3

x x2 = x+1 x

Gii

K 0 x 1, Ta t t = x+1 x th x x2 = t2 1

2, phng trnh tr thnh bc 2 vi n

l t

1 + t2 1

3= t t2 3t+ 2 = 0 t = 1; t = 2

TH1 t = 2 x+1 x = 2 (VN)TH2 t = 1 x+1 x = 1 x = 0;x = 12

Gii cc phng trnh sau

1/(HVKTQS-1999)

3x 2 +x 1 = 4x 9 + 23x2 5x+ 22/

2x+ 3 +x+ 1 = 3x+ 2

2x2 + 5x+ 3 16

3/

4x+ 3 +

2x+ 1 = 6x+

8x2 + 10x+ 3 164/(CSPHN-2001)

x 2x+ 2 = 2x2 4 2x+ 2

Th l xong cc v d c bn ri by gi ta xt n cc v d m cn s bin i kho lo mtcht v c s quan st nh gi mi c th a v dng c bn t n ph c.

II-t n ph a v phng trnh tch

Xut pht t 1 s hng ng thc c bn khi t n ph:x3 + 1 = (x+ 1)(x2 x+ 1)x4 + 1 = (x2 2x+ 1)(x2 +2x+ 1)x4 + x2 + 1 = (x4 + 2x2 + 1) x2 = (x2 + x+ 1)(x2 x+ 1)4x4 + 1 = (2x2 2x+ 1)(2x2 + 2x+ 1)

Ch : Khi t n ph xong ta c gng a v nhng dng c bn nh sauu+ v = 1 + uv (u 1)(v 1) = 0au+ bv = ab+ vu (u b)(v a) = 0Phng trnh ng cp bc hai ax2 + bxy + cy2 = 0 at2 + bt+ c = 0 vi t = x

y

Li ly Bi 1 trn 1 ln na

Gii

Gii phng trnh 1 +2

3

x x2 = x+1 x

Nhn xt: Ta thy (x)2 + (

1 x)2 = 1(**), m t phng trnh u ta rt c mt cn thc

qua cn thc cn li

Gii

x = 3

1 x 32

1 x 3 . Do nu t t =

1 x x = 3t 32t 3

Thay vo (**) ta bin i thnh t(t 1)(2t2 4t+ 3) = 0 t = 0; t = 1 hay x = 0;x = 1 l nghimca phng trnh.2

Page 2


Ta xt v d sau

Bi 2: Gii phng trnh 3x+ 1 + 3

x+ 2 = 1 + 3

x2 + 3x+ 2

Gii

Ta thy (x+ 1)(x+ 2) = x2 + 3x+ 2t u = 3

x+ 1; v = 3

x+ 2

PT u+ v = 1 + uv (u 1)(v 1) = 0Gii tip ta c x = 0;x = 12

Ta xt v d sau, kh ging bi trn nhng kh hn.

Bi 3: Gii phng trnh 3x2 + 3x+ 2( 3

x+ 1 3x+ 2) = 1

Nhn xt: Cch lm bi ny cng kh ging nhng phi tht k bn VP v ta tch VPthnh biu thc "lin quan" n biu thc n ph.

Gii

Li gii: Phng trnh cho tng ng vi(x+ 1) (x+ 2) + 3x2 + 3x+ 2( 3x+ 1 3x+ 2) = 0

Ta t 3x+ 1 = a; b = 3x+ 2, khi phng trnh tng ng

a3 + b3 ab(a+ b) = 0 (a+ b)(a b)2 = 0 a = b 3x+ 1 = 3x+ 2 x = 3

2

Th li thy x = 32tha mn. Vy phng trnh c nghim duy nht x = 3

22

V d tng t

Bi 4: Gii phng trnh (x+ 2)(

2x+ 3 2x+ 1) +2x2 + 5x+ 3 1 = 0

Gii

K

x 3

2x 1

x 1

t

2x+ 3 = ax+ 1 = b

a; b 0

x+ 2 = a2 b2

2x2 + 5x+ 3

1 = a2 2b2Nn PT (a2 b2)(a 2b) + ab = a2 2b2 (a2 b2)(a 2b) + b(a+ b) (a2 b2) = 0. V a+ b > 0 nn ta chia 2 v cho a+ b (a b)(a 2b) (a 2b) = 0 (a 2b)(a b 1) = 0 Vi a = b+ 1 2x+ 3 = x+ 1 + 1 (VN) Vi a = 2b 2x+ 3 = 2x+ 1 x = 1

2(TMK)

Vy phng trnh c nghim S =

{1

2

}

Page 3


Bi tp nghGii cc phng trnh sau1/(x+ 5x+ 2)(1 +x2 + 7x+ 10) = 3

2/(x+ 1 +

x 2)(1x2 x 2) = 3

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

4/

3x2 18x+ 25 +4x2 24x+ 29 = 6x x2 4

Bi 5: Gii phng trnh2 +x

2 +

2 +x

+2x

2

2x =

2

Gii

Thot nhn ta a ra nh gi rt d thy 2 +x+ 2x = 4

Nn ta t

2 +x = a;

2x = b

Ta c ab =

4 x; a2 + b2 = 4Ta vit li phng trnh nh sau:a22 + a

+b22 b =

2

a22 a2b+ b22 + ab2 = 2(2 b2 + a2 ab) 2(a2 + b2 + ab 2) ab(a b) = 2(a b) 2(ab+ 2) = (a b)(ab+ 2). a2 + b2 = 4V ab+ 2 6= 0 nn a b = 2 a2 + b2 2ab = 2 ab = 1 4 x = 1Nn x = 3Vy phng trnh c nghim S = 32.

Bi 6: Gii phng trnh (13 4x)2x 3 + (4x 3)5 2x = 2 + 816x 4x2 15Nhn xt: D thy rng (2x 3)(5 2x) = 16x 4x2 15, nhng cn cc nh thc ngoi cn takhng th biu din ht theo 1 n ph c, ta t 2 n ph v c a v phng trnh tch.

Gii

Li gii: K3

2 x 5

2t u =

2x 3 u2 = 2x 3; 2u2 + 3 = 4x 3

v =

5 2x v2 = 5 2x; 2v2 + 3 = 13 4x u2 + v2 = 2;uv = 16x 4x2 15(1) PT (2v2 + 3)u+ (2u2 + 3)v = 2 + 8uv = u2 + v2 + 8uv 2uv(u+ v) + 3(u+ v) = (u+ v)2 + 6uv (u+ v 3)(2uv u v) = 0TH1 : u+ v = 3

16x 4x2 15 = 72(VN)

TH2 : u+ v = 2uv 16x 4x2 15 = 1 x = 2 (Tha K)Vy phng trnh cho c nghim duy nht x = 22

Bi 7: Gii phng trnh x2 +x+ 1 = 1 (*)

Gii

Page 4


tx+ 1 = t; t 0

PT(*) (t2 1)2 + t = 1 t(t 1)(t2 + t 1) = 0TH1 Vi t = 0 th x = 1.TH2 Vi t = 1 th x = 0.

TH3 Vi t =1 +5

2th x =

152

2

Ta t lm kh vi kiu bi trn ln mt t nh, nng bc ly tha, ta xt v d sau

Bi 8: Gii phng trnh x4 +x2 + 3 = 3

Gii

n gin ha, ta t x2 = a, a 0PT a2 +a+ 3 = 3, ta s tch a v phng trnh tch nh sau: a2 (a+ 3) + (a+a+ 3) = 0 (a+a+ 3)(aa+ 3 + 1) = 0

V a 0 a+a+ 3 > 0 (VN)Ta c a+ 1 =

a+ 3

a2 + a 2 = 0 a = 1(a 0) nn x = 12

Bi 9: Gii phng trnh (x2 + 2)2 + 4(x+ 1)3 +x2 + 2x+ 5 = (2x 1)2 + 2

( thi chn i tuyn 10 THPT chuyn Lng Vn Chnh-Ph Yn)

Nhn xt: Bi ny c ly tha bc cao nht l 4, v c c cn bc 2 nn ta s c nhm cc biuthc ly tha ging trong cn c th t n ph.

Gii

x4 + 4x2 + 4 + 4(x3 + 3x2 + 3x+ 1) +x2 + 2x+ 5 = 4x2 4x+ 3 (x2 +2x)2 +8(x2 +2x)+x2 + 2x+ 5+5 = 0 (Cng on nhm li th ny cng rt quan trng)

t t =x2 + 2x+ 5, t 2 t2 5 = x2 + 2x

Ta vit li PT cho tng tng vi (t2 5)2 + 8(t2 5) + t+ 5 = 0 t4 2t2 + t 10 = 0 (t 2)(t3 + 2t2 + 2t+ 5) = 0V t 2 nn t3 + 2t2 + 2t+ 5 > 0Ta c t = 2 x2 + 2x+ 5 = 2Vy x = 12

Bi 10: Gii phng trnhx2 2x+ 5 +x 1 = 2

Gii

t:t =x 1, vi x 1, t 0 t2 = x 1

Phng trnh cho vit li:

(x 1)2 + 4 = 2x 1Tr thnh:

t4 + 4 = 2 t(t 2)

t4 t2 + 4t = 0

Page 5


V t [0; 2] nn t3 t+ 4 > 0Vy t = 0 x = 12

Bi 11: Gii phng trnh (4x2 + 1)x+ (y 3)5 2y = 0

Gii

iu kin y 52.

t a = 2x v b =

5 2y (b 0) ta c phng trnh vit li thnha3 + a

2+ (b3 + b)

2= 0 a = b

Hay 2x =

5 2y x = 5 4y2

2. Vy x =

5 4y22

l nghim ca phng trnh.

Nhn xt. Mt li gii tht p phi khng ! Chc cc bn s thc mc rng lm sao m ta li cth t c n ph nh trn.

Trc tin ta s t

5 2y = b y 3 = 5 b2

2 3 = (b

2 + 1)

2

(y 3)5 2y = (b2 + 1) b

2

By gi ta mun (4x2 + 1)x =a (a3 + 1)

2 (4x2 + 1) .2x = a3 + a 8x3 + 2x = a3 + a a = 2xT ta c c cch t n ph nh li gii 2

Bi 12: Gii phng trnh

x+ 2

2 1 = 33(x 3)2 + 39(x 3)

Gii

iu kin x 2 t t = 39 (x 3) th ta c x = t3 + 279

x+ 2

2=

t3 + 45

18; 3

3(x 3)2 = t2

3.

Phng trnh cho tr thnht3 + 45

18 1 = t

2

3+ t

t3 + 45

2= t2 + 3t+ 3 (1)

Ta c t2 + 3t+ 3 =

(t+

3

2

)2+

3

4> 0 nn phng trnh (1) tng ng vi

t3 + 45

2= (t2 + 3t+ 3)2

2t4 + 11t3 + 30t2 + 36t 27 = 0(2t 1)(t+ 3)(t2 + 3t+ 9) = 0 t = 1

2; t = 3

Vi t = 12th x =

t3 + 27

9=

217

72

Page 6


Vi t = 3 th x = t3 + 27

9= 0

Cc nghim trn tha mn iu kin ca bi ton. Vy phng trnh c hai nghim x = 0 v

x =217

722.

Bi 13: Gii phng trnh 5 3x 5x+ 3 5

x 3x = 8

Gii

Phng trnh cho tng ng vi: 53

5x6 + 3

5

3x4 = 8

5 15x6 + 3 15x4 = 8t:y =

15x2 vi y 0 ta c:

5y3 + 3y2 8 = 0 (y 1)(5y2 + 8y + 8) = 0 y 1 = 0 y = 1Do ta c:

15x2 = 1 x2 = 1 x = 1.

Vy: tp nghim ca phng trnh cho l:S = {1; 1}2.

Bi 14: Gii phng trnh5x4 7

5x2

+6

x= 0

Gii

K x 6= 0. Ta c phng trnh cho tng ng vi5x4 7

5x2

+6

5x5

= 0 5x9 7 5x3 + 6 = 0()

t:y =5x3, y 6= 0, phng trnh (*) tr thnh:

y3 7y + 6 = 0 (y 1)(y2 + y 6) = 0

y = 1y = 2y = 3

5x3 = 1

5x3 = 2

5x3 = 3

x = 1x = 2 34x = 3 39

Vy tp nghim ca phng trnh cho l{

1; 2 3

4;3 39}2Bi 15: Gii phng trnh

4x 1 +4x2 1 = 1

Gii

K

{4x 1 04x2 1 0 x

1

2

Bnh phng hai v phng trnh cho, ta c:(4x 1) + (4x2 1) + 2(4x 1)(4x2 1) = 1 2(4x 1) (4x2 1) = 3 4x2 4x = 4 (2x+ 1)2t y = 2x+ 1 4x 1 = 2y 3, 4x2 1 = y2 2yPhng trnh tr thnh2

(2y 3)(y 2) = 4 y2

{

4 y2 04(2y 3)(y 2)y = (4 y2)2

Page 7


2 y 2[y 2 = 04(2y 3)y = (y + 2)2(y 2)

2 y 2[y = 2

y3 6y2 + 8y 8 = 0 y = 2

Hm s G(y) = y3 6y2 + 8y 8 ly gi tr m trn ton min [2; 2]Do ta c 2x+ 1 = 2 x = 1

2

Vy phng trnh c nghim duy nht x =1

22


2x 1 + x2 3x+ 1 = 0 (D-2006)

Gii

t t =

2x 1 x = t2 + 1

2PT t4 4t2 + 4t 1 = 0 (t 1)2(t2 + 2t 1) = 0* Vi t = 1 x = 1*Vi t =

2 1 x = 222

Bi 17: Gii phng trnh 2x2 6x 1 = 4x+ 5

Gii

K x 3

11

2;x 3 +

11

2

t t =

4x+ 5 x = t2 5

4PT t4 22t2 8t+ 27 = 0 (t2 + 2t 7)(t2 2t 11) = 0i chiu iu kin ta tm c nghim ca phng trnh x = 12;x = 2 +32

Nhn xt: i vi nhng bi c dngax+ b+cx2+dx+e = 0 th cch gii l t

ax+ b = t,

sau a v phng trnh bc 4, dng ng nht thc phn tch nhn t. Nhng c 1 s bikhng gii c bng cch , ta s nhc li vn ny phn sau.

Bi 18: Gii phng trnh (x+ 3x+ 2)(x+ 9

x+ 18) = 168x

i vi nhng bi m khi phn tch thnh cc nh thc hoc tam thc ta thng nhm cnghim hu t kh p, vy cn i vi nhng nghim v t?

Ta xt bi ton sau:

Bi 19: Gii phng trnh (x 2)x 12x+ 2 = 0Nhn xt: Ta thy trong cn c

x 1, nn ta s c gng thm bt v tch s c mt phng

trnh theo n mi

Gii

Page 8


tx 1 = t, t 0

Ta bin i phng trnh nh sau : [(x 1) 1]x 12[(x 1)2]2 = 0 t3 2t2 t+ 22 = 0Phng trnh ny ta bm my khng c nghim hu t, nhng bn no tinh mt t s thyt = 0.4142......?Nhn vo s ny kh quen nh, n chnh l

2 1

p dng s Horner, ta phn tch c nh sau :(t+ 12)(t2 t2) = 0*TH1 Vi t =

2 1 x 1 = 2 1 x = 4 22

*TH2 t2 t2 = 0, v ch nhn t > 0

Ta c t =1 +

1 + 4

2

2 x =

(1 +

1 + 4

2

2

)2+ 12

III- t n ph a v phng trnh ng cp bc hai, bc ba.

Bi 20: Gii phng trnh 2(x2 + 2) = 5x3 + 1 ( ngh Olympic 30/4/2007)

i vi bi ton ny u tin ta phn tch nhn t trong cn x3 + 1 = (x + 1)(x2 x + 1) ri c bin i v tri thnh tng hoc hiu ca hai tha s trong cn.

Gii

Ta bin i nh sau 2(x2 + 2) = 2(x2 x+ 1) + 2(x+ 1)Ta t

x2 x+ 1 = a;x+ 1 = b

PT 2a2 + 2b2 = 5abn y gii ra c 2 nghim t =

1

2; t = 2 vi t = (

a

b)

Vy x =537

22

Sau y l mt s bi tp tng tGii PT1/2(x2 3x+ 2) = 3x3 + 82/2x2 + 5x 1 = 7x3 13/10x3 + 8 = 3(x2 x+ 6)

4/10x3 + 1 = 3(x2 + 2)

Ngoi ra cc bn vn c th sng to thm cc PT bng cc ng thc ti nu trn s rtth v y, c mt phng trnh p ta phi chn h s a, b, c sao cho PT at2 + bt + c = 0 c"nghim p" l c, bn hy th xem.

V d bi ny chng hn 4x2 22x+ 4 = x4 + 1Cng th sc vi bi ton sau nh, bi ny kh hn so vi cc v d ti nu trn


5x2 14x+ 9x2 x 20 = 5x+ 1 (HSG Qung Ngi 2012)

Gii

K x 5, chuyn v bnh phng ta c :2x2 5x+ 2 = 5(x2 x 20)(x+ 1)

n y li gp 1 vn na l ta khng th tm c hai s , sao cho(x2 x 20) + (x + 1) = 2x2 5x + 2 nn ta khng th t a = x2 x 20; b = x+ 1 nh

Page 9


cc v d trn c.

Nhng li thy x2 x 20 = (x 5)(x+ 4)PT 2x2 5x+ 2 = (x2 4x 5)(x+ 4)Ta th li ln na v tm c , tha mn, ta bin i li PT nh sau 2(x2 4x 5) + 3(x+ 4) = 5(x2 4x 5)(x+ 4)t a =

x2 4x 5; b = x+ 4

PT 2a2 + 3b2 = 5abT ta c a = b; a =

3

2b

Vi a = b x = 5 +

61

2(x 5)

Vi a =3

2b x = 8;x = 7

4

i chiu vi iu kin ta nhn x = 8;x =5 +

61

2l nghim ca phng trnh.2

BI TPGii cc phng trnh sau:

1/x2 + x 6 + 3x 13x2 6x+ 19 = 0 S: x = 23

341

2

2/ 3x2 + 4x 5 +x 311x2 + 25x+ 2 = 0 S: x = 21

161

2

3/

7x2 + 25x+ 19x2 2x 35 = 7x+ 2 S: S ={

61 +

11137

18; 3 + 2

7

}

Bi 22: Gii phng trnh 3x2 2x 2 = 630

x3 + 3x2 + 4x+ 2

Nhn xt:Bi ny hi khc mt cht so vi nhng bi trn l biu thc trong cn khngc dng hng ng thc, v vy ta xem nh mt phng trnh hu t v nhm nghim.

K 3x2 2x 2 0 x 1

7

3;x 1 +

7

3 : x3 + 3x2 + 4x+ 2 = (x+ 1)3 + (x+ 1) = (x+ 1)(x2 + 2x+ 2)

Gii

Ta vit li PT nh sau 3(x2 + 2x+ 2) 8(x+ 1) = 630

(x+ 1)(x2 + 2x+ 2)

n y d ri, ta t a =x2 + 2x+ 2; b =

x+ 1 nn PT vit li nh sau

3a2 8b2 = 630ab

p s : x = 232

Bi 23: Gii phng trnh (x2 6x+ 11)x2 x+ 1 = 2(x2 4x+ 7)x 2

Gii

Li gii: K x 2tx2 x+ 1 = a;x 2 = b vi a, b 0

Ta biu din cc biu thc ngoi cn theo a v b nh saux2 6x+ 11 = (x2 x+ 1)2 + (x 2)2

Page 10


x2 4x+ 7 = (x2 x+ 1)2 + (x 2)2S dng ng nht thc ta gii cx2 6x+ 11 = a2 5b2 v x2 4x+ 7 = a2 3b2

Phng trnh cho tng ng vi(a2 5b)a = 2(a2 3b2)b a3 2a2b 5ab2 + 6b3 = 0 t3 2t2 5t+ 6 = 0 vi t = a

b t 0

(t 1)(t 3)(t+ 2) = 0TH1 Vi t = 1 a = b

x2 x+ 1 = x 2 (VN)

TH2 Vi t = 3 a = 2bx2 x+ 1 = 3x 2

x = 56 (Tha mn K)TH3 Vi t = 2 0 nn phng trnh v nghim.

Vy S ={

5 +

6; 56}2Nhn xt: Ci kh dng bi ny l ta phi bin i biu thc trong cn sao cho ph hp vi

bn ngoi tm c , thch hp, cc bn cng c th t sng to cc PT kiu ny bng cchlm ngc li l t PT bc 2 nghim p ri chn cc tam thc v nh thc thch hp s c cmt bi ton hay.

IV-n ph khng trit i vi nhiu PT v t, khi khng biu din hon ton c theo n ph th c mt cch l xembin mi l n, bin c l tham s.Dng ton ny gi l n ph khng hon ton.

*Ni dung phng phpa phng trnh cho v dng phng trnh bc hai vi n l n ph hay l n ca phng trnh cho.a phng trnh v dng sau

f(x)Q(x) = f(x) + P (x)x khi :

tf(x) = t; t 0. Phng trnh vit li thnh t2 t.Q(x) + P (x) = 0

n y chng ta gii t theo x. Cui cng l gii quyt phng trnhf(x) = t sau khi n gin

ha v kt lun.

Ta xt v d sau hiu r hn.

Bi 24: Gii phng trnh x2 + 3x+ 1 = (x+ 3)x2 + 1 (HQG-2001)

Nhn xt: Ta thy trong cn c x2 + 1, ta t t =x2 + 1. Ta s khng rt x theo t m coi x

l tham s. Tht vyPT t2 (x+ 3)t+ 3x = 0Ta c = (x+ 3)2 12x = (x 3)2 = x 3 t = 3; t = x+ 3TH1 t = x+ 3 (VN)TH2 t = 3 x = 222

Bi 25: Gii phng trnh x2 +(3x2 + 2)x = 1 + 2x2 + 2

Gii

Phng trnh tng ng vi

Page 11


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

x2 + 3x 1 (x+ 2)x2 + 2 = 0t t =

x2 + 2; (t 2), phng trnh vit li thnh

t2 (x+ 2) t+ 3x 3 = 0 c = (x 4)2Nn phng trnh c 2 nghim l

t = x 1 x2 + 2 = x 1{x 12x 1 = 0 h ny v nghim.

t = 4 x2 + 2 = 4 x2 = 14 x = 14Vy phng trnh c hai nghim l x =

14 v x = 142

Bi 26: Gii phng trnh (3x+ 1)

2x2 1 = 5x2 + 32x 3

Gii

t t =

2x2 1; (t 0)Phng trnh vit li thnh 2t2 (3x+ 1)t+ x2 + 3

2x 1 = 0

= (x 3)2 suy ra phng trnh c hai nghim l

t = 2x 1 2x2 1 = 2x 1x

1

2x2 2x+ 1 = 0

x = 1

t = x+ 2 2x2 1 = x+ 2{x 2x2 4x 5 = 0 x = 1;x = 5

Vy S = {1; 5; 1}2

Nhn xt: Thng thng sau khi t n ph th ta vit phng trnh cho li thnh

t2 (3x+ 1) t+ 3x2 + 32x 2;nhng bi ton ny li c s khc bit l ta s vit phng trnh ny

li thnh 2t2 (3x+ 1) t + x2 + 32x 1. Chng ta quan tm ti liu h s trc t2 c p tc ta

mong mun phi l bnh phng ca mt s hoc mt biu thc, v iu ny s quyt nh ti ligii s ngn gn hay phc tp. c th iu chnh c h s trc t2 sao cho p cc bn c

th lm nh sau mt2 (3x+ 1) t+ (5 2m)x2 + 32x+m 3 = 0 co.

= (3x+ 1)24m[(5 2m)x2 + 3

2x+m 3

]= (8m2 20m+ 9)x2+(6 6m)x+(4m2 + 12m+ 1)

Ta xt tip ca bng cch gii phng trnh sau

2

(8m2 20m+ 9) (4m2 + 12m+ 1) = 6 6m m = 2 chnh l h s m ta cn tm.

Bi 27: Gii phng trnh 3(

2x2 + 1 1) = x (1 + 3x+ 82x2 + 1)Gii

Phng trnh tng ng vi 3(

2x2 + 1 1) = x (1 + 3x+ 82x2 + 1) 3x2 + x+ 3 + (8x 3)2x2 + 1 = 0

Page 12


t

2x2 + 1 = t (t 1), phng trnh vit li thnh3t2 + (8x 2) t 3x2 x = 0 c = (10x 3)2Nn phng trnh c hai nghim

t = x 2x2 + 1 = xx 02x2 + 1 = x2

9

h ny v nghim.

t = 1 3x 2x2 + 1 = 1 = 3x

x

1

37x2 6x = 0

x = 0

Vy phng trnh c nghim l x = 02.

Bi 28: Gii phng trnh 3x3 13x2 + 30x 4 = (6x+ 2)(3x 1)3Gii

K x 43

;x 13

Ta bin i nh sau 3x3 13x2 + 30x 4 = (x2 3x+ 2)(3x 4) + 2(6x+ 2)Nu x 1

3VT

4

36x+ 2

3x 4 = (x 1)2 x = 3

Vi t = x 22

6x+ 2

3x 4 =x 2

2{x 23x3 16x2 + 4x 24 = 0()

Gii phng trnh (*) ta c nghim gn ng x 5, 36278, cc bn c th s dng phng phpCardano tnh chnh xc nhng n qu di v phc tp nn ta khng cp.Vy phng trnh cho c 2 nghim x = 3 x 5, 362782

Bi tp Gii cc PT sau:1/6x2 10x+ 5 (4x 1)6x2 6x+ 5 = 02/(x+ 3)

10 x2 = x2 x 12 (H Dc-1999)

3/2(1 x)x2 + 2x 1 = x2 2x 1 (H Dc 1997)4/(4x 1)x2 + 1 = 2x2 + 2x+ 15/2(1 x)x2 + x+ 1 = x2 3x 16/(x+ 1)

x2 2x+ 3 = x2 + 1 (Ch thm bt c chnh phng).

7/(4x 1)x3 + 1 = 2x3 + 2x+ 1

Page 13


Bi 29: Gii phng trnh 2

2x+ 4 + 4

2 x = 9x2 + 16

Bi ny thot nhn th ch c dng iu ging mt phng trnh a v n ph khng hon ton.Nhng n chnh l mt phng php gii quyt rt hay cho bi ton ny.

Gii

Li gii: K |x| 2Bnh phng 2 v ta c :4(2x+ 4) + 16

2(4 x2) + 16(2 x) = 9x2 + 16

8(4 x2) + 162(4 x2) = x2 + 8xn y bn no tinh , s quan st c c 2 v c dng hng ng thc, v c a v a2 = b2

Tht vy thm 16 vo 2 v ta c (2

2(4 x2) + 4)2 = (x+ 4)2, n y th rt d dng ri nh,nhng mc ch ca ta l a v n ph khng hon ton.Ta vit li PT 8(4 x2) + 162(4 x2) = x2 + 8x, t t = 2(4 x2) 4t2 + 16t x2 8x = 0

Gii phng trnh trn theo n t ta c t1 =x

2; t2 =

x2 4

V K |x| 2 nn t2 khng tha iu kin.Vi t =

x

2th

2(4 x2) = x2

x = 4

2

3(Tha mn K) 2

Bi 30: Gii phng trnh (3x+ 2)

2x 3 = 2x2 + 3x 6

Gii

Li gii: iu kin x 32

t t =

2x 3; t 0 t2 + 3 = 2xTa s thm bt theo n ph a v phng trnh theo t v x l tham s.PT t2 (3x+ 2)t+ 2x2 + x 3 = 0Ta c = 9x2 + 12x+ 4 4(2x2 + x 3) = (x+ 4)2Nn ta gii c t = 2x+ 3 hoc t = x 1TH1 Vi t = 2x+ 3

2x 3 = 2x+ 3 (VN)

TH2 Vi t = x 1

2x 3 = x 1(x 1) x = 2 (Tha K)Vy phng trnh c nghim duy nht x = 22

Bi 31: Gii phng trnh 4x+ 1 1 = 3x+ 21 x+1 x2

Nhn xt: Mi u khi gp bi ton ny ti thy n kh d bi thy s xut hin cax+ 1

v

1 x, nn ti t n ph a v h phng trnh nhng khi nhn k li th 3x khng thbiu din hon ton theo n ph c b tc.

Gii

Page 14


Ta gii bi ny nh sau: t

1 x = t

PT 3t2 (2 +1 x)t+ 4(x+ 1 1) = 0Ta tnh = (2 +

1 x)2 48(x+ 1 1), ta thy khng c dng chnh phng, mu cht bi

ton ny nm ch 3x.Ta s tm v sao cho:3x+ 1 = (

1 x)2 + (1 + x)2, s dng ng nht h s ta d dng tm c = 1; = 2

PT t2 (2 +x+ 1)t 2(x+ 1) + 4x+ 1 = 0Ta c = 9x+ 13 12x+ 1 = 9(x+ 1) 12x+ 1 + 4 = (3x+ 1 2)2Phn tip theo xin dnh cho bn c.2Bi ton ny khng d mt cht no i vi nhng ai khng nm k cch gii cng nh bin i,vn y l phi tinh tch 3x thnh hai dng c biu thc nh trong cn, n y bi tonmi thc s c gii quyt.

Bi 32: Gii phng trnh 2(2

1 + x2 1 x2)1 x4 = 3x2 + 1

Gii

Li gii: iu kin 1 x 1t a =

1 + x2; b =

1 x2 3x2 + 1 = 2(1 + x2) (1 x2) = 2a2 b2

Khi phng trnh tr thnh2(2a b) ab = 2a2 b2 2a2 + a(b 4) + 2b b2 = 0Ta c a = (b 4)2 8(2b b2) = (3b 4)2Nn ta suy ra a =

b

2hoc a = 2 b

TH1 Vi a =b

2 21 + x2 = 1 x2 (VN)

TH2 Vi a = 2 bx2 + 1 = 21 x2

x2 + 1 +1 x2 = 2 2 + 21 x4 = 4 x = 0Vy S = {0}2

S dng h s bt nh

Bi 33: Gii phng trnh 2x2 11x+ 21 3 34x 4 = 0 (Hc sinh gii quc gia -1995, bngA)Bi ny c cch gii rt hay v gn bng Bt ng thc Cauchy, mt cch gii bng n ph cngrt sng to.Li gii: Ta cn tm a, b, c sao cho:2x2 11x+ 21 = a(4x 4)2 + b(4x 4) + c 2x2 11x+ 21 = 16ax2 + (4b 32a)x+ (16a 4b+ c)ng nht h s ta thu c a =

1

8; b = 7

4; c = 12

Ta vit li PT nh sau:1

8(4x 4)2 7

4(4x 4) + 12 34x 4 = 0

t u = 3

4x 4, khi PT tr thnhu6 14u3 24u+ 96 = 0 (u 2)2(u4 + 4y3 + 18u+ 24) = 0D thy u4 + 4y3 + 18u+ 24 = 0 (VN) v:*Nu u 0 th u6 14u3 24u+ 96 > 0*Nu u > 0 th u4 + 4y3 + 18u+ 24 > 0Vy u = 2 x = 32

Page 15



1 xx+ 1 + 31 x2 = 3 x

K 1 x 1Ta tm ; sao cho x+ 3 = (1 x)2 + (x+ 1)2 x+ 3 = ( )x+ + Gii ra ta c = 2; = 1Ta vit li phng trnh thnh 1 + x+ 2(1 x) 21 x+x+ 1 31 x2 = 0t u =

1 + x; v =

1 x v u, v 0, phng trnh tr thnh

u2 + 2v2 2v + u 3uv = 0 u2 + (1 3v)u+ 2v2 2v = 0 = (1 3v)2 4(2v2 2v) = (v + 1)2Nn u = 2v hoc u = v 1 u = 2v x+ 1 = 21 x x = 5

3

u = v 11 x

1

24x2 = 3

Nn phng trnh c 2 nghim x =5

3;x =

3

2

Vy S =

{5

3;

3

2

}2

Tuy nhin phng php dng h s bt nh ny ch gii quyt c mt s lp bi v trongphng trnh v t dng bi ny cng khng nhiu, ta cng xt 2 v d thot u nhn th cngtng t nhng khng th gii quyt bng cch nh trn c, phn ny ca dng a v phngtrnh tch nhng ti mun a ra y gip ta linh hot khi gii ton ch khng phi ci my nh.


1 x = x+ 6 31 x2 + 51 + x

Ta s lm cch nh trn nh, biu din x+ 6 = (

1 x)2 + (x+ 1)2Gii ra ta c =

1

2; =

7

2, thay vo PT u nhng ta khng nhn c chnh phng.

Li gii: t a =

1 + x v b =

1 xPT 2x+ 2 + 1 x+ 51 + x 31 x2 41 x+ 3 = 0 2a2 + b2 3ab+ 5a 4b+ 3 = 0By gi ta s c nhm sao cho t c nhn t chung, thng l nhm v dng a = b (a b)(2a b) + 3(a b) + (2a b) + 3 = 0 (a b+ 1)(2a b+ 3) = 0

TH1 a+ 1 = b x+ 1 + 1 = 1 x 2x+ 1 = (2x+ 1);x 1

2

x =

3

2TH2 2a+ 3 = b (PTVN)

V d tng t sau xin dnh cho bn c

Bi 36: Gii phng trnh 4 + 2

1 x = 3x+ 5x+ 1 +1 x2

Page 16


p s: Phng trnh c 3 nghim S =

{0;

24

25;

3

2

}2

t n ph a v h phng trnh.

Ta s tip tc vi 1 phng php lm l t n ph a v h, ch ny kh "di hi" vnhiu bi ton s c gii quyt rt gn bng phng php ny

Dng 1. Phng trnh c dng xn + a = b nbx a

Cch gii: t y = nbx a khi ta c h i xng loi II{

xn by + a = 0yn bx+ a = 0

Ta xt bi ton sau

Bi 37: Gii phng trnh x3 + 1 = 2 3

2x 1 (H Dc-1996)

Gii

t y = 3

2x 1 y3 = 2x 1Ta c h PT sau

{x3 + 1 = 2yy3 + 1 = 2x

y l h i xng loi II, tr v theo v ta c:x3 y3 = 2(y x) (x y)(x2 + y2 + xy + 2) = 0x = y 32x 1 = x x3 2x+ 1 = 0 (x 1)(x2 + x 1) = 0Vy x = 1;x =

152

Ta c x2 + y2 + xy + 2 =(x+

y

2

)2+

3y2

4+ 2 > 0,x, y

Vy PT cho c 3 nghim 2.

Phng trnh c dng:na f(x) + mb+ f(x) = cCch gii: t u = n

a f(x); v = mb+ f(x)

Ta c h sau

{u+ v = c

un + vm = a+ b

Bi 38: Gii phng trnh 4x+ 8 + 4

x 7 = 3

Gii

t u = 4x+ 8 0 u4 = x+ 8 x = u4 8

v = 4x 7 0 v4 = x 7 x = v4 + 7

Ta c h:

u+ v = 3

u, v 0u4 v4 = 15

v = 3 u(u2 v2)(u2 + v2) = 15u, v 0

v = 3 u(u v)(u+ v)(u2 + v2) = 15u, v 0

v = 3 u0 u 3(u v)(u2 + v2) = 5

Page 17


{

0 u 3(2u 3) [u2 + (3 u)2] = 5

{0 u 3(2u 3)(2u2 6u+ 9) = 5

{

0 u 34u3 18u2 + 36u 32 = 0

{0 u 3u = 2

{

4x+ 8 = 2

4x 7 = 1

{x+ 8 = 16

x 7 = 1 x = 8

Vy phng trnh cho c nghim duy nht x = 82.

Bi tp Gii phng trnh1/ 3x+ 34 3x 3 = 14

2/ 4

97 x+ 4x = 53/ 3x+ 2 +

x+ 1 = 3

4/ 4

18 x+ 4x 1 = 35/ 4

17 x8 32x8 1 = 1

Bi 39: Gii phng trnh 2 3

3x 2 + 36 5x = 8 (A-2009)

Gii

t u = 3

3x 2; v = 6 5x 0{u3 = 3x 2v2 = 6 5x

5u3 + 3v2 = 5(3x 2) + 3(6 5x) = 8(1)Mt khc ta li c 2u+ 3v 8 = 0(2)T (1) v (2) ta c h sau:{

5u3 + 3v2 = 82u+ 3v = 8

5u3 + 3(

8 2u3

)2= 8

15u3 + 4u2 32u+ 40 = 0Phng trnh c nghim duy nht u = 2Nn 3

3x 2 = 2 x = 22


1 +

1 x2[

(1 + x)3 (1 x)3] = 2 +1 x2 (Olympic 30/4/2011)Nhn xt: Bi ton ny nhn vo c v kh phc tp nhng (

1 + x)2 + (

1 x)2 = 2.

Gii

Li gii: K 1 x 1t

1 + x = a;

1 x = b vi a, b 0 a2 + b2 = 2 (*)Ta c h sau{

a2 + b2 = 2(1)1 + ab(a3 b3) = 2 + ab(2)

Ta c 1 + ab =1

2(2 + 2ab) =

1

2(a2 + b2 + 2ab) do (*)

1 + ab = 12

(a+ b) v a, b 0

Page 18


Vy t PT(2) ta c12

(a+ b)(a b)(a2 + b2 + ab) = 2 + ab

12

(a2 b2) = 1

Kt hp vi (1) ta c h sau

{a2 b2 = 2a2 + b2 = 2

Cng v ta c 2a2 = 2 +

2

a2 = 1 + 12

1 + x = 1 + 12

x = 122

Nhn xt: bi ton ny tng ca ta l thay h s bng n t phng trnh th nht ca hc th gii quyt bi ton c d dng hn. Sau y l mt v d nh tng t.

Bi 40b: Gii phng trnhx+ 1 + x+ 3 =

1 x+ 31 x2

Gii

t

{u =x+ 1 0

v =

1 x 0

Phng trnh cho tr thnh

{u2 + u+ 2 = v + 3uv

u2 + v2 = 2

Thay 2 = u2 + v2 vo phng trnh u ta c 2u2 + u+ v2 = v+ 3uv 2u2 + (1 3v)u+ v2 v = 0Ta c = (v + 1)2. n y cc bn c th gii quyt d dng 2.

Bi 41: Gii phng trnh (x+ 5)x+ 1 + 1 = 3

3x+ 4

Gii

Li gii: K x 1t a =

x+ 1; b = 3

3x+ 4

x = a2 1 v 3a2 + 1 = b3

Thay vo phng trnh ta c h sau{(a2 + 4)a+ 1 = b

3a2 + 1 = b3

Cng v theo v ta c a3 + 3a2 + 4a+ 2 = b3 + bn y quan st k mt cht, ta bin i nh sau (a+ 1)3 + (a+ 1) = b3 + bXt hm s c trng f(t) = t3 + tTa c f (t) = 3t2 + 1 > 0, vy hm s ng bin. f(a+ 1) = f(b)

Nnx+ 1 + 1 = 3

3x+ 4

t u =x+ 1; v = 3

3x+ 4

Ta c h sau

Page 19


{u+ 1 = v

v3 3u2 = 1 {

u = v 1v3 3u2 = 1

S dng php th ta c v3 3(v 1)2 = 1 v3 3v2 + 6v 4 = 0 (v 1)(v2 v + 4) = 0Phng trnh c nghim duy nht v = 1 33x+ 4 = 1 x = 1Vy phng trnh c nghim duy nht x = 12

Phng trnh dngax+ b = cx2 + dx+ e

Ta gp cc dng bi ton nhax+ b = cx + d v mt s v d nu trn bng cch bnh

phng bc 4 v ng nht h s tm c nghim, nhng i vi nhng bi ton khng dngc phng php th sao? Chng ta cng lm r vn .

Xt v d sau

Bi 42: Gii phng trnh 2x2 6x 1 = 4x+ 5

Gii

Li gii: K x 54

Ta bin i phng trnh nh sau4x2 12x 2 = 24x+ 5 (2x 3)2 = 24x+ 5 + 11.t 2y 3 = 4x+ 5 ta c h phng trnh sau:{

(2x 3)2 = 4y + 5(2y 3)2 = 4x+ 5

(x y)(x+ y 1) = 0

Vi x = y 2x 3 = 4x+ 5 x = 2 +3.Vi x+ y 1 = 0 y = 1 x x = 122

Nhn xt: Chc cc bn ang ngc nhin v khng bit ti sao ta c th t nh vy, khngphi l on m u. Phng php ny rt hu dng vi ai hc qua o hm l c th d dngt c.Bi ton c dng nh sau

Dng 1:ax+ b = cx2 + dx+ e, (a 6= 0, c 6= 0, a 6= 1

c)

Xt f(x) = cx2 + dx+ e f (x) = 2cx+ dGii PT f (x) = 0, khi bng php t

ax+ b = 2cy + d, ta s a c v h i xng loi II

tr mt s trng hp c bit.C th thy r rng qua v d trn, ta xt v d tip theo

Bi 43: Gii phng trnh x2 4x 3 = x+ 5Lm nhp: Xt f(x) = x2 4x 3 f (x) = 2x 4Gii f (x) = 0 x = 2

Gii

Li gii: K x 27;x 2 +7tx+ 5 = y 2 (y 2)2 = x+ 5

Page 20


Ta bin i phng trnh u lix+ 5 = (x 2)2 7

Thay y 2 vo PT u ta thu c h sau{(x 2)2 = y + 5(y 2)2 = x+ 5Tr v theo v ta c (x y)(x+ y 3) = 0

TH1 : x = y x+ 5 = x 2;x 2

[x = 1

2(5 +

29)

x = 12(529)

TH2 : 1 x =x+ 5;x 1

[x = 1x = 4

i chiu vi iu kin ta nhn x = 1;x = 12

(5 +

29)

Vy PT cho c nghim S =

{1; 1

2(5 +

29)

}

Bi 44: Gii phng trnh x2 + 5 +

3x+ 1 = 13x

Nhn xt. Lm tng t ta vit li phng trnh nh sau3x+ 1 = 4x2 + 13x 5 v t f(x) = 4x2 + 13x 5

Ta c f (x) = 8x + 13 nu ta gii ra v t bng phng php tng t nh trn s khng thuc h i xng loi II.

Gii

Li gii. K x 13

t

3x+ 1 = (2y 3); y 32

Ta c h phng trnh sau

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

Tr v theo v ta c (x y)(2x+ 2y 5) = 0 Vi x = y x = 15

97

8

Vi 2x+ 2y 5 = 0 x = 11 +

73

8

Vy tp nghim ca phng trnh l S =

{1597

8;11 +

73

8

}2

Ngoi cch ny, cc bn vn c th t

3x+ 1 = t, ri bin i phng trnh thnh4x2 10x t2 + t+ 6 = 0 c = (2t 1)2Nhn xt. Ta thy cch gii bi ton ny khc so vi v d trn v a v h "gn i xng" loiII nhng vn gii c mt cch d dng. Dng ton ny c dng nh sau:ax + b = r(ux + v)2 + dx + e v tha mn

{u = ar + d

v = br + e

Cch gii. t uy + v =ax+ b khi ta c h

{uy + v = r(ux+ v)2 + dx+ e

ax+ b = (uy + v)2

Ta vit li phng trnh trn nh sau

3x+ 1 + (2x 3)2 x 4 = 0. D dng ta kim tra ccc h s u tha mn, nhng khi t

3x+ 1 = 2y 3 th ta thu c h phng trnh khng

d dng gii mt cht no. Ta chuyn v v i du s a v h gn i xng gii c nh bi

Page 21


ton trn. Chc cc bn ang thc mc l khi no th dng o hm khi no th dng cch ti vanu. Tht s l kt hp c 2 cch y. o hm p dng c khi h s d = 0, cc bn c th ddng kim tra, cn nu khng c th dng cch thm bt nh trn.

Bi tp: Gii cc phng trnh sau1/

2x 1 + x2 3x+ 1 = 02/

2x+ 15 = 32x2 + 32x 203/x 1 + x2 x 3 = 0

4/x2 x 20041 + 16032x = 2004 (HSG Bc Giang 2003-2004)5/

9x 5 = 3x2 + 2x+ 36/x2 =

2 x+ 2

Dng 2:ax+ b =

1

ax2 + cx+ d(a 6= 0) v tha mn b+ ad = a

2c

2

(1 +

c

2

)Cch gii: Xt hm s f(x) =

1

ax2 + cx+ d f (x) = 2

ax+ c = 0 x = ac

2ta a v h i

xng quen thuc.

V d: Gii phng trnh 3x2 + x 296

=

12x+ 61

36

Lm nhp:f(x) = 3x2 + x 296 f (x) = 6x+ 1 = 0 x = 1

6

Gii

t

12x+ 61

36= y +

1

6, y 1

6 12x+ 61

36= y2 +

1

3y +

1

36 12x+ 61 = 36y2 + 12y + 1 3y2 + y = x+ 5Mt khc t phng trnh u ta c 3x2 + x 29

6= y +

1

6 3x2 + x = y + 5

Nn ta c h sau

{3x2 + x = y + 5

3y2 + y = x+ 5

Tr v theo v ta c (x y)(3x+ 3y + 2) = 0 x = y y = 3x+ 23

Vi x = y 3y2 = 5 x = y =

5

3; y 1

6

Vi y = 3x+ 23 3x2 + x = 3x+ 2

3+ 5 9x2 + x 13 = 0

x = 3

126

9T y tm c y v kt lun nghim 2

Dng 3: 3ax+ b = cx3 + dx2 + ex+m, (a 6= 0, c 6= 0, a = 1

c)

Cch gii: Xt hm s cx3 + dx2 + ex + m, ta gii phng trnh o hm cp hai bng khng.

f (x) = 6cx+ 2d = 0 x = d3c

.

Sau bng php t 3ax+ b = y +

d

3cta a c v h i xng.

V d 1: Gii phng trnh 3

3x 638

=x3

3 3

2x2 +

9

4x

Page 22


Lm nhp: Ta c f (x) = 2x 3 = 0 x = 32

Gii

t 3

3x 638

= y 32 3x 63

8= y3 9

2y2 +

27

4y 27

8 12x 18 = 4y3 18y2 + 27y

Mt khc t phng trnh u ta c c 12y 18 = 4x3 18x2 + 27x, ta c h sau{12x 18 = 4y3 18y2 + 27y12y 18 = 4x3 18x2 + 27x

Gii h ny khng cn kh khn2.

Dng 4: 3ax+ b = cx3 + dx2 + ex+m, (a 6= 0, c 6= 0, a 6= 1

c)

Cch gii: Cng tng t nh trn : Xt hm s f(x) = cx3 + dx2 + ex + m, gii phng trnh

f (x) = 0 6cx+ 2d = 0 x = d3c

.

Khi cng bng cch t 3ax+ b = 3cy + d, ta a v h i xng.

V d 2: Gii phng trnh 3

81x 8 = x3 2x2 + 43x 2 (THTT T6/2001)

Lm nhp: Ta c f (x) = 6x 4 x = 23

Gii

t 3

81x 8 = 3y 2 3x = y3 2y2 + 43y, bin i tng t nh trn ta c h{

3y = x3 2x2 + 43x

3x = y3 2y2 + 43y

(x y)(x2 + xy + y2 2x 2y + 133

= 0

V (x2 + xy + y2 2x 2y + 133

=1

2(x+ y)2 +

1

2(x 2)2 + 1

2(y 2)2 + 1

3> 0

Nn x = y thay vo phng trnh ta gii tip tc2.Nhn xt: Tuy dng bi ny vn gii c cch dng hm s, nhng y cng l mt cch rt huhiu gii quyt dng ton ny. Ta cng n vi mt s bi ton tng t xut hin trong cc kthi.Gii cc phng trnh sau:1/x3 + 3x2 3 33x+ 5 = 1 3x ( ngh Olympic 30/4/2009)2/x3 15x2 + 78x 141 = 5 32x 9 (Olympic 30/4/2011)3/8x3 4x 1 = 36x+ 1


2 1 x+ 4x = 14

2

Gii

iu kin 0 x 2 1t

{2 1 x = u

4x = v

{

0 u

2 10 v 4

2 1

Nh vy ta c h

u+ v =14

2u2 + v4 =

2 1

u =

14

2 v(

14

2 v

)2+ v4 =

2 1

Page 23


T phng trnh th hai ta c: (v2 + 1)2 =

(14

2+ v

)2 v2 v + 1 1

4

2= 0

v =1

44

2 3

2(Tha K). Nn x =

1

44

2 3

2

4

2


1 x2 =(

2

3x

)2Gii

iu kin{1 x2 0x 0

{1 x 1x 0 0 x 1

t u =x, v =

2

3x vi u 0, v 2

3Ta c h phng trnh

1 x2 = 1 u4(2

3x

)2= v2

Do ta c h{u+ v =

2

31 u4 = v2

{u+ v =

2

3u4 + v4 = 1

u+ v =

2

3(u2 + v2)

2 2u2.v2 = 1 u+ v =

2

3[(u+ v)2 2u.v]2 = 1

u+ v =

2

3(4

9 2u.v

)2 2u2.v2 = 1

{

u+ v = 23

2u2.v2 169u.v 65

81= 0

u+ v =

2

3

u.v =8194

18u+ v =

2

5

u.v =8 +

194

18

Nn u, v l nghim ca phng trnhy2 23y + 8

194

18= 0(a)

y2 23y +

8 +

194

18= 0(b)

Ch c (a) l c nghim nn

y =1

6

(2

2(2

194 6))

Do {u1 = y1v1 = y2

{u2 = y2v2 = y1

V u 0 nn ta chn u = 16

(2 +

2(2

194 6))

x = 16

(2 +

2(2

194 6))

Vy phng trnh c nghim duy nht

x =1

9

(2 +

2(

194 6) +

97

2

)2

Page 24


Nhn xt: Bi ny thc s l kh, phc tp khng ch i hi s sng to linh hot trong ccht n ph m khi ta gii cc phng trnh bc 2 my tnh khng bm ra s c m i hi ta phivng k nng tnh ton ch khng phi lc no cng da vo my tnh.

Bi 47: Gii phng trnh 4x2 11x+ 10 = (x 1)2x2 6x+ 2

Nhn xt: Bi ny khi c ta ngh ngay n cch gii bng n ph khng hon ton bngcch t

2x2 6x+ 2, ri thm bt VT nhng ta khng nhn c chnh phng, ta gii bi

ny bng cch a v h phng trnh n ph khng hon ton.

Gii

PT (2x 3)2 + x+ 1 = (x 1)(x 1)(2x 3) x 1t u = 2x 3; v = (x 1)(2x 3) x 1Ta c h phng trnh

{u2 + x+ 1 = (x 1)vv2 + x+ 1 = (x 1)u

Tr v theo v ta c u2 v2 = (x 1)(v u) (u v)(u+ v + x 1) = 0 u = v u2 + x+ 1 = (x 1)u (2x 3)2 + x+ 1 = (x 1)(2x 3) 2x2 6x+ 7 = 0 phng trnh v nghim. u+ v + x 1 = 0 2x 3 +2x2 6x+ 2 + x 1 = 0

2x2 6x+ 2 = 4 3xx

4

37x2 18x+ 14 = 0

h ny v nghim.

Vy phng trnh cho v nghim.2Nhn xt: Cch gii ca bi ton ny pht trin ln t 1 cch lm nu trnfn(x) + b = a n

af(x) b, trong a, b R

Ta d on f(x) = (2x+ c)2. n y ta ng nht h s tm c4x2 + 4cx+ c2 + (11 4c)x+ 10 c2 = (x 1)(x 1)(2x+ c) (11 4c)x 10 + c2b = (11 4c)x+ 10 c2. i chiu vi bi ton ng nht h s suy ra x = 3

Ta xt tip v d sau, kh nhnh hn 1 cht

Gii phng trnh 3x3 6x2 3x 17 = 3 39(3x2 + 21x+ 5)Nhn xt: Cng ging nh bi ton trn nhn vo biu thc ta c th d on f(x) = (3x + c) f(x)x+ c. y ta chn f(x) = 3x+ c

PT 27x3 54x2 27x+ 153 = 27 39(3x2 + 21x+ 5)Tuy nhin n y nu ta p dng cch cn bng h s th c l phc tp. Ta hy ch n nuchng ta tm ra biu thc f(x) ph hp th biu thc b cn tm s c bc cao nht l 2. V cbit hn khi ta p dng af(x) b cho biu thc trong cn th h s bc 2 trong biu thc b ginguyn ch thay i bc nht v hng t t do. V af(x) = a(3x + c) ch cho cc hng t bc nht3ax v h s t do ac . Ta thy 27x2 s chc chn c trong biu thc b. Vy biu thc bc 2 trongf(x) = 81x2. Mt khc khi khai trin f(x) th h s ca hng t bc 2 l 27x2c, vy c = 3. Hayta d on f(x) = 3x 3Ta phn tch kim tra(3x 3)3 + (27x2 108x 126) = 27 39(3x2 + 12x+ 5) = 27 327(3x 3) (27x2 108x 126)Ta t u = 3x 3; v = 327(3x 3) (27x2 108x 126)

Page 25


Ta c h phng trnh sau

{u3 + (27x2 108x 126) = 27vv3 + (27x2 108x 126) = 27u

Phn tip theo xin dnh cho bn c.2Cng bng cch tng t nh trn bn c cng c th gii quyt bi ton tng t nh sauBi tp: Gii phng trnh1/ x3 x2 10x 2 = 37x2 + 23x+ 122/7x2 13x+ 8 = 2x2 3x(1 + 3x 3x2)3/8x2 13x+ 7 =

(x+

1

x

)3

3x2 2

Phng php lng gic ha.

Nu bi ton cha a2 x2 c tht x = |a| sin t vi pi

2 t pi

2hocx = |a| cos t vi 0 t pi

Nu bi ton cha x2 a2 c tht x =

|a|sin t

vi t [pi

2;pi

2

]\ {0}

Hoc x =|a|

cos tvi t [0; pi] \

{pi2

} Nu bi ton cha a2 + x2 c th: t:x = |a| tan t vi t

(pi

2;pi

2

)Hoc x = |a| cot t vi t (0; pi). Nu bi ton cha

a+ x

a x hoca xa+ x

c th: tx = a cos 2t.

Nu bi ton cha(x a) (b x) c th t x = a+ (b a) sin2t.Li th ca phng php ny l a phng trnh ban u v mt phng trnh lng gic cbn bit cch gii nh phng trnh ng cp, i xng ... V iu kin nhn hoc loi nghimcng d dng hn rt nhiu. V lng gic l hm tun hon nn ta ch t iu kin cc biuthc lng gic sao cho khi khai cn khng c du tr tuyt i, c ngha l lun dng.

Bi 48: Gii phng trnh x3 +

(1 x2)3 = x2(1 x2)K 1 x 1T iu kin ca bi ton ta t n ph x = cos t, khi

1 x2 = |sin|

Ch cn chn m 0 pi khi 1 cos = x 1 v sin 0 v |sin| = sinPT cho bin i c v dng :cos3+ sin3 =

2cossin

(cos+ sin) (1 cossin) = 2cossint u = cos+ sin =

2 sin

(+

pi

4

)Do 0 x pi pi

4 + pi

4 5pi

4

2

2 sin

(+

pi

4

) 1, ta c 1

2 u 2

Phng trnh i s vi n u c dng :

u

(1 u

2 12

)=

2u2 1

2

u3 +2u2 3u2 = 0 (u2) (u2 + 22u+ 1) = 0

Page 26


u = 2u = 2 + 1u = 2 1 < 2

TH1 :u =

2 sin(+

pi

4

)=

2 sin(+

pi

4

)= 1 = pi

4+ k2pi , k Z

TH2 :u =

2 sin(+ pi

4

)= 12 v cossin = u

2 12

= 12

Khi cos , sin l nghim ca phng trnh

X2 (12)X + 12 = 0 X = 12(

2 1) (2 + 3)2

Do sin 0 cho nn cos =12

(2 1) (2 + 3)2

Vy pt c nghim : x =12

(2 1) (2 + 3)2

, x=

2

22

Bi 49: Gii phng trnh 2x2 +

1 x+ 2x1 x2 = 1

Gii

K x [1; 1]t x = cos t, t [0; pi]Phng trnh tng ng 2 cos2 t+

2 sin

t

2+ 2 sin t cos t = 1

cos 2t+ sin 2t = 2 sin t2

cos(

2t pi4

)= sin t

2

cos(

2t pi4

)= cos

(t

2+pi

2

)

2t pi4 = t2 + pi2 + k2pi

2t pi4

= t2 pi

2+ k2pi

t = pi2 + k4pi3t = pi

10+k4pi

5Da vo iu kin nghim ca phng trnh ta nhn 2 nghim l

x = cospi

2;x = cos

7pi

10

Vy S =

{cos

7pi

10; 0

}2


1 +

1 x2 = x(1 + 21 x2)

Gii

Page 27


iu kin:1 x2 0 1 x 1t x = sin t vi t

[pi

2;pi

2

]Khi phng trnh c dng:

1 +

1 sin2t = sin t

(1 + 2

1 sin2t

) 1 + cos t = sin t (1 + 2 cos t)

2 cos t2

= sin t+ sin 2t 2 cos t2

= 2 sin3t

2cos

t

2

2 cos t2

(12 sin 3t

2

)= 0

cos t2 = 0sin

3t

2=

2

2

t =

pi

6

t =pi

2

x =1

2

x = 1

Vy tp nghim ca phng trnh cho l S =

{1

2; 1

}2

Bi 51: Gii phng trnhx2 + 1 +

x2 + 1

2x=

(x2 + 1)2

2x(1 x2)

Gii

iu kin

{x 6= 1x 6= 0

t x = tan t; t (pi

2;pi

2

)\{pi

4; 0}

Khi x2 + 1 = tan2t+ 1 =1

cos2t x2 + 1 = 1

cos t

sin 2t =2 tan t

1 + tan2t=

2x

x2 + 1 x

2 + 1

2x=

1

sin 2t

cos 2t =1 tan2t1 + tan2t

=1 x2x2 + 1

sin 2t. cos 2t = 2x (1 x2)

(x2 + 1)2

sin 4t = 4x (1 x2)

(x2 + 1)2 2

sin 4t=

(x2 + 1)2

2x (1 x2)Khi phng trnh c bin i v dng

1

cos t+

1

sin 2t=

2

sin 4t 4 sin t. cos 2t+ 2 cos 2t = 2

2 sin t. cos 2t = 1 cos 2t 2 sin t. cos 2t = 2sin2t (cos 2t sin t) sin t = 0 (1 2sin2t sin t) sin t = 0 (sin t+ 1) (2 sin t 1) sin t = 0 sin t = 1

2 t = pi

6 x = 1

3.

Vy nghim duy nht ca phng trnh cho l:x =132.

Bi 52: Gii phng trnh 5 + 3

1 x2 = 8x6 + 8(1 x2)3

Gii

iu kin x [1; 1]t x = sin t;pi

2 t pi

2Phng trnh cho tng ng vi 5 + 3

1 x2 = 8 (x6 + (1 x2)3)

5 + 3 cos t = 8(sin6 t+ cos6 t) sin6 t+ cos6 t = 58

+3

8cos 4t

Page 28


cos 4t = cos t

t = k2pi3t =

k2pi

5

Kt hp vi iu kin ca t, ta gii c x = 0 x = sin 2pi52

MT S BI TON CHN LC


7x+ 1 3x2 x 8 + 3x2 8x+ 1 = 2

Gii

t a = 3

7x+ 1; b = 3

8 + x x2; c = 3x2 8x+ 1Ta c h sau

{a+ b+ c = 2

a3 + b3 + c3 = 8.

{(a+ b+ c)3 = 8

a3 + b3 + c3 = 8 (a+ b)(b+ c)(c+ a) = 0

a = b x = 1 x = 9 b = c x = 1 c = a x = 0 x = 1Vy phng trnh c nghim S = {1; 1; 0; 9}2

Bi tp tng ng 3

3x+ 1 + 3

5 x+ 32x 9 34x 3 = 0

Bi 2: Gii phng trnh15

2(30x2 4x) = 2004(30060x+ 1 + 1)

Gii

PT (30x2 4x) = 4008(30060x+ 1 + 1)t y =

30060x+ 1 + 1

15 15y 1 = 30060x+ 1

15y2 + 2y = 2004xMt khc t phng trnh u ta c 30x2 4x = 4008y 15x2 2x = 2004yTa c h phng trnh

{15x2 2x = 2004y15y2 + 2y = 2004x

Tr v theo v ta c (x y)(15(x+ y) + 2002) = 0 Vi x = y 15x 1 = 30060x+ 1 x = 0 x = 2006

15Vi x = 0 th phng trnh u v nghim.

Vi 15(x+ y) + 2002 = 0. Ta c 30060x+ 1 0 y =

30060x+ 1 + 1

15 1

15

Nn x+ y 130060

+1

15> 0

Vy 15(x+ y) + 2002 = 0 v nghim2.


1 x = x+ 3 + 31 x+1 x2

Gii

Page 29


t x = cos t; t [0; pi]PT 41 + cos t = cos t+ 3 + 31 cos t+ sin t 42 cos t

2= cos t+ 3 + 3

2 sin

t

2+ 2 sin

t

2cos

t

2

42 cos t2

= 4 2 sin2 t2

+ 3

2 sint

2+ 2 sin

t

2cos

t

2

2 cos t2

(2

2 sin t2

)+ 2 sin2

t

2 32 sin t

2 4 = 0

2 cos t2

(2

2 sin t2

)+

(sin

t

2 22

)(2 sin

t

2+

2

)= 0(

sint

2 22

)(2 sin

t

2 2 cos t

2+

2

)= 0

Vi sin t2

= 2

2 (PTVN)

Vi 2 sin t2 2 cos t

2+

2 = 0

sin t2 cos t

2= 1

2

sin(t

2 pi

4

)= 1

2

t

2 pi

4= pi

6+ k2pi

t

2 pi

4=

7pi

6+ k2pi

t =

pi

6

t =17

6pi(l)

i chiu vi iu kin ca t, phng trnh c nghim duy nht x = cospi

6=

3

22

Bi 4: Gii phng trnh (x3 3x+ 1)x2 + 21 + x4 3x2 + x = 21(Trch bi vit ca anh L Phc L)

Gii

t t =x2 + 21 > 0

(x2 + 21) (x3 3x+ 1)x2 + 21 (x4 2x2 + x) = 0 t2 (x3 3x+ 1)t (x4 2x2 + x) = 0 = (x3 3x+ 1)2 + 4.(x4 2x2 + x) = x6 2x4 + 2x3 + x2 2x+ 1 = (x3 x+ 1)2Suy ra phng trnh c 2 nghim l:

t = x3 3x+ 1) (x3 x+ 1)

2[t = x(1)t = x3 2x+ 1(2)

[

x2 + 21 = xx < 0

[x2 + 21 = x2

x < 0Phng trnh ny v nghim.

(2) x2 + 21 = x3 2x+ 1 x2 + 21 5 = x3 2x 4

x2 4

x2 + 21 + 5= (x 2)(x2 + 2x+ 2)

x = 2x+ 2x2 + 21 + 5

= x2 + 2x+ 2(3)

Xt (3) V T |V T | 0x. Nn ta suy ra c iu kin xcnh l 0 < x < 1PT (x(x+ 1))2 + (x x)2 = (x2 + 1)x(x+ 1)(1 x) (2)t u = x(x+ 1); v = 1 x (vi u,v>0) th u+ v = x2 + 1. Ta c th vit (2) di dngu2 + v2 = (u+ v)

uv (uv)2(u+ v +uv) = 0

u = v (V u+ v +uv > 0)Vi u = v x(x+ 1) = 1 x x2 + 2x 1 = 0 PT c 2 nghim x = 12. i chiu iu kinch c x = 1 +2 tha mn.Vy phng trnh c nghim duy nht x = 1 +22.

Bi 6: Gii phng trnh 10x2 + 3x+ 1 = (6x+ 1)x2 + 3

Gii

t u = 6x+ 1; v =x2 + 3

M ta biu din VT theo n ph c nh sau 10x2 + 3x+ 1 =1

4(6x+ 1)2 + (x2 + 3) 9

4

Vy ta vit li phng trnh1

4u2 + v2 9

4= uv (u 2v)2 = 9 u 2v = 9

Vi u 2v = 3 1 + 6xx2 + 3 = 3 3x 1 = x3 + 3{

3x 1 0x2 + 3 = (3x 1)2 x = 1

Vi u 2v = 3 3x+ 2 = x2 + 3{

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

x =

7 34

Vy phng trnh c 2 nghim l x = 1 x =

7 34

2

PHP T N PH GII H PHNG TRNH

Bi ton 1 Gii h phng trnh{ 2 (x y) (1 + 4xy) = 3 (1)

x2 + y2 = 1 (2)

Gii

t : {x = siny = cos

Khi , phng trnh (2) tha mn vi mi .

Page 31


Phng trnh (1) tng ng vi phng trnh

2(sin cos)(1 + 2 sin 2) =

3

2.

2 sin( 450).2(

1

2+ sin 2

)=

3

4 sin( 450)(sin 2 + sin 300) =

3

8 sin( 450). sin( + 150) cos( 150) =

3

4 cos( 150)[cos 600 cos(2 300)] =

3

2 cos( 150) 4 cos( 150). cos(2 300) =

3

2 cos(3 450) =

3

[ = 650 + k1200

= 350 + l1200 (k, l Z)

Vy h cho c su nghim nh sau:

{x1 = sin 65

0

y1 = cos 650{

x2 = sin 1850

y2 = cos 1850{

x3 = sin 3050

y3 = cos 3050{

x4 = sin 850

y4 = cos 850{

x5 = sin 350y5 = cos 35

0{x6 = sin 205

0

y6 = cos 2050 2

Bi 2 Gii h phng trnh: {x+ y + x2 + y2 = 8xy(x+ 1)(y + 1) = 12

Gii

Bin i h tr thnh {x(x+ 1) + y(y + 1) = 8x(x+ 1).y(y + 1) = 12

t: {u = x(x+ 1)v = y(y + 1)

Khi , h cho tr thnh: {u+ v = 8 =u.v = 12

{u = 2v = 6{u = 6v = 2

Trng hp 1 {u = 2v = 6

{x2 + x 2 = 0y2 + y 6 = 0

[(1; 2), (1;3)(2; 2), (2;3)

Page 32


Trng hp 2 {u = 6v = 2

{x2 + x 6 = 0y2 + y 2 = 0

[(2; 1), (3; 1)(2;2), (3;2) 2

Bi 3 Gii h phng trnh {(x y)(x2 y2) = 3(x+ y)(x2 + y2) = 15

Gii

Bin i h cho ta thu c {x3 + y3 xy(x+ y) = 3x3 + y3 + xy(x+ y) = 15

t {u = x3 + y3

v = xy(x+ y)

H cho tr thnh {u+ v = 15u v = 3

{u = 9v = 6

Khi , ta c: {x3 + y3 = 9xy(x+ y) = 6

{x+ y = 3xy = 2

{x = 1y = 2{x = 2y = 1

2

Bi 4 Gii h phng trnh (2x+ y)2 5(4x2 y2) + 6(2x y)2 = 0

2x+ y +1

2x y = 3

Gii

iu kin: 2x y 6= 0t {

u = 2x+ yv = 2x y

H cho tr thnh

{u2 5uv + 6v2 = 0u+

1

v= 3

u

v= 3

u+1

v= 3

u

v= 2

u+1

v= 3

{u = 2v =2v2 3v + 1 = 0

{u = 2v = 1{u = 1

v =1

2

Trng hp 1 {2x+ y = 22x y = 1

x =

3

4

y =1

2

Page 33


Trng hp 2 {2x+ y = 1

2x y = 12

x =

3

8

y =1

4

p s: (x; y) =

(3

4;1

2

),

(3

8;1

4

)2

Bi 5 Gii h phng trnh x+

1

y+x+ y 3 = 3

2x+ y +1

y= 8

Gii

iu kin y 6= 0x+

1

y 0

x+ y 3 0()

Bin i h cho tr thnh x+

1

y+x+ y 3 = 3

x+1

y+ x+ y 3 = 5

t u =x+

1

yv =x+ y 3

vi

{u 0v 0 ()

Khi h cho tr thnh

{u+ v = 3 =u2 + v2 = 5

{u+ v = 3uv = 2

{u = 2v = 1{u = 1v = 2

tha mn ()

Trng hp 1

{u = 2v = 1

x+

1

y=

1

4x+ y 3 = 1

{x = 3y = 1{x = 5y = 1

tha mn ()

Trng hp 2

{u = 1v = 2

x+

1

y= 1

x+ y 3 = 4

{x = 410y = 3 +

10{

x = 4 +

10

y = 310tha mn ()

p s: (x; y) = (3; 1), (5;1), (410; 3 +10), (4 +10; 310)2

Page 34


Bi 6 Gii h phng trnh {4log3(xy) = 2 + (xy)log3 2

x2 + y2 3x 3y = 12Gii

iu kin: xy > 0t: u = log3(xy) xy = 3uKhi , h cho tr thnh {

xy = 3x2 + y2 3(x+ y) = 12

{xy = 3(x+ y)2 3(x+ y) 18 = 0

{x+ y = 6xy = 3{x+ y = 3xy = 3

(v nghim)

{x = 3 +

6

y = 36{x = 36y = 3 +

6

p s: (x; y) = (3 +

6; 3

6), (3

6; 3 +

6) 2

Bi 7 Gii h phng trnh {x2 + y2 + x+ y = 4x(x+ y + 1) + y(y + 1) = 2

Gii

H cho tng ng vi h sau{x2 + y2 + x+ y = 4x2 + y2 + x+ y + xy = 2 xy = 2

t S = x+ y; P = xy(S2 4P ) S2 = x2 + y2 + 2xy x2 + y2 = S2 2PVy {

S2 2P + S = 4S2 P + S = 2

P = 2[S = 0S = 1

Trng hp 1 {x+ y = 0xy = 2

{x =

2

y = 2{x = 2y =

2

Trng hp 2 {x+ y = 1xy = 2

{x = 1y = 2{x = 2y = 1

Page 35


p s: (x; y) = (

2;

2), (

2;

2), (1;2), (2; 1)Bi 8 Gii h phng trnh sau {

x2 + y2 + xy = 4y 1x+ y =

y

x2 + 1+ 2

Gii

H cho tng ng vi x2 + 1

y+ x+ y = 4 =

x+ y =y

x2 + 1+ 2

t u =x2 + 1

y, v = x+ y. H phng trnh c dng

{u+ v = 4

v =1

u+ 2

Gii h trn ta thu c u = 1, v = 3.

Vi

{u = 1v = 3

x2 + 1

y= 1

x+ y = 3

{x = 1y = 2{x = 2y = 5

p s: (x; y) = (1; 2), (2; 5) 2Bi 9 Gii h phng trnh

3

x2 + y2 1 +2y

x= 1

x2 + y2 2xy

= 4

Gii

iu kin: xy 6= 0t a = x2 + y2 1, b = x

yvi ab 6= 0

H cho tr thnh { 3a

+2

b= 1

a 2b = 3

{a = 1b = 1{a = 9b = 3

Trng hp 1 {a = 1b = 1

{x = 1y = 1{x = 1y = 1

Trng hp 2 {a = 9b = 3

{x = 3y = 1{x = 3y = 1

Page 36


p s: (x; y) = (1;1), (1; 1), (3; 1), (3;1) 2Bi 10 Gii h phng trnh sau {

x2 + xy 3x+ y = 0x4 + 3x2y 5x2 + y2 = 0

Gii

Xt x = 0 y = 0. Vy (0; 0) l mt nghim ca h.Xt x 6= 0, chia hai v ca phng trnh u cho x, hai v ca phng trnh th hai cho x2, ta ch phng trnh sau x+

y

x+ y = 3

x2 +y2

x2+ 3y = 5

(x+

y

x

)+ y = 3(

x+y

x

)2+ y = 5

t z = x+y

x, ta thu c h {

z + y = 3z2 + y = 5

Gii h ny, ta c: z = 2, y = 1 hoc z = 1, y = 4.Gii trng hp u c x = y = 1, trng hp th hai v nghim.

p s: (x; y) = (0; 0), (1; 1) 2

Bi 11 Gii h phng trnh sau {x

3x+ y +

5x+ 4y = 512

5x+ 4y + x 2y = 35

Gii

iu kin 3x+ y 0, 5x+ 4y 0.t u =

3x+ 4y; v =

5x+ 4y. Suy ra x 2y = 2(3x+ y) (5x+ 4y) = 2u2 v2.

H cho tr thnh{u+ v = 512v + 2u2 v2 35 = 0

{u = 5 v2(5 v)2 v2 + 12v 35 = 0

{u = 5 vv2 8v + 15 = 0

Trng hp 1 {v = 3u = 2

{

5x+ 4y = 93x+ y = 4

{x = 1y = 1

Trng hp 2 {v = 5u = 0

{

5x+ 4y = 253x+ y = 0

x = 25

7

y =75

7

p s (x; y) = (1; 1),

(25

7;75

7

)2

Bi 12 Gii h phng trnh {x2y + 2x2 + 3y 15 = 0x4 + y2 2x2 4y 5 = 0

Gii

Page 37


H phng trnh cho tng ng vi h sau{(x2 1)(y 2) + 4(x2 1) + 4(y 2) = 5(x2 1)2 + (y 2)2 = 10

t u = x2 1, v = y 2H tr thnh

{u2 + v2 = 10uv + 4(u+ v) = 5

{

(u+ v)2 2uv = 10uv + 4(u+ v) = 5

{u+ v = 10uv = 45{u+ v = 2uv = 3

{u = 3v = 1{u = 1v = 3

Trng hp 1 {u = 3v = 1

Khi , c hai nghim ca h l: (x; y) = (2; 1) v (x; y) = (2; 1)Trng hp 2 {

u = 1v = 3

Khi , h c nghim l: (x; y) = (0; 5).

p s: (x; y) = (2; 1), (2; 1), (0; 5) 2Bi 13 Gii h phng trnh sau{

2x 1 y (1 + 22x 1) = 8y2 + y

2x 1 + 2x = 13

Gii

iu kin: x 12. t t =

2x 1 vi t 0. H phng trnh tr thnh{

t y (1 + 2t) = 8y2 + yt+ t2 = 12

{t y 2ty = 8 (1)(t y)2 + 3ty = 12 (2)

T (1) v (2), suy ra: 2(t y)2 + 3 (t y) = 0 t y = 0 t y = 32

Vi t = y, ta c: t = y = 2. Khi :

2x 1 = 2 x = 52.

Vy h c nghim l

(5

2; 2

).

Vi y = t+3

2, c 4t2 + 6t 13 = 0 t = 3 +

61

4( do t 0). Khi :

t =3 +61

4

y =

3

2+3 +61

42x 1 = 3 +

61

4

y =

3 +

61

4

x =43 361

16

p s: (x; y) =

(5

2; 2

),

(43 361

16;3 +

61

4

)2

Bi 14 Gii h phng trnh sau xy + 2 = 3

2

y + 2 (x 2)x+ 2 = 74

Page 38


Gii

iu kin: x 2, y 2t u =

x+ 2, v =

y + 2 vi u, v 0 ()

H cho tr thnh u2 v = 7

2(1)

v2 + 2 (u2 4)u = 14

(2)

T (1) v (2), thu c: (u2 7

2

)2+ 2u3 8u = 1

4 u4 + 2u3 7u2 8u+ 12 = 0 (u 1) (u 2) (u2 + 5u+ 6) = 0

u = 1 u = 2Vi u = 1 thay vo (1) c v = 5

2, khng tha ().

Vi u = 2 thay vo (1) c v =1

2, tha ().

Vy h c nghim l (x; y) =

(2;

7

4

)2.

Bi 15 Gii h phng trnh {2x2 x(y 1) + y2 = 3yx2 + xy 3y2 = x 2y

Gii

Xt y = 0 x = 0.Xt y 6= 0. t t = x

y x = ty. H cho tr thnh{

y2(2t2 t+ 1) = y(3 t) (1)y2(t2 + t 3) = y(t 2) (2)

T (1) v (2) c:

3t3 7t2 3t+ 7 = 0

t = 1t = 1t =

7

3

H cho c nghim (x; y) = (0; 0), (1; 1), (1; 1),(

7

43;

3

43

)2

Bi 16 Gii h phng trnh {1 + x+ xy = 5y1 + x2y2 = 5y2

Gii

Vi y = 0, h v nghim.Vi y 6= 0, h c dng

x+1

y+x

y= 5

x2 +1

y2= 5

x+

1

y+ x.

1

y= 5(

x+1

y

)2 2x1

y= 5

Page 39


t x+1

y= u v x.

1

y= v, h tr thnh:

{u2 2v = 5u+ v = 5

{u = 5v = 10{u = 3v = 2

Vi {u = 5v = 10

x+

1

y= 5

x.1

y= 10

H v nghim

Vi {u = 3v = 2

x+

1

y= 3

x.1

y= 2

{x = 2y = 1{x = 1

y =1

2

p s: (x; y) = (2; 1),

(1;

1

2

)2

Cu 17: Gii h phng trnh:

{x2 + 1 + y (y + x) = 4y (1)(x2 + 1) (y + x 2) = y (2) (I)

Gii

+) Do y = 0 khng l nghim ca h nn: (I)

x2 + 1

y+ y + x = 4

x2 + 1

y(y + x 2) = 1

+) t

u =x2 + 1

yv = x+ y

. H tr thnh:

{u+ v = 4u (v 2) = 1

{u = 4 v(4 v) (v 2) = 1

{u = 1v = 3

x2 + 1

y= 1

x+ y = 3

{x = 1y = 2{x = 2y = 5

Vy h cho c 2 nghim: (1; 2) , (2; 5)2


{2x2y4 + 2xy2 y4 + 1 = 2 (32 x) y2 (1)x y2 + x = 3 (2) (I)

Gii

+) (2)x y2 = 3 x { x 3x2 + y2 7x+ 9 = 0

+) (1) 2(xy2 + y2 + 1) (xy2 y2 + 1) = 2 (32 x) y2(xy2 + y2 + 1) (xy2 y2 + 1) = 2 [1 (32) y2 + xy2] ()

Page 40


+) t

{u = xy2 + 1v = y2

. Phng trnh (*) tr thnh:(u+ v) (u v) = 2 [u (32) v]

{u (32) v 0u2 v2 = 4[u (32) v]2

{u (32) v 03u2 8 (32)uv + (45 242) v2 = 0 () (II)

Ta thy v = 0 khng l nghim ca h (II) nn:

() 3(uv

)2 8 (32) u

v+(45 242) = 0

uv = 3uv

= 5 82

3

xy2 + 1

y2= 3

xy2 + 1

y2= 5 8

2

3

xy2 + 1

y2= 3 (Do u (32) v)

xy2 + 1 = 3y2 (x 3) y2 + 1 = 0 ( ) T (1) ta c: y2 = x2 + 7x 9 thay vo (***) ta c:

(x 3) (x2 + 7x 9) + 1 = 0 x3 + 10x2 30x+ 28 = 0 x = 2x = 4 +2(loi)x = 42

Vi x = 2 y2 = 1 y = 1 Vi x = 42 y2 = 1 +2 y =

1 +

2Vy h phng trnh cho c 4 nghim:

(x; y) ={

(2; 1) ; (2;1) ;(

42;

1 +

2)

;(

42;

1 +

2)}2


4xy + 4 (x2 + y2) +

3

(x+ y)2= 7 (1)

2x+1

x+ y= 3 (2)

(I)

Gii

+) iu kin: x+ y 6= 0

+) (I)

3(x+ y)2 + (x y)2 + 3

(x+ y)2= 7

x+ y +1

x+ y+ x y = 3

3

(x+ y +

1

x+ y

)2+ (x y)2 = 13

x+ y +1

x+ y+ x y = 3

+) t a = x+ y +1

x+ y(|a| 2) ; b = x y ta c h:

{3a2 + b2 = 13 (3)a+ b = 3 (4)

T (4) ta c: b = 3 a thay vo (3):

3a2 + (3 a)2 = 13 4a2 6a 4 = 0[a = 2

a =1

2(loi)

Vi a = 2 b = 1 t ta c h:{x+ y + 1

x+y= 2

x y = 1 {x+ y = 1x y = 1

{x = 1y = 0

+) Vy h cho c nghim duy nht l (1; 0)2

Page 41



x2 + y2 =

1

5(1)

4x2 + 3x 5725

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

Gii

+) (I){

5 (x2 + y2) = 1

4x2 + 3x+ 3xy + y =57

25

2 (x2 + y2) =

10

25

2x2 2y2 + 3x+ 3xy + y = 4725

+) Ta c: 2x2 2y2 + 3x+ 3xy + y = 4725 (2x y) (x+ 2y) + (2x y) + (x+ 2y) = 47

25+) t 2x y = a, 2x+ y = b ta c h:{

a2 + b2 = 1

ab+ a+ b =47

25

{

(a+ b)2 2ab = 12ab+ 2 (a+ b) =

94

25

{

2ab = (a+ b)2 1(a+ b+ 1)2 =

144

25

a+ b =

7

5

ab =12

25

()a+ b = 17

25

ab =132

25

()

+) Ta thy h (**) v nghim, cn h (*) c hai nghim l: (a; b) =

(3

5;4

5

),

(4

5;3

5

)Tng ng ta c: (x; y) =

(2

5;1

5

),

(11

25;2

5

)+) Vy h cho c hai nghim: (x; y) =

(2

5;1

5

),

(11

25;2

5

)2

Nhn xt: Bi ny xut pht t h i xng

{x2 + y2 = 1xy + x+ y = 47

25

. Sau khi thay x, y tng ng bi

2x y, 2x + y th bi ton tr nn phc tp i hi ngi gii phi c nhng bin i kho lo c kt qu nh trn.


{x4 2x = y4 y (1)(x2 y2)3 = 3 (2) (I)

Gii

+) t x+ y = a, x y = b, 3 = c3

x =

a+ b

2

y =a b

2+) T (2) ta c: (ab)3 = c3 ab = c+) Ta c: x4 y4 = (x y) (x+ y) (x2 + y2) = ab

[(a+ b

2

)2+

(a b

2

)2]=ab

2(a2 + b2)

+) Mt khc: 2x y = a+ b a b2

=a+ 3b

2=a+ c3b

2

+) Khi (1) tr thnh:ab

2(a2 + b2) =

a+ c3b

2 c (a2 + b2) = a+ c3b

Page 42


+) T ta c h:

{c (a2 + b2) = a+ c3bab = c

(II)

c(a2 +

c2

a2

)= a+

c4

a ca4 + c3 = a3 + ac4 (ca 1) (a3 c3) = 0

[a =

1

ca = c

Suy ra h (II) c hai nghim l: (a; b) = (c; 1) ,

(1

c; c2)

Vi{a = cb = 1

x =

c+ 1

2=

3

3 + 1

2

y =3

3 12

Vi{a =

1

cb = c2

x =

1

2

(1

c+ c2

)=

1 + c3

2c=

23

3

y =1

2

(1

c c2

)=

1 c32c

= 13

3

+) Vy h cho c hai nghim: (x; y) =

(3

3 + 1

2;

3

3 12

),

(23

3; 1

3

3

)2

Page 43

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