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7/28/2019 Mt s phng php gii phng trnh v t

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I HC THI NGUYN

TRNG I HC KHOA HC

TRNH HNG UYN

MT S PHNG PHPGII PHNG TRNH V T

LUN VN THC S TON HC

Chuyn ngnh: PHNG PHP TON S CPM s: 60.46.40

Ngi hng dn khoa hc:GS. TSKH. NGUYN VN MU

THI NGUYN - NM 2011

1S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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Mc lc

M u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Chng 1. Phng php gii phng trnh v t . . . . . . . . . . 5

1.1. Phng php hu t ha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2. Phng php ng dng cc tnh cht ca hm s . . . . . . . . . . 241.3. Phng php a v h i xng . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.4. Phng trnh gii bng phng php so snh . . . . . . . . . . . . . . . 32

Chng 2. Mt s phng php gii phng trnh v t chatham s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.1. S dng phng php bin i tng ng . . . . . . . . . . . . . . . . 402.2. S dng phng php t n ph. . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3. S dng nh l Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.4. S dng phng php iu kin cn v . . . . . . . . . . . . . . . . . . 432.5. S dng phng php hm s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Chng 3. Mt s cch xy dng phng trnh v t. . . . . 483.1. Xy dng phng trnh v t t cc phng trnh bit cchgii. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.2. Xy dng phng trnh v t t h phng trnh. . . . . . . . . . . 523.3. Dng hng ng thc xy dng cc phng trnh v t . . 533.4. Xy dng phng trnh v t da theo hm n iu. . . . . . . 553.5. Xy dng phng trnh v t da vo hm s lng gic v phngtrnh lng gic. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.6. Xy dng phng trnh v t t php "t n ph khng tonphn". . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.7. Xy dng phng trnh v t da vo tnh cht vect. . . . . . . 603.8. Xy dng phng trnh v t da vo bt ng thc . . . . . . . . 613.9. Xy dng phng trnh v t bng phng php hnh hc . . 65Kt lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68



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Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69



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M u

Phng trnh v t l mt lp bi ton c v tr c bit quan trngtrong chng trnh ton hc bc ph thng. N xut hin nhiu trong cck thi hc sinh gii cng nh k thi tuyn sinh vo i hc. Hc sinh phii mt vi rt nhiu dng ton v phng trnh v t m phng phpgii chng li cha c lit k trong sch gio khoa. l cc dng tonv phng trnh v t gii bng phng php a v h (i xng hockhng i xng), dng phng php t n ph khng ton phn, dngn ph lng gic, . . . .

Vic tm phng php gii phng trnh v t cng nh vic xy dngcc phng trnh v t mi l nim say m ca khng t ngi, c bit l

nhng ngi ang trc tip dy ton. Chnh v vy, p ng nhu cuging dy v hc tp, tc gi chn ti "Mt s phng php giiphng trnh v t" lm ti nghin cu ca lun vn. ti nhm mtphn no p ng mong mun ca bn thn v mt ti ph hp msau ny c th phc v thit thc cho vic ging dy ca mnh trong nhtrng ph thng. Lun vn c hon thnh di s hng dn trc tipca NGND. GS.TSKH. Nguyn Vn Mu. Tc gi xin c by t lngbit n chn thnh v su sc i vi ngi thy ca mnh, ngi nhittnh hng dn, ch bo v mong mun c hc hi thy nhiu hn na.

Tc gi xin chn thnh cm n qu thy c trong Ban gim hiu, Phngo to i hc v sau i hc Trng i hc Khoa hc, i hc ThiNguyn, cng qu thy c tham gia ging dy kha hc to mi iukin, gip tc gi trong sut qu trnh hc tp v nghin cu tcgi hon thnh kha hc v hon thnh bn lun vn ny.

Lun vn gm phn m u, ba chng, phn kt lun v danh mc

ti liu tham kho.



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Chng 1 trnh by h thng cc phng php gii c bn lp ccphng trnh v t.

Chng 2 trnh by phng php gii v bin lun phng trnh v tc cha tham s.

Chng 3 trnh by mt s cch xy dng phng trnh v t mi.

Mc d c gng rt nhiu v nghim tc trong qu trnh nghin cu,nhng do thi gian v trnh cn hn ch nn kt qu t c tronglun vn cn rt khim tn v khng trnh khi thiu xt. V vy tc gimong nhn c nhiu kin ng gp, ch bo qu bu ca qu thy c,

cc anh ch ng nghip lun vn c hon thin hn.Thi Nguyn 2011

Trnh Hng Uyn



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Chng 1

Phng php gii phng trnh vt

1.1. Phng php hu t ha

Nhn chung gii phng trnh v t ta thng quy v phng trnhhu t gii. Ta thng dng cc phng php sau y a ccphng trnh v t v phng trnh hu t m ta c th gi cc phngphp ny l "hu t ho".

1.1.1. S dng cc php bin i tng ngNi dung chnh ca phng php ny l lu tha hai v vi s m ph

hp.Mt s php bin i tng ng thng gp.

[1]. 2n

f(x) = 2n

g(x)

f(x) = g(x)f(x) 0g(x) 0

[2]. 2n

f(x) = g(x) khi v ch khi

f(x) = g2n(x)g(x) 0

[3]. 2n+1

f(x) = g(x) f(x) = g2n+1(x).V d 1.1. Gii phng trnh

2x + 1 = 3x + 1. (1.1)



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Gii. Ta c

(1.1)

3x + 1 02x + 1 = (3x + 1)

2

x 13

9x2 + 4x = 0

x 13

x = 0, x = 49

x = 0, x =

4

9

(loi).

Vy nghim ca phng trnh l x = 0.

Nhn xt 1.1. Phng trnh trn c dng tng qut

f(x) = g(x). Khigp dng phng trnh ny, ta s dng bin i sau.

f(x) = g(x)

g(x) 0f(x) = g2(x)

V d 1.2. Gii phng trnh

1 + 23

x x2 = x + 1 x. (1.2)

Gii. iu kin

x x2 0x 01 x 0

0 x 1.

gii phng trnh ny, ta thng ngh n vic bnh phng hai vkhng m ca mt phng trnh c phng trnh tng ng.

(1.2) 2(x x2

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

x x2 = 0

2

x x2 = 3

x x2 = 04x2 4x + 9 = 0 (v nghim)

Suy ra x = 1 hoc x = 0.Kt hp vi iu kin bi ra, ta c x = 0; x = 1 l nghim phng

trnh.



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Nhn xt 1.2. Dng tng qut ca phng trnh trn l

f(x) +g(x) = h(x). Khi gp dng phng trnh ny ta bin itng ng nh sau

f(x) +

g(x) =

h(x)

f(x) 0g(x) 0f(x) + g(x) + 2

f(x)g(x) = h(x)

V d 1.3 (Hoc sinh gii quc gia nm 2000). Gii phng trnh4 310 3x = x 2. (1.3)

Gii. Ta c(1.3)

x 24 310 3x = x2 4x + 4

x 24x x2 = 310 3x

2 x 4x4 8x3 + 16x2 + 27x 90 = 0

2 x 4(x 3)(x3 5x2 + x + 30) = 0

2 x 4(x 3)(x + 2)(x2 7x + 15) = 0

x = 3.

Vy x = 3 l nghim ca phng trnh.

1.1.2. Thc hin php nhn lin hp n gin vic tnh ton

Ta bit nu x = x0 l nghim ca phng trnh f(x) = 0 th iu c ngha l

x0 Dff(x0) = 0

Nu x = a l nghim ca a thc P(x) th P(x) = (x a)P1(x), trong P1(x) l a thc vi deg P1 = deg P 1.

Nu x0 l mt nghim ca phng trnh f(x) = 0 th ta c th aphng trnh f(x) = 0 v dng (x x0)f1(x) = 0 v khi vic giiphng trnh f(x) = 0 quy v phng trnh f1(x) = 0.



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3(2 +

x

2) = 2x +

x + 6. (1.4)

Gii. iu kin

x 2 0x + 6 0 x 2.

Ta thy x = 3 l mt nghim ca phng trnh cho. Nhn xt rngkhi x = 3 th x 2 v 4x + 6 l nhng s chnh phng. Do ta tmcch a phng trnh cho v dng (x 3)f1(x) = 0.

Bin i phng trnh v dng sau 2(x3)+(x + 6 3x 2) = 0.Vn cn li ca chng ta l phi phn tch

x + 6

3

x

2 = 0

c tha s (x 3). Ta c (x + 6) 9(x 2) = 8(x 3), iu ny gipta lin tng n hng ng thc a2 b2 = (a + b)(a b). Ta bin i

x + 6 3x 2 = (

x + 6 3x 2)(x + 6 + 3x 2)

x + 6 + 3

x 2=

8(x 3)x + 6 + 3

x 3

Suy ra phng trnh cho tng ng vi phng trnh

(x 3)

2 8x + 6 + 3

x 2

= 0.

n y ta ch cn gii phng trnh

2 8x + 6 + 3

x 2 = 0

hayx + 6 + 3x 2 = 4.

Phng trnh ny c mt nghim x =11 35

2, x =

11 + 3

5

2

Vy phng trnh c nghim l x = 3 v x =11 35

2, x =

11 + 3

5

2

Nhn xt 1.3. Qua v d trn ta thy kh cn thc ta c th s dnghng ng thc

an bn = (a b)(an1 + an2b + + abn2 + bn1).



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Ta gi hai biu thc a b v an1 + an2b + + abn2 + bn1 l ccbiu thc lin hp ca nhau. Nn phng php trn thng c gi tt

l phng php nhn lin hp.V d 1.5 ( thi ngh Olympic 30-4 THPT Thi Phin, Nng).Gii phng trnh

1 + 3

x

4x +

2 + x 1 = 0. (1.5)

Gii. iu kin

x 02 + x 04x +

2 + x

= 0

x 0.

Ta c

(1.5) 1 + 3x 4x 2 + x = 0 3x 2 + x = 4x 1 (3x 2 + x)(3x + 2 + x) = (4x 1)(3x + 2 + x) 8x 2 = (4x 1)(3x + 2 + x)

(4x

1)(3

x +

2 + x

2) = 0

4x 1 = 03

x +

2 + x = 2

x =1

416x2 28x + 1 = 0

Gii h tuyn hai phng trnh trn, ta c

x =1

4, x =

7 358

, x =7 + 3

5

8l nghim cn tm.

1.1.3. t n ph

Ni dung ca phng php ny l t mt biu thc cha cn thcbng mt biu thc theo n mi m ta gi l n ph, ri chuyn phngtrnh cho v phng trnh vi n ph va t. Gii phng trnh theon ph tm nghim ri thay vo biu thc va t tm nghim theon ban u.

Vi phng php ny ta tin hnh theo cc bc sau.



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Bc 1. Chn n ph v tm iu kin xc nh ca n ph.y l bc quan trng nht, ta cn chn biu thc thch hp t

lm n ph. lm tt bc ny ta cn phi xc nh c mi quan hca cc biu thc c mt trong phng trnh. C th l, phi xc nhc s biu din tng minh ca mt biu thc qua mt biu thc khctrong phng trnh cho.

Bc 2. Chuyn phng trnh ban u v phng trnh theo n phva t v gii phng trnh ny.

Thng thng sau khi t n ph th nhng phng trnh thu c lnhng phng trnh n gin hn m ta bit cch gii.

Bc 3. Gii phng trnh vi n ph bit xc nh nghim caphng trnh cho.

Nhn xt rng, c rt nhiu cch t n ph. Ta s m t mt scch t n ph qua v d sau y.


1 +2

3x x

2 =

x +

1

x. (1.6)

iu kin

x 01 x 0 0 x 1

Phn tch. Ta c th la chn cc cch chn n ph nh sau.Cch 1. Ta nhn thy

x x2 c th biu din qua x + 1 x nh

vo ng thc(

x +

1 x)2 = 1 + 2

x x2. (1.7)

C th nu ta t x + 1 x = t, t 0 th x x2 = t2

12 Khi phng trnh cho tr thnh phng trnh bc hai vi n t l

1 +t2 1

3= t hay t2 3t + 2 = 0 suy ra t = 1, t = 2.

Vi t = 1, ta c

x +

1 x = 1 hay 2x x2 = 0, suy ra x = 0 hocx = 1.

Vi t = 2, ta c

x +

1

x = 2 v nghim.Vy x = 0, x = 1 l nghim phng trnh.



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Ta nhn thy cch gii trn da theo mi lin h l ng thc (1.7).Ngoi ra, ta c th ta c th to ra mi quan h khc gia cc i tng

tham gia phng trnh theo cch sau.Cch 2. T phng trnh cho ta c th rt ra c mt cn thc theobiu thc cha cn cn li l

x =

3

1 x 32

1 x 3 Do , nu ta t

1 x = t, t 0.

Khi ta c

x =3t 32t 3 V t ng thc

(

x)2 + (

1 x)2 = x + 1 x = 1 (1.8)ta thu c phng trnh t(t 1)(2t2 4t + 3) = 0 c nghim t = 0 vt = 1, hay x = 1, x = 0 l nghim ca phng trnh cho.Cch 3. Nhn xt rng phng trnh cho ch cha tng v tch ca haibiu thc cha cn v chng tho mn (1.8), do ta c th t

x = a,

1 x = b, a 0, b 0. T phng trnh cho kt hp vi (1.8) ta ch phng trnh.

1 +2

3ab = a + b

a2 + b2 = 1y l h i xng loi I. Gii h ny ta thu c nghim ca phng

trnh l x = 0, x = 1.

Tip tc nhn xt, ta thy ng thc (1.8) gip ta lin tng n ngthc lng gic sin2 + cos2 = 1. iu ny dn n cch gii sau.

Cch 4. t x = sin2 t, vi t

0;

2

(iu ny hon ton hp l v

x

[0; 1] nn ng vi mi gi tr ca x xc nh duy nht mt gi tr ca

t).Khi , ta c

(1.6) 1 + 23

sin t. cos t = sin t + cos t

3(1 sin t) +

(1 sin t)(1 + sin t)(2 sin t 3) = 0 1 sin x[31 sin x (3 sin2x)1 + sin x] = 0

Suy ra sin t = 1 hoc 3

1

sin t = (3

2sin t)

1 + sin t

hay sin t(4 sin2 8sin t + 6) = 0 suy ra sin t = 0.



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Vy x = 1; x = 0 l nghim.

Nhn xt 1.4. Qua v d trn ta thy c nhiu cch t n ph giiphng trnh v t. Tuy nhin t nh th no cho ph hp v cho cchgii hay l tu thuc vo kinh nghim pht hin ra mi quan h c thgia cc i tng tham gia trong phng trnh. Sau y l mt s dngton v mt s cch t n ph thng dng.

Dng 1. Phng trnh dng F( n

f(x) ) = 0, vi dng ny ta tn

f(x) = t (nu n chn th phi c iu kin t 0 ) v chuyn v

phng trnh F(t) = 0. Gii phng trnh ny ta tm c t, tip theo suy

ra x t phng trnhn

f(x) = t.Ta thng gp phng trnh c dng nh sau af(x)2 + bf(x) + c = 0.

V d 1.7 ( thi ngh Olympic 30-4 Trng THPT Chuyn Chu VnAn, Ninh Thun). Gii phng trnh

2x2 + 5x 1 = 7

x3 1.Gii. iu kin x3

1

0

x

1.

Khai trin phng trnh cho nh sau

(1.7) 3(x 1) + 2(x2 + x + 1) = 7

(x 1)(x2 + x + 1).Ta nhn thy x = 1 khng l nghim ca phng trnh nn chia c hai vca phng trnh trn cho x 1, ta c phng trnh

3 + 2x2 + x + 1

x

1

7

x2 + x + 1

x

1

= 0.

t

x2 + x + 1

x 1 = t, t 0. Khi ta c phng trnh

2t2 7t + 3 = 0 suy ra t = 3 hoc t = 12

Vi t = 3, ta c x2 8x + 10 = 0 hay x = 4 6 (tha mn iukin).

Vi t =1

2

, ta c 4x2 + 3x + 5 = 0 (v nghim).

Vy nghim ca phng trnh l x = 4 6.



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x2 +

x2 + 11 = 31. (1.9)

Gii. t

x2 + 11 = t, t 0. Ta c

(1.9) t2 + t 42 = 0

t = 6t = 7 (loi)

Vi t = 6 ta c

x2 + 11 = 6 suy ra x = 5 l nghim.

Vy nghim ca phng trnh l x = 5.V d 1.9. Gii phng trnh

x 1 + 3x 1 2 = 0

Gii. iu kin x 1 0 x 1.t t = 6

x 1, t 0 phng trnh cho tr thnh

t3 + t2 2 = 0 hay (t 1)(t2 + 2t + 2) = 0 suy ra t = 1.

Vi t = 1 ta c6

x 1 = 1 suy ra x = 2 tha mn iu kin.Vy x = 2 l nghim ca phng trnh.Nhn xt 1.5. Cc phng trnh c cha cc biu thc

n1

f(x), n2

f(x), n3

f(x), . . . . . . . . . nn

f(x) th ta gii phng trnh

bng cch t t = n

f(x) trong n l bi s chung nh nht ca cc sn1, n2, . . . . . . . . . nn.

Dng 2. Trong phng trnh c cha f(x) g(x) v f(x).g(x).Khi gp phng trnh dng ny ta t

f(x) g(x) = t sau bnhphng hai v ta s biu din c nhng i lng cn li qua t v chuyn

phng trnh ban u v phng trnh bc hai i vi t.


3 + x +

6 x = 3 +

(3 + x)(6 x).

Gii. iu kin3 + x

0

6 x 0 3 x 6.



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t

3 + x +

6 x = t, t 0 suy ra

t2 = 9 + 2(3 + x)(6 x). (1.10)p dng bt ng thc Cauchy ta c 2

(3 + x)(6 x) 9 nn t

(1.10), suy ra 3 t 32.Phng trnh cho tr thnh t = 3 +

t2 92

hay t2 2t 3 = 0c nghim t = 3 (tha mn). Thay vo (1.10), ta c phng trnh

(3 + x)(6 x) = 0 c x = 3, x = 6 l nghim.

V d 1.11. Gii phng trnh2x + 3 +

x + 1 = 3x + 2

2x2 + 5x + 3 16.

Gii. iu kin

2x + 3 0x + 1 0 x 1.

t

2x + 3 +

x + 1 = t, t 0, suy ra

t2 = 3x + 2

(2x + 3)(x + 1) + 4. (1.11)

Khi phng trnh cho tr thnh t = t2 20 hay t2 t 20 = 0 suyra t = 5 (do t 0).

Thay t = 5 vo (1.11), ta c

(1.11) 21 3x = 2

2x2 + 5x + 3

21 3x 0441 126x + 9x2 = 8x2 + 20x + 12

1 x 7x2

146x + 429 = 0 x = 3.V d 1.12. Gii phng trnh

x3

35 x3(x + 3

35 x3) = 30.Gii. t x + 3

35 x3 = t suy ra t3 = 35 + 3x 335 x3(x + 335 x3)

hay

x 3

35 x3 = t3

353t (1.12)



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(v t=0 khng l nghim ca phng trnh).Phng trnh cho tr thnh

t3 353t

t = 30 suy ra t = 5.

Thay vo (1.12), ta c

(1.12) x 3

35 x3 = 6 x3(35 x3) = 216 x6 35x3 + 216 = 0 x = 2 hoc x = 3.

Dng 3. Phng trnh dng a n

f(x) + b 2n

f(x)g(x) + c n

g(x) = 0 (Vig(x) = 0).

gii phng trnh dng ny ta chia hai v phng trnh cho n

g(x)

v t 2n

f(x)

g(x)= t, t 0 ta c phng trnh bc hai i vi n t c

dng at2 + bt + c = 0.


10

x3 + 8 = 3(x2 x + 6). (1.13)

Gii. iu kin x3 + 8 0 x 2.Ta c

(1.13) 10

(x + 2)(x2 2x + 4) = 3[(x + 2) + (x2 2x + 4)].

Chia hai v cho x2 2x + 4 (do x2 2x + 4 0 vi mi x).Ta c phng trnh

10

x + 2

x2 2x + 4 = 3[x + 2

x2 2x + 4 + 1]

t

x + 2

x2 2x + 4 = u, u 0 phng trnh vi n u c dng

3u2 10u + 3 = 0 hay u = 3 hoc u = 13

Vi u = 3, ta c

x + 2x2 2x + 4 = 3, hay 9x219x +34 = 0 v nghim.



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Vi u =1

3ta c

x + 2

x2 2x + 4 =1

3

hay x2 11x 14 = 0. Suy ra x = 11 1772 (tha mn).Vy phng trnh cho c nghim x =

11 1772


5

x3 + 1 = 2(x2 + 2). (1.14)

Gii. iu kin x3 + 1

0

x

1. Ta c

(1.14) 5

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

2 x + 1x2 x + 1 5

x + 1

x2 x + 1 + 2 = 0 (do x2 x + 1 > 0 vi mi x).

t

x + 1

x2 x + 1 = t vi t 0 ta c phng trnh

2t2

5t + 2 = 0 suy rat = 2 hoc t =

1

2Vi t = 2 ta c

x + 1

x2 x + 1 = 4 hay4x2 5x + 3 = 0 phng trnh v nghim.

Vi t =1

2ta c

x + 1

x2 x + 1=

1

4hay x2

5x

3 = 0

x =

5 372

Vy nghim ca phng trnh l x =5 37

2Dng 4. p(x)f(x) + g(x)

f(x) + h(x) = 0.

Vi dng phng trnh ny ta c th t

f(x) = t, t 0.Khi ta c phng trnh theo n t lp(x)t2 + g(x)t + h(x) = 0, ta gii phng trnh ny theo t, xem x l

tham s (ta tm c t theo x) nn ta gi dng ny l dng n ph khngtrit .



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2(1

x)x2 + 2x

1 = x2

6x

1.

Gii. iu kin x2 + 2x 1 0.t

x2 + 2x 1 = t, t 0 ta c phng trnh

t22(1x)t4x = 0. y l phng trnh bc hai n t ta coi x l thams c = (x + 1)2, do phng trnh ny c hai nghim t = 2, t = 2x.

Vi t = 2 ta c

x2 + 2x 1 = 2 hay x2 + 2x 5 = 0 suy rax = 1 6.Vi t =

2x ta c

x2 + 2x

1 =

2x hay x 0

3x2

2x + 1 = 0h

ny v nghim.Vy phng trnh cho c hai nghim tha mn l x = 1 6.

V d 1.16 ( thi ngh Olympic 30-4 THPT Bc Liu). Gii phngtrnh

x2 + (3

x2 + 2)x = 1 + 2

x2 + 2.

Gii.t

x2 + 2 = t, t

0 ta c x2 = t2

2 nn phng trnh cho tr

thnht2 (2 + x)t 3 + 3x = 0 hay t = 3 hoc t = x 1.Vi t = 3 ta c

x2 + 2 = 3 suy ra x = 7.

Vi t = x 1 ta c x2 + 2 = x 1suy ra

x 1 0x2 + 2 = x2 2x + 1 (v nghim).

Vy nghim ca phng trnh l x = 7.

1.1.4. Phng php a v h khng i xngPhng trnh c dng sauA( n

f(x)+ m

g(x))+B n

f(x) m

g(x)+C = 0 vi (af(x)bg(x) = D)trong A, B, C , D, a, b l cc hng s.

t

n

f(x) = u

m

g(x) = v

Khi phng trnh cho tr thnh h phng trnh "hu t".A(u v) + Buv + C = 0aun + bvm = D



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2(

x

1

3

x + 1) + 3

x

1. 3

x + 1 + 7 = 0.

Gii iu kin x 1.t

x 1 = u, u 0

3

x + 1 = v

Khi ta c h

2(u v) + 3uv + 7 = 0u v = 2

Rt u t phng trnh th hai ca h ta c u = v 2 thay vophng trnh u ta c

3v2

6v

+ 3 = 0 suy rav

= 1 tha mn.Vy x + 1 = 1 khng tha mn iu kin.Phng trnh v nghim.

V d 1.18. Gii phng trnh3

24 + x +

12 x = 6.Gii. iu kin 12 x 0 x 12.

t 3

24 + x = u,

12

x = v suy ra u

3

36, v

0.

Ta c

u + v = 6u3 + v2 = 36

v = 6 uu3 + (6 u)2 = 36

v = 6 uu(u2 + u 12) = 0

Suy ra u = 0; u = 4; u = 3 l nghim tho mn iu kin.T y ta c x = 24; x = 88; x = 3.Vy phng trnh c nghim x = 24; x = 88; x = 3.


x + 7

x = 1.

Gii iu kin x 0.t

u3 = x + 7v2 = x

suy ra

u 37v 0

Ta c h

u v = 1u3 v2 = 7

Rt v t phng trnh th nht ca h ta cv = uv thay vo phng trnh th hai ca h ta c u3(u1)2 = 7

hay u3

u2 + 2u

8 = 0 suy ra (u

2)(u2 + u + 4) = 0. Phng trnh

ny c nghim u = 2 vy v = 1.



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19

Tr v tm x ta gii

u3 = 8v2 = 1

suy ra x + 7 = 8x = 1

suy ra x = 1 (tha mn).

Vy x = 1 l nghim ca phng trnh.Mt s v d khc


x +

1 x 2

x(1 x) 2 4

x(1 x) = 1.

Gii iu kin

x 01 x 0

t 4x = u41 x = v vi u 0, v 0.

T iu kin v t phng trnh cho ta c hu4 + v4 = 1u2 + v2 2uv + 1 2u2v2 = 0 hay

u4 + v4 = 1(u v)2 + (u2 v2)2 = 0

hoc

u v = 0u2 v2 = 0u4 + v4 = 1

suy ra

u = v

u4 = v4 =1

2Tr v tm x ta c

x = 12

1 x = 12

suy ra x =1

2


8x + 1 +

3x 5 = 7x + 4 + 2x 2Gii iu kin x 5

3.

t u =

8x + 1, v =

3x

5, z =

7x + 4, t =

2x

2 viu, v , z , t khng m.

T cch t v phng trnh cho ta thu c hu + v = z+ tu2 v2 = z2 t2

T phng trnh th hai ca h ta thu c (u+v)(uv) = (z+t)(zt).Li do u + v > 0 v u 0, v 0, u, v khng ng thi bng 0, ta thu

c u v = z t kt hp vi phng trnh u ca h suy ra u = z.T ta c

8x + 1 =

7x + 4 suy ra x = 3. (tha mn).

Vy x = 3 l nghim ca phng trnh.



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20

1.1.5. Phng php lng gic ha

- Nu phng trnh cha

a2

x2 t x =

|a

|sin t vi

2 x

2hoc x = |a| cos t vi 0 x .- Nu phng trnh cha

x2 a2 t x = |a|

sin tvi t

2; 0) hoc

t (0; 2

t x = |a|

cos tvi t

0;

2

hoc t

2

;

- Nu phng trnh cha

x2 + a2 t x = |a| tan t vi t

2

;

2

- Nu phng trnh cha a + x

a

xhoc

a xa + x

t x = a cos2t.

- Nu phng trnh cha

(a x)(b x) t x = a + (b a)sin2 t.V d 1.22. Gii phng trnh

1 +

1 x2 = x(1 + 2

1 x2).Gii. iu kin |x| 1.

t x = sin t vi 2

t 2

Khi (1.22) 1 + cos t = sin t(1 + 2 cos t)

2cost

2= sin t + sin 2t = 2sin

3t

2cos

t

2

cos t2

(

2sin3t

2 1) = 0

cost

2= 0

cos

3t

2 =

1

2

Suy ra t = (2k + 1) (k Z) hoc t = 6

+ k4

3hoc t =

5

6+ k

4

3kt hp iu kin ta c t =

6

Vy x = sin

6=

1

2


x3 3x = x + 2. (1.15)



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21

Gii. iu kin x 2.Vi x < 2 phng trnh khng xc nh.Vi x > 2 ta c x

3

3x = x + x(x2

4) > x > x + 2.Vy gii phng trnh ta ch cn xt t [2;2].t x = 2 cos t vi t [0; ]. Khi

(1.15) cos3t = cos t2

t =

k4

5

t =k7

7

Kt hp vi iu kin ta c t =4

5, t =

4

7, t = 0.

Vy x = 2, x = 2 cos4

5, x = 2 cos

4

7


x2 + 1 +x2 + 1

2x=

(x2 + 1)2

2x(1

x2)

(1.16)

Gii. iu kin x = 0 v x = 1.t x = tan t vi t

2;

2

, t = 0 v t =

4

Ta c

(1.16) 1cos t

+1

sin2t=

2

sin4t

1cos t

(1 +1

2sin t 1

2sin t. cos2t) = 0

2sin t. cos2t + cos 2t

1 = 0

2sin t(1 2sin2 t) 2sin2 t = 0 sin t(1 sin t 2sin2 t) = 0

sin t = 0sin t = 1sin t =

1

2

t =

2+ k2

t =

6 + k2(k Z)



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22

Kt hp vi i kin ca t ta c t =

6

Vy phng trnh c nghim x = tan

6=

1

3 1.1.6. Phng php s dng nhiu hn mt n ph


4x2 + 5x + 1 2

x2 x + 1 = 9x 3. (1.17)

Gii. iu kin 4x2 + 5x + 1 0x2

x + 1

0

t4x2 + 5x + 1 = a

2

x2 x + 1 = b a 0, b 0.Khi

(1.17) a2 b2 = 9x 3 (a b)(a + b 1) = 0

a b = 0a + b

1 = 0

x =

1

3a b = 9x 32a = 9x 2

x =1

3x = 0

x =56

65

Vy nghim phng trnh l x =1

3; x = 0; x =

56

65


2(x2 3x + 2) = 3

x3 + 8. (1.18)

Gii. iu kin x3 + 8 0 x 2t u =

x2

2x + 4; v =

x + 2, u

0, v

0.

Ta c u2 v2 = x2 3x + 2.



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23

Lc

(1.18)

2(u2

v2) = 3uv

(2u + v)(u 2v) = 0 u = 2v (do 2u+v >0).

Tm x ta giix2 2x + 4 = 2x + 2 x26x + 4 = 0 x = 313(tha mn).

Vy phng trnh c nghim x = 3 +

13; x = 3 13.V d 1.27. Gii phng trnh

x = 3 x4 x + 5 x4 x + 3 x5 x.Gii. iu kin 0 x 3.

t

3 x = a; 4 x = b, 5 x = c; a,b,c 0 ta c h phngtrnh.

ab + bc + ca = 3 a2ab + bc + ac = 4 b2ab + bc + ac = 5 c2

hay

(a + c)(a + b) = 3(b + c)(b + a) = 4(c + a)(c + b) = 5

Suy ra (a + b)(b + c)(c + a) = 215.

Vy

a + b = 2

3

5

b + c = 2

5

3

c + a =

15

4

a =23

4

15

b =17

4

15

c =7

4

15

suy ra x =671

240l nghim

ca phng trnh cho.


3x + 1 + 3

5 x + 32x 9 34x 3 = 0.Gii. t 3

3x + 1 = a; 3

5 x = b; 32x 9 = c.

Suy ra a3 + b3 + c3 = 4x 3.Khi phng trnh cho tr thnh(a + b + c)3 = a3 + b3 + c3 hay (a + b)(b + c)(c + a) = 0 suy ra a = b

hoc a =

c hoc b =

c.

Gii ra ta c nghim phng trnh l x = 3; x = 4; x = 85



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24

1.2. Phng php ng dng cc tnh cht ca hms

1. Nu hm s y = f(x) n iu thc s, lin tc trn tp D thphng trnh f(x) = k vi k l hng s, nu c nghim x = x0 th lnghim duy nht ca phng trnh.

2. Nu hm s y = f(x) n iu trn tp D v u(x), v(x) l cc hms nhn cc gi tr thuc D th f(u(x)) = f(v(x)) u(x) = v(x).

Mt s lu khi s dng phng php hm s.Vn quan trng khi s dng phng php hm s l chng ta phi

nhn ra c hm s n iu v "nhm hoc tnh c nghim caphng trnh vic ny c th nh my tnh".

pht hin c tnh n iu ca hm s ta cn nm vng cc tnhcht.

Nu hm s y = f(x) ng bin hoc nghch bin trn D th khi .- Hm s y = n

f(x) ng bin hoc nghch bin trn D.

- Hm s y =1

f(x)vi f(x) > 0 nghch bin hoc ng bin trn D.

- Hm s y = f(x) nghch bin hoc ng bin trn D.- Tng ca cc hm ng bin hoc nghch bin trn D l hm s ng

bin hoc nghch bin trn D.- Tch ca cc hm s dng ng bin hoc nghch bin trn D l mt

hm ng bin hoc nghch bin trn D.

V d 1.29. T tnh n iu ca cc hm s y = x + 3, y = 3 x vy = 2

x nu nm c tnh cht trn ta pht hin ra ngay cc hm s

y = 3x + 3 + x + 3 + x l ng bin trn tp xc nh.Hm s y =

6

3 x +

8

2 x ng bin trn tp xc nh ca n.

Hm s y =1

x + 3+

3 x nghch bin trn tp xc nh ca n.


5x3 1 + 3

2x

1 + x = 4.



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25

Nhn xt 1.6. Quan st v tri ca phng trnh, t tnh ng binnghch bin ca hm bc nht v tnh cht n iu ca hm s nu

trn, ta thy v tri ca phng trnh l hm ng bin trn tp xc nh.V phi ca phng trnh l hm hng nn ta s dng tnh n iu cahm s gii bi ton.

Gii. iu kin 5x3 1 0 x 13

5

Ta c f(x) =15x2

2

5x3 1 +2

3 3

(2x 1)2 + 1 vi mi x 1

3

5; +

nn hm s ng bin trn 135; +

M f(1) = 4 tc x = 1 l mt nghim ca phng trnh.Ta chng minh l nghim duy nht ca phng trnh tht vy.- Nu x > 1 th f(x) > f(1) = 4 suy ra phng trnh v nghim.

- Nu13

5 x < 1 th f(x) < f(1) = 4 suy ra phng trnh v nghim.

Vy phng trnh c nghim duy nht x = 1.

V d 1.31 ( thi ngh Olympic 30-4 Trng Chuyn L Qu n

B Ra -Vng Tu). Gii phng trnh3

6x + 1 = 8x3 4x 1. (1.19)Gii. Ta c

(1.19) 6x + 1 + 36x + 1 = (2x)3 + 2xXt hm s f(t) = t3 + t l s ng bin trn R.

Vy 3

6x + 1 = 2x suy ra 8x3

6x = 1. Nhn xt nu

|x

|> 1 th

4x3 3 > 1, suy ra |8x3 6x| = 2|x|(4x2 3) > 2.Nn nghim nghim ca phng trnh cho phi thuc [1;1].t x = cos t, t [0; ] khi phng trnh cho tr thnh.4cos3 t 3cos t = 1

2 cos3t = 1

2 t =

9+ k

2

3, k R.

Vy nghim phng trnh cho l x = cos

9, x = cos

5

9, x = cos

7

9


2x3 + 3x2 + 6x + 16 4 x = 23.



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26

Gii.iu kin

2x3

+ 3x2

+ 6x + 16 04 x 0 2 x 4.Phng trnh cho c dng f(x) = 2

3.

Trong f(x) =

2x3 + 3x2 + 6x + 16 4 x.f(x) =

3(x2 + x + 1)2x3 + 3x2 + 6x + 16

+1

2

4 x > 0 vi mi x (2;4)nn hm s ng bin trn [2;4].

M f(1) = 2

3, t ta c x = 1 l nghim duy nht ca phng

trnh.V d 1.33. Gii phng trnh

3

x + 2 3

2x2 + 1 =3

2x2 3x + 1.Gii. Bin i phng trnh cho v dng

3

x + 2 + 3

x + 1 =3

2x2 +3

2x2 + 1.

Xt hm s f(t) = 3

t + 3

t + 1, ta c phng trnh f(x + 1) = f(2x2).V f(t) = 3t + 3t + 1 lin tc v ng bin trn tp xc nh nnf(x + 1) = f(2x2) 2x2 x 1 = 0 x = 1, x = 1

2

Vy phng trnh c hai nghim l x = 1, x = 12

1.3. Phng php a v h i xng

1.3.1. Phng trnh dng

na+

f(

x) +

nb f(x) = ct u = n

a + f(x), v = n

b f(x)

Nh vy ta c hu + v = cun + vn = a + b

l h i xng loi I.


57 x + 4x + 40 = 5. (1.20)

Gii. iu kin 40 x 57.



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27

t u = 4

57 x v v = 4x + 40 u 0, v 0. Ta c

(1.20) u + v = 5

u4 + v4 = 97

u + v = 5[(u + v)2 2uv]2 2u2v2 = 97

u + v = 52(uv)2 100uv + 528 = 0

u + v = 5

uv = 6uv = 44.

u + v = 5uv = 6u + v = 5uv = 44.

(v nghim)

u = 2v = 3u = 3v = 2.

Vi u = 3 v v = 3 ta c4

57 x = 24

x + 40 = 3suy ra

57 x = 16x + 40 = 81

suy ra x = 41

Vi u = 3 v v = 2 ta c h57 x = 81x + 40 = 81

suy ra x = 24.

Vy x = 41 hoc x = 24 l nghim phng trnh.

Nhn xt 1.7. Xt v mt no cch dng n ph di dng ny cv ngc vi vic ta thng lm "Chuyn thnh bi ton nhiu n nhiuphng trnh hn bi ton ban u". y do tnh cht phc tp ca biton m ta nh chu "thit" v s lng tc lm tng s n v phngtrnh nhng li c ci c bn l chuyn c t bi ton kh v bi tond hn.


3

7 + tan x +3

2 tan x = 3.



29/70

28

Gii.iu kin x =

2+ k.

t

u = 37 + tan xv = 3

2 tan x suy ra

u3 = 7 + tan xv3 = 2 tan x

T phng trnh cho v t cng thc t n ph ta thu c hu + v = 3u3 + v3 = 9

hay

u + v = 3(u + v)[(u + v)2 3uv] = 9 suy ra

u = 2, v = 3 hoc u = 1, v = 2.Tr v tm x gii

H 7 + tan x = 82

tan x = 1

tan x = 1

x =

4

+ k.

H

7 + tan x = 12 tgx = 8 tan x = 6 = tan x = + l

(k, l Z).

1.3.2. Phng trnh dng n

ax + b = r(ux + v)n + dx + e

a = 0, u = 0, r = 0 Vi cc h s tha mn

u = ar + dv = br + e

Cch gii. t n

ax + b = uy + v.

Sau a v h

(uy + v)n = 1r

(ux + v) dr

x er

(ux + v)n =1

r(uy + v) d

rx e

r

y l h i

xng loi II c gii bng cch tr v vi v ca hai phng trnh trongh c mt phng trnh tch.

V d 1.36 (Tp ch ton hc tui tr s 303). Gii phng trnh

2x + 15 = 32x2

+ 32x 20. (1.21)Gii. iu kin x 15

2

Bin i phng trnh cho tr thnh

2x + 15 = 2(4x + 2)2 28.t n ph

2x + 15 = 4y + 2 (4y + 2)2 = 2x + 15 (4y + 2 0).

Khi

(1.21) (4x + 2)2 = 2y + 15.



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Vy ta c h phng trnh

(4x + 2)2 = 2y + 15(4y + 2)2 = 2x + 15

y l h i xng loi hai.Gii h trn bng cch tr v vi v ca hai phng trnh ta c

nghim l x =1

2, x =

9 22116


3

3x 5 = 8x3

36x2

+ 53x 25. (1.22)Gii.

(1.22) 33x 5 = (2x 3)3 x + 2.

t 3

3x 5 = 2y 3 suy ra (2x 3)3 = 3x 5, khi ta c h phngtrnh.

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

3)3 = 2x

3 + x

2

Gii h trn bng cch tr v vi v ca hai phng trnh trong h sau th tr li phng trnh u ta c phng trnh

(x 2)(8x2 20x + 11) = 0 suy ra x = 2

x =5 3

4

Vy nghim phng trnh cho l x = 2, x =5 3

4

1.3.3. Phng trnh dng (f(x))n + b = an

af(x) bCch gii. t n

af(x) b = t ta c h

(f(x))n + b = attn + b = af(x)

y l

h i xng loi II.


x2 4 = x + 4.

Gii. iu kin x 4.



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t

x + 4 = t th ta c h sau

x2 = t + 4t2 = x + 4

Tr v vi v ca phng trnh th nht cho phng trnh th hai trongh ta c x2 t2 = t x hay (x t)(x + t + 1) = 0 suy ra t = x hoct = 1 x(t 0).Vi t = x ta c x2 = x + 4 hay x2 x 4 = 0 suy ra x = 1

17

2so

snh iu kin x = t 0 ta c nghim phng trnh l x = 1 +

17

2

Vi t = 1 x khi phng trnh x2 = 3 x hay x2 + x 3 = 0 suy

ra x = 1

13

2 so snh iu kin ta c x = 1

13

2 Vy nghim phng trnh l x =

1 +

17

2; x =

1 132

V d 1.39 ( thi hc sinh gii tnh Thi Nguyn Lp 10 nm 2011).Gii phng trnh sau

x3 + 1 = 2 3

2x 1.Gii. t y = 3

2x 1 kt hp vi phng trnh cho ta c h

x

3

+ 1 = 2yy3 + 1 = 2x tr v vi v ca phng trnh th nht cho phngtrnh th hai ca h ta c (x y)(x2 + y2 + xy + 2) = 0 hay x = y (dox2 + y2 + xy + 2 > 0 vi mi x, y).

Thay li c x3 2x + 1 = 0 suy ra x = 1, x = 1

5

2

1.3.4. Phng trnh dng x = a +

a +

x

Cch gii. t a + x = t phng trnh cho tng ng vix = a + tt = a +

x

l h i xng loi II.


x = 2007 +

2007 +

x.

Gii. iu kin x 0. t 2007 + x = t. Ta c h phng trnhx = 2007 + tt = 2007 + x



32/70

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Ly phng trnh u tr i phng trnh th hai v vi v ta cx t = t x hay (t x)(t + x + 1) = 0 suy ra x = t.Khi ta c phng trnh xx2007 = 0 suy ra x = 8030 + 280294

(do x 0).

1.3.5. Phng trnh dng n

ax + b = r(ux + v)n + e.

Vi a = 0, u = 0, r = 0.Vi cc h s tha mn

u = arv = br + e

Cch gii t nax + b = uy+v. Sau a v h(uy + v)

n

=1

r (ux + v) e

r

(ux + v)n =1

r(uy + v) e

rh ny c gii nh h i xng loi II bng cch tr v vi v ca haiphng trnh trong h c mt phng trnh tch.


4x + 928

= 7x2 + 7.

Gii. iu kin x 94

Phng trnh cho tng ng vi

4x + 9

28= 7(x +

1

2)2 7

4

Kim tra a =1

7, b =

9

28, r = 7, e = 7

4, u = 1, v =

1

2(tha mn).

t y +1

2

= 4x + 9

28

Ta c h

(x +1

2)2 =

1

7(y +

1

2) +

1

4

(y +1

2)2 =

1

7(x +

1

2) +

1

4y l h i xng loi II tr v vi v sau rt y theo x th vo

phng trnh u gii ra ta c nghim phng trnh l

x =6 50

14, x =

49 399784



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1.4. Phng trnh gii bng phng php so snh

1.4.1. p dng tnh cht n iu ca hm s gii phngtrnh v t

Ta p dng tnh nghch bin ca hm y = ax khi 0 < a < 1 v ngbin khi a>1 gii phng trnh cha cn.


4

1 x + 4x = 1.

Gii. iu kin 0 x 1.T iu kin suy ra4

x x, 41 x 1 x 1 4x.Vy 4

1 x + 4x 1.

Du bng xy ra khi v ch khi x = 0, x = 1.p s x = 0, x = 1 l nghim ca phng trnh.


5 4

x + 8

1 5x = 1.

Gii. iu kin 0 x 15

T iu kin suy ra5 4

x 5x, 81 5x 1 5x 1 5 4x.Vy 5 4

x + 8

1 5x 1.

Du bng xy ra khi x = 0.

p s x = 0.

1.4.2. S dng bt ng thc lu tha gii mt s phngtrnh v t

Mt s bt ng thc lu .1- Vi A,B > 0 ta cn

A + n

B

2 n

A + B

2

(*).

Du bng xy ra khi v ch khi A = B.



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33

2- Vi A, B, C > 0 ta cn

A + n

B + n

C

3 nA + B + C

3

(**).

Du ng thc xy ra khi v ch khi A = B = C.Ta chng minh cc bt ng thc trnVi a, b > 0, n, m N, ta c bt ng thcam+n + bm+n 1

2(am + bm)(an + bn).

Tht vy bt ng thc trn tng ng viam+n + bm+n ambn + anbm (am bm)(an bn) 0 (hin nhin

ng).

Ta can + bn

2 1

22(a + b)(an1 + bn1) 1

2n(a + b)n.

t an = A, bn = B ta c bt ng thc (*). Ta chng minhan + bn + cn

3a + b + c

3

n, a, b, c > 0.

Xt P = an + bn + cn +a + b + c

3

nhay P

2a + b

2n + 2c +

a+b+c3

2n 4

a + b + c + a+b+c3

4n =

4a + b + c

3

nVy

an + bn + cn

3a + b + c

3

nt an = A, bn = B, cn = C.Bin i bt ng thc choA + B + C

3

n

A + n

B + n

C

3 n

hayn

A + n

B + n

C

3 n

A + B + C

3

Bt ng thc (**) c chng minh.V d 1.44. Gii phng trnh

4x2 + x 4 +

6 4x2 x = 2.

Gii. iu kin

4x2 + x 4 06 4x2 x 0

p dng bt ng thc (*) ta c

4x2 + x 4 + 6 4x2 x 222 = 2.34S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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Du bng xy ra khi4x2 + x 4 = 6 4x2 x hay 4x2 + x 4 = 6 4x2 x suy ra

x = 1, x = 5

4

Vy nghim phng trnh l x = 1, x = 54


2

x +

3 2x = 3.

Gii. iu kin 0

x

3

2p dng bt ng thc (**) ta c

x +

x +

3 2x 3

x + x + 3 2x3

= 3.

Du bng xy ra khi v ch khi

x =

3 2x suy ra x = 1.Vy nghim ca phng trnh l x = 1.

1.4.3. S dng mt s bt ng thc quen thuc so snh ccv ca phng trnh v t

1-Bt ng thc Cauchy.Vi mi b s (xi, yi) ta lun c bt ng thc sau

(ni=1

xiyi)2 (

ni=1

x2i )(ni=1

y2i )

Du ng thc xy ra khi b s (xi) v (yi) t l nhau, tc l tn ticp s thc , khng ng thi bng 0, sao cho

xi + yi = 0 vi mi i = 1, 2, 3, . . . np dng cho cc s a,b,c,d ta c(ac + bd)2 (a2 + b2)(c2 + d2).Du ng thc xy ra khi

a

c=

b

d

2-Bt ng thc trung bnh cng trung bnh nhn

Cho n s dng x1, x2, . . . , xn ta cx1 + x2 + . . . + xn

n nx1x2 . . . xn

Du ng thc xy ra khi x1 = x2 =

= xn



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35

p dng cho hai s dng a, b ta ca + b

2 ab. Du ng thc xy

ra khi a = b.

V d 1.46 ( thi ngh Olympic 30-4 THPT Chuyn Bn tre). Giiphng trnh

2

2x + 1

+

x =

x + 9.

Gii. iu kin

x + 1 > 0x 0x + 9 0

x 0.

p dng bt ng thc Cauchy cho hai cp s saua = 2

2, b =

x + 1, c =

1x + 1

, d =

x

x + 1, khi

22x + 1

+

x2

=

2

21

x + 1+

x + 1

x

x + 1

2 (8 + x + 1)

1x + 1

+x

x + 1

Suy ra 22x + 1

+ x 9 + x.

Du bng xy ra khi2

2x + 1

=1

x + 1

x

x + 1suy ra

x =1

7

Vy x =1

7l nghim phng trnh.

V d 1.47 ( thi ngh Olympic 30-4 Trng Chuyn Thng Long Lt Lm ng). Gii phng trnh

4x 1 + 48x 3 = 4x4 3x2 + 5x.

Gii. iu kin 8x 3 0 x 38

Vi iu kin trn ta chia c hai v ca phng trnh trn cho x ta c4x 1

x+

4

8x 3x

= 4x3 3x + 5.Theo bt ng thc trung bnh cng v trung bnh nhn ta c



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4x 1

x 4x 1 1

2x= 2 v

4

8x 3x

8x 3 + 1 + 1 + 14x

= 2

Do

4x

1

x +

4

8x

3

x 4.ng thc xy ra khi x =

1

2

Mt khc 4x3 3x + 5 = 4x3 3x + 1 + 4 = (2x 1)2(x + 1) + 4 4vi x 3

8

ng thc xy ra khi x =1

3

Vy phng trnh c nghim l x =1

2 V d 1.48. Gii phng trnh

x +

2 x2 = x2 + 1x2

Gii. iu kin 2 x2 0 2 x 2, x = 0.Ta c x +

2 x2 2, x2 + 1

x2 2

Hai v phng trnh bng nhau khi v ch khi x = 1.p s x = 1.


2x 1 = x3 2x2 + 2x.

Gii. iu kin 2x 1 0 x 12

Bin i phng trinh cho thnh

2x 1x

= 1 + (x

1)2.

Ta c2x 1

x= 2

1(2x 1)

2x 21 + 2x 1

2.2x= 1.

Cn 1 + (x 1)2 1.Hai v bng nhau khi v ch khi x = 1p s x = 1.


2x

2 1 + x2x 1 = 2x2

.



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Gii. iu kin

2x2 1 02x 1 0 x

1

2

Bin i phng trnh cho thnh 2x2

1x

+ 2x 1x

= 2

Tng t v d trn ta c

2x 1x

1 v

2x2 1x2

1.

Vy

2x 1x

+

2x2 1

x2 2.

Du ng thc xy ra khi v ch khi x = 1p s x = 1.

V d 1.51 ( thi ngh Olympic 30-4 Trng Nguyn Bnh KhimQung Nam). Gii phng trnh

x4 + 2006x3 + 1006009x2 + x 2x + 2007 + 1004 = 0. (1.23)

Gii. iu kin 2x + 2007 0 x 20072

.

Phng trnh cho tng ng vi phng trnh

(1.23)

x2(x2 + 2x.1003 + 10032) +

1

2

(2x + 2007

2x + 2007 + 1) = 0

x2(x + 1003)2 + 12

(

2x + 2007 1)2 = 0

x(x + 1003) = 02x + 2007 1 = 0 x = 1003

Vy x = 1003 l nghim ca phng trnh.V d 1.52 ( d b Olympic 30-4 Chuyn Hng Vng). Gii phngtrnh

4x =

30 +

1

4

30 +

1

4

30 +

1

4

x + 30.

Gii. iu kin x > 0.

t1

4

30 +

1

4

30 + x = u, u 0 ta thu c h

4x =

30 +

1

4

30 + u

4u =

30 +

1

4

30 + x

Gi sx u suy ra 4u =

30 + 1430 + x

30 + 1430 + u = 4x.



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Vy x = u v ta thu c phng trnh 4x =

30 +

1

4

30 + x

t v = 1430 + x ta thu c h

4x = 30 + v4v = 30 + xGi sx v suy ra 4v = 30 + x 30 + v = 4x.Gi sx v suy ra 4v = 30 + x 30 + v = 4x.Vy x = v ta thu c phng trnh

4x =

30 + x hay

x > 016x2 x 30 = 0 suy ra x =

1 +

1921

32

Vy nghim phng trnh l x =1 +

1921

32V d 1.53. Gii phng trnh

x2 + 3x 2 + x + 3 = 2.Gii.

iu kinx2 + 3x 2 0

x + 3 0 1 x 2.Suy ra

x + 3 2. Vy x2 + 3x 2 + x + 3 2.

Du bng xy ra khi v ch khi x = 1.V d 1.54. Gii phng trnh

x +

1 x + 1 x = 2.

Gii. iu kin 0 x 1.Ta c

x x , du bng xy ra khi v ch khi x = 0, x = 1.

1 x 1 x, du bng xy ra khi v ch khi x = 0, x = 1.1 + x 1, du bng xy ra khi x = 0.Suy ra x + 1 + x + 1 x 2.Du ng thc xy ra khi v ch khi x = 0.Vy x = 0 l nghim ca phng trnh.

1.4.4. S dng tnh cht vc t

Tnh cht 1.1. |a + b| |a| + |b|.Du bng xy ra khi hai vect a v b cng hng, iu ny xy ra khi

a = kb vi mi s thc dng k.



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Dng phng trnh

f2(x) + A2 +g

2(x) + B2 = h2(x) + C2.

Vi

f(x) + g(x) = h(x)A + B = C

t

a = (f(x); A)b = (g(x); B)

Suy ra a +b = (f(x) + g(x); A + B) = (h(x); C) (A, B, C c th l ccbiu thc cha x).

Phng trnh cho tr thnh |a + b| = |a| + |b|.Du ng thc xy ra khi v ch khi hai vect a,b cng hng hay

a = kb vi mi s thc dng k.V d 1.55 (Tuyn tp olympic 2003). Gii phng trnh

x2 8x + 816 +

x2 + 10x + 267 =

2003.

Gii. t

(4 x; 202) = a(5 + x; 11

2) = b

suy ra a + b = (9; 31

2).

Ta c |a| = x2 8x + 816; |b| = x2 + 10x + 267; |a + b| = 2003.Phng trnh cho tr thnh |a + b| = |a| + |b|.Du ng thc xy ra khi v ch khi hai vc t a,b cng hng iu

ny tng ng vi

a = kb vi mi s dng k. Khi 4 x = 2011

(5 + x). Gii ra ta c

x = 5631

Vy nghim ca phng trnh l x = 56

31



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Chng 2

Mt s phng php gii phngtrnh v t cha tham s

2.1. S dng phng php bin i tng ng

2.1.1. Phng trnh dng

f(x, m) =

g(x, m)

Cch gii. Ta thng bin i phng trnh trn v hg(x, m) c ngha v g(x, m) 0f(x, m) = g2(x, m)

Nhn xt 2.1. Khng cn t iu kin f(x, m) 0V d 2.1. Gii bin lun phng trnh

x2 1 x = m.

Gii. Ta c

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

x m2mx = m2 1

Ta xt cc trng hp.

Vi m = 0 h v nghim do c mt phng trinh th hai trong h vnghim.

Vi m = 0 h c nghim khi v ch khi phng trnh 2mx = m2 1c nghim tha mn x m iu ny xy ra khi m

2 + 1

2m m hay

m2 12m

0 suy ra m 1 hoc 1 m 0Vy m 1 hoc 1 m 0 l gi tr cn tm.



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2.1.2.Phng trnh dng

f(x, m) +

g(x, m) =

h(x, m)

Cch gii. Bin i phng trnh trn tng ng vi h

f(x, m) c ngha v f(x, m) 0g(x, m) c ngha v g(x, m) 0f(x, m) + g(x, m) + 2

f(x, m)g(x, m) = h(x, m)

V d 2.2. Tm m phng trnh sau c nghimx2 + 3x 2 =

2m + x x2.

Gii. Ta c

x2 + 3x

2

0

x = m + 1 hay1

x

2

x = m + 1Do iu kin phng trnh c nghim l 1 m + 1 2 hay

0 m 1

2.2. S dng phng php t n ph

2.2.1. Mt s dng thng gp

Cch t n ph trong trng hp bi ton phng trnh v t chatham s ging nh cch t n ph trong trng hp phng trnh v tkhng cha tham s c trnh by chng trc. Sau y l mt sv d minh ha.

V d 2.3. Vi gi tr no ca m phng trnh sau c nghim?

2(x2 2x) +

x2 2x 3 m = 0.Gii. iu kin x2

2x

3

0.

t x2 2x 3 = t vi t 0.Khi phng trnh cho c dng2(x22x3)+x2 2x 3m+6 = 0 hay f(t) = 2t2+tm+6 = 0.Phng trnh cho c nghim khi f(t) = 0 c t nht mt nghim

khng m. iu ny tng ng vi.

a.f(0) 0

0a.f(0) 0S2

0hay

6 m 0

8m 47 06 m 01

2 0



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Vy phng trnh c nghim khi v ch khi m 6.

2.2.2. t n ph a v h phng trnh i xngV d 2.4. Vi gi tr no ca a th phng trnh sau c nghim3

1 x + 31 + x = a.Gii. t

3

1 x = u3

1 + x = vsuy ra u3 + v3 = 2.

Khi phng trnh cho c chuyn v h.

u3 + v3 = 2u + v = a

hay(u + v)(u2 + v2 uv) = 2u + v = a

suy raa(u2 + v2 uv) = 2u + v = a

a. Nu a = 0 th h v nghim.b. Nu a = 0 th h c bin i nh sau

u2 + v2 uv = 2a

u + v = ahay

(v + u)2 3uv = 2

au + v = a

suy ra

u + v = a

uv =1

3(a2 2

a)

Phng trnh cho c nghim khi v ch khi

a2 413

(a2 2a

) 0 hay 8 a3

3a 0 suy ra 0 < a 2.

Vy phng trnh c nghim khi 0 < a

2.

2.3. S dng nh l Lagrange

nh l Lagrange Cho hm s f(x) lin tc trn [a; b] v tn ti ohm trn (a; b) lun tn ti mt s c (a; b) sao cho

f(c) =f(b) f(a)

b aT ta c th s dng nh l Lagrange thc hin yu cu t ra chophng trnh l.

Bi ton. Chng minh phng trnh c nghimT nh l Lagrange nu f(a) = f(b) th tn ti c (a; b) sao chof(c) =

f(b) f(a)b a = 0 hay phng trnh f

(x) = 0 c nghim thuc

(a; b).

Vy p dng c kt qu trn vo vic chng minh phng trnhf(x) = 0 c nghim trong (a; b) iu quan trng nht l nhn ra cnguyn hm F(x) ca hm f(x). C th l ta thc hin cc bc sau.



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Bc 1. Xc nh hm s F(x) kh vi lin tc trn [a; b] v tha mnF(x) = f(x) v F(a) = F(b).

Bc 2. Khi tn ti mt s c (a; b) sao cho f(c) = f(b) f(a)b a .Vy phng trnh f(x) = 0 c nghim x = c.

V d 2.5. Chng t rng vi a + 3b = 27, phng trnh

2(6x b)x + 1 = a

lun c t nht mt nghim dng.

Gii. Bin i phng trnh v dng6x b = a2

x + 1hay a

2

x + 1 6x + b = 0.

Xt hm s F(x) = a

x + 1 3x2 + bx kh vi lin tc trn (0;+)c F(x) =

a

2

x + 1 6x + b

v F(3) F(0) = (2a 27 + 3b) a = a + 3b 27 = 0.Khi tn ti c (0; 3) sao choF(x) =

F(3) F(0)3

0

haya

2

c

+ 1

6c + b = 0 tc l phng trnh cho lun c t nht mt nghim c (0; 3).

2.4. S dng phng php iu kin cn v

Phng php iu kin cn v thng t ra kh hiu qu cho ccdng ton tm iu kin ca tham s .

Dng 1: Phng trnh c nghim duy nht.Dng 2: Phng trnh c nghim vi mi gi tr ca tham s.Dng 3: Phng trnh nghim ng vi mi x D. Khi ta cn thc

hin cc bc:Bc 1: t iu kin cc biu thc trong phng trnh c ngha.Bc 2: Tm iu kin cn cho h da trn vic nh gi hoc tnh i

xng ca h.Bc 3: Kim tra iu kin trong bc ny cn c mt s k nng

c bn.



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V d 2.6. Tm m phng trnh sau c nghim duy nht

1 x2 + 2

31 x2 = m.

Gii. iu kin cn.Nhn xt rng phng trnh c nghim x0 th cng nhn x0 lm

nghim.Do phng trnh c nghim duy nht th iu kin l x0 = x0

thay vo phng trnh ta c m = 3.iu kin .Vi m = 3 khi phng trnh c dng

1

x2 + 2 3

1

x2 = 3.

V1 x2 1

3

1 x2 1 suy ra1 x2 + 2 31 x2 3.

Do phng trnh c nghim khi v ch khi

1 x2 = 13

1 x2 = 1 hayx = 0.

Vy phng trnh c nghim duy nht khi v ch khi m = 3.

V d 2.7. Tm m phng trnh sau nghim ng vi mi x 0.

x2 + 2x m2 + 2m + 4 = x + m 2.Gii.

iu kin cn.Gi s phng trnh cho c nghim vi mi x 0 th x = 0 cng

l mt nghim ca phng trnh khi thay x = 0 vo phng trnh tac

m2 + 2m + 4 = m

2 hay m 2 0m2 + 2m + 4 = (m 2)2

suy ra

m = 3.iu kin .Vi m = 3 khi phng trnh cho c dng

x2 + 2x + 1 = x + 1 phng trnh ny lun ng vi mi x 0.Vy m = 3 l gi tr cn tm.

2.5. S dng phng php hm s

Kin thc cn nh



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Cho hm s y = f(x) lin tc trn D.Phng trnh f(x) = g(m) vi m l tham s c nghim x D khi v

ch khi minD f(x) g(m) maxD f(x).Bt phng trnh f(x) g(m) c nghim x D khi v ch khi

minD

f(x) g(m).Bt phng trnh f(x) g(m) nghim ng vi mi x D khi v ch

khi maxD

f(x) g(m).Bt phng trnh f(x) g(m) c nghim x D khi v ch khi

maxD

f(x) g(m).Bt phng trnh f(x) g(m) c nghim ng vi mi x D khi v

ch khi minD

f(x) g(m).Nhn xt 2.2. Trong trng hp khng tn ti min

Df(x), max

Df(x) th

ta s lp bng bin thin ca hm s y = f(x) trn D. Sau cn c vobng bin thin kt lun cho bi ton.

Dng ton thng gp l bi ton tm gi tr ca tham s m sao chophng trnh cha tham s m c nghim ta lm nh sau.

1. Bin i phng v dng f(x) = g(m).2. Tm max

Df(x); min

Df(x) (nu c). Nu f(x) khng t gi tr max

Df(x)

hoc minD

f(x) th ta phi tnh gii hn v lp bng bin thin ca hm s

y = f(x) trn D.3. T bng bin thin suy ra gi tr m cn tm.

Ch 2.1. Trong trng hp phng trnh cha cc biu thc phc tpta c th t n ph.

- t t = (x) ((x) l hm s thch hp c mt trong f(x)).- T iu kin rng buc ca x D ta tm iu kin t K.- Ta a phng trnh v dng f(t) = h(m).- Lp bng bin thin y = f(t) trn K.- T bng bin thin suy ra kt lun ca bi ton.

V d 2.8 (Tuyn sinh i hc Cao ng khi A nm 2004). Tm m phng trnh sau c nghim.

m(1 + x2

1 x2 + 2) = 2

1 x4 + 1 + x2

1 x2.



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Gii. iu kin 1 x 1.t

1 + x2 1 x2 = t vi x [1;1].

Ta c t =x

1 + x2

+x

1 x2 = x

1

1 + x2+ 1

1 x2

.t = 0 suy ra x = 0 vi x [1;1] ta c bng bin thin.

x 1 0 1t 0 +t

2 0

2

Nh vy 0 t 2.Phng trnh (2.11) tng ng vi phng trnh sau

m(t + 2) = 2 t2 + t, t [0; 2] hay m = t2 + t + 2

t + 2

t m = g(t) = t +3 4t + 2

, t [0; 2] khi g(t) = 1 + 4(t + 2)2

g(t) = 0 suy ra t = 0 ta c bng bin thin

t 0

2

g(t) 0 g(t) 1

2 1

Nh vy phng trnh c nghim khi m [2 1;1].V d 2.9. "i hc cao ng khi A nm 2007". Tm m phng trnhsau c nghim thc.

3x 1 + mx + 1 = 24

x2 1.Gii.

iu kin x 1.Phng trnh cho tng ng vi 3( 4

x 1x + 1

)2 + m = 2 4

x 1x + 1

t t = 4

x 1x + 1

vi x 1.

Ta c 0 x

1

x + 1 = 1 2

x 1 < 1 suy ra 0 t < 1.



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Xt phng trnh theo n t vi t [0; 1) c dng3t2 + m = 2t hay m = 3t2 + 2t, t [0;1).Xt hm s f(t) = 3t

2

+ 2t, t [0; 1)Bng bin thin

t 0 13

1

f(t) 01

3 1

Nh vy phng trnh cho c nghim vi m (1; 13

]



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Chng 3

Mt s cch xy dng phng trnhv t

Phng trnh v t c nhiu dng v nhiu cch gii khc nhau. Ngigio vin ngoi vic nm c cc dng phng trnh v cc phng phpgii chng cn phi bit xy dng ln cc ton khc nhau lm ti liuging dy. Trong phn ny tc gi xin trnh by mt s cch xy dng lncc phng trnh v t, hy vng s em li nhng iu b ch.

3.1. Xy dng phng trnh v t t cc phngtrnh bit cch gii.

Con ng sng to ra nhng "phng trnh v t" l da trn c scc phng php gii c trnh by. Ta tm cch "che y" v bini i mt cht t du i bn cht, sao cho phng trnh thu c dnhn v mt hnh thc v mi quan h gia cc i tng tham gia trongphng trnh cng kh nhn ra th bi ton cng kh. Ta tm hiu mt s

cch xy dng sau

3.1.1. Xy dng phng trnh v t t phng trnh bc hai.

T phng trnh dng at2 + bt + c = 0 ta thay th t =

f(x) ta snhn c mt phng trnh v t t n ph a v phng trnh bchai gii.

V d 3.1. T phng trnh 2t2

7t + 3 = 0 ta chn t =x2 + x + 1

x 1



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49

ta c phng trnh v t sau.

2x2 + x + 1

x 1 7

x2 + x + 1

x 1 + 3 = 0

hoc bin i bi ton tr nn kh hn bng cch nhn c hai vca phng trnh trn vi x 1 ta c phng trnh sau

3(x 1) + 2(x2 + x + 1) = 7

x3 1

t phng trnh ny ta xy dng ln mt bi ton gii phng trnh v tnh sau.

Bi ton 3.1. ( thi ngh Olympic 30/4/2007) Gii phng trnh

2x2 + 5x 1 = 7

x3 1

Hng dn 3.1. Phng trnh ny c gii bng phng php av dng phng trnh bc hai nh phng trnh ban u xy dng.

Mt s dng phng trnh sau c gii bng phng php t n pha v dng phng trnh bc hai.

Dng 1. ax + b +

cx + d = 0 t

cx + d = t khi x =t2 d

cta

thu c mt phng trnh bc hai at2 + ct + bc ad = 0.Dng 2. A(

a + x +

a x) + Ba2 x2 = C.

t t =

a + x +

a x suy ra t2 = 2a + 2a2 x2.Ta thu c phng trnh bc hai At + B

t2 2a

2 = C.

Dng 3. A(x + x + a) + B(x2 + x + 2xx + a) + C = 0.t t = x +

x + a suy ra t2 = x2 + x + a + 2x

x + a.

Ta thu c phng trnh bc hai At + B(t2 a) + C = 0.Dng 4. A(x +

x2 + a) + B(x2 + x

x2 + a) + C = 0.

t t = x +

x2 + a khi t2 = 2x2 + 2x

x2 + a + a hay

x2 + x

x2 + a =t2 a

2

Cui cng ta thu c phng trinh bc hai At + B

t2

a

2 + C = 0.



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50

By gi mun to ra cc phng trnh v t mi ta c th thay thA,B,C,a,b,c bng cc s "hoc cc biu thc" theo mun l ta s c

cc dng phng trnh v t c gii theo phng php t n ph av phng trnh bc hai.

V d 3.2. ( Cho dng 2.)Chn A = 1, B = 2, C = 4, a = 1 ta c bi ton sau.

Bi ton 3.2. Gii phng trnh phng trnh

1 x + 1 + x + 21 x2 = 4.Hng dn 3.2. iu kin 1 x 1.

t

1 x + 1 + x = t, t 0 suy ra t2 = 2 + 21 x2 .Ta thu c phng trnh t2 + t 6 = 0 suy ra t = 2.Thay th tr li ta c

1 x + 1 + x = 2 suy ra x = 0.

3.1.2. Xy dng phng trnh v t t phng trnh tch.

Phng trnh dng au + bv = ab + uv. Ta c au + bv = ab + uv hay(u b)(v a) = 0 suy ra u = b, v = a.

Chn u, v l bng cc biu thc cha cn a, b bng cc s thc chotrc ta s xy dng c cc phng trnh v t.

V d 3.3. chn a = 1, b = 5, u =

x 1, v = x2 + 1. Ta thu cphng trnh

x 1 + 5

x2 + 1 x

3 x2 + x 1 = 5

v ta c bi ton sau.

Bi ton 3.3. Gii phng trnh

x 1 + 5

x2 + 1

x3 x2 + x 1 = 5

Hng dn 3.3. Nghim ca phng trnh l nghim ca phng trnhl nghim ca mt trong hai phng trnh

x2

+ 1 = 1 hoc x 1 = 5.



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Thc ra ta ang xy dng phng trnh v t da trn phng trnhc dng tch. Kh hn mt cht ta c th xy dng t phng trnh cha

nhiu tch (u a)(v b)(w c) = 0 gn cho u, v , w cc biu thc chacn. Gn cho a,b,c l cc s thc thm ch c th l cc biu thc chacn. Bin i i mt cht l ta s c c cc phng trnh v t "p"hay "khng" ph thuc vo vic ta c kho chn hay khng.

3.1.3. Xy dng phng trnh v t t mt s dng phngtrnh v t c gii theo phng php bin i tng tng.

Xy dng phng trnh v t t phng trnh dng

A +

B =

C +

D

Gn cc biu thc cha x cho A, B, C , D ta s c cc phng trnh vt c gii bng cch bnh phng hai v

V d 3.4. Gn A = x + 3, B = 3x + 1, C = 4x, C = 2x + 2 ta c biton gii phng trnh v t sau.

Bi ton 3.4. Gii phng trnhx + 3 +

3x + 1 = 2

x +

2x + 1

Hng dn 3.4. gii phng trnh ny khng kh nhng hi phctp mt cht.

Phng trnh ny s n gin hn nu ta chuyn v phng trnh3x + 1 2x + 2 = 4x x + 3

Bnh phng hai v ta c phng trnh h qu6x2 + 8x + 2 = 4x2 + 12x suy ra x = 1 th li thy x = 1 lnghim phng trnh.

Tng t ta cng c mt s dng sauDng 1: Phng trnh

A =

B.

Dng 2:Phng trnh

A = B

B 0A = B2

Dng 3:

A +

B =

C

A 0B

0A + B + 2AB = C



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Dng 4: 3

A + 3

B = 3

C suy ra A + B + 3 3

AB( 3

A + 3

B) = C

i vi dng ny thng s dng php th: 3

A + 3

B = 3

C ta c

phng trnh A + B + 33

ABC = CT cc dng ton ny gn cho A, B, C cc biu thc cha x ta s c

cc phng trnh v t tuy nhin mc kh hay d ph thuc vo vicchn cc biu thc cho A, B, C sao cho sau khi lu tha hai v ln ta thuc mt phng trnh c th gii c.

3.2. Xy dng phng trnh v t t h phng trnh.

3.2.1. Xy dng t h i xng loi II

Ta i xt mt phng trnh i xng loi II sau(x + 1)2 = y + 2(y + 1)2 = x + 2

vic gii h ny th n gin.

By gi ta s bin h trn thnh phng trnh bng cch t y = f(x)sao cho phng trnh th hai trong h lun ng vy y =

x + 2 1, khi

thay vo phng trnh u ca h ta c phng trnh(x + 1)2 = (x + 2 1) + 1 hay x2 + 2x = x + 2T c bi ton.


x2 + 2x =

x + 2

Hng dn 3.5. Ta li tin hnh t nh trn v a v gii h i xng

loai II.Bng cch tng t xt h tng qut dng bc hai

(x + )2 = ay + b(y + )2 = ax + b

ta s xy dng ln mt phng trnh dng sau.T phng trnh th hai trong h ta c y + =

ax + b khi thay

vo phng trnh u tin ca h ta c phng trnh

(x + )2 =a

ax + b + b a

()

Tng t cho bc cao hn (x + )n = anax + b + b a

(**)



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gii phng trnh ny ta li t y + = n

ax + b a v hnh ban u xy dng.

Vy c mt phng trnh v t gii bng cch a v h i xngloi II, ta ch vic chn , , a, b ph hp vi mc kh d ca bi ton.Sau xy dng phng trnh dng khai trin, hc sinh mun gii cphng trnh dng ny th phi bit vit phng trnh v dng phngtrnh (*) hoc (**) gii.

V d 3.5. Ta xy dng bi ton nh sauChn = 2, = 3, a = 4, b = 5Ta c phng trnh(2x 3)2 = 24x + 5 + 11 hay 4x2 12x 2 = 24x + 5 suy ra

2x2 6x 1 = 4x + 5Khi ta c mt bi ton mi.

Bi ton 3.6. Gii phng trnh v t

2x2 6x 1 = 4x + 5

Hng dn 3.6. Hc sinh phi bit bin i dng khai trin ny vphng trnh (2x 3)2 = 24x + 5 + 11

Sau t 2y 3 = 4x + 5 c h phng trnh.(2x 3)2 = 4y + 5(2y 3)2 = 4x + 5 suy ra (x y)(x + y 1) = 0

Vi x = y khi 2x 3 = 4x + 5 suy ra x = 2 + 3Vi x + y 1 = 0 khi y = 1 suy ra x = 1 2

3.3. Dng hng ng thc xy dng cc phngtrnh v t

3.3.1. T nhng nh gi bnh phng A2 + B2 0.Ta xy dng nhng phng trnh dng

A2 + B2 = 0 suy ra

A = 0B = 0

V d t phng trnh (

5x

1

2x)2 + (

9

5x

2)2 +

x

1 = 0

khai trin ra ta c phng trnh



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4x2 + 12 +

x 1 = 4x5x 1 + 49 5x)Khi ta c th xy dng bi ton.


4x2 + 12 +

x 1 = 4x5x 1 + 49 5x)

Hng dn 3.7. Khi mun gii bi ton trn ta bin i a v phngtrnh trc khi khai trin v gii l tt nht. Sau p dng nh gi nh trnh by.

3.3.2. T hng ng thc (A B)2 = 0 suy ra A = B

V d chn A = 1, B =

4x

x + 3ta c phng trnh

(1

4x

x + 3)2 = 0 khai trin ra ta c phng trnh

1 +4x

x + 3= 2

4x

x + 3

Nhn hai v phng trnh vi x + 3 ta c phng trnhx + 3 +

4xx + 3

= 4x ta c bi ton.


x + 3 +4x

x + 3= 4x

3.3.3. Xy dng phng trnh v t t hng ng thc sau.

Ta c

(A + B + C)3 = A3 + B3 + C3 + 3(A + B)(B + C)(C + A)

Khi (A + B + C)3 = A3 + B3 + C3 khi (A + B)(B + C)(C + A) = 0iu ny xy ra khi A = B hoc B = C hoc A = CTa c th xy dng bi ton nh sau.Gn A = 37x + 2010, B = 3x2 + 2011, C = 3x2 7x + 2012



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Nh vy A3 + B3 + C3 = 2011. T ta c bi ton gii phng trnhnh sau.


3

7x + 2010 3

x2 + 2011 +3

x2 7x + 2012 = 3

2011

Hng dn 3.8. Bi ta tha mn (A + B + C)3 = A3 + B3 + C3. Nn

nghim ca phng trnh l nghim ca h tuyn

A = BB = CA = C

Trong

A,B,C l cc biu thc nh chn.

Tng t ta c th xy dng nhiu bi ton theo cch ny!

3.4. Xy dng phng trnh v t da theo hm niu.

3.4.1. Xy dng phng trnh v t da theo tnh cht ca hmn iu.

Da vo kt qu "nu hm y = f(x) n iu thf(x) = f(t) x = t" ta c th xy dng c nhng phng trnh v

t.

V d 3.6. Xut pht t hm n iu y = f(x) = 2x3 + x2 + 1 mix 0 ta xy dng phng trnh.

f(x) = f(

3x 1) hay 2x3 + x2 + 1 = 2(3x 1)3 +(3x 1)2 + 1Rt gn ta c phng trnh 2x3 + x2 3x + 1 = 2(3x 1)3x 1.

Ta c bi ton sauBi ton 3.10. Gii phng trnh

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

Hng dn 3.9. Bi ton c gii theo phng php hm s.

V d 3.7. T hm s ng bin trn R, f(t) = t3 + t v t phngtrnh f( 3

7x2 + 9x

4) = f(x + 1)

Ta xy dng c bi ton sau.



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x3

4x2

5x + 6 =

37x2 + 9x 4Hng dn 3.10. Bi ton c gii theo phng php s dng phngphp hm s. Gii phng trnh ta c nghim phng trnh l nghim caphng trnh.

(x + 1) = 3

7x2 + 9x 4 suy ra x = 5 hoc x = 1

5

2

3.4.2. Xy dng phng trnh v t da vo cc c lng ca

hm n iu. d s dng v kt hp nhiu c lng ta xy dng mt s c lng

nh sau bng cch s dng tnh n iu ca hm s ta nhn c[1] 1 x 1 x 1Hm s f(x) =

x 1 x l hm n iu tng trn [0; 1]

Nn ta c 1 = f(0) f(x) f(1) = 1[2] 1 4x 41 x 1

Hm s f(x) =4

x 4

1 x 1 l hm tng trn [0;1]Nn ta c 1 = f(0) f(x) f(1) = 1[3] 1 4x 1 x 1Hm s f(x) = 4

x 1 x l hm tng trn [0;1]

Nn ta c 1 = f(0) f(x) f(1) = 1[4] 0

x + 3

2 +

1 x 1

Hm s f(x) =

x + 3

2 + 1 xl hm tng trn [

3;1]

Nn ta c 0 = f(3) f(x) f(1) = 1[4] 0

4

x + 15

2 + 4

1 x 1

Hm s f(x) =4

x + 15

2 + 4

1 x l hm tng trn [15; 1]Nn ta c 0 = f(15) f(x) f(1) = 1[6] 3 +

x 1 +

1 x 1

Hm s f(x) =

3 + x 1 + 1 x l hm tng trn on [0;1]57S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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Suy ra f(x) f(1) = 1[7]

x + 3 1 +

1 x 1

Hm s f(x) = x + 3

1 + 1 x l hm tng trn [3;1]suy ra f(x) f(1) = 1S dng tnh cht nghch bin ca hm s m y = ax vi c s 0 a 1

ta nhn c[8]

x +

1 x 1

Ta c

x x hoc 1 x 1 xsuy ra

x +

1 x x + (1 x) = 1

Du ng thc t c khi x = 0 hoc x = 1Cng hai hay nhiu cc c lng c bn chng ta thu c cc phng

trnh cha cn sau.


x + 4

x + 6

x = 3 +

1 x + 41 x + 61 x.Hng dn 3.11. iu kin 0 x 1

Phng trnh cho tng ng vi.

(x 1 x) + (4

x 4

1 x) + (6

x 6

1 x) = 3S dng cc c lng c bn ta thu c v tri nh hn hoc bng

3, v du ng thc xy ra khi v ch khi x = 1p s x = 1


x + 4

x + 6

x + 3 4

1 x = 3.Hng dn 3.12. iu kin 0

x

1

Phng trnh cho tng ng vi(

x + 4

1 x) + ( 4x + 41 x) + ( 6x + 41 x)V tri ln hn hoc bng 3. Du ng thc xy ra khi v ch khi x = 1.p s x = 1.


x + 3

2 + 1 x +

3x + 1

2 + 41 x



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Hng dn 3.13. iu kin 13

x 1V tri nh hn hoc bng 2. Du bng xy ra khi x = 1.p s x = 1.Nhn cc c lng c bn dng ta thu c cc phng

trnh cha cn sau.


2x 1 44x 3 66x 5 = x3

Hng dn 3.14. iu kin x 5

6 Chia hai v cho x3 = 0 ta thu c

2x 1x

4

4x 3x

6

6x 5x

= 1

V tri nh hn hoc bng 1. Du ng thc xy ra khi v ch khi x = 1.p s x = 1.

3.5. Xy dng phng trnh v t da vo hm s

lng gic v phng trnh lng gic.T cng thc lng gic n gin cos3t = sin t, ta c th to ra c

nhng phng trnh v t.T cos3t = 4cos3 t 3cos t ta c phng trnh v t 4x3 3x =

x2

x2 1 (1)Nu thay x trong phng trnh (1) bi

1

xta s c phng trnh v t

kh hn 4 3x2

= x2

x2

1 (2).Nu thay x trong phng trnh (1) bi x 1 ta s c phng trnh vt kh 4x3 12x2 + 9x 1 = 2x x2(3).

Tng t nh vy t cc cng thc sin3x, sin4x . . . ta cng c th xydng phng trnh v t theo kiu lng gic!

V d 3.8. T phng trnh lng gic1

cos t+

1

sin t= 2

2 v t ng

thc lng gic sin2 t +cos2 t = 1 suy ra sin t =

1 cos2 t thay th cos tbi x, ta s c mt phng trnh v t nh sau 1x + 11 x2 = 22.



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Vy ta c bi ton gii phng trnh v t c gii theo phng phpt n ph lng gic nh sau.


1

x+

1x2 + 1

= 2

2

Hng dn 3.15. Khi phng trnh ny c gii theo phng phpt n ph lng gic.

V d 3.9. T phng trnh cos3 t + sin3 t = 2cos t sin t thay th cos tbi x ta c phng trnh v t x3 +

(1 x2)3 = x2(1 x2).

V ta c bi ton gii phng trnh v t


x3 +

(1 x2)3 = x

2(1 x2)

Hng dn 3.16. Phng trnh ny c gii theo phng php t nph lng gic.

V d 3.10. T phng trnh 5+3sin t = 8(cos6 t +sin6 t) thay th cos tbi x, ta c phng trnh v t. 5 + 3

1 x2 = 8[x6 + (1 x2)3]. V

ta c bi ton.

Bi ton 3.18. Gii phng trnh 5 + 3

1 x2 = 8[x6 + (1 x2)3]Hng dn 3.17. T iu kin

|x|

1 ta t x = cos t; t

[0; ] v tathu c.

5 + 3 sin t = 8(sin6 t + cos6 t) 3sin t = 8(1 3sin2 t cos2 t) 3sin t = 3 24 sin2 t cos2 t sin t = 1 8sin2 t cos2 t = 1 2sin2 2t = cos4t cos4t = cos

2

t

.

T phng trnh ny ta s tm c t, sau suy ra c x.



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3.6. Xy dng phng trnh v t t php "t nph khng ton phn".

Ta xt bi ton xy dng lp phng trnh dng At2 + Bt + C = 0,trong t l biu thc cha cn ca x, cn A, B, C l cc biu thc hu tcha x, sao cho = B2 4AC lun lun l mt biu thc chnh phng.

Thng cho tin ta chn BA

= f(x) + g(x) cnC

A= f(x)g(x) khi

t = g(x) hoc t = f(x).

V d 3.11. Ta chn t =

x2 + 2, f(x) = 3, g(x) = x

1.

Ta c bi ton gii phng trnh v t nh sau.Bi ton 3.19. Gii phng trnh

x2 + (3

x2 + 2)x = 1 + 2

x2 + 2

Hng dn 3.18. gii bi ny hc sinh phi bit bin i v dng(x2+2)(2 + x)x2 + 23 + 3x = 0 sau t t = x2+ 2 th phng

trnh cho tr thnh phng trnh n t l t2 (2 + x)t 3 + 3x = 0 rigii phng trnh ny.

C th nhm nghim hoc tnh , c nghim t = 3, t = x 1.Vi t = 3 th

x2 + 2 = 3 suy ra x = 7.

Vi t = x 1(x 1) v nghim.Nh vy vi cch xy dng nh trn ta c th c nhiu bi ton. Gii

phng trnh v t bng phng php n ph khng ton phn.

3.7. Xy dng phng trnh v t da vo tnh chtvect.

Tnh cht |a + b| |a| + |b| du "=" xy ra khi a,b cng hng.Ta xy dng nh sau.

t

a = (f(x); A)b = (g(x); B)

suy ra a + b = (f(x) + g(x); A + B).

Khi ta c phng trnh

f2(x) + A2 +

g2(x) + B2 =

(f(x) + g(x))2 + (A + B)2



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V d 3.12. chn

a = (4 x; 202)b = (5 + x; 10

2)

suy ra a + b = (9;31

2).

Vy |a| = x2 8x + 816, |b| = x2 + 10x + 267

v |a + b| = 2003.Ta xy dng c phng trnh nh sau|a + b| = |a| + |b| x2 8x + 816 + x2 + 10x + 267 = 2003

Bi ton 3.20 (Thi olympic 30-4 nm 2003). Gii phng trnh

x2 8x + 816 +

x2 + 10x + 267 = 2003.

3.8. Xy dng phng trnh v t da vo bt ngthc

3.8.1. Bt ng thc "tam thc bc (xem [2], trang 25).

Vi mi > 1 ta c x + 1 x vi mi x 0 .Du ng thc xy ra khi x = 1.Chn =2010 ta c phng trnh

x2010 + 2009 =

2010x v ta c

bi ton.


x2010 + 2009 =

2010x.

Hng dn 3.19. S dng bt ng thc "tam thc bc " ta s cx = 1 l nghim.

V d 3.13. C th to ra cc bi ton kh hn bng cch thay x phngtrnh trn bng cc biu thc cha x.V d thay x bng (x2 1)2 v = 3 ta s c phng trnh

(x2 1)3 + 2 = 3(x2 + 1) hay x6 3x4 + 3x2 + 1 = 3(x2 1).Ta c bi ton sau.


x6 3x4 + 3x2 + 1 = 3(x2 1).Hng dn 3.20. Gp bi ton ny hc sinh cn bit bin i a vphng trnh ban u xy dng p dng bt ng thc th bi ton



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tr nn d dng. p dng bt ng thc trn tm ra c nghim phngtrnh l khi du ng thc xy ra x2

1 = 1 suy ra x =

2.

Nh vy c th thay x bng cc biu thc khc ca x ta s c ccphng trnh kh hay d tu thuc vo ta chn biu thc thay th cho x.

3.8.2. Xy dng phng trnh v t da vo bt ng thc giatrung bnh cng v trung bnh nhn.

S dng bt ng thc gia trung bnh cng v trung bnh nhn tanhn c

[1]

2x

1

x 1V

2x 1

x 2x 1 + 1

2x= 1

[2]4

4x 3x

1

V4

4x 3x

4x 3 + 1 + 1 + 14x

= 1 du ng thc xy ra khi x = 1

[3]6

6x 5x

6x 5 + 1 + 1 + 1 + 1 + 1

6x

= 1 du ng thc xy ra

khi x = 1[4]

x +

2 x 2

V

x +

2 x 2

x + 2 x2

= 2 du ng thc xy ra khi x = 1

[5] 4

x + 4

2 x 2V 4

x + 4

2 x 2 4

x + 2 x

2= 2 du ng thc xy ra khi x = 1

[6] x +

2 x2 2 vx + 2 x2 |x| + 2 x2 = x2 + 2 x2 2

x2

+ 2 x2

2= 2

du ng thc xy ra khi x = 1[7] x + 4

2 x4 2 v

x + 4

2 x4 |x| + 42 x4 =

x2 + 4

2 x4

=4

x4 + 4

2 x4 2 4

x4 + 2 x42

= 1

du ng thc xy ra khi x = 1

[8] x +6

2 x6

2 v



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63

x + 6

2 x6 |x| + 62 x6 6

x6 + 6

2 x6 2 6

x6 + 2 x62

= 1

du ng thc xy ra khi x = 1.Ta gi y l cc c lng c bn. C th to ra c rt nhiu ccc lng c bn theo cch ny.

Ta xy dng cc phng trnh v t nh sauCch 1. Cng hai hay nhiu cc c lng c bn


2x

1 + 4

4x

3 + 6

6x

5 = 3x

Hng dn 3.21. iu kin x 56

.

Phng trnh cho tng ng vi2x 1

x+

4

4x 3x

+6

6x 5x

= 3

V tri nh thua hoc bng 3. Du ng thc xy ra khi v ch khix = 1.

Bi ton 3.24. Cng cc c lng c bn nh sau.

(

x + 4

1 x) + ( 4

x4 + 4

1 x) + ( 6x + 41 x) = 3

Vit li

x + 4

x + 6

x + 3 4

1 x = 3.Khi ta c bi ton sau.

Bi ton 3.25.

x + 4

x + 6

x + 3 4

1 x = 3

Hng dn 3.22. V tri phng trnh ln hn hoc bng 3. Du ngthc xy ra khi x = 1. Vy x = 1 l nghim phng trnh.Cch 2. Nhn cc c lng c bn ta c cc phng trnhcha cn.

V d nhn cc c lung c bn ta c bi ton gii phng trnh.

2x 1. 44x 3. 66x 5 = x3



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Hng dn 3.23. Vi iu kin x 56 Sau chia c hai v cho x ta

thu c.

2x 1x

.44x 3

x.

66x 5x

= 1

V tri nh thua hoc bng 1. Du ng thc xy ra khi x = 1. Nhn ccc lng c bn ta c bi ton.


2x

1

x(x + 1 x) = 1

Hng dn 3.24. V1

x +

1 x 1;

2x 1x

1 suy ra v trinh hn hoc bng 1. Du ng thc xy ra khi v ch khi x = 1. p sx = 1 l nghim phng trnh.

3.8.3. Xy dng phng trnh da theo bt ng thc Cauchy.

T bt ng thc (AB + CD)2 (A2 + C2)(B2 + D2) du ng thcxy ra khi "AD = BC"

V d 3.14. Ta chn cc cp s A = 2

2, B =

x + 1, C =1

x + 1,

D =

x

x + 1Ta xy dng phng trnh

(221

x + 1 + x + 1

x

x + 1)2

(8 + x + 1)1

x + 1 +

x

x + 1

suy ra2

2x + 1

9 + xT y ta i xy dng bi ton gii phng trnh v t sau.


x +

2

2x + 1

=

9 + x



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Hng dn 3.25. Nh vy hc sinh c th bin i phng trnh pdng bt ng thc Cauchy cho hai cp s A, C v B, D nh trn rt

ra du bng xy ra khi 22x + 1

= 1x + 1

xx + 1

suy ra x =1

7

Vy x =1

7l nghim phng trnh.

V d 3.15. T cc cp s A = 3

2, B =

2x + 1, C =

x

2x + 1,

D =

x + 1

2x + 1Tng t nh trn ta c th xy dng bi ton gii phng trnh v t

sau.


3

2x2x + 1

+

x + 1 =

19 + 2x

Hng dn 3.26. p dng bt ng thc Cauchy cc cp s A,C v B,Dnh trn ta c du ng thc xy ra hay nghim phng trnh l nghim

ca phng trnh3

22x + 1

=

x

x + 1iu kin x 0. Bnh phng hai

v gii phng trnh ta c x =17 +

433

4l nghim.

3.9. Xy dng phng trnh v t bng phng php

hnh hcCho tam gic ABC c ng AD l ng phn gic trong ca gc A,

A = 2, m AD, t AM = x khi nu M BC th M BM C

=AB

AC

V d 3.16. Xt tam gic vung ABC vung ti A c AB = 4, AC = 3.Gi AD l phn gic trong ca A. Trn AD ly im M, t AM = x.Trong tam gic AMC c CM2 = x2 32x + 9.

Trong tam gic AMB c BM2 = x2

4

2x. Ta xy dng bi ton nhsau.



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Bi ton 3.29 (D b Olympic 30-4 THPT Chuyn Tin Giang nm2007). Gii phng trnh

x2 3x2 + 9 +

x2 4x2 + 16 = 5.

Hng dn 3.27. Khi vi x < 0 th

x2 3x2 + 9 > 3x2 42x + 16 > 4

phng trnh v nghim.Vi x > 0 ta cCM + BM = x2 3x

2 + 9 +x2 4x

2 + 16 BC = 5.

Du bng xy ra khi M trng vi D hay M chia BC theo t sk =

4

3 x = AD = 12

2

7

V d 3.17. Cho tam gic ABC c AB = 3, AC = 4, A = 120o, AD lng phn gic trong gc A. Ly M AD, t AM = x

Khi ta c BM =

x2 3x + 9, CM = x2 4x + 16,BC = 9 + 16 2.3.4.

1

2 = 37.Ta xy dng phng trnh x2 3x + 9 + x2 4x + 16 = 37. Ta

c bi ton.

Bi ton 3.30. Gii phng trnhx2 3x + 9 +

x2 4x + 16 =

37

Hng dn 3.28. Lp lun tng t nh hai bi ton trn ta c x < 0phng trnh v nghim.

Vi x > 0 ta cBM + CM =

x2 3x + 9 + x2 4x + 16 BC = 37.

Du bng xy ra khi M trng vi D hay M chia BC theo t sM B

M C=

AB

AC=

3

4

x2 3x + 9

x2 4x + 16 =3

4 x = 12

7

V d 3.18. Cho tam gic ABC c AB = 5, AC = 4, A = 60o, AD lng phn gic trong gc A. Ly M

AD, t AM = x

Khi ta c BM =

x2 53x + 25, CM = x2 43x + 16,67S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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BC =

25 + 16 2.4.5.1

2=

21.

Ta xy dng phng trnh

x2 53x + 25 +

x2 43x + 16 = 21. Ta c bi ton.Bi ton 3.31. Gii phng trnh

x2 5

3x + 25 +

x2 4

3x + 16 =

21

Hng dn 3.29. Vi x < 0 ta c x2 5

3x + 25 > 5x2 43x + 16 > 4

phng

trnh v nghim.Xt x > 0 khi du bng ca phng trnh xy ra khiM B

M C=

5

4hay 4

x2 53x + 25 = 5

x2 43x + 16.

Bnh phng hai v gii ra ta c nghim phng trnh l x =20

3

9



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Kt lun

Lun vn Mt s phng php gii phng trnh v t gii quytc nhng vn sau:

- H thng cc phng php gii cc phng trnh v t.- Trnh by cc phng php gii v bin lun phng trnh v t ccha tham s.

- a ra cc phng php xy dng phng trnh v t mi.Kt qu ca lun vn gp phn nng cao cht lng dy v hc Ton

trng ph thng trong giai on hin nay.



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Ti liu tham kho

[1] Nguyn Vn Mu, 1993, Phng php gii phng trnh v bt phngtrnh, NXB Gio Dc.

[2] Nguyn Vn Mu, 2004, Bt ng thc, nh l v p dng, NXBGio Dc.

[3] Cc chuyn phng php gii phng trnh bt phng trnh trnmng Internet.

[4] Nguyn Vn Mu, 2002, a thc i s v phn thc hu t, NXBGio Dc.

[5] Tp ch ton hc tui tr.[6] Cc tuyn tp thi Olympic 30-4, NXB i Hc S Phm.

[7] Nguyn V Lng, 2008, H phng trnh phng trnh cha cn, NXBi Hc Quc Gia H Ni.

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