4 . 3 . 1 N h a n d a n g v a p h i f d n g p h a p g i a i .
He bac hai vcli hai an x va y la | ^^^^J + 1 '''A t w'^" t ^ (*)
• • \ b2xy + C2y'^+ d2X + C2y h-
Mot so trirSng hop dac biet (doi xi'mg loai 1, loai 2, dS,ng cap...) da difdc xet (1 cac phan t n t d c . K h i cac t m l i chat dac biet khong con t h i he (*) d\trtc giai theo m o t so do chnng se dudc t r i n h bai t r o n g cac bai toan 67 d t r a n g 2GG, bai toan 68 6 t r a n g 267. T i i y nhien phirong phap nay khong i)hai la t o i uu.
N h i n chung c:ac dang thitcJng gap dcu dita t r c n nigt vai dac t h u cua dang bfic hai. Nen biet khai thac cac tfnh chat dac biet do t a se t i m ditrJc Idi giai ngiln ggn.
B a i t o a n 67. Gidi he phitcing trinh G i a i .
{5 2 + y 2 + a; - 2y = 2
• ^ + y ^ + 2(x + y) = 11.
K h i X = 0 t h i he viet lai | _,_ 2y - 11 "SliiC-ni.
K h i X 7^ 0. Dat y = a x , thay vao he da cho t a ditOc x 2 + ft'-^x^ + X - 2 a x = 2
I x^ + d^x^ + 2x + 2 a x = 11
( l + a 2 ) : ; ; 2 4 . ( i _ 2 r k ) x = 2 ( l + a ^ ) x ' ^ + 2 ( l + a ) x = 11.
C (1 + f v ^ ) z + (1 - 2fv)x = 2
at x^ = i t a ditdc he (1 + a'-^)^ + 2(1 + a ) x = 11 T a tfnh cac d i n h th^ic
D = l + a^ I--2a l + a^ 2 +2a
1 + fv2 2
1 + ^ 2 11
y = ax, x^ = z.
= (1 + a'){2 + 2a - 1 + 2a) = (1 + a'^){Aa + 1). 2 \
= (l + a 2 ) . 9 . r - ' l ; , 2 1 - 2a
11 2 + 2 a 4 + 4a - 11 + 22a = 26a - 7. * ' !? -aJ
VI DX 7^ 0, Va nen neu 4a 4- 1 = 0 t h i D = 0, he v6 nghiem. T i e p theo t a xet a K h i do
O r 26a - 7
D 4 a + r ' D (l + a 2 ) ( 4 a + l ) ' Dieu kien x!^ = z cho ta phUdng t r i n h
<^ 81(1 + a^) = (26a - 7)(4a + 1) a = 2
81 26a - 7
( 4 a + 1 ) 2 (l + a 2 ) ( 4 a + l )
<^81a2 + 81 = 104a2 - 2a - 7 4* 23a2 - 2a - 88 = 0
a = -- 44 23' V6i a = 2, t a diMc { J = 2.
Vdi a = - — , t a dildc 44 2 3 '
X =
y =
9 9
44 - 153 ~ 2 3 \4
23 17 / 4 4 \
23 17 17'
Vay he da cho c6 hai nghieui (x; y) = ( 1 ; 2 ) , (x; y) = J^! J^)"
L i f u y. B a i toan 6 7 se dUdc giai nhanh hdn neu t a n h i n t h a y dUdc : L a j ' hai phitdng t r i n h cua he trir nhau, t a se t h u dUdc m o t phitdng t r i n h bac nhat theo hai an x va y, tit day r i i t y theo x, giai bang phifdng phap the.
4 . 3 . 2 S a n g t a c c a c h e b a c h a i t o n g q u a t . ' ' V i d u 1 . Xet hai s6x = 3vdy = 0. Khi do { ""1 + y' " + ^y = - 3
[ x ' ' - x y + y ' ' + X - 2y = 1^- Vay ta thu ditOc mot he bac hai tong quat, he do ch&c ch&n c6 mot nghiem
"dep" la {x;y) = {3;0). Ta CO bai todn sau day. . ,.
B a i t o a n 6 8 . Gidi h$phmng trinh { ""l^~ H'[ ,o — xy + y + Zy — i^-
G i a i .
• K h i .T = 0 t h i he viet lai | ^2 ^ 2y = 12^ " ^ ^ nghiem.
• K h i X 7^ 0. Dat y = ax, thay vao he da cho t a dilOc (1 +a2)x2 + 2( Q - 2) i = - 3 ( l - a + a 2 ) x 2+ ( l - 2 a ) x = 1 2 . Dat = z t a dvTdc he
•+ .1
(1 + Q 2 ) 2 + 2 ( Q - 2 ) X = - 3
{l-a + a^)z+{l-2a)x=12 y = ax, x^ — z.
Ta t i n h cac d i n h thiic
1 + ^2 2a - 4
1 - a + 1 _ 2Q - 3 2 ( Q - 2 )
12 1 - 2a 1 + a2 - 3 l - a + a^ 12
= - 4 a ^ + 7a2 - 8a + 5.
= - 1 8 a + 45.
= ISa^ - 3a + 15.
K h i D = 0, tufc la a = 1 t h i he v6 nghiem. Tiep theo chi xet a ^ 1. Diou kicu x^ = z cho t a phiMng t r i n h de xac dinh a
X =
z D
^ (I5a2 - 3a + 15)^ = ( - 4 a ^ + 7a - 8a + 5) (45 - 18a)
<^153a'* + 216a^ + 360a = 0
<^a (I53a^ + 216a2 + 360) = 0 <^ a = 0 a = - 2 .
K h i a = 0 t h i D = 5, Z)^ = 15, suy r a x = ^ = 3 ^ y 0.
K h i a = - 2 t h i D = 81, D^, = 81, suy T& x = 1 y = - 2 . He d a cho c6 h a i nghiem | ^ = Q v a | ^ ^ I 2
Lufu y. V i c a c phudng t r i n h d a thi'rc bac khong q u a 4 luon giai diWc n c n vdi phudng phap d a t r i n h b a y 6 tren t a luon giai dUdc c a c he bac h a i tong quat.
V i d u 2. Tic mot he. dang cap bac hai, hang each tinh tien nghiem, ta se thu dUdc mdt he bac hai tSng qudt. Xet he I " + ^ " o ~ f ^dt u = x + 2, V = y - 3. Khi do
x^ + 4 x + 4 + xy - 3 x + 22/ - 6 + 2y2 - 12?/ + 18 = 4 3x2 + 12x + 12 - xy + 3 x - 22/ + 6 - y2 + 62/ - 9 = 1
^ / x2 + 22/2 + a;y + x - 1 0 ? / = - 1 2 , .?•
^ \2 - 2/2 - x y + 15x + 4 y = - 8 . ^ fa thu dUdc bdi todn sau.
Bai t o a n 6 9 . Giai he phudng trinh [ 2y2 + xy + x - lOy = - 1 2
^ • ^ \2 - y2 - xy + + 4 y = _ 8 . Htfdng d a n . Dat x = w, - 2 va y = ?; + 3, t a dUdc he dang cap bac hai
I 3u2 -uv-v^ = \. ' T f ^ > ' LuM y- Phep d f t t x = u - 2 v a y = i; + 3 dUdc t i m ra n h u sau : Ta dat j ; = u + a va y = t; + 6, vdi a, 6 t i m sau. K h i do, thay vao phitdng t r i n h thit nhat ciia ho, t a dudc
+ 2n.u + a2 + 2t;2 ^ 4^,; + 2^2 + uv + bu + av + ab + M + a - 10?; - 106 = - 1 2 . De t h u dutdc phUdng t r i n h dang cap bac hai t h i dieu kien la
/ 2a + 6 + 1 = 0 ^ r a = - 2
\6 + a - 1 0 = 0 ^ 1 6 = 3.
Vay ta dat X = u - 2 va y = ?; + 3. Ngoai ra ta c6 the lam nhanh hdn nhiT
sau: Lan lUdt dao ham hai v c phUdiig trinh t h i i nhat thco bicn x (xeni y la hkng so), theo bien y (xem x l a hang so) ciia he t a dUdc
t r 2 x + y + l = 0 ^ r x = - 2
\y + X - 10 = 0 ^ \ = 3.
Tit do C O phep dat x = u - 2 va y = ?; + 3. • Bai t o a n 70. Giai he phUdng trinh ,, 1 i
1 x 2 + 3y2 + 4xy - 18x - 22y + 31 = 0 (1)
\ 2 x 2 + 4 y 2 + 2 x y + 6 x - 4 6 y + 1 7 5 = 0. (2)
Hwdng d i n . Lan htdt dao ham hai vc phitdng trinh thi't nhat theo bicn x (xeni y l a hang so), theo bien y (xem x la hang so) ciia he t a dUdc
r 2x + 4y - 18 = 0 ^ / x = - 5
„ \x + 6 y - 2 2 = 0 ^ \ = 7. -3
• v a y , thitc hien phep doi bien x = - 5 + iz, y = 7 + t;, t a ditdc ' '' '
Jl ju"^+ 3v^+ 4uv = 1 (3)
2u2 + 4?;2 + 2?iu = 1. (4) ' .f MAO (11)
He nay l a he dang cap, c6 t h e giai theo each t h o n g thucJng, n h u n g I m i y la tr(t hai phUdng t r i n h (3) va (4) v ^ theo ve t a c6 ngay u'^+ v'^-2uv = 0 ^ u = y
\ = V =
T a c o h e ( 7/ ) o{ 8 w! = l 2\/2
- 1 Vay ( / ) CO nghiem 2v/2'
5; —i + 7
B a i t o a n 71 ( D l n g h i O l y m p i c 3 0 / 0 4 / 2 0 1 1 ) . Giai he phuang trinh a.2 + 4 y 2 _ 4^ + 1 2 y+ l l = 0
x2 + 4y2 - 2a;y - X + 4y - 12 = 0.
H i f d n g d a n . N c u tinh tao nhhi nhan thi thay ugay rang day la bai toan dS '
3.7; — 23 '
lay hai phitdng trinh trijf nliau t a ditdc y = —, the vac phitdng trinh thii nhat ciia he dUdc phUdng trinh bac 4 va may man la phiTdng trinh nay CO t 6 i hai nghiem "dep" x = 1 va x = 4.
B a i t o a n 7 2 . Giai he phudng trinh ( H\+ f 2^," ' + 2 / + 2 = 0 (1)
^ \x^ + y^ + x + y-4 = 0. (2) H i f d n g d a n . T a c6 (1) o y'^ - ( x + l)y - 2x2 + 5^,- - 2 = 0. T a coi day la phudng trinh b^c h a i v d i an y, con x la thara so, c6 5 = x + 1 va
P = -2x2 + 5a; _ 2 ^ - ( x - 2) (2x - 1) = {-x + 2) (2x - 1 ) .
T a chon P = {-x + 2) (2x - 1) de ( - x + 2) + (2x - 1) - x + 1 - S. Vay (1) CO h a i nghiem yi = 2 - x, y2 — 2x - 1. D a n den
( y = 2-x { y = 2x-l
l x 2+ 2/ 2+ x + y - 4 = 0 ' \ + i/ +x + y -i=-0.
( 4 13^
He CO hai nghiem (x; ?/) = ( 1 ; 1 ) , (x; ?/) = - 7 ; - " T • L i f u y. De bien doi 2x2 + x y - _ 5^ + y + 2 = 0 ^\^^^
(x + y - 2) (2x - y - 0) = 0
thu dUdc y = 2 - X , y = 2x - 1 bang each tach, them bdt,... ro rang kho khS"
hdn nhieu so vdi phan tich - 2 x 2 ^ 53. _ 2 thanh (2 - x ) (2x - 1).
B a i t o a n 7 3 . Giai h$ phuang trinh / - - 2 ^ + 2y = - 3
U r - 2 x y + 2x = - 4 .
p a p s 6 . He CO nghiem l ^ - 2^) = + 1- ^ + l ) ^ ( 1' ^1 + l ) •
^ , , ^ ^ , , f x2 - 2xy + 2y + 15 = 0 B a i t o a n 74. Giai he phuang tnnh | _ ^^^^ + ^2 ^ 5 ^ 0
D a p so. He CO n g h i § m (x; y) l a {2^2 + 1; 3 ^ 2 + l ) ; ( - 2 \ / 2 + i ; - 3 ^ 2 + 1).
4,4 P h i f d n g p h a p d u n g t i n h ddn d i e u c u a h a m so
T i n h ddn dion c.iia ham so la mot cong cu hfru hic\ dfi sang tac va giai phUdng t r i n h , van de nay da diTdc t r i n h bay ci bai 1.3 : PhUdng phap dua phUdng t r i n h ve phirdng t r i n h ham (5 t r a n g 15). Trong bai nay t a se khai thac t i n h (Idu dicu ham so dc giai ho phUdng t r i n h . M o t so van dc ve pluidng phap giai da CO a t r a n g 15 nen k h o n g neu r a d day m a t a se di vao n h i i n g vi du, bai toan cv the.
t o a n 7 5 Gidi he / W I T ^ + ^1+7) = 1 (1) / n
B a i t o a n 7 5 . Giai h^ ^ ^^^^ - 2xy + 1 = 4xy + 6x + 1. (2)
G i a i . Dieu kien 6x - 2xy + 1 > 0. T a c6 1
- 2 / + \/y2 + 1 .
y + \/y2 + 1
(1) ^ X + v/x2 + l = -y + V'y2 + 1 ^ fix) = fi-y), vdi f{t) = t + ^/W^.
-1 > 0, do do fit) dong bien tren R.
M a (1) I t i o n CO nghia vdi m o i x G R, y G R va (1) <^ fix) = fi-y) nen (1) ^ X = - y . T h e vao (2), t a dudc
x V 6 x + 2x2 + 1 = -4x2 + 6x + 1
< ^ x \ / 2 x 2 + 6x + 1 = 2x2 + 6x 4-1 - g^2
x\ 25x2
(^2x2 + 6x + 1 - I) = x/2x2 + 6 x 4- 1 = 3x (3)
\/2x2 + 6x 4-1 = -2x. (4) [ ^ c 6 ( 3 ) ^ { 2 ^ i 6 x + l = 9x2 = =
[ T a c 6 ( 4 ) . . { 2 < t 6 x . l = 4 x 2 ^ ^ = zl^^ .
| | y he ( / ) CO hai nghiem (x; y) - (1; - 1 ) ; (x; y) = ( ^ ; )•
B a
I
^ • x ' - r e n - ' i ' 1 7 3 7 ^ 4 - 4 ( 2 x 4 - 1 ) = y y ^ + 3y (1) (i) m toan 76. Giai he + y) (2x - y) + 4 = - 6 x - 3y. (2)
G i a i . Dicu kieii .x > > 1. Ta c6:
o
(2) <^ (x + y){2x - ?;) + 4 + 4(x + jj) + (2x - ?y) = 0
^{x + y + l){2x-y + 4) = 0
7 . ^\/ = 2x + 4 (do tir dieu kien suy r a x + y + l > - > 0 ) . Thay vao (1), t a dUdc : ' V
\ / 3 ^ ^ + 2x - 8 = v / 2 ^ + 3 ^
<:ằ2 (3x - 1) + v/ 3 x - 1 = 2 (2x + 3) + V2x + 3. (3) Xet ham so / (i) = 2f + t vdi ( > 0. K h i do
(3) 4^ / ( V s T ^ ) - / (y2irr3).
M a /'(O = 4t + 1 > 0 Vf > 0 nen ham so / dong bicn trcn [0; +oo), do do / ( v^ 3 x ^ ) = / ^ VSx^ = V2F+3 ^ X = 4 =^ y = 12.
Vay he phirong t r i n h ( / ) c6 nghiem duy nhat (x, y) = (4,12).
B a i t o a n 77. Gidi he phMng trinh j^s + ^s^^.^'l^o ^ ^ (2) Y tifdng. Rat ti.r nhien t a nhin vao tiifng phiWng t r i n h dc danh gia vdi muc dich t i m moi quan he gifra hai bien. Txi (1) t a thay rang 2 ve l a 2 da thiic doc lap ciia 2 bien x, y va ciing bac. N h u vay viec ap dung phitdng phap sut dung t i n h ddn dicu c6 cd hoi thanh cong rat cao. Va day cQng la liic chung ta dung tdi k y thuat he so bat dinh. Dau ticn, t a chon mot da thi'tc bat k i lam chuan d (1). De thay nen chon da thiic ben ve trai v i nhin no ddn gian hdn. Vdi y tulcing do t a dUdc ham so dac trUng f{t) = t^ + t-2, nhuT vay viec ciia chiing t a can lam do la phan tich :
y^ + 3y^+Ay = g\y) + g{y)-2.
R6 rang g{y) c6 dang g{y) = y + h t i t day t a khai trien va dong nhat he so dUdc 6 = 1 . N h u vay t a c6 phitdng trinh + x - 2 = (y + 1)^ + (y + 1) - 2- T6i day t h i y titcfng giai bai toan da ditdc hoan thien.
G i a i . T i t (1) t a c6
x3 + x - 2 = ( y + l ) ' + (y + l ) - 2 . (3) Xet ham so f{t) = + i - 2, i G R. Ta c6 f'{t) = 3^^ + 1 > 0, V( G R. Suy ra
/ dong bien tren E. Vay (3) / ( x ) = f{y + 1) <^ x = y + 1. T h e vao (2) : x^ + (x - 1)^ + 1 = 0 x^ + x^ - 3x2 + 3x = 0
^ x ( x ^ + x^ - 3x + 3) = 0 ^ [ _ 3^ ^ 3 : / 3\ 3
x'* + x2 - 3x + 3 = .T^ + (^x - - J + ^ > 0 nen (4) vo nghiem. Vay (/) c6 liem duy nhat {x;y) = (0,1).
f (41/2 + 1) + 2 (x2 + 1) = 6
B a i t o a n 78. G.dz/.e | ,2^ (2 + 2 y v T l ) = x + (2) (^) G i a i . Dieu kien : x > 0. Neu x = 0 t h i t i t (1), t a c6 0 = 6 (sai'). Vay gia sii x> 0, chia ca hai ve ciia (2) cho x^, t a dUdc
22/(1 + ^/VTT) = i ( ' i + yjTi^ (3)
Xet ham so f (t) = t (l + V T + F ) , vdi t G M, phitdng trinh (3) viet lai thanh /(2y) = / ( i ) . T a c6 f'{t) = 1 + ^ T T ? + ^ i = =
/(2y) = / ( - ) t a CO 2y = - . Thay vao (1) t a dUdc
> 0, V i G 1 , do do t i l
X X
.x^ + X + 2 (x2 + 1) - 6 = 0 4=> x^ + X - 6 = - 2 (x^ + l ) V ^ . (4) p t cac ham so 5 (x) = x^ + x - 6, / i (x) = - 2 (x^ + l ) ^ x , vdi x G (0; +00).
cd g'{x) = 3x2 + 1 > 0 va /^/(^•^ = _ 2 f2 x v ^ + < 0, Vx > 0.
V 2v/x J
/ay g{x),h{x) ddn dicu ngUdc chieu trcn (0;+00) va g{l) - h{\) nen (4) cd nghiem duy nhat x = 1, suy ra y = ^. Do do (/) cd nghiem (x, y) = ^ 1 , ^ B a i t o a n 79. Giai he phudng trinh
r ( \ / ^ 2 ^ - 3 x 2 y + 2) ( v / V T T + l ) = 8 x 2 , / (1)
. x2y - X + 2 = 0. (2) (I)
G i a i . Vdi x = 0 hoSc y = 0 t h i thay vao he (I) dan tdi v6 l i . Gia sii x 0 va 2/ 7^ 0. Phitdng t r i n h (1) tUdng dildng vdi
^^^^tizl^'^ + ^4y2 = 8x2y3 ^ ^^^-^=^'y + ^ ^ 2x2y
^ \ / x 2 + l - 4x2y + X = 2x2y v/4y2 + 1 - 2x2y
<!=>\/x2 + 1 + X = 2x2y (V4y2 + 1 + 1)
1
X -o + 1 + 1 = 2y J{2yf + 1 + 1 ) (3)
Xet ham so f{t) = t (^fi?T\ l ) c6 /'(O = 1 + + 1 + , > 0 neri
/ dong bien tren R. Tir (3) ta c6 / ( - I = / (2y) <^ - = 2y <^ 2x1^ = 1. Thf>
\ / X
vao (2), ta c6 : 2x'^y - 2 x + 4 = 0 < ^ x - 2 x + 4 = 0 < ^ x = 4=>?/ = ^.
8 Ket luan : He c6 nghiem duy nhat (x; y) = 4; - .
V 8 / B a i toan 80. Giai he X^ - 3X2 + 2 = ^
(2) (/)
Y t i f d n g . Chung ta lai bat dau t i m toi t i i cai ddn gian t d i phiic tap. Tvi (]) de y rang ta da c6 dang g{x) - h{y) nhit mong muon, n h u vay y tu6ng dting t i n h ddn dieu de xet ham dac trUng da xuat hien. Se tot hdn neu g{x), h{y) la ham da thiic. Vay ta thijf binh phudng de loai bo can thirc :
(1) ^ (x^ - 3 x 2+ 2 f = y3 +32/2.
Cong viec tiep theo la t i m ham dac tritng. De thay h{y) = + 3y2 la lira chon tot v i day la ham so ddn gian va dong bicn tren [0; + 0 0 ) . Ta se c6 g d n g
phan tich {x^ - 3x2 ^ 2f = q^{x) + 372(x). Dong nhat he so se t i m diTcic q{x) = x2 - 2x - 2. Suy ra x2 - 2x - 2 = y (chu y dieu kien c6 nghiem la x^ - 3x2 + 2 = (x - i) ( ^ 2 _ 2x - 2) > 0 x2 - 2x - 2 > 0, do X > 2). Nhmig cau hoi dat ra la, vice khai tridn va dong nhat he so v6i (.r^ - 3x2 _|_ 2)2 j^jj/^
phiic tap. Lai chii y rang ham so dac trUng khong phai la duy nhat. Lieu co mot ham so nao ddn gian hdn ? Vay dieu tU nhien la ta se d i t i m each d a t an phu : mot ham chiia can nao do dg khong phai luy thita. De y rang
(1) <^ x^ - 3x2 + 2 = ^ ^ - 3x2 + 2 = + 3
N h u vay ta se dat a = \/y~+3 => y = - 3, y-^y + 3 = {a^ - 3)o = - 3ô
Ham dac t r u n g se la f{t) = - "it. Do do can phan tich x^ - 3 x 2 + 2 = 5^ ( x) - 3 g( x ) . De thay r^ng g{x) = x + 6. Tir do
x^ - 3x2 + 2 = (x + _ 3(^ ^ ^)
^x^ - 3x2 + 2 = x^ + 36x2 ^ ^3^2 _ 3^^. ^ ^3 _ 3^
r ng nhat he so ta dUdc
- 3 = 36
0 = 36^ - 3 6 = - 1 . Do do 2 = 63 - 36
Vt.f
(x - 1)^ - 3(x - 1) = i^/y + zf - 3\ / ^ T 3
ham dac trUng /(() = - 3i la ham dong bien tren [1; + 0 0 ) . N h u vay y tiffing da ro rang.
G i a i . Dieu kien
I x > 2 ( x>2 f ^ o
y2 + 8 y> 0 [ y G ( - o o; - 8] U [ 0 ; + c x ) ) ^ ^ Ta CO
(1) ^ x^ - 3x2 + 2 = ^ (3. _ 1)3 _ 3(^ _ 1) {^/yT^f - 3\fyT^- Ta CO s/y + ^ > ^3 > 1, x - 1 > 1. Xet ham dac t n m g f{t) = - 3t, Vi > 1 CO / ' ( < ) = 3*2 - 3 > 0, V<> 1. Suy ra ham so / dong bien tren [1; + 0 0 ) , do do tit /(x - 1) = /(VyT^) ta CO X - 1 = sfyT^ ^y = x^ -2x-2. The vao (2) ta dudc
9(x - 2) - y2 ^. 8y <^ 9(x - 2) = (x2 - 2x - 2f + 8(x2 - 2x - 2)
^x"^ - 4x3 + 8x2 - 17x + 6 = 0
<^(x - 3)(x3 - x2 + 5x - 2) = 0 <^ x = 3
x^ - x2 + 5x - 2 = 0.
Xet Q(x) = x'^ - x2 + 5x - 2 c6 Q'(x) = 3x2 - 2x + 5 > 0, Vx € g^y j a y la mot ham dong bicn tren R. Lai c6 x > 2 Q(x) > Q(2) = 13 > 0, suy ra phUdng t r i n h Q{x) = 0 vo nghiem. Vay he phildng t r i n h (/) c6 nghiem duy nhat (x;y) = ( 3; l ) .
N h a n xet 1. NhUng bai toan tren da cho thay c6 nhieu each de dUa hai vi cua mot phuang trinh ve ham dac triing. Tuy nhien mot so bai loan kho hdn sc. dai hoi phai bicn doi cdc phuang trinh cua he de' tim ra ham dac trUng.
B a i toan 81. Giai he phudng trinh | + 3y) - 1
I x(y* - 2) = 3.
j G i a i . Vdi x = 0, the vao (I) thay v6 H. G i a sii x 7^ 0, tit (I) ta c6 2 + 3y = —
2 = -
X
, vdi fit) = t^ + St.
Co fit) = 3*2 + 3 > 0 , V t e K , suy ra / la ham dong bien tren M , do (j^
y = —. T h a y vao phUdng trinh thu: nhat cua he, t a dUdc
X
3 1 '
x ^ ( 2 + - ) = 1 ô - 2 x ^ + 3x2 - 1 = 0 < ^ x - , x = -l.
X ^
TM lai ta thay he ( I ) c6 hai nghiem (x; y) = ( - 1 ; - 1 ) , (x; y) = ( ^ ; 2).
L i f u y . Bang each dat t = -,t?i dua vc ho doi xilfng loai hai theo t va y.
B a i t o a n 8 2 . Gidi he phUcJng trinh x2 + l
(1)
31og2(x + 22/ + 6) = 21og2(x + y + 2) + l . (2)
G i a i . D i l u k i e n ( | \ " * o l ^ n ° ^ ^ t ham s6 : f{t) = e\t +l),te [0, +
\^x + y + Z > yj.
V i / ' ( / , ) = e^{t + 1) + e* > 0, V^, > 0 nen / la ham dong bien t r e n [0; + o o ) .
do (1) <^ e^'(x2 + 1) = ey\y^ + \ ) ^ / ( x ^ ) = f{y^) ^x^ = y^^x=±y.
00).
'0
(2) log2 [ ( x + 2y + 6)^] = logs [2(x + y + 2f
^ (x + 2y + 6)^ = 2(x + y + 2f.
• Neu x — y t h i t h a y vao (3), t a ditdc
(3x + 6 f = 2(2x + 2)2.
Theo dieu k i e n t a c6 x > - 1 . L a i c6
(3x + 6)^ - 2(2x + 4)^ = (x + 2)^(27x + 46) > 0 (3x + 6)^ > 2(2x + 4 f . D o do (3x + 6)^ > 2(2x + 4)^ > 2(2x + 2 ) ^ suy ra (4) v6 nghiem.
• Neu X = - y , t h a y vao (3), t a dUdc :
( - X + 6 f = 2(2)2 < - > ( 6 - x ) ^ = 8 ^ 6 - x = 2 < ^ x = 4=J>j/ = - 4 . Vay he (*) da cho c6 nghiem d u y nhat la (x, y) = (4, - 4 ) . I B a i t o a n 8 3 . Gidi cdc h$ phUdng trinh sau :
? ( 8 x- 3 ) \ / 2 ^ ^ - y- 4 y 3 = o I 4x2 _ 8x + 2y^ + y2 _ 2y + 3 = 0
..3/
(3)
(4)
rx-*(3-; + 55:
I xy{y'^ + 3y ) = 64
+ 3) = 12 + 51X
f 2x2_^x'^ + 4x - 1 = 2x^(2 - y)^^ - 2y
\VFT2= -^14 - X y / S ^ ^ + 1 ( ^x + y + l + ^/¥Ty = 5
\ + xy + 4 + >/?y2 + xy + 4 = 1 2 a) -y* -2 = Zx-Zy
+ yr=^ - Z^2y - y2 + 2 = 0.