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CHUYN :
PHNG TRNH - HPHNG TRNH
THNG 6/2012
Din n MATHSCOPE
Niels Henrik Abel(1802-1829)
Gerolamo Cardano(1501-1576)
variste Galois(1811-1832)
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Din n MATHSCOPE
PHNG TRNH
H PHNG TRNHCh bin: Nguyn Anh Huy
26 - 7 - 2012
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Mc lcLi ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Cc thnh vin tham gia chuyn . . . . . . . . . . . . . . . . . . . . . . . . 8
1 I CNG V PHNG TRNH HU T 10
Phng trnh bc ba . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Phng trnh bc bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Phng trnh dng phn thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Xy dng phng trnh hu t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Mt s phng trnh bc cao . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2 PHNG TRNH, H PHNG TRNH C THAM S 32
Phng php s dng o hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Phng php dng nh l Lagrange - Rolle . . . . . . . . . . . . . . . . . . . . . . 42
Phng php dng iu kin cn v . . . . . . . . . . . . . . . . . . . . . . . . . 46Phng php ng dng hnh hc gii tch v hnh hc phng . . . . . . . . . . . . . 55
Hnh hc khng gian v vic kho st h phng trnh ba n . . . . . . . . . . . . . 76
Mt s bi phng trnh, h phng trnh c tham s trong cc k thi Olympic . . . 81
3 CC PHNG PHP GII PHNG TRNH 93
Phng php t n ph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Mt s cch t n ph c bn . . . . . . . . . . . . . . . . . . . . . . . . . . 93
t n ph a v phng trnh tch . . . . . . . . . . . . . . . . . . . . . . . 94t n ph a v phng trnh ng cp . . . . . . . . . . . . . . . . . . . . 101Phng php t n ph khng hon ton . . . . . . . . . . . . . . . . . . . . 103Phng php s dng h s bt nh . . . . . . . . . . . . . . . . . . . . . . . 108t n ph a v h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . 109Phng php lng gic ha . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
Phng php bin i ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
Phng php dng lng lin hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
Phng php dng n iu hm s . . . . . . . . . . . . . . . . . . . . . . . . . . 138
Phng php dng bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
Mt s bi ton chn lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
3
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4 PHNG TRNH M-LOGARIT 158L thuyt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Phng php t n ph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Phng php dng n iu hm s . . . . . . . . . . . . . . . . . . . . . . . . . . 166
Phng php bin i ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
Bi tp tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
5 H PHNG TRNH 177Cc loi h c bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
H phng trnh hon v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
Phng php t n ph trong gii h phng trnh . . . . . . . . . . . . . . . . . . 206
Phng php bin i ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
Phng php dng n iu hm s . . . . . . . . . . . . . . . . . . . . . . . . . . 222
Phng php h s bt nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231K thut t n ph tng - hiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
Phng php dng bt ng thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
Tng hp cc bi h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258H phng trnh hu t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258H phng trnh v t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
6 SNG TO PHNG TRNH - H PHNG TRNH 297
Xy dng mt s phng trnh c gii bng cch a v h phng trnh . . . . 297S dng cng thc lng gic sng tc cc phng trnh a thc bc cao . . . . 307
S dng cc hm lng gic hyperbolic . . . . . . . . . . . . . . . . . . . . . . . . . 310
Sng tc mt s phng trnh ng cp i vi hai biu thc . . . . . . . . . . . . . 312
Xy dng phng trnh t cc ng thc . . . . . . . . . . . . . . . . . . . . . . . . 318
Xy dng phng trnh t cc h i xng loi II . . . . . . . . . . . . . . . . . . . 321Xy dng phng trnh v t da vo tnh n iu ca hm s. . . . . . . . . 324Xy dng phng trnh v t da vo cc phng trnh lng gic. . . . . . . . 328
S dng cn bc n ca s phc sng to v gii h phng trnh. . . . . . . 331S dng bt ng thc lng gic trong tam gic . . . . . . . . . . . . . . . . 338S dng hm ngc sng tc mt s phng trnh, h phng trnh. . . . . 345
Sng tc h phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
Kinh nghim gii mt s bi h phng trnh . . . . . . . . . . . . . . . . . . . . . 353
7 Ph lc 1: GII TON BNG PHNG TRNH - H PHNG TRNH 362
8 Ph lc 2: PHNG TRNH V CC NH TON HC NI TING 366
Lch s pht trin ca phng trnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 366C my cch gii phng trnh bc hai? . . . . . . . . . . . . . . . . . . . . . 366Cuc thch chn ng th gii ton hc . . . . . . . . . . . . . . . . . . . . 368Nhng vinh quang sau khi qua i . . . . . . . . . . . . . . . . . . . . . . . 372
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Tu s mt s nh ton hc ni ting . . . . . . . . . . . . . . . . . . . . . . . . . 376Mt cuc i trn bia m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376Ch v l sch qu hp! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376Hai gng mt tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377Sng hay cht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
9 Ti liu tham kho 381
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Li ni uPhng trnh l mt trong nhng phn mn quan trng nht ca i s v c nhng ng
dng rt ln trong cc ngnh khoa hc. Sm c bit n t thi xa xa do nhu cu tnhton ca con ngi v ngy cng pht trin theo thi gian, n nay, ch xt ring trong Tonhc, lnh vc phng trnh c nhng ci tin ng k, c v hnh thc (phng trnh hu t,phng trnh v t, phng trnh m - logarit) v i tng (phng trnh hm, phng trnhsai phn, phng trnh o hm ring, ...)
Cn Vit Nam, phng trnh, t nm lp 8, l mt dng ton quen thuc v cyu thch bi nhiu bn hc sinh. Ln n bc THPT, vi s h tr ca cc cng c gii tchv hnh hc, nhng bi ton phng trnh - h phng trnh ngy cng c trau chut, trthnh nt p ca Ton hc v mt phn khng th thiu trong cc k thi Hc sinh gii, thii hc.
c rt nhiu bi vit v phng trnh - h phng trnh, nhng cha th cp mt
cch ton din v nhng phng php gii v sng to phng trnh. Nhn thy nhu cu cmt ti liu y v hnh thc v ni dung cho c h chuyn v khng chuyn, Din nMathScope tin hnh bin son quyn sch Chuyn phng trnh - h phng trnh mchng ti hn hnh gii thiu n cc thy c gio v cc bn hc sinh.
Quyn sch ny gm 6 chng, vi cc ni dung nh sau: Chng I: i cng v phng hu t cung cp mt s cch gii tng qut phngtrnh bc ba v bn, ngoi ra cn cp n phng trnh phn thc v nhng cch xy dngphng trnh hu t. Chng II: Phng trnh, h phng trnh c tham s cp n cc phng phpgii v bin lun bi ton c tham s ,cng nh mt s bi ton thng gp trong cc k thiHc sinh gii. Chng III: Cc phng php gii phng trnh ch yu tng hp nhng phngphp quen thuc nh bt ng thc, lng lin hp, hm s n iu, ...vi nhiu bi tonm rng nhm gip bn c c cch nhn tng quan v phng trnh.Chng ny khng cp n Phng trnh lng gic, v vn ny c trong chuyn Lng gic ca Din n.
Chng IV: Phng trnh m logarit a ra mt s dng bi tp ng dng ca hms logarit, vi nhiu phng php bin i a dng nh t n ph, dng ng thc, hm niu, ... Chng V: H phng trnh l phn trng tm ca chuyn . Ni dung ca chng
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bao gm mt s phng php gii h phng trnh v tng hp cc bi h phng trnh haytrong nhng k thi hc sinh gii trong nc cng nh quc t. Chng VI: Sng to phng trnh - h phng trnh a ra nhng cch xy dng mt bihay v kh t nhng phng trnh n gin bng cc cng c mi nh s phc, hm hyperbolic,hm n iu, . . .
Ngoi ra cn c hai phn Ph lc cung cp thng tin ng dng phng trnh, h phngtrnh trong gii ton v v lch s pht trin ca phng trnh.
Chng ti xin ng li cm n ti nhng thnh vin ca Din n chung tay xy dngchuyn . c bit xin chn thnh cm n thy Chu Ngc Hng, thy Nguyn Trng Sn,anh Hong Minh Qun, anh L Phc L, anh Phan c Minh v h tr v ng gp nhng kin qu gi cho chuyn , bn Nguyn Trng Thnh v gip ban bin tp kim tra ccbi vit c mt tuyn tp hon chnh.
Nim hi vng duy nht ca nhng ngi lm chuyn l bn c s tm thy nhiu iub ch v tnh yu ton hc thng qua quyn sch ny. Chng ti xin n nhn v hoan nghnhmi kin xy dng ca bn c chuyn c hon thin hn. Mi gp xin vui lngchuyn n [email protected]
Thnh ph H Ch Minh, ngy 11 thng 7 nm 2012
Thay mt nhm bin sonNguyn Anh Huy
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Cc thnh vin tham gia chuyn
hon thnh c cc ni dung trn, chnh l nh s c gng n lc ca cc thnh vin cadin n tham gia xy dng chuyn :
Ch bin: Nguyn Anh Huy (10CT THPT chuyn L Hng Phong - TP HCM)
Ph trch chuyn : Nguyn Anh Huy (10CT THPT chuyn L Hng Phong - TP HCM),
Nguyn An Vnh Phc (TN Ph thng Nng khiu- TP HCM)
i cng v phng trnh hu t: Hunh Phc Trng (THPT Nguyn Thng Hin TP HCM), Phm Tin Kha (10CT THPT chuyn L Hng Phong - TP HCM)
Phng trnh, h phng trnh c tham s: thy Nguyn Trng Sn (THPT Yn M A Ninh Bnh), V Trng Hi (12A6 THPT Thi Phin - Hi Phng), nh V Bo Chu(THPT chuyn L Qu n - Vng Tu), Hong B Minh ( 12A6 THPT chuyn Trni Ngha - TP HCM), Nguyn Hong Nam (THPT Phc Thin - ng Nai), Ong Th
Phng (11 Ton THPT chuyn Lng Th Vinh - ng Nai) Phng php t n ph: thy Mai Ngc Thi (THPT Hng Vng - Bnh Phc), thy
Nguyn Anh Tun (THPT L Qung Ch -H Tnh), Trn Tr Quc (11TL8 THPTNguyn Hu - Ph Yn), H c Khnh (10CT THPT chuyn Qung Bnh), on ThHo (10A7 THPT Long Khnh - ng Nai)
Phng php dng lng lin hp: Ninh Vn T (THPT chuyn Trn i Ngha -TPHCM) , inh V Bo Chu (THPT - chuyn L Qu n, Vng Tu), on ThHa (THPT Long Khnh - ng Nai)
Phng php dng bt ng thc: Nguyn An Vnh Phc (TN Ph thng Nng khiu-TP HCM), Phan Minh Nht, L Hong c (10CT THPT chuyn L Hng Phong - TPHCM), ng Hong Phi Long (10A10 THPT Kim Lin H Ni), Nguyn Vn Bnh(11A5 THPT Trn Quc Tun - Qung Ngi),
Phng php dng n iu: Nguyn Anh Huy (10CT THPT chuyn L Hng Phong- TP HCM), Hong Kim Qun (THPT Hng Thi H Ni), ng Hong Phi Long(10A10 THPT Kim Lin H Ni)
Phng trnh m logarit: V Anh Khoa, Nguyn Thanh Hoi (i hc KHTN- TPHCM), Nguyn Ngc Duy (11 Ton THPT chuyn Lng Th Vinh - ng Nai)
Cc loi h c bn: Nguyn Anh Huy (10CT THPT chuyn L Hng Phong - TP HCM)
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H phng trnh hon v: thy Nguyn Trng Sn (THPT Yn M A Ninh Bnh),Nguyn Anh Huy (10CT THPT chuyn L Hng Phong TP HCM), Nguyn nh Hong(10A10 THPT Kim Lin - H Ni)
Phng php bin i ng thc: Nguyn nh Hong (10A10 THPT Kim Lin - HNi), Trn Vn Lm (THPT L Hng Phong - Thi Nguyn), Nguyn c Hunh (11
Ton THPT Nguyn Th Minh Khai - TP HCM) Phng php h s bt nh: L Phc L (i hc FPT TP HCM), Nguyn Anh Huy,
Phan Minh Nht (10CT THPT chuyn L Hng Phong TP HCM)
Phng php t n ph tng - hiu: Nguyn Anh Huy (10CT THPT chuyn L HngPhong TP HCM)
Tng hp cc bi h phng trnh: Nguyn Anh Huy (10CT THPT chuyn L Hng PhongTP HCM), Nguyn Thnh Thi (THPT chuyn Nguyn Quang Diu ng Thp), Trn
Minh c (T1K21 THPT chuyn H Tnh H Tnh), V Hu Thng (11 Ton THPTNguyn Th Minh Khai TP HCM)
Sng to phng trnh: thy Nguyn Ti Chung (THPT chuyn Hng Vng Gia Lai),thy Nguyn Tt Thu (THPT L Hng Phong - ng Nai), Nguyn L Thu Linh (10CTTHPT chuyn L Hng Phong TP HCM)
Gii ton bng cch lp phng trnh: Nguyn An Vnh Phc (TN Ph thng Nng khiu-TP HCM)
Lch s pht trin ca phng trnh: Nguyn An Vnh Phc (TN Ph thng Nng khiu-TP HCM), Nguyn Hong Nam (THPT Phc Thin - ng Nai)
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C I: I CNG V PHNG TRNHHU T
PHNG TRNH BC BA
Mt s phng php gii phng trnh bc ba
Phng php phn tch nhn t:Nu phng trnh bc ba ax3 + bx2 + cx + d = 0 c nghim x = r th c nhn t(x
r) do
c th phn tch
ax3 + bx2 + cx + d = (x r)[ax2 + (b + ar)x + c + br + ar2]
T ta a v gii mt phng trnh bc hai, c nghim l
b ra b2 4ac 2abr 3a2r22a
Phng php Cardano:Xt phng trnh bc ba x3 + ax2 + bx + c = 0 (1).Bng cch t x = y a
3, phng trnh (1) lun bin i c v dng chnh tc:
y3 +py + q = 0(2)
Trong : p = b a2
3, q = c +
2a3 9ab27
Ta ch xt p, q= 0 v p = 0 hay q = 0 th a v trng hp n gin.t y = u + v thay vo (2), ta c:
(u + v)3 +p(u + v) + q = 0
u3 + v3 + (3uv + p)(u + v) + q = 0 (3)
Chn u, v sao cho 3uv +p = 0 (4).Nh vy, tm u v v, t (3) v (4) ta c h phng trnh:u
3 + v3 = qu3v3 = p
3
27
Theo nh l Viete, u3 v v3 l hai nghim ca phng trnh:
X
2
+ qX p3
27 = 0(5)
t =q2
4+
p3
27
10
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11
Khi > 0, (5) c nghim:u3 = q
2+
, v3 = q
2
Nh vy, phng trnh (2) s c nghim thc duy nht:
y = 3
q2
+
+ 3
q2
Khi = 0, (5) c nghim kp: u = v = 3q2
Khi , phng trnh (2) c hai nghim thc, trong mt nghim kp.
y1 = 23
q
2, y2 = y3 =
3
q
2
Khi < 0, (5) c nghim phc.Gi u30 l mt nghim phc ca (5), v30 l gi tr tng ng sao cho u0v0 =
p
3.
Khi , phng trnh (2) c ba nghim phn bit.y1 = u0 + v0
y2 = 12
(u0 + v0) + i
3
2(u0 v0)
y3 = 12
(u0 + v0) i
3
2(u0 v0)
Phng php lng gic ho - hm hyperbolic:Mt phng trnh bc ba, nu c 3 nghim thc, khi biu din di dng cn thc s lin quann s phc. V vy ta thng dng phng php lng gic ho tm mt cch biu din
khc n gin hn, da trn hai hm s cos v arccosC th, t phng trnh t3 + pt + q = 0 () ta t t = u cos v tm u c th a () vdng
4cos3 3cos cos3 = 0
Mun vy, ta chn u = 2p
3v chia 2 v ca () cho u
3
4 c
4cos3 3cos 3q2p
.
3p
= 0 cos3 = 3q2p
.
3p
Vy 3 nghim thc l
ti = 2
p3
. cos
1
3arccos
3q
2p.
3p
2i
3
vi i = 0, 1, 2.
Lu rng nu phng trnh c 3 nghim thc th p < 0 (iu ngc li khng ng) nn cngthc trn khng c s phc.Khi phng trnh ch c 1 nghim thc v p = 0 ta cng c th biu din nghim bng cng
thc hm arcosh v arsinh:
t =2|q|
q.
p3
cosh
1
3.arcosh
3|q|2p
.
3p
nu p < 0 v 4p3 + 27q2 > 0.
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t = 2
p
3. sinh
1
3.arsinh
3q
2p.
3
p
nu p > 0
Mi phng php trn u c th gii quyt phng trnh bc ba tng qut. Nhng mc chca chng ta trong mi bi ton lun l tm li gii ngn nht, p nht. Hy cng xem quamt s v d:
Bi tp v dBi 1: Gii phng trnh x3 + x2 + x = 1
3
Gii
Phng trnh khng c nghim hu t nn khng th phn tch nhn t. Trc khi ngh ticng thc Cardano, ta th quy ng phng trnh:
3x3 + 3x2 + 3x + 1 = 0
i lng 3x2+3x+1 gi ta n mt hng ng thc rt quen thuc x3+3x2+3x+1 = (x+1)3.Do phng trnh tng ng:
(x + 1)3 = 2x3hay
x + 1 = 32xT suy ra nghim duy nht x =
11 + 3
2
.
Nhn xt: V d trn l mt phng trnh bc ba c nghim v t, v c gii nh kho lobin i ng thc. Nhng nhng bi n gin nh th ny khng c nhiu. Sau y ta s i
su vo cng thc Cardano:
Bi 2: Gii phung trnh x3 3x2 + 4x + 11 = 0
Gii
t x = y + 1 . Th vo phng trnh u bi, ta c phng trnh:y3 + 1.y + 13 = 0
Tnh = 132
+
4
27 .1
3
=
4567
27
0p dng cng thc Cardano suy ra:
y =3
13 +4567272
+3
13 4567272
Suy ra x =3
13 +4567272
+3
13 4567272
+ 1.
Nhn xt: V d trn l mt ng dng c bn ca cng thc Cardano. Tuy nhin cng
thc ny khng h d nh v ch c dng trong cc k thi Hc sinh gii. V th, c l chngta s c gng tm mt con ng hp thc ha cc li gii trn. l phng php lnggic ho. u tin xt phng trnh dng x3 + px + q = 0 vi p < 0 v c 1 nghim thc:
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13
Bi 3: Gii phng trnh x3 + 3x2 + 2x 1 = 0Gii
u tin t x = y 1 ta a v phng trnh y3 y 1 = 0 (1). n y ta dng lng gicnh sau:
Nu |y| 0 khng kh chng minh phng trnh c nghim duy nht:
Bi 5: Gii phng trnh x3 + 6x + 4 = 0
Gii
tng: Ta s dng php t x = k
t 1t
a v phng trnh trng phng.
php t ny khng cn iu kin ca x, v n tng ng k(t2 1) xt = 0. Phng trnh
trn lun c nghim theo t.Nh vy t phng trnh u ta c
k3
t3 1t3
3k3
t 1
t
+ 6k
t 1
t
+ 4 = 0
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Cn chn k tho 3k3 = 6k k = 2Vy ta c li gii bi ton nh sau: Li gii:
t x =
2
t 1
t
ta c phng trnh
22t3 1t3+ 4 = 0 t6 1 + 2t3 = 0 t1,2 = 31 3
2
Lu rng t1.t2 = 1 theo nh l Viete nn ta ch nhn c mt gi tr ca x l
x = t1 + t2 =
2
3
1 + 3
2+ 3
1 3
2
. 2
Bi 6: Gii phng trnh 4x3 3x = m vi |m| > 1
GiiNhn xt rng khi |x| 1 th |V T| 1 < |m| (sai) nn |x| 1. V vy ta c th tx =
1
2
t +
1
t
.
Ta c phng trnh tng ng:1
2
t3 +
1
t3
= m
T :t =
3
m
m2 1 x = 12
3
m +
m2 1 + 3m
m2 1 .Ta chng minh y l nghim duy nht.
Gi s phng trnh c nghim x0 th x0 [1, 1] v |x0| > 1. Khi :4x3 3x = 4x30 3x0
hay(x x0)(4x2 + 4xx0 + 4x20 3) = 0
Xt phng trnh:4x2 + 4xx0 + 4x
2
0 3 = 0
c = 12 12x20 < 0 nn phng trnh bc hai ny v nghim.
Vy phng trnh u bi c nghim duy nht l
x =1
2
3
m +
m2 1 + 3
m m2 1
.
Bi tp t luynBi 1: Gii cc phng trnh sau:a) x3 + 2x2 + 3x + 1 = 0b) 2x3 + 5x2 + 4x + 2 = 0c) x3 5x2 + 4x + 1 = 0
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d) 8x3 + 24x2 + 6x 10 36 = 0Bi 2: Gii v bin lun phng trnh:
4x3 + 3x = m vi m RBi 3: Gii v bin lun phng trnh:
x3 + ax2 + bx + c = 0
PHNG TRNH BC BN
[1] Phng trnh dng ax4 + bx3 + cx2 + bkx + ak2 = 0 (1)
Ta c
(1) a(x4 + 2x2.k + k2) + bx(x2 + k) + (c 2ak)x2 = 0 a(x2 + k)2 + bx(x2 + k) + (c 2ak)x2 = 0
n y c hai hng gii quyt:Cch 1: a phng trnh v dng A2 = B2:Thm bt, bin i v tri thnh dng hng ng thc dng bnh phng ca mt tng, chuyncc hng t cha x2 sang bn phi.Cch 2: t y = x2 + k y kPhng trnh (1) tr thnh ay2 + bxy + (c 2ak)x2 = 0
Tnh x theo y hoc y theo x a v phng trnh bc hai theo n x.
V d: Gii phng trnh: x4 8x3 + 21x2 24x + 9 = 0 (1.1)
Cch 1:(1.1) (x4 + 9 + 6x2) 8(x2 + 3) + 16x2 = 16x2 21x2 + 6x2 (x2 4x + 3)2 = x2
x2 4x + 3 = xx2 4x + 3 = x
x2 5x + 3 = 0x2 3x + 3 = 0
x =5 13
2
x = 5 + 132
Cch 2:
(1.1) (x4 + 6x2 + 9) 8x(x2 + 3) + 15x2 = 0 (x2 + 3)2 8x(x2 + 3) + 15x2 = 0t y = x2 + 3. (1.1) tr thnh: y2 8xy + 15x2 = 0 (y 3x)(y 5x) = 0
y = 3x
y = 5x
Vi y = 3x: Ta c x2 + 3 = 3x: Phng trnh v nghim
Vi y = 5x: Ta c x2 + 3 = 5x x2 5x + 3 = 0
x =5 13
2
x =
5 +
13
2
Vy phng trnh (1.1) c tp nghim: S =
5 +
13
2;
5 132
Nhn xt: Mi phng php gii c li th ring. Vi cch gii 1, ta s tnh c trc tip m
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khng phi thng qua n ph. Vi cch gii 2, ta s c nhng tnh ton n gin hn v t bnhm ln.
Bi tp t luynGii cc phng trnh sau:1) x4 13x3 + 46x2 39x + 9 = 02) 2x4 + 3x3 27x2 + 6x + 8 = 03) x4 3x3 6x2 + 3x + 1 = 04) 6x4 + 7x3 36x2 7x + 6 = 05) x4 3x3 9x2 27x + 81 = 0
[2] Phng trnh dng (x + a)(x + b)(x + c)(x + d) = ex2 (2) vi ad = bc = m
Cch 1: a v dng A2 = B2(2) (x +px + m)(x2 + nx + m) = ex2(ad = bc = m, p = a + d, n = b + c)
x2 +p + n
2x + m n p
2x
x2 +
p + n
2x + m +
n p2
x
= ex2
x2 +p + n
2x + m
2=
n p2
2+ e
x2
Cch 2: Xt xem x = 0 c phi l nghim ca phng trnh khng.Trng hp x = 0:
(2) x + mx
+px + mx
+ n = et u = x +
m
x. iu kin: |u| 2|m|
(2) tr thnh (u + p)(u + n) = e. n y gii phng trnh bc hai theo u tm x.
V d: Gii phng trnh: (x + 4)(x + 6)(x 2)(x 12) = 25x2 (2.1)
Cch 1:(2.1) (x2 + 10x + 24)(x2 14x + 24) = 25x2
(x2 2x + 24 + 12x)(x2 2x + 24 12x) = 25x2
(x2 2x + 24)2 = 169x2
x2 2x + 24 = 13xx2 2x + 24 = 13x
x2 15x + 24 = 0x2 + 11x + 24 = 0
x = 8x = 3
x =
15
129
2Cch 2:
(2.1) (x2 + 10x + 24)(x2 14x + 24) = 25x2Nhn thy x = 0 khng phi l nghim ca phng trnh.
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x = 0 : (2.1)
x +24
x+ 10
x +
24
x 14
= 25
t y = x +24
x |y| 46. (2.1) tr thnh:
(y + 10)(y 14) = 25 (y + 11)(y 15) = 0
y = 11y = 15
Viy
= 11
: Ta c phng trnh:x +
24
x= 11 x2 + 11x + 24 = 0
x = 3x = 8
Vi y = 15: Ta c phng trnh:
x +24
x= 15 x2 15x + 24 = 0 x = 15
129
2
Phng trnh (2.1) c tp nghim S =
3; 8; 15
129
2;
15 +
129
2
Nhn xt: Trong cch gii 2, c th ta khng cn xt x = 0 ri chia m c th t n phy = x2 + m thu c phng trnh bc hai n x, tham s y hoc ngc li.
Bi tp t luyn
Gii cc phng trnh sau:1) 4(x + 5)(x + 6)(x + 10)(x + 12) = 3x2
2) (x + 1)(x + 2)(x + 3)(x + 6) = 168x2
3) (x + 3)(x + 2)(x + 4)(x + 6) = 14x2
4) (x + 6)(x + 8)(x + 9)(x + 12) = 2x2
5) 18(x + 1)(x + 2)(x + 5)(2x + 5) =19
4x2
[3] Phng trnh dng (x + a)(x + b)(x + c)(x + d) = m (3) vi a + b = c + d = p
Ta c (3) (x2 + px + ab)(x2 +px + cd) = mCch 1:
(3)
x2 + px +
ab + cd
2
+ab cd
2 x2 +px +
ab + cd
2
ab cd
2 = m
x2 + px +ab + cd
2
2= m +
ab cd
2
2Bi ton quy v gii hai phng trnh bc hai theo x.Cch 2:
t y = x2 +px iu kin: y p2
4. (3) tr thnh:(y + ab)(y + cd) = m
Gii phng trnh bc 2 n y tm x.
V d: Gii phng trnh: x(x + 1)(x + 2)(x + 3) = 8 (3.1)
Cch 1:
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Ta c(3.1) (x2 + 3x)(x2 + 3x + 2) = 8
(x2 + 3x + 1 1)(x2 + 3x + 1 + 1) = 8
(x2 + 3x + 1)2 = 9
x2 + 3x + 1 = 3
x2 + 3x + 1 = 3
x2 + 3x 2 = 0x2 + 3x + 4 = 0
x = 3 172
Cch 2:
(3.1) (x2 + 3x)(x2 + 3x + 2) = 8t y = x2 + 3x y 9
4(3.1) tr thnh:
y(y + 2) = 8
y2 + 2y
8 = 0
y = 2
y = 4(loi) y = 2
Vi y = 2: Ta c phng trnh:
x2 + 3x 2 = 0 x = 3
17
2
Phng trnh (3.1) c tp nghim: S =
3 + 17
2;3 17
2
Bi tp t luynGii cc phng trnh sau:1. (x + 2)(x + 3)(x 7)(x 8) = 1442. (x + 5)(x + 6)(x + 8)(x + 9) = 40
3.
x +1
4
x +
3
5
x +
1
20
x +
4
5
=
39879
400004. (6x + 5)2(3x + 2)(x + 1) = 355. (4x + 3)2(x + 1)(2x + 1) = 810
Nhn xt: Nh dng (2), ngoi cch t n ph trn, ta c th t mt trong cc dngn ph sau: t y = x2 +px + ab t y = x2 +px + cd t y =
x +
p
2
2 t y = x2 +px +
ab + cd
2
[4] Phng trnh dng (x + a)4 + (x + b)4 = c (c > 0) (4)
t x = y a + b2
. (4) tr thnh:y +
a b2
4+
y a b
2
4= c
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S dng khai trin nh thc bc 4, ta thu c phng trnh:
2y4 + 3(a b)2y2 + 2
a b2
4= c
Gii phng trnh trng phng n y tm x.
V d: Gii phng trnh: (x + 2)4 + (x + 4)4 = 82 (4.1)
t y = x + 3. Phng trnh (4.1) tr thnh:(y + 1)4 + (y 1)4 = 82 (y4 + 4y3 + 6y2 + 4y + 1) + (y4 4y3 + 6y2 4y + 1) = 82 2y
4
+ 12y2
80 = 0 (y2
4)(y2
+ 10) = 0 y2 = 4 y = 2Vi y = 2, ta c x = 1Vi y = 2, ta c x = 5Vy phng trnh c tp nghim: S = {1; 5}
Bi tp t luynGii cc phng trnh sau:
1. (x + 2)4 + (x + 8)4 = 2722. (x +
2)4 + (x + 1)4 = 33 + 12
2
3. (x + 10)4 + (x 4)4 = 285624. (x + 1)4 + (x 3)4 = 90
[5] Phng trnh dng x4 = ax2 + bx + c (5)
a (5) v dng A2
= B2
:
(5) (x2 + m)2 = (2m + a)x2 + bx + c + m2
Trong , m l mt s cn tm.Tm m f(x) = (2m + a)x2 + bx + c + m2 c = 0. Khi , f(x) c dng bnh phng camt biu thc.
Nu 2m + a < 0 : (5) (x2 + m)2 + g2(x) = 0 (vi f(x) = g2(x))
x2 + m = 0
g(x) = 0
Nu2m + a > 0 : (5) (x2 + m)2 = g2(x) (vi f(x) = g2(x)) x2 + m = g(x)
x2 + m = g(x)
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V d: Gii phng trnh: x4 + x2 6x + 1 = 0 (5.1)
Ta c:(5.1) x4 + 4x2 + 4 = 3x2 + 6x + 3 (x2 + 2)2 = 3(x + 1)2
x2 + 2 =
3(x + 1)
x2 + 2 =
3(x + 1)
x2 3x + 2 3 = 0x2 +
3 + 2 +
3 = 0
x =
3
4
3 52
x =
3 +
4
3 52
Phng trnh (5.1) c tp nghim: S =
3
4
3 52
;
3 +
4
3 52
Bi tp t luynGii cc phng trnh sau:1. x4 19x2 10x + 8 = 02. x4 = 4x + 13. x4 = 8x + 74. 2x4 + 3x2 10x + 3 = 05. (x2 16)2 = 16x + 1
6. 3x4
2x2
16x 5 = 0Nhn xt: Phng trnh dng x4 = ax + b c gii theo cch tng t.Phng trnh = 0 l phng trnh bc ba vi cch gii c trnh by trc. Phngtrnh ny c th cho 3 nghim m, cn la chn m sao cho vic tnh ton l thun li nht. Tuynhin, d dng nghim m no th cng cho cng mt kt qu.
[6] Phng trnh dng af2(x) + bf(x)g(x) + cg2(x) = 0 (6)
Cch 1: Xt g(x) = 0, gii tm nghim v th li vo (6).
Trng hp g(x) = 0: a
f(x)
g(x)
2+ b.
f(x)
g(x)+ c = 0
t y =f(x)
g(x), gii phng trnh bc hai ay2 + by + c = 0 ri tm x.
Cch 2: t u = f(x), v = g(x), phng trnh tr thnh
au2 + buv + cv2 = 0 (6)Xem (6) l phng trnh bc hai theo n u, tham s v. T tnh u theo v.
V d: Gii phng trnh: 20(x 2)2 5(x + 1)2 + 48(x 2)(x + 1) = 0 (6.1)
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t u = x 2, v = x + 1. Phng trnh (6.1) tr thnh:
20u2 + 48uv 5v2 = 0 (10u v)(2u + 5v) = 0
10u = v
2u = 5v
Vi 10u = v, ta c: 10(x
2) = x + 1
x =
7
3 Vi 2u = 5v, ta c: 2(x 2) = 5(x + 1) x = 1
7
Vy phng trnh (6.1) c tp nghim: S =
7
3; 1
7
Nhn xt: Nu chn y =
f(x)
g(x)Vi f(x) v g(x) l hai hm s bt k (g(x) = 0), ta s to
c mt phng trnh. Khng ch l phng trnh hu t, m cn l phng trnh v t.
Bi tp t luynGii cc phng trnh sau:1. (x 5)4 12(x 2)4 + 4(x2 7x + 10)2 = 02. (x 2)4 + 3(x + 3)4 4(x2 + x 6)2 = 03. 4(x3 1) + 2(x2 + x + 1)2 4(x 1)2 = 04. 2(x2 x + 1)2 + 5(x + 1)2 + 14(x3 + 1) = 05. (x 10)4 15(x + 5)4 + 4(x2 5x 50)2 = 0
[7] Phng trnh bc bn tng qut ax4 + bx3 + cx2 + dx + e = 0 (7)
Phn tch cc hng t bc 4, 3, 2 thnh bnh phng ng, cc hng t cn li chuyn sang vphi:
(7) 4a2x4 + 4bax3 + 4cax2 + 4dax + 4ae = 0 (2ax2 + bx)2 = (b2 4ac)x2 4adx 4ae
Thm vo hai v mt biu thc 2(2ax2 + bx)y + y2 (y l hng s) v tri thnh bnh phngng, cn v phi l tam thc bc hai theo x:
f(x) = (b2 4ac 4ay)x2 + 2(by 2ad)x 4ae + y2
Tnh y sao cho v phi l mt bnh phng ng. Nh vy, ca v phi bng 0. Nh vy taphi gii phng trnh = 0. T ta c dng phng trnh A2 = B2 quen thuc.
V d: Gii phng trnh x4 16x3 + 66x2 16x 55 = 0 (7.1)
(7.1) x4
16x3
+ 64x2
= 2x2
+ 16x + 55 (x2 8x)2 + 2y(x2 8x) + y2 = (2y 2)x2 + (16 16y)x + 55 + y2
Gii phng trnh = 0 (8 8y)2 (55 + y2)(2y 2) = 0 tm c y = 1, y = 3, y = 29.Trong cc gi tr ny, ta thy gi tr y = 3 l thun li nht cho vic tnh ton.
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Nh vy, chn y = 3, ta c phng trnh:
(x2 8x + 3)2 = 4(x 4)2
x2 8x + 3 = 2(x 4)x2 8x + 3 = 2(x 4)
x2 10x + 11 = 0x2 6x 5 = 0
x = 3 14x = 5 14
Phng trnh (7.1) c tp nghim S = 3 + 14;3 14; 5 + 14;5 14 Nhn xt: V d trn cho ta thy phng trnh = 0 c nhiu nghim. C th chn y = 1nhng t ta c phng trnh (x2 8x + 1)2 = 56 th khng thun li lm cho vic tnh ton,tuy nhin, kt qu vn nh nhau.Mt cch gii khc l t phng trnh x4 + ax3 + bx2 + cx + d = 0 t x = t a
4, ta s thu c
phng trnh khuyt bc ba theo t, ngha l bi ton quy v gii phng trnh t4 = at2 + bt + c.
Bi tp t luyn
1. x4 14x3 + 54x2 38x 11 = 02. x4 16x3 + 57x2 52x 35 = 03. x4 6x3 + 9x2 + 2x 7 = 04. x4 10x3 + 29x2 20x 8 = 05. 2x4 32x3 + 127x2 + 38x 243 = 0
PHNG TRNH DNG PHN THC[1] Phng trnh cha n mu c bn
t iu kin xc nh cho biu thc mu. Quy ng ri gii phng trnh.
V d: Gii phng trnh: 12 x +
x
2x 1 = 2 (1.1)
iu kin: x= 2; x
=
1
2.
(1.1) 2x 1 + x(2 x)(2 x)(2x 1) = 2 2x 1 + 2x x
2 = 2(4x 2 2x2 + x)
3x2 6x + 3 = 0 x = 1(tha iu kin)Vy phng trnh (1.1) c tp nghim S = {1}
[2] Phng trnh dng x2 + a2x2
(x + a)2= b (2)
Ta c:
(2) x ax(x + a)
2 + 2x. axx + a
= b
x2
x + a
2+ 2a.
x2
x + a+ a2 = b + a2
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t y =x2
x + a. Gii phng trnh bc hai theo y tm x.
V d: Gii phng trnh: x2 + 9x2
(x + 3)2= 7 (2.1)
iu kin: x=
3.
(2.1)
x 3xx + 3
2+ 6.
x2
x + 3= 7
x2
x + 3
2+ 6.
x2
x + 3= 7
t y =x2
x + 3. Ta c phng trnh
y2 + 6y 7 = 0 y = 1
y = 7
Nu y = 1: Ta c phng trnh x2 = x + 3 x = 1
13
2 Nu y = 7: Ta c phng trnh x2 + 7x + 21 = 0 (v nghim)Vy phng trnh (2.1) c tp nghim: S =
1 +
13
2;
1 132
Nhn xt: Da vo cch gii trn, ta c th khng cn phi t n ph m thm bt hngs to dng phng trnh quen thuc A2 = B2
Bi tp t luyn
Gii cc phng trnh sau:
1. x2 +4x2
(x + 2)2= 12
2. x2 +25x2
(x + 5)
2= 11
3. x2 +9x2
(x 3)2 = 14
4.25
x2 49
(x 7)2 = 1
5.9
4(x + 4)2+ 1 =
8
(2x + 5)2
[3] Phng trnh dng x2 + nx + a
x2 + mx + a+
x2 + qx + a
x2 +px + a= b (3)
iu kin:x2 + mx + a = 0x2 +px + a = 0
Xt xem x = 0 c phi l nghim phng trnh khng.
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Trng hp x = 0:
(2) x +
a
x+ n
x +a
x+ m
+x +
a
x+ q
x +a
x+p
= b
t y = x +a
x. iu kin: |y| 2
|a| ta c phng trnh
y + ny + m
+ y + qy +p
= b
Gii phng trnh n y sau tm x.
V d: Gii phng trnh: x2 3x + 5
x2 4x + 5 x2 5x + 5x2 6x + 5 =
1
4(3.1)
iu kin: x = 1, x = 5.x = 0 khng phi l nghim ca phng trnh.
Xt x = 0 :
(3.1) x +
5
x 3
x +5
x 4
x +
5
x 5
x +5
x 6
= 14
t y = x +5
x |y| 25, y = 6. Phng trnh (3.1) tr thnh:
y 3y 4
y 5y 6
=
1
4 2
y2 10y + 24=
1
4 y2
10y + 16 = 0
y = 2 (loi)y = 8T ta c phng trnh
x +5
x= 8 x2 8x + 5 = 0 x = 4
11
Vy phng trnh (3.1) c tp nghim: S =
4 +
11;4 11 Nhn xt: Cc dng phng trnh sau c gii mt cch tng t: Dng 1: mx
ax2 + bx + d+
nx
ax2 + cx + d= p
Dng 2: ax2 + mx + cax2 + nx + c
+ pxax2 + qx + c
= b
Bi tp t luyn
Gii cc phng trnh sau:1)
4x
4x2 8x + 7 +3x
4x2 10x + 7 = 1
2)2x
2x2 5x + 3 +13x
2x2 + x + 3= 6
3)3x
x2 3x + 1+
7x
x2 + x + 1=
4
4)x2 10x + 15x2 6x + 15 =
4x
x2 12x + 55)
x2 + 5x + 3
x2 7x + 3 +x2 + 4x + 3
x2 + 5x + 3= 7
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Tng ktQua cc dng phng trnh trn, ta thy phng trnh hu t thng c gii bng mt trongcc phng php:[1.] a v phng trnh tch[2.] t n ph hon ton[3.] t n ph a v h phng trnh[4.] a v ly tha ng bc (thng l dng A2 = B2)[5.] Chia t v mu cho cng mt s[6.] Thm bt to thnh bnh phng ngTuy nhin, c mt s dng phng trnh c nhng phng php gii c trng. Nhng phngtrnh ny s c trnh by c th hn nhng phn khc.
Bi tp tng hp phn phng trnh hu t
Phn 1:1) x3 3x2 + 18x 36 = 02) 8x2 6x = 1
23) x3 4x2 4x + 8 = 04) x3 21x2 + 35x 7 = 05) x3 6x2 + 8 = 0Phn 2:1) 6x5 11x4 11x + 6 = 02) (x2
6x)2
2(x
3)2 = 81
3) x4 + (x 1)(3x2 + 2x 2) = 04) x4 + (x + 1)(5x2 6x 6) = 05) x5 + x2 + 2x + 2 = 06) (x2 16)2 = 16x + 17) (x + 2)2 + (x + 3)3 + (x + 4)4 = 2
8) x3 +1
x3= 13
x +
1
x
9)
x 1
x 2
+
x 1x
2
2
=40
9
10) x(3 x)x 1
x +
3 xx 1
= 15
11)1
x+
1
x + 2+
1
x + 5+
1
x + 7=
1
x + 1+
1
x + 3+
1
x + 4+
1
x + 6
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XY DNG PHNG TRNH HU TBn cnh vic xy dng phng trnh t h phng trnh, vic xy dng phng trnh t nhngng thc i s c iu kin l mt trong nhng phng php gip ta to ra nhng dngphng trnh hay v l. Di y l mt s ng thc n gin.
4.1 T ng thc (a + b + c)3 = a3 + b3 + c3 + 3(a + b)(b + c)(c + a) (1) :
V d: Gii phng trnh: (x 2)3 + (2x 4)3 + (7 3x)3 = 1 (1.1)
Nhn xt: Nu t a = x 2, b = 2x 4, c = 7 3x. Khi ta c phng trnh: a3 + b3 + c3 =(a + b + c)3. T ng thc (1), d dng suy ra (a + b)(b + c)(c + a) = 0. T , ta c li gii:(1) (x 2)3 + (2x 4)3 + (7 3x)3 = [(x 2) + (2x 4) + (7 3x)]3
(3x 6)(3 x)(5 2x) = 0 x = 2x = 3
x =5
2
Vy phng trnh (1.1) c tp nghim: S =
2;3;5
2
Vi bi ton trn, cch t nhin nht c l l khai trin ri thu v phng trnh bc ba. Tuynhin, vic khai trin c th khng cn hiu qu vi bi ton sau:V d: Gii phng trnh: (x2 4x + 1)3 + (8x x2 + 4)3 + (x 5)3 = 125x3 (1.2)
4.2 T mnh 1a +1b
+ 1c
= 1a + b + c
(a + b)(b + c)(c + a) = 0 (2) :
V d: Gii phng trnh: 1x 8 +
1
2x + 7+
1
5x + 8=
1
8x + 7(2.1)
iu kin: x = 8, x = 72
, x = 85
, x = 78
T bi ton (2), ta c:
(2.1) (x 8 + 2x + 7)(x 8 + 5x + 8)(2x + 7 + 5x + 8) = 0 x =
1
3x = 0
x = 157
Phng trnh c tp nghim: S =
1
3; 0; 15
7
4.3 T ng thc a3 + b3 + c3 3abc = (a + b + c)(a2 + b2 + c2 ab bc ca) (3) :
V d: Gii phng trnh 54x3 9x + 2 = 0 (3.1)Ta tm cch vit v tri ca phng trnh di dng x3 + a3 + b3 3abx. Nh vy th a, b l
nghim ca h phng trnh a3 + b3 =
2
54
a3.b3 =1
183
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a3, b3 l nghim ca phng trnh
t2
2
54t +
1
182= 0 t = 1
54
2 a = b = 1
3
2
Khi phng trnh cho tng ng vi
(x + a + b) x2 + a2 + b2 a x bx ab = 0
x +2
3
2
x2 2
3
2x +
1
18
= 0
x =2
3
2
x =1
3
2
Vy (3.1) c tp nghim S =
23
2;
1
3
2
4.4 T bi ton Nu xyz = 1 v x + y + z = 1x + 1y + 1z th (x 1)(y 1)(z 1) = 0 (4) :
V d: Gii phng trnh: 110x2 11x + 3 =
1
2x 1 +1
5x 3 + 10x2 18x + 7 (4.1)
iu kin: x = 35
, x = 12
Nhn xt: Nu t a = 2x 1, b = 5x 3, c = 110x2 11x + 3
Th ta c: abc = 1 v a + b + c =1
a+
1
b+
1
cT (3.1) (2x 1 1)(5x 3 1)
1
10x2 11x + 3 1
= 0
x = 1
x =4
510x2 11x + 2 = 0
x = 1
x =4
5
x =11 41
20
Phng trnh (3.1) c tp nghim: S =
1;
4
5;
11 4120
Qua cc v d trn, ta c th hnh dung c bn vic s dng ng thc xy dng phngtrnh. Hi vng da vo vn hiu bit v kh nng sng to ca mnh, bn c c th to ranhng phng trnh p mt v c o hn na. Sau y l mt s bi tp t luyn t ccng thc khc.
Bi tp t luyn
1. (x 2)6 + (x2 5x + 4)3 = (2x2 + 3)3 + (5 9x)32. (x2
2x + 3)5 + (8x
x2 + 7)5 = (9x + 5)5 + (5
3x)5
3. (x35x+4)2(127x5x2)+(x3+2x2 +7)(7x2 +12x9) = (x3+7x2+7x5)2(2x2+5x+3)4. (x 5)4(4x x2) + (2x + 7)4(x2 3x + 12) = (x2 2x + 7)4(12 + x)5.
(x2 5x + 3)3(4 9x)(x2 12x + 7) +
(x2 + 4x 1)3(9x 4)(x2 3x + 3) +
(7x 4)3(12x 7 x2)(x2 + 3x 3) = 7
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6.1
x2 + 4x + 3+
1
x2 + 8x + 15+
1
x2 + 12x + 35+
1
x2 + 16x + 63= 5
7. (x2 8x + 5)7 + (7x 8)7 = (x2 x 3)7
MT S PHNG TRNH BC CAONh ton hc Abel chng minh rng khng c cng thc nghim tng qut cho phngtrnh bc cao (> 4). y cng khng phi l dng ton quen thuc ph thng. V th bivit ny ch cp n mt s phng trnh bc cao c bit, c th gii bng bin i s cp.
Bi 1: Gii phng trnh x5 x4 x3 11x2 + 25x 14 = 0
GiiPhng trnh cho tng ng
(x5 2x4) + (x4 2x3) + (x3 2x2) + (9x2 + 18x) + (7x 14) = 0(x 2)(x4 + x3 + x2 9x + 7) = 0 x = 2 x4 + x3 + x2 9x + 7 = 0 ()
Xt (*) ta c(
)
(x4 + x3 + x2
9x + 6) + 1 = 0
(x4 x3 + 2x3 2x2 + 3x2 3x 6x + 6) + 1 = 0 (x 1)2(x3 + 3x + 6) + 1 = 0 (v nghim)
Vy phng trnh c tp nghim S = {2} 2
Bi 2: Gii phng trnh x6 7x2 + 6 = 0 ()
Gii
R rng ta khng th on nghim ca phng trnh ny v bc cao v h s xu. Mt cch tnhin ta t
6 = a. Lu rng ta hi vng c th a (*) v phng trnh bc hai theo a, do
ta phn tch 7 = a2 + 1. Cng vic cn li l gii phng trnh ny:t
6 = a, khi
() x6 x2(a2 + 1) + a = 0 a2x2 a + x2 x6 = 0 ()
() c: = 1 + 4x2
(x6
x2
) = (2x4
1)2
nn c nghim:a1 = x2
a2 =1 x4
x2
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Vy ()
x2 =
6
1 x4 =
6x2
x = 4
6
x =
5
2
3
2
Vy phng trnh c tp nghim S = 4
6;
5
2 3
2 . 2Bi 3: Gii phng trnh x6 15x2 + 68 = 0 ()
Gii
Do x = 0 khng tho (*) nn x = 0. Vit li (*) di dng
x3 +
68
x3=
15
x x3 + 2
17
x3=
17 2x
t a =
17 > 0 ta c phng trnh
x3 +2a
x3=
a2 2x
x2a2 2a x6 2x2 = 0 ()
Coi (**) l phng trnh n a ta tm c nghim a = x2 (loi do a > 0)a =
2 + x4
x2
Vy ta c
17 =2 + x4
x2 x4
17x2 + 2 = 0
x2 =
17 3
2 x =
17 32
Kt lun: (*) c tp nghim S =
17 32
2
Bi 4: Chng minh phng trnh x5 5x4 + 30x3 50x2 + 55x 21 = 0 c nghim duynht
x = 1 +5
2 5
4 +5
8 5
16
Gii
t f(x) = x5 5x4 + 30x3 50x2 + 55x 21.Ta c f(x) = 5x4 20x3 + 90x2 100x + 55 = 5(x2 2x + 3)2 + 10(2x 1)2 > 0 x, do phng trnh f(x) = 0 c khng qu 1 nghim.Ta s chng minh nghim l x = 1 + 5
2 54 + 58 516:
tx = 1 +
5
2 5
4 +5
8 5
16
5
2x =5
2 +5
4 5
8 +5
16 2 x + 5
2x = 2
5
2 1 x + 1 = 5
2(2 x) (x + 1)5 = 2(2 x)5
Khai trin biu thc trn, sau rt gn, ta c phng trnh:
x5 5x4 + 30x3 50x2 + 55x 21 = 0
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Vy ta c iu phi chng minh. 2
Bi 5: Chng minh phng trnh sau c 7 nghim thc:
g(x) = x9 9x7 + 3x6 + 27x5 18x4 27x3 + 27x2 1 = 0 ()
Gii
t f(x) = x3 3x + 1 th g(x) = f(f(x)). Ta s tm nghim ca g(x) = f(x).f(f(x)): Nghim ca f(x) l: f(x) = 0 3x2 3x = 0 x = 1. tm nghim ca f(f(x)) ta tm nghim ca f(x) = 1 v f(x) = 1: f(x) = 1 x3 3x + 2 = 0 x {2; 1} f(x) = 1 x3 3x = 0 x {0; 3}Nh vy tp nghim ca phng trnh g(x) = 0 l {2; 3; 1;0;1; 3}.
Suy ra g(x) c ti a 7 nghim. Li c:
g(x) khi x g(2) = 3 > 0 g(x) c 1 nghim trong (; 2)g(3) = 1 < 0 g(x) c 1 nghim trong (2; 3)g(1) = 19 > 0 g(x) c 1 nghim trong (3; 1)g(0) = 1 < 0 g(x) c 1 nghim trong (1;0)g(1) = 3 > 0 g(x) c 1 nghim trong (0; 1)g(3) = 1 < 0 g(x) c 1 nghim trong (1; 3)g(x) + khi x + g(x) c 1 nghim trong (3; +)
Nh vy g(x) = f(f(x)) c 7 nghim thc. 2
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C II: PHNG TRNH, H PHNGTRNH C THAM S
PHNG PHP S DNG O HM
L thuyti vi bi ton tm iu kin ca tham s phng trnh f(x) = g(m) c nghim min D tada vo tnh cht: phng trnh c nghim khi v ch khi hai th ca hai hm s y = f(x)v y = g(m) ct nhau. Do bi ton ny ta tin hnh theo cc bc sau:Bc 1: Lp bng bin thin ca hm s y = f(x) .Bc 2: Da vo bng bin thin ta xc nh m ng thng y = g(m) ct th hm sy = f(x).Ch : Nu hm s y = f(x) lin tc trn D v m = Min
Df(x) , M = Max
Df(x) th phng
trnh: k = f(x) c nghim m k MSau y l mt s bi tp v d:
Bi tp v d
Bi 1:Tm m phng trnh
x2 + x + 1 x2 x + 1 = m c nghim.
Gii
Xt hm s f(x) =
x2 + x + 1 x2 x + 1 vi x R cf(x) =
2x + 1
2
x2 + x + 1 2x 1
2
x2 x + 1Ta s tm nghim ca f(x):
f(x) = 0 2x + 12
x2 + x + 1 2x 1
2
x2 x + 1 = 0
(2x + 1)
x2 x + 1 = (2x 1)
x2 + x + 1
x +1
2
2 x 1
2
2+
3
4
=(x 1
2)2
x +1
2
2+
3
4
x +1
2
2=
x 1
2
2 x = 0
Th li ta thy x = 0 khng l nghim ca f(x). Suy ra f(x) khng i du trn R, mf(0) = 1 > 0 f(x) > 0x R . Vy hm s f(x) ng bin trn R.Mt khc: lim
x+f(x) = lim
x+
2xx2 + x + 1 +
x2 x + 1 = 1 v limx f(x) = 1.
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Da vo bng bin thin ta thy phng trnh cho c nghim khi v ch khi m (1;1) 2
Bi 2: Tm m phng trnh 4
x2 + 1 x = m c nghim
Gii
KX: x 0Xt hm s f(x) = 4x2 + 1 x vi x 0 ta c f(x) lin tc trn [0;+)Li c: f(x) =
x
2 4
(x2 + 1)3 1
2
x 0.
Suy ra hm s f(x) nghch bin trn [0;+)Mt khc: lim
x+f(x) = 0
Do phng trnh cho c nghim khi v ch khi m (0;1] 2 . Nhn xt: i khi ta phi tm cch c lp m a phng trnh v dng trn.
Bi 3: Tm m phng trnh sau c nghim:4x4 13x + m + x 1 = 0 (*)
Gii
Ta c:
() 4
x4 13x + m + x1 = 0
1 x 0x4 13x + m = (1 x)4
1 x
4x3 6x2 9x = 1 m
Xt hm s f(x) = 4x3 6x2 9x vi x 1
Ta c: f(x) = 12x2 12x 9, f(x) = 0 x = 32
x = 12
.
Da vo bng bin thin suy ra phng trnh c nghim khi v ch khi m 32
. 2
Bi 4: Tm m phng trnh sau c nghim: x
x +
x + 12 = m(
5 x + 4 x)
Gii
KX: x [0;4].Khi phng trnh tng ng vi
(x
x +
x + 12)(
5 x 4 x) = m
Xt hm s f(x) = (x
x +
x + 12)(
5 x 4 x) lin tc trn on [0; 4].Ta c: f(x) =
3
2
x +
1
2
x + 12
1
2
4 x 1
2
5 x
> 0 x [0; 4]Vy f(x) l hm ng bin trn [0;4].Suy ra phng trnh c nghim khi v ch khi 2
3(
5
2) m 12 2 .
Bi 5: Tm m h sau c nghim: ()
2x2
1
2
45x3x2 mxx + 16 = 0
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Gii
Ta c:
()
x2 4 + 5x3x2 mxx + 16 = 0
x [1; 4]3x2 mxx + 16 = 0
m = 3x2 + 16
x
x
Xt f(x) =3x2 + 16
x
xvi x [1; 4]. Ta c:
f(x) =6x2
x 3
2
x(3x2 + 16)
x3=
3
x(x2 16)2x3
0x [1; 4]
Nh vy m = f(x) nghch bin trn [1; 4], do f(4) m f(1) 8 m 19Vy h c nghim khi v ch khi m [8; 19] 2
Nhn xt: Khi gp h phng trnh trong mt phng trnh ca h khng cha tham sth ta s i gii quyt phng trnh ny trc. T phng trnh ny ta s tm c tp nghim(i vi h mt n) hoc s rt c n ny qua n kia. Khi nghim ca h ph thuc vonghim ca phng trnh th hai vi kt qu ta tm c trn.
Bi 6: Tm m h sau c nghim:
72x+x+1 72+
x+1 + 2007x 2007
x2 (m + 2)x + 2m + 3 = 0
GiiTa c:
72x+x+1 72+
x+1 + 2007x 2007 72+
x+1(72x+2 1) 2007(1 x) ()
Nu x > 1 72+x+1(72x2 1) > 0 > 2007(1 x).Suy ra (*) v nghim.
Nu x 1 72+x+1(72x2 1) 0 2007(1 x). Suy ra (*) ng .
Suy ra h c nghim khi v ch khi phng trnh x2 (m + 2)x + 2m + 3 = 0 c nghim vi
x [1;1], hay phng trnh m =x2
2x + 3
x 2 c nghim vi x [1;1].Xt hm s f(x) =
x2 2x + 3x 2 vi x [1;1], c
f(x) =x2 4x + 1
(x 2)2 = 0 x = 2
3
Da vo bng bin thin suy ra h c nghim khi v ch khi m 2 3.
Bi 7: Tm m h phng trnh sau c nghim: () x y + m = 0y +
xy = 2
Gii
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KX: xy 0. T h ta cng c y = 0.
()
x y + m = 0
xy = 2 y
x y + m = 0
x =y2 4y + 4
y
y 2
m = y y2 4y + 4
y=
4y 4y
(y 2)
Xt hm s f(y) =4y
4
y (y 2) ta c f(y) =
4
y2 > 0 y = 0, suy ra hm s f(y) ng bintrn cc khong (; 0) v (0; 2].Mt khc, lim
yf(y) = 4, lim
y0+f(y) = ; lim
y0f(y) = +
Suy ra h c nghim khi v ch khi m (; 2] (4;+) 2
Bi 8: Tm m phng trnh c ng hai nghim phn bit:
x4 4x3 + 16x + m + 4
x4 4x3 + 16x + m = 6 ()
Gii
Ta c:() 4
x4 4x3 + 16x + m = 2 m = x4 + 4x3 16x + 16
Xt hm s f(x) = x4 + 4x3 16x + 16 vi x R. Ta c:
f(x) = 4x3 + 12x2 16; f(x) = 0
x = 1x = 2
Da vo bng bin thin suy ra phng trnh c hai nghim phn bit khi v ch khi m < 27.
Bi 9: Tm m phng trnh m
x2 + 2 = x + m () c ba nghim phn bit.Gii
T (*) ta c: () m = xx2 + 2 1
Xt hm s: f(x) =x
x2 + 2
1
vi x R.
Ta c: f(x) = 2 x2 + 2x2 + 2(
x2 + 2 1)2
; f(x) = 0 x = 2 .
Da vo bng bin thin phng trnh c nghim khi v ch khi 2 < m < 2 2
Bi 10: Tm m phng trnh: mx2 + 1 = cos x () c ng mt nghim x
0;
2
Gii
Ta thy (*) c nghim th m 0. Khi
mx2 + 1 = cos x m = cos x 1x2
2m =sin2
x
2x2
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Xt hm s f(t) =sin2t
t2vi t
0;
4
Ta c
f(t) =2t2 sin t cos t 2tsin2t
t4=
2sin t(t cos t sin t)t3
=sin2t(t tan t)
t3< 0 t
0;
4
Suy ra hm s f(t) nghch bin trn (0;
4).
Da vo bng bin thin suy ra phng trnh c ng mt nghim trn (0; 4 ) khi v ch khi82
< 2m < 1 12
< m < 422
Bi 11: Tm m h phng trnh c ba cp nghim phn bit:
()
3(x + 1)2 + y m = 0x +
xy = 1
Giiiu kin xy 0Ta c
()
3(x + 1)2 + y = m
xy = 1 x
3(x + 1)2 + y = m
y =x2 2x + 1
x
y 1
m = 3(x + 1)2 + x2 2x + 1
x
m 3 = 3x2
+ 6x +
x2
2x + 1
x
Xt hm s: f(x) = 3x2 + 6x +x2 2x + 1
x(x 1) ta c
f(x) =6x3 + 7x2 1
x2= 0
x = 1x =
12
x =1
3
Da vo bng bin thin ta thy h phng trnh c ba nghim phn bit khim [4; 15
4] [20
3;12] 2
Nhn xt: Khi t n ph ta phi tm min xc nh ca n ph v gii quyt bi ton nph trn min xc nh va tm. C th:* Khi t t = u(x)(x D), ta tm c t D1 v phng trnh f(x, m) = 0 (1) tr thnhg(t, m) = 0 (2). Khi (1) c nghim x D (2) c nghim t D1.* tm min xc nh ca t ta c th s dng cc phng trnh tm min gi tr (v min xcnh ca t chnh l min gi tr ca hm u(x)).* Nu bi ton yu cu xc nh s nghim th ta phi tm s tng ng gia x v t, tc l mi
gi tr t D1 th phng trnh t = u(x) c bao nhiu nghim x D.
Bi 12: Tm m phng trnh m(
x 2 + 2 4x2 4) x + 2 = 2 4x2 4 c nghim
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Gii
KX; x 2Ta thy x = 2 khng l nghim ca phng trnh nn ta chia hai v phng trnh cho 4
x2 4:
m
4
x 2x + 2
+ 2
4
x + 2
x
2
= 2 ()
t t = 4x + 2
x 2(t > 1). Khi (*) tr thnh: m1
t+ 2
t = 2 m = t2 + 2t
2t + 1.
Xt hm s f(t) =t2 + 2t
2t + 1(t > 1) ta c f(t) =
2t2 + 2t + 2
(2t + 1)2> 0 t > 1.
Vy hm s f(t) ng bin trn (1;+).Vy phng trnh c nghim khi v ch khi m > 1 2 Nhn xt: Trong cc bi ton trn sau khi t n ph ta thng gp kh khn khi xc nhmin xc nh ca t . tm c iu kin ca n ph t, chng ta c th dng cng c hm
s, bt ng thc, lng gic ha. . .
Bi 13: Tm m phng trnh log23 x +
log23 x + 1 2m 1 = 0 c nghim trn
1; 33
Gii
t t =
log23 x + 1. Do x
1; 33
1 t 2Phng trnh cho tr thnh: t2 + t = 2m + 2Xt hm s f(t) = t2 + t vi 1 t 2, ta thy f(t) l hm ng bin trn [1; 2]
Vy phng trnh c nghim khi v ch khi 2 2m + 2 6 0 m 2 2Bi 14: Xc nh m h sau c 2 nghim phn bit
log3(x + 1) log3(x 1) > log34log2(x
2 2x + 5) mlog(x22x+5)2 = 5
Gii
iu kin : x > 1. T bt phng trnh th nht ca h ta c: log3
x + 1
x 1 > log32 x (1; 3).t t = log2(x2 2x + 5)(t (2;3))v phng trnh th hai ca h tr thnht +
m
t= 5 t2 5t = m
T cch t t ta c: Vi mi gi tr t (2; 3) th cho ta ng mt gi tr x (1; 3). Suy ra h c2 nghim phn bit khi v ch khi phng trnh t2 5t = m c 2 nghim phn bit t (2; 3).Xt hm s f(t) = t2 5t vi t (2; 3). Ta c f(t) = 2t 5; f(t) = 0 t = 5
2. Da vo bng
bin thin ta c, h phng trnh c 2 nghim phn bit khi v ch khi m (6; 254
) 2
Bi 15:( thi H khi B - 2004) Tm m phng trnh c nghim:
m(
1 + x2
1 x2 + 2) = 2
1 x4 +
1 + x2
1 x2 ()
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Gii
iu kin: 1 x 1Trc tin, ta nhn thy rng: (1 x2) (1 + x2) = 1 x4 v 1 x2 + 1 + x2 = 2 nn ta c phpt n ph nh sau:t t =
1 + x2 1 x2
Phng trnh cho tr thnh:
m (t + 2) = t2 + t + 2 t2 + t + 2
t + 2= m (1)
Do
1 + x2
1 x2 t 0Mt khc: t2 = 2 21 x4 2 t 2Ta xt hm s: f(t) =
t2 + t + 2t + 2
, t 0; 2, ta c:f(t) = t2 4t(t + 2)2
0
Vy hm f(t) nghch bin trn on
0;
2
. M hm s lin tc trn
0;
2
nn phng trnh
cho c nghim x khi phng trnh (1) c nghim t 0; 2.iu ny tng ng vi:min f(t) m max f(t) t
0;
2
f
2 m f(0)
Vy cc gi tr m cn tm l
2 1 m 1 2.
Bi 16: Tm m phng trnh sau c nghim:
xx + x + 12 = m 5 x + 4 xGii
Cng ging nh nhng bi ton trc, bi ny ta ngh ngay l phi a bi ton v dngf(x) = m ri s dng tng giao gia 2 th v suy ra iu kin m.Ta gii bi ton nh sau:iu kin: 0 x 4
Phng trnh cho tng ng vi:x
x +
x + 125
x +
4
x
= m
t f(x) =xx + x + 12
5 x + 4 x x [0; 4]Tuy nhin, nu n y ta kho st hm s ny th c v kh phc tp v di dng. V vy tas gii quyt theo mt hng khc. rng, ta c tnh cht sau:
Vi hm s y =f(x)
g (x). Nu y = f(x) ng bin v y = g (x) nghch bin th y =
f(x)
g (x)ng
bin.Ta vn dng tnh cht trn nh sau:Xt hm s g (x) = x
x +
x + 12 ta c:
g(x) = 32x + 1
2
x + 12> 0 nn hm g(x) l ng bin.
Xt hm s h(x) =
5 x + 4 x ta c:h(x) = 1
2
5 x 1
2
4 x < 0 nn hm h(x)l nghch bin
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Vy, theo tnh cht trn ta c hm y =g (x)
h (x)ng bin x [0; 4].
Do phng trnh f(x) = m c nghim khi v ch khif(0) m f(4) 2 15 12 m 12 2.Bi 17: Gii v bin lun phng trnh sau theo m:
x2 2x + m2 = |x 1| m (1)
Gii
Xt x 1 ta c:
(1)
x2 2x + m2 = x 1 m
x 1 m 0x2 2x + m2 = (x 1 m)2
x 1 + m
2mx = 2m + 1
Nu m = 0: h trn v nghim Nu m = 0 x = 2m + 1
2m
Ta c x 1 + m 2m + 12m
1 + m 2m2 + 1
2m 0 m
2
2 0 < m 0
Kt hp iu kin trn ta c 0 < m 0, x nn chia hai v ca phng trnh cho x2 x + 1 ta c:x2 + x + 1
x2 x + 12
x2 + x + 1
x2 x + 1 = m
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t t =x2 + x + 1
x2 x + 1 ,1
3 t 3
Phng trnh trn tr thnh: t2 t = my l mt phng trnh bc hai n gin nn vic kho st xin dnh cho bn c.iu lu y l iu kin ca t. Thc cht y ta tm gi tr ln nht v nh nht
ca biu thcx2 + x + 1
x
2
x+ 1
(c th dng phng php min gi tr).
Bi 19: Tm m h sau c nghim thc:x +
1
x+ y +
1
y= 5
x3 +1
x3+ y3 +
1
y3= 15m 10
Gii
Cch 1:
Nhn vo h ny, ta thy ngay c hng i l phi t n ph v gia cc i lng x + 1xv x3 +
1
x3dng nh c mi lin h vi nhau. Vi tng , ta c php t nh sau:
t
a = x +
1
x
b = y +1
y
, (|a| 2, |b| 2).
Ta c
x3 +1
x3=
x +
1
x
3 3
x +1
x
= a3 3a
y3
+
1
y3 = y + 1y3
3y + 1y = b3 3bKhi , h cho tr thnh:
a + b = 5
a3 + b3 3 (a + b) = 15m 10
a + b = 5
ab = 8 m
D thy a, b l nghim ca phng trnh X2 5X + 8 m = 0 X2 5x + 8 = m (1) Xthm s f(X) = X2 5X+ 8, |X| 2, ta c:f (X) = 2X 5 f (X) = 0 X = 5
2
T , k bng bin thin v ch rng h cho c nghim thc khi v ch khi phng trnh(1) c nghim |X| 2. Ta tm c: 7
2 m 2 m 22.
Cch 2:Ta nhn thy rng phng trnh th nht khng c cha tham s nn ta s xut pht tphng trnh ny. Khai trin phng trnh ny ra, ta c:
x3 y3 + 4 (x y) = 9y + 8 3x2 + 6y2 x3 + 3x2 + 4x = y3 + 6y2 + 13y + 8
(x + 1)3
+ (x + 1) = (y + 2)3
+ (y + 2)
Xt hm s f(t) = t3 + t, d thy l hm s ny ng bin nn
f(x + 1) = f(y + 2) x + 1 = y + 2 x = y + 1
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T y, thay x = y + 1 vo phng trnh th hai ta c:
15 + 2y y2 = 2m +
4 y2
(5 y) (y + 3)
4 y2 = 2m
n y tng r, ta ch cn chuyn v tng giao gia hai th.
Bi 20: Tm m h sau c nghim thc: x3 + (y + 2) x2 + 2xy = 2m 3x2 + 3x + y = m
Gii
tng: h ny ta quan st thy bi ton cn cha r ng li no v c hai phngtrnh trong h u cha n tham s m. V vy i n hng gii quyt tt ta nn bt uphn tch hai v tri trong hai phng trnh trong h. C th ta c:
x3 + (y + 2) x2 + 2xy = x3 + yx2 + 2x2 + 2xy = x2 (x + y) + 2x (x + y) = (x + y)
x2 + 2x
Mt khc:
x2 + 3x + y = x2 + 2x + x + y
R rng bc phn tch ny ta tm ra li gii cho bi ton ny chnh l t n ph. Li gii:
t: a = x2 + 2x 1
b = x + y
ta c h phng trnh
a + b = m
ab = 2m 3
a2 3 = (a + 2) m (1)b = m a
T phng trnh (1) trong h ta c:a2 3a + 2
= m (2)
H cho c nghim khi v ch khi phng trnh (2) c nghim a 1.Xt hm s: f(x) =
x2 3
x + 2
vi x
1
n y ta ch cn lp bng bn thin. Cng vic tip theo xin dnh cho bn c.
Bi tp t luyn
Bi 1: Tm m phng trnh tan2x + cot2x + m(cot x + tan x) = 3 c nghimBi 2: Tm m phng trnh
x +
x + 9 = 9x x2 + m c nghimBi 3: Tm m phng trnh
3 + x +
x + 6 18 + 3x x2 = m c nghimBi 4: Tm m phng trnh x3 4mx2 + 8 = 0 c 3 nghim phn bit.
Bi 5: Tm m phng trnh x3 + 3x2 + (3 2m) x + m + 1 = 0 c ng mt nghim lnhn 1.Bi 6: Tm m phng trnh sau c ng 2 nghim thc phn bit:4x2 2mx + 1 = 38x3 + 2x
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PHNG PHP DNG NH L LAGRANGE-ROLLE
L thuyt
nh l Rolle: : Nu f(x) l hm lin tc trn on [a; b], c o hm trn khong (a; b)
v f(a) = f(b) th tn ti c (a; b) sao cho f(c) = 0.T ta c 3 h qu: H qu 1: Nu hm s f(x) c o hm trn (a; b) v f(x) c n nghim (n l s nguyndng ln hn 1) trn (a; b) th f(x) c t nht n - 1 nghim trn (a; b). H qu 2: Nu hm s f(x) c o hm trn (a; b) v f(x) v nghim trn (a; b) th f(x)cnhiu nht 1 nghim trn (a; b). H qu 3: Nu f(x)c o hm trn (a; b) v f(x) c nhiu nht n nghim (n l s nguyndng) trn (a; b) th f(x) c nhiu nht n + 1 nghim trn (a; b).Cc h qu trn vn ng nu cc nghim l nghim bi (khi f(x) l a thc) v cho ta
tng v vic chng minh tn ti nghim cng nh xc nh s nghim ca phng trnh, vnu nh bng mt cch no ta tm c tt c cc nghim ca phng trnh th ngha lkhi phng trnh c gii.
nh l Lagrange: : Nu f(x)l hm lin tc trn on [a; b], c o hm trn khong
(a; b)th tn ti c (a; b) sao cho f(c) = f(b) f(a)b a .
Sau y l mt s ng dng ca hai nh l trn:
Bi tp v d
Dng nh l Lagrange - Rolle bin lun phng trnh:
Bi 1: Chng minh rng phng trnh a cosx + b cos2x + c cos3x = 0 lun c nghim vimi a,b,c.
Gii
Xt f(x) = a sin x +b sin2x
2 +c sin3x
3ta c f(x) = a cosx + b cos2x + c cos3xM f(0) = f() = 0 x0 (0; ) : f(x0) = 0, suy ra iu phi chng minh. 2
Bi 2: Cho a + b + c = 0. Chng minh rng phng trnh sau c t nht 4 nghim thuc[0; ]: a sin x + 9b sin3x + 25c sin5x = 0
Gii
chng minh f(x) c t nht n nghim ta chng minh F(x) c t nht n + 1 nghim vi F(x)
l mt nguyn hm ca f(x) trn (a; b) (c th phi p dng nhiu ln)Xt hm s f(x) = a sin x b sin3x c sin5x, ta c:
f(x) = a cos x 3b cos3x 5c cos5x; f(x) = a sin x + 9b sin3x + 25c sin5x
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Ta c f(0) = f(
4) = f(
3
4) = f() = 0
Suy ra x1 (0; 4
), x2 ( 4
;3
4), x3 ( 3
4; ) : f(0) = f(x1) =
f(x2) = f(x3) = 0
x4 (x1; x2), x5 (x2; x3)|f(x4) = f(x5) = 0m f(0) = f() = 0, suy ra iu phi chng minh. 2
Bi 3: Chng minh phng trnh x5 5x + 1 = 0 c ng ba nghim thc.Gii
t f(x) = x5 5x + 1 th f(2) = 21 < 0, f(0) = 1 > 0, f(1) = 3 < 0, f(2) = 23 > 0nn t y suy ra phng trnh cho c ba nghim thcGi s c nghim th t ca phng trnh. p dng nh l Rolle ta c:f(x) = 0 5x4 5 = 0. Phng trnh ny c hai nghim thc nn suy ra mu thun.Vy ta c iu phi chng minh 2.
Bi 4: Cho a thc P(x) v Q(x) = aP(x) + bP(x) trong a, b l cc s thc, a =0.
Chng minh rng nu Q(x) v nghim th P(x) v nghim.
Gii
Ta c deg P(x) = deg Q(x)V Q(x) v nghim nn deg Q(x) chn. Gi sP(x) c nghim, v deg P(x) chn nn P(x) ct nht 2 nghim. Khi P(x) c nghim kp x = x0 ta c x0 cng l mt nghim ca P(x) suy ra Q(x) c
nghim. Khi P(x) c hai nghim phn bit x1 < x2: Nu b = 0 th hin nhin Q(x) c nghim. Nu b =0 : Xt f(x) = eab xP(x) th f(x)c hai nghim phn bit x1 < x2 v
f(x) =a
bea
bxP(x) + e
a
bxP(x) =
1
bea
bx[aP(x) + bP(x)] =
1
bea
bxQ(x)
V f(x) c hai nghim suy ra f(x) c t nht 1 nghim hay Q(x) c nghim.Tt c trng hp u mu thun vi gi thit Q(x) v nghim. Vy khi Q(x) v nghim th
P(x) v nghim2
Bi 5: Gi s phng trnh sau c n nghim phn bit:a0x
n + a1xn1 + ... + an1x + an = 0, (a0 = 0)
Chng minh (n 1)a21 > 2na0a2Gii
t a0xn + a1xn1 + ... + an1x + an = f(x)Nhn xt f kh vi v hn trn R nn suy ra
f(x) c n 1 nghim phn bit.f(x) c n 2 nghim phn bit.. . . . .f[n2](x) =
n!
2a0x
2 + (n 1)!a1x + (n 2)!a2 c 2 nghim phn bit.
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Nhn thy f[n2](x) c > 0 nn ((n 1)!a1)2 2n!a0(n 2)!a2 > 0Suy ra iu phi chng minh. 2
Bi 6: (VMO 2002) Xt phng trnhni=1
1
i2x 1 =1
2, vi n l s nguyn dng. Chng
minh rng vi mi s nguyn dng n, phng trnh nu trn c mt nghim duy nht ln
hn 1; k hiu nghim l xn
Gii
Xt fn(x) =ni=1
1
i2x 1 1
2, ta c: fn(x) lin tc v nghch bin trn (1;+)
M limx1+
fn(x) = +, limx+
fn(x) = 12
fn(x) = 0 c mt nghim duy nht ln hn 1. 2
Dng nh l Lagrange -Rolle gii phng trnh:
Bi 7: Gii phng trnh 3x + 5x = 2.4x ()Gii
Nhn xt: x = 0; x = 1 l nghim ca phng trnh (*).Gi x0 l nghim khc ca phng trnh cho. Ta c:
3x0 + 5x0 = 2.4x0 5x0 4x0 = 4x0 3x0 ()
Xt hm s f(t) = (t + 1)x0 tx0, ta c () f(4) = f(3)V f(t) lin tc trn [3; 4] v c o hm trong khong (3; 4), do theo nh l Rolle tn tic (3; 4) sao cho
f(c) = 0 x0[(c + 1)x01 cx01]=0
x0 = 0
x0 = 1(loi)
Vy phng trnh (*) c tp nghim S = {0; 1}. 2
Bi 8: Gii phng trnh 5x 3x = 2x ()Gii
Nhn xt: x = 0; x = 1 l nghim ca phng trnh (2). Gi x0 l nghim khc ca phngtrnh cho, ta c:
5x0 5x0 = 3x0 3x0 (2a)Xt hm s: f(t) = tx0 tx0, khi : (2a) f(5) = f(3)V f(t) lin tc trn [3;5] v c o hm trn (3; 5), do theo nh l Lagrange lun tn tic (3; 5) sao cho
f(c) = 0 x0(cx01 1)=0 x0 = 0
x0 = 1
(loi)
Vy phng trnh (*) c tp nghim S = {0; 1} 2
Bi 9: Gii phng trnh: 3x + 2.4x = 19x + 3 ()
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Gii
Ta c() 3x + 2.4x 19x 3 = 0
Xt hm s: y = f(x) = 3x + 2.4x 19x 3 ta c: f(x) = 3x ln 3 + 2.4x ln 4 19 ta cf(x) = 3x(ln 3)2 + 2.4x(ln 4)2 > 0,
x
R hay f(x) v nghim, suy ra f(x) c nhiu nht 1
nghim, suy ra f(x) c nhiu nht 2 nghim.M f(0) = f(2) = 0 do (*) c ng hai nghim x = 0, x = 2 2
Bi 10: Gii phng trnh: (1 + cos x)(2 + 4cosx) = 3.4cos x ()
Gii
t t = cos x, (t [1; 1])
Ta c: () (1 + t)(2 + 4t) = 3.4t (1 + t)(2 + 4t) 3.4t = 0Xt hm s: f(t) = (1 + t)(2 + 4t) 3.4t ta c:f(t) = 2 + 4t + (t 2)4t ln 4, f(t) = 2.4t ln 4 + (t 2)4tln24Li c: f(t) = 0 t = 2 + 2
ln 4, suy ra f(t) c nghim duy nht.
Suy ra f(t) c nhiu nht hai nghim, ngha l f(t) c nhiu nht ba nghim.
Mt khc d thyf(0) = f(1
2) = f(1) = 0, do f(t) c ba nghim t = 0,
1
2, 1.
Kt lun: Nghim ca phng trnh (*) l: x =
2+ k2, x =
3+ k2, x = k2, k Z 2
Bi 11: Gii phng trnh 3cosx 2cosx = 2cosx 2cos x ()Gii
Xt hm f(t) = tcos t cos , f(t) = cos (tcos1 1)Ta nhn thy f(3) = f(2) v f(x) kh vi trn [2;3] nn p dng nh l Lagrange ta c:
c [2; 3] : f(c) = f(3) f(2)1
cos (ccos1 1) = 0
T ta suy ra nghim ca phng trnh (*) l x =
2+ k,x = k2(k
Z) 2
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PHNG PHP DNG IU KIN CN V
L thuyt
Bi ton:
Cho h phng trnh (hoc h bt phng trnh) cha tham s c dng:
(I)
f(x, m) = 0
x Dxm Dm
hoc (II)
f(x, m) 0
x Dxm Dm
Trong x l bin s, m l tham s, Dx, Dm l min xc nh ca x v m.Yu cu t ra: ta phi tm gi tr ca tham s m h (I) hc (II) tha mn mt tnh chtno .
Phng php gii:
Bc 1 (iu kin cn): Gi s h tha mn tnh cht P no m u bi i hi. Khi, da vo c th ca tnh cht P v dng ca phng trnh ta s tm c mt rng buc no i vi tham s m v rng buc y chnh l iu kin cn c tnh cht P. iu c nghal: nu vi m0 khng tha mn rng buc trn th chc chn ng vi m0, h khng c tnh cht P.
Bc 2 (iu kin ): Ta tm xem trong cc gi tr ca m va tm c, gi tr no lmcho h tha mn tnh cht P. bc ny ni chung ta cng ch cn gii nhng h c th khngcn tham s. Sau khi kim tra, ta s loi i nhng gi tr khng ph hp v nhng gi tr cnli chnh l p s ca bi ton.
Nh vy, tng ca phng php ny kh r rng v n gin. Trong rt nhiu bi ton vbin lun th phng php ny li th hin u th r rt. Tuy nhin, thnh cng ca phngphp cn nm ch ta phi lm th no pht hin iiu kin cn mt cch hp l v chniu kin mt cch ng n.
Bi tp v d
S dng tnh i xng ca cc biu thc c mt trong bi ton
Bi 1: Tm m phng trnh sau c nghim duy nht
4
x + 4
1 x + x + 1 x = m (1)
Gii
iu kin cn:Gi s (1) c nghim duy nht x = D thy nu (1) c nghim x = th (1) cng c nghim x = 1 . V nghim l duy nht
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nn = 1 = 12
Thay =1
2vo (1) ta tm c m =
2 + 4
8.
iu kin :Gi sm =
2 + 4
8, khi (1) c dng sau:
4
x +4
1 x + x + 1 x =
2 +
4
8 (2)
Theo bt ng thc AM-GM ta c:
x +
1 x
2 v 4
x + 4
1 x 4
8
Do (2) x = 1 x x = 12
.
Vy (1) c nghim duy nht th iu kin cn v l m =
2 + 4
8 2
Bi 2: Tm a v b phng trnh sau c nghim duy nht
3(ax + b)2 + 3(ax b)2 + 3a2x2 b2 = 3b (1)Gii
iu kin cn:Gi s (1) c nghim duy nht x = x0, khi d thy x = x0 cng l nghim ca (1). Do t gi thit ta suy ra x0 = 0. Thay x0 = 0 vo (1) ta c :
3
b2 =3
b
b = 0
b = 1
iu kin : Khi b = 0, (1) c dng:
3
a2x2 +3
a2x2 +3
a2x2 = 0 a2x2 = 0Do (1) c nghim duy nht khi v ch khi a = 0 Khi b = 1, (1) c dng:
3
(ax + 1)2 +
3
(ax 1)2 +
3
a2x2 1 = 1 ()t u = 3ax + 1; v = 3ax 1, ta thy:
()
u3 v3 = 2u2 + uv + v2 = 1
u v = 2u2 + uv + v2 = 1
u = 1
v = 1
ax + 1 = 1
ax 1 = 1 ax = 0
Vy (*) c nghim duy nht khi v ch khi a = 0Tm li, phng trnh (1) c nghim duy nht th iu kin cn v l
a = 0; b = 0b = 1
2
Bi 3: Tm m h sau c nghim duy nht:
7 + x +
11 x 4 = m
4 310 3m7 + y +
11 y 4 = m
4 310 3m
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Gii
iu kin: 7 x, y 11; 7427 m
10
3Tr theo v hai phng trnh ta c:
x + 7 11 x =
y + 7
11 y
Xt hm s: f(t) = t + 7 11 t; 7 t 11 ta c:f (t) =
1
2
t + 7+
1
2
11 t > 0 Vy hm s ng bin, suy ra: f(x) = f(y) x = y.Thay vo mt trong hai phng trnh ca h ta c:
7 + x +
11 x 4 = m
4 310 3m ()
iu kin cn:Ta thy l nu x0 l mt nghim ca phng trnh th 4 x0 cng l nghim ca phng trnh.Nn h cho c nghim duy nht khi v ch khi
x0 = 4 x0 x0 = 2
Thay vo phng trnh (*) ta c:4 310 m = m 2 ()
Gii phng trnh (**) ta tm c m = 3. iu kin :
Vi m = 3, ta thu c h phng trnh:7 + x + 11 x = 67 + y + 11 y = 6
V x = y nn ta ch vic gii phng trnh
7 + x +
11 x = 6 x = 2Vy m = 3 l gi tr cn tm h cho c nghim duy nht. 2
Bi 4: Tm a,b h sau c nghim duy nht:
xyz+ z = a
xyz2 + z = b
x2 + y2 + z2 = 4
Gii
iu kin cn:Gi s(x0; y0; z0) l nghim ca h phng trnh cho th (x0; y0; z0) cng l nghim. Do
tnh duy nht nn x0
=
x0; y
0=
y0
x0
= y0
= 0 Thay tr li vo h , ta c:
z0 = a
z0 = bz20 = 4
T y ta suy ra a = b = 2 hoc a = b = 2 iu kin :
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Xt hai trng hp sau: Nu a = b = 2: Khi h c dng:
xyz+ z = 2 (1)
xyz2 + z = 2 (2)
x
2
+ y
2
+ z
2
= 4 (3)
Ly (1) (2) ta c xyz(1 z) = 0, t (1) li c z= 0 do xy(1 z) = 0* Nu x = 0 z = 2 y = 0* Nu y = 0 z = 2 x = 0
* Nu z = 1 x2 + y2 = 3xy = 1 .
H trn c nghim (x1; y1) = (0;0).V vy ngoi nghim (0, 0, 2), h cn c nghim khc(x1; y1; 1) do h khng c nghim duy nht. Trng hp ny khng tha mn.
Nu a = b = 2:Khi h c dng:
xyz = 2xyz22 + z = 2x2 + y2 + z2 = 4
Tin hnh lm nh trng hp trn ta i n:* Nu x = 0 z = 2 y = 0* Nu y = 0 z = 2 x = 0* Nu z = 1
x2 + y2 = 3xy = 3Ta thy t h phng trnh trn, ta suy ra x2 + y2 < 2|xy| nn h v nghim.Vy trong trng hp ny h c duy nht nghim (x; y; z) = (0, 0, 2)Vy iu kin cn v h phng trnh cho c nghim duy nht l a = b = 2 2
S dng im thun li
Bi 5: Tm a phng trnh sau nghim ng vi mi x:
log2
a2x2 5ax2 + 6 a = log2+x2 3 a 1 ()Gii
iu kin cn:Gi s (*) ng vi mi x. Vi x = 0 ta c log2
6 a = log2 (3
a 1)
Li c: 1 a 6
a 1 < 3
a 1 + 6 a = 3 a {2; 5}
iu kin :
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Nu a = 2 th () log2(2 12x2) = log2+x2 2 ()R rng (**) khng ng vi mi x, v log2(2 12x2) c ngha th phi c 12x2 < 2 Nu a = 5 th () log2 1 = log2+x2 1 (lun ng)Vy a = 5 l iu kin cn v (*) ng vi mi x 2
Bi 6: Tm a h phng trnh n (x; y) c nghim vi mi b:2bx + (a + 1)by2 = a2(a 1)x3 + y2 = 1Gii
iu kin cn:Gi s h c nghim vi mi b, thay b = 0 ta c
a2 = 1(a 1)x3 + y2 = 1Do iu kin cn l a = 1 iu kin :
Nu a = 1: ta c h
2bx + 2by2 = 1y2 = 1Khi b >
1
2h v nghim. Vy trng hp ny loi.
Nu a = 1: ta c h2bx = 12x3 + y2 = 1
H trn lun c nghim (x; y) = (0; 1)Vy a = 1 l iu kin cn v h phng trnh c nghim vi mi b 2
Bi 7: Tm a h phng trnh n (x; y) c nghim vi mi b:
(x2 + 1)a + (b2 + 1)y = 2
a + bxy + x2
y = 1
Gii
iu kin cn:Gi s h c nghim vi mi b, thay b = 0 ta c
()
(x2 + 1)a = 1
a + x2y = 1
a = 0; x2y = 1
x2 + 1 = a + x2y = 1 a {0; 1}
iu kin :
Nu a = 0 : ta c
(b2 + 1)y = 1 (1)bxy + x2y = 1 (2)
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Nu b = 0 b2 + 1 = 1 nn t (1) c y = 0, nhng khng tho (2). Vy trng hp ny loi.
Nu a = 1: ta c
x2 + (b2 + 1)y = 1bxy + x2y = 0H trn lun c nghim x = y = 0.Vy a = 1 l iu kin cn v h cho c nghim vi mi b 2
Bi 8: Tm iu kin ca a,b,c,d,e,f hai phng trnh n (x; y) sau l tng ng:ax2 + bxy + cy2 + dx + ey + f = 0 (1)x2 + y2 = 1 (2)Gii
iu kin cn:
Ta thy (x; y) = (0; 1) , (1;0) , 12 ; 12 , 12; 12 l nghim ca (2). Do (1)cng phi c cc nghim trn.
Nh vy
c + e + f = c e + f = a + d + f = a d + f = 0a + b + c +
2d +
2e + 2f
2=
a + b + c 2d 2e + 2f2
= 0
Gii h trn ta tm c iu kin cn ca bi ton l ()b = d = e = 0a = c = f = 0
iu kin :
D thy vi (*) th (2) trng vi (1).Vy (*) l iu kin cn v (1) (2) 2
Bi 9: Cho phng trnh x3 + ax + b = 0 (1)Tm a, b phng trnh trn c ba nghim phn bit x1 < x2 < x3 cch u nhau.
Gii
iu kin cn:
Gi s phng trnh (1) c 3 nghim khc nhau x1, x2, x3 tha gi thit x1 + x3 = 2x2Theo nh l Viete vi phng trnh bc 3 ta c: x1 + x2 + x3 = 0 3x2 = 0 x2 = 0Thay x2 = 0 vo (1) ta c b = 0 iu kin :Gi sb = 0 , khi (1) tr thnh:
x3 + ax = 0 x(x2 + a) = 0 (2)
Ta thy (2) c 3 nghim phn bit nu a < 0. Khi cc nghim ca (2) l
x1 = ax2 = 0
x3 =a
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Cc nghim trn cch u nhau nn iu kin cn v (1) c nghim tha mn bi lb = 0, a < 0 2
Bi 10: Cho phng trnhx3 3x2 + (2m 2) x + m 3 = 0
Tm m phng trnh c ba nghim x1, x2, x3 sao cho x1 0 khi x1 < x < x2Suy ra f(1) > 0 hay m 5 > 0 m < 5. iu kin :
Gi sm < 5.Do limx
f(x) = nn tn ti < 1 m f() < 0Li c: f(1) = m 5 > 0 v f(x) lin tc nn ta c < x1 < 1 sao cho f(x1) = 0Ta c: f(0) = m 3 < 0 ( do m < 5)Vy tn ti 1 < x2 < 0 sao cho f(x2) = 0Mt khc, do lim
x+f(x) = + nn phi c > 0 sao cho f() > 0.
T , tn ti x3 m 0 < x3 < sao cho f(x3) = 0.Nh vy, phng trnh f(x) = 0 khi m < 5 c 3 nghim x1, x2, x3 tha mn x1 < 1 < x2 0 (vi x1 < < x2 < < x3)
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Phng trnh f(x) = 3x2 + 2ax + b = 0 (1) c hai nghim dng.Nh vy
= a2 3b > 0P =
b
3> 0
S =2a
3> 0
a2 > 3b
b > 0
a < 0
()
b > 0
a < 3b
(*) l iu kin cn ca bi ton. iu kin :Gi s a, b tho mn (*) th r rng phng trnh 3x2 + 2ax + b = 0 c 2 nghim dng0 < <
Suy ra hm s f(x) = x3 + ax2 + bx + c c cc i ti x = v cc tiu ti x = .Do > 0 nn tm c x1 (c; ) sao cho f() < f(x1) < f() phng trnh f(x) = f(x1)c ba nghim dng. t c = f(x1) th phng trnh x3 + ax2 + bx + c = 0 c 3 nghimdng.
Vy (*) l iu kin cn v tn ti c sao cho phng trnh x3 + ax2 + bx + c = 0 c 3nghim dng.T , ta suy ra: b 0 hoc a 3b l iu kn cn v sao cho phng trnhx3 + ax2 + bx + c = 0 c khng qu 2 nghim dng. 2
Bi tp t luyn
Bi 1: Tm m hai phng trnh sau l tng ng x2 + (m2 5m + 6) x = 0 v x2 +2 (m
3) x + m2
7m + 12 = 0
Bi 2: Tm m phng trnh sau c nghim duy nht:|x + m|2 + |x + 1| = |m + 1|Bi 3: Tm m phng trnh sau c nghim duy nht:
x + 3 +
6 x
(3 + x) (6 x) = m
Bi 4: Tm a h sau c ng mt nghim:
x2 + y2 1 a x + y 1 = 1x + y = xy + 1
Bi 5: Tm a h sau c ng mt nghim:
x2 + 3 + |y| = a
y2 + 5 + |x| =
x2 + 5 +
3
a
Bi 6: Cho h phng trnh (x + y)4 + 13 = 6x2y2 + m
xy
x2 + y2
= m
Tm m h c nghim duy nht.
Bi 7: Tm s thc m sao cho h:
x3 my3 = 1
2(m + 1)2
x3 + mx2y + xy2 = 1c nghim (x; y) tho x + y = 0
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PHNG PHP NG DNG HNH HC GII TCHV HNH HC PHNG
L thuyt
Trong mt phng ta Oxy, cho cc vector: u = (x1, y1), v = (x1, y2) khi ta c:
|u + v | |u | + |v |
(x1 + x2)2 + (y1 + y2)2
x21 + y21 +
x22 + y22
Du bng xy ra khi v ch khi hai vector u v v cng hng x1x2
=y1y2
= k 0.
Vi hai vector u , v bt k trong khng gian th
u .v = |u | . |v | . cos(u , v ) |u | . |v |
(nh l cosin trong tam gic) Cho a, b, c l ba cnh ca tam gic ABC v A l gc nh A ca tam gic khi :
a2 = b2 + c2 2bc cos A
Cho tam gic ABC c ba gc nhn v im M ty trong mt phng khi gi T lim nhn cc cnh BC, CA, AB di cng mt gc 1200 th vi mi im M trn mt
phng ta c:MA + MB + MC T A + T B + T C
T c gi l im Torricelli ca tam gic ABC.
Sau y l mt s dng ton thng gp:
Kho st h phng trnh cha dng tuyn tnh v phn tuyn tnh
Bi 1: Bin lun s nghim ca h
x2 + y2 = rax + by = c (r>0) Phn tch: H cho gm phng trnh ng trn (C) v phng trnh ng thngd. Nh vy ta cn kho st s giao im ca (C) v d. Cch 1: Lp cng thc tnh khong cch t gc to n d. Bin lun s giao im cang thng v ng trn bng cch so snh khong cch vi bn knh ca (C).
Cch 2: Tm di mt phng P hoc min gc Q nh nht cha (C). Bin lun v tr cang thng d i vi cc min phng . Nu pht hin ng thng i qua mt im trongng trn th h phng trnh lun c nghim phn bit.
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Bi 2: Bin lun s nghim ca h phng trnh
(x p)2 + (y q)2 = rax + by = c (r>0) Phn tch: t u = x p; v = y p a v bi ton 1.
Bi 3: Bin lun s nghim ca h phng trnh px2 + qy2 = r
ax + by = c(p, q, r > 0)
Phn tch: H cho gm phng trnh ng elip (E) v phng trnh ng thng d.Nh vy ta cn kho st s giao im ca d v (E). S dng php co - dn bin (E) thnhng trn (C) v bin d thnh d, ta a v kho st s im chung ca d v (C). Ta bitrng s im chung khng ln hn 2.
Phng php: t u = x; v =
q
py h cho tr thnh
()u
2 + v2 =r
p
au +
b
p
qv
= c
T y lm tip nh bi 1. Nu pht hin ng thng i qua mt im nm trong (E) th hlun c 2 nghim phn bit.
Bi 4: Bin lun s nghim ca h phng trnh
p(x k)2 + q(y h)2 = r
ax + by = c
(p, q, r > 0)
Phng php: t u = x k; v = y h a v bi ton 3.
Bi 5: Bin lun s nghim ca h phng trnh vi p, q,r > 0; k = 1:px2 + kxy + qy2 = rax + by = c Phng php: Bin i phng trnh u v dng mu2 + nv2 = r a v bi ton 4.
Bi 6: Kho st tnh cht nghim ca h
x2 + y2 = rax + by = c (r > 0) Phng php: Coi mi nghim ca h l mt im vi to l cp s . Bin i hphng trnh v iu kin ca h thnh ngha hnh hc. iu kin bi ton thng lin quann mt s tnh cht nh tp hp im thuc phn chung ca cc na mt phng hoc mintrn, min elip hoc khong cch gia 2 im, tch v hng ca 2 vect c lp ra t ccim , s o gc to bi 2 vect hoc 2 ng thng.
Bi 7: Kho st tnh cht nghim ca h phng trnh
px2 + qy2 = rax + by = c (p, q, r > 0)
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Phng php: t u = px; v = qy h cho tng ngu2 + v2 = rau + bv = c
Trong a =
a
p ; b =b
q. Bi ton a v bi 6.
a phng trnh v t mt n sang h phng trnh hai n
Bi 1: Kho st nghim ca phng trnh
a bx2 = kx + m vi a, b > 0
Phng php: t y = a
b x2 khi ta c h
()
x2 + y2 =
a
bkx by = my 0
Bi ton a v kho st nghim ca h ().
Bi 2: Kho st nghim ca phng trnh a b(x + c)2 = kx + m vi a, b > 0
Phng php: t z = x + c; y =
a2 + bz2 v a v bi 1.
Bi 3: Kho st nghim ca phng trnh a sin t + b cos t = c vi t [; ]
Phng php: t x = sin t; y = cos t ta c h
()
ax + by = c
x2 + y2 = 1
Bi ton a v kho st nghim ca h () vi x, y tho iu kin xc nh.
Bi 4: Kho st s nghim ca phng trnh p
a + bx + q
c + dx = m vi bd < 0; q = 1
Phng php: Coi b > 0 d < 0. t y =
a
b+ x; z =
cd
x ta c h
()
y2 + z2 =a
b
c
dpby + qdz = my 0
z 0
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Bi ton a v kho st s nghim ca h () gm cung trn AnB nm trong min gc P xcnh bi cc min x 0, y 0 v ng thng d.
Bi tp v d
Cc bi tp v h tuyn tnh v phn tuyn tnh
Bi 1: Bin lun s nghim ca h
(x 1)2 + (y 2)2 = 5x + my = m + 1
Gii
t u = x 1; v = y 2 ta c h cho tng ng:u2 + v2 = 5u + mv = m
(C)
O
(0,-1)
H trn gm phng trnh ng trn (C) : x2 + y2 = 5 tm l gc ta v bn knh l 5v phng trnh ng thng d : x + my + m = 0 ph thuc m. Hn na, ng thng d luni qua im c nh A(0; 1) nm trong (C) nn d lun ct (C) ti 2 im phn bit vi mim.2
Bi 2: Tm m h c nghim x > 0; y > 1:
()x2 + y2 = 42x y = mGii
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d3 d2 d1
= 1
x
1
n
A( 3 , 1)
M(m
2; 0)
M2 M1
B(0; 2)
O
iu kin ca bi ton xc nh min gc P vi x > 0; y > 1. () gm phng trnh ng trn(C) : x2 + y2 = 4 v phng trnh ng thng d : 2x y = m ph thuc m. Phn chung ca(C) v P l cung AnB c cc u mt A(
3;1) v B(0;2). Yu cu bi ton l tm m d v
cung AnB c t nht 1 im chung.
Phng trnh ng thng d1, d2 i qua A, B v cng phng vi d l 2x y = 23 1 v2x y = 2. Cc ng thng ct Ox ti M1(
3 1
2; 0) v M2(1;0). Cc im nm
trong (C) nn giao im th hai ca d1, d2 vi (C) thuc na mt phng P1 = {(x; y) : y < 0}.V vy d1, d2 ch c mt im chung vi AnB l A v B.
Gi Q l di mt phng bin d1, d2, khi Q cha AnB. ng thng d ct on thngM1M2 ti M(
m
2; 0) khc M1, M2.
T suy ra 2 < m < 23 1 l tp hp cc gi tr m tho gi thit. 2
Bi 3: Tm m h c 2 nghim phn bit tho x1x2 + y1y2 > 0:
()x2 + y2 = 4x + y = m
Gii
d
H
O
B
Gi s h c hai nghim phn bit l (x1, y1) v (x2, y2), khi ta c: Gi A(x1; y1), B(x2; y2),
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th do OA.
OB = OA.OB. cos
OA,
OB
= x1x2 + y1y2
Nn suy ra:cos
OA,OB
> 0 AOB < 2
Hn na do (x1, y1) v (x2, y2) u l nghim ca h nn cc im A v B u nm trn ngtrn (C) : x2 + y2 = 4 v ng thng d : x + y = m, nh vy AB l mt dy ca ng trn(C), ng thi AB cng chnh l ng thng d.Nh vy h c nghim tha iu kin x1x2 + y1y2 > 0 th AB phi ct ng trn (C) tihai im phn bit, hn na phi tha mn AOB
2 22 > |m| > 2 2
Bi 4: Tm m h c 2 nghim phn bit tho (x1 x2)2 + 4(y1 y2)2 = 3:
(
)
x2 + 4y2 = 16
x my = m
Gii
H () gm phng trnh elip (E) : x2
16+
y2
4= 1 v ng thng d : x my = m ph thuc m.
-1
O
Ta c ng thng d lun i qua im c nh A(0; -1) hn na A nm trong (E). Vy d lunct (E) ti 2 im phn bit, do h () c 2 nghim phn bit vi mi m.Gi B(x1; y1), D(x2; y2) v t u = x; v = 2y (1). Cng thc (1) xc nh mt php dn vi h
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s 2 bng trc dn Ox. Php dn bin cc im B, D thnh B(u1; v1), D(u2; v2) v h ()tr thnh:
()u2 + v2 = 162u mv = 2m
To cc im B, D l nghim ca (). H gm phng trnh ng trn (C) v phng
trnh ng thng d
l nh ca d qua php dn .
Q'
P'
-2
O
Mt khcBD2 = (u1u2)
2 + (v1 v2)2 = (x1 x2)2 + 4(y1 y2)2 = 3Ngha l d ct (C) ti 2 im B, D sao cho BD =
3 hn na (C) c tm l gc ta nn
d2O/d +BD2
4 = R2 = 16 Suy ra dO/d
=
61
2 , do d = |2m
|4 + m2 nn tm c m = 1223 2Bi 5: Tm m h c 2 nghim phn bit v (x1 x2)2 + (y1 y2)2 t gi tr nh nht:
()x2 + y2 = 42x + my = m + 2
Gii
(C)
d
BM
O
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H () gm phng trnh ng trn (C) : x2 + y2 = 4 v ng thng d ph thuc vom. Gi A, B l giao im ca d v (C) th ta cn tm m AB2 t gi tr nh nht. DoM(1; 1) d m v M nm trong (C) nn d lun ct (C) ti 2 im phn bit A, B.Ta c: AB Min d OM ().Gi u = (m; 2) l vect ch phng ca d th
() u .OM = 0 2 m = 0 m = 2Do vi m = 2 th h c hai nghim phn bit tha (x1x2)2 +(y1y2)2 t gi tr nh nht.
Bi 6. Bin lun s nghim ca h phng trnh sau:|x| + 2 |y| = 4(x 2a) (y a) = 0Gii
Xt h phng trnh: |x| + 2 |y| = 4 (1)(x 2a) (ya) = 0 (2)Ta thy rng cc im tha mn (1) l bn cnh ca hnh thoi ABCD trong :
A((4;0); B (0;2); C(4;0); D (0; 2)ng thi, cc im tha mn (2) nm trn hai ng thng d1 : x = 2a v d2 : y = a. Snghim ca h hai phng trnh (1) v (2) chnh l s giao im ca bn cnh hnh thoi vi
hai ng thng ni trn.Trc tin ta tm xem khi no 3 ng thng x = 2a; y = a v |x| + 2 |y| = 4 ng quy. Gi(x0; y0) l im chung ca ba ng thng ny, khi ta c h sau:
x0 = 2a
y0 = a
|x0| + 2 |y| = 4
x0 = 2a
y0 = a
4 |a| = 4
|a
|= 1
x0 = 2; y0 = 1
x0 = 2; y0 = 1
-4
-2
4
2
y = 1
x = 2
D
B
CO
y = a
x = 2a
-4
-2
4
2
D
B
CO
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Da vo th trn ta c kt lun sau:
Nu |a| > 2: H (1), (2) v nghim.
Nu |a| = 2: H (1), (2) c hai nghim.
Nu |a| < 2 v |a| = 1: H (1), (2) c 4 nghim.
Nu |a| = 1: H (1), (2) C 3 nghim.
a phng trnh v t v h phng trnh
Bi 7: Bin lun s nghim ca phng trnh
m9 x2 x + 5m = 0 (1)
Gii
t y =
5 x2 ta c h
()
x2 + y2 = 9
x my = 5my 0
H () gm phng trnh ng thng d ph thuc m v na ng trn (C) xc nh bi hbt phng trnh x2 + y2 = 9y 0Ta c A(0; 5) d m do A nm ngoi (C)
(C)
d
-3 35m
M2 M1OM
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Cc im M1(3; 0), M2(3;0) l u mt ca ng knh ca (C). Cc tia AM1, AM2 l binca min gc P. ng thng d ct M1M2 ti M(5m; 0) nn vi m tho 3 5m 3 3
5
m 3
5th d v na trn ca (C) c duy nht 1 giao im. Trng hp ny phng trnh c 1
nghim.Vi cc gi tr cn li ca m th d v na trn ca (C) khng c giao im nn phng trnh
v nghim. 2
Bi 8: Bin lun theo m nghim ca phng trnh sau:
4 x2 = mx + 2 m ()
Gii
Ta bit rng s nghim ca phng trnh () l s giao im ca hai ng y = mx + 2 m
v y = 4 x2.Hn na v y =
4 x2 x2 + y2 = 4, y 0 nn th ca y = 4 x2 l na ng trn
tm l gc ta bn knh bng hai v nm pha trn trc honh.Cn y = mx + 2 m l mt h ng thng lun i qua im c nh A(1; 2). Ta nhn thyc hai tip tuyn vi ng (C) : y =
4 x2, mt l ng thng y = 3 song song vi trc
honh, v tip tuyn AD.
2
-2 21O
B C
E
Gi B(2, 0) v C(2, 0) l hai u mt ca ng knh BOC, gi s m1, m2, m3, m4 tngng l h s gc ca cc ng thng AC, AD, AB, AE th ta c cc iu sau:
m1 = tan ACO = 2.
m2 =
tanDCO =
tanEAD =
tan2OAE =
4
3
.
m3 = tan ABO = 23
.
m4 = 0.
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T suy ra:
1. Phng trnh () c hai nghim
0 < m
2
3
2 m
4
3
2. Phng trnh () c mt nghim
m >
2
3m < 2m = 0
m = 43
3. Phng trnh () v nghim 4
3 < m < 0 2
Bi 9: Bin lun s nghim ca phng trnh sau theo a:a
9 x2 + x
x a
3
= 0
Gii
t y =
9 x2 x2 + y2 = 9Phng trnh (4) tng ng:
x2 + y2 = 9, y 0
(ay + x)
x a3 = 0D thy phng trnh th nht biu din phn pha trn trc honh ca ng trn tm l gcta bn knh l 3 cn phng trnh th hai biu din hai ng thng x = a3 v y = 1
a.x
(nu a = 0).S nghim ca h chnh l s giao im ca hai ng thng vi na ng trn. Ta ch cnxt khi a > 0 (v khi a < 0 ta c kt qu tng t v khi a = 0 th (4) x = 0 v lc hc 1 nghim).
Thy rng ng thng y = 1a
x nn n i qua O v v vy lun ct na ng trn ti mt
im. Ta ch cn quan tm n v tr tng i ca x = a
3 vi ng trn v ng thngy =
1
ax.
Do y 0 nn nu ng trn v hai ng thng ng quy th phi c x = a3 > 0 vy = 1
ax < 0. iu ny l v l. Do s khng c trng hp ba ng ny ng quy. Nh
th nhn vo th ta s thu c kt lun:
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x = a 3
a > 3
y = -1
ax
3
O
x = a 3
0 < a < 3
y = -1
ax
3
O
Do ta c kt lun sau:
Phng trnh (4) c mt nghim
a = 0
|a| > 3
Phng trnh (4) c hai nghim 0 < |a| < 3 2
Bi 10: Bin lun s nghim ca phng trnh:
9 2x x2 = x + m (()
Gii
Quan st mt cht ta thy bi ny c cu trc hi khc cc bi trn nhng tht ra chng nhnhau:
()
10 (x + 1)2 = x + mt z = x + 1; y =
10 z2 ta c h:
y2 + z2 = 10
y
z = m
1
y 0
Bi ton a v bin lun s im chung ca ng thng y = z+ m 1 v na ng trny2 + z2 = 10y 0n y tng t nh Bi 1, xin dnh cho bn c gii quyt tip phn cn li ca bi ton.
Bi 11: Bin lun s nghim ca phng trnh
2sin t (m + 3) cos t = m 1, t
3;
2
3
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Gii
t x = sin t; y = cos t v a iu kin ca bi ton thnh:
x2 + y2 = 1 (i)
2x (m + 3)y = m 1 (ii)3
2 x 1 (iii)
12 y
1
2(iv)
Do hai iu kin (iii) v (iv) nn ta ch c