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TRAN STUNG---- & ----
TI LIU N THI I HC CAO NG
Na m 2011
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Trn STng H phng trnh nhiu n
Trang1
1. H phng trnh bc nht hai n
a x b y c a b a b
a x b y c 2 2 2 21 1 11 1 2 2
2 2 2
( 0, 0) + =
+ + + =
Gii v bin lun:
Tnh cc nh thc:a b
Da b
1 1
2 2
= , xc b
Dc b
1 1
2 2
= , ya c
Da c
1 1
2 2
= .
Ch :gii h phng trnh bc nht hai n ta c thdng cc cch gii bit nh:phng php th, phng php cng i s.
2. H phng trnh bc nht nhiu nNguyn tc chung gii cc h phng trnh nhiu n l khbt na v ccphng trnh hay h phng trnh c sn t hn. kh bt n, ta cng c th dng ccphng php cng i s, phng php th nhi vi h phng trnh bc nht hai n.
Bi 1. Gii cc h phng trnh sau:a)
x y
x y
5 4 3
7 9 8
=
=b)
x y
x y
2 11
5 4 8
+ =
=c)
x y
x y
3 1
6 2 5
=
=
d)( )
( )x y
x y
2 1 2 1
2 2 1 2 2
+ + =
=e)
x y
x y
3 216
4 35 3
112 5
+ =
=
f)x y
y
3 1
5x 2 3
=
+ =
Bi 2. Gii cc h phng trnh sau:
a)x y
x y
1 818
5 451
=
+ =
b)x y
x y
10 11
1 225 3
2
1 2
+ = +
+ =
+
c)x y x y
x y x y
27 327
2 345 48
1
2 3
+ = +
=
+
d)x y
x y
2 6 3 1 5
5 6 4 1 1
+ + = + =
e)x y x y
x y x y
2 9
3 2 17
+ = + + =
f)x y x y
x y x y
4 3 8
3 5 6
+ + = + =
Bi 3. Gii v bin lun cc h phng trnh sau:a)
mx m y m x my
( 1) 1
2 2
+ = + + =
b)mx m y
m x m y ( 2) 5
( 2) ( 1) 2
+ = + + + =
c)m x y m m x y m
( 1) 2 3 1
( 2) 1
+ = + =
d)m x m y m x m y m
( 4) ( 2) 4
(2 1) ( 4)
+ + = + =
e)m x y m
m x y m m 2 2( 1) 2 1
2
+ =
= +f)
mx y m x my m
2 1
2 2 5
+ = + + = +
Bi 4. Trong cc h phng trnh sau hy:i) Gii v bin lun. ii) Tm m Z h c nghim duy nht l nghim nguyn.
a)m x y m
m x y m m 2 2( 1) 2 1
2
+ =
= +b)
mx yx m y m
1
4( 1) 4
= + + =
c)mx yx my m
3 3
2 1 0
+ = + + =
I. H PHNG TRNH BC NHT NHIU N
Xt D Kt qu
D 0 H c nghim duy nht yxDD
x yD D
;
= =
Dx 0 hoc Dy 0 H v nghimD = 0Dx = Dy = 0 H c v snghim
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H phng trnh nhiu n Trn STng
Trang2
Bi 5. Trong cc h phng trnh sau hy:i) Gii v bin lun.ii) Khi h c nghim (x; y), tm h thc giax,yc lp i vi m.
a)mx y m
x my m2 1
2 2 5
+ = + + = +
b)mx m y
m x my6 (2 ) 3
( 1) 2
+ = =
c)mx m y m
x my( 1) 1
2 2
+ = + + =
Bi 6. Gii v bin lun cc h phng trnh sau:a)
ax y bx y3 2 5
+ = + =
b)y ax b
x y2 3 4 = =
c)ax y a bx y a2
+ = + + =
d)a b x a b y a
a b x a b y b( ) ( )
(2 ) (2 )
+ + = + + =
e) ax by a bbx ay ab
2 2
2
+ = ++ =
f)ax by a b
bx b y b
2
2 4
=
=Bi 7. Gii cc h phng trnh sau:
a)
x y zx y z
x y z
3 1
2 2 5
2 3 0
+ = + =
=b)
x y zx y zx y z
3 2 8
2 6
3 6
+ + =+ + =
+ + =c)
x y zx y z
x y z
3 2 7
2 4 3 8
3 5
+ = + + =
+ =
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Trn STng H phng trnh nhiu n
Trang3
1. H gm 1 phng trnh bc nht v 1 phng trnh bc hai T phng trnh bc nht rt mt n theo n kia.
Th vo phng trnh bc hai a v phng trnh bc hai mt n. S nghim ca h tu theo s nghim ca phng trnh bc hai ny.
2. Hi xng loi 1
H c dng: (I) f x yg x y
( , ) 0
( , ) 0
= =
(vif(x, y) = f(y, x) v g(x, y) = g(y, x)).
(C ngha l khi ta hon v giax vy thf(x, y) v g(x, y) khng thay i).t S = x + y, P = xy.a h phng trnh (I) v h (II) vi cc n l Sv P. Gii h (II) ta tm c S v P.
Tm nghim (x, y) bng cch gii phng trnh: X SX P 2 0 + = .
3. Hi xng loi 2
H c dng: (I) f x yf y x( , ) 0 (1)
( , ) 0 (2)
= =
(C ngha l khi hon v giax vy th (1) bin thnh (2) v ngc li). Tr (1) v (2) v theo v ta c:
(I) f x y f y x f x y( , ) ( , ) 0 (3)
( , ) 0 (1)
= =
Bin i (3) v phng trnh tch:(3) x y g x y ( ). ( , ) 0 = x y
g x y( , ) 0
= =
.
Nh vy: (I)
f x y
x y
f x yg x y
( , ) 0
( , ) 0
( , ) 0
= = = =
.
Gii cc h trn ta tm c nghim ca h (I).Ch : Vi cc h phng trnh i xng, nu h c nghim x y0 0( ; ) th y x0 0( ; )
cng l nghim ca h. Do nu h c nghim duy nht th x y0 0= .
4. Hng cp bc hai
H c dng: (I)a x b xy c y d
a x b xy c y d
2 21 1 1 1
2 22 2 2 2
+ + =
+ + =.
Gii h khix = 0 (hocy = 0). Khix 0, t y kx= . Th vo h (I) ta c h theo kvx. Khx ta tm c phngtrnh bc hai theo k. Gii phng trnh ny ta tm c k, t tm c (x; y).
II. H PHNG TRNH BC HAI HAI N
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H phng trnh nhiu n Trn STng
Trang4
Bi 1. Gii cc h phng trnh sau:a) x y
x y
2 24 8
2 4
+ =+ =
b) x xyx y
2 24
2 3 1
= =
c) x yx y
2( ) 49
3 4 84
=+ =
d) x xy y x y x y
2 23 2 3 6 0
2 3
+ + + = =
e)x y
xy x y 3 4 1 0
3( ) 9
+ = = +
f)x y
xy x y 2 3 2
6 0
+ = + + + =
g) y x x
x y
2 4
2 5 0
+ =+ =
h)x y
x y y2 22 3 5
3 2 4
+ =
+ =i)
x y
x xy y 2 22 5
7
=
+ + =
Bi 2. Gii v bin lun cc h phng trnh sau:a)
x y
x y m2 26 + =
+ =
b)x y m
x y x2 2 2 2
+ =
+ =c)
x y
x y m2 23 2 1 =
+ =
Bi 3. Trong cc h phng trnh sau:i) Tm s nguyn m h c nghim duy nht l nghim nguyn.ii) Khi h c nghim (x, y) , tm h thc giax, yc lp vi m.
a)
mx y m
x my a
2 1
2 2 1
+ = + + = b)
mx y m
x my m
3
2 1
+ = + = +
c)x y m
x y m2 4
2 3 3
= + = +
d)x y
y x m2 5
2 10 5
+ = = +
Bi 4. Gii cc h phng trnh sau:a)
x xy y
x y xy x y 2 211
2( ) 3
+ + =
+ + = b)
x y
x xy y 2 24
13
+ =
+ + =c)
xy x y
x y x y 2 25
8
+ + =
+ + + =
d)
x y
y x
x y
13
6
6
+ =
+ =
e) x x y y x y xy
3 3 3 3 17
5
+ + =+ + =
f)x x y y
x xy y
4 2 2 4
2 2
481
37
+ + =
+ + =
Bi 5. Gii cc h phng trnh sau:a)
x xy y
x y y x 2 21
6
+ + =
+ = b)
x y
x x y y
2 2
4 2 2 4
5
13
+ =
+ =c)
x y y x
x y
2 2
3 3
30
35
+ =
+ =
d)x y
x y x y
3 3
5 5 2 2
1 + =
+ = +e)
x y xy
x y x y
2 2
4 4 2 2
7
21
+ + =
+ + =f)
x y xy
x y x y 2 211
3( ) 28
+ + =
+ + + =
Bi 6. Gii cc h phng trnh sau:
a)
x yxy
x yx y
2 2
2 2
1( ) 1 5
1( ) 1 49
+ + =
+ + =
b) ( )
y x x y
x yx y
2 2
2 2
2 2
( 1) 2 ( 1)
11 24
+ = +
+ + =
c)
x yx y
x yx y
2 2
2 2
1 14
1 14
+ + + =
+ + + =
d)
x y
x y
x yxy
2 2
2
31 11
( )(1 ) 6
+ = + +
+ + =
e)x y y x y x xy
y xxy xy x y
2 22 2 6
1 4
+ + + =
+ + + =
f)
xyxy
x yxy
14
1( ) 1 5
+ =
+ + =
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Trn STng H phng trnh nhiu n
Trang5
Bi 7. Gii v bin lun cc h phng trnh sau:a)
x y xy m
x y m2 2 3 2
+ + =
+ = b)
x y m
x y xy m m 2 2 21
2 3
+ = +
+ = c)
x y mxy x y m ( 1)( 1) 5
( ) 4
+ + = + + =
Bi 8. Gii cc h phng trnh sau:a)
x x y
y y x
2
2
3 2
3 2
= +
= +b)
x y x y
y x y x
2 2
2 2
2 2
2 2
= +
= +c)
x x y
y y x
3
3
2
2
= +
= +
d)
yx y
xx
y xy
3 4
3 4
=
=
e)
yy
x
xx
y
2
2
2
2
23
23
+=
+ =
f)
x yy
y xx
2
2
12
12
= +
= +
g)x x y
y y x
3
3
3 8
3 8
= +
= +h)
xy x y
xy y x
2
2
8( 1)
8( 1)
+ =
+ = i)
x yx
y x
y
2
2
32
32
+ =
+ =
k)
2 2
2 2
91 2 (1)
91 2 (2)
+ = +
+ = +
x y y
y x x
Bi 9. Gii v bin lun cc h phng trnh sau:a)
x x my
y y mx
2
2
3
3
= +
= +b)
x y m m
y x m m
2 2
2 2
(3 4 ) (3 4 )
(3 4 ) (3 4 )
=
= c)
xy x m y
xy y m x
2
2
( 1)
( 1)
+ =
+ =
Bi 10. Tm m h phng trnh sau c nghim duy nht:
a) x y m y xy m x
2 2
2 2 + =
+ =b) xy x m y
xy y m x
2
2( 1)( 1)
+ = + =
c)
m
x y y
my x
x
22
22
2
2
= + = +
Bi 11. Gii cc h phng trnh sau:a)
x xy y
x xy y
2 2
2 2
3 1
3 3 13
+ =
+ =b)
x xy y
x xy y
2 2
2 2
2 4 1
3 2 2 7
+ =
+ + =c)
y xy
x xy y
2
2 2
3 4
4 1
=
+ =
d)x xy y
x xy y
2 2
2 2
3 5 4 38
5 9 3 15
+ =
=e)
x xy y
x xy y
2 2
2 2
2 3 9
4 5 5
+ =
+ =f)
x xy y
x xy y
2 2
2 2
3 8 4 0
5 7 6 0
+ =
=
Bi 12. Gii v bin lun cc h phng trnh sau:a)
x mxy y m
x m xy my m
2 2
2 2( 1)
+ + =
+ + =b)
xy y
x xy m
2
2
12
26
=
= +c)
x xy y m
y xy
2 2
2
4
3 4
+ =
=
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H phng trnh nhiu n Trn STng
Trang6
Cc h phng trnh i s tng qut thng rt kh gii v khng th nu ra phngphp chung gii chng. y xin nu ra mt s phng php c th la chn thchhp.
1. Phng php th: T phng trnh n gin nht ca h hoc t phng trnh tchtm cch rt mt n theo n kia, ri th vo phng trnh cn li. Gii phng trnh ny.S nghim ca h tu thuc s nghim ca phng trnh ny.2.t n ph: Bin i cc phng trnh c tht n ph, ri chuyn v h cbn.3.Phng php nh gi: Tiu kin ca n, xt trng hp xy ra du "=" btng thc.4. Phng php iu kin cn v :5. Phng php hm s: Chn hm s thch hp, ri s dng tnh n iu ca hm s.
Cho hm sy = f(x) ng bin (hoc nghch bin) trn khong (; ). Khi , vi mia, b (; ) ta c: f(a) = f(b) a = b.
Ch : Cc h phng trnh hon v vng quanhx f yy f zz f x
( )
( )
( )
= =
=, thng sdng tnh n
iu ca hm schng minh x = y = z. Xt tnh n iu hm sf(t). Chng t x < y, x > y, khng xy ra. T suy ra x = y = z. Thvo h cho gii tm x, y, z.
Vn 1: Phng php th
Bi 1. Gii h phng trnh sau:x y y x y
x y x y
2
2
1 ( ) 4 (1)
( 1)( 2) (2)
+ + + =
+ + =
Dthy y 0. HPT[ ]y y y x y x y 4 ( ) ( 2) + + = [ ]y x2
(3 ) 0 = y x3= Nghim: (1; 2), (2; 5).
Bi 2. Gii h phng trnh sau:2 2
2 2
3 (1)
1 1 4 (2)
+ = + + + =
x y xy
x y
(2) x y x y xy xy xy 2 2 2 2 22 ( 1).( 1) 14 2 ( ) 4 11+ + + + = + + + = (3)
t xy = p.p
pp p p
pp p
22
311
(3) 2 4 11 353 26 105 0
3
= + + = =+ =
(1) ( )x y xy 2
3 3+ = + p = xy =35
3 (loi) p = xy = 3 x y 2 3+ =
1/ Vixy
x yx y
33
2 3
=
= = + =2/ Vi
xyx y
x y
33
2 3
=
= = + =
Vy h c hai nghim l: ( ) ( )3; 3 , 3; 3
III. H PHNG TRNH DNG KHC
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Trn STng H phng trnh nhiu n
Trang7
Bi 3. Gii h phng trnh sau:x x y
x yx
2
2
( 1) 3 0
5( ) 1 0
+ + = + + =
(D 2009)
V x 0 nn HPTx y
x
x yx
2
2
31
5( ) 1 0
+ =
+ + =
x y
x
xx2
31
4 62 0
+ =
+ =
xxx y x y
1 11
1 21
22
= = + = + =
. Nghim:3
(1;1), 2;2
.
Bi 4. Gii h phng trnh sau:x x y xy y
x y x y
3 2 2 36 9 4 0 (1)
2 (2)
+ =
+ + =
Ta c: (1) x y x y 2( ) ( 4 ) 0 = x yx y4
= =
Vi x = y: (2) x = y = 2
Vi x = 4y: (2) x y32 8 15; 8 2 15= = Bi 5. Gii h phng trnh sau:
+ + =
+ + + =
x x y
x x y xy x
2
3 2 2
5 9
3 2 6 18
H y x x
x x x x+
2
4 3 2
9 5
4 5 18 18 0
=
+ =
y x xxx
x
2
9 51
3
1 7
= = = =
x yx y
x y
x y
1; 33; 15
1 7; 6 3 7
1 7; 6 3 7
= = = =
= = + = + =
Bi 6. Gii h phng trnh sau:x y xy
x y
2 0
1 4 1 2
=
+ =
H PT( ) ( )x y x y
x y
2 0
1 4 1 2
+ =
+ =
x y
x y
2 0
1 4 1 2
=
+ =
x y
y
4
4 1 1
= =
x
y
2
1
2
= =
Bi 7. Gii h phng trnh sau:xy
x yx y
x y x y
2 2
2
21 (1)
(2)
+ + =
+ + =
iu kin: x y 0+ > .
(1) x y xy x y
2 1( ) 1 2 1 0 + = + x y x y x y 2 2( 1)( ) 0+ + + + = x y 1 0+ =
(v x y 0+ > nn x y x y 2 2 0+ + + > )
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H phng trnh nhiu n Trn STng
Trang8
Thay x y1= vo (2) ta c: x x21 (1 )= x x2 2 0+ = x yx y
1 ( 0)
2 ( 3)
= = = =
Vy h c 2 nghim: (1; 0), (2; 3).Bi 8. Gii h phng trnh sau:
( )x y xy
x y
3 3
2 2
3 4 (1)
9 (2)
=
=
T(2): x y xy 2 2 9 3= = .
Khi: xy 3= , ta c: x y3 3 4 = v ( )x y3 3. 27 =
Suy ra: ( )x y3 3; l cc nghim ca phng trnh: X X X2 4 27 0 2 31 = =
Vy nghim ca H PT l x y3 32 31, 2 31= + =
hoc x y3 32 31, 2 31= = + .
Khi: xy 3= , ta c: x y3 3 4 = v ( )x y3 3. 27 =
Suy ra: x y3 3; ( ) l nghim ca phng trnh: X X PTVN 2 4 27 0 ( )+ + = Bi 9. Gii h phng trnh sau:
( )2
3 2 (1)
2 8 (2)
=
=
x y xy
x y
iu kin : x y x y . 0 ;
Ta c: (1) x y xy x y x y 23( ) 4 (3 )( 3 ) 0 = =y
x y hay x 33
= =
Vi x y3= , thvo (2) ta c : y y y y 2 6 8 0 2 ; 4 + = = =
H c nghim x xy y
6 12;
2 4
= = = =
Viy
x3
= , thvo (2) ta c : y y23 2 24 0 + = V nghim.
Kt lun: h phng trnh c 2 nghim l:x xy y
6 12;
2 4
= = = =
Bi 10.Gii h phng trnh sau:x y y x
y x
3 3
2 2
4 16 (1)
1 5(1 ) (2)
+ = +
+ = +
T(2) suy ra y x2 2 5 4= (3). Thvo (1) c:
( )yx x y y x 2 23 3 5 . 16+ = + x x y x 3 2 5 16 0= xx xy2
0
5 16 0
= =
Vi x 0= y2 4= y 2= .
Vi x xy2 5 16 0= xyx
2 16
5
= (4). Thvo (3) c:
x
xx
22
216
5 45
= x x x x 4 2 4 2
32 256 125 100+ =
x x4 2124 132 256 0+ = x2 1= x yx y
1 ( 3)1 ( 3)
= = = =
.
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Trang9
Vy h c 4 nghim: (x; y) = (0; 2) ; (0; 2); (1; 3); (1; 3)Bi 11.Gii h phng trnh sau:
xy y
xy x
2
2
4 8
2
=
= +
Nu xy 4 th HPT
xy y
xy x
2
2
4 8 (1)
2 (2)
=
= +
T(2) x 0, x2 2 vx
yx
22 +=
Thay vo (1) ta c:x
xx
22
2 22 4 8 +
+ =
x x2 2( 2)( 1) 0 = x 2=
H c nghim (x; y) l: ( ) ( )2; 8 , 2; 8
Nu xy < 4 th x2 2< .
HPT xy yxy x
2
24 8
2 =
= + xx
x
2
22 24 2 8 + =
x22(2 ) 0 = x2 2= (loi)
Kt lun: Nghim (x; y) ca h: ( ) ( )2; 8 , 2; 8 Bi 12.Gii h phng trnh sau:
x y x y x x
xy x x
2 2
2
( 1)( 1) 3 4 1 (1)
1 (2)
+ + + = +
+ + =
T(2) x 0 vx
yx
2 11
+ = . Thay vo (1) ta c:
x xx x x x
x x
2 22 21 1 3 4 1
+ = +
x x x2 ( 1)( 2) 0 + = xx
1
2
= =
(v x 0)
Nghim (x; y):5
(1; 1), 2;2
Bi 13.Gii h phng trnh sau:xy x y x y
x y y x x y
2 22 (1)
2 1 2 2 (2)
+ + =
=
iu kin x 1, y 0 x + y > 0.(1)
x y x y ( )( 2 1) 0+ =
x y2 1= +
(3)
Thay (3) vo (2) ta c: y y y y y y (2 1) 2 2 2(2 1) 2+ = +
( )y y( 1) 2 2 0+ = y = 2 x = 5Nghim (x; y): (5; 2)
Bi 14.Gii h phng trnh sau:y x x
y x xy x y
2
2 2
(5 4)(4 ) (1)
5 4 16 8 16 0 (2)
= +
+ + =
T(1) y x x2 25 16 16= + + .
Thay vo (2) ta c: y xy y 22 4 8 0 = yy x02 4 = = +
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H phng trnh nhiu n Trn STng
Trang10
Vi y = 0 x x25 16 16 0 + + = x
x
4
54
=
=
Vi y x2 4= + x x x2 2(2 4) 5 16 16+ = + + x = 0 y = 4.
Kt lun: Nghim (x; y):
4
(0; 4), (4; 0), ; 05
.Bi 15.Gii h phng trnh sau:
xyx y
x y
x y x y x
2 2
3
816 (1)
3 3 2 1 (2)
+ + =
+ + + + =
Dthy x + y > 0.
T(1) x y xy x y
2 4( ) 16 2 1 0
+ = + x y x y x y 2 2( 4) 4( ) 0 + + + + =
x y 4 0+ = x y 4+ =
Thay vo (2) ta c: x x3 2 7 3 2+ = x x x3 22 9 14 5 0 + = x y1 7
2 2
= =
Nghim (x; y):1 7
;2 2
.
Bi 16.Gii h phng trnh sau: x x y y x y x y
4 2 2
2 2
4 6 9 0 (1)
2 22 0 (2)
+ + =
+ + =
T(2)x
yx
2
2
22
2
=
+. Thay vo (1) ta c:
xx x
x
224 2
2
224 3 0
2
+ =
+
xx x
x
2 22 2
2 2
16( 4)( 4) 0
( 2)
+ =
+
x x x x 2 6 4 2( 4)( 4 20 64) 0 + + =
x yx y
x y
x y
2 ( 3)
2 ( 3)
2 ( 5)
2 ( 5)
= = = =
= = = =
Bi 17.Gii h phng trnh sau:x y x y
x y x y
3 (1)
2 (2)
=
+ = + +(B - 2002)
iu kin:x y
x y0
(3)0
+
(1) ( ) x yx y x y x y
3 61 01
= = = +. Thay vo (2) ta c:
x y
x y
1
3 1,
2 2
= =
= =
.
Bi 18.Gii h phng trnh sau:x y
x y
y x3
1 1(1)
2 1 (2)
=
= +
(A - 2003)
iu kin xy 0. Ta c: (1) x yx yxyxy
1( ) 1 0
1
= + = =
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Trang11
Trng hp 1:
x y
x y x y x y
x x x x x
x y
3 2
1
1 5
22 1 ( 1)( 1) 0
1 5
2
= = = = + = =
= + + =
= =
Trng hp 2:yxy yx
xy x
x x x VN x
33 4
1 11
22 11 2 0 ( )
= = = = + = + + + =
Kt lun: Nghim (x; y):1 5 1 5 1 5 1 5
(1;1), ; , ; ;2 2 2 2
+ +
.
Bi 19.Gii h phng trnh sau:x x y y
x y
3 3
2 2
8 2
3 3( 1)
= +
= +
(DB A 2006)
H PT x y x y
x y
3 3
2 2
3( ) 6(4 ) (1)
3 6 (2)
= +
=.
Th(2) vo (1) ta c: x y x y x y 3 3 2 23( ) ( 3 )(4 ) = + x x y xy 3 2 212 0+ =
xx yx y
0
3
4
= = =
.
Nghim (x; y):6 6 6 6
(3;1), ( 3; 1), 4. ; , 4. ;13 13 13 13
.
Bi 20.Gii h phng trnh sau:
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Vn 2: Phng php t n ph
Bi 1. Gii h phng trnh sau:x x y y
x y x y
4 2 2
2 2
4 6 9 0
2 22 0
+ + =
+ + =
HPT x yx y x
2 2 2
2 2( 2) ( 3) 4
( 2 4)( 3 3) 2 20 0 + =
+ + + =. t u x
v y
2 23
= =
.
HPT u uu vv vuv u v
2 2 2 040 24( ) 8
= =+ = = =+ + = . Nghim (2; 3), ( 2; 3),( 2; 5), ( 2; 5)
Bi 2. Gii h phng trnh sau:x y y
x y x y
3 3 3
2 2
8 27 18
4 6
+ =
+ =
HPTx
y
x xy y
33 3(2 ) 18
3 32 . 2 3
+ =
+ =
. t a = 2x; b =y3 . HPT a b
ab3
1 + = =
H cho c nghim:3 5 6 3 5 6
; , ;4 43 5 3 5
+ +
Cch 2: D thy y 0. HPT x y y
x y xy y
3 3 3
2 2 3
8 27 18
4 6
+ =
+ = x y x y xy 3 3 2 28 27 18(4 6 )+ = +
(*)
t t xy= . (*) t t t2(2 3)(4 42 9) 0+ + = t
t
32
21 9 5
4
=
=
Bi 3. Gii h phng trnh sau:x y y x y
x y x y
2
2
1 ( ) 4
( 1)( 2)
+ + + =
+ + =
Dthy y 0. HPT
xy x
y
xy x
y
2
2
12 2
1( 2) 1
++ + =
+ + =
.t
xu
yv y x
2 1
2
+ = = +
.
HPT u vuv
2
1
+ = =
uv
1
1
= =
x
y
y x
2 11
2 1
+=
+ =
x
y
1
2
=
=hoc
x
y
2
5
=
=
Bi 4. Gii h phng trnh sau: + + =
+ + + =
x x y
x x y xy x
2
3 2 2
5 9
3 2 6 18
HPT x x x y x x x y
2
2
2 3 9
( 2 )(3 ) 18
+ + + =
+ + =. t u x x
v x y
2 2
3
= += +
.
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Trn STng H phng trnh nhiu n
Trang13
HPT u vuv
9
18
+ = =
u vu v
6, 3
3, 6
= = = =
. Nghim:
x yx y
x y
x y
1; 3
3; 15
1 7; 6 3 7
1 7; 6 3 7
= = = =
= = + = + =
Bi 5. Gii h phng trnh sau:xy x y x y xy y 2 2 2
1 71 13
+ + =+ + =
(B - 2009)
Dthy y 0. HPT
xx
y y
xx
y y
2
17
113
+ + =
+ =
.tu x
yx
vy
1= +
=
.
HPTu v
u v27
13
+ =
= u
v5
12
= =
hoc uv
4
3
= =
. Nghim1
1; , (3;1)3
.
Bi 6. Gii h phng trnh sau:2 2
3 3
2 1
2 2
y x
x y y x
=
=
HPT ( )( )x y y x y x x x y xy y 3 3 2 2 3 2 2 32 2 2 2 2 5 0 = + + = Khi y 0= th h VN.
Khi y 0 , chia 2 vcho y3 0 ta c:x x x
y y y
3 2
2 2 5 0
+ + =
tx
t y= , ta c : t t t t 3 22 2 5 0 1+ + = =
y xx yx yy2 1 11 = = = = = =
Bi 7. Gii h phng trnh sau:y
xx yx
x yy
2 2
2 2
32 1
1
4 22
+ = +
+ + =
iu kin: x y x y 2 20, 0, 1 0 +
tx
u x y v y2 2 1;= + = . H PT trthnh: u v u v
u v u v
3 2 3 2
1 1 (1)1 4 22 21 4 (2)
+ = + = + + = =
Thay (2) vo (1) ta c:v
v vvv v
23
3 21 2 13 21 0 7
21 42
=+ = + =
=
Nu v = 3 th u = 9, ta c H PT:
x yx xx y
xy yx y
y
2 22 21 9 3 310
1 13 3
+ = = = + = = = = =
Nu v 72
= th u = 7, ta c H PT:
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H phng trnh nhiu n Trn STng
Trang14
y yx y x yx
x yy x x
2 2 2 2 2 24 41 7 853 537 7
2 22 14 142
53 53
= = + = + = = = = =
So snh iu kin ta c 4 nghim ca H PT.
Bi 8. Gii h phng trnh sau:2 2
2 2
1 4
( ) 2 7 2
x y xy y
y x y x y
+ + + =
+ = + +
Th PT y 0 . Khi ta c:
xx y
x y xy y y
y x y x y x x y
y
2
2 2
2 2 22
14
1 4.
( ) 2 7 2 1( ) 2 7
++ + = + + + =
+ = + + + + =
tx
u v x y
y
2 1,
+= = + ta c h:
u v u v v u
v uv u v v 2 2
4 4 3, 1
5, 92 7 2 15 0
+ = = = = = = = + =
Vi v u3, 1= = ta c h:x yx y x y x x
y x x yx y y x
2 2 2 1, 21 1 2 0
3 2, 53 3
= = + = + = + = = = =+ = = .
Vi v u5, 9= = ta c h: x y x y x x x y y x y x
2 2 21 9 1 9 9 46 0
5 5 5
+ = + = + + = + = = =
, h VN
Kt lun: H cho c hai nghim: (1; 2), ( 2; 5) .Bi 9. Gii h phng trnh sau:
=++++
=++++
011)1(
030)2()1(
22
3223
yyyxyx
xyyyxyyx
H PT xy x y x y x y xy x y xy x y
2 2 2( ) ( ) 30
( ) 11
+ + + =+ + + + =
xy x y x y xy xy x y xy x y
( )( ) 30
( ) 11
+ + + = + + + + =
tx y uxy v
+ = =
. HPTuv u v uv u v
( ) 30
11
+ = + + =
uv uv uv u v
(11 ) 30 (1)
11 (2)
= + + =
. T(1)uvuv
5
6
= =
Vi uv = 5 u v 6+ = . Gii ra ta c cc nghim (x; y) l: 5 21 5 21;2 2
+
v
5 21 5 21;
2 2
+
Vi uv = 6 u v 5+ = . Gii ra ta c cc nghim (x; y) l: (1;2) v (2;1)
Kt lun: H PT c 4 nghim: (1;2), (2;1),5 21 5 21
;2 2
+
,5 21 5 21
;2 2
+
Bi 10.Gii h phng trnh sau:x y x y
x y x y
2 2
2 2
3 4 1
3 2 9 8 3
+ + =
=
HPT x x y y
x x y y
2 2
2 2
3 4 1
3( 3 ) 2( 4 ) 3
+ + =
+ =
.t u x x
v y y
2
2
3
4
=
= +
.
HPT u vu v
1
3 2 3
+ = =
. Nghim (x; y):3 13 3 13
; 0 , ; 42 2
.
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Bi 11.Gii h phng trnh sau:x y x y
x y x y 2 2 2 2
2
1 3
+ = +
+ + =
iu kin: x y x y 0, 0+ > .t:u x y
v x y
= +
=
ta c h:
u v u v u v uv
u v u v uv uv
2 2 2 2
2 ( ) 2 4
2 23 3
2 2
= > + = +
+ + + + = =
u v uv
u v uv uv
2
2 4 (1)
( ) 2 23 (2)
2
+ = +
+ + =
.
Th(1) vo (2) ta c:uv uv uv uv uv uv uv
2
8 9 3 8 9 (3 ) 0+ + = + + = + =.
Kt hp (1) ta c:uv
u vu v
04, 0
4
= = =
+ =(vi u > v).
T ta c: x = 2; y = 2.(thok)Kt lun: Vy nghim ca h l: (x; y) = (2; 2).
Bi tng t:x y x y
x y x y 2 2 2 2
2
4
+ =
+ + =
Bi 12.Gii h phng trnh sau:xy x y
x y x y 2 2
3 2 16
2 4 33
=
+ =
HPTx y x y
x y2 2( 1)( 2) ( 1) ( 2) 21
( 1) ( 2) 38
=
+ =.t u x
v y1
2
= =
.
HPTuv u v
u v2 2( ) 21
38
+ =
+ =. Nghim (x; y): ( ) ( )3 3; 2 3 , 3 3; 2 3 + + .
Bi 13.Gii h phng trnh sau:x y xy
x y
xx y
2 2
2
34( ) 7
( )
12 3
+ + + = +
+ = +
HPTx y x y
x y
x y x y x y
2213 ( ) 13
13
+ + + = + + + + = +
.tu x y
x yv x y
1= + +
+ =
(vi u 2 )
HPT u vu v
2 23 13
3
+ =+ =
u
v uv
2( 2)
1
= =
x y xx y
y
x y
12 1
0
1
+ + = = + = =
.
Bi 14.Gii h phng trnh sau:
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H phng trnh nhiu n Trn STng
Trang16
=+
=+
358
152
33
22
yx
xyyx
H PTxy x y
x y3 32 (2 ) 30
(2 ) 35
+ =
+ =.t
u xv y
2 = =
. H PTuv u v
u v3 3( ) 30
35
+ =
+ = u v
u v2; 3
3; 2
= = = =
Nghim (x; y): 3(1;3), ; 22 .
Bi 15.Gii h phng trnh sau:x y x y
x y
2 1 1
3 2 4
+ + + =
+ =
H PT x y x y x y x y
2 1 1
(2 1) ( ) 5
+ + + =
+ + + + =.t u x y v x y 2 1 0, 0= + + = + .
H PTu v u v
u v loai u v2 21 2, 1
1, 2 ( )5
= = = = = + =
xy
2
1
= =
.
Bi 16.Gii h phng trnh sau:
=++
=++
64
9)2)(2(
2 yxx
yxxx
H PT x x x y
x x x y
2
2
( 2 )(2 ) 9
( 2 ) (2 ) 6
+ + =
+ + + =.t u x x
v x y
2 2
2
= += +
Bi 17.Gii h phng trnh sau:
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Trang17
Vn 3: Phng php nh gi
Bi 1. Gii h phng trnh sau:x z z a
y x x b
z y y c
3 2
3 2
3 2
9 27( 1) ( )
9 27( 1) ( )
9 27( 1) ( )
= = =
Cng (a), (b), (c) ta c: x y z d 3 3 3( 3) ( 3) ( 3) 0 ( ) + + = + Nu x > 3 th t(b) suy ra: 3 9 ( 3) 27 27 3y x x y= + > >
t(c) suy ra: 3 9 ( 3) 27 27 3z y y z= + > > (d) khng tho mn+ Tng t, nu x < 3 th t(a) 0 < z < 3 0 < y 0, y > 0, z > 0.
Khng mt tnh tng qut, gi sx y y z y z 1 1+ + .
Ta li c: z x y x 1 1= + + = x y z x x = y = z.
x x 1 0 = ( )
x
25 1
4
+= . Nghim x = y = z =
( )2
5 1
4
+.
Bi 3. Gii h phng trnh sau:x
yx
yz
y
zx
z
2
2
2
2
2
2
2
1
2
1
2
1
= + =
+
= +
Nu x = 0 th y = 0, z = 0 H c nghim (x; y; z) = (0; 0; 0) Nu x 0 th y > 0, z > 0 x > 0.
Ta c:x x
y x
xx
2 2
2
2 2
21
= =
+
. Tng tta suy ra c: y x z y x = y = z
x
xx
2
2
2
1=
+ x = 1. Nghim (0; 0; 0), (1; 1; 1).
Bi 4. Gii h phng trnh sau:xy
x x yx x
xyy y x
y y
2
3 2
2
23
2
2 92
2 9
+ = +
+ + = + +
Cng hai phng trnh, vtheo v, ta c:xy xy
x yx x y y
2 2
3 2 23
2 2
2 9 2 9+ = + + + (*)
Ta c: x x x3 2 232 9 ( 1) 8 2 + = + , y y y2 23 32 9 ( 1) 8 2 + = +
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VT (*) xy xy
xy xy x y 2 22 2
2 22 2
+ = +
Du "=" xy ra x yx y
1
0
= = = =
. Nghim: (0; 0), (1; 1).
Bi 5. Gii h phng trnh sau:y x xx y y
3
33 4
2 6 2 = + +
=
y x x
x y y
3
3
3 4
2 6 2
= + +
= y x x
x y y
2
2
2 ( 1) ( 2) (1)
2 2( 1) ( 2) (2)
= +
= +
Ddng thy (x; y) = (2; 2) l mt nghim ca h.Nu x > 2 th t(1) y < 2. Nhng t(2) x 2 v y 2 cng du Mu thun.Nu x < 2 th cng suy ra iu mu thun tng t.Vy h c nghim duy nht x = y = 2.
Bi 6. Gii h phng trnh sau:xy x
xy y
2
2
10 20 (1)
5 (2)
= = +
HD : Rut ra yyy
yx +=
+=
55 2
C si202 x theo (1) 202 x suy ra x,y
T(2) yyy
yx +=
+=
55 2.
p dng BT C-si ta c: 525
+= yyx x2
20 . M theo (1) th x2
20 .
Do x2 20= x yx y
2 5 ( 5)
2 5 ( 5)
= =
= =
Bi 7. Gii h phng trnh sau:
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Vn 4: Phng php sdng phng trnh h qu
Bi 1. Gii h phng trnh sau:x yy z
z x
1 (1)
2 (2)
3 (3)
+ =+ =
+ =
Cng 3 phng trnh, vtheo v, ta c: x y z 3 (4)+ + = T(4) v (1) z = 2; t(4) v (2) x = 1; t(4) v (3) y = 0.Thli Nghim (x; y; z): (1; 0; 2).
Bi 2. Gii h phng trnh sau:xy x y yz y z zx z x
2 1
2 7
2 2
= + += + +
= + +
xy x y yz y z
zx z x
2 1
2 7
2 2
= + += + +
= + +
x yy z
z x
(2 1)(2 1) 3
(2 1)(2 1) 15
(2 1)(2 1) 5
= =
=
Nhn cc phng trnh trn, vtheo v, ta c:
x y z2 2 2(2 1) (2 1) (2 1) 225 = x y z a x y z b
(2 1)(2 1)(2 1) 15 ( )
(2 1)(2 1)(2 1) 15 ( )
= =
Trng hp (a) x
yz
2 1 1
2 1 3
2 1 5
= =
=
x
yz
1
2
3
==
=
Trng hp (b) xy
z
2 1 1
2 1 3
2 1 5
= =
=
xy
z
0
1
2
==
=
Thli Nghim (x; y; z): (1;2;3), ( 0; 1; 2) .Bi 3. Gii h phng trnh sau:
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Vn 5: Phng php hm s
Bi 1. Gii cc h phng trnh sau:y
x
x x x
y y y
2 1
2 1
2 2 3 1
2 2 3 1
+ + = +
+ + = +
tu xv y
1
1
= =
. HPTv
u
u u
v v
2
2
1 3
1 3
+ + =
+ + =
u vu u v v 2 23 1 3 1+ + + = + + + f u f v ( ) ( )= vi tf t t t 2( ) 3 1= + + + .
Ta c: tt t
f t
t
2
2
1( ) 3 ln 3 0
1
+ + = + >+
f(t) ng bin.
u v= ( )uu u u u u 2 231 3 log 1 0+ + = + + = (2)
Xt hm s: ( )g u u u u 23( ) log 1= + + g u( ) 0 > g(u) ng bin.M g(0) = 0 nn u = 0 l nghim duy nht ca (2).Nghim: (1; 1).
Bi 2. Gii h phng trnh sau:x x y y
x y
3 3
8 4
5 5 (1)
1 (2)
=
+ =
T(2) x y8 41, 1 x y1, 1 .
Xt hm s f t t t t 3( ) 5 , [ 1;1]= f t t t 2( ) 3 5 0, [ 1;1] = < f(t) nghch bin
trn [1; 1].Do : T(1) f(x) = f(y) x = y.
Thay vo (2) ta c: x x8 4 1 0+ = x y41 5
2
+= =
Bi tng t: x x y y
x y
3 3
6 6
3 3
1
=
+ =
Bi 3. Gii h phng trnh sau:
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Vn 6: Gii phng trnh bng cch a v h phng trnh
Bi 1. Gii phng trnh sau: 3 18 1 2 2 1++ = x xt x xu v3 12 0; 2 1+= > = .
Ta c h
u vu v u v
u uv u u v u uv v
3 3
33 2 2
01 2 1 2
2 1 01 2 ( )( 2) 0
= > + = + =
+ =+ = + + + =
x
x 2
0
1 5log
2
= + =
Bi 2. Gii phng trnh sau: x x3 31 2 2 1+ = t y x3 2 1= . Ta c h x y
y x
3
3
1 2
1 2
+ =
+ = x y x y xy 2 2( )( 2) 0 + + + = x y=
x x3 2 1 0 + = x
x
1
1 5
2
= =
.
Bi 3. Gii phng trnh sau: x x32 3 2 3 6 5 8 0 + = t u x v x v 3 3 2, 6 5 , 0 (* )= = .
Ta c h:u v
u v3 22 3 8
5 3 8
+ =
+ =
uv
u u u3 2
8 2
3
15 4 32 40 0
=
+ + =
u
v2
4
= =
x = 2.
Bi 4. Gii phng trnh sau:
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Vn 7: H phng trnh cha tham s
Bi 1. Tm m h phng trnh:( )
2 2
2 2
2
4
+ =
+ =
x y x y
m x y x yc ba nghim phn bit.
H PT
m x m x m
xy
x
4 2
2
2
( 1) 2( 3) 2 4 0 (1)
2
1
+ + =
+=+
.
+ Khi m = 1: H PTx
VNxy
x
2
2
2
2 1 0
( )2
1
+ = +
=+
+ Khi m 1.t t = x2 , t 0 . Xt f t m t m t m 2( ) ( 1) 2( 3) 2 4 0 (2)= + + =
H PT c 3 nghim phn bit (1) c ba nghim x phn bit
(2) c mt nghim t = 0 v 1 nghim t > 0 ( )f
mmS
m
(0) 0... 22 3
01
= == >
.
Bi 2. Tm m h phng trnh sau c nghim: 11 3
+ =
+ =
x y
x x y y m(D 2004)
t u x v y u v , ( 0, 0)= = . H PTu v u v
uv mu v m3 31 1
1 3
+ = + = =+ = .
S: m1
0
4
.
Bi 3. Tm m h phng trnh sau c nghim duy nht: y x my xy
2 (1)
1 (2)
= + =
T (1) = x y m2 , nn (2) = y my y 22 1 = +
y
m yy
1
12
(vy 0)
Xt ( ) ( )= + = + >f y y f y y y21 1
2 ' 1 0
Da vo BTT ta kt lun c h c nghim duy nht >m 2 .Bi 4. Tm m h phng trnh sau:
a)
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Trn STng H phng trnh nhiu n
Trang23
Bi 1. Gii cc h phng trnh sau:a)
=+
=+
22
22
xy
yxb)
x y x y
x y
. 3
1 1 4
+ = + + + =
c)
x y
y x xy
x xy y xy
71
78
+ = +
+ =
d)
x y
y x
x y xy2 2
5
2
21
+ =
+ + =
e)x y
y x
x y xy
2 23
3
+ = + =
f)( )x y x y
x y x y
3 3
2 2
7
2
=
+ = + +
g) x yxy
1 1 43
9
+ = =
h) x y
xy
3 3 4
27
+ =
=
Bi 2. Gii cc h phng trnh sau:a)
xy x y
x y x y xy2 23
6
+ =
+ + + =b) x xy y
x y xy
2 2 3
1
+ + =+ + =
c) x y
x y2 2
1 1 1
2
5
+ =
+ =
d)( ) ( )
x y x y
x x y y y
2 2 4
1 1 2
+ + + =
+ + + + =
Bi 3. Gii cc h phng trnh sau:a)
x y x
y x y
2 2
2 2
2 3 2
2 3 2
=
= b)
x y x
y x y
3
3
2 2
2 2
= + +
= + +
c)x y x
y x y
2 2
2 2
2 3 4
2 3 4
= +
= +d)
x y x y
y x y x
2 2
2 2
2 2
2 2
= +
= +
e)
yx
yx
yx
2
2
2
12
1
=
=
f)
yx
y
xyx
2
2
2
2
1
1
11
=
+
= +Bi 4. Gii cc h phng trnh sau:
a)x xy y
x xy y
2 2
2 2
6 2 56
5 49
=
=b)
x xy y
x xy y
2 2
2 2
2 3 15
2 8
+ + =
+ + =
c)x xy y
x xy y
2 2
2 2
2 3 9
2 2 2
+ + =
+ + =d)
x xy y
x xy y
2 2
2 2
2 4 1
3 2 2 7
+ =
+ + =
e)x y
x y
5 5
3 3
1
1
+ =
+ =
f)x y
x y
2 2
3 3
1
1
+ =
+ =
IV. BI TP N
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H phng trnh nhiu n Trn STng
g)x y
x x y y
2 2
4 2 2 4
5
13
+ =
+ =h)
x y
x y
4 4
6 6
1
1
+ =
+ =Bi 5. Gii cc h phng trnh sau:
a)x xy y x y
x xy y x y
2 2
2 2 2
3( )
7( )
+ =
+ + =
b)xy x y x y
x y y x x y
2 22
2 1 2 2
+ + =
=
c)x y x y
x y x y
2 2
2 2
( )( ) 13
( )( ) 25
+ =
+ =d)
x x y x y
x y x xy
4 3 2 2
3 2
1
1
+ =
+ =
e)
x y x y xy xy
x y xy x
2 3 2
4 2
5
4
5(1 2 )
4
+ + + + =
+ + + =
f)x x y x y x
x xy x
4 3 2 2
2
2 2 9
2 6 6
+ + = +
+ = +
g)x x x y
x x y
2
2
( 2 )(3 ) 18
5 9 0
+ + =
+ + =h)
x y x y
x y x y
2 2
2 2
( )( ) 3
( )( ) 15
=
+ + =Bi 6. Tm m cc h phng trnh sau c nghim:
a)xy x y m
x y x y2 2( 1)( 1)
8
+ + =
+ + + =b)
x xy y
x xy y m
2 2
2 2
1
3 2
+ =
+ =
c)x y
x y y x x y m
1 1 3
1 1 1 1
+ + + =
+ + + + + + + =d) x y m
x y m
1 2
3
+ + =
+ =
e)
+=
=
mxyx
yxy
26
12
2
2
f)
x yx y
x y mx y
3 3
3 3
1 15
1 1 15 10
+ + + =
+ + + =
Bi 7. Tm m cc h phng trnh sau c nghim duy nht:
a)y m x
x m y
2
2
( 1)
( 1)
+ = +
+ = +b)
mx y
y
my x
x
22
22
2
2
= +
= +
c)x y m
x xy
2 0
1
=
+ =
d)y x x mx
x y y my
2 3 2
2 3 2
4
4
= +
= +
e)
=+
=+
)1(
)1(
2
2
xayxy
yaxxy (K cn v ) f)
+=+
+=+
axy
ayx2
2
)1(
)1( (K cn v )
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