Upload
chuong-gio
View
231
Download
0
Embed Size (px)
Citation preview
7/28/2019 Mt s phng php gii phng trnh v t
1/70
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
7/28/2019 Mt s phng php gii phng trnh v t
2/70
1
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
2S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
3/70
2
Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
4/70
3
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.
4S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
5/70
4
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
5S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
6/70
5
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)
6S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
7/70
6
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.
7S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
8/70
7
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.
8S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
9/70
8
V d 1.4. Gii phng trnh
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).
9S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
10/70
9
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.
10S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
11/70
10
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.
V d 1.6. Gii phng trnh
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.
11S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
12/70
11
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.
12S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
13/70
12
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.
13S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
14/70
13
V d 1.8. Gii phng trnh
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.
V d 1.10. Gii phng trnh
3 + x +
6 x = 3 +
(3 + x)(6 x).
Gii. iu kin3 + x
0
6 x 0 3 x 6.
14S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
15/70
14
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)
15S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
16/70
15
(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.
V d 1.13. Gii phng trnh
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.
16S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
17/70
16
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
V d 1.14. Gii phng trnh
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 .
17S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
18/70
17
V d 1.15. Gii phng trnh
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
18S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
19/70
18
V d 1.17. Gii phng trnh
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.
V d 1.19. Gii phng trnh3
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.
19S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
20/70
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
V d 1.20. Gii phng trnh
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
V d 1.21. Gii phng trnh
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.
20S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
21/70
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
V d 1.23. Gii phng trnh
x3 3x = x + 2. (1.15)
21S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
22/70
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
V d 1.24. Gii phng trnh
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)
22S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
23/70
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
V d 1.25. Gii phng trnh
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
V d 1.26. Gii phng trnh
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.
23S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
24/70
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.
V d 1.28. Gii phng trnh3
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
24S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
25/70
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.
V d 1.30. Gii phng trnh
5x3 1 + 3
2x
1 + x = 4.
25S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
26/70
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
V d 1.32. Gii phng trnh
2x3 + 3x2 + 6x + 16 4 x = 23.
26S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
27/70
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.
V d 1.34. Gii phng trnh4
57 x + 4x + 40 = 5. (1.20)
Gii. iu kin 40 x 57.
27S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
28/70
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.
V d 1.35. Gii phng trnh
3
7 + tan x +3
2 tan x = 3.
28S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
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.
29S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
30/70
29
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
V d 1.37. Gii phng trnh
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.
V d 1.38. Gii phng trnh
x2 4 = x + 4.
Gii. iu kin x 4.
30S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
31/70
30
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.
V d 1.40. Gii phng trnh
x = 2007 +
2007 +
x.
Gii. iu kin x 0. t 2007 + x = t. Ta c h phng trnhx = 2007 + tt = 2007 + x
31S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
32/70
31
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.
V d 1.41. Gii phng trnh
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
32S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
33/70
32
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.
V d 1.42. Gii phng trnh
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.
V d 1.43. Gii 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.
33S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
34/70
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
7/28/2019 Mt s phng php gii phng trnh v t
35/70
34
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
V d 1.45. Gii phng trnh
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
35S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
36/70
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
36S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
37/70
36
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.
V d 1.49. Gii phng trnh
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.
V d 1.50. Gii phng trnh
2x
2 1 + x2x 1 = 2x2
.
37S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
38/70
37
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.
38S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
39/70
38
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.
39S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
40/70
39
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
40S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
41/70
40
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.
41S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
42/70
41
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
42S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
43/70
42
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.
43S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
44/70
43
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.
44S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
45/70
44
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
45S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
46/70
45
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.
46S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
47/70
46
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.
47S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
48/70
47
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
]
48S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
49/70
48
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
49S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
50/70
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.
50S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
51/70
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.
51S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
52/70
51
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
52S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
53/70
52
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.
Bi ton 3.5. Gii phng trnh
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
(**)
53S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
54/70
53
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
54S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
55/70
54
4x2 + 12 +
x 1 = 4x5x 1 + 49 5x)Khi ta c th xy dng bi ton.
Bi ton 3.7. Gii phng trnh
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.
Bi ton 3.8. Gii phng trnh
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
55S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
56/70
55
Nh vy A3 + B3 + C3 = 2011. T ta c bi ton gii phng trnhnh sau.
Bi ton 3.9. Gii phng trnh
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.
56S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
57/70
56
Bi ton 3.11. Gii phng trnh
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
7/28/2019 Mt s phng php gii phng trnh v t
58/70
57
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.
Bi ton 3.12. Gii phng trnh
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
Bi ton 3.13. Gii phng trnh
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.
Bi ton 3.14. Gii phng trnh
x + 3
2 + 1 x +
3x + 1
2 + 41 x
58S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
59/70
58
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.
Bi ton 3.15. Gii phng trnh
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.
59S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
60/70
59
Vy ta c bi ton gii phng trnh v t c gii theo phng phpt n ph lng gic nh sau.
Bi ton 3.16. Gii phng trnh
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
Bi ton 3.17. Gii phng trnh
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.
60S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
61/70
60
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
61S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
62/70
61
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.
Bi ton 3.21. Gii phng trnh
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.
Bi ton 3.22. Gii phng trnh
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
62S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
63/70
62
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
63S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
64/70
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
Bi ton 3.23. Gii phng trnh
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
64S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
65/70
64
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.
Bi ton 3.26. Gii phng trnh
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.
Bi ton 3.27. Gii phng trnh
x +
2
2x + 1
=
9 + x
65S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
66/70
65
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.
Bi ton 3.28. Gii phng trnh
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.
66S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
67/70
66
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
7/28/2019 Mt s phng php gii phng trnh v t
68/70
67
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
68S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
69/70
68
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.
69S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn
7/28/2019 Mt s phng php gii phng trnh v t
70/70
69
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.