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8/3/2019 Chuyen de He Phuong Trinh Dai So
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NHOM THC HIEN
Dng Minh Ch(Bin tp)
V Hoi BoHunh Nguyn Minh T
Phm Anh ThH Ngc Trm
Nguyn Bch Vi
L Hu Tr(Kthut)
Trn Ch ThinNguyn Phc HinNguyn Thanh ViTrn Th Anh Th
Nguyn Th Qu Chi
Gio Vin: Hunh Ch Ho
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Mng ton h phng trnh l mng ton khng kh
lm, nhng n li c nhiu dng v nhiu cch gii khc
nhau, thi gian gn y (t 2002 n 2010) cc thi Cao
ng v i hc thng cho m nh im l nm 2010
thi khi A v khi B c cc cu ny, a s cc hc sinh
u bngv cha tng luyn gii loi ny.
Nay nhm mnh son quyn chuyn H phngtrnh nhm gip cc bn c ti liu tham kho v n tp.
Trong cun chuyn ny nhm mnh chia ra tng loi
cc bn s thun tin hn trong vic tham kho.
Trong quyn chuyn ny cc bn sc gp mt
s cch gii hay, ngn gn v chnh xc. Mc d c gng
khi bin son, nhng khng th trnh khi vi thiu st,
mong cc bn gp v thng cm quyn chuyn ny
c hon thin hn.
Nhm 2
vntoanhoc.com
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& &Muc Luc
Trang
Phn 1 CC H PHNG TRNHTHNG THNG
1. H phng trnh bc nht hai n...4
2. H phng trnh bc nht ba n9
3. H phng trnh i xng loi I.16
4. H phng trnh i xng loi II245. H phng trnh ng cp bc hai30
Phn 2 CC H PHNG TRNH KHC
1. H phng trnh bc cao hai n..36
2. H phng trnh v t......40
3. H phng trnh khng mu mc...47
4. H phng trnh dng phng php hnh hc vect...56
5. H phng trnh trong cc k thi.60
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1Cac He Phng Trnh
Thong Thng
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1. H PHNG TRNH BC NHT HAI N
1/ Dng:
1 1 1
2 2 2
a x b y c
a x b y c+ =+ =
Cch gii:php th, php cng...
Php th:
V d 1: ( )2 1
12 3
+ =
=
x y
x y
Gii
( )1 2 1 2
12 3 2(1 2 ) 3
1 71 21 2
5 51
15 5
x y x y
x y y y
xx y
yy
= =
= =
= == =
=
Vy ( )7 1
; ;5 5
x y x y
= = =
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Php cng:
V d 2:2 2 5 2 2 5
3 6 2 6 12
x y x y
x y x y
+ = + =
= =
Cng 2 v phng trnh li, ta c:4 17
174
y
y
=
=
Gii h phng trnh (1) ta c27
4x
=
Vy ( )27 17
; ;4 4
x y x y
= = =
2/ Gii v bin lun phng trnh:
Bc 1: Tnh cc nh thc:
1 1
1 2 2 1
2 2
a ba b a b
a bD = =
(gi l nh thc ca h)
1 1
1 2 2 1
2 2
x
c bc b c b
c bD = =
(gi l nh thc ca x)1 1
1 2 2 1
2 2
y
a ca c a c
a cD = =
(gi l nh thc ca y)
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Bc 2:Bin lun:
NuD 0 th h c nghim duy nht:
;yx
x yDD
D D= =
NuD = 0 vDx 0 v Dy0 th h v nghim
NuD = Dx= Dy= 0 th h c v s nghim
hoc v nghim.
V d 1: Gii h phng trnh:
5 2 9
4 3 2
x y
x y
=
+ =
5 215 8 23
4 3Ta c: D
= = + =
9 2
27 4 232 3
xD
= = + =
5 9
10 36 464 2
yD
= = + =
23 46
1 223 23
;yx
x yDD
D D
= = = == =
Vy nghim ca h phng trnh l (1;2)
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V d 2: Gii v bin lun h phng trnh:
1
2
mx y m
x my
+ = +
+ =
Gii
2
2
1 1 ( 1)( 1)1
1 12 ( 1)( 2)
2
12 1 1
1 2
x
y
mm m m
m
mm m m m
m
m mm m m
D
D
D
= = +
+= + = +
+= =
=
=
=
Bin lun:NuD 0 m 1 th h phng trnh
c nghim duy nht l
2
11
1
mx
m
ym
+ = + = +
NuD = 0 m = 1 th:
Khi m = 1 ta c h phng trnh:2
22
x yx y
x y
+ = + =
+ =
H c v s nghim: ( )0 00
2
=
=
x xx
y x
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Khi m = 1 ta c h phng trnh:
0
2
0
(VN)2
x y
x y
x y
x y
+ =
=
=
=
3/ Gii h phng trnh bng phng php t n ph:
V d 3: Gii h phng trnh:
( )3( ) 9 2( )
22( ) 3( ) 11
x y x y
x y x y
+ + =
+ =
Gii
( )3( ) 2( ) 9
22( ) 3( ) 11
x y x y
x y x y
+ = + =
t u = x + y; v = x y
Ta c h phng trnh:
3 2 9 12 3 11 3
1 1
3 2
u v uu v v
x y x
x y y
= = = =
+ = =
= =
Vy h phng trnh c nghim ( 1; 2)
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2. H PHNG TRNH BC NHT BA N
1/ Cc phng php chung: Nguyn tc chung gii h phng trnh nhiu n
vn l bin i h phng trnh cho thnh nhng htng ng hoc nhng h phng trnh h qu d gii
hn, (trong c nhng phng trnh vi sn ngy cngt). t c iu ny ta thng dng:-Phng php cng i s-Phng php th
Nu c dng php bin i khng tng ng thcn phi th li cc gi tr tm c ca n.V d 1: Gii h phng trnh :
2 2
3 3
3
( ) 5
9
x y z
I x y z
x y z
+ =
+ = + =
Gii
2 2
3 23
2 2
3 2
3 3
( ) ( ) 2 5 9 5 2
( ) 3 ( ) 9 9 527 9 9
2
3 3
2 9 5 2 9 5
23 5 2 0 0 13
+ = + =
+ = = + + = =
+ = + =
= = + = = = =
x y z x y z
I x y xy z z z xy
x y xy x y z z z z z z
x y z x y z
xy z z xy z z
z z z z z z
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Vi 0z= ta c :
0
0
0
x y
xy
z
+ =
= =
H ny c nghim ( )0;0;0
Vi 1z= ta c :
3
2
1
x y
xy
z
+ =
= =
Gii h ta c hai nghim : ( ) ( )1;2;1 , 2;1;1
Vi 23z= ta c :
2
1
32
3
x y
xy
y
+ =
=
=
H c hai nghim :3 6 3 6 2 3 6 3 6 2
; ; , ; ;3 3 3 3 3 3
+ + .
Vy h (I) c 5 nghim :
( ) ( ) ( )3 6 3 6 2 3 6 3 6 2
0;0;0 , 1;2;1 , 2;1;1 , ; ; , ; ;3 3 3 3 3 3
+ +
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V d 2: Gii h phng trnh :
( )
12
III 20
15
xy
yz
zx
=
= =
Nh
n xt: Nghin cu cch gii. Hy th gii h ny bngphng php th.
Tuy nhin c th nhn xt v du ca , ,x y z ramt cch gii cch.
R rng , ,x y z khc 0 v cng du.
Gii
Nhn v vi v ca ba phng trnh ta c mtphng trnh; kt hp vi hai trong ba phng trnh cho
ta c h:
( )2
12
20
3600
xy
yz
xyz
=
=
=
Ta d thy rng ( ) ( )III IV
V ( )2 60
360060
xyzxyz
xyz
== =
Nn h ( )IV tng ng vi hai h :
( )
12
V 20
60
xy
yz
xyz
=
= =
v ( )
12
VI 20
60
xy
yz
xyz
=
= =
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Gii ( )V : Thay 12xy = vo phng trnh th bat a c
5=z ; thay 20yz= vo phng trnh th ba ta c 3=x .Thay 3, 5= =z vo phng trnh th ba ta c 4=y .
H ( )V c nghim: ( ).3;4;5
Gii ( )IV : Tng t h ( )IV c nghim ( )3; 4; 5 . Vy h ( )III c hai nghim: ( ) ( )3;4;5 , 3; 4; 5 .
2/ p dng h thc Vit i vi phng trnh bc ba:
H thc Vit i vi phng trnh bc ba c phtbiu nh sau :
NH L:Nu phng trnh 3 2 0+ + + =ax bx cx d c
ba nghim 1 2 3, ,x x th :
1 2 3
1 2 2 3 3 1
1 2 3
+ + =
+ + =
=
bx x x
a
cx x x x x x
a
dx x x
a
Ngc li, nu ba s 1 2 3, ,x x x tha mn cc ng thc :
1 2 3 1
1 2 2 3 3 1 2
1 2 3
x x x S x x x x x S
x x x P
+ + = + + = =
th chng l ba nghim ca phng trnh :
3 21 2 0 + =x S x S x P
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Theo nh l trn y c thgii h phng trnh ba n
1 2 3 1
1 2 2 3 3 1 2
1 2 3
+ + =
+ + =
=
x x x S
x x x x x x S
x x x P
bng cch gii mt phng trnh bc ba
3 21 2 0 + =x S x S x P
V d 3: Gii h phng trnh :
1 2 3
1 2 2 3 3 1
1 2 3
1
10
8
x x x
x x x x x x
x x x
+ + =
+ + = =
GiiTheo nh l Vit , ,x y z l ba nghim ca phng trnh
3 2 10 8 0x x x+ + = Ta thy 1=x l mt nghim ca phng trnh trn. Do
phng trnh trn c vit thnh : 2( 1)( 2 8) 0x x x + =
Gii phng trnh ny ta c : 1 2 31, 4, 2x x x= = =
V h phng trnh ny i xng i vi , ,y z nn h c
3! 6= nghim sau :
( ) ( ) ( ) ( ) ( ) ( )1; 4;2 , 1;2; 4 , 2; 4;1 , 4;1;2 , 2;1; 4 , 4;2;1
V d 4:Gii h phng trnh :
2 2 2
4
( ) 22
18
x y z
II x y z
xyz
+ + =
+ + = =
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GiiBin i h ny thnh h c dng ( )I . Bnh phng hai v
phng trnh th nht ri tr tng v vi phng trnh thhai ta c :
4 4
2( ) 4 hay 3
18 18
+ + = + + =
+ + = + + = = =
x y z x y z
xy yz zx xy yz zx
xyz xyz
Theo nh l Vit, , ,y z l ba nghim ca phng trnh :3 24 3 18 0x x x+ + =
Ta c th thy 2x = l mt nghim. Do :( )( )22 6 9 0x x x + + =
Gii phng trnh ny ta c : 1 2 32, 3x x x= = =
Vy h c ba nghim: ( ) ( ) ( )2; 3; 3 , 3;2; 3 , 3; 3;2
V d 5: Gii h phng trnh :
( ) 2 2 2
3 3 3
2
III 6
8
x y z
x y z
x y z
+ + =
+ + = + + =
Gii
( ) ( ) ( )
( ) ( ) ( )
2
2 2 2 2 2 2 2 2 2
2
III 2 6
8
x y z
x y z xy yz zx
x y z x y z xy x y yz y z zx z x
+ + =
+ + + + =
+ + + + + + + + + =
2
2 2 2
2
( ) 2( ) 6
( )( ) ( )( ) 3 8
x y z
x y z xy yz zx
x y z x y z xy yz zx x y z xyz
+ + =
+ + + + = + + + + + + + + + =
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2
1
2
x y z
xy yz zx
xyz
+ + =
+ + = =
Vy , ,y z l ba nghim ca phng trnh bc ba2 2 2 0t t t+ = hay 2( 2)( 1) 0t t+ =
Phng trnh ny c nghim l : 1 2 31, 1, 2t t t= = =
Do h phng trnh c 6 nghim l hon v ca ba s1,1, 2 :
( 1;1; 2), ( 1; 2;1), (1; 1; 2), (1; 2; 1), ( 2; 1;1), ( 2;1; 1)
BITPRNLUYN
Gii cc h phng trnh :2 2 2
2 2 2 2 2 2
2 2 2 3 3 3
( ) 2 2 1
a) ( ) 3 b) 3 c) 9
( ) 4 4 1
+ = + = + + + =
+ = + = + + + = + = + = + + + =
y z x zx xy x x y z
z x y xy yz y x y z
x y z yz zx z x y z
2
2
2
1 1 13
6 ( ) 13( )( )1 1 1
d) ( )( ) 2 e) 3 f) 3 ( ) 5
( )( ) 3 6 ( ) 51
1
+ + =
+ =+ + =
+ + = + + = + = + + = + =
=
x y z x y z yz x y z x x
y z x y y y z x zxxy yz zx
z x y z z z x y xy
xyz
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3. H PHNG TRNH I XNG LOI I
TM TT L THUYTV PHNG PHP GII TON:
1/ Dng tng qut ca hi xng loi I:nh ngha: Hi xng loi I l h cha 2 n x,y
m khi ta thay i vai tr x,y cho nhau th h phng trnhkhng thay i.
( )
( )
, 0
, 0
f x y
g x y
=
=, trong
( ) ( )
( ) ( )
, ,
, ,
f x y f y x
g x y g y x
=
=
Phng php gii tng qut:
i) Bc 1: t iu kin (nu c)ii) Bc 2: tS = x + y;P = xy (viS2 4P) .
Khi , ta a h v h mi chaS,P.
iii) Bc 3: Gii h mi tmS,P. ChnS,Ptha mn
S2 4P.
iiii) Bc 4: ViS,Ptm c thx,y l nghim caphng trnh:
X2
SX + P = 0
( nh l Vit o)*Ch :
i) Cn nh:2 2 2
3 3 3
2
3
x y S P
x y S SP
+ =
+ =
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ii) i khi ta phi t n ph:
( )
( )
u u x
v v x
=
=v
S u v
P uv
= +
=
iii) C nhng h phng trnh trthnh
hi xng loi I sau khi ta t n ph.
2/ Mt s v d minh ha:
V d 1:Gi h phng trnh sau:
2 2
2 2 7 8x y xyx y x y
+ + = + + + =
(1)
GII
t:S x y
P xy
= +=
, vi S2 4P.
Khi , h (1) trthnh:
( )
22
2 22
2
2
2
77
2 7 82 8
77
36 0
2
= =
+ = + =
= =
= = =
P SS P
S S SS P S
P SP S
SS S
S
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Vi: 3 2S P= = . Khi , x v y l nghim ca phng
trnh: 2 3 2 0X X + =
1
21
2 21
x
yX
X x
y
= ==
= = =
Vi: 2 3S P = = . Khi , x v y l ngim ca
phng trnh: 2 2 3 0X X+ =
1
31
3 31
= = =
= = =
x
yX
X xy
Vy h cho c 4 nghim
(x,y) = (1;2), (2;1), (1;3), (3;1).
V d 2: Gii h phng trnh:
2 22 2
1 1 5
1 19
x yx y
x yx y
+ + + = + + + =
( H Ngoi Thng TPHCM - Khi A,D nm 1997)
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GII
t:
2
2
22
22
1 12
1 1 2
u x x ux x
v y y vy y
= + + =
= + + =
Khi , h (1) trthnh:
( )
2 2
2
5
13
5
2 13
5
6
u v
u v
u v
u v uv
u v
uv
+ =
+ =
+ =
+ =
+ =
=
u, v l nghim ca phng trnh: X2 5X+ 6 = 0
3
2
2
3
3
2
= =
= =
=
=
X
X
u
v
u
v
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Trng hp 1: u = 2; v = 3
12
13
1 1
3 5 3 5
2 2
xx
y
yx x
y y+
+ =
+ =
= =
= =
Trng hp 2: u = 3; v = 2
1 3
12
3 5 3 5
2 2
1 1
xx
yy
x x
y y
+
+ =
+ =
= =
= =
Vy h cho c 4 nghim (x,y) l
3 5 3 5 3 5 3 51; , 1; , ;1 , ;1
2 2 2 2
+ +
.
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3/ iu kin tham s hi xng loi I c nghim:
Phng php gii tng qut:
i) Bc 1: t iu kin (nu c).
ii) Bc 2: tS = x + y; P = xy vi iu kin caS,Pv S2 4P(*).
iii) Bc 3: Thayx,y biS,Pvo h phng trnh.Gii h tmS,Ptheo m, ri tiu kin (*) tm m.
(vi m l tham s)
V d 3: Tm iu kin m h phng trnh sau cnghim:
( )
4 1 4
13
+ =
+ =
x y
x y m
GII
t:4 0
1 0
=
=
u x
v y
Khi , h (1) trthnh:
2 2
4
3 5
u v
u v m
+ = + =
4
21 32
u v
muv
+ =
=
Suy ra u,v l nghim (khng m) ca phng trnh:
2 21 34 0 (*)2
mX X
+ =
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Theo , h (1) c nghim Pt (*) c 2 nghim khng m.
3 13' 0 01320 7.
21 3 300
2
m
P mm
S
Vy 13 73
m l gi tr cn tm.
V d 4: Tm m h phng trnh sau c nghim thc:
1(1)
1 3
x y
x x y y m
+ =
+ =
GII
iu kin: 0; y 0x
Khi :
( ) ( )3 3
11
1 3 1 3
+ =+ =
+ = + =
x yx y
x x y y m x y m
t: ( )2
0; P= 0 4S x y xy S P = +
H phng trnh trthnh:
3
1 1
3 1 3
= =
= =
S S
S SP m P m
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H (1) c nghim thc2 4
1 4 10 0 m
0 40
S Pm
Pm
S
Vy1
04
m l gi tr cn tm.
BI TP RN LUYN
Bi 1: Gii h phng trnh: 2 219
133
x y xy
x y xy
+ + =
+ + =.
Bi 2: Gii h phng trnh:2 2
2 2
1 14
1 14
x yx y
x yx y
+ + + = + + + =
.
Bi 3: Tm m h phng trnh c ng 2 nghim thc
phn bit.2 2
2
2 1
4
( )
( )
y m
x y
+ = + + =
Bi 4: Tm m h phng trnh sau c nhgim thc:2 2
4 4 104 4
( )( )
x y x yy x y m
+ + + =+ + =
Bi 5:Tm m h phng trnh c nghim thcx > 0,y > 0:
2 2
1x xy y m
x y xy m
+ + = + + =
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4. H PHNG TRNH I XNG LOI II
1/ nh ngha:
H phng trnh i xng loi II l h cha hai nx,y mkhi ta thay i vai trx,y cho nhau th phng trnh ny tr
thnh phng trnh kia ca h.*Ch :Nu 0 0( ; )y l nghim ca h th 0 0( ; )y x cng l nghim ca h.
2/ Cc dng ca h phng trnh i xng loi II:
DDnngg 11::( , ) 0
( , ) 0
f x y
f y x
=
=(i v trx vy cho nhau th
phng trnh ny trthnh phng trnh kia).
Phng php gii chung:Tr v vi v hai phng trnh v bin i v dng
phng trnh tch s.
Kt hp mt phng trnh tch s vi mt phng trnhca h suy ra nghim ca h
V d1:Gii h phng trnh sau:2
2I 2( ) 2x y
y y x
= =
Nhn xt: Nu thay ng thix biy vy bix thphng trnh th nht s trthnh phng trnh th hai vngc li.
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Gii
Tr tng v hai phng trnh trong h, ta c
( )( ) 2( ) ( )
( )( 1) 0
0
1 0
y x y x y x y
x y x y
x y
x y
+ =
+ =
=
+ =
Do , h phng trnh cho tng ng vi:
2
0(Ia)
2
x y
x x y
=
=hoc
2
1 0(Ib)
2
x y
x y y
+ =
=
Gii h (Ia) ta c nghim (0;0), (3;3).
Gii h (IIa) ta c nghim:
1 5 1 5 1 5 1 5; , ;2 2 2 2
+ +
Vy h phng trnh c 4 nghim l
(0;0), (3;3),1 5 1 5 1 5 1 5
; , ;2 2 2 2
+ +
DDnngg 22::( , ) 0
( , ) 0
f x y
g x y
= = (trong ch c 1 phng trnh
i xng loi I)
Cch gii:a phng trnh i xng v dng tch,giiy theox ri th vo phng trnh cn li.
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V d 2: Gii h phng trnh:2
1 1(1)
2 1 0 (2)
x yx y
x xy
= =
Giiiu kin: 0; y 0x . Khi :
1(1) ( ) 1 0 1
x yx y
xy yx
= + = =
Vix = y th (2) 2 1 0 1x x = =
Vi1
y = th (2) v nghim
Vy h phng trnh c 2 nghim phn bit (1;1), (1;1).
3/ Mt s bi tp v phng trnh i xng loi II :
V d 3: Gii h phng trnh:2
2
3 2
3 2
x y
y y x
=
=
Gii
Tr v theo v ca hai phng trnh, ta c:2 2 3 3 2 2
0(x-y)(x+y-1) 01 0
x y x y y x
x yx y
+ =
= = + =
Vy h phng trnh cho tng ng vi:2 3 2
(I)0
x y
x y
=
=hoc
2 3 2(II)
1 0
x y
x y
=
+ =
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Gii (I):2 ( 5) 03 2
( ) 0 5x xx x x
I x y x yx yx y
= = = = = =
==
Gii (II):2 23 2(1 ) 2 0
( ) 1 1
1 2
2 1
x x x x x
II y x y x
x x
y y
= =
= =
= =
= =
Vy h phng trnh c bn nghim(0;0), (5;5), (1;2), (2;1).
V d 4: Gii h phng trnh2 3 4 4 (1)
2 3 4 4 (2)
x y
y x
+ + =
+ + =
Gii
iu kin:
34
23
42
x
y
.
Ly(1) tr (2) ta c:
( ) ( )2 3 2 3 4 4 0
(2 3) (2 3) (4 ) (4 )
02 3 2 3 4 42 1
( ) 02 3 2 3 4 4
x y y x
x y y x
x y y x
x yx y y x
y
+ + + =
+ +
+ =+ + + +
+ = = + + + +
Thayx = y vo (1), ta c:2 3 4 4 7 2 (2 3)(4 ) 16x x x x x+ + = + + + =
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22
39 02 2 5 12 9 11
9 38 33 09
= + + =
+ = =
xxx x x
x x x
Vy h phng trnh c 2 nghim phn bit
( ) ( )11 11
; 3;3 , ;9 9
x y
= .
V d 5: Gii h phng trnh
2
2
2
2
12
12
yy
x
xx
y
+=
+ =
Giiiu kin: , 0>x y Khi , h phng trnh cho tng ng
2 2
2 22 1 (1)2 1 (2)
yx y
xy x
= += +
Ly (1) tr (2) v theo v ta c:
( ) ( )
2 ( )
( ) 2
(3)
0 2 0
m
=
+ + = + + >
=
xy x y y x
x y xy x y xy x y
x y
Thay (3) vo (1) ta c:3 2
3 2
2
0
2 1
2 1 0
( 1)(2 1) 0 1x
x x
x x
x x x x
>
= +
=
+ + = =14243
Vy h phng trnh c nghim duy nht(x;y) = (1;1).
(thok)
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BI TP RN LUYN
Bi 1:Gii h phng trnh:
) )
) )
22
2
2
321
3 12
4 43 3
4 43 3
x yy x yx
a by y xy x
y
y yx y x y
xc d
x xy x y x
y y
+ = + = +
+ = + + =
= = = =
Bi 2: Tm a h sau c nghim duy nht:
2
2
( 1)
( 1)
x xy a y
y xy a x
+ =
+ =
Bi 3: Chng minh rng vi 0a th phng trnh
sau c nghim duy nht:
22
22
2
2
ax y
y
ay x
x
= + = +
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5. H PHNG TRNH NG CPBC HAI
1/ nh ngha:Biu thcf(x; y) gi l phng trnh ng cp bc 2nu
f(mx; my) = m2f(x; y)
H phng trnh ng cp bc hai c dng:
( )
( )
,
,
=
=
f x y a
g x y b
Trong :f(x; y) vg(x; y) l phng trnh ng cp bc 2;
vi a v b l hng s.
2/ Cch gii:
Xt x = 0 thay vo h kim tra.
Vix 0 ta ty = xtthay vo h ta c:
( )
( )
( )
( )
2
2
, 1,
, 1,
= =
= =
f x xt a x f t a
g x xt b x g t b
Sau , chia 2 v ca 2 phng trnh vi nhau ta c:( ) ( ) ( )1, 1, *=
af t g t
b
Gii phng trnh (*) ta tm c t.
Thtvo h ta tm c (x; y).
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3/ Cc v d:
V d 1: Gii h phng trnh sau:( )
( )2 2
2 2
2 3 121
2 14
+ + =
=+
x y xy
yx y
Gii
D thyx = 0 khng l nghim ca h phng trnh
Vix 0 ta ty = xt. Khi h phng trnh trthnh:
( )
( )
( )
2 22 2 2 2
2 2 2 2 2 2
22
2
3 2 122 3 12
2 4 14 4 2 14
k : 4 2 03 2 62
4 2 7 2 2
+ + = + + =
+ + = + + =
+ + + + = + +
x t t x x t x t
x x t x t x t t
t tt t
t t t
Khi (2) 21
3 2 02
= + = =
tt t
t(tha)
Khi t = 1 th vo h ta c (x; y) = ( )2 2;
Khi t = 2 th vo h ta c (x; y)= (1; 2), (1; 2)
Vy nghim ca h l:(x; y) = ( )2 2; , (1; 2), (1; 2)
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V d 2: Tm m h phng trnh sau c nghim:
( )
2 2
2 21
+ + =
+ + =
x xym y m
x m xy my m
Gii
D thyx = 0 khng l nghim ca h phng trnh
Vix 0 ta ty = xt. Th vo h phng trnh ta c
( )
( )
( )
( )( )
2 22 2 2 2
2 2 2 2 2 2
22
2
1
1 1
011 1 0
1 11
+ + = + + =
+ + = + + =
=+ + = + =
=+ + +
x t tm mx x tm x t m
x m x t x t m m x t m tm t m
tt tmm t t
m tt m tm t
Khi t = 0 th ( )2
k : 00
=
=
x mm
y
Khi (1m)t= 1 ( )2
2
1*
2 3 2
=
= +
xy
m
my m m
V2
2 732 3 2 2 084
m m m + = + >
nn (*) c ngha 1m
Vy vi 1m th h phng trnh trn c nghim.
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V d3: Cho h phng trnh sau:2 2
2
4
3 4
+ =
=
x xy y m
y xy
Chng minh h phng trnh lun lun c nghim m .
GiiKhix = 0 khng l nghim ca h phng trnh.
Vix 0 ta ty = xt. Khi h phng trnh trthnh
( )
( )
( )
2 22 2 2 2
2 2 2 2 2
22
2
4 14
4 3 4
k : 3 04 1*3 4 3
+ = + =
+ = =
+ =
x t t mx x t x t m
x t x t x t t
tt t m
t t
Khi ( ) ( ) ( ) ( )24 16 3 4 0* ** + =m t m t
Vi m = 4 th (**) c dng 4 4 0 1 + = =t t (tho)
Vi m 4 th (**) c dng:
( ) ( )
24 16 3 4 0 + =m t m t
Vi2
2 40 1289 80 192 3 03 9
m m m = + = + >
Vy h phng trnh lun lun c nghim m .
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BI TP RN LUYN
Bi 1: Gii cc h phng trnh sau:
)
)( )
)
)
)
2 2
2 2
2 2
2 2 2
2 2 2
2 2
2 2
2 2
2 2
3 2 11
2 3 17
5 2
10
2 2 2
2 2 1
2 2
2 2 2 0
2 3 3
2 2
x xy ya
x xy y
x y xyb
y x y
x y x xc
x y x y xy
x y xy y xd
x y y
x y x xy yd
x y x y
+ + =
+ + =
+ =
+ =
+ + =
+ =
+ + + + =
=
+ = + +
+ = +
Bi 2: Tm gi tr ca m phng trnh c nghim:
2 2
2 2
3 2 11
2 3 17
+ + =
+ + = +
x xy y
x xy y m
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2Cac He Phng Trnh
Khac
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1. H PHNG TRNH BC CAO HAI N
H phng trnh bc cao hai n l h gm hai phngtrnh hai n trong c t nht mt phng trnh c bc lnhn 1.
Phng php chung:Cc h phng trnh bc cao thng kh gii v khng
th nu ra tng phng php cu th gii.Do phngphp thng c s dng l chuyn chng v h phngtrnh bc hai bng mt trong hai phng php:
Phng php bin i tng ng.Phng php t n ph.
Cc dng h phng trnh thng gp:H phng trnh i xng loi 1H phng trnh i xng loi 22/ Mt s v d minh ha:
V d 1:
( )2 2
4 4
31
17
x xy y
x y
+ + =
+ =
GiiKhi (1) trthnh:
2 2 2
2 2 2 2 2 2 2 2
2 2
2 2 2 2 2 2
2
3 ( ) 3
( ) 17 2 (3 ) 17 2
( ) 3 ( ) 39 6 17 2 6 8 0
( ) 31
22
4
x y xy x y xy
x y x y xy x y
x y xy x y xyxy x y x y x y xy
x y xyx y
xyxy
xy
+ + + = +
+ = + = +
+ = + + = + + = + + + =
+ = ++ =
= =
=
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Vi1
2x y
xy
+ ==
x,y l nghim ca phng trnh:
1 2
1 2
2 1 2
2 1
12 0
2
x x
y y
tt t
t
= = = =
= =
=
Vi
=
=+
2
1
xy
yx
x,y l nghim ca phng trnh:
3 42
3 4
1 212 0
2 2 1
x xtt t
t y y
= = = + = = = =
Vy h phng trnh c bn cp nghim l
(1;2), (2; 1), (1; 2), (2;1)
V d 2:Gii cc h phng trnh sau:
) ( )3 3 2 126+ =+ =
x yax y
Gii
t:x y S
xy P
+ =
=; iu kin 042 PS
Khi :
( )3 3
2 2 21
3( ) 3 ( ) 26 3 26
x y S S
Px y xy x y S SP
+ = = =
= + + = =
Vi S = 2 ; P = 32 1 3
3 3 1
x y x y
xy x y
+ = = = = = =
Vy h phng trnh c hai cp nghim l (1;3),(3; 1)
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) 2 2 3 34
( )( ) 280(2)
+ =
+ + =
x yb
x y x y
Gii
( ) 2 3
2
2 [( ) 2 ][( ) 3 ( )] 280
(16 2 )(64 12 ) 2803
3( ) 40 93 0 31
3
x y xy x y xy x y
xy xy
xy
xy xyxy
+ + + =
==
+ = =
Vixy = 3 ta c:
=
=+
3
4
xy
yxx;y l nghim ca phng trnh:
2
x 1
y 3t 1t 4t 3 0
t 3 x 3
y 1
= == + = = =
=
Vi3
31=xy ta c:
=
=+
331
4
xy
yx
x;y l nghim ca phng trnh:
03
3142 =+ tt (v nghim)
Vy h c hai cp nghim l (1;3),(3;1)
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BI TP RN LUYN
Bi 1: Gii h phng trnh:
)
)
)
)
3 3
4 4 2 2
2 22 2
2 2
4 4
7( ) 2
1
12
1 14
1 14
3
17
= =
+ =
+ =
+ + + =
+ + + =
+ + =
+ =
x ya
xy x y
x yb
x y x y
x yx y
c
x yx y
x xy yd
x y
Bi 2: Cho h phng trnh:
4 4 2 212
x y m
y x y
+ =
+ =
Tm m h phng trnh c ng hai nghim.
Bi 3: Tm m h sau c nghim duy nht:3 2 2
3 2 2
7
7
x y x mx
y x y my
= +
= +
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2. H PHNG TRNH V T
NHNG PHNG PHP THNG DNG
GII H PHNG TRNH V T:
1/ Khcn thc a h cho v h hu t:
Mt vi nh l khi kh cn thc:
2 2
22
2 12 1
2 12 1
2 12 1
22
2
( ) ( ) ( ) ( ) 0
( ) 0( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) 0( ) ( )
( ) ( )
( ) 0( ) 0
( ) ( )(
++
++
++
= =
= =
= =
> >
<
Vy h (IV) c nghim khi1
0;4
m
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4/ Phng php nh gi:
V d 5:Gii h phng trnh
1 (V)2 1xy y y
xy y y
+ = =
Gii
iu kin: 0 y 1, x 0, xy y 0 (*)
D thyy = 0 khng phi l nghim ca h.
Viy 0, (*) x 1.
Suy ra
1
1
1
1
xy y y
y y y
y
x
+
+ =
=
=
Thay (x;y) = (1;1) vo h, ta thy tha h.
Vy nghim ca h (V) l (x;y) = (1;1)
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BI TP RN LUYN
Bi 1: Gii cc h phng trnh sau:
a)2 2 3
3 1 3 1 4
x y xy
x y
+ =
+ + + =
b)2 1 1
3 2 4
x y x y
x y
+ + + =
+ =
c)2 2 2 22 1 2 1
12 0
x x y x x y
xy y
+ + + + = +
=
d)2 2
3 2 2
4 3
4 0
x y x y
x x y
+ + =
=
Bi 2:
Tm cc gi tr ca m h phng trnh sau c nghim
1 2
2 1
y m
y m
+ + =
+ + =
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3. H PHNG TRNH KHNG MU MC
Khi nim:
H phng trnh khng mu mc l h phng trnh
khng c cu trc (dng) c th, do cng khng c cchgii tng qut. Phi ty vo tng h phng trnh m c
cch gii ph hp.
Mt s cch gii cbn:
Phng php th,
Phng php t n s ph,
Phng php cng,
Phng php dng tnh n iu ca hm s,
Phng php dng bt ng thc,
Phng php nh gi,
Phng php a v h phng trnh cng bc
(ng cp).
Sau y l mt s v d c th cho cc phng php:
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1/ Phng php th:
V d 1: Gii h phng trnh sau:
( )
( )
2
2 2
6 3 1 1
1 2
x xy x y
x y
+ + =
+ =
GiiTa bin i (1) thnh phng trnh bc hai theo n x:
( )26 1 3 1 0x y x y+ + =
Ta tnh bit s delta ca phng trnh trn:
( ) ( ) ( )2 2
1 3 24 1 3 5y y y = =
Ta tm dc nghim l 1 12 3
yx x
= =
Th 13
x = vo (2) 2 23
y =
Th 12
yx
= vo (2)
3 44 5
1 0
y x
y x
= =
= =
Vy nghim ca h l:
( ) ( )
1 2 2 1 2 2, ; , ;
3 3 3 3
3 4; 1;0 , ;
4 5x y
=
V d 2: Gii h phng trnh sau:
( )( ) ( )
( )
2 2
2
1 1 3 4 1 1
1 2
x y x y x x
xy x x
+ + + = +
+ + =
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GiiD thyx = 0 khng tha mn phng trnh (2).
Vi x 0, t (2) ta c2 1
1x
yx
+ = . Thay vo (1) ta c:
( )( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( )
2 22 2 2 2
3 2
2
3 4 11 1
1 2 1 1 3 1
1 2 2 1 1 3 1
2 2 1 0 1 2 do 0
x xx x x x x x x x
x x
x x x x x x
x x x x x x
+ = + =
+ =
+ = = =
Vi 1 1x y= = , Vi 52
2x y= =
Vy h c nghim l ( ) ( )5
21;1 , 2;;x y
=
V d 3: Gii h phng trnh:( )
( )
2 2 1
2
2 7
3
x x y
xy x y
+ + =
+ =
Gii
T ( ) ( )3
2 11
xy x
+ =
+, thay vo (1) ta c:
( )( )( )4 3 2 22 5 2 7 2 0 1 2 2 3 1 0
12 3 17 3 17
1 2 4 43 172 1 1 17 1 174
2 23 174
x x x x x x x x
x
xx x
x x
x y yy y
x
+ + = + + =
== +
= == = + = = = +
= =
=
( ) ( )3 17 1 17 3 17 1 17
1;2 , 2; 1 , ; , ;4 2 4 2
S
+ + +=
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BI TP RN LUYN
Gii cc h phng trnh sau:
) )( )( )
) )
2 2
3 32 2
2 2 2
2 2 2
33 2 16
2 9 2 32 3 33
3 4 7 4 9 02 2 2 1 0 4 3
x xy yxy x x ya b
x y x y xyx y x y
xy y x y x yc dxy y x y x y x y
+ = =
= ++ =
+ + = =+ + = + = +
2/ Phng php t n s ph:
V d 4: Gii h phng trnh sau:
( )
( )( )( )
2
2
1 4I
1 2
x x y y
x x y
y
y
+ +
+ +
+ =
=
GiiD thyy = 0 khng tha h (I), nn ta c:
( )
( )
2
2
14
I1
2 1
x
y
xx y
y
x y
+
++
+ + =
=
t
2
1, 2u v x yxy= = + + , ta c: 12 1
1u v u
uv v
+ = == =
Khi , suy ra:
221 1 1 21
2 532 1
xx yy x
yx yy x
x y
+= = == +
= =+ =
+ =
Vy nghim ca h l: ( ) ( ) ( )1;2 , 2;5;x y = .
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V d 5: Gii h phng trnh sau:
( )( )
( )
2 22
3
II1
4 4 7
2 3
yxy xx y
xx y
+ +
+ =+
+ =+
Giiiu kin:x + y 0. Khi :
( )
( ) ( )( )
2 2
2
3
II1
3 7
3
x y x yx y
x y x yx y
+
+
+ + =+
+ + =+
t 1u x yy
= + ++
(iu kin: 2u ), v x y=
( )( ) ( )
2 2
22II
2 133 131
loai3 3 3 132
u vv uu v
uu v u u
= == + =
= + = + =&
Suy ra:1
2 10
1
x y xx y
yx y
=
+ + =+
= =
Vy h c mt nghim duy nht ( ) ( )1;0;yx =
BI TP RN LUYNGii cc h phng trnh sau:2 2 2
3 32 2 2
2 2
52 2 2) b)
( 1) ( 1) 352 2 1
4 2 1 1) d)
( 1) ( 1) 2 3 2 4
x y xyx y x xa
x yx y x y xy
x y x y x y x yc
x x y y y x y
+ + =+ + =+ + + = + =
+ + + = + + + =+ + + + = + =
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3/ Phng php cng:
V d 6: Gii h phng trnh sau:
1 1 4
6 4 6
x y
x y
+ + =
+ + + =
Gii
iu kin: 1, 1y
Cng v tr v theo v ca hai phng trnh, ta c:
( )*6 1 4 1 4
6 1 4 1 6
x x y y
x x y y
+ +
+ + + + =
+ + + + =
t5
6 1 6 1u ux x x x= + =+ + + +
54 1 4 1v
vy y y y= + =+ +
Khi h (*) trthnh
10 6 1 55 35 5 5 42 4 1 5
u vx xu x
v yy yu v
+
+
+ = + + == =
= =+ = + =
Vy nghim ca h l ( ) ( )3;4;yx =
V d 7: Gii h phng trnh sau:
( )
( )
2 2
2 2
1
2
91 2
91 2
x y y
y x x
+ = +
+ = + +
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Gii
iu kin: , 2y >
Ly (1) tr (2) ta c:
( )
2 2 2 2
2 22 2
2 2
2 2
0 , 2
91 91 2 2
2 291 91
10
2 291 91
x y
x y y x y x
x y y xy x
y xx y
x yx y x y
y xx y
x y
> >
+
+
+ + = + +
= +
+ ++ +
+ + + + =
+ ++ +
=
144444444424444444443
Th y= vo phng trnh (1), ta c:
( )( )
( ) ( )
2 2 2 2
2
2
2
0 , 2
91 91 10 1 9
91
91
2 2
9 33 3
2 120
1 13 3 1 0
2 110
3 3x y
x x x x x x
x xx x
xx
x x
xx
x y
> >
+ = + + = +
= + +
++ +
+ =
++ +
= =
1444444442444444443
Vy h c m nghim duy nht: ( ) ( )3;3;yx =
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BI TP RN LUYN
Gii cc h phng trnh sau:
2 2
2 3 2
4 2
13 3
1) b)
3( ) 2812 5
52 6 2
4) c)5
(1 2 ) 2 3 24
+ + + = + + =
+ + + =
+ + =
+ + + + = + =
+ + + = = +
x x yx y xy y
ax y x y
x yy
xx y x y xy xy y x yyb
x y xy x x x y x y
3/ Phng php dng bt ng thc:
V d 8: Gii h phng trnh sau:
1 1 1 6
9
x y z
x y z
+ + + + + =
+ + =
Giiiu kin: , , 1x y z
p dng bt ng thc Bunyakovsky, ta c:
( ) ( )2
1. 1 1. 1 1. 1 3 36x y z x y z + + + + + + + =
Suy ra: 1 1 1 6x y z + + + + +
ng thc xy ra 3x y z = = = tha mn phng trnh th
hai ca h.Vy h c mt nghim duy nht ( ) ( )3;3;3;;y zx =
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V d 9: Gii h phng trnh sau:
2
2
3
4 2
4
6 4 2
2
1
3
1
4
1
xy
x
yz
y y
z
z z z
= +
=
+ +
=+ + +
Gii
V2
2
20
1
xy
x=
+nn xy ra hai trng hp sau:
Viy = 0, khi x = y = z = 0Vy ( ) ( )0;0;0;;y zx = l mt nghim ca h phng trnh.
Viy > 0, khi x > 0,z>0.
D thy 2 21 2x x+ nn2
2
2hay
1x
x y xx
+
.
Theo BT Cauchy, ta c:2
4 2 4 2 234 2
31 3 . .1 3 hay
1y
y y y y y y z yy y
+ + = + +
T phng trnh th 3 ca h suy ra x z . Vy x y z x ,iu ny xy ra x y z = = .
Thay vo phng trnh u ta c 1y z= = = (tho)
Vy nghim ca h l ( ) ( )( )0;0;0 1;1;1;;y zx =
BI TP RN LUYNGii cc h phng trnh sau:
) )( 1) ( 1) 2 4 1 4 1 4 1 9
61 1
x y y x xy x y z a b
x y z x y y x xy
+ = + + + + + =
+ + = + =
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4. SDNG VCT GII
H PHNG TRNH
Mt vi btng thc vectthng dng:
Trong mt phng hoc trong khng gian cho hai vc t ,ur ura b ;
khi ta c:
( )| | | | | | 1+ +r r r ra b a b
Du " = " xy ra * :+ =r r r ra b k a kb hoc
mt trong hai vc tbng0r
( )| | | | | | 2
r r r ra b a b
Du " = " xy ra * : =r r r ra b k a kbhoc
mt trong hai vc tbng 0r
( )| | . | | . 3 r r r ru v u v
Du " = " xy ra * : =r r r ra b k a kbhoc
mt trong hai vc tbng 0r
( ). | | . | | 4r r r ru v u v
Du " = " xy ra * :+ =r r r ra b k a kb hoc
mt trong hai vc tbng 0r
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V d 1: Gii h phng trnh sau:
( )2 2 2
2011 2011 2011
I
3
3
3
+ + =
+ + =
+ + =
x y z
x y z
x y z
Gii
Xt ( ; ; ), (1;1;1)u x y z v= =ur ur
Khi ta c:
2 2 2
2 2 2
3
1 1 1 3
= + + =
= + + =
ur
ur
x y z u
v
M 3 3. . .= = = =+ +ur ur ur ur ur uru v x y z u v u v
Vy uru cng phng vi urv 01 1 1x y z = = >
0x y z = = >
Kt hp vi (I) ta cx = y = z = 1 l nghim ca h.
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V d 2: Gii h phng trnh sau:
2 2
2
2 2
( )
2
3 8 8 8 2 4 2
+ = +
+ + =
+ + + = + +
x y y x z
x x y yz
x y xy yz x z
Gii
2 2 2 2
( ) ( ) 0( 1) (2 1) 0
4( ) 4( ) ( 1) (2 1)
+ + + = + + + =
+ + + = + + +
x x y y y z
x x y z
x y y z x z
Xt: ( ; ); ( ; ); ( 1;2 1)a x y b x y y z c x z = = + + = + +ur ur ur
2 2. 0; . 0;4 = = =
ur ur ur ur ur ura b a c b c
Nu th 10 0;2
= = = = ur ura x y z
Nu 0 th vur r ur ur
ba c cng phng 2 = urur
c b
Xt 2 trng hp 2 2v= = ur urur ur
c b c b
Ta cx = 0; y = 12; z = 1
2
Vy h c 2 nghim 1 1 10;0; v 0; ;2 2 2
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BI TP RN LUYN
Bi 1: Gii h phng trnh sau:
2( ) 1 3 2 ( ) 1
2
2 1
x y x y x y x yx y
x y
+ + + + = + +
+
=
Bi 2: Chng minh rng h phng trnh sau v nghim:
4 4 4
2 2 2
1
2 7
x y z
x y z
+ + =
+ + =
Bi 3: Gii h phng trnh sau:
2 2 2
2009 2009 2009
3
3
3
x y z
x y z
x y z
+ + =
+ + =
+ + =
Bi 4: Gii h phng trnh sau:
1 2 2008
1 2 2008
20091 1 ... 1 2008
2008
20071 1 ... 1 2008
2008
x x x
x x x
+ + + + + + =
+ + + =
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5. H PHNG TRNH
TRONG CC K THI
V d 1: Gii h phng trnh
( )2 2 21 7
,
1 13
xy x yx y
x y xy y
+ + =
+ + =
(Tuyn sinh i hc, cao ng khi B 2009)
GiiD thyy = 0 khng l nghim nn h cho tng ngvi
22
2
11 77
1 1
13 13
xx xxy yy y
x x
x xy y y y
+ + =+ + =
+ + = + =
Suy ra2
1 120 0x x
y y+ + + =
151
5
12
xyx
yx y
+ = + =
=
(h v nghim)
141 4
3
x yxy
x y
+ =+ =
=
Vy trong trng hp ny, h c hai nghim
(x;y) =1
13
;
; (3;1).
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V d 2: Gii h phng trnh:
( )( )
2 3 2
4 2
5
4 ,5
1 24
x y x y xy xy
x y
x y xy x
+ + + + =
+ + + =
(Tuyn sinh i hc, cao ng khi A 2008)Gii
H phng trnh cho tng ng
( )
2 2
22
5( )
45
4
x x xy x y xy
x y xy
+ + + + = + + =
Suy ra2 2 2 2
( ) ( )x y xy x y x y+ + + = + 2 2( )( 1 ) 0x y x y xy + + =
Vi
2 3
2
3
50
40 5
254
16
xx y
x yxy
y
=+ =+ =
= =
Vi
2
2
1
121 0 33
22
xx yx y xy
yxy
=+ = + = =
=
Vy nghim ca h l (x;y) = 3 35 25
;4 16
,3
1;2
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V d 3: Tm m h phng trnh c nghim thc:
3 33 3
1 15
1 115 10
x yx y
x y m
x y
+ + + = + + + =
(Tuyn sinh i hc, cao ng khi D 2007)
Gii
t ( )1 1
; 2, 2x a y b a bx y
+ = + =
H phng trnh cho trthnh
3 3
5 5
83( ) 15 10
a b a b
ab ma b a b m
+ = + =
= + + =
Do a, b l nghim ca phng trnh
( )2 5 8 0 *X X m + =
H phng trnh cho c nghim khi v ch khi PT (*) c
hai
nghimX1, X2 tha 1 22, 2x x
Vy7
2 224
m m
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V d 4: Gii h phng trnh:
1 22
2 33
2002 11
1 1
2
1 1
2
1 1
2
...
x
x
x
xx
xx
xx
= +
= +
= +
(Olympic 30/4/2002 ti THPT Chuyn L Hng Phong)
Gii
Nhn xt: Nu ( )1 2 2002, ,...,x x x l nghim th 1 2 2002, ,...,x x x
phi cng du v khc 0. ng thi ( )1 2 2002, ,...,x x x cng
l nghim, nn ta ch cn xt vi 1 2 2002, ,...,x x x dng.
Theo bt ng thc Cauchy: ( ) ( )1
2 1,2,...,2002 1,+ =ii
x ix
T cc phng trnh trong h v (1), ta c:
( )2 2 1 2hayi ix x
Mt khc cng cc phng trnh trong h th:
1 2 20021 2 2002
1 1 1
... ...x x x x x x+ + + = + + + ( )3
T (2) v (3) ta c: 1 2 2002 1...x x x= = = =
Kt qu: H c 2 nghim 1 2 2002
1 2 2002
1
1
...
...
x x x
x x x
= = = =
= = = =
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TUYN SINHI HC, CAONG
Tuyn sinh i hc, cao ng khi B 2002:
Gii h phng trnh3
2
y x y
x y x y
=
+ = + +
Kt qu: ( ) ( )3 1
; 1;1 , ;2 2
x y =
.
Tuyn sinh i hc, cao ng khi A 2003:
Gii h phng trnh3
1 1
2 1
x yx y
y x
=
= +
Kt qu:
( ) ( )1 5 1 5 1 5 1 5
; 1;1 , ; , ;2 2 2 2
x y + +
=
.
Tuyn sinh i hc, cao ng khi D 2004:
Tm m phng trnh sau c nghim:
( )1
,1 3
x yx y
x x y y m
+ =
+ =
Kt qu:1
04
m
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Tuyn sinh i hc, cao ng khi A 2006:
Gii h phng trnh3
1 1 4
x y xy
x y
+ =
+ + + =
Kt qu: ( ) ( )1;1;x y =
Tuyn sinh i hc, cao ng khi B 2008:
Gii h phng trnh ( )4 3 2 2
2
2 2 9,
2 6 6
x x y x y xx y
x xy x
+ + = +
+ = +
Kt qu: ( )17
; 4;4
x y =
Tuyn sinh i hc, cao ng khi D 2009:
Gii h phng trnh( )
( )( )2
2
1 3 0;5
01
x x yx y
x yx
+ + =
+ = +
( ) ( )3
; 1;1 , 2;2
x y =
Tuyn sinh i hc, cao ng khi A 2010:
Gii h phng
trnh( ) ( )
( )2
2 2
4 1 2 5 2 0;
4 2 3 4 7
x x y yx y
x y x
+ + =
+ + =
Kt qu: ( )1
; ;22
x y =
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OLYMPIC 30/4
Olympic 30/4/1998 ti THPT Chuyn
L Hng Phong TP H Ch Minh:
Gii h phng trnh( )( )
( )
2
2
2
0 126 2
26 3
ax bx cbx cx a
cx ax b
=
+ ++ + =
+ + =
Hng dn: Cng (1), (2) v (3) v theo v ta c:
( )( )2 1 0 0a b c x x a b c+ + + + = + + = (V 2 1 0x x>+ + )
( )1 1c
x xa
= = (khng tho (2), (3) nn loi)
ax c = thay vo (2) tm ra a v bin lun tm b, c.
Kt qu: ( ) ( ) ( )1;4; 3 , 13;0; 13 , 26; 26;0 l 3 b s cn tm.
Olympic 30/4/2000 ti THPT Chuyn
L Hng Phong TP H Ch Minh:
Gii h phng trnh
53 2 4
42
53 2
42
yy x
xy x
= +
+ = +
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Hng dn:
Tm iu kin, sau vit li thnh:
( )
( )
5
42
1 21
1 23 2
y xx y
x y
+
=
+ =
1 2 15
42x y y x =
+
Tm y theox sau th vo (2) ta s tm dc nghim cah.
Kt qu: ( )5 2 6 5 2 6
; ;27 9
x y + +
=
Olympic 30/4/2005 ti THPT ChuynL Qu n TP Nng:
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
22 2 2 2
22 2 2 2
22 2 2 2
3 1 1
3 1 2
3 1 3
x y z x x y z
y z x y y z x
z x y z z x y
=
=
=
+ + +
+ + +
+ + +
Hng dn: Chia lm 2 trng hp:
Trng hp 1:xyz= 0
Trng hp 2:xyz 0
Kt qu: H c nghim l:
( ) ( ) ( ) ( )9 3 9 5 5
0 0 0 0 0 0 113 4 11 6 4
; ; , ; ; , ; ; , , ; ; , ; ;vx y z x y z
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Olympic 30/4/2007 ti Hu:
Gii h phng trnh2 2
2
816
xyx y
x y
x y x y
+ + = + + =
Kt qu: ( ) ( ) ( )2;2 3;7; ,x y =
Olympic 30/4/2008 ti THPT Chuyn
L Hng Phong TP H Ch Minh:
Gii h phng trnh 2 2 23 3 3
7
371
x y z
x y z
x y z
=+
+ = + =
Kt qu: ( ) ( ) ( ); 9;10;12 10;9;12,x y =
Olympic 30/4/2009 ti THPT Chuyn
L Hng Phong TP H Ch Minh:
Gii h phng trnh
3 2 3
3 2 3
3 2 3
2 3 18
2 3 18
2 3 18
x y y
y y z z
z z x x
+ = +
+ = + + = +
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Hng dn: t 3 2( ) 2 3 18f t t t = + v 3( )g t t t = + th
phng trnh c vit li:( ) ( )
( ) ( )
( ) ( )
f x g y
f y g z
f z g x
=
= =
Gi s ( )max , ,x y z = th ( ) ( )( ) ( )x y g x g yx z g x g z
Do hm sng bin( ) ( )
( ) ( )
x f y
x f z
Kt qu: H c nghim duy nht ( ) ( )2;2;2; ;x y z =
Olympic 30/4/2010 ti THPT Chuyn
L Hng Phong TP H Ch Minh:
Gii h phng trnh
3 3
1 19
1 1 1 11 1 18
x y
x yx y
+ = + + + =
Hng dn: t3 3
1 1;a b
x y= =
Kt qu: ( ) 1;1 1
; , ;18 8
x y =