Upload
truong-tan-dat
View
469
Download
0
Embed Size (px)
DESCRIPTION
Tài liệu rất hay và đầy đủ về các phương pháp giải hệ phương trình thông qua giải chi tiết 410 bài hệ.
Citation preview
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
Nguyn Minh TunSinh vin K62CLC - Khoa Ton Tin HSPHN
TUYN CHN 410 BI HPHNG TRNH I S
BI DNG HC SINH GII V LUYN THI I HC - CAONG
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
H Ni, ngy 9 thng 10 nm 2013
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
nMc lc
Li ni u 4
1 Mt s phng php v cc loi h c bn 5
1.1 Cc phng php chnh gii h phng trnh . . . . . . . . . . . . . . . . . . 51.2 Mt s loi h c bn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Tuyn tp nhng bi h c sc 7
2.1 Cu 1 n cu 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Cu 31 n cu 60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.3 Cu 61 n cu 90 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.4 Cu 91 n cu 120 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2.5 Cu 121 n cu 150 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
2.6 Cu 151 n cu 180 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
2.7 Cu 181 n cu 210 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
2.8 Cu 211 n cu 240 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.9 Cu 241 n cu 270 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
2.10 Cu 271 n cu 300 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
2.11 Cu 301 n cu 330 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
2.12 Cu 331 n cu 360 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
2.13 Cu 361 n cu 390 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
2.14 Cu 391 n cu 410 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
Ti liu tham kho 228
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
nLi ni u
H phng trnh i s ni chung v h phng trnh i s hai n ni ring l mt phnquan trng ca phn i s ging dy THPT . N thng hay xut hin trong cc k thi hcsinh gii v k thi tuyn sinh i hc - Cao ng.
Tt nhin gii tt h phng trnh hai n khng phi n gin . Cn phi vn dng ttcc phng php, hnh thnh cc k nng trong qu trnh lm bi. Trong cc k thi i hc, cu
h thng l cu ly im 8 hoc 9.
y l mt ti liu tuyn tp nhng kh dy nn ti trnh by n di dng mt cun schc mc lc r rng cho bn c d tra cu. Cun sch l tuyn tp khong 400 cu h c sc,t n gin, bnh thng, kh, thm ch n nh v kinh in. c bit, y hon ton lh i s 2 n. Ti mun khai thc tht su mt kha cnh ca i s. Nu coi Bt ng thc3 bin l phn p nht ca Bt ng thc, mang trong mnh s uy nghi ca mt ng hong thH phng trnh i s 2 n li mang trong mnh v p gin d, trong sng ca c gi thnqu lm say m bit bao g si tnh.
Xin cm n cc bn, anh, ch, thy c trn cc din n ton, trn facebook ng gp vcung cp rt nhiu bi h hay. Trong cun sch ngoi vic a ra cc bi h ti cn lng thmmt s phng php rt tt gii. Ngoi ra ti cn gii thiu cho cc bn nhng phng phpc sc ca cc tc gi khc . Mong y s l mt ngun cung cp tt nhng bi h hay chogio vin v hc sinh.
Trong qu trnh bin son cun sch tt nhin khng trnh khi sai st.Th nht, kh nhiubi ton ti khng th nu r ngun gc v tc gi ca n. Th hai : mt s li ny sinh trongqu trnh bin son, c th do li nh my, cch lm cha chun, hoc trnh by cha p dokin thc v LATEX cn hn ch. Tc gi xin bn c lng th. Mong rng cun sch s honchnh v thm phn s. Mi kin ng gp v sa i xin gi v theo a ch sau y :
Nguyn Minh TunSinh Vin Lp K62CLC
Khoa Ton Tin Trng HSP H NiFacebook :https://www.facebook.com/popeye.nguyen.5
S in thoi : 01687773876Nick k2pi, BoxMath : Popeye
http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.5http://%20https//www.facebook.com/popeye.nguyen.55/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
nChng 1Mt s phng php v cc loi h cbn
1.1 Cc phng php chnh gii h phng trnhI. Rt x theo y hoc ngc li t mt phng trnh
II. Phng php th1. Th hng s t mt phng trnh vo phng trnh cn li2. Th mt biu thc t mt phng trnh vo phng trnh cn li3. S dng php th i vi c 2 phng trnh hoc th nhiu ln.
III. Phng php h s bt nh1. Cng tr 2 phng trnh cho nhau2. Nhn hng s vo cc phng trnh ri em cng tr cho nhau.3. Nhn cc biu thc ca bin vo cc phng trnh ri cng tr cho nhau
IV. Phng php t n ph
V. Phng php s dng tnh n iu ca hm s
VI. Phng php lng gic ha
VII. Phng php nhn chia cc phng trnh cho nhau
VIII. Phng php nh gi1. Bin i v tng cc i lng khng m2. nh gi s rng buc tri ngc ca n, ca biu thc, ca mt phng trnh3. nh gi da vo tam thc bc 24. S dng cc bt ng thc thng dng nh gi
IX. Phng php phc ha
X. Kt hp cc phng php trn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
6 Mt s phng php v cc loi h c bn
1.2 Mt s loi h c bnA. H phng trnh bc nht 2 n
I. Dng ax + by= c (a2 + b2 = 0)
ax + by=c (a2
+ b2
= 0)II. Cch gii1. Th2. Cng i s3. Dng th4. Phng php nh thc cp 2
B. H phng trnh gm mt phng trnh bc nht v mt phng trnh bc hai
I. Dng ax2 + by2 + cxy+ dx + ey+ f= 0
ax + by=c
II. Cch gii:Th t phng trnh bc nht vo phng trnh bc hai
C. H phng trnh i xng loi II. Du hiui vai tr ca xvycho nhau th h cho khng iII. Cch gii:Thng ta s t n ph tng tch x + y = S, xy=P (S2 4P)
D. H phng trnh i xng loi II
I. Du hiui vai tr ca xvycho nhau th phng trnh ny bin thnh phng trnh kiaII. Cch gii:Thng ta s tr hai phng trnh cho nhau
E. H ng cpI. Du hiu
ng cp bc 2
ax2 + bxy+ cy2 =d
ax2 + bxy+ cy2 =d
ng cp bc 3ax3 + bx2y+ cxy2 + dy3 =e
ax3 + bx2y+ cxy2 +dy3 =e
II. Cch gii:Thng ta s t x= ty hoc y = tx
Ngoi ra cn mt loi h na ti tm gi n l bn ng cp, tc l hon ton c th av dng ng cp c .Loi h ny khng kh lm, nhng nhn nhn ra c n cn phikho lo sp xp cc hng t ca phng trnh li. Ti ly mt v d n gin cho bn c
Gii h :
x3 y3 = 8x + 2yx2 3y2 = 6
Vi h ny ta ch vic nhn cho v vi v s to thnh ng cp. V khi ta c quynchn la gia chia c 2 v cho y3 hoc t x= ty
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
nChng 2Tuyn tp nhng bi h c sc
2.1 Cu 1 n cu 30
Cu 1
(x y) (x2 + y2) = 13(x + y) (x2 y2) = 25
Gii
D dng nhn thy y l mt h ng cp bc 3, bnh thng ta c nhn cho ln ri chia 2v cho x3 hoc y3. Nhng hy xem mt cch gii tinh t sau y:Ly(2) (1)ta c : 2xy(x y) = 12 (3)Ly(1) (3)ta c : (x y)3 = 1 x= y + 1V sao c th c hng ny ? Xin tha l da vo hnh thc i xng ca h. Ngon lnh
ri. Thay vo phng trnh u ta c
(y+ 1)2 + y2 = 13
y = 2y = 3
Vy h cho c nghim (x; y) = (3; 2), (2;3)
Cu 2 x3 8x= y3 + 2y
x
2
3 = 3 (y2
+ 1)
Gii
nh sau : Phng trnh 1 gm bc ba v bc nht. Phng trnh 2 gm bc 2 v bc 0(hng s).R rng y l mt h dng na ng cp. Ta s vit li n a v ng cpH cho tng ng :
x3 y3 = 8x + 2yx2
3y2 = 6
Gi ta nhn cho hai v a n v dng ng cp
6 x3 y3 = (8x + 2y) x2 3y2 2x (3y x) (4y+ x) = 0
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
8 Tuyn tp nhng bi h c sc
TH1 : x= 0thay vo (2) v nghimTH2 : x= 3ythay vo (2) ta c:
6y2
= 6 y= 1, x= 3y= 1, x= 3TH3 : x= 4ythay vo (2) ta c:
13y2 = 6
y=
6
13, x= 4
6
13
y=
6
13, x= 4
6
13
Vy h cho c nghim :(x; y) = (3; 1), (3;1),4
6
13 ;6
13
,
46
13 ;6
13
Cu 3
x2 + y2 3x + 4y= 13x2 2y2 9x 8y = 3
Gii
khi nhn 3 vo PT(1) ri tr i PT(2) s ch cn y . Vy
3.P T(1) P T(2) y2 + 4y= 0
y= 0 x=
372
y= 4 x= 3
7
2
Vy h cho c nghim : (x; y) =
3 7
2 ; 0
,
37
2 ;4
Cu 4
x2 + xy+y2 = 19(x y)2x2 xy+y2 = 7 (x y)
Gii
Nhn xt v tri ang c dng bnh phng thiu, vy ta th thm bt a v dng bnhphng xem sao. Nn a v (x y)2 hay (x+ y)2. Hin nhin khi nhn sang v phi ta schn phng n u
H cho tng ng
(x y)2 + 3xy= 19(x y)2(x y)2 + xy= 7 (x y)
t x y = a vxy= b ta c h mi
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 9
b= 6a2
a2 + b= 7a
a= 0, b= 0a= 1, b= 6
x y = 0xy= 0
x y = 1xy= 6
x= 0, y= 0x= 3, y= 2x= 2, y = 3
Vy h cho c nghim :(x; y) = (0; 0) , (3;2)(2;3)
Cu 5
x3 + x3y3 + y3 = 17x +xy+ y= 5
GiiH i xng loi I ri. No problem!!!
H cho tng ng (x + y)3 3xy(x + y) + (xy)
3 = 17(x + y) +xy = 5
t x + y= avxy= bta c h mi
a3 3ab + b3 = 17a +b= 5
a= 2, b= 3a= 3, b= 2
x + y= 2xy= 3
x + y= 3xy= 2
x= 2, y= 1x= 1, y= 2
Vy h cho c nghim (x; y) = (1; 2), (2; 1)
Cu 6
x(x + 2)(2x + y) = 9x2 + 4x + y = 6
Giiy l loi h t n tng tch rt quen thuc
H cho tng ng
(x2 + 2x) (2x +y) = 9(x2 + 2x) + (2x + y) = 6
t x2 + 2x= av2x + y = b ta c h mi ab= 9a + b= 6
a= b = 3
x2 + 2x= 32x + y = 3
x= 1, y = 1x= 3, y= 9
Vy h cho c nghim (x; y) = (1; 1), (3;9)
Cu 7
x +y xy= 3x + 1 +
y+ 1 = 4
GiiKhng lm n g c c 2 phng trnh, trc gic u tin ca ta l bnh phng ph skh chu ca cn thc
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
10 Tuyn tp nhng bi h c sc
(2) x + y+ 2 + 2
xy+ x + y+ 1 = 16
M t (1) ta c x + y= 3 + xynn
(2) 3 + xy+ 2 + 2xy+ xy+ 4 = 16 xy= 3 xy= 9x +y = 6 x= y = 3Vy h cho c nghim (x; y) = (3; 3)
Cu 8
x + 5 +
y 2 = 7x 2 + y+ 5 = 7
Giii xng loi II. Khng cn g ni. Cho 2 phng trnh bng nhau ri bnh phng tungte ph s kh chu ca cn thciu kin : x, y 2T 2 phng trnh ta c
x + 5 +
y 2 = x 2 +
y 5
x + y+ 3 + 2
(x + 5)(y 2) =x + y+ 3 + 2
(x 2)(y+ 5)
(x + 5)(y
2) = (x
2)(y+ 5)
x= y
Thay li ta c
x + 5 +
x 2 = 7 x= 11
Vy h cho c nghim : (x; y) = (11;11)
Cu 9 x2 + y2 +
2xy= 8
2
x +y = 4
Gii
H cho c v l na i xng na ng cp, bc ca PT(2) ang nh hn PT(1) mtcht. Ch cn php bin i bnh phng (2) s va bin h tr thnh ng cp va ph bbt i cniu kin : x, y 0H cho
2(x2 + y2) + 2xy= 16x +y+ 2xy= 16 2 (x2 + y2) =x + y x= yThay li ta c : 2
x= 4 x= 4
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 11
Vy h cho c nghim (x; y) = (4; 4)
Cu 10
6x
2
3xy+x= 1 yx2 + y2 = 1
Gii
Mt cch trc gic khi nhn thy h cha tam thc bc 2 l th xem liu c phn tch cthnh nhn t hay khng ? Ta s th bng cch tnh theo mt n c chnh phng haykhng. Ngon lnh l PT(1) xp nh tin.Phng trnh u tng ng (3x 1)(2x y+ 1) = 0Vi x=
1
3 y= 2
2
3
Vi y= 2x + 1 x2 + (2x + 1)2 = 1
x= 0, y = 1
x= 45
, y=3
5
Vy h cho c nghim (x; y) =
1
3;2
2
3
, (0, 1),
4
5;3
5
Cu 11 x 2y xy= 0
x
1 +
4y
1 = 2
Gii
Phng trnh u l dng ng cp riiu kin x 1, y1
4T phng trnh u ta c :
x +
y
x 2y = 0 x= 4yThay vo (2) ta c
x 1 + x 1 = 2 x= 2
Vy h cho c nghim (x
;y
) =
2;
1
2
Cu 12
xy+ x +y = x2 2y2x
2y yx 1 = 2x 2y
Gii
iu kin : x
1, y
0
Phng trnh u tng ng
(x + y) (2y x + 1) = 0
x= yx= 2y+ 1
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
12 Tuyn tp nhng bi h c sc
Vi x= yloi v theo iu kin th x, yphi cng duVi x= 2y+ 1 th phng trnh 2 s tng ng
(2y+ 1)
2y y
2y = 2y+ 2
2y(y+ 1) = 2y+ 2 y= 2 x= 5
Vy h cho c nghim (x; y) = (5; 2)
Cu 13
x + 1 +
y+ 2 = 6x +y = 17
Gii
iu kin x, y 1H cho tng ng
x + 1 + y+ 2 = 6(x + 1) + (y+ 2) = 20t
x + 1 =a 0,y+ 2 =b 0. H cho tng ng
a + b= 6a2 +b2 = 20
a= 4, b= 2a= 2, b= 4
x= 15, y = 2x= 3, y= 14
Vy h cho c nghim (x; y) = (15; 2), (3;14)
Cu 14
y2 = (5x + 4)(4 x)y2 5x2 4xy+ 16x 8y+ 16 = 0
Gii
Phng trnh 2 tng ng
y2 + (5x + 4)(4 x) 4xy 8y= 0 2y2 4xy 8y = 0
y= 0y= 2x + 4
Vi y= 0th suy ra : (5x + 4) (4 x) = 0 x= 4
x= 45
Vi y= 2x + 4th suy ra (2x + 4)2 = (5x + 4)(4 x) x= 0Vy h cho c nghim (x; y) = (4; 0),
4
5; 0
, (0; 4)
Cu 15
x2 2xy+x +y = 0x4 4x2y+ 3x2 + y2 = 0
Gii
H cho tng ng
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 13
x2 + y=x(2y 1)(x2 + y)
2+ 3x2 (1 2y) = 0 x
2(2y 1)2 + 3x2(2y 1) = 0 x2(2y 1)(2y 4) = 0
x= 0, y = 0
y=1
2(L)
y= 2, x= 1 2Vy h cho c nghim (x; y) = (0; 0), (1; 2), (2; 2)
Cu 16 x + y+ xy(2x +y) = 5xyx + y+ xy(3x
y) = 4xy
Gii
P T(1) P T(2) xy(2y x) =xy
xy= 0x= 2y 1
Vi xy= 0 x + y= 0 x= y = 0Vi x= 2y 1
(2y 1) +y+ (2y 1)y(5y 2) = 5(2y 1)y y= 1, x= 1
y=
9
41
20 , x= 1 +
41
10
y=9 +
41
20 , x=
41 1
10
Vy h cho c nghim(x; y) = (0; 0), (1; 1),
1 +
41
10 ;
9 4120
,
41 1
10 ;
9 +
41
20
Cu 17 x2 xy+y2 = 32x3 9y3 = (x y)(2xy+ 3)
Gii
Nu ch xt tng phng trnh mt s khng lm n c g. Nhng 2 ngi ny b rngbuc vi nhau bi con s 3 b n. Php th chng ? ng vy, thay 3 xung di ta s ra mtphng trnh ng cp v kt qu p hn c mong iTh 3 t trn xung di ta c
2x3
9y3 = (x
y) x2 +xy+ y2 x3 = 8y3 x= 2y
(1) 3y2 = 3 y = 1, x= 2Vy h cho c nghim (x; y) = (2; 1), (2;1)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
14 Tuyn tp nhng bi h c sc
Cu 18
x + y+
x y = 1 +
x2 y2x +
y = 1
Gii
iu kin :x y 0Phng trnh u tng ng
x +y 1 = x y x + y 1 x + y= 1
x y = 1
x=
1 yx=
1 +y
T
1 y+ y= 1y+ 1 +
y= 1
y= 0, x= 1y= 1, x= 0(L)
y= 0, x= 1
Vy h cho c nghim (x; y) = (1; 0)
Cu 19
2x y = 1 +x(y+ 1)x3 y2 = 7
Gii
iu kin : x(y+ 1) 0T (2) d thy x >0 y 1(1)
x
y+ 1 2x +y+ 1 = 0 x= y + 1 (y+ 1)3 y2 = 7 y = 1, x= 2Vy h cho c nghim (x; y) = (2; 1)
T cu 20 tr i ti xin gii thiu cho cc bn mt phng php rt mnh gii quyt gn p rt nhiu cc h phng trnh hu t. gi h s bt nh(trong y ti s gi n bng tn khc : UCT). S mt khong hn chc v d din t trn vn phng php ny
Trc ht im qua mt mo phn tch nhn t ca a thc hai bin rt nhanh bng mytnh Casio. Bi vit ca tc gi nthoangcute.
V d 1 : A= x2 + xy 2y2 + 3x + 36y 130Thc ra y l tam thc bc 2 th c th tnh phn tch cng c. Nhng th phn tchbng Casio xem .Nhn thy bc ca x v y u bng 2 nn ta chn ci no cng cCho y = 1000ta c A= x2 + 1003x 1964130 = (x + 1990) (x 987)Cho 1990 = 2y 10 v 987 = y 13A= (x + 2y
10)(x
y+ 13)
V d 2 : B= 6x2y 13xy2 + 2y3 18x2 + 10xy 3y2 + 87x 14y+ 15
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 15
Nhn thy bc ca x nh hn, cho ngay y = 1000B= 5982x2 12989913x + 1996986015 = 2991 (2x 333)(x 2005)Cho 2991 = 3y 9 ,333 =
y 13
, 2005 = 2y + 5
B= (3y 9)2x y
1
3
(x 2y 5) = (y 3)(6x y+ 1) (x 2y 5)
V d 3 : C=x3 3xy2 2y3 7x2 + 10xy+ 17y2 + 8x 40y+ 16Bc ca xvynh nhauCho y = 1000ta c C=x3 7x2 2989992x 1983039984Phn tch C= (x 1999)(x + 996)2Cho 1999 = 2y 1v996 =y 4C= (x 2y+ 1) (x + y 4)2
V d 4 : D= 2x2y2 + x3 + 2y3 + 4x2 +xy+ 6y2 + 3x + 4y+ 12Bc ca xvynh nhauCho y = 1000ta c D= (x + 2000004) (x2 + 1003)Cho 2000004 = 2y2 + 4v1003 =y + 3D= (x + 2y2 + 4) (x2 +y+ 3)
V d 5 : E=x3y+ 2x2y2 + 6x3 + 11x2y xy2 6x2 7xy y2 6x 5y+ 6Bc ca ynh hn
Cho x = 1000 ta c E = 1998999y2
+ 1010992995y + 5993994006 =2997 (667y+ 333333) (y+ 6)
o ha E=999 (2001y+ 999999) (y+ 6)Cho 999 =x 1, 2001 = 2y+ 1, 999999 =x2 1E= (x 1) (y+ 6) (x2 + 2xy+y 1)
V d 6 : F = 6x4y+ 12x3y2 + 5x3y 5x2y2 + 6xy3 +x3 + 7x2y+ 4xy2 3y3 2x2 8xy+3y2 2x + 3y 3Bc ca y nh hnCho x= 1000ta c F= 5997y3 + 11995004003y2 + 6005006992003y+ 997997997Phn tch F= (1999y+ 1001001) (3y2 + 5999000y+ 997)Cho 1999 = 2x 1, 1001001 =x2 + x + 1, 5999000 = 6x2 x, 997 =x 3F = (x2 + 2xy+ x y+ 1) (6x2y xy+ 3y2 +x 3)
Lm quen c ri ch ? Bt u no
Cu 20
x2 +y2 =1
5
4x2 + 3x5725
= y(3x + 1)
Gii
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
16 Tuyn tp nhng bi h c sc
Li gii gn p nht ca bi trn l
25.P T(1) + 50.P T(2) (15x + 5y 7)(15x + 5y+ 17) = 0
n y d dng tm c nghim ca h : (x; y) = 25
;1
5 ,11
25
; 2
25
Cu 21
14x2 21y2 6x + 45y 14 = 035x2 + 28y2 + 41x 122y+ 56 = 0
Gii
Li gii gn p nht ca bi ny l
49.P T(1) 15.P T(2) (161x 483y+ 218)(x + 3y 7) = 0V n y cng d dng tm ra nghim (x; y) = (2;3), (1; 2)
Qua 2 v d trn ta t ra cu hi : V sao li th ? Ci nhm thnh nhn t th ti khngni bi t hn cc bn c n trn ri. V sao y l ti sao li ngh ra nhng hng skia nhn vo cc phng trnh, mt s tnh c may mn hay l c mt phng php. Xin tha chnh l mt v d ca UCT. UCT l mt cng c rt mnh c th qut sch gn nh tonb nhng bi h dng l hai tam thc. Cch tm nhng hng s nh th no. Ti xin trnhby ngay sau y. Bi vit ca tc gi nthoangcute.
Tng Qut:
a1x2 +b1y
2 + c1xy+ d1x + e1y+ f1 = 0a2x
2 +b2y2 + c2xy+ d2x + e2y+ f2 = 0
Gii
Hin nhin nhn xt y l h gm hai tam thc bc hai. M nhc n tam thc th khngth khng nhc ti mt i tng l . Mt tam thc phn tch c nhn t hay khngphi xem x hoc y ca n c chnh phng hay khng. Nu h loi ny m t ngay mtphng trnh ra k diu th chng ni lm g, th nhng c hai phng trnh u ra rtk cc th ta s lm nh no. Khi UCT s ln ting. Ta s chn hng s thch hp nhn vomt (hoc c hai phng trnh) p sao cho chnh phng.
Nh vy phi tm hng s k sao cho P T(1) +k.P T(2)c th phn tch thnh nhn tt a= a1+ ka2, b= b1+ kb2, c= c1+kc2, d= d1+kd2, e= e1+ke2, f=f1+ kf2S kl nghim ca phng trnh sau vi a = 0
cde+ 4abf=ae2 +bd2 +f c2
D vng c hn mt cng thc gii h phng trnh loi ny. Tc gi ca n kh xutsc !!!. Th kim chng li v d 21 nha= 14 + 35k, b=
21 + 28k, c= 0, d=
6 + 41k, e= 45
122k, f=
14 + 56k
S ks l nghim ca phng trnh
4(14+35k)(21+28k)(14+56k) = (14+35k)(45122k)2+(21+28k)(6+41k)2 k= 1549
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 17
Nh vy l P T(1) 1549
.P T(2)hay 49.P T(1) 15.P T(2)Mt cht lu l khng phi h no cng y cc hng s. Nu khuyt thiu phn no thcho hng s l 0. Ok!!Xong dng ny ri. Hy lm bi tp vn dng. y l nhng bi h ti tng hp t nhiu
ngun.
1.
x2 + 8y2 6xy+ x 3y 624 = 021x2 24y2 30xy 83x + 49y+ 585 = 0
2.
x2 + y2 3x + 4y= 13x2 2y2 9x 8y= 3
3.
y2 = (4x + 4)(4 x)y2 5x2 4xy+ 16x 8y+ 16 = 0
4. xy 3x 2y= 16x2
+ y2
2x 4y= 335.
x2 + xy+ y2 = 3x2 + 2xy 7x 5y+ 9 = 0
6.
(2x + 1)2 + y2 + y= 2x + 3xy+x= 1
7.
x2 + 2y2 = 2y 2xy+ 13x2 + 2xy y2 = 2x y+ 5
8.
(x 1)2 + 6(x 1)y+ 4y2 = 20x2 + (2y+ 1)2 = 2
9. 2x2 + 4xy+ 2y2 + 3x + 3y 2 = 0
x2 + y2 + 4xy+ 2y= 0
10.
2x2 + 3xy= 3y 133y2 + 2xy= 2x + 11
11.
4x2 + 3y(x 1) = 73y2 + 4x(y 1) = 3
12.
x2 + 2 =x(y 1)y2 7 =y(x 1)
13.
x2 + 2xy+ 2y2 + 3x= 0xy+ y2 + 3y+ 1 = 0
Cu 22
x3 y3 = 352x2 + 3y2 = 4x 9y
Gii
Li gii ngn gn cho bi ton trn l
P T(1) 3.P T(2) (x 2)3 = (y+ 3)3 x= y + 5
Thay vo (2) ta d dng tm ra nghim(x; y) = (2;3), (3;2)Cu hi t ra y l s dng UCT nh th no ? Tt nhin y khng phi dng trn nari. Trc ht nh gi ci h ny
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
18 Tuyn tp nhng bi h c sc
- Bc ca xvyl nh nhau- Cc bin x,yc lp vi nhau- Phng trnh mt c bc cao hn PT(2)Nhng nhn xt trn a ta n tng nhn hng s vo PT(2) P T(1) +a.P T(2)ac v dng hng ng thc A3 =B3P T(1) +a.P T(2) x3 + 2ax2 4ax y3 + 3ay2 + 9ay 35 = 0Cn tm asao cho v tri c dng (x + )3 (y+ )3 = 0
Cn bng ta c :
3 3 = 353= 2a32 = 4a
a= 3= 2= 3
VyP T(1) 3.P T(2) (x 2)3 = (y+ 3)3OK ?? Th mt v d tng t nh
Gii h: x3 + y3 = 91
4x2
+ 3y2
= 16x
+ 9y
Gi : P T(1) 3.P T(2) (x 4)3 = (y+ 3)3
Cu 23
x3 + y2 = (x y)(xy 1)x3 x2 + y+ 1 =xy(x y+ 1)
Gii
Hy cng ti phn tch bi ton ny. Tip tc s dng UCTnh gi h :-Bc ca x cao hn bc ca y-Cc bin x,y khng c lp vi nhau-Hai phng trnh c bc cao nht ca x v y nh nhauV bc x ang cao hn bc y v bc ca y ti 2 phng trnh nh nhau nn ta hy nhn tungri vit li 2 phng trnh theo n y. C th nh sau :
y2 (x + 1) y (x2 + 1) +x3 + x= 0y2x y (x2 +x 1) +x3 x2 + 1 = 0
By gi ta mong c rng khi thay x bng 1 s no vo h ny th s thu c 2 phngtrnh tng ng. Tc l khi cc h s ca 2 phng trnh s t l vi nhau . Vy :
x + 1
x =
x2 + 1
x2 + x 1= x3 + x
x3 x2 + 1 x= 1
Rt may mn ta tm c x = 1. Thay x = 1 li h ta c 2 (y2 y+ 1) = 0y2 y+ 1 = 0 2.P T(2) P T(1)s c nhn tx 1
C th l (x 1) (y2 (x + 3) y+x2 x 2) = 0TH1 :x= 1thay vo th v nghim
TH2: Kt hp thm vi PT(1) ta c h mi : y2 (x + 3) y+x2 x 2 = 0 (3)x3 +y2 x2y+ x + xy2 y = 0
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 19
Nhn xt h ny c c im ging vi h ban u l bc y nh nhau. Vy ta li vit li htheo n y v hi vng n s li ng vi x no . Tht vy, l x =1
2. Tip tc thay n
vo h v ta s rt ra :
2P T(2) P T(1) (2x + 1) y2 (x 1) y+ x2 x + 2TH1 : x= 1
2 y=5 3
5
4TH2 : Kt hp vi (3) ta c
y2 (x 1) y+x2 x + 2 = 0y2 (x + 3) +x2 x 2 = 0
Vi h ny ta ch vic tr cho nhau s ra y= 1 x2 + 2 = 0(V nghim)Vy h cho c nghim :(x; y) =
1
2
;5 + 3
5
4 ,
1
2
;5 35
4
Cu 24
2 (x +y)(25 xy) = 4x2 + 17y2 + 105x2 + y2 + 2x 2y= 7
Gii
Hnh thc bi h c v kh ging vi cu 23
Mt cht nh gi v h ny- Cc bin x v y khng c lp vi nhau- Bc cao nht ca x 2 phng trnh nh nhau , y cng vyVi cc c im ny ta th vit h thnh 2 phng trnh theo n x v y v xem liu h cng vi x hoc y no khng. Cch lm vn nh cu 23. Vit theo x ta s khng tm c y,nhng vit theo y ta s tm c x = 2 khin h lun ng. Thay x = 2 vo h ta c
21y2 42y+ 21 = 0y2 2y+ 1 = 0 P T(1) 21P T(2) (x 2)
2y2 + 2xy+ 4y 17x 126 = 0
TH1 : x= 2 y= 1
TH2 :
2y
2
+ 2xy+ 4y 17x 126 = 0x2 + y2 + 2x 2y 7 = 0H ny c cch gii ri nh ??3.P T(2) P T(1) (x y+ 5)2 + 2x2 + x + 80 = 0(V nghim)Vy h cho c nghim : (x; y) = (2; 1)
Tip theo chng ta s n vi cu VMO 2004.
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
20 Tuyn tp nhng bi h c sc
Cu 25
x3 + 3xy2 = 49x2 8xy+y2 = 8y 17x
Gii
Li gii ngn gn nht ca bi trn l :
P T(1) + 3.P T(2) (x + 1) (x + 1)2 + 3(y 4)2 = 0n y d dng tm ra nghim (x; y) = (1;4), (1;4)
Cu hi c t ra l bi ny tm hng s nh th no ? C rt nhiu cch gii thch nhngti xin trnh by cch gii thch ca ti :tuzki:Lm tng t theo nh hai cu 23 v 24 xem no. Vit li h cho thnh
3xy2 + x3 + 49 = 0y2 + 8(x + 1)y+ x2 17x= 0
Mt cch trc gic ta th vi x= 1. V sao ? V vi x= 1phng trnh 2 s khng cnphn y v c v 2 phng trnh s tng ng. Khi thay x= 1h cho tr thnh
3y2 + 48 = 0y2 16 = 0
Hai phng trnh ny tng ng. Tri thng ri !! Vy x =1chnh l 1 nghim cah v t h th hai ta suy ra ngay phi lm l P T(1) + 3.P T(2). Vic cn li ch l phntch nt thnh nhn t.
Tip theo y chng ta s n vi mt chm h d bn ca tng trn. Ti khng trnhby chi tit m ch gi v kt qu
Cu 26
y3 + 3xy2 = 28x2 6xy+y2 = 6x 10y
Gi : P T(1) + 3.P T(2) (y+ 1) (3(x 3)2 + (y+ 1)2) = 0Nghim ca h : (x; y) = (3;1), (3;1)
Cu 27
6x2y+ 2y3 + 35 = 05x2 + 5y2 + 2xy+ 5x + 13y= 0
Gi : P T(1) + 3.P T(2) (2y+ 5)
3
x +
1
2
2+
y+
5
2
2= 0
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.1 Cu 1 n cu 30 21
Cu 28
x3 + 5xy2 = 352x2 5xy 5y2 + x + 10y 35 = 0
Gi : P T(1) + 2.P T(2) (x 2) (5(y 1)2
+ (x + 3)2
) = 0
Cu 29
x3 + 3xy2 = 6xy 3x 49x2 8xy+y2 = 10y 25x 9
Gi : P T(1) + 3.P T(2) (x + 1) ((x + 1)2 + 3(y 5)2) = 0
im qua cc cu t cu 23 n cu 29 ta thy dng nh nhng cu h ny kh c bit.Phi c bit th nhng h s kia mi t l v ta tm c x = hay y = l nghim cah. Th vi nhng bi h khng c c may mn nh kia th ta s lm nh no. Ti xin giithiu mt phng php UCT rt mnh. C th p dng rt tt gii nhiu bi h hu t (kc nhng v d trn). l phng php Tm quan h tuyn tnh gia x v y. V ta skhng ch nhn hng s vo mt phng trnh m thm ch nhn c mt hm f(x)hay g(y)vo n. Ti s a ra vi v d c th sau y :
Cu 30
3x2 + xy 9x y2 9y= 02x3 20x x2y 20y= 0
Gii
Bi ny nu th nh cu 23, 24, 25 u khng tm ra ni x hay y bng bao nhiu l nghim cah. Vy phi dng php dng quan h tuyn tnh gia x v y. Quan h ny c th xy dng
bng hai cch thng dng sau :- Tm ti thiu hai cp nghim ca h- S dng nh l v nghim ca phng trnh hu t
Trc ht ti xin pht biu li nh l v nghim ca phng trnh hu t :Xt a thc : P(x) =anx
n + an1xn1 + ....+a1x + a0a thc c nghim hu t
p
q p l c ca a0 cn q l c ca an
OK ri ch ? By gi ta hy th xy dng quan h theo cch u tin, l tm ti thiu haicp nghim ca h ( Casio ln ting :v )D thy h trn c cp nghim l (0;0v(2;1)Chn hai nghim ny ln lt ng vi ta 2 im, khi phng trnh ng thng quachng s l : x + 2y= 0 x= 2y
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
22 Tuyn tp nhng bi h c sc
Nh vy quan h tuyn tnh y l x= 2y. Thay li vo h ta c 9y (y+ 1) = 020y (y+ 1) (y 1) = 0
Sau ta chn biu thc ph hp nht nhn vo 2 phng trnh. y s l 20 (y 1) .P T(1) + 9.P T(2)Nh vy
20 (y 1) .P T(1) + 9.P T(2) (x + 2y) 18x2 + 15xy 60x 10y2 80y = 0TH1 : x= 2ythay vo (1)TH2 : Kt hp thm vi PT(1) na thnh mt h gm hai tam thc bit cch giiNghim ca h :
(x; y) = (0; 0), (2;
1), (10; 15),
15 1452
; 11
145 ,
15 +
145
2 ; 11 +
145
S dng cch ny chng ta thy, mt h phng trnh hu t ch cn tm c mt cpnghim l ta xy dng c quan h tuyn tnh v gii quyt bi ton. y chnh l uim ca n. Bn c th vn dng n vo gii nhng v d t 23 n 29 xem. Ti th lmcu 25 nh : Cp nghim l(1;4), (1;4)nn quan h xy dng y l x = 1. Thay livo h v ta c hng chn h s nhn.
Tuy nhin cch ny s chu cht vi nhng bi h ch c mt cp nghim hoc nghim qul khng th d bng Casio c. y l nhc im ln nht ca n
No by gi hy th xy dng quan h bng nh l nh.
Vi h ny v phng trnh di ang c bc cao hn trn nn ta s nhn avo phng trnhtrn ri cng vi phng trnh di. V bc ca x ang cao hn nn ta vit li biu thc saukhi thu gn di dng mt phng trnh bin x. C th l
2x3 + (3a y) x2 + (ay 9a 20) x y (ay+ 9a + 20) = 0()Nghim ca (*) theo nh l s l mt trong cc gi tr
1,12
,y2
,y, ....Tt nhin khng th c nghim x= 1
2hayx = 1c. Hy th vi hai trng hp cn li.
* Vi x= y thay vo h ta c 3y2 18y= 0y3 40y= 0
Khi ta s phi ly(y2 40).P T(1) 3(y 6).P T(2). R rng l qu phc tp. Loi ci ny.* Vi x= ythay vo h ta c
y2 = 03y3 = 0
Khi ta s ly 3y.PT(1) +P T(2). Qu n gin ri. Khi biu thc s l
(x + y)
2x2 + 6xy 3y2 + 27y+ 20 = 0Cch s hai rt tt thay th cch 1 trong trng hp khng tm ni cp nghim. Tuy nhinyu im ca n l khng phi h no dng nh l cng tm c nghim. Ta phi bit kt
hp nhun nhuyn hai cch vi nhau. V hy th dng cch 2 lm cc cu t 23 n 29 xem.N s ra nghim l hng s.
Lm mt cu tng t na. Ti nu lun hng gii.
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 23
2.2 Cu 31 n cu 60
Cu 31
x2y2 + 3x + 3y 3 = 0x2y 4xy 3y2 + 2y x + 1 = 0
Gii
P T(1) (y 1).P T(2) (x + y 1) 3y2 +xy 2y+ 2 = 0TH1 : x= 1 y. No problem !!!Th2 :
3y2 + xy 2y+ 2 = 0x2y 4xy 3y2 + 2y x + 1 = 0
y li l h c bit, ta tm c x= 3l nghim ca h. Thay vo v rt ra kt qu
PT(1) + PT(2) (x 3) (xy 1) = 0
Vy h cho c nghim (x; y) = (0; 1), (1; 0)
Bi vit v phng php UCT hay cn gi l h s bt nh kt thc y. Qua hn chccu ta thy : s dng phng php UCT nng cao (tm quan h tuyn tnh gia cc n) lmt phng php rt mnh v rt tt gii quyt nhanh gn cc h phng trnh hu t. Tuynhin nhc im ca n trong qu trnh lm l kh nhiu. Th nht : tnh ton qu tru b
v hi no. Hin nhin ri, dng quan h tuyn tnh kh, sau cn phi nhc cng phntch mt a thc hn n thnh nhn t. Th hai, nu s dng n mt cch thi qu s khinbn thn tr nn thc dng, my mc, khng chu my m suy ngh m c nhn thy l laou vo UCT, c khc g lao u vo khng ?
Mt cu hi t ra. Liu UCT c nn s dng trong cc k thi, kim tra hay khng ? Xintha, trong nhng VMO, cng lm tng ca h l dng UCT dng c bn, tc l nhnhng s thi. UCT dng c bn th ti khng ni lm g ch UCT dng nng cao th tt nhtkhng nn xi trong cc k thi. Th nht mt rt nhiu thi gian v sc lc. Th hai gy khkhn v c ch cho ngi chm, h hon ton c th gch b ton b mc d c th bn lmng. Vy nn : CNG NG LM RI MI DNG NH !! :D
y c l l bi vit ln nht m ti km vo trong cun sch. Trong nhng cu tip theoti s ci nhng bi vit nh hn vo. n xem nh. Nhng cu tip theo c th cn mt scu s dng phng php UCT. Vy nn nu thc mc c quay tr li t cu 20 m xem. Tmthi gc li , ta tip tc n vi nhng cu tip theo.
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
24 Tuyn tp nhng bi h c sc
Cu 32
x5 + y5 = 1x9 + y9 =x4 + y4
Gii
Nhn thy r rng y l loi h bn ng cp. Ta nhn cho hai v vi nhau c
x9 + y9 = (x4 + y4)(x5 + y5) x4y4(x + y) = 0
TH1 : x= 0 y= 1TH2 : y= 0 x= 1TH3 : x= ythay vo (1) r rng v nghimVy h cho c nghim (x; y) = (1; 0), (0; 1)
Cu 33
x3 + 2xy2 = 12y8y2 +x2 = 12
Gii
Li thm mt h cng loi, nhn cho hai v cho nhau ta c
x3 + 2xy2 =y(8y2 +x2) x= 2y
Khi (2) s tng ng12y2 = 12 y= 1, x= 2
Vy h cho c nghim (x; y) = (2; 1), (2;1)
Cu 34
x2 +y2 +
2xy
x + y = 1
x + y = x2 y
Gii
iu kin : x + y >0R rng khng lm n c t phng trnh (2). Th bin i phng trnh (1) xem
(1) (x + y)2 1 + 2xyx +y
2xy= 0
(x + y+ 1)(x + y 1) 2xy(x + y 1)x + y
= 0
C nhn t chung ri. Vi x + y= 1thay vo (2) ta c
1 = (1 y)2 y y= 0, y = 3
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 25
Gi ta xt trng hp cn li. l x + y+ 1 = 2xy
x +y
x +y+ 1 = 1 x2 y2 x2 + y2 + x + y= 0
R rng sai v t iu kin cho ngay x +y >0Vy h cho c nghim (x; y) = (1; 0), (2;3)
Cu 35
x3 y3 = 3(x y2) + 2x2 +
1 x2 3
2y y2 + 2 = 0
Gii
iu kin :1 x 1, 0 y 2Thng th bi ny ngi ta s lm nh sau. phng trnh (1) mt cht
(1) x3 3x= (y 1)3 3(y 1)
Xt f(t) =t3 3tvi1 t 1th f(t) = 3t2 3 0Suy ra f(t)n iu v t suy ra x= y 1thay vo (2)Cch ny n. Tuy nhin thay vo lm vn cha phi l nhanh. Hy xem mt cch khc rt mim m ti lm
(2) x2 +
1 x2 + 2 = 3
2y y2 f(x) =g(y)
Xt f(x)trn min [1;1]ta s tm c 3 f(x) 134Ta li c : g(y) = 3
y(2 y) 3 y+ 2 y
2 = 3
Vyf(x) g(y). Du bng xy ra khi y= 1x= 1, x= 0 Thay vo phng trnh u ch c cp(x; y) = (0; 1)l tha mn
Vy h cho c nghim (x; y) = (0; 1)
Cu 36
x3 3x= y3 3yx6 + y6 = 1
Gii
D thy phng trnh (1) cn xt hm ri, tuy nhin f(t) =t33tli khng n iu, cn phib thm iu kin. Ta s dng phng trnh (2) c iu kin. T (2) d thy 1 x, y 1.Vi iu kin r rng f(t)n iu gim v suy ra c x= yThay vo (2) ta c
2x6
= 1 x
= 1
62Vy h cho c nghim :(x; y) =
16
2;
16
2
,
1
6
2; 1
6
2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
26 Tuyn tp nhng bi h c sc
Cu 37
x3(2 + 3y) = 1x(y3 2) = 3
Gii
Nhn thy x= 0khng l nghim. H cho tng ng
3y+ 1 = 1
x33
x+ 2 =y3
y = 1x
Thay li (1) ta c
2x3 + 3x2 1 = 0
x= 1 y= 1x=
1
2y = 2
Vy h cho c nghim :(x; y) = (1;1),
1
2; 2
Cu 38
x2 + y2 + xy+ 1 = 4yy(x +y)2 = 2x2 + 7y+ 2
Gii
S dng UCT s thy y= 0l nghim ca h. Thay li v ta s c
2P T(1) +P T(2) y(x + y+ 5)(x +y 3) = 0 y= 0x= 5 y
x= 3 y
Vi y= 0thay li v nghimVi x= 5 ykhi phng trnh (1) s tng ng
(y+ 5)2 +y2
y2
5y+ 1 = 4y
V L
Tng t vi x= 3 ycng v nghimVy h cho v nghim
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 27
Cu 39
x +y x y= y2
x2 y2 = 9
Gii
iu kin : y min{x}Ta khng nn t n tng hiu v vn cn st li
y
2s lm bi ton kh khn hn. Mt cch
trc gic ta bnh phng (1) ln. T (1) ta suy ra
2x 2
x2 y2 = y2
4
n y nhn thy
x2 y2 theo (2) bng 3. Vy suy ra
2x 6 = y2
4 y2
= 8x 24Thay vo (2) ta c
x2 8x + 15 = 0 x= 3 y= 0(T M)x= 5 y= 4(T M)
x= 5 y= 4(T M)Vy h cho c nghim (x; y) = (3; 0), (5; 4), (5;4)
Cu 40
x y+ 1 =52
y+ 2(x 3)x + 1 = 34
Gii
iu kin : x, y 1Khng tm c mi quan h c th no. Tm thi ta t n d nhnt
x + 1 =a
0,
y+ 1 =b
0. H cho tng ng
a2 1 b= 5
2
b2 1 + 2a(a2 4) = 34
Ta th b=7
2 a2 t (1) vo (2) v c :
72 a
22+ 2a(a
2
4) 1
4 = 0
a= 3 b= 112
(L)
a=
2
b=
1
2(L)
a= 1 b= 52
(T M)
a= 2 b= 12
(L)
x= 0
y = 34
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
28 Tuyn tp nhng bi h c sc
Vy h cho c nghim : (x; y) =
0;34
Cu 41
(x2 + xy+ y2)
x2 + y2 = 185
(x2 xy+ y2)
x2 + y2 = 65
Gii
Thot nhn qua th thy y l mt h ng cp bc 3 r rng. Tuy nhin nu tinh ta emcng 2 phng trnh cho nhau s ch cn li x2 + y2Cng 2 phng trnh cho nhau ta c
2(x2 +y2)
x2 +y2 = 250 x2 + y2 = 5Khi thay li h ta c
(25 +xy).5 = 185(25 xy).5 = 65
xy= 12x2 + y2 = 25
x= 3, y = 4x= 4, y = 3x= 3, y= 4x= 4, y= 3
Vy h cho c nghim (x; y) = (3; 4), (4; 3), (3;4), (4;3)
Cu 42
y
x+
x
y =
7xy
+ 1
x
xy+y
xy= 78
Gii
iu kin : xy 0H cho tng ng
x + yxy
=7 +
xy
xyxy(x +y) = 78
t x + y= a,xy=b. H cho tng ng
a b= 7ab= 78
a= 13b= 6
a= 6b= 13 (L)
x +y = 13xy= 36
x= 9, y= 4x= 4, y= 9
Vy h cho c nghim (x; y) = (9; 4), (4; 9)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 29
Cu 43
x3 y3 = 9x2 + 2y2 x + 4y= 0
Gii
Dng UCTP T(1) 3.P T(3) (x 1)3 = (y+ 2)3 x= y + 3
n y d dng tm nghim (x; y) = (1;2), (2;1)
Cu 44
8x3y3 + 27 = 18y3
4x2y+ 6x= y2
Gii
y l mt h hay. Ta hy tm cch loi b 18y3 i. V y = 0khng l nghim nn (2) tngng
72x2y2 + 108xy= 18y3
n y tng r rng ri ch ? Th 18y3 t (1) xung v ta thu c
8x3y3
72x2y2
108xy+ 27 = 0
xy= 3
2
xy=
21
9
5
4
xy=21 + 9
5
4
Thay vo (1) ta s tm c y v x
y= 0(L)
y= 3
8(xy)3 + 27
18 = 3
2
5 3 x= 1
4
35
y= 3
8(xy)3 + 27
18 =
3
2 3 +
5x=
1
4 3 +
5Vy h cho c nghim :(x; y) =
1
4
35 ;3
2
5 3 ,1
4
3 +
5
;3
2
3 +
5
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
30 Tuyn tp nhng bi h c sc
Cu 45
(x +y)
1 +
1
xy
= 5
(x2 + y2)
1 +
1
x2y2
= 9
Gii
iu kin : xy= 0Ta c nhn ra . H tng ng
x +y+1
x+
1
y = 5
x2 + y2 + 1
x2+
1
y2 = 9
x +
1
x
+
y+
1
y
= 5
x +1
x
2+
y+
1
y
2= 13
x +
1x
= 2, y+1y
= 3
x +1
x= 3, y+
1
y = 2
x= 1, y =
3 52
x=3 5
2 , y= 1
Vy h cho c nghim : (x; y) =
1;
3 52
,
35
2 ; 1
Cu 46
x2 + y2 + x + y = 18x(x + 1)y(y+ 1) = 72
Gii
Mt bi t n tng tch cng kh n gint x2 + x= a, y2 + y= b. Ta c
a + b= 18ab= 72
a= 12, b= 6
a= 6, b= 12
x2 + x= 6y2 +y = 12
x
2
+ x= 12y2 +y = 6
x= 2, x= 3y = 3, y = 4
x= 3, x= 4y = 2, y = 3
Vy h cho c c thy 8 nghim
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 31
Cu 47
x3 + 4y= y3 + 16x1 +y2 = 5(1 + x2)
Gii
H cho tng ng x3 16x= y (y2 4)y2 4 = 5x2
Nh vy phng trnh (1) s l
x3 16x= 5x2y x= 0, y = 2
y=x2 16
5x
Trng hp 2 thay vo (2) s l(x2 16)2
25x2 4 = 5x2
x2 = 1
x2 = 6431
x= 1, y = 3x= 1, y= 3
Vy h cho c nghim (x; y) = (0; 2), (0;2), (1;3), (1;3)
Cu 48 x +y2 x2 = 12
y
x
y2 x2 = 12
Gii
iu kin : y2 x2 x
y2 x2 sinh ra t vic ta bnh phng (1). Vy th bm theo hng xem. T (1)
ta suy tax2 + y2 x2 + 2x
y2 x2 = (12 y)2
y2 + 24 = (12 y)2 y= 5
Thay vo (2) ta c x
25 x2 = 12 x= 3, x= 4i chiu li thy tha mnVy h cho c nghim (x; y) = (3; 5), (4; 5)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
32 Tuyn tp nhng bi h c sc
Cu 49
x4 4x2 + y2 6y+ 9 = 0x2y+x2 + 2y 22 = 0
Gii
nu t x2 = ath h cho bin thnh h tam thc bc 2 ta hon ton bit cchgii. C th y s l
P T(1) + 2.P T(2) (x2 + y)2 2(x2 + y) 35 = 0
TH1 : x2 +y = 7 x2 = 7 ythay (2) ta c
(7 y)y+ 7 y+ 2y 22 = 0
y = 3 x= 2y = 5 x= 2
TH2 : x2
+y = 5 x2
= 5 y. Hon ton tng t thay (2) s cho y v nghimVy h cho c nghim : (x; y) = (2; 3), (2;3), (2;5), (2;5)
Cu 50
x2 +y+ x3y+ xy+ y2x= 54
x4 +y2 + xy(1 + 2x) = 54
Giiy l cu Tuyn sinh khi A - 2008. Mt cch t nhin khi gp hnh thc ny l ta tin hnhnhm cc s hng liH cho tng ng
(x2 + y) +xy+ (x2 +y)xy= 5
4
(x2 + y)2 +xy = 54
n y hng i r rng. t x2 +y = a, xy= b ta c
a + b + ab= 54
a2 + b= 54
a= 0, b=
5
4
a= 12
, b= 32
x2 + y= 0xy= 5
4
x2 +y = 12
xy= 32
x= 3
5
4, y= 3
25
16
x= 1, y = 32
Vy h cho c nghim (x; y) =
3
5
4; 3
25
16
,
1;3
2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 33
Cu 51
x2 + 1 + y(y+x) = 4y(x2 + 1)(x + y 2) =y
Gii
H gn nh ch l cu chuyn ca x2 + 1v x+y. Tuy nhin ychen vo khin h tr nnkh chu. Hy dit yi . Cch tt nht l chia khi m y = 0khng phi l nghim cah. H cho tng ng
x2 + 1
y + x + y 2 = 2
x2 + 1
y (x + y 2) = 1
Hng i r rng. t x2 + 1
y =a, x +y 2 =b
H cho tr thnh a +b= 2ab= 1
a= 1b= 1
x2 + 1 =yx + y= 3
x= 1, y= 2x= 2, y = 5
Vy h cho c nghim (x; y) = (1; 2), (2;5)
Cu 52 y+ xy2 = 6x2
1 +x2y2 = 5x2
Gii
Loi h ny khng kh. tng ta s chia bin v phi tr thnh hng sNhn thy x= 0khng l nghim. H cho tng ng
y
x2+
y2
x = 6
1
x2+y2 = 5
y
x
1
x+y
= 6
1
x+ y
2
2 yx
= 5
t y
x=a,
1
x+ y=b. H tr thnh
ab= 6b2 2a= 5
a= 2b= 3
y = 2x1
x+ y= 3
x= 1, y = 2
x=1
2, y = 1
Vy h cho c nghim (x; y) = (1; 2),
1
2; 1
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
34 Tuyn tp nhng bi h c sc
Cu 53
x2 + 2y2 =xy + 2y2x3 + 3xy2 = 2y2 + 3x2y
Gii
mt cht y l h bn ng cp. Nu ta vit li nh sau x2 + 2y2 xy= 2y2x3 + 3xy2 3x2y= 2y2
T ta c
2y2(x2 + 2y2 xy) = 2y 2x3 + 3xy2 3x2y 4y (y x) x2 xy+y2 = 0TH1 : y= 0 x= 0
TH2 : x= y = 0TH3 : x= y thay vo (1) ta c
2y2 = 2y
x= y = 0x= y = 1
Vy h cho c nghim (x; y) = (0; 0), (1; 1)
Cu 54 2x2y+y3 = 2x4 + x6
(x + 2)y+ 1 = (x + 1)2
Gii
iu kin : y 1Khai thc t (1). C v nh l hm no . Chn chia cho ph hp ta s c mc ch, ys chia cho x3 v x= 0khng l nghim ca h. PT(1) khi s l
2y
x+ y
x3
= 2x + x3 yx
=x y = x2
Thay vo (2) ta s c
(x + 2)
x2 + 1 = (x + 1)2 (x + 2)2 x2 + 1 = (x + 1)4 x= 3, y= 3(T M)x= 3, y = 3(T M)
Vy h cho c nghim : (x; y) = (3;3)
Ta s n mt cu tng t n
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 35
Cu 55
x5 + xy4 =y10 +y64x + 5 +
y2 + 8 = 6
Gii
iu kin : x 54
Thy y= 0khng l nghim ca h. Chia 2 v ca (1) cho y5 ta c
x
y
5+
x
y =y5 + y x
y =y x= y2
Thay vo (2) ta c
4x + 5 +
x + 8 = 6 x= 1 y = 1
Vy h cho c nghim (x; y) = (1;1)
Cu 56
xy+ x + 1 = 7yx2y2 + xy+ 1 = 13y2
Gii
y l cu Tuyn sinh khi B - 2009. Cc gii thng thng nht l chia (1) cho y , chia (2)choy2 sau khi kim tra y= 0khng phi l nghim. Ta s c
x +x
y+
1
y= 7
x2 +x
y+
1
y2 = 13
x +1
y+
x
y = 7
x +1
y
2 x
y = 13
a + b= 7a2 b= 13
a= 4, b= 3a= 5, b= 12
x +1
y = 4
x= 3y x +1y = 5x= 12y
x= 1, y=
1
3x= 3, y= 1
Vy h cho c nghim : (x; y) =
1;1
3
, (3; 1)
Tip tc ta n thm mt cu tuyn sinh na
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
36 Tuyn tp nhng bi h c sc
Cu 57
x4 + 2x3y+ x2y2 = 2x + 9x2 + 2xy= 6x + 6
Gii
tht k nu ta th kho lo xyln (1) s ch cn li phng trnh n x. D s l bc 4nhng liu th n nhiu. H vit li
x4 + 2x2(xy) +x2y2 = 2x + 9
xy=6x + 6 x2
2
T (1) s tng ng
x4
+ x2
(6x + 6 x2
) +6x + 6 x
2
22
= 2x + 9 x= 4x= 0 y=17
4V L
Vy h cho c nghim (x; y) =4;17
4
Cu 58
3
1 +x +
1 y = 2x2 y4 + 9y= x(9 +y y3)
Gii
iu kin : y 1Khng lm n g c t (1). Xt (2). 1 to th (2) c th phn tch c thnh
(x y) (9 x y3) = 0
x= yx= 9 y3
Vi x= y thay vo (1) ta s c
3
1 +y+
1 y= 2 a + b= 2a3 +b2 = 2
b 0 a= 1, b= 1a= 1 3, b= 3 + 3
a=
3 1, b= 3 3 y= 0y= 63 11
y= 63 11
Vi x= 9 y3 thay vo (1) ta s c3
10 y3 +
1 y = 2
Ta c3
10 y3 +
1 y 3
9> 2
Vy h cho c nghim : (x; y) = (0; 0), (63 11;63 11), (63 11;63 11)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.2 Cu 31 n cu 60 37
Cu 59
xy+
1 y= y2
y
x 1y = 1
Gii
iu kin : x 1, 0 y 1Thot nhn bi ton ta thy nh lc vo m cung nhng cn thc. Tuy nhin ch vi nhngnh gi kh n gin ta c th chm p bi tonVit li phng trnh (2) nh sau
2
y
x 1 = y 1
T iu kin d thy V T 0 V PDu bng xy ra khi x= y = 1Vy h cho c nghim (x; y) = (1; 1)
Cu 60
x
17 4x2 +y
19 9y2 = 317 4x2 +
19 9y2 = 10 2x 3y
Gii
iu kin :172 x
172
,193 y
193
Bi ton ny xut hin trn thi th ln 2 page Yu Ton hc v ti l tc gi ca n. tng ca n kh n gin, ph hp vi 1 thi tuyn sinh x
17 4x2 lin quan n 2xv17 4x2, y
19 9y2 lin quan n 3yv19 9y2.
V tng bnh phng ca chng l nhng hng s. y l c s ta t nt 2x +
17 4x2 =a, 3x +
19 9y2 =b. H cho tng ng
a + b= 10a2 17
4 +
b2 196
= 3
a= 5, b= 5a= 3, b= 7
TH1 :
2x +17 4x2 = 53y+
19 9y2 = 5
x= 1
2x= 2
y=5 13
6
TH2 :
2x +
17 4x2 = 33y+
19 9y2 = 7 (Loi)
Vy h cho c nghim :(x; y) =
1
2;5 +
13
6
1
2;5 13
6
2;
5 +
13
6
2;
5 136
V y l tng gc ca n. Hnh thc n gin hn mt cht
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
38 Tuyn tp nhng bi h c sc
2.3 Cu 61 n cu 90
Cu 61
x
5 x2 +y
5 4y2 = 15 x2 +5 4y
2 =x 2y
Nghim : (x; y) = (1;1),
2;12
Cu 62
x3 xy2 + y3 = 14x4 y4 = 4x y
Gii
R rng l mt h a v c dng ng cp bng cch nhn cho v vi v. Tuy nhin, biny nu s dng php th tt ta s a v mt kt qu kh p mtPhng trnh (2) tng ng
4x(x3 1) =y(y3 1)n y ta rt x3 1vy3 1t (1). C th t (1) ta c
x3 1 =y3 y2xy3 1 =xy2 x3
Thay tt c xung (2) v ta thu c
4xy2(y x) = xy(x2 y2)
x= 0y = 0x= y4y= y + x
y= 1x= 1x= y = 1
y= 13
25, x=
33
25
Vy h cho c nghim (x; y) = (0; 1), (1; 0), (1; 1),
13
25;
33
25
Cu 63
x +
x2 y2x
x2 y2 +
xx2 y2x +
x2 y2 =
17
4
x(x +y) +
x2 + xy+ 4 = 52
Gii
iu kin : x =
x2 y2, x2 y2 0, x2 +xy+ 4 0Hnh thc bi h c v kh khng b nhng nhng tng th l ht. Ta c th khai thc c2 phng trnh. Pt(1) c nhiu cch x l : ng cp, t n, lin hp. Ti s x l theo hngs 3. (1) khi s l
x +
x2 y22
x2 (x2 y2) +
x
x2 y22
x2 (x2 y2) =17
4 2 (2x
2 y2)y2
=17
4 y= 4x
5
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.3 Cu 61 n cu 90 39
Tip tc khai thc (2). D thy t
x2 +xy+ 4 =t 0th (2) tr thnh
t2 + t= 56
t= 7t= 8(L) x
2 +xy = 45
Kt hp li ta c y= 4
5x
x2 + xy= 45
x= 5, y = 4x= 5, y= 4x= 15, y= 12x= 15, y= 12
Vy h cho c nghim : (x; y) = (5;4), (5; 4), (15; 12), (15;12)
Cu 64x +y+x
y = 2
y+ xy x= 1Gii
iu kin : x, y 0,y min{x},x min{y}Khng tm c mi lin h g t c hai phng trnh, ta tin hnh bnh phng nhiu ln ph v ton b cn thc kh chu. Phng trnh (1) tng ng
2x + 2
x2 y= 4
x2 y= 2 x x2 y= x2 4x 4 4x y= 4Lm tng t phng trnh (2) ta s c : 4x
4y =
1. Kt hp 2 kt qu li d dng tm
c x,yVy h cho c nghim : (x; y) =
17
12;5
3
Cu 65
x + 2xy
3
x2 2x + 9 =x2 + y
y+ 2xy
3y2 2y+ 9=y2 +x
Gii
Hnh thc ca bi h l i xng. Tuy nhin biu thc kh cng knh v li nhn xt thyx= y = 1l nghim ca h. C l s nh giCng 2 phng trnh li ta c
x2 + y2 = 2xy
1
3
x2 2x + 9 + 1
3
y2 2y+ 9
T ta nhn xt c nghim th xy 0v l 3
t2 2t+ 9 2nn ta nh gix2 + y2 2xy
1
2+
1
2
(x y)2 0
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
40 Tuyn tp nhng bi h c sc
Du bng xy ra khi (x; y) = (1; 1)
Cu 66
6
x
y 2 = 3x y+ 3y2
3x +
3x y= 6x + 3y 4
Gii
iu kin : y= 0, 3x y, 3x +3x y 0Phng trnh (1) khi s tng ng
6x
2y=y3x y+ 3y2
2 (3x
y)
y3x y 3y
2 = 0
3x y = y
3x y =3y
2
TH1 :
3x y= y. T y suy ra y 0v3x= y2 + ythay tt c vo (2) ta c
2
y2 + y y= 2 y2 + y+ 3y 4 2y2 + 7y 4 = 0y 0 y = 4 x= 4
TH2 :
3x y=3y2
. T y suy ra y 0v3x= 9y2
4 + ythay tt c vo (2) ta cng s tm
c y =8
9x=
8
9Vy h cho c nghim (x; y) = (4;4),
8
9;8
9
Cu 67
(3 x)2 x 2y2y 1 = 03
x + 2 + 2
y+ 2 = 5
Gii
iu kin : x 2, y12
Phng trnh (1) tng ng
(2 x)2 x +2 x= (2y 1)
2y 1 +
2y 1 f(2x 1) =f(
2y 1)
Vi f(x) =x3 +xn iu tng. T suy ra
2 x= 2y 1 x= 3 2ythay vo (2)ta c
3
5 2y+ 2
y+ 2 = 5
a + 2b= 5a3 + 2b2 = 9
a= 1, b= 2
a=3 654
, b= 23 + 658
a=
65 3
4 , b=
23 658
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.3 Cu 61 n cu 90 41
y= 2
y=233 + 23
65
32
y=233 2365
32
Vy h cho c nghim
(x; y) = (1;2),
23
65 18516
;233 2365
32
23
65 + 185
16 ;
233 + 23
65
32
S dng tnh n iu ca hm s cng l mt hng kh ph bin trong gii h phng trnh.Ch cn kho lo nhn ra dng ca hm, ta c th rt ra nhng iu k diu t nhng phngtrnh khng tm thng cht no
Cu 681 +xy+ 1 +x + y= 2
x2y2 xy= x2 + y2 + x + y
Gii
iu kin : xy 1, x + y 1Mt cht bin i phng trnh (2) ta s c
x2y2 + xy= (x +y)2 + x + y (xy x y)(xy+ x + y+ 1) = 0
x +y =xyx +y = xy 1
TH1 : xy=x + ythay vo (1) ta c2
1 +xy = 2 xy= 0 x= y = 0TH2 : x + y = xy 1thay vo (1) ta c
1 +xy+xy= 2(V L)
Vy h cho c nghim : (x; y) = (0; 0)
Cu 69
x + 3x yx2 + y2
= 3
y x + 3yx2 +y2
= 0
Gii
Ti khng nhm th bi ton ny xut hin trn THTT, tuy nhn hnh thc ca h kh pmt v gn nh nhng khng h d gii mt cht no. Hng lm ti u ca bi ny l phcha. Da vo tng h kh i xng ng thi di mu nh l bnh phng ca Moun mta s dng cch ny. Hng gii nh sau
PT(1)+i.PT(2) ta s c
x + yi +3(x yi) (xi + y)
x2 + y2 = 0
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
42 Tuyn tp nhng bi h c sc
t z=x +yikhi phng trnh tr thnh
z+3z iz|z|2 = 3 z+
3z izz.z
= 3 z+3 iz
= 3
z= 2 + iz= 1 i
Vy h cho c nghim (x; y) = (2; 1), (1;1)Hnh thc ca nhng bi h ny kh d nhn thy. Th lm mt s cu tng t nh.
Cu 70
x +5x + 7
5y
x2 +y2 = 7
y+7
5x 5yx2 + y2
= 0
Cu 71
x + 5x yx2 + y2
= 3
y x + 5yx2 +y2
= 0
Cu 72
x +16x 11y
x2 +y2 = 7
y 11x + 16yx2 + y2
= 0
Cu 73
(6 x)(x2 + y2) = 6x + 8y(3 y)(x2 +y2) = 8x 6y
Gi : Chuyn h cho v dng
x +6x + 8y
x2 +y2 = 6
y+8x 6yx2 + y2
= 3
Nghim : (x; y) = (0; 0), (2;1), (4; 2)
Phc ha l mt phng php kh hay gii h phng trnh mang tnh nh cao. Khngch vi loi h ny m trong cun sch ti s cn gii thiu mt vi cu h khc cng s dngphc ha kh p mt.
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.3 Cu 61 n cu 90 43
Cu 74
4x2y2 6xy 3y2 = 96x2y y2 9x= 0
Gii
y l mt bi ton cng kh p mt. Thy x= 1l nghim ca h . Ta suy raP T(1) +P T(2) (x 1)(4y2(x + 1) + 6xy 9) = 0
TH1 : x= 1 y= 3TH2 : 4y2(x + 1) + 6xy 9 = 0V x= 0khng l nghim. Suy ra 4y2x(x + 1) + 6x2y 9x= 0(*)V sao nhn xvo y. UCT chng ? Ti ch gii thiu cho cc bn UCT nng cao thi chti ch dng bao gi. L do ch n gin ti mun xut hin 6x2y 9x= y2 t (2) thiVy (*)
4y2x(x + 1) +y2 = 0
y2(2x + 1)2 = 0
TH1 : y= 0v nghimTH2 : x= 1
2 y= 3, y= 3
2
Vy h cho c nghim : (x; y) = (1; 3),1
2; 3
,
1
2;3
2
Cu 75
x2
(y+ 1)2+
y2
(x + 1)2 =
1
2
3xy=x + y+ 1
Gii
iu kin x, y= 1Bi ton ny c kh nhiu cch gii. Ti xin gii thiu cch p nht ca bi nyp dng Bt ng thc AMGMcho v tri ca (1) ta c
V T 2xy(x + 1)(y+ 1)
= 2xy
xy+ x + y+ 1=
2xy
xy+ 3xy =
1
2
Du bng xy ra khi (x; y) = (1; 1),13 ;13
Cu 76
3y2 + 1 + 2y(x + 1) = 4y
x2 + 2y+ 1y(y x) = 3 3y
Gii
iu kin : x2 + 2y+ 1
0
Khng lm n g c t (2). Th bin i (1) xem sao. PT(1) tng ng
4y2 4y
x2 + 2y+ 1 +x2 + 2y+ 1 =x2 2xy+ y2
2y
x2 + 2y+ 12
= (x y)2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
44 Tuyn tp nhng bi h c sc
x2 + 2y+ 1 = 3y xx2 + 2y+ 1 =x + y
C v hi o nh ? Nhng mt cht th (1) c vc dng ca cc hng ng thc nn tangh n hng ny
By gi x l hai trng hp kia th no ? Chc bnh phng thi. Tt qu ! Phng trnh sch cn li xyv ym nhng ci th (2) c cTH1 :
x2 + 2y+ 1 = 3y x
3y xx2 + 2y+ 1 = 9y2 6xy+ x2
3y x6xy= 9y2 2y 1xy= y2 + 3y 3(2)
x= 1, y= 1(T M)
x=415
51 , y=
17
3(T M)
TH2 :
x2 + 2y+ 1 =x +y
x +y 0x
2
+ 2y+ 1 =x
2
+ 2xy+ y
2
x + y 02xy=
y2 + 2y+ 1
xy= y2 + 3y 3 x= 1, y= 1
x=
41
21 , y = 7
3(L)
Vy h cho c nghim : (x; y) = (1; 1),
415
51;
17
3
Nh chng ta bit. Tam thc bc hai c kh nhiu ng dng trong gii ton v h cngkhng phi l ngoi l. Ch vi nhng nh gi kh n gin : t iu kin ca tamthc c nghim m ta c th tm ra cc tr ca cc n. T nh gi v gii quyt nhngbi ton m cc phng php thng thng cng b tay. Loi h s dng phng php ny
thng cho di hai dng chnh. Th nht : cho mt phng trnh l tam thc, mt phngtrnh l tng hoc tch ca hai hm f(x) v g(y). Th hai : cho c 2 phng trnh u lphng trnh bc hai ca 1 n no . Hy th lt qua mt chm h loi ny nh.
Cu 77
x4 + y2 =698
81x2 + y2 + xy 3x 4y+ 4 = 0
GiiHnh thc ca h : mt phng trnh l tam thc bc hai mt c dng f(x) +g(y)v mt skh khng b. Ta hy khai thc phng trnh (2) bng cch nh gi Vit li phng trnh (2) di dng sau
x2 + (y 3)x + (y 2)2 = 0()y2 + (x 4)y+ x2 3x + 4 = 0()
(*) c nghim th x 0 (y 3)2 4(y 1)2 0 1 y73
(**) c nghim th y 0 (x 4)4 4(x2 3x + 4) 0 0 x 43
T iu kin cht ca hai n gi ta xt (1) v c mt nh gi nh sau
x4 + y2
4
3
4+
7
3
2=
697
81 0 V Pnn v nghimVy h cho c nghim : (x; y) = (0; 1), (1;1)
Cu 86
x3(4y2 + 1) + 2(x2 + 1)
x= 6
x2y(2 + 2
4y2 + 1) =x +
x2 + 1
Gii
iu kin : x 0Hnh thc ca bi h r rng l kh rc ri. Tuy nhin, (2) nu ta chia c 2 v cho x2th s c lp c xvyv hi vng s ra c iu g.
Nhn thy x= 0khng l nghim. Chia 2 v ca (2) cho x2
ta c
2y+ 2y
4y2 + 1 = 1
x+
1
x
1
x2+ 1
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
48 Tuyn tp nhng bi h c sc
R rng 2 v u c dng f(t) =t +t
t2 + 1v hm ny n iu tng. Vy t ta suy ra
c 2y= 1
xthay vo (1) ta c
x31x2 + 1
+ 2(
x2
+ 1)
x
= 6
x3 + x + 2(x2 + 1)x= 6R rng v tri n iu tng vi iu kin ca x. Vy x= 1l nghim duy nht
Vy h cho c nghim : (x; y) =
1;1
2
Cu 877x + y+ 2x + y = 52x + y+ x y= 2
Gii
y l cu trong VMO 2000-2001. Khng hn l mt cu qu khiu kin : y min{2x;7x}Xut hin hai cn thc vy th t
7x + y = a ,
2x +y = bxem
Nhng cn x yth th no ? Chc s lin quan n a2, b2. Vy ta s dng ng nht thc
x y = k(7x + y) +l(2x + y) k= 35 , l= 85Vy h cho tng ng
a + b= 5
b +3a2
5 8b
2
5 = 2
a, b 0
a=15 77
2
b=
77 5
2
7x + y=151 1577
2
2x + y=51 577
2
x= 10 77y =
11 772
Vy h cho c nghim : (x; y) =
10
77;
11
77
2
Mt cch khc cng kh tt. t
7x + y = a,
2x + y= bv ta xy dng mt h tm sau a + b= 5a2 b2 = 5x
a +b= 5a b= x b=
5 x2
Thay vo (2) v ta c5 x
2 + x y= 2 x= 2y 1
n y thay li vo (2) v ta cng ra kt qu
Mt v d tng t ca bi ny
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.3 Cu 61 n cu 90 49
Cu 88
11x y y x= 17
y x + 6y 26x= 3
Nghim : (x; y) = 37
20;81
10
Cu 89
3x
1 +
1
x + y
= 2
7y
1 1
x +y
= 4
2
Gii
y l cu trong VMO 1995-1996. Mt tng kh p mt m sng toiu kin : x, y 0, x + y >0H cho tng ng
1 + 1
x +y =
23x
1 1x + y
=4
27y
1
x + y =
13x2
2
7y
1 = 1
3x+
2
27y
1x + y
=
1
3x 2
2
7y
1
3x+
2
27y
1x + y
= 1
3x 8
7y 21xy= (x + y)(7y 3x)
(y 6x)(7y+ 4x) = 0 y= 6xThay vo phng trnh u ta c
1 + 1
7x=
23x
x= 11 + 4
7
21 y = 22
7 +
87
Mt cch khc c th s dng trong bi ny l phc ha. N mi xut hin gn ytx= a >0,y= b >0. Ta c h mi nh sau
a +
a
a2 + b2 =
23
b ba2 + b2
=4
27
P T(1) +i.P T(2) (a +bi) + a bia2 + b2
= 2
3+
4
27
i
t z=a +biphng trnh cho tr thnh
z+1
z =
23
+4
27
i z a, b x, y
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
50 Tuyn tp nhng bi h c sc
Vy h cho c nghim : (x; y) =
11 + 4
7
21 ;
22
7 +
87
Bi h ny c kh nhiu d bn phong ph. Ti xin gii thiu cho cc bn
Cu 90
x
3
1 +
6
x + y
=
2
y
1 6
x +y
= 1
Nghim : (x; y) = (8; 4)
2.4 Cu 91 n cu 120
Cu 91
x
1 12
y+ 3x
= 2
y
1 +
12
y+ 3x
= 6
Nghim : (x; y) = (4 + 2
3; 12 + 6
3)
Cu 92
10x
1 +
3
5x + y
= 3
y
1 3
5x +y
= 1
Nghim : (x; y) =
2
5; 4
Cu 93
4
x
1
4+
2
x +
y
x +y
= 2
4
y
1
42
x +
y
x + y
= 1
Tip theo ta n mt vi v d v s dng phng php lng gic ha trong gii h phng trnh
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 51
Cu 94
x
1 y2 +y1 x2 = 1(1 x)(1 +y) = 2
Gii
iu kin :|x| 1,|y| 1iu kin ny cho ta tng lng gic ha. t x= sina, y= sinbvi a, b
2;
2
Phng trnh u tng ng
sinacosb + sinbcosa= 1 sin(a + b) = 1 a + b= 2
Phng trnh (2) tng ng
(1
sina)(1 +sinb) = 2
(1
sina)(1 +cosa) = 2
a=
2
a= 0 b=
b=
2
x= 1, y= 0(L)x= 0, y= 1
Vy h cho c nghim : (x; y) = (0; 1)
Cu 95 2y=x(1 y2)3x
x3 =y(1
3x2)
Gii
Thot nhn ta thy c v h ny cng xong, ch c g khi vit n di dng xy2 =x 2yx3 3x2y= 3x y
a n v dng ng cp, nhng ci chnh y l nghim n qu l. Vy th hng khcxem. Vit li h cho sau khi xt
x= 2y1 y2y =
3x x31 3x2
Nhn biu thc v phi c quen thuc khng ? Rt ging cng thc lng gic nhn i vnhn ba ca tan. Vy tng ny rat x= tanvi
2;
2
. T PT(2) ta s c
y=3tan tan3
1
3tan2
= tan 3
M nh th theo (1) ta s c
x= 2tan3
1 tan23 = tan 6
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
52 Tuyn tp nhng bi h c sc
T suy ra
tan = tan 6 = k5 =
2
5 ;
5; 0;
5;2
5
Vy h cho c nghim : (x; y) = tan 25
;tan6
5 ,tan
5
;tan3
5 , (0; 0)
Lm mt bi tng t nh.
Cu 96
y =3x x31 3x2
x=3y y31 3y2
S dng phng php lng gic ha trong gii h phng trnh cn phi nm r cc hngng thc, ng thc, cng thc lng gic, v cn mt nhn quan tt pht hin mt biuthc no ging vi mt cng thc lng gic.
Cu 97 x3y(1 +y) +x2y2(2 +y) +xy3 30 = 0x2y+x(1 +y+y2) +y
11 = 0
Gii
y l mt h kh mnh nhng hay. Nhn vo 2 phng trnh ta thy cc bin "kt dnh" vinhau kh tt v hng s c v nh ch l k ng ngoi. Vy hy vt hng s sang mt bn vthc hin bin i v tri. H phng trnh cho tng ng
xy(x + y)(x +y+ xy) = 30xy(x + y) +x + y+ xy= 11
n y tng r rng. t a= xy(x + y), b= xy + x + yv h cho tng ng
ab= 30a + b= 11
a= 5, b= 6a= 6, b= 5
xy(x +y) = 5xy+ x + y= 6
xy(x +y) = 6xy+ x + y= 5
TH1 :
xy(x + y) = 6xy+ x + y= 5
xy= 2x + y = 3
xy= 3x + y = 2
(L)
x= 2, y= 1x= 1, y= 2
TH2 :
xy(x + y) = 5xy+ x + y= 6
xy= 5
x + y = 1 (L) xy= 1x + y = 5
x= 5 212 , y = 5 + 212
x=5 +
21
2 , y=
5212
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 53
Vy h cho c nghim : (x; y) = (1; 2), (2; 1),
521
2 ;
5 212
Tc gi ca n rt kho lo trn nhiu ln cch t n tng tch vo mt h, gy nhiu kh
khn cho ngi lm
Cu 98
sin2x +
1
sin2x+
cos2y+
1
cos2y =
20y
x + ysin2y+
1
sin2y+
cos2x +
1
cos2x=
20x
x + y
Gii
Bi ton xut hin trong VMO 2012-2013. Hnh thc bi h c s khc l khi c c hmlng gic chen chn vo. Vi kiu h ny nh gi l cch tt nhtTa s cng hai phng trnh vi nhau v s chng minh V T 210 V Pp dng Bt ng thc Cauchy Schwarzcho v phi ta c
20y
x + y+
20x
x + y
2
20y
x + y+
20x
x +y
= 2
10
Gi ta s chng minh : V T
2
10tc l phi chng minhsin2x +
1
sin2x+
cos2x +
1
cos2x
10
V T =
sin x 1
sin x
2+
22
+
cos x 1
cos x
2+
22
1
sin x+
1
cos x (sin x + cos x)
2+
2
22
Hin nhin ta c sinx + cosx
2nn1
sin x+
1
cos x (sin x + cos x) 4
sin x + cos x
2 42
2 =
2
VyV T 2 + 8 = 10. Tng t vi bin yv ta c iu phi chng minhng thc xy ra khi x= y =
4+ k2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
54 Tuyn tp nhng bi h c sc
Cu 99
x
xx= yy+ 8yx y= 5
Gii
iu kin : x, y 0 h ny cho mt phng trnh n gin qu. Th thng ln (1) chng ? Khng nn ! Bin i1 to ri hy th. Hng bin i kh n gin l lm ph v cn thcPhng trnh (1) tng ng
x(x 1) = y(y+ 8) x(x 1)2 =y(y+ 8)2
n y thc hin th x= y + 5 ln (1) v ta c
(y+ 5)(y+ 4)2 =y(y+ 8)2
y= 4
x= 9
Vy h cho c nghim (x; y) = (9; 4)
Cu 100
1x
+y
x=
2
x
y + 2
y
x2 + 1 1 = 3x2 + 3Gii
iu kin : x >0, y= 0R rng vi iu kin ny th t (2) ta thy ngay c nghim th y >0Phng trnh (1) tng ng
x + y
x =
2 (
x +y)
y
x +y = 0(L)
y= 2x
Vi y= 2xthay vo (2) ta c
2xx2 + 1 1 = 3x2 + 3 2x3x2 + 1 = 2x x2 + 1 = 2x2x 3
R rng v tri n iu tng v v phi n iu gim nn phng trnh ny c nghim duynht x=
3 y = 23
Vy h cho c nghim (x; y) = (
3; 2
3)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 55
Cu 101
y= x3 + 3x + 4x= 2y3 6y 2
Gii
Hnh thc bi h kh gn nh nhng cng khin nhiu ngi phi lng tng. Nhn xtx= y = 2l nghim. Ta tin hnh tch nh sau
y 2 = (x + 1)2(x 2)x 2 = (y+ 1)2(y 2)
n y nhn cho v vi v ta c
2(y 2)2(y+ 1)2 = (x + 1)2(x 2)2
D thy V T 0 V P. y ng thc xy ra khi x= y = 2
Cu 102
x3 xy2 + 2000y= 0y3 yx2 500x= 0
Gii
D dng a c v h ng cp. Nhng ta bin i mt to n ti u.
H cho tng ng
x (x2 y2) = 2000yy(x2 y2) = 500x 500x
2(x2 y2) = 2000y2(x2 y2)
x= yx= yx= 2yx= 2y
Thay li vi mi trng hp vo (1) v ta c
y = 0, x= 0
y = 1010
3 , x= 20103y = 10
10
3 , x= 20
10
3
Vy h cho c nghim : (x; y) = (0; 0),
20
10
3;10
10
3
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
56 Tuyn tp nhng bi h c sc
Cu 103
3
x2 + y2 1+ 2y
x= 1
x2 +y2 + 4x
y = 22
Gii
tng t n ph r rng. t x2 + y2 1 =a , yx
=b . H cho tng ng
3
a+ 2b= 1
a +4
b = 21
a= 7, b=
2
7
a= 9, b=1
3
x2 +y2 = 82x= 7y
x2 +y2 = 10x= 3y
y= 4
2
53, x= 14
2
53x= 3, y = 1
Vy h cho c nghim : (x; y) = (3;1)14253 ;4253
Cu 104
x +1
y+
x + y 3 = 3
2x + y+1
y = 8
Giiiu kin : y= 0, x +1
y 0, x + y 3
tng t n ph cng kh r rng.
t
x +1
y =a 0,x + y 3 =b 0. H cho tng ng
a + b= 3a2 + b2 = 5
a= 1, b= 2
a= 2, b= 1
x +1
y = 1
x + y
3 = 4
x +1y
= 4
x + y 3 = 1
x= 410, y= 3 + 10x= 4 +
10, y = 3
10
x= 3, y= 1x= 5, y= 1
Vy h cho c nghim : (x; y) = (3; 1), (5;1)(4 10;310)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 57
Cu 105
x3(2 + 3y) = 8x(y3 2) = 6
Gii
y l mt cu kh ging cu s 37Nghim : (x; y) = (2;1), (1;2)
Cu 106
2x2y+ 3xy= 4x2 + 9y7y+ 6 = 2x2 + 9x
Gii
Bi ny nu li ngh c th dng mn v th thn chng y vo PT(1). Nhng hy dng UCT y s tt hn.Nhn thy y = 3l nghim (ci ny gi li nh, ti khng gii thch na), thay y = 3vo hta c
2x2 + 9x 27 = 027 2x2 + 9x= 0
Nh vy hng ca ta s cng hai phng trnh ban u li v nhn ty 3s xut hin. Vy
P T(1) +P T(2) (3 y) 2x2 + 3x 2 = 0
n y d dng gii ra (x; y) =2;16
7
,
1
2;1
7
,
3(3
33)4
; 3
Cu 107
x2 + 3y= 9y4 + 4(2x 3)y2 48y 48x + 155 = 0
Giiy l mt cu kh hc, khng phi ai cng c th d dng gii n c.Th 3y= 9 x2 t (1) xung (2) ta c
y4 + 8xy2 12y2 16(9 x2) 48x + 155 = 0
y4 + 8xy2 + 16y2 12(y2 + 4x) + 11 = 0
y2 + 4x= 1y2 + 4x= 11
TH1 :
y2 + 4x= 11
9 x23
2
+ 4x= 11
x4
18x2 + 36x
18 = 0
x4 = 18(x 1)2
x2 32x + 32 = 0x2 + 3
2x 32 = 0
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
58 Tuyn tp nhng bi h c sc
x=
3
2
18 1222
y= 12
2 6
36 24212
x=32
18 122
2 y=12
2 6
36 242
12
TH2 :y2 + 4x= 1
9 x2
3
2+ 4x= 1 x4 18x2 + 36x + 72 = 0
x2 6x + 12 x2 + 6x + 6 = 0 x= 3 3 y = 1 23Vy h c c thy 6 nghim nh trn
Mt thc mc nh l TH2 v sao x4 18x2 + 36x+ 72 = (x2 6x+ 12)(x2 + 6x+ 6). Tchnhn t kiu g hay vy ? Casio truy nhn t chng ? C th lm. Nhng thc ra phng trnhbc 4 c cch gii tng qut bng cng thc Ferrari. i vi v d trn ta lm nh sau
x4 18x2 + 36x + 72 = 0 x4 2ax2 + a2 = (18 2a) x2 36x + a2 72Ta phi tm asao cho v phi phn tch c thnh bnh phng. Nh th ngha l
182 = (18 2a) a2 72 a= 9Nh vy
x4 18x2 + 36x + 72 = 0 (x2 + 9)2 = 9(2x 1)2 (x2 6x + 12)(x2 + 6x + 6) = 0
Chi tit v gii phng trnh bc 4 cc bn c th tm d dng trn google. Gi ta tip tc cc
bi h. Tip theo l mt chm h s dng tnh n iu ca hm s kh d nhn.
Cu 108
x +
x2 + 1
y+
y2 + 1
= 1
y+ yx2 1 =
35
12
Gii
iu kin : x2 >1Khng th lm n c g t (2). T (1) ta nhn xt thy hai hm ging nhau nhng chngli dnh cht vi nhau, khng chu tch ri. Vy ta dt chng ra. Php lin hp s gip taPhng trnh (1) tng ng
x +
x2 + 1
y+
y2 + 1
y2 + 1 y
=
y2 + 1 y x +
x2 + 1 = y +
y2 + 1
Tch c ri nhng c v hai bn khng cn ging nhau na. Khoan !! Nu thay y2 = (y)2th sao nh. Qu tt. Nh vy c hai v u c dng f(t) =t+
t2 + 1v hm ny n iu
tng. T ta rt ra x=
y
Thay li vo (2) ta cy+
yy2 1 =
35
12
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 59
y thc ra l mt phng trnh kh kh chu. Thot tin khi thy loi ny ta s bnh phng2 v ln. iu kin bnh phng l y >0khi ta c
y2 + 2y2
y2 1+
y2
y2
1
= 35
122
y4 y2 + y2
y2
1
+ 2y2
y2 1=
35
122
n y kh r rng . t y2
y2 1 =t >0v phng trnh tng ng
t2 + 2t
35
12
2= 0
t=
49
12(L)
t=25
12
y2
y2 1 =25
12
y=
5
4
y= 53
i chiu iu kin bnh phng ch ly 2 gi tr dng.
Vy h cho c nghim : (x; y) =54;54 ,53;53
Cu 109
(4x2 + 1)x + (y 3)5 2y= 04x2 + y2 + 2
3 4x= 7
Gii
iu kin : y5
2 , x 3
4 Vit li phng trnh (1) nh sau
(4x2 + 1)x= (3 y)
5 2y (4x2 + 1)2x= (6 2y)
5 2y f(2x) =f
5 2y
Vi f(t) =t3 + tl hm n iu tng. T ta c 2x=
5 2y x 0thay vo (2) ta c
4x2 +
5
2 2x2
2+ 2
3 4x= 7
Gi cng vic ca ta l kho st hm s v tri trn 0;34v chng minh n n iu gim.
Xin nhng li bn cVi hm s v tri n iu gim ta c x=
1
2l nghim duy nht y = 2
Vy h cho c nghim : (x; y) =
1
2; 2
Hy k mi tng quan gia cc biu thc trong mt phng trnh va ta s t mc ch
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
60 Tuyn tp nhng bi h c sc
Cu 110
y3 + y = x3 + 3x2 + 4x + 21 x2 y= 2 y 1
Gii
iu kin : 0 y 2,1 x 1Phng trnh (1) tng ng
y3 +y = (x + 1)3 + (x + 1) y = x + 1
Thay vo (2) ta c 1 x2 1 +x= 1 x 1
Phng trnh ny khng qu kh. t t =
1 +x+
1 x 1 x2 = t2 2
2 . Thay vo
phng trnh ta c
t2 22
=t 1
t= 0t= 2
1 x + 1 +x= 01 x + 1 +x= 2 x= 0, y = 1
Vy h cho c nghim :(x; y) = (0; 1)
Nhng bi ny thng s nng v gii phng trnh v t hn.
Cu 111x + 1 + x + 3 + x + 5 = y 1 + y 3 + y 5
x + y+x2 + y2 = 80
Gii
iu kin : x 1, y 5Phng trnh u c dng
f(x + 1) =f(y 5)Vi f(t) =
t +
t+ 2 +
t+ 4l hm n iu tng. T ta c y= x + 6thay vo (2) ta
c
x + x + 6 + x2 + (x + 6)2 = 80 x= 55 72 y= 55 + 5
2
Vy h cho c nghim : (x; y) =
5
5 72
;5
5 + 5
2
y ti a ra mt s cu h s dng tnh n iu ca hm s kh n gin. Ni l ngin v t mt phng trnh ta nhn thy ngay hoc mt cht bin i nhn ra dng cahm cn xt. Ti s cn gii thiu kh nhiu nhng bi cn bin i tinh t nhn ra dnghm, nhng cu sau ca cun sch.
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 61
Cu 112
x + 4
32 x y2 = 34
x +
32 x + 6y = 24
Gii
iu kin : 0 x 32C v y l mt h khc rc ri khi xut hin cn bc 4. Ta s dng cc nh gi giiquyt ci h nyCng 2 phng trnh cho nhau ta c
x +
32 x + 4x + 432 x= y2 6y+ 21
Hin nhin ta c : V P 12Gi ta tin hnh nh gi v tri. p dng bt ng thcCauchySchwarzcho v tri ta c
x +
32 x
(1 + 1)(x + 32 x) = 8
4x + 432 x (1 + 1)(x +32 x) 4VyV T V PDu bng xy ra khi (x; y) = (16; 3)
Ti cn mt cu tng ging bi ny nhng hi kh hn mt cht. Bn c c th gii n
Cu 113
2x + 2 4
6
x
y2 = 2
2
42x + 26 x + 22y = 8 + 2
Nghim : (x; y) = (2;
2)
Cu 114
x2(y+ 1)(x + y+ 1) = 3x2 4x + 1xy+x + 1 =x2
GiiBi ny c l khng cn suy ngh nhiu. C th y+ 1 ln (1) coi saoNhn thy x= 0khng l nghim. Phng trnh (2) tng ng
x(y+ 1) =x2 1 y+ 1 = x2 1
x
Thay ln (2) ta s c
x(x2 1)x +x2 1
x = 3x2 4x + 1
x= 2 y= 52
x= 1 y = 1Vy h cho c nghim : (x; y) = (1;1),
2;5
2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
62 Tuyn tp nhng bi h c sc
Cu 115
4xy+ 4(x2 +y2) + 3
(x + y)2 = 7
2x + 1
x + y = 3
Giiiu kin : x + y= 0y l mt bi h khng n gin cht no. Tuy nhin ta c mt nhn xt kh tt sau y :
a(x2 +y2) +bxy =k(x + y)2 + l(x y)2
Gi hy phn tch4x2 + 4y2 + 4xy= k(x + y)2 + l(x y)2Cn bng h s ta thu c : 4x2 + 4y2 + 4xy= 3(x + y)2 + (x y)2Nh vy tng s l t n ph tng-hiu chng ? Cng c c s khi2x= x + y + x y. Nhvy tng s b l th. Bin i h thnh
3(x + y)2 + (x y)2 + 3
(x + y)2 = 7
x + y+ 1
x +y+x y= 3
ng vi t ngay. mt cht 3(x+ y)2 + 3
(x + y)2 = 3
x + y+
1
x + y
2 6. Nh vy
cch t n ca ta s trit hn.t x + y+
1
x + y =a, x y = b ta thu c h mi
b2 + 3a2 = 13a + b= 3|a| 2
a= 2, b= 1
a= 12
, b=7
2(L)
x + y+
1
x + y = 2
x y = 1
x + y= 1x y= 1
x= 1y= 0
Vy h cho c nghim (x; y) = (1; 0)
OK cha ? Tip tc thm mt cu tng t nh
Cu 116
x2 +y2 + 6xy 1(x y)2 +
9
8= 0
2y 1x y +
5
4= 0
Gii
iu kin : x =yH cho tng ng
2(x + y)2
(y x)2
1
(y x)2 +9
8= 0y x + 1
y x
+ (x + y) +5
4= 0
2(x +y)2 y x + 1y x2
+25
8 = 0y x + 1
y x
+ (x + y) +5
4= 0
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.4 Cu 91 n cu 120 63
t x + y= a, y x + 1y x =b, |b| 2ta c h mi
a + b=
5
42a2 b2 = 25
8
a=
5
4b= 5
2
y+ x=
5
4y
x=
2
y+ x= 5
4
y x= 12
x=13
8 , y=
3
8x=
7
8, y=
3
8
Vy h cho c nghim : (x; y) =
7
8;3
8
,
13
8;3
8
Ti s a thm 2 cu na cho bn c luyn tp
Cu 117
3(x2 + y2) + 2xy+ 1
(x y)2 = 20
2x + 1
x y = 5
Nghim : (x; y) = (2; 1),
410
3 ;
10 3
3
,
4 +
10
3 ;
3 103
Cu 118
(4x2 4xy+ 4y2 51)(x y)2 + 3 = 0(2x 7)(x y) + 1 = 0
Th ng no mt cht xem v sao li a c v ging 3 cu trn ?
Nghim :(x; y) = 5
3
2
;1 +
3
2 ,5 +
3
2
;1 3
2
Cu 119
2x
2 +x 1y
= 2
y y2x 2y2 = 2
Giiiu kin : y
= 0
Phng trnh (2) tng ng vi1
y x 2 = 2
y2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
64 Tuyn tp nhng bi h c sc
t a=1
yta chuyn h v
2x
2
+x a= 22a2 + a x= 2
x= 1, a= 1x= 1, a= 1
x=132
, a= 3 12
x=
3 1
2 , a=
1 32
Vy h cho c nghim : (x; y) = (1;1),13
2 ; 1 3
Cu 120
4x2 + y4 4xy3 = 14x2 + 2y2 4xy= 2
Gii
Hnh thc kh gn nh nhng cng rt kh chi. Mt cht tinh ta nhn thy y2 = 1 lnghim ca h. Thay vo v ta rt ra
P T(1) P T(2) y4 4xy3 2y2 + 4xy+ 1 = 0 (y2 1)(y2 4xy 1) = 0
Vi y= 1thay vo (2) ta tm c x= 0hoc x= 1Vi y= 1thay vo (2) ta tm c x= 0hoc x= 1Vi y2 = 4xy+ 1. Khng cn ngh nhiu, th tru b vo cho nhanh !!!
Ta rt ra x=y2 1
4y thay vo (2) ta c
4
y2 1
4y
2+ 2y2 + 1 y2 = 2 5y4 6y2 + 1 = 0
y= 1 x= 0y= 1 x= 0y= 1
5 x= 1
5
y= 1
5 x=
1
5
Vy h cho c nghim :(x; y) = (1; 1), (1;1), (0;1), (0;1),
15
; 15
,
1
5;
15
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.5 Cu 121 n cu 150 65
2.5 Cu 121 n cu 150
Cu 121
x4 + x3y+ 9y = y3x + x2y2 + 9xx(y3 x3) = 7
Gii
Khng cn bit T quc ni u, chin phng trnh u
P T(1) (x y)(x(x + y)2 9) = 0Vi x= y kt hp vi (2) r rng khng thaCn li ta kt hp thnh mt h mi
x (y3 x3) = 7x(x + y)2 = 9
y l mt bi ton kh quen thuc v hp dn tng xut hin trn bo THTT, cch lmph bin nht vn l "tru b"
Trc ht c nh gi x >0v rt ra y = 3
x3 +7
x. Thay xung ta c
x
x +
3
x3 +
7
x
2= 9 x3 + 2x 3
x6 + 7x2 + 3
x(x4 + 7)2 = 9
t v tri l f(x). Ta c
f(x) = 3x2 + 2
3
x6 + 7x2 + 6x6 + 14x2
3 3
(x6 + 7x2)2
+
1
3.9x8 + 70x4 + 49
3
x2(x4 + 7)4>0
Vyf(x) = 9c nghim duy nht x= 1 y= 2Vy h cho c nghim : (x; y) = (1; 2)
Tip theo ti xin gii thiu cho cc bn mt s cu h s dng Bt ng thc Minkowskigii. Bt ng thc Minkowski l mt bt ng thc khng kh v cng thng c dng,bt ng thc cp n vn di ca vect trong khng gian m sau ny hc sinh quen
gi n l bt ng thc V ectorVi hai vectu ,v bt k ta lun c
|u |+ |v | |u +v |Nu ta ha 2 vecto ny ta s thu c
a12 +b12 +
a22 + b22
(a1+ a2)2 + (b1+ b2)
2
ng thc xy ra khi (a1, a2)v(b1, b2)l 2 b t ly l mt h qu hay dng trong gii h
Th khi no nhn vo mt bi h ta c th ngh n s dng Bt ng thc Minkowski. Thngkhi nhn thy tng hai cn thc m bc ca biu thc trong cn khng vt qu 2 th ta cth chn hng ny. Ti s nu 3 v d bn c hiu r hn
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
66 Tuyn tp nhng bi h c sc
Cu 122
3x + 4y= 26x2 + y2 4x + 2y+ 5 +
x2 + y2 20x 10y+ 125 = 10
Gii
tng s dng hin r ri. Bc u tin ta lm l phn tch biu thc trong cnthnh tng cc bnh phng . V tri ca (2) khi s l
(x 2)2 + (y+ 1)2 +
(x 10)2 + (y 5)2
Tuy nhin nu ta s dng Bt ng thc Minkowskingay by gi th n s l
V T
(x 2 +x 10)2 + (y+ 1 +y 5)2
Khng phi 10 na m l mt biu thc kh phc tp. Khi ta phi xem li cch vit cc
bnh phng ca mnh nu l hng s v phi th khi cng vo ta phi lm trit tiu n i. Vy cn phi vitnh sau
V T =
(x 2)2 + (y+ 1)2+
(10 x)2 + (5 y)2
(x + 2 + 10 x)2 + (y+ 1 + 5 y)2 = 10
Ok ri. ng thc xy ra khi 10 x
x 2 =5 yy+ 1
3x 4y = 10Kt hp (1) d dng gii ra (x; y) = (6; 2)
Nh ta thy, s dng khng kh. Tuy nhin ci kh y chnh l ngh thut i du vsp xp cc hng t ca bnh phng ta t c mc ich
Cu 123
x2 2y2 7xy= 6x2 + 2x + 5 +
y2 2y+ 2 =
x2 + y2 + 2xy+ 9
Gii
Xt phng trnh (2) ta c
V T =
(x + 1)2 + 22 +
(y 1)2 + 12
(x + y)2 + 32 =V P
ng thc xy ra khi x + 1 = 2(y 1) x= 2y 3Thay vo (1) v ta d dng gii ra (x; y) =
5
2;1
4
, (1;1)
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.5 Cu 121 n cu 150 67
Cu 124
2x2 + 6xy+ 5y2 + 5 =
2x2 + 6xy+ 5y2 + 14x + 20y+ 25x4 + 25y2 2 = 0
Gii
By gi nu chuyn cn sang v tri, hng s sang v phi l cht d. Mu cht y l g ?S 5 chng ? ng vy, ta phn tch 5 =
32 + 42 s dng bt ng thc Minkowski. Tuy
nhin cc i du v sp xp s hng nh th no. Ci ta phi quan tm n v phi chn la cho ph hp. y s l
V T =
(x + y)2 + (x + 2y)2 +
42 + 32
(x + y+ 4)2 + (x + 2y+ 3)2 =V P
ng thc xy ra khi x + y
4 =
x + 2y
3 x= 5y
Thay vo (2) v ta d dng gii ra (x; y) =
1;15
,1;15
Cu 125
2y(x2 y2) = 3xx(x2 +y2) = 10y
Gii
Mt h a v dng ng cp r rng. Tuy nhin, ta hy x l s b h ny loi mt strng hpT (2) d thy x.yphi cng du, m nu th (1) x2 y2Trc ht x= y = 0l mt nghim ca hNhn cho 2 phng trnh cho nhau ta c
20y2(x2 y2) = 3x2(x2 +y2) (x 2y)(2y+ x)(5y2 3x2) = 0
V x v y cng du nn nn t y ta suy ra x= 2yhoc x=
5
3y
n y ch vic thay vo (1). Xin nhng li cho bn cVy h cho c nghim :(x; y) = (0; 0), (2; 1), (2,1),
4
30375
6 ;
4
135
2
,
4
30375
6 ;
4
135
2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
68 Tuyn tp nhng bi h c sc
Cu 126
7 +x +
11 y = 67 +y+
11 x= 6
Gii
Cng 2 phng trnh cho nhau ta c
7 +x +
11 x +
7 +y+
11 y= 12
p dng bt ng thc Cauchy Schwarzcho v tri ta c
V T
(1 + 1)(7 + x + 11 x) +
(1 + 1)(7 +y+ 11 y) = 12Du bng xy ra khi (x; y) = (2; 2)
Cu 127
2x2y2 + x2 + 2x= 22x2y x2y2 + 2xy= 1
Gii
Bin i 1 t, h cho tng ng
2x2y2 + (x + 1)2 = 3
2xy(x + 1) x2
y2
= 1 (xy+ x + 1)2 = 4
xy= 1 xxy= 3 x
Vi xy= 1 xthay vo (1) ta c
2(1 x)2 + x2 + 2x= 2
x= 0(L)
x=2
3 y=1
2
Vi xy= 3 xthay vo (2) ta c
2(x + 3)2
+ x2
+ 2x= 3 x=
8
3y =
1
8x= 2 y = 12
Vy h cho c nghim : (x; y) =
2
3;1
2
,
8
3;1
8
,
2;1
2
Nguyn Minh Tun - K62CLC Ton Tin - HSPHN. My facebook : Popeye Nguyn
5/25/2018 Tuy n ch n 410 h ph ng tr nh v c c ph ng ph p gi i h ph ng tr nh
NguynM
inhTu
n
2.5 Cu 121 n cu 150 69
Cu 128
(x 1)(y 1)(x + y 2) = 6x2 + y2 2x 2y 3 = 0
Gii
Bi ny tng t n ph r rngt x 1 =a, y 1 =b ta a v h sau
ab(a +b) = 6a2 +b2