Phương pháp đặc biệt giải toán THPT sử dụng phương pháp điều kiện cần và đủ NXB hà nội 2004

Cuo'n 1: Sir dung phuong phap luong giac hoa giai Toan Cuon 2: Sii dung phuong phap vecto giai Toan Cuon 3: Su dung cac phep bien hinh giai Toan Cuon 4: Su dung phuong phap toa do giai

T h s T o a n hoc - K s T i n hoc L E H O N G DUG - C h i i bieii

Cuo'n 1: Sir dung phuong phap luong giac hoa giai Toan

Cuon 2: Sii dung phuong phap vecto giai Toan

Cuon 3: Su dung cac phep bien hinh giai Toan

Cuon 4: Su dung phuong phap toa do giai Toan

Cuon 5: Sir dung phuong trinh tham so Duong thiing, Duong tron,

Eh'p va Hypebol giai Toiin

Cuon 6: Sir dung phuong phap dat in phu giai Toan

Cuon 7: Su dung phuong phap di6u kien can va du giai Toan

Cuon 8: Su dung phuong phap ham so va do thi giai Tosin

Cuon 9: Sir dung gidfi han giai Toan

Cuon 10: Su dung dao ham giai Toan

Cuon 11: Sir dung may ti'nh giai Toan

Muc tieu cua bp tai lieu tham khdo nay la cung cap cho cue Thdy, Co gido nipt bp bdi gidng cintyen sdu cd chat hd/ng va cho cac em hpc sinli Trung hpc phu thong yeu thicli mdn Todn mot bp tai lieu hoc tap bo ich

Bp tai lieu ditac viet tren mot tit tudng liodn todn mdi me, c6 tinh sit

pham, CO tinh long h(/p cao, tan dung ditcrc day di'i the manh ciia cac

phuong phdp ddc biet de giai Todn

Bp tai lieu nay chac cliaii phu hpp vdi nhieu ddi titpiig ban dpc tit cdc Thdy, Co gido den cdc em Hpc sinh lap 10, 11, 12 va cdc em chud'n bi dit thi mdn Todn Tot nghiep PTTH hodc vuo cdc Trudng Dai hpc

Cuon

Sir uiJNG iMiifoi\ iMiAi' niv.ii KIEN CAIM V A nt o i i i ro/iiv

r//;'c/ ///a//// 3 chi'i de:

Chii de I: Sir diiiig phiraiig phap dieu kien can va dii giJi bai loan ve tinh

chat duy nhat nghiem

Chii del: Siidung phuang phap dieu kien can va du gitii bai loan ve tinh

chat nghiem

Chi'i de3: Sir dung phuang phtip dieu kien can va du giiii bai toiin ve tinh

chat tham so

inieii id (III lic't plncang phap gidi cho 9 dang loan tan dung dupe day du

ihe inanli cita phii'o'ng phap dlcit kien can vd dt'i

Tdi Cling xin bay id tai day long biet on sdu sac td'l sif gli'ip do' dong

vien llnli than cita lial nginyi Thdy ind Idl rat ini/c kinli Irpng, gdin GS.TS

Trail Maiili Titan ngtiyen Fhd Gidiii doc Trung Tdiii KHTN & CNQG, Nhd

gido I fit lit Ddo Thieii Klidi iigityen Hleu trudng TrUdng FTTH Hd Ndl

-Ainslerdaiii

Citdl ciing, cito dii dd rat cd gang, ninfng thai khd Irdnh klidi iihuiig

thlcit sol bdi nhifng hleu blel vd kinli nghlein con liaii cite, id't inong nlidn

dupe iiltilng y kien dong gop c/tiy bdu cita ban dpc gdn xa Mpl y kien dong

gap xIn lien he tiri:

Dia chi: Nhom lac gia Cu Mon - Nha sach Toan TMPT Cir Mon

So 20 - Ngo 86 - Duong To Ngoc Van - Quan Tay Ho - Ha Noi Dien thoai: (04) 7196671

bat phuang tiinh 10

Bai toan 2 Giai bai toan duy nhat nghiem cho he phuang trlnh,

he bat phuang trlnh 33

C H U D ^ 2

SU DUNG PHUONG PHAP DIEU KIEN CAN VA DU (ilAI BAI TOAN VE TINH CHAT N(,HIEM Bai toan 1 Giai bai toan ve tinh chat cac nghiem

cho phuang trlnh 66

Bai toan 2 Giai bai toiin ve tap nghiem 77 Bai toan 3 Giai bai toan ve phuang trlnh ha qua 92

Bai toan 4 Giai bai toan vi hai phuang trlnh tirang duang 95

Bai toan 5 Su dung do thi 126

CIIU Y)t 3

Sir DUNG PHUONG PHAP DIEU KIEN CAN VA DU (ilAI BAI TOAN VE TINH CHAT THAM SO Bai toan 1 Phuang trlnh nghiem diing vai gia tri

xiic dinh ciia tham so 102

Bai toan 2 He nghiem diing vol gia tri xac dinh cua tham so 105

TAI LIEU THAM KHAO 110

C H U D E I

GIAI BAI TOAN V E TINH DUY NHAT NGHIEM

M 6 D A U

Trong chu de nay se minh hoa each sir dung phuong phap dieu kien can va dii giai bai toan duy nhat nghiem cho phuong trinh, bat phuong trinh, he phuong trinh va he bat phuong trinh dugc chia thanh hai dung:

Dgiigl: Giai bai toan duy nhat nghiem cho phuong trinh, "bat

phuong trinh chua tham so

Dang 2: Giai bai toan duy nhat nghiem cho he phuong trinh, he bat

phuong trinh chua tham so

CO iii>hiem duy nlid't"

ta lliirc hien theo cac budfc sau:

Btioc 1: Dat dieu k i e n de cac bieu thirc trong (1) c6 nghla

BiioT 2: Dieu kien can: Gia sir (1) c6 nghiem la x = x,„ k h i do:

a Dua tren tinh chat doi x u n g ciia cac bieu thiic glai t i c h

trong (1), ta di khang dinh k h i do x = cpCx,,) cung la nghiem cua (1)

b D o do, de he c6 nghiem duy nhii't can c6:

x„ = (p(x„) =^ Gia t r i ciia x,, (2)

c Thay (2) vao (1) ta xac dinh dugc dieu k i e n can cho

tham so m de (1) c6 nghiem duy nhat, gia sur meD,„

Bu(>c3: Dieu kien dir V o i meD,„, ta d i k i e m tra lai tinh duy nhat

nghiem cho (1)

T h o n g thuong trong buoc nay, ta chi phai xet cac phirong

trlnh, bat phuong trinh cu the (thuong la k h o n g c6 tham so

hoac neu c6 t h l da duoc don gian d i nhieu) Ket qua ciia

budrc nay cho phep ta loai d i k h o i tap D,„ cac gia t r i k h o n g

thich hop ciia m

Buoc 4: Ket hop ba buoc giai tren ta t u n duoc dap so

II V i DU MINH HOA

Trudc tien chung ta m i n h hoa cac v l du sir dung tinh chat ham chSn

de xac d i n h dieu kien can, tuc la xuat phat tii' nhan xet:

• Gia su phuong trinh c6 nghiem x,, khang d i n h r i n g no cung se

m x ^ - 2 ( m - l ) x f , + m - 1 = 0

o m( - X,,)' - 2 ( m - 1)( - X,,)' + m - 1 = 0 tuc la - x„ cQng la nghiem cua phuang trinh

V a y de phuong trinh c6 nghiem duy nhat dieu kien la:

- x„ = x„ x„ = 0

K h i do:

(1) c ^ m - 1 = 0 » m = 1

D o c h i n h la dieu k i e n can de phuong trinh c6 nghiem duy nhat

Dieu kien du: V o l m = 1, ta c6:

x"* = 0 » X = 0 la nghiem duy nhat cua phuong trinh

Vay, v 6 i m = 1 phuong trinh c6 nghiem duy nhat

2 Nhu viiy, de tim dieu kien cua tham so sao ciio phuong trinh triJng

phuong:

ax"* + bx^ + c = 0 (1)

CO ngliiem duy nhat, bang phuong phap dieu kien can va dii dugc thirc

hien Iheo Ciic buoc:

Biiocl: Dieu kien can:

Giai sir (1) c6 ngliiem x,„ suy ra - x„ cQng la nghiem ciia

phuong trlnh Vay de phuong tiinh c6 nghiem duy nhat

dieu kien la:

- x„ = x„ <=> x„ = 0

Khi do:

(l)c:>c = 0

Do chinh la dieu kien can de phuong tiinh c6 nghiem duy nhat

Bia'rc 2: Dieu kien dir Thirc hien viec thu lai vdi c = 0

Vi du 2: Tim m de phuong tiinh sau c6 nghiem duy nhat:

V l - x ^ + 2 N / I - X 2 = m (1)

Gicii

Dieu kien can: Nhan xet rang neu phuong trlnh c6 nghiem x,„ thl cQng

nhan - x,, lam nghiem

Do do phuong trlnh c6 nghiem duy nhat thl dieu kien can la

x„ = - x„ » x„ = 0

Khi do:

1 + 2 = m « m = 3

Do chinh la dieu kien can de phuong tiinh c6 nghiem duy nhat

Dieu kien dir V6i m = 3, khi do phuong tiinh c6 dang:

Vay, phuong tiinh c6 nghiem duy nhat khi va chi khi m = 3

Vi du 3: Tun m de bat phuong tiinh co nghiem duy nhat:

V x 2 - 2 m < m x ' ^ (1)

Gicii Dieu kien rein: Nhan xet rang neu phuong tiinh c6 nghiem x,„ thl cung

nhan - x„ lam nghiem

Vay (1) CO nghiem duy nhat khi

X„ = - X „ <::i>X„ = 0

Khi do:

( ! ) « V - 2 m <Oc:>m = 0

Do chmh la diSu kien can de phuong tiinh c6 nghiem duy nhat

Dieu kien dir Gia su m = 0, khi do (1) c6 dang:

4^ < 0 c : > x = 01a nghiem duy nhat ciia bat phuong tiinh

Vay, voi m = 0 bat phuong tiinh c6 nghiem duy nhat

Vi du 4: Tim m de bat phuong tiinh sau c6 nghiem duy nhat:

log 2 , ( 2 - V x - + r ) > ( m - 1 ) ' (1)

Gicii Dieu kien can: Gia sir (1) c6 nghiem la x = x„ suy ra - x„ cung la

nghiem cua (1)

Vay (1) CO nghiem duy nhat khi x„ = - x„ <=> x„ = 0 • Thay x„ = 0 vao (1), ta duoc:

log , 1 > (m - 1 )^ « (m - 1)- < 0 <=> m = 1

Do chinh la dieu kien can de phuong tiinh c6 nghiem duy nhat

Dic'u kien dir Voi m = 1, khi do (1) c6 dang:

log2(2-Vx- + l) >C » 2 - Jx^ + 1 > 1 o VX- + 1 <1

<=> x- + 1 < 1 <=> x^ < 0 <=> X = 0 la nghiem duy nhat

Vay, voi m = 1 bat phuong tiinh c6 nghiem duy nhat

Vi du 5: Tim m de bat phuong tiinh sau c6 nghiem duy nhat:

Gidi Dieu kien can: Gia sir (1) c6 nghiem la x = x„ suy ra - x„ cDng la

nghiem cua (1)

V a y ( 1 ) C O nghiem d u y nhat k h i

Thay x„ = 0 vao ( I ) , ta duoc:

1 > 1 + m^ « m^ < 0 » m = 0

D o c h i n h la dieu k i e n can de phuong trinh c6 nghiem duy nhat

Dieu kien du: Gia sir m = 0, k h i do (1) c6 dang:

^ > 1 » 2'^' < I Ixl < 0 o X = 0 la nghiem duy nhat

Vay, voi m = 0 phuong i i l n h cc nghiem duy nhat

V i d i i 6: T i m m de phuong trinh:

V m - c o s x = c o s 2 x (1)

CO n g h i e m d u y nhat thuoc ( - - , - )

Gicii

Dieu kien van: Gia su (1) c6 nghiem la x = x,„ tiic la:

V' " - c o s x , | = cos2x„ ^m-co.s(-x„) = cos2{ - x„)

^ - x,| cung la nghiem CLia (1)

la n g h i e m duy nhat ciia phuong t r i n h

Vay, v o i m = 2 phuong trinh c6 n g h i e m duy nhat

V i d u 7: T i m m de phuong t r i n h sau c6 nghiem duy nhat:

X - - 2mcosA + 2 = 0

Gicii Dieu kien can: N h a n xet rang neu phuong t r i n h c6 n g h i e m x,„ t h i cung

D o c h i n h la dieu kien can de phuong trinh c6 n g h i e m d u y nhat

Dieu kien dii: V o i m = 1, k h i do phuong trinh c6 dang:

x^ - 2co.sx + 2 = 0 <x> x ' = 2(cosx - 1)

V i :

VT = x^ > 0 ] lo)>xl<l •

VP = 2(co,sx-l) < 0

D o do phuong t r i n h c6 nghiem k h i va chi k h i

2 ( c o s x - l ) = 0 Vay, phuong t r i n h c6 nghiem duy nhat k h i va chi k h i m = 1

Chii y: N h u vay, thong qua viec danh gia t i n h ch5n ciia cac bieu thirc

giiii tich trong phuong t r i n h , bat phuong t r i n h chiing ta da thitc hien

duoc yeu cau " Tim dieu kien ciia tliam so de phifcfng trinh, bat phmng trinh c6 nghiem duy nhat"

Cac V I d u tiep theo vSn voi yeu cau nliir trcn xong de t i m dieu kien can chung ta su dung cac phep bien d o i dai so de thong qua nghiem x„

lam xua't hien nghiem (p(x„)

V i d i i 8: T i m m de phuong t r i n h :

x"* + m x ' + 2inx^ + m x + 1 = 0 (1)

CO n g h i e m duy nhat

Gidi

N h a n xet r i n g x = 0 k h o n g phai la nghiem ciia phuong t r i n h

Dieu kien can: G i a i sii (1) c6 n g h i e m x^^^O, suy ra

x^ + m x,^ + 2 m xfi + mx,, + 1 = 0

1 s

1 + m — +2m~ +m~ + - L =o

+ m + 2 m + m — + 1 = 0

tiic la — cung la nghiem cua phuofng trinh

V a y de phucfng trinh c6 nghiem duy nhat dieu kien la:

2 2

<=> ( X - 1 )^(2x^ + 3x + 2) = 0 <=> X = 1 la nghiem duy nhat

V a y , m = - ^ phuang trinh c6 nghiem duy nhat

Chii y:

1 Yeu cau tien hoan loan c6 the dugc thuc hien bang phuong phap dat

an phu, cu the:

Nhan xet r i n g x = 0 khong phai la nghiem ciia phuang trinh Chia ca

hai ve cua phuang trinh cho x V O , ta dugc:

gt nghiem thoa man Itl > 2 - De nglu ban doe ticlani

N h u vay, de t i m dieu kien cua tham so sao cho phuang trinh hoi y:

Bia'fc 2: Dieii kien vein:

Giai su (1) C O nghiem x,,, suy ra — cung la nghiem c i a

phuang trinh Vay de phuang trinh c6 nghiem duy nhat dieu kien la:

1

— = x,| » X|| = ±1 => Gia tri tham so

I Do chinh la dieu kien can de phuang tiinli c6 nghiem duy nhat

BIIOC 3: Dieu kien dir Thuc hien viec t h u lai

du 9: T\m m de phuang trinh:

Isinx - ml + Icosx - ml = v 2 (1)

C O dung mot nghiem thugc (0, - )

Gicii

Dieu kien ran: Gia su (1) c6 nghiem la x = x,, suy ra

Isinx,, - ml + Icosx,, - ml = yfz

<=> lcos( ^ - x„) - ml + lsin( ^ - x„) - ml = V2

- - X|, cung la nghiem ciia (1)

V a y (1) C O nghiem duy nhat k h i

Thay x„ = ^ vao (1), la duac:

(1) « Isinx - N/2 I + Icosx - V2 I = N/I

o sinx + cosx - -Jl - gicii tmmg ti( nhinren

V a y , vdi m = 0 hoac m = >/2 phuong trlnh c6 d i i n g 1 n g h i e m thuoc

=> y - X|| cung la nghiem cua (1)

Vay (1) C O nghiem duy nhat k h i

A p d u n g bat dang thiic Cosi, ta dugc:

V T = -y/igx + ^cot gx >2^^/tgx.ycoTgx^ = 2

Do do:

x e ( ( ) , " )

(2) « > ^/tgx - T c o t g x = ! <=> tgX = 1

4

la n g h i e m d u y nhat cua phuang t r l n h

V a y , voi m = 0 phuang trlnh c6 nghiem duy nhat

V i d i i 1 1 : T i m m de phuong trlnh sau c6 nghiem duy nhat;

Gicii Die II kien can: Gia su phuang trlnh (1) c6 nghiem la x = x,, suy ra

2 - x,| cung la n g h i e m cua (1)

V a y (1) C O nghiem duy nhat k h i x„ = 2 - x„ » x„ = 1

Thay x„ = I vao {!) ta dugc m - 4 •

D o c h i n h la dieu k i e n can de phuang t n n h c6 n g h i e m duy nhat

Dieu kien di'i

Do do:

<=> X = 1 la nghiem duy nhat cua phuong trlnh

Viiy, voi m = 4 phuang trinh c6 nghiem duy nhat

Vi du 12: T u n a, b, c de phuang trinh sau c6 nghiem duy nhat:

=> a + b - X|| cung la nghiem cua (1)

Vay (1) CO nghiem duy nha't khi

, a + b

X|, = a + b - x„ <=> x„ = •

Thay x„ = vao (1), ta dugc:

c = la - bl

D o chinh la dieu kien can de phuang trinh c6 nghiem duy nhat

Diet! kien di'i

Gia sir c = la - bl, k h i do (1) c6 dang:

Ix - al + Ix - bl = la - bl « Ix - al + Ix - bl = l(x - a) - (x - b)l

• Neu a ;^ b (ta gia su k h i do a < b), khi do:

(2) <^ a < X < b, tuc la (2) khong c6 nghiem duy nhat

• Neu a = b, k h i do:

( 2 ) c : > ( x - a ) - < 0

o X = a la nghiem duy nhat ciia phirong trinh

V a y , voi c = 0 va a = b phuong trinh c6 nghiem duy nhat

Chii y: Bai toan tren la dang tong quat va phuang phap duqc ap dung de

giiii cho mot lap cac bai toan g o m 1 va 2 tham so (thong thuong cac bai

thi dai hoc chi g o m mot tham so) Cac em hoc sinh can nam vilng cac

buoc thuc hien de ap dung trong m 6 i bai toan cu the

Vi du 13: T i m m de phuong trinh sau c6 nghiem duy nhat:

Gicii

Oicu kien can: Giai sii (1) c6 nghiem x,„ ta c6:

mx„(2 - x„) = lx„ - 11 « m [ 2 - (2 - x„)](2 - x„) = 1(2 - x„) - 11

tiic la 2 - x„ cung se la nghiem cua phuang trinh

Vay de phuang trinh c6 nghiem duy nhat dieu kien la

2 - x„ = x„ o x,| = 1

K h i do:

(1) « m = 0

D o chinh la dieu kien can de phuang trinh c6 nghiem duy nhat

Dieu kien di'i: V o i m = 0, ta c6:

(1) <=> Ix - 11 - 0 « X = 1 la nghiem duy nhat cua phuang trinh

Vay, vdi m = 0 phuang trinh c6 nghiem duy nhat

Vi du 14: T i m m de phuang trinh sau c6 nghiem duy nhat:

1

3IX-21 = 2m - 1

Gicii Dieu kien cc'in: Gia su phuang trinh c6 nghiem la x = x,, suy ra

=> 4 - X|, cung la nghiem cua (1)

Vay phuang trinh c6 nghiem duy nhat k h i x„ = 4 - x„ « x„ = 2

Thay x„ = 1 vao pliuang trinh, ta duac m = 1

D o chinh la dieu kien can de phuang trinh c6 nghiem duy nhat

Dieu kien du: Gia su m = 1, khi do phuang trinh c6 dang:

— ! — = 1 Ci> 3'" = 1 » Ix - 21 = 0 <=> X = 2 la nghiem duy nlia't

Vay, voi m = 1 phuang trinh c6 nghiem duy nhat

Vi du 15: T i m m de phuong trinh sau c6 nghiem duy nhat:

2 m x , 2 - x , ^ 3 l x - l l + ,ii_

Giiii Dieu kien cc'iu: Giai su phuong trinh c6 nghiem x,,, ta c6:

2 n i x „ ( 2 - x „ ) _ ^Ix,,-!! + j-j^ < ^ 2 " i ( 2 - x „ ) | 2 - ( 2 - X | , ) | _ 3 l ( 2 - X o ) - l l + tiic la 2 - X|, cLiiig se la nghiem cua phuong trinh

V a y de pliuang trinh c6 nghiem duy nhat dieu kien la

D o chfnh la dieu kien con de phuong trinh c6 nghiem duy nhat

Dic'ii kien di'i

V i du 16: T i m m de phuang trinh sau c6 nghiem duy nhat:

=> m^ - 4 m - x„ cung la nghiem cua (1)

V a y ( 1 ) CO nghiem duy nhat k h i

( x „ + l ) ' + {x„ + 3)'* = 2 m o ( - x „ - l ) ' + ( - x „ - 3 ) - ' = 2 m

« [3 + ( - x„ - 4)]-' + [1 + ( - x„ - 4)]^ = 2 m tire la - x„ - 4 cQng la nghiem cua phuang trinh

Vay de phuang trinh c6 nghiem duy nhat dieu kien l i i :

- X o- 4 = x„<=>x„= - 2

K h i do:

(1) « ( - 2 + 1)-^ + ( - 2 + 3 ) ' = 2 m « m = 1

D o chinh la dieu kien can de phuang trinh c6 nghiem duy nhat

Dieu kien clii: V o i m = 1, ta c6:

(1) » ( x + l ) ' + (x + 3)'' = 2 (2) Dat t = X + = X + 2, suy ra :

<=>x + 2 = 0<::^x = - 2 la nghiem duy nhat

V a y , m = 1 phuang trinh c6 nghiem duy nhat

Clu'i y:

1 Nliu vay, de tim dieu kien cua tham so sao cho phuang trinh:

CO nghiem duy nhat, bing phuang phiip di^u kien ciin va du dugc thuc

hien iheo cac buac:

Bum-1: DieII kien can:

Giai su (1) c6 nghiem x„, suy ra - X n - a - b cung la

nghiem ciia phuang trinh Vay de phuang trinh c6 nghiem

duy nhii't dieu kien la:

- x„ - a - b = x„

a + b

( I )

=> Gia tri tham so

Do chinh la di<!u kien can de phuang tilnh c6 nghiein duy

nhat

Biioc- 2: Dieu kien dir Thuc hien viec thu lai

2 Yeu cau tien hoan toan c6 the duac thuc hien bang phuang phap dat

rin phu, cu the:

Dat I = x + = x + 2, suy ra:

( x „ - l ) ( x „ + l)(x„ + 3)(x„ + 5) = m

« ( - x„ + 1 ) ( x „ - - ] ) ( - x„ - 3)( - x„ - 5) = m

o [5 + ( - X, - 4)][3 + ( - ^, - 4)][ 1 + ( - x<, - 4)][ - 1 + ( - x, - 4)] = m tuc la - x,i - 4 cung la nghiem cua phuang trinh

Vay de phuang trinh c6 nghiem duy nhat dieu kien la:

- x „ - 4 = x „ o x „ = - 2

K h i do:

( l ) » ( - 2 - l ) ( - 2 + l ) ( - 2 + 3 ) ( - 2 + 5) = m o m = 9

Do chinh la dieu kien can de phuang trinh c6 nghiem duy nhat

Dieu kien du: V d i m = 9, ta c6:

( l ) » ( x - l ) ( x + l ) ( x + 3)(x + 5) = 9 (x- + 4x - 5)(x' + 4x + 3) = 9

Dat t = x^ + 4x - 5, dieu kien t> - 9, suy ra x" + 4x + 3 = t + 8

K h i do phuang trinh tren c6 dang:

t(t + 8) = 9 » t^ + 8t - 9 = 0 «

x + 4x - 6 = 0 +4x + 4 = 0

1 = 1

1 = - 9

+ 4 x - 5 = l x^ + 4 x - 5 = - 9

x = - 2 ± V l O

x = -2 tuc la phuang trinh khong c6 nghiem duy nhat

Vay, khong ton tai m de phuong trinh c6 nghiem duy nhat

Bum-1: i)ieu kien can: Giai su (1) c6 nghiem x,„ suy ra - x,, - a - b

cung la nghiem ciia phuang trinh

Vay de phuomg trinh c6 nghiem day nhat dieu kien la:

a + b

( I )

a b = x,| o x,| =

-Gia tri tham so

Do chinh la dieu kien can de phuong trinh c6 nghiem duy nhat

Bif()c 2: Dieu kien dir Thuc hien viec thu lai

2 Yeu cau tien hoan toiin c6 the' ducfc thuc hien bang phuang phap dat

an phu, cu the:

Viet lai phuang trinh du6i dang:

(x' + 4x - 5)(x^ + 4x + 3) = m

Dat t = X' + 4x - 3, dieu kien t > - 9 suy ra x^ + 4x + 3 = t -i- 8

Khi do phuong trinh tren c6 dang:

t(t + 8) = m « f ( t ) = t ' + 8 t - m = 0 (2)

Phuang trinh (1) co nghiem duy nhat

<=> (2) CO nghiem thoa man t, < t, = - 9

Vay, khong ton tai m de phuong trinh c6 nghiem duy nhat

Vi dii 19: T i m m de phuong trinh sau c6 nghiem duy nhat:

tiic la 2 - X|, cung la nghiem ciia (1)

Vay ( 1) CO nghiem duy nhat khi

Vx = V2 - x x = 1 Ja nghiem duy nhat

Vay, m = 2 phuong trinh c6 nghiem duy nha't

Chii y:

1 Nhu vay, de t l m dieu kien ciia tham so sao cho phuang trinh:

Vx + a + V b - x = c

(1)

CO nghiem duy nhat, bang phuang phap dieu kien can va du duoc thuc

hien iheo cac bu6c:

BiiocJ: Dien kien can:

Giiii sir (1) c6 nghiem x„, suy la

<=> , / b - ( - x „ - a + b) + ^/a + (-v„ - a + b) = c

<=> ^a + (-x„ - a + b) + ^ b - ( - x „ - a + b) = c tuc la - x„ - a + b cung la nghiem ciia phuang trinh Vay

de phuang trinh c6 nghiem duy nhat dieu kien la:

b - a

- X|, a + b = x„ <=> x„ = y

-=> Gia tri tham so

Do chinh la dieu kien ciin de phuang trinh c6 nghiem duy nhat

Bui/c 2: Dien kien dir Thuc hien viec thu lai

2 Yeu cau tren hoan toan c6 the duac thuc hien bang cac each khac,

cu the:

Ccicli I: Phuang phap ddr an pint:

x + a , dieu kien u, v > 0

V = V b - x Khi do phuang trinh dugc chuyen thanh he:

u + V = c

u + V = a + b

do chinh la he dx loai I ma chiing ta da biet each giiii

Ci'icli 2: Phuang pluip ham so:

Xct ham so y = V x + a + V b - x l i e n tap D = [ - a, b ] , tCr do xac d i n h :

• D a o ham l o i giai phuang trlnh y ' = 0

• Bang bien thien

K h i do phuang trlnh c6 nghiem d u y nhat k h i va c h i k h i duang thang

y = c cat phan d o thj h a m so tren D tai m o t d i e m d u y nhat

Cacli 3: Pliii'(Hi}> plidp liMiiy, i^icic lioci

3 D e nghj ban doc m a rong cho phuang trlnh :

Tuc la k h i do - 1 - x,, cung la nghiem cua (1)

Vay ( 1 ) CO nghiem d u y nhat k h i

X(| — 1 X(| X|| — —

V a i X|| = - ^ , ta dugc:

( 1 ) « iog^^(J4Tl+^-l + 5) = a « a = 1

V a y a = 11a dieu kien can de phuang trlnh c6 nghiem d u y nhat

Dicii kicn dir V d i a = 1, phirong trlnh (1) c6 dang :

4x-^+4x + l = 0 2 Vay, a = 1 phuang trlnh c6 nghiem d u y nhat

Chu

y-1 T r o n g phan xac d i n h dieu kien can ta c6 the su d u n g :

• Bat dang thilc Bunhiacopski n h u sau:

2 V 4 ^ 2Vx + 5

« Vx + 5 = V4 - X <=> X = - ^

Being bien thien:

Dat I = 2sinx dieu kien Itl < 2

K h i do phuong trinh c6 dang:

It - l l + l t - a l = b

K h i do (1) CO diing 2 nghiem phan biet thuoc [0, 2n)

<=> (2) CO nghiem duy nhat thuoc [ - 2, 2 ]

Dicti kien can

Gia sir (1) c6 nghiem la t = t,, suy ra

lt„ - 11 -i- ll„ - al b

c : > l ( l + a - t „ ) - a l + l ( l + a - t „ ) - l l = b

=> 1 + a - t|, cung la nghiem ciia (2)

V a y (2) CO nghiem duy nhat k h i

D o c h i n h la dieu kien can de phuong trinh c6 nghiem duy nha't

Dicii kien di'i:

V d i b = la - 11, k h i do (2) c6 dang:

I t - 1 1 + I t - a l = l a - I I

o l t - l l + l t - a l = l ( t - l ) - ( t - a ) l

c ^ ( t - l ) ( t - a ) < 0 (3)

• Neu a?tl (ta gia sir k h i do a < 1), k h i do:

(3) o a < t < 1, tuc la (3) khong c6 nghiem duy nhat

^~ 6

Vay voi b = 0 va a = 1 co dung 2 nghiem phan biet thuoc [0, In)

I I I B A I T A P D ^ N G H I Biii tap 1: T u n m de cac phuong trinh sau c6 nghiem duy nhat:

Hiii tap 6: T i m m de cac phuang trinh sau c6 nghiem duy nhat

Voi yeu cau:

" Tim dicn klcn ciia tham so(i;id sir la m) dc he pliuang trinh, / t f hat phiioiiii trinh:

f ( x , y , m ) < 0

g( x , y , m ) > 0

CO ii^liieru day nhat"

ta thuc hien theo cac budc sau:

Biio'c J: Dat dieu kien de cac bieu ihiic irong (1) c6 nghla

Bum- 2: Dieu kien can: Gia sir (1) c6 nghiem la (x,,, y,,) khi do:

a Dua tren tinh chat doi xung cua cac bieu thiic giai tich trong (1), ta di khang dinh khi do ((p,(x,|, y,,), (pjCx,,, y,,)) cung la nghiem cua (1)

b Do do, de he CO p.ghiem duy nhat Clin c6:

Biioc3: Dieu kien dir V o i meD,,,, ta di kiem tra lai tinh duy nhat

Biioc 4: Ket hop ba buac giai tren ta tim dugc dap so

Gia tri ciia (x,„ y,,)

33

Chii y: V a i cixc he mot an bai toan duac tbi/c hien dua tren phirong phap

da biet trong bai toan 1

Dicii kicii cciii:

Nhan xet nlng neu he c6 nghiem (x„ y„) thi (y^, x„) cQng la nghiem

cua lie, do do he co nghiem duy nhat khi:

Khi do, he c6 dang:

2xf| = m

2x„ = 6

Do chfnh ia dieu kien ciin de he c6 nghiem duy nhat

DicH kien dir

Voi m = 18, ta duac:

m = 18

+ y- = 18 fx + y = 6

x + y = 6 XV = 9 <=> X = y = 3 la nghiem duy nhat

Vay, vdi m = 18 he phuang trinh c6 nghiem duy nhat

Chii y:

1 Nhu vay, de tim dieu kien cua tham so sao cho he phuang trinh doi

xiTng loai I va ioai II c6 nghiem duy nhat ta thuc hien theo cac bu6c:

Biioc I: Dieu kien can

• Nhan xet rang, neu he c6 nghiem (x,„ y„) thi (y,„ x„)

cung la nghiem ciia he, do do he c6 nghiem duy nha't

x„ = y,, (**)

• Thay (**) vao he ta duac gia tri ciia tham so Do chinh

la dieu kien can de he c6 nghiem duy nhat

Bitoc 2: Dieii kien dii

2 Yeu cau tren hoan toan c6 the dugc thuc hien bang nhung phuang

phap khac, cu the:

Cckli J: Sir dung phuang plidp cluing ciia he doi xiing loai I

Bien doi he phuang trinh v6 dang:

He CO nghiem duy nhat

<r> (1) CO nghiem duy nhat o A',,, = 0 < = > m - 1 8 = 0 < = > m = 1 8

Khi do he c6 nghiem x = y = 3

Vay, vai m = 18 he phuang trinh c6 nghiem duy nhat

Cdcli 2: Siiditng plutang plidp the

Bien doi he ve dang:

x~ + ( 6 - x ) " = m <=> 2x^-12x + 36 i.i-0 (2)

y = 6 - x

He CO nghiem duy nhat

<=> phuang trinh (2) c6 nghiem duy nhat

« A',|, = 0 » m - 18 = 0 « m = 18

K h i do he c6 nghiem x = y = 3

Vay, vai m = 18 he phuang trinh c6 nghiem duy nhat

Cdch 3: Sit dung phuang phap do thi

Nhan xet rang vdi m<0, he v6 nghiem, do do ta xet v6i m > 0

Ta c6:

• Phirang trinh (1) la du6ng tron (C) c6 tarn 0 ( 0 , 0), ban kinh R ^

yfm

• Phuang trinh (2) la duang thang (d)

He CO nghiem duy nhat

<=> (d) tiep xiic v6i (C)

<=> d ( 0 , (d)) = R » = Vi^ « m = 18

Vi + i

Khi do, he c6 nghiem x = y = 3

Vay, vai m = 18 he phuang trinh c6 nghiem duy nhat

Ccuh 4: Si'cclungphucfngplidp luang gidc hoa

Nhan xet r i n g vai m < 0, he v6 nghiem, do do ta xet v 6 i m > 0

T u phuang trlnh t h i i nhat ciia he:

Thay (3) vao phucrng trlnh t h u hai ciia he, ta duoc:

Vim sint + 4m cost = 6 «

« s i n ( t + ^ ) = ^

He CO nghiem duy nhat

phifong trlnh (4) c6 nghiem duy nhat tren tap [0, 2n)

Vay, voi m = 18 he phuang trlnh c6 nghiem duy nhat

V i d u 2: T u n m de he phuang trlnh sau c6 nghiem duy nhat:

log2(x + y) = m logjCxy) log2(xy) = 6-xy

Dial kien (dir

N h a n xet rSng neu he c6 nghiem (x,„ y,,) t h i cQng c6 nghiem (y,,, x„),

do do he c6 nghiem duy nhat t h i x„ = y„

|log2(xy) = 6-xy [xy = 4 [xy = 4

Vay, v 6 i m = 1 he c6 nghiem duy nhat (2, 2)

la nghiem cua he, do do he c6 nghiem duy nhat k h i :

D o chfnh la dieu k i e n can de he c6 nghiem duy nhat

ieu kien di'i

• V o i m = 1, ta duac:

^ j ^ ^ | x y + (x + y) = 3 [xy(x + y) = 2

khi do X + y va xy la nghiem ciia phuang trlnh:

Nhan thay he luon c6 hai cap nghiem (0, 1) va ( 1 , 0)

-la nghiem duy nhat ciia he

Vay, v d i m = 1 hoac m = - he da cho c6 n g h i e m duy nhat

Chii y: Nhu da tha'y trong v i d u tren, chiing ta c6 the thuc hien bang

k h i do S, P la n g h i e m ciia phuong trinh:

<=> X, y la n g h i e m cua f(u)—u —u+m+1 = 0 (1)

» X, y la n g h i e m ciia g(u)=u^-(m+1 )u+1 =0 (2)

I S

He CO n g h i e m d u y nhat

(1) v6 nghiem & (2) c6 nghiem kep

<=> (2) v6 nghiem & (1) c6 nghiSiii kep

(i) & (2) CO nghiem kep u „

Dicii kien can: N h a n xet rang neu he co nghie m (x,,, y,,) t h i cung

n g h i e m (y,„ x„), do do he co nghiem duy nhat t h i x,, = y,,

^hii y: V i d u tren, chiing ta c6 the thuc hien bang each sau:

Viet lai he phuong trinh d u d i dang:

He CO nghiem duy nhat

» ( 1 ) CO nghiem kep duong

Vay, v o i m = logj y thoa man dieu k i c n dau bai

V I (UI 5: Tim m de he bat phuong trinh sau c6 nghiem d u y nhat:

_ yfx+yjy = 1

X + y < m

Gicii

D i e u k i e n x, y > 0

Die II kien can:

G i a su he c6 nghiem (x,„ y,,) =:> (y,„ x,,) cQng la nghiem cua he V a y

he CO nghiem d u y nhat t h i dieu kien can la: x„ = y„

K h i do he (I) c6 dang:

2 ^ = ^ = ^ m > i

2x < m 2

V a y m > ^ la dieu k i e n can de he c6 nghiem d u y nhat

Dien kien dir

Vdfi m > ^ , ta xet hai t i u o n g hop :

Tnfd-iii!, lu/p 2: V d i m > ^ , he (I) c6 dang:

^/^ + ^/y =1 t=VI JVy = i - t

^ CO v 6 so gia t r i y thoa m a n => he k h o n g c6 nghiem d u y nhat

Vav, m = - la dieu kien can va du de he c6 nghiem duy nhat

2 •

Chu y: V i d u tren, chung ta c6 the thuc hien bang cdc each sau:

Truoc het, dat:

Trudc het can c6

Vay, vdi = ^ h? c6 nghiem duy nhat

Cc/c// 2; Si'fdiing phifc/ngphdp do thi

Triayng lu/p 1: V 6 i m<0 thi (III) c6 v6 so nghiem

Tnayng lu/p 2: V o i m > 0

Goi X , va X j Ian lugt la tap nghiem ciia (1) va (2), ta c6:

• X | la tap cac diem trong doiui thang A B ciia duong tliang (d): u + v - ! = 0

" X j la tap cac diem trong hinh tron (C) c6:

Vay, voi m = ^ he c6 nghiem duy nhat

Vi du 6: Tun m de he bat phuong trinh sau c6 nghiem duy nhat:

X + y + ^2xy + m > 1

x + y < 1

Gidi

Die It kien can: Gia sir he c6 nghiem (x,„ y„) ^ (y,,, x„) cung la nghiem

ciia he Vay he c6 nghiem di^y nhat thi dieu kien can la: x,, = y,,

Vay m>~ ^ la dieu kien can de he c6 nghiem duy nha't

Pic'ii kien dir V d i m> - - , ta xet hai tru5ng hop :

fi-iiviig lu/p 7: V d i m = - - , he (I) c6 dang:

x + y < 1

He (II) CO nghiem duy nhat x = y = ^ , v i k h i do dirong thang

X + y - 1 = 0 tiep xiic voi dudng trong ( C ) : (x - 1)^ + (y - 1)^ = ^

=> CO v6 so gia tri y thoa man => he khong c6 nghiem duy nhat

Vay, m = - ^ la dieu kien can va dii de he c6 nghiem duy nhat

Chu y: Co the sir dung phuong phap do thi de giai v i du tren bang viec

bie'n doi tuong duong he ve dang:

7 2 x y + 111 > l - ( x + y)

X + y < 1

x + y < 1 2xy + iii > [ l - ( x + y ) f

Goi X | va X j Ian luot la tap nghiem ciia (1) va (2) Ta c6:

• X , la tap cac diem trong phan matphang phia dudi dudng thang (d): x + y - 1 = 0

4-?

X 2 la tap c a c d i e m tiong hinh tron ( C ) c 6 :

Vay, vtfi m = - - he c6 nghiem duy nhat

Vi till 7: T u n m de he sau c6 nghiem duy nhat:

X = - y +1)1 (*)

y = - X + m

Gicii

Dieu kien can: Nhaii xet rang, neu he c6 nghiem (x,,, y,,) t h i cung c6

nghiem (y,,, x„), do do he c6 nghiem duy nhat i h i

D o chinh la dieu kien can de he c6 nghiem duy nhat

Dieu kien dir Vdfi m = 1, he c6 dang:

Nghiem thoa man he va la nghiem duy nhat

Vay, vcfi m - 1 he c6 nghiem duy nhat

Vi dH 8: T u n m de; cac he sau c6 nghiem duy nhat:

( x - i ) ^ + ( y + i ) - < m (x + l ) ^ + ( y - l ) 2 < m

Giiii

Dieu kien can: Gia sir he c6 nghiem (x,„ y„), -suy ra (y,„ x„) cQng la nghiem

cua he Vay de he c6 nghiem duy nhat thl dieu kien can la x„ - y„

Nhan xet l i n g x = y = 0 thoa man he (II)

Vay, he c6 nghiem duy nhat k h i m = 2

Vi du 9: Tim m de he sau c6 nghiem duy nhat:

x"+(y +1)" < 111 (x + l)^+y^ < m

Gicii

Dieu kien can:

Gia sir he c6 nghiem (x,„ y„), suy ra (y,„ x„) cung la nghiem cua he

Vay de he c6 nghiem duy nhat thi dieu kien can la x„ - y„

K h i do:

xr, + ( x „ + l ) - < m » 2 x ? , + 2 x „ - m + 1 < 0 (1)

Ta can (1) phai c6 nghiem duy nhat

<=>A = 0 < = > m = - Vay dieu kien can de he c6 nghiem duy nhat la ' i i = ^ •

Dieu kien du: V o i m = ^ , he c6 dang:

x + ( y + 1) < (X M)^+y^ < ^

-(ID

x ' + (y + + (X + + y ' < l 2 x ' + 2x + 2 y ' + 2y + 1 < 0

«(xV^ + ^ ) ^ + (yV^ + ^ ) ^ < 0 « x = y = - i Nhan xet rang x = y = - ^ thoa man he (II)

Vay, he c6 nghiem d u y nhat k h i m = ^

45

Chii y: Ta c6 the su dung phuang phap do thi de thuc hien v i du tren, cu

the:

Goi X , va X2 Ian lugt la tap nghiem ciia (1) va (2) Ta c6:

• X | la tap cac diem trong hinh tron:

Vay he c6 nghiem duy nhat khi (C,) tiep xilc vdi (C,)

<=> I , ! = R, + R, I L = 2 <=> m = -

Vay, vai m = - thoa man dieu kien dau bai

VI du 10: Tun m de he phuang trlnh c6 nghiem duy nhat:

xy + x^ = m ( y - l )

xy + y = m(x - 1 )

Gicii

Die It kien can:

Nhan xet rang: neu he c6 nghiem (x,„ y„) thi cQng c6 nghiem (y,,, x,,),

do do he c6 nghiem duy nhat thi

y = - 8 - x

72 = 0

<=> x = y = 2 la nghiem duy nhat

V a y , vdi m = 8 he c6 nghiem duy nhat

Vi (111 11: Ti'ii m de he sau c6 nghiem duy nhat:

x-^ = y 2 + 7 x ^ - m x (1)

y-' = x^ +7y" - my (2)

Gicii OU'ii kien can: Nhan xet rang: neu he c6 nghiem (x,„ y„) thi cung c6

nghiem (y,„ x„), do do he c6 nghiem duy nhat thi

bai phirang trinh (**) v6 nghiem do:

Vay, voi m > 16 he c6 nghiem duy nhat x = y = 0

Vi du 12: Tun m de he sau c6 nghiem duy nhat:

x(4y' - 3 ) = in

y(4x^ - 3 ) = 111

Gidi

Dieu kien can:

Nhan xet rang: neu he c6 nghiem (x,„ y„) thi cung co nghiem (y,„ x,,)

do do he c6 nghiem duy nhat thl x„ = y„ K h i do:

(1) <» x„(4xr, - 3) = m o 4xf) - 3x„ = m

Do x„ duy nhat nen (1) phai c6 nghiem duy nhat « hnl > 1

Dieu kien du:

x(4y- - 3 ) = 111

X = y x(4y" - 3 ) = m 4xy + 3 = 0

111

x + y = - - ^

xy = - l

( I V ,

Nhan xet rang ( I V ) luon c6 2 nghiem phan biet, do vay he (I) khong

the CO nghiem duy nhat

Vay, khong ton tai m de he c6 nghiem duy nhat

nghiem ciia he Vay de he c6 nghiem duy nhat thl dieu kien can la

Do chinh la difiu kien can de he co nghiem duy nhat

Dieu kien clii : V o l in - 2 he co dang;

( x - 2 ) ' + y - - = 2

x - + ( y - 2 ) 2 = 2

^ (X - 2)- + y ' + x ' + (y - 2)- = 4 <» (X - + (y - = 0

<=> X = y = 1

Nhan xet rang x = y = 1 thoa man he (II)

Vay, he co nghiem duy nhat khi m = 2

Chii y: Chung ta co the sir dung phuong phap do thi de thuc hien v i du

• X , la tap cac diem tren dudng tron (Cj) co:

Vi du 14: T i m a de he bat phucfng trinh sau c6 nghiem duy nhat:

Vx + 1 + ^ <-d

4 I + Vx < a

Gidi

Dieu kien x, y > 0

Dieu kien can:

Gia sir he c6 ngliiem (x,,, y„) =:> x,„ y ;i> 0 tCi' do:

• => neu a < 1 thi he vo nghiem

Vay a >1 la dieu kien can de he c6 nghiem duy nhat

Dieu kien dir Xet hai trudng hop:

V o i a = 1, he CO dang:

+ 1 + < 1

V o i a > 1, xet cac cap nghiem ciia he c6 x = 0 he co dang:

Bien doi bii't phuong trinh (1) ve dang:

<z> X = y = 0 la nghiem duy nhat cua he

yjy + l <i\

y < ( a - i r

7 « 0 < y < m i n { ( a - 1 ) ' , a ^ - 1 }

y < a ' ' - l

=> CO v6 so gia tri y thoa man => he khong c6 nghiem duy nhat

Vay, voi a = 1 he c6 nghiem duy nhat x = y = 0

Chu y: V o i each lap luan tuong tii nhu tren ta c6 the giai duoc bai toan

voi yeu cdu:

" Tim a de he but phmyng trinh sau c6 nghiem "

Khi do dieu kien la a > l

V i du 15: T i m m d6 he phuong trinh sau c6 nghiem duy nhat:

I g x l g y + lg X = m ( l g y - l ) ( I )

(D

I g x l g y + l g - y = i n ( l g x - l )

Gidi

Dieu kien x,y > 0

Dieu kien can: Nhan xet ring neu he c6 nghiem (x,,, y,,) thl cung c6

ngiiiem (y,,, x,,), do do he c6 nghiem duy nhat thl x,, = y„ K h i do;

TrCr tCing ve he phuong trinh, ta duoc :

Ig-x - Ig-y = - 8(igx - Igy) (Igx - lgy)(lgx + Igy + 8) = 0

Vay, voi m = 8 he c6 nghiem duy nhat

Vi du 16: T i m m de he sau c6 nghiem duy nhat:

Do do he t o nghiem duy nhal thi:

X(i = 2 - x , )

• y ( ) = 2 - y „ o x „ = y „ = 1

Voi x,| = y„ = 1 ta suy ra m = 4

Dicii kieii du: V o i m = 4, he c6 dang:

j l x + l l + l y - 3 l = 4 rx = y = l

| l y + 1 1 + I X - 3 1= 4 [ x = y = 3 •

Vay, khong ton tai m de he c6 nghiem duy nhat

Vi (ill 17: Tim a de he sau c6 nghiem duy nhat:

[ i i x ~ + a = V + I

<

I x l + y - = 1

Gicii

Dicii kien can: Nhan xet n\ng: neu he co nghiem (x,,, y,,) tiVi ( - x,,, y,,)

Cling [i\m cua he

Do do he co nghiem duy nhat thi: x„ = ~ x,, <=> x„ = 0

Voi x„ - 0 ta suy ra:

la nghiem duy nhat

Vay voi a = 2 he co nghiem duy nhat

£)iai kien can:

Nhan xet rang neu he co nghiem (x,,, y„) thi cung co

( - x,„ y„) K h i do de he co nghiem duy nhat \h :

x„ = - x„ <=> x„ = 0 Voi x,| = 0, ta dugc:

Vay, voi m = 6 he co nghiem duy nhat

V i d u l 9 : Cho he phuong tilnh:

ax^ + a - 1 = y - I s i i i x I tg~x + y^ = l

Tim a de he phirong trlnh co nghiem duy nhat

Gidi

£>ieu kien can:

Nhan xet rang neu he co nghiem (x,,, y„) thi cung co

(•- X|„ y„) K h i do de he co nghiem duy nhat la

x„ = - x„ <=> x„ = 0 Voi x„ = 0, ta duoc:

Dicn kien di'i

la nghiem duy nhat ciia he

Vay, voi a = 2 he co nghiem duy nhii't

Vi du 2 0 : Tim a de he sau co nghiem duy nhat:

ax'^+I sin 2x I + a = y + 1

1 tg6x I + 2 y - = 2 'j

(jiiii

Dicii kien can:

Nhan xet rang: neu he co nghiem (x,„ y„) ihi ( - x,„ y„) cung la

nghiem ciia he

Do do he co nghiem duy nhat thi: x„ = - x„ o x„ = 0

Voi x„ = 0 ta suy ra:

P Vay, voi a = 2 he co nghiem duy nhat

Vi du 2 1 : Tim m de he sau co nghiem duy nhat:

2 ' ' ' ' - 2 > ' = y - 1 x 1 ( 1 1 1 + 1 )

' 1X + y = 111 2

Gidi Dieu kien can: Nhan xet rang neu he CT nghiem (x,„ y„) suy ra ( - x„, y„) cQng la nghiem

Vay de he co nghiem duy nhat thi x„ = - x,, <=> x„ = 0

Do chinh la dieu kien can de he nghiem duy nhat

f^ieu kien du: Gia su m - 0, khi do ht co dang:

2'-^'-2>' = y - l x l

X - -f y - 0

, 1 x 1

<=> x l = 2 > ' + y (3) x*^+v = 0 ( 4 )

5 5