Sang Tao Va Giai Phuong Trinh He Phuong Trinh Bat Phuong Trinh P4

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.[r]

B a i t o a n 23 G^â he phuang trinh [ IP^f^lÚi^g^X^^ ZlUl

B a i t o a n 24 Giọ he ph^Cng trinh { + g + ^Z = I + 8

V i d u 12 Xet mot phu&ng trinh bac ba nao do : 8x^ + 6x = VS Ta bien đi

thdnh phuang trinh, ch&ng han 8x^ + x'^ + bx = \/3 - x + x'^ Trong vi phdi,

thay X bdi y ta dUdc Sx^ + + 5x = \/3 - y + ý^ "Tron " x vd y vdo hai vi

de phicang trinh dU0c "sinh dong" hdn, chang han

<^8x^ - 2x2 + 4x = 8y^ - 2y2 + 4ỵ (1)

Ham so f{t) = 8t^ - 2t^ + At c6 fit) = - 4t + 4 > 0, V i G R nen dong

bien tren R, do do tvr (1) c6 / ( x ) = / ( y ) , hay y = x Thay vao he d a cho, ta

/

V i d u 13 Ham so y = logi x la ham so ngU0c cua ham so y =^ [

-(0; + 0 0 ) Do do ta co hdi toan saụ

tren

B a i t o a n 26 Gidi he phuang trinh (a)

/ 1 \ ( 4 ) •

ciia ham / ( x ) = j , do do do t h i cua hai ham nay doi xiing nhau qua

difdiig phan giac ciia goc plian t u t h i i nhat y = x B d i vay (x, y) la nghiem ciia (1) k h i va chi k h i x = y, nghia la (1) | ^ ^ x = y = ^

C a c h 2. De (x; y) la nghiem ciia he t h i x > 0 va y > 0 Khong m a t t i n h tong

quat, gia sii" y > x D a t y = tx, khi do i > 1 Thay vao (a) dudc

(c)

X = V4; = â, vdi a = Q V 0 < a < ^ (do t>\)

Tren (0; + 0 0 ) , xet hai ham so / ( x ) = x, ry(x) = ậ De thay ham / dong bien

con ham g nghich bien, do do (c) c6 khong qua mot nghiem Thay x = - vao

V 2 ' 2 J la nghiem duy nhat ciia hẹ

V i d u 14 Chon phuang trinh (x + y)2 + (x - 1)2 + (y _ i ) 2 ^

Tii day ta se tao ra rat nhieu he doi xiing loai hai, chang han

x^=x^ + y + C

Tic he (2) nay, neu y = x thi r^ - r^ - r - C = 0, do do ta nen chon^ C sao

cho he phuang trinh c6 nghiem "dep" vd bai toan khong qua kho Chang han

chon C = 0 thi thdy ngay rang he phuang trinh (2) nhan (0; 0) Idm nghieni

vd ta duac hdi toan sau

4.1.4 C a c b a i t o a n r e n l u y e n v a n a n g c a o

C a c bai t o a n ve he doi xiJng loai 1 va c a c bai t o a n difdc dufa ve he

doi xiJng loai 1

B a i t o a n 2 8 Giai he phuang trinh | ^^^^=2

nghiem ciia he da cho la (x; y) = (1; 1)

B a i t o a n 29 ( D H - 2 0 1 2 A ) Gidi he phuang trinh '{) •r'X

Vay nghiem cua he la ; - 0 ; ;

B a i t o a n 30 Gidi he phuang trinh (^" + ^2/+ ^ = 5

B a i t o a n 38 ( D e nghi O l y m p i c 3 0 / 0 4 / 2 0 0 6 ) Gidi phuang trinh

G i a i Dieu kion cosx > 0, siiiy > 0 Lay (1) t n r (2) then v6, t a diMo

log2 (1 + 3cosx) + log3 ( c o s i ) = log2 (1 + 3siri2/) + logg ( s m y ) (3)

Tren khoang (0; + 0 0 ) , xet ham so f{t) = logall + 3f.) + loggf K h i do

Vay ham so / dong bien tren khoang (0; + 0 0 ) T t f (3) t a c6

/ (cos x) - f (sin y) <^ cos x = sin y

Thay vao (1) t a ditdc

G i a i Dieu kien | y ^ 2 ^^^y { W = 2 khong la nghiem ciia he phiTdng

trinh da cho Lay (1) trir (2), t a duoc

\/x2 + 91 Vy^ToI = ^ y 2 + 2/^

-y-x x^-y^

v/x2 + 91 = v ^ x - 2 + x^

+ x + y

+ x + y > 0

251

Vay (3) X = 3 He phiTdng trinh c6 nghiem duy nhat x = y = 3

B a i toan 41 (De thi c h i n h thiJc O l y m p i c 3 0 / 0 4 / 2 0 0 6 ) Giai he phuang

Dat t = ^ , ta dUdc | ^3 ~ } Lay (1) trfr (2) thoo vo, ta dildc

t^-y^-3{y-t)=0^{t- y){e + ty+ y'^ + 2>) =Q ^ t = y

Thay vao (1), ta diTdc

Theo bai toan 3 d trang 119, suy ra cac nghiem cua (4) la cos—, cos-g-i

c o s - ^ Vay tat ca cac nghiem cua (3) la 2 c o s ^ , 2 c o s - ^ , 2cos ^ Nghiem

4.2 H e C O yeu to dang cap '2M I

4.2.1 H e C O chiia m o t phu-dng t r i n h dSng cap (tiJc l a thu4n

/ x2 - 2xy + 3y2 = 9

\-Bai toan 42 ( H V N H - 2 0 0 1 ) Gidi he (1)

13xy + 15y^ = 0 (2) Giai

Ngu X = 0 t h i thay vao (2) dudc y = 0, thay vao (1) thay khong thoa man

Xet X ^ 0 Chia ca hai ve ciia (2) cho x^ ta dUdc

4.2.2 H e C O h a i phrfdng t r i n h b a n d a n g cap bac h a i

a^x"^ + b-ixy + C 2 y ^ = d-^

Dua ve dang phttCng trinh da xet 6 muc 4.2.1 T i l hai phudng trinh ban dan^

cap bac hai nay, ta tao ra mot phitdng trinh dang cap bac hai nhu sau

dx {aax^ + h'lxy + C2y'^) = d-^ (aix^ + bixy + ciy^)

B a i toan 44 ( D H Q G T p H o C h i M i n h - 1 9 9 8 ) Cho he phMng trinh

3x'^ + 2xy + y^ = 11 + 2xy + 3y2 = 17 + m (A) a) Gidi he khi m = 0

-• K h i /, = 2, thay vao (1) ta dUdc l l x ^ = 11 ^ x = ± 1 Vay ( x ; y ) = (1;2)

(x; y) = (-1; -2) la nghiem ciia he (A)

• K h i i = - 7 , thay vao (1) dUdc ^ x ^ = l l < ^ x 2 = ^ < ^ x =

4 / \

^4v/3 5v/3\ , / 4v/3 5v/3\

, (x; y) = Vay (x; y) = ^ , ^

Khi /ft = 0 he (/I) CO bon nghiem

la nghiem ciia he

Bai toan 45 Cho cdc so thUc x,y thod man dieu kien x ^ - x y + y^ = 3 Tim gid tri Idn nhat vd gid tri nho nhat cua bieu thiic G = x^ + xy - 2y^

Hu'd'ng dan Goi T la tap gia t r i ciia G Ta com khi va chi khi he sau

a) Chiing minh rhng vdi moi a he da cho ludn c6 nghiem

b) Tim a de he c6 nghiem duy nhat

la nghiem ciia (*) vdi m o i a

b) T i r kgt cjua cau a) suy ra : K h o n g t o n t a i a de he (*) c6 nghiem d u y nhat

B a i t o a n 4 9 Xnc dinh cnc gin tri ciln a de he snu c.6 nghiem duy nhat

r x'^ + 2xy + 2f < a (1)

\ Axy -y^ <a (2)

G i a i B a t phUdng t r i n h (1) v i l t lai (x + yf + y"^ < a Vay neu a < 0 t h i h?

da cho vo nghiem T i e p theo xet a > 0 N h a n xet rang k h i do (x; y) = (0; 0)

I (x; y) = {s/a; 0) la nghiem ciia h ^ Suy ra ngu a > 0 t h i h § c6 i t n h a t hai

Tigp theo t a c h i i n g m i n h , vdi a < t h i he ( I ) c6 nghiem T h a t vay, goi

(a^o; ijo) la nghiem ciia he ( H ) , k h i do

Ltfu y Digm mau chot cua Idi giai la tim ra dufdc bat phvtdng trinh (3) Vj^y

(3) dUUc tim l a nliU the nao ? Vdi m < 0, ta c6

62 + 4

2 ( 1 - 6 2 ) + 2

4.2.4 H e d a n g cap bo p h a n Khi gap he dang cap bo phan, ta thudng sii dung phep the de tao ra he dang cap Nhieu k h i da biet chfic dua dUdc ve he dang cap t h i ta lai khong lam

dieu nay ma giai luon bKng each dat x = ty hoftc y = tx

+ y^ = 1

t^y + 2xy2 + y3 = 2

Bai toan 53 Giai he phuang trinh | ^2

Giai Ho phUdng trinh da cho vict lai

x2y + 2xy2 + y3 = 2 ^ \ + y^ - x^y - 2xy2 = 0 (2)

• K h i y 7^ 0 Ta co

.7:3+y3 = l (3) ' - V - f - ] " - 2 f - ) + 1 - 0 (4)

B a i i t o a n 54 Gidi he phuang trinh { "^^'^^ = + Vs/v

G i a i Dion kion x > 0, y > 0

• Xet 2/ = 0, khong thoa mail he phvtdng trinh

• Xet y T^O, dat = t^, dieu kien i > 0 va t 7^ 1 He t r d thanh

kien ta duoc nghiem ciia he phiTdng trinh la (x; y) = (9; 4)

B a i t o a n 55 ( D H - 2 0 1 1 A ) Gidi he phuang trinh

X

4 3 2 6 3

5 x - - + — - 2 x - - = 0 X X < S ^ 3 x - - + ^ = 0

<^3x4 - 6x2 + 3 = 0 <^ 3 (x^ - 1)^ = 0 x^ = 1 <j=! X = ± 1

Vay (x; y) = (1; 1), (x; y) = (-1; -1) la nghiem ciia he

Tru-dng hdp 2 : x^ + = 2 Thay vao (1) ta dudc

5x2y - 4xy2 + 3y^ - (x^ + y^){x + y) = 0

<i =>5x2y - 4xy2 + 3y^ - x^ - y^ - x^y - xy^ = 0

<;^4x2y - 5xy2 + 2y^ - x^ = 0

Neu y = 0 t h i tiJt (3) suy ra x = 0, mau thuan vdi x^ + y2 ^ 2, vay tiep theo

ta chi xet y 7^ 0 K h i do chia ca hai ve ciia (3) cho y^ ta diWc

(3)

4.T;2 5X „ X ^ ( 3 < » ^ + 2 - - ^ = 0 < ^

• Xet y = 0 Thay vao he ta dUdc

(.T; y) = (1; 0) la nghiem ciia he da cho

• Xet y ^ 0 Dat x = ty K h i do

B a i t o a n 57 Gtwi, he phuang trinh | '^y^^^^Jzl^

G i a i Xet y = 0 x = 0 Xet y 7^ 0 Dat / = - <^ x ty He da c l

Chia (1) cho (2) vc theo ve ta duoc

^^_:±±l = i z i ^ 3 / 3 _ 7 / 2 _ 3 / + 7 = 0.=

/2 + / - 3 / - 2

/ = - 1

' = 3

-He d a clio c6 b o n n g h i e m ( 0 ; 0 ) , (1; 1), (-1; 1),

B a i t o a n 5 8 Gidi he phMng trinh

3 ^ 3 _ 3 ^ (1)

X + iJ x'+y' = l (2)

G i a i N h i n vac he t a de d a n g n g h i den viec t h e so 1 t i t p h u d n g t r i n h (2)

vao i)h\mng t r i n h (1) dfi r o phirclng t r i n h t h u a n n h a t ( d a n g cap) D i o n kioii

x + yj^O.T^ CO (2) 4=> (x2 + y2)2 = 1 T h 6 vao (1) t a d u d c :

i3x '-y')ix + y)={x' + y'f

^ S x - i -y' + 3x^y - xy^ = x* + y'+ 2x^y\

Giai Dieu kien x > 0, y > 0, x + y =;>^ 0 De thiy x = 0, y = 0 khong thoa

man he, vay chi can xet x > 0 va j/ > 0 Dat u = > 0, v = ^ > 0 Khi

<^4t;u^ + 4i;^ - - uv^ = 2u^v + uw^

^2vu^ + 4w^ - - 2uu2 = 0 (2uu^ - u^) + (4^^ - 2uv^) = 0

^u^{2v - u) + 2u2(2u - u) <t4- (u^ + 2v^){2v - u) = 0 u = 2t)

Thay = 2?' vao (1) ta (hruc

4.2.5 Phufdng phap sang tac bai toan mdi

V i du 1 V6i { J = 1 to CO I + ^ _9 2/ = 2x, nen khi

giai he, ta se di(a ve mot phMng tnnh hac ba theo i — hon niia phuong

X

trinh nay c6 "nghiem, dep" t = 2 Ta c6 bdi todn sau

Bai toan 64 Giai he phuang trinh

.3 = 6

2x'^ + 2/3 ^ 10

x^y - 3xy2 + x^ = - 9

Giai Khi x = 0, he da cho tr6 thanh | g _r_^g Vay he khong c6 nghiem

dang (0; y) Tiep theo xet x 7^ 0 TiJt he da cho, ta co

Bai toan 65 Giai he phUdng trinh | ^ 4 ^ ^4 + J_~3y ^ Q

Hirdtng dan Khi y = 0, he da cho trd thanh | ^4 ^ ~1 Q <^ X = - 1 Vay

nghiem co dang (x;0) ciia he la (-1;0) Tiep theo xet y ^ 0 Dat x = ty,

thay vao ho ta dudc

ty^ - 2iy.y2 = - 1 <io^o J ((^ _ 2t) = - 1

C h u y 2 Viec sang tdc cdc he phUdng trinh duac gidi bang each dit.a ve cdc

phicang trinh dang cap bdc hai nhic cdc hai todn 60, 6 1 , 62 se duac trinh bay

a muc 4.6.2 d trang 311 Sau day ta se tiinh bay mot cdcli sung tdc klidc cuug

khd nhanh chong cho cdc he phiMng trinh dang nay {xem vi du 3 )

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 trang 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 • ^ + y ^ + 2(x + y) = 11 2 + y 2 + a; - 2y = 2

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

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

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)

V6i a = 2, t a diMc { J = 2

44 Vdi a = - — , t a dildc

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

X — xy + y + X — 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

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^ = \ 'Tf^ > '

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

\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)

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 ta c6 ngay u'^+ v'^-2uv = 0 ^ u = y

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 '

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)

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

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

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 = =

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

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 trinh 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

/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)

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 trinh (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

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 = ^

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 l a

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 = - 3«

Ham dac t r u n g se la f{t) = - "it Do do can phan tich

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

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

Co fit) = 3*2 + 3 > 0,Vt 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 « - 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

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

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 :

2 P h u ' d n g p h a p g i a i Xet he lap ba an (*), vdi / la ham so CO tap xac

dinh la D, tap gia trj la T va T C D , ham so / dong bien tron T

C a c h 1: Doan nghiem roi chiing minh he c6 nghiem d u y nhat T h u d n g de cluing minh he c6 nghiem duy nhat ta cong ba phudng trinh ciia he ve theo

vc, sau do suy ra x = y = z

C a c h 2: Tir T C D suy ra / ( x ) , / ( / ( x ) ) va / ( / ( / ( x ) ) ) thuoc D D i (x; y; z)

la nghiem ciia he thi x e T Neu x > / ( x ) thi do / tang tren T nen ta c6

/ ( x ) > / ( / ( x ) ) Vay / ( / ( x ) ) > / ( / ( / ( x ) ) ) D o do

^ > / ( x ) > / ( / ( x ) ) > / ( / ( / ( x ) ) ) = x

Dieu mau thuan nay chiing to khong the co x > / ( x ) T u d n g tir cung khong thg CO X < / ( x ) D o do / ( x ) = x Viec giai he (*) dUdc quy ve giai phUdng trinh / ( x ) = x H d n nfra ta c6 :

vdi A, B,C >\ f{y), f{z) > 0 {xem bdi todn 101 d trang 286) hoQ.c

C h u y 4 Khi ham f khong thod cdc dieu kien da noi 3 phan phuang phdp

gidi thi ta phdi c6 nhUng each xii li khdc, chang han xem bdi todn 90 d trang

280, bdi todn 109 d trang 292, bdi todn 99 d trang 284, bdi todn 107 d trang

Vay tap gia t r i cua ham / ( x ) la T - [ v ^ ; +oo

[ l ; + o c nen / dong bien tren [ ^ ; + o o )

Theo phan phitdng phap giai, he da cho

viet hii I ^ f ^ ^ I y = 2 He c6 nghiem duy nhat la (2; 2; 2)

C a c h 2: Cong ba phiTdng trinh ciia he ve theo ve ta dUtJc

(x - 2)^ + (y - if + (z - 2)^ = 0 (4)

Ta CO (2; 2; 2) la mpt nghiem ciia he Ta se chiing minh (2; 2; 2) la nghiem

duy nhat ciia he

• Neu X > 2 thi tir (1) ta c6

2/^ - 8 = 6x(x - 2) > 0 y > 2

278

tr y > 2 va tir (2) ta c6 ; ^; i :

^3 _ 8 = %y{y - 2) > 0 2 > 2 *

Vay 0 = (x - 2f + (y - 2)^ + (2 - 2)^ > 0 Day la dieu vo Ii

• Neu 0 < X < 2 (ta CO ngay x > 0 vi theo (3) t h i x^ = 6(2 - 1)^ + 2 > 0) thi

tif (3) suy ra

62(2 - 2 ) = X ^ - 8 < 0 = ^ 0 < 2 < 2 Ket hop vdi (2) suy ra 0 < y < 2 Vay 0 = (x - 2)^ + (y - 2)^ + (2 - 2)^ < 0

Day la dieu vo l i Vay x = 2, tir (1) ta c6 y = 2, thay y = 2 vao (2) ta c6

^ = 2 Vay (2; 2; 2) la nghiem duy nhat ciia he

C h u y 5 Doi vdi he lap ba an thi c6 mot sai lam rat tinh vi, kho phdt hien

do la sai lam: Do x, y, 2 cd vai trb nhu nhau nen khong mat tinh tSng qudt

gid sti X > y > z " Thuc ra x, y, 2 hodn vi vbng quanh nen phdi xet hai thii

Hufdng dan Xet ham so / ( x ) = sinx c6 tap xac dinh la R va tap gia t r i la

[ - 1 ; 1], / dong bien tren [ - 1 ; 1] He da cho viet lai

Ta chiing minh ditdc x = / ( x ) va | y = ^fi^ | ^ Z ^f(^y Xet phvtdng

trinh x = sinx tren [ - 1 ; 1] Xet ham so ^(x) = x - sinx tren [ - 1 ; 1] Ta c6:

g'{x) = 1 - cos X > 0, Vx e [ - 1 ; 1]

Ham g dong bien tren [ - 1 ; 1], 5(0) = 0 Do do x = 0 la nghiem duy nhat ciia

phudng trinh x = sinx tren [ - 1 ; 1] Vay (0;0; 0) la nghiem duy nhat ciia he

Bai toan 87 Giai he phUdng trinh nrf

x^ - 3.T2 + 6 x - 6 + ln(x2 - 3 x + 3) = y

y3 - 3y2 + 6y - 6 + \n{y'^ -2,y + i) = z

2^ - 32^ + 62 - 6 + ln(22 - 32 + 3) = X

279

G i a i X e t ham so /(.x) = - Sx^ + 6x - 6 + hi(x2 - 3x + 3) Ham so nay c6

tap xac dinh la M va ,

Vay f{x) dong bien tren E He da cho viet lai: < y = /(x) Tuong t i t nhit

cac v i du trutdc t a dUdc:

1 x = / W

Tiep theo t a giai phifcing t r i n h f{x) = x <^ / ( x ) - x = 0 D a t

h{x) = / ( x ) - X =^ li{x) = / ' ( x ) - 1 > 0,Vx G R

Vay li{x) dong bion tren R Hdn nifa /i(2) = 0 Do do /i(x) = 0 x = 2 Do

do (*) I ^ ^ ^ ^ ^ He da cho c6 mot nghiem duy n h i t la (2; 2; 2)

B a i t o a n 88 Gidi he phiiOng trinh

G i a i De thay he da cho tiTdng dudng vdi fiz) trong do

Vay ham s6 /(x) dong bien tren cac khoang

1 \I 1 \ 1 -cx); - -

Vay tap gia t r i ciia /(x) la R Tkp xac dinh ciia ham so / l a con, thitc su cua

tap gia t r i ciia ham so / nen t a khong the ap dung each giai n h u da t r i n h

bay trong phan phiTdng phap giai X e t phudng t r i n h x^^ - 3x = y(3x^ - 1) V i

X = ±—r^ khong thoa man phudng t r i n h nay nen de x la nghiem ciia phUdng

v 3

1 x^ - 3x trinh nay t h i x khac ±—7=, k h i do y = - — 5 — - Do do t a dat x = t a n a , vdi

»iai Dg (x; y; z) la nghiem ciia he da cho t h i dieu kien la x, y, z nho hdn 6

p da cho tUdng dirdng vcli • " •

(1) (2) (3)

hay ,, ,

( \ogM -y) = fix) m

log3(6 ^ z] = f{y} (2) vdi fix) = ^

\olti6 - x) = f(z) (3) Vx^ -2x + 6

Ta C O fix) = > 0,Vx < 6, suy ra fix) la han,

(.T2 - 2.r + 6) Vx2 - 2.T 4- 6

tang, con g(x) = log3(6 - x) la ham giiim vdi x < 6 Neu ix;y;z) la mot

nghiem ciia he phiMng t r i n h ta chiing minh x = y = z Khong mat tinh t6ng

quat gia siif x = max(.r, y, z) t h i c6 2 trirdng hop:

TrvfSng hdpl x>y> z Do fix) tang nen fix) > f{y) > f{z), suy ra

N' M - log3(6 -y)> log3(6 - z) > log3(6 - x) .^ij,

Do giani nen suy ra <

6 - y < 6 - 2 < 6 - a ; < s ^ a ; < 2 < y = i > a ; = y = 2

Tru'dng hdp 2: x > z > y. Tildng t i t nhir tren suy ra x = y = z Phuong

t r i n h g{x) = f{x) c6 nghiem duy nhat x = 3 Vay h f c6 nghiem duy nhat la

( x ; y ; z ) - ( 3 : 3 ; 3 )

r x 2 ( x + l ) = 2 ( y 3 - x ) + l

B a i t o a n 92 Giai he phuang trinh I y'^{y + 1) = 2 ( 2 ^ - y) + 1

I z'^iz + l) = 2{x^-z) + l

r x3 + x2 + 2x = 2y3 + 1 f fix) = gi^y) (1)

Hifdng d a n Ta c6 I y3 + y2 ^ 2y = 2z^ + 1 hay <^ / y) - g z) (2 vdi

I ? + 22 + 2 z - 2 x 3 + 1 I f\z)=9{x) (3)

<7(f) = 2^3 + 1 va f{t) = + ^2 + 2( Do dang cln'mg minh diwc / va g la nhOng

ham dong bien tren M Gia sii rang (x; y; 2) la nghiem ciia h$ va khong giam

t6ng quat, c6 the coi x > y K l i i do t i t (1) va ( 2 ) suy ra

r l o g 5 X = : l o g 3 ( 4 + y ^ )

B a i t o a n 94 Gidi he phuang trinh < logg y = log3(4 + y/z)

I log5^ = log3(4 + \ / i ) ; 282

B a i t o a n 95 Gidi he phuang trinh

B a i t o a n 96 Gidi he phuong trinh

B a i t o a n 97 Gidi he phuang trinh

B a i t o a n 98 ( D H - 2 0 1 0 A - P h a n C h u n g ) Gidi he phuang trinh

Dat w = v / 3 ^ ^ K h i do { u;2 ^ 2u = 3 Neu u < 1 thi

[ u2 + 2i; = 3

w'^=3-2u>l=^w>l=^v'^ = 3-2w<l

=^ti < 1 = 3 - 2u > 1 =^ u > 1,

;Den day ta gap m a u thuan, vay khong the c6 u < 1 TitOng tU, triTdng h0p

u > 1 cung khong t h i xay ra Vay u = 1 Suy ra u = = i f = 1 Do do x = 2 '

y = 2, thoa man d i e u kien

283

Lixtu y Viec t i m ra 2x = \/5 — 2y bang t i n h ddn dieu cua ham so ditdc tien

hanh bang phUdng phap he so bat dinh n h u sau : Ta se b a t ' d a u phan tich t i t

(1) PhUdng t r i n h nay c6 x, y tach bict ucn kha nang diing ddn dieu la cao

Vay t a bien doi phUdng t r i n h ve dang

Ta hy vong r^ng f{t) = x + ^ chinh la ham dac tritng ma t a can t i m Vay

can phai phan tich {4x^ + l ) x = 4x^ + x = + R6 rang p(x) c6

dang mx + n, dung he so bat dinh t a t h u diWc m = 2; n = 0 =^ p{x) = 2x

Nhir vay ham so dac t n t n g chinh la /(«) = y + 2'

B a i t o a n 99 ( D e n g h i cho k y t h i chon hoc s i n h gioi c a c trvfdng

C h u y e n k h u vulc D u y e n H a i v a D o n g B a n g B a c B o ) Gidi h$ phiCOng

T i t do, theo d i n h l i Lagrange t a c6

\x-y\ \fiy)-f{z)\ \f'{0\\{y-z)\<\y-z\

K h i do g'{x) = 1 + ^ sin > 0 Vay ham s6 g d6ng bi^n ma

= 0 nen he phitong t r i n h c6 nghi^m duy nhat x = y = z = ~

B a i t o a n 100 ( D e nghi cho k y t h i chon h o c s i n h gioi c a c trtfdng

C h u y e n k h u vu-c D u y e n H a i va D 6 n g B ^ n g B i c B o ) Gidi he phuang

trinh sau vdi a e (^\\

ax^ + 1= xy

a y 2 + 1 = yz az"^ + 1= xz

C)' 10.1 n.feo,} 11:

G i a i T i t he suy ra xyz ^ 0 va x,y,z cung dau Ta thay ngu ix;y;z) la

nghi$m ciia he t h i ( - x ; -y; -z) cung la nghieni cua he

• Xet trifdng hdp x, y, z cung dudng He da cho titdng ditdng vdi he sau

Vay / (t) la ham dong bien tren [2^; +oo) Ham <? ( 0 = < la ham dong bien

tren [2v^; +oo) Hg da cho t r d thanh