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

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

LE null T R i

PIIlTO]\ P H A P D A C B I E T G I A I
TRU]\ HOC P I I O T I l 6 ] \

Str D U N G PHU^dNG P H A P
D I E U K I E N CAN VA D U

GIAI TOAN

NHA XUAT BAN HA

NOI


G l 6 l T H I E U CHUNG
Kill trciii tigng gic'ri lliieu tai ban doc bp tai lieu:

i*iiiJOi\ i>iiAi» »Ac

BI£:T

Gi\i

roAi\
I UUI\ H O C IMIO Tll6l\
do Tliac

sTTodn

Jipc Le Hong

Di'tc cliii

bien.

D p t a i l i e u g o m I J tap:

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\ i M i A i ' niv.ii

KIEN

CAIM V A

nt

oi.ii

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
E-mail:
Ud iipl, ngdy 1 llidng 8 iidiit 2004
LE HONG DirC

GlOl THifiU CHUNG

MUC LUC
CHIJ DE 1

SU DUNG PHl/ONCi PHAP DIEU KIEN CAN VA DU
(JIAI BAI TOAN VE TINH CHAT DUY NHAT NGHIEM
Bai toan 1.
Bai toan 2.

Giai bai toan duy nha't nghiem cho phuang trlnh,
bat phuang tiinh
Giai bai toan duy nhat nghiem cho he phuang trlnh,
he bat phuang trlnh

10
33

CHUD^2

SU DUNG PHUONG PHAP DIEU KIEN CAN VA DU
(ilAI BAI TOAN VE TINH CHAT N(,HIEM
Bai toan 1.
Bai
Bai
Bai
Bai

toan 2.
toan 3.
toan 4.
toan 5.

Giai bai toan ve tinh chat cac nghiem

cho phuang trlnh
Giai bai toiin ve tap nghiem
Giai bai toan ve phuang trlnh ha qua
Giai bai toan vi hai phuang trlnh tirang duang
Su dung do thi

66
77
92
95
126

CIIU Y)t 3

(ilAI BAI
TOAN PHAP
VE TINH
CHAT
SO DU
Sir DUNG
PHUONG
DIEU
KIENTHAM
CAN VA

Bai toan 1.
Bai toan 2.

Phuang trlnh nghiem diing vai gia tri
xiic dinh ciia tham so

He nghiem diing vol gia tri xac dinh cua tham so

TAI LIEU THAM KHAO

102
105
110


CHU D E I

Str DUNG PHU'dNG P H A P DIEU KIEN CAN VA DU
GIAI BAI TOAN V E TINH DUY NHAT NGHIEM
M 6 DAU
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.


=

—

=

—

=

BAI TOAN 1

[|

GiAi BAI T O A N D U Y N H A T N G H I E M

C H O PHL/ONG TRINH, B A T PHUaNG TRINH

•
I

V l (111 I :

•

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

mx^-2(m-l)xf,

V a i yeu cau:

fix, m) >() (liodcf(x,
CO iii>hiem duy

hat

o

BiioT 2:

(1) c ^ m - 1 = 0 » m =

Dieu

kien du: V o l m = 1, ta c6:
x"* = 0 »

trong (1), ta d i khang d i n h k h i do x = cpCx,,) cung la
n g h i e m cua (1).
D o do, de he c6 n g h i e m d u y nhii't can c6:
(2)

T h a y (2) vao (1) ta xac d i n h dugc dieu k i e n can cho
t h a m so m de (1) c6 n g h i e m duy nhat, gia sur meD,„.

Dieu

kien dir. V o i meD,„, ta d i k i e m tra l a i tinh d u y nhat

Chii y:
1.

Y e u ciiu tren hoan toan c6 the duoc thuc hien bang phuong phap dat

an phu, cu the:
Dat t = x% v o i t > 0, phuong t r l n h c6 dang:

Buoc 4:

phai xet cac phirong
la k h o n g c6 tham so
nhieu). K e t qua ciia
D,„ cac gia t r i k h o n g

Ket hop ba buoc giai tren ta t u n duoc dap so.

Tru'd'ni^ hop I. V o i m = 0
(2) » 2 t - l = 0 «

t = ^-«

^ ' " ^ ^

Phuong t r i n h c6 hai n g h i e m phan biet.
Trifdiii^

hop 2. V o i m * 0.

Phuong t r i n h (1) c6 n g h i e m d u y nhat

II. V i DU M I N H H O A

2{m--i)

T r u d c tien c h u n g ta m i n h hoa cac v l du sir d u n g t i n h chat h a m chSn
de xac d i n h dieu k i e n can, tuc la xuat phat tii' nhan xet:
•

G i a su phuong t r i n h c6 n g h i e m x,, khang d i n h r i n g no c u n g se
nhiin - x,, l a m n g h i e m

•

V a y de' phuong t r i n h c6 n g h i e m d u y nhat dieu k i e n la:
- x„ = x„ »

(2)

f(t) = m t ^ - 2 ( m - l ) t + m - 1 = 0.

n g h i e m cho (1).
T h o n g thuong trong buoc nay, ta c h i
t r l n h , bat phuong t r i n h cu the (thuong
hoac neu c6 t h l da duoc d o n gian d i
budrc nay cho phep ta loai d i k h o i tap
thich hop ciia m .

X = 0 la n g h i e m duy nhat cua phuong t r i n h .

V a y , v 6 i m = 1 phuong t r i n h c6 n g h i e m d u y nhat.

x„ = (p(x„) =^ G i a t r i ciia x,,.

Bu(>c3:

1.

D o c h i n h la dieu k i e n can de phuong t r i n h c6 n g h i e m d u y nhat.

Dua tren t i n h chat d o i x u n g ciia cac bieu thiic glai t i c h

c.

x„ = 0.

K h i do:

Dieu kien can: Gia sir (1) c6 n g h i e m la x = x,„ k h i do:

b.

m ( - X,,)' - 2 ( m - 1)( - X,,)' + m - 1 = 0

- x„ = x„

nlid't"

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

a.

=0

V a y de phuong t r i n h c6 n g h i e m d u y nhat dieu k i e n la:

ta lliirc hien theo cac budfc sau:
1:

+ m - 1

tuc la - x„ cQng la n g h i e m cua phuang t r i n h .

(1)

m) <0)

(1)

kien can: G i a i sU (1) c6 n g h i e m X,,, suy ra

Dieu

" Tim dieit kieii ciia tluim so (i>id sii Id m) dc plurc/iii^ trinh,
plni(/i\^ triiili:

=0.

Gidi

I. PHir()N(J P H A P

Btioc

T i m m de phuong t r i n h :

inx'-2(m-l)x- + m - I

x„ = 0.

(2) CO n g h i e m t,<0 = t,
•<0

S<0
P-0

m-1
m

V a y , v o i m - 1 phuong t r i n h c6 n g h i e m duy nhat.

o m = l .
= 0


2. Nhu viiy, de tim dieu kien cua tham .so sao ciio phuong trinh triJng
phuong:
ax"* + bx^ + c = 0

Tun m de bat phuong tiinh co nghiem duy nhat:
V x 2 - 2 m
(1)

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

hien Iheo Ciic buoc:
Biiocl:

Vi du 3:

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.

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

(l)c:>c = 0.

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

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.

+ 2 N / I - X 2 =m.

4^

Vi du 4:

(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

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

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

Tim m de phuong tiinh sau c6 nghiem duy nhat:
Vl-x^

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.

Khi do:

•

Thay x„ = 0 vao (1), ta duoc:
log , 1 > (m - 1 )^ « (m - 1)- < 0 <=> m = 1.

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:

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

VI:

4r-

(1)

( ! ) « V - 2 m <Oc:>m = 0.

Khi do:

Vi du 2:

^

- Jx^ + 1 > 1 o

VX- + 1 <1

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

+2N/I^

9

<3.

Vay, voi m = 1 bat phuong tiinh c6 nghiem duy nhat.
Vi du 5:

x~ <

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

Do do phuong tiinh c6 nghiem khi va chi khi
Ci>x = 0.
x2

=1

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

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

V i d u 7:

V a y ( 1 ) C O n g h i e m d u y nhat k h i

T i m m de p h u o n g t r i n h sau c6 n g h i e m duy nhat:
X - - 2mcosA + 2 = 0.

Gicii

T h a y x„ = 0 vao ( I ) , ta duoc:
1 > 1 + m^ «

m^ < 0 »

Dieu

m = 0.

D o c h i n h la d i e u k i e n can de p h u o n g t r i n h c6 n g h i e m d u y nhat.
Dieu

2'^' < I

> 1»

Ixl < 0 o

- x,, l a m n g h i e m .

Do do p h u o n g t r i n h c6 n g h i e m duy nhat t h i di6u k i e n can la

kien du: G i a sir m = 0, k h i do (1) c6 dang:
^

kien can: N h a n xet rang neu p h u o n g t r i n h c6 n g h i e m x,„ t h i c u n g

nhan

x„=

X = 0 la n g h i e m duy nhat.

( D o

Dieu
=cos2x.

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

-).

CLia

giiii t i c h t r o n g p h u o n g t r i n h , bat phuong t r i n h c h i i n g ta da thitc h i e n
duoc yeu cau " Tim
trinh c6 nghiem

duy

dieu kien ciia tliam so de phifcfng

trinh,

bat

phmng

nhat"

Cac V I d u tiep theo vSn v o i yeu cau nliir trcn x o n g de t i m dieu k i e n

2

can c h u n g ta su d u n g cac phep bien d o i dai so de t h o n g qua n g h i e m x„

kien dii: V d i m = 2 , k h i do (1) c6 dang:
=

C0S2X

(2)

lam xua't h i e n n g h i e m (p(x„).
V i d i i 8:

Ta CO nhan xet:
= V2-C0SX

T i m m de p h u o n g t r i n h :

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

> 1

(1)

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

VP = c o s 2 x < l

Gidi

D o do:
fVT = l

^^^^

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

Chii

thu6c(-^,^).

VT

0

V a y , p h u o n g t r i n h c6 n g h i e m d u y nhat k h i va c h i k h i m = 1.

D o chfnh la d i e u k i e n can de phuong t r i n h (1) c6 n g h i e m d u y nhat

yJl-COSK

<

•

2(cosx-l) = 0

ci>x„ = 0.

V m - c o s O = cosO <=> V m - l = 1 < » m = 2 .

Dieu

lo)>xl
D o do p h u o n g t r i n h c6 n g h i e m k h i va c h i k h i

(1).

T h a y x„ = 0 vao (1), ta dugc:

2

> 0

VP = 2 ( c o , s x - l )

V a y ( I ) CO n g h i e m d u y nhat k h i

x„ = - x „

= x^

]

^ m - c o . s ( - x „ ) = cos2{ - x„)

- x,| c u n g la n g h i e m

1.

Vi:

kien van: G i a su (1) c6 n g h i e m la x = x,„ tiic la:

^

-2m + 2 = 0 o i n =

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

VT

V ' " - c o s x , | = cos2x„

= 0

kien dii: V o i m = 1, k h i do phuong t r i n h c6 dang:

(1)

Gicii
Dieu

« x „

D o c h i n h la dieu k i e n can de p h u o n g t r i n h c6 n g h i e m d u y nhat.

T i m m de p h u o n g t r i n h :
Vm-cosx

X|,

K h i do:

V a y , v o i m = 0 p h u o n g i i l n h cc n g h i e m duy nhat.
V i d i i 6:

-

VP

l ^ ^ U

IVP = 1

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

cosx = l

2

[2cos2x-I = l

"^'"f'f'

«cosx=l

la n g h i e m d u y nhat ciia p h u o n g t r i n h .

«

x = 0

Dieu

kien can: G i a i s i i (1) c6 n g h i e m x^^^O, suy ra
x^ + m x,^ + 2 m xfi + mx,, +

1=0

V a y , v o i m = 2 p h u o n g t r i n h c6 n g h i e m d u y nhat.

1s


1

+ m — +2m~

X„

+ -L

+m~

K h i do phuang t r i n h c6 dang:

=o

r(t) = t ' + m t + 2 m - 2 = 0.

^•i

v2

(2)

Phuang t r i n h (1) c6 n g h i e m d u y nhat »
+ m

+ 2m

+ m —

+1=0

phuang t r i n h (2) c6 d i i n g

gt n g h i e m thoa m a n Itl > 2 - De nglu ban doe

ticlani.

N h u vay, de t i m dieu k i e n cua tham so sao cho p h u a n g t r i n h h o i
tiic la —

c u n g la n g h i e m cua phuofng t r i n h .

y:
ax'* + b x ' + cx" + bx + a = 0, v a i a

V a y de phucfng t r i n h c6 n g h i e m d u y nhat dieu k i e n la:
—

•

= X „ « X o

1 + m + 2 m + m + 1 = 0 <=> m = - ^ .

Biioc I:

N h a n xet l i i n g x = 0 k h o n g phai la n g h i e m cua phuang
trinh.

Bia'fc 2:

Dieii kien vein:
G i a i su (1) C O n g h i e m x,,, suy ra —

(!)<=> 1 - m + 2 m - m + 1 = 0, v6 n g h i e m .
m =

- ^

1
—

nhat.
- - , ta c6:
2

(l)<::>x'- - x ' - x - - - x +
2
2

I
1= 0<=>2x'-x'-2x--x

+ 2 = 0

<=> ( X - 1 )^(2x^ + 3x + 2) = 0 <=> X = 1 la n g h i e m d u y nhat.

cung la n g h i e m c i a

phuang t r i n h . V a y de phuang t r i n h c6 n g h i e m d u y nhat
dieu k i e n la:

la dieu k i e n can de phuong t r i n h c6 n g h i e m d u y

Dieu kien dir. V o i m =

= x,| »

X|| = ±1 => G i a t r i t h a m so.

D o chinh la dieu kien can de phuang t i i n l i c6 nghiem duy nhat.

BIIOC 3:
du 9:

Dieu kien dir. Thuc hien viec t h u l a i .
T\m m de phuang t r i n h :
Isinx - ml + Icosx - ml = v 2 .

V a y , m = - ^ p h u a n g t r i n h c6 n g h i e m d u y nhat.
Chii

(1)

hien theo cac bu6c:

±l.

V o i x„ = - 1, ta dugc:

Vay,

0

6 n g h i e m d u y nhat, bang phuang phap dieu k i e n can va du dugc thuc

V a i x„ = 1, ta du-gc:
(1) o

•

=

5i

(1)

C O d u n g m o t n g h i e m thugc (0, - ) .

y:

1. Y e u cau t i e n hoan loan c6 the dugc thuc h i e n bang phuong phap dat
an p h u , c u the:
N h a n xet r i n g x = 0 k h o n g phai la n g h i e m ciia phuang t r i n h . Chia ca
hai ve cua p h u a n g t r i n h cho x V O , ta dugc:
x^ + m x + 2 m + m . -

+ -i^ = 0

X

<^{x~ + ~)

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

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

- X|, cung la n g h i e m ciia (1).

X

J
Dat t = X + - , dieu k i e n ltl>2.
X

Suy ra x" + ~

Dieu kien ran: G i a su (1) c6 n g h i e m la x = x,, suy ra

X*

+ m ( x + - ) + 2 m = 0.

X

Gicii

= t^ - 2.

V a y (1) C O n g h i e m d u y nhat k h i
7t

2

C„

» X „

=

Tt

4•

T H l / V!
17


T h a y x„ = ^ vao (1), la dugc:

T h a y x„ = ^ vao (1), la duac:
I s i n - - m i + l c o s - - ml = V2 « > l ^
4
4
2
,

V2 ,

<=> Im - —

V2

I = —

2

2

-ml + l

^ - mi = V2
2

m = 0

<=>

DieII kien dir.
V o i m - 0, k h i do (1) c6 dang:

kien di't

^

Vai m = 0
xe((),^)

(1)

Isinxl + Icosxl = •s/2

o

V2 sin(x + - )
4

= V2

o

X + - ^ - + 2k7t o
4
2

c?"

•»

o

=2

D o do:
xe((),")

(2) « >

« > X = ^ la n g h i e m d u y nhat.

^/tgx -

Tcotgx

= ! <=> tgX = 1

4

la n g h i e m d u y nhat cua phuang t r l n h .
V a y , v o i m = 0 p h u a n g t r l n h c6 n g h i e m d u y nhat.

Isinx - N/2 I + Icosx - V2 I = N/I

sinx + cosx -

-Jl - gicii tmmg ti(

Vi dii 11:

T i m m de phuong t r l n h sau c6 n g h i e m d u y nhat;

nhinren.

khocing (0, ^ ) .

^ + ^42^ + 4^ +

^Jl^K

= in.

(1)

Gicii

DieII kien can: G i a su phuang t r l n h (1) c6 n g h i e m la x = x,, suy ra
2 - x,| c u n g la n g h i e m cua (1).

T i m m de p h u a n g t r l n h c6 n g h i e m d u y nhat thuoc (0, ^ ) :
^tgx-m

+ ^cot g x - i i i

=2.

(1)

Gicii

V a y (1) C O n g h i e m d u y nhat k h i
x„ = 2 - x„ »

x„ = 1.

T h a y x„ = I vao { ! ) . ta dugc m - 4.

Dieu kien can: G i a sir (1) c6 n g h i e m la x = x,,, tiic la:
V'gX() - m

+

^cot gX() - m

<=^ A / c o t g ( ^ - X | ) ) - i n

+

=

2

^tg(^-x„)-ni

=> y - X|| cung la n g h i e m cua (1).
V a y (1) C O n g h i e m d u y nhat k h i
X||

_
—

Tt
— — X,,

2
18

(2) ,

X = - + 2k7t
4

V a y , v d i m = 0 hoac m = >/2 p h u o n g t r l n h c6 d i i n g 1 n g h i e m thuoc

V i d u 10:

= 2

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

sin(x + - ) = 1
4

V o i m = V2
(1) «

+ Vcotgx

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

sinx + cosx = V2

xe((),?)

•

m = 0.

thuoc k h o i i n g (0, ^ ) .

k h o i i n g (0, ^ ) .

•

- "11 = 2 <=> 2 V l - m = 2 «

D o c h i n h la d i e u k i e n can de phuong t r l n h (1) c6 n g h i e m d u y nha^t

m = V2

D o c h i n h \i\u Icien can de phuong t i l n h c6 d i i n g 1 n g h i e m thuoc

Dicii

tg ^ - m + ^cot g

<=>

_
7t
X|| — — .

•

D o c h i n h la d i e u k i e n can de phuang t n n h c6 n g h i e m d u y nhat.
Dieu kien di'i

= 2

V o i m = 4, k h i do (1) c6 dang:

ill

+ ifl^

= 4.

+ V7 +

(2)

A p d u n g bat dang thCrc B u n h i a c o p x k i , ta dugc:
+ V 2 ^

< 2 va ^

+

ifl^ < 2

4

1 n


Vi du 13:

Do d o :

T i m m de p h u o n g t r i n h sau c6 n g h i e m d u y nhat:
mx(2

- X)

= ix - II.

(1)

Gicii
Oicu kien can:
<=> X = 1 la n g h i e m d u y nhat cua phuong t r l n h .

mx„(2 - x„) = lx„ - 11 «

V i i y , v o i m = 4 phuang t r i n h c6 n g h i e m d u y nhat.
V i d u 12:

V a y de phuang t r i n h c6 n g h i e m d u y nhat dieu k i e n la

|X|,

K h i do:
(1) «

G i a sii (1) c6 n g h i e m la x = x,, suy la

- al +

IX|,

2 - x„ = x„ o

(1)

Gicii
Dieu kien can:

m = 0.

Dieu kien di'i: V o i m = 0, ta c6:
(1) <=> Ix - 11 - 0 «

=> a + b - X|| cung la n g h i e m cua (1).

X = 1 la nghiem. d u y nhat cua phuang t r i n h .

V a y , v d i m = 0 phuang trinh c6 n g h i e m d u y nhat.
V i du 14:

V a y (1) CO n g h i e m d u y nha't k h i

T i m m de phuang t r i n h sau c6 n g h i e m d u y nhat:
1

,
a+b
= a + b - x„ <=> x„ =
•

Thay x„ =

x,| = 1.

D o c h i n h la dieu k i e n can de phuang t r i n h c6 nghiem. d u y nhat.

- bl = c

<=> l(a + b - x„) - al + l(a + b - x„) - bl = c

X|,

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

tiic la 2 - x„ cung se la n g h i e m cua phuang t r i n h .

T u n a, b, c de phuang t r i n h sau c6 n g h i e m d u y nhat:
l x - a l + l x - b l = c.

G i a i sii (1) c6 n g h i e m x,„ ta c6:

3IX-21

= 2 m - 1.

Gicii

vao (1), ta dugc:

Dieu kien cc'in: Gia su phuang t r i n h c6 n g h i e m la x = x,, suy ra

c = la - bl.
D o c h i n h la dieu k i e n can de phuang t r i n h c6 n g h i e m d u y nhat.

=> 4 - X|, cung la n g h i e m cua (1).

Diet! kien di'i

V a y phuang t r i n h c6 n g h i e m d u y nhat k h i

G i a .sir c = la - bl, k h i do (1) c6 dang:
Ix - al + Ix - bl = la - bl «
(X

•

x„ = 4 - x„ «

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

- a)(x - b) < 0

(2)

N e u a ;^ b (ta gia su k h i do a < b), k h i do:

T h a y x„ = 1 vao p l i u a n g t r i n h , ta duac m = 1.
D o c h i n h la d i e u k i e n can de phuang t r i n h c6 n g h i e m d u y nhat.
Dieu kien du: G i a su m = 1, k h i do phuang t r i n h c6 dang:

(2) <^ a < X < b, tuc la (2) k h o n g c6 n g h i e m d u y nhat.
•

N e u a = b, k h i d o :
(2)c:>(x-a)-<0
o

X = a la n g h i e m d u y nhat ciia phirong t r i n h .

V a y , v o i c = 0 va a = b phuong t r i n h c6 n g h i e m d u y nhat.
Chii y: Bai toan tren
g i i i i cho mot l a p cac
thi dai hoc chi g o m
buoc thuc hien de ap

la dang tong quat va phuang phap duqc ap d u n g de
bai toan g o m 1 va 2 t h a m so (thong t h u o n g cac bai
m o t tham so). Cac e m hoc sinh can n a m v i l n g cac
d u n g trong m 6 i bai toan cu the.

x„ = 2.

— ! — = 1 Ci> 3'"

= 1»

Ix - 21 = 0 <=> X = 2 la nghiem duy nlia't.

V a y , v o i m = 1 phuang t r i n h c6 n g h i e m d u y nhat.
Vi du 15:

T i m m de phuong t r i n h sau c6 n g h i e m d u y nhat:
2mx,2-x,

^ 3 l x - l l

+

,ii_

Giiii
Dieu kien cc'iu: G i a i su phuong t r i n h c6 n g h i e m x,,, ta c6:
2nix„(2-x„)

_

^Ix,,-!!

+

j-j^ < ^

2"i(2-x„)|2-(2-X|,)| _

tiic la 2 - X|, cLiiig se la n g h i e m cua phuong t r i n h .

3l(2-Xo)-ll

+


V a y de p l i u a n g t r i n h c6 n g h i e m d u y nhat dieu k i e n la
2-X„

=

X„C:>X„=

1.

{)int
•

kien

dii

Voi m = 2

K h i d o phuang t r i n h c6 dang:
Bcinouli

2'" = m + 1 « 2 ' " + m ( l - 2 ) = 1

o

( l ) c : > l x + 2l + lx + 2l = 0 < » x = - 2

"m = 0

la n g h i e m d u y nhat ctia phuang t r i n h .

111 = i

D o c h f n h la dieu k i e n con de phuong trinh c6 n g h i e m d u y nhat.

•

Voi m =

Dic'ii kien di'i
•

( l ) < = > l x + - I + I X + 1|=
2
2

V 6 i m = 0, ta c6:
1 = 3'' ^" «

"

2

Ix - 11 = 0 <=> X = 1 la n g h i e m d u y nhat.

V 6 i m = 1 , ta c6:

- ( --hi"^'"'-'"^ > v6 n g h i e m .
4

V a y , v a i m = 2 p h u a n g t r i n h c6 n g h i e m d u y nhat.
V i du 17:

T i m m de phuang t r i n h :
( x + l ) ' + (x + 3r = 2 m .

Ta CO nhan x e l sau:

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

x(2-x)= - x ' + 2x= 1 - ( X -

1)'< 1

V T = 2^*'-^'<2.

l x - l l > 0 = ^ VP = 3 ' ' - " + 1 > 3 " + 1 = 2 .

-x-+2x = f

VP = 1

IX

-

»

X = 1 la n g h i e m d u y nhat.

11= 0

Ix - m^ + 3ml + Ix + m l = m^ - 3 m + 2.

l(m^ - 4 m - x„) - n r + 3 m I + l( m^ - 4 m - x„) + ml = c

=> m^ - 4 m - x„ cung la n g h i e m cua (1).
V a y ( 1 ) CO n g h i e m d u y nhat k h i
111"
111

( - 2 + 1)-^ + ( - 2 + 3 ) ' = 2 m «

m = 1.

Dieu kien clii: V o i m = 1, ta c6:
(1) » ( x + l ) ' + (x + 3)'' = 2.

(2)

= X + 2, suy ra :

x + l = t - l

x+3

4ll1
V
/ , V
1
—
vao (1), ta dugfc:

m^ - 3 m + 2 = Im" - 2ml<=>

-2.

- 4in

111" -

1 hay X|| = —

= x„<=>x„=

Dat t = X +

x„ = m - - 4 m - X,, <=> x„ = —
rr,

-Xo-4

D o c h i n h la dieu k i e n can de phuang t r i n h c6 n g h i e m d u y nhat.

lx,| - m " + 3ml + lx„ + m l = m^ - 3 m + 2

A

V a y de phuang t r i n h c6 n g h i e m d u y nhat dieu k i e n l i i :

(1) «

G i a sir (1) c6 n g h i e m la x = x„ suy ra

~>

[3 + ( - x„ - 4)]-' + [1 + ( - x„ - 4)]^ = 2 m

Khi do:

Gicii

«

«

tire la - x„ - 4 cQng la n g h i e m cua phuang t r i n h .

T i m m de phuang t r i n h sau c6 n g h i e m d u y nhat:

Dieu kien can:

G i a i sir (1) c6 n g h i e m x,,, suy ra

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

V a y , v o i m = 0, m = 1 phuang t r i n h c6 n g h i e m d u y nhat.
V i du 16:

Gicii
Dieu kien can:

V a y p h u a n g t r i n h (2) tuang d u a n g v a i :
VT = 1

(1)

= t +

r

Khi do:

i i r - 2in > 0

(2)

in^ - 3iii + 2 = m" - 2m

o t^(t^ + 6) = 0 »

111^ - 2iii < 0

<=>x + 2 = 0<::^x = - 2 la n g h i e m d u y nhat.

111^

-3m+ 2 =

-111"

+2m

D o c h i n h la d i e u k i e n can de phuang t r i n h c6 n g h i c m d u y nhat.

o (t - I ) ' + (t + 1 = 2 »

2t'' + 12t' = 0

t= 0

V a y , m = 1 phuang t r i n h c6 n g h i e m d u y nhat.


Vi dii IX:

Clu'i y:
1.

Trni m de phuang trinh:
(x - l ) ( x + l ) ( x + 3)(x + 5) = m.

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

+ a)-* + (X + b)'* = c.

(1)

CO nghiem duy nhat, bing phuang phiip di^u kien ciin va du dugc thuc
hien iheo cac buac:

CO nghiem duy nhat.
Giai
Dic'ii kien can: Giai su (1) c6 nghiem x,,, suy ra

DieII kien can:

Bum-1:

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

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„ + 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.

- x„ - a - b = x„

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

a +b

-x„-4 = x„ox„=
(I)

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

Do chinh la diDieu kien dir. Thuc hien viec thu lai.

Do chinh la dieu kien can de phuang trinh c6 nghiem duy nhat.
Dieu kien du: V d i m = 9, ta c6:

2. Yeu cau tien hoan toan c6 the duac thuc hien bang phuang phap dat
rin phu, cu the:
Dat I = x +
|x + l
X+3

-2.

Khi do:

=> Gia tri tham so.
nhat.
Biioc- 2:

(1)

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

= x + 2, suy ra:

Dat t = x^ + 4x - 5, dieu kien t> - 9, suy ra x" + 4x + 3 = t + 8.
Khi do phuang trinh tren c6 dang:

= t-l
= t+r

t(t + 8) = 9 »

t^ + 8t - 9 = 0

«

1=1
1 = -9

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

K h i do:
x

(1) » ( t - l ) V ( t + l ) ' = 2 m » 2 t ' ' + 12t' + 2 = 2m
o

t"* + 6t- + 1 - m = 0.

(2)

+4x + 4 = 0

x = -2

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

Khi d o :
(2) <=> f(u) = u ' + 6u + 1 - m = 0.
Phuang trinh (1) c6 nghiem duy nhat

<=>

x = -2±VlO

tuc la phuang trinh khong c6 nghiem duy nhat.

Dat u = t", u > 0.

»

+ 4x - 6 = 0

(3) CO nghiem u, < 0 = U2
fS <0

<=> m = 1.

P =0

Vay, vai m = 1 phuang trinh c6 nghiem duy nhat.

(3)

Chu y:

1.

Nhu vay, de' t i m dieu kien cua tham. so sao cho phuang trinh.
(X

+ a)(x + b)(x + c)(x + d) = m, v6i a + b = c + d

(1)

CO nghiem duy nhat, bang phuang phap dieu kien can va du duac thuc
hien theo cac buac:
Bum-1:

i)ieu kien can: Giai su (1) c6 nghiem x,„ suy ra - x,, - a - b
cung la nghiem ciia phuang trinh.


Dien kien dir. V 6 i m - 2, ta c6:

Vay de phuomg trinh c6 nghiem day nhat dieu kien la:
a - b = x,| o

x,| = -

a +b
liunhiiiccipxki

4=(

(I)

Gia tri tham so.

V2-7 )

Vx = V 2 - x

Do chinh la dieu kien can de phuong trinh c6 nghiem duy nhat.
Bif()c 2:

+

<

(1 + l ) ( x + 2 - x ) - 4

x = 1 Ja nghiem duy nhat.

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

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:

Chii y:
1

Nhu vay, de tlm dieu kien ciia tham so sao cho phuang trinh:

(1)

V x + a + V b - x = c.

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

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

hien iheo cac bu6c:

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

BiiocJ:

Khi do phuong trinh tren c6 dang:

Dien kien can:
Giiii sir (1) c6 nghiem x„, suy la

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

(2)

Phuang trinh (1) co nghiem duy nhat
<=> , / b - ( - x „ - a + b) + ^/a + (-v„ - a + b) = c

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

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

A'> 0

16 + m > 0

f(-9) = 0 <=>

9 - m = 0 v6 nghiem.

S/2 < - 9

-4<-9

tuc la

- x„ - a + b cung la nghiem ciia phuang trinh. Vay

de phuang trinh c6 nghiem duy nhat dieu kien la:
b-a

Vay, khong ton tai m de phuong trinh c6 nghiem duy nhat.
Vi dii 19:

- X|, -

T i m m de phuong trinh sau c6 nghiem duy nhat:
N/X

+

V2 -

= m.

X

(i)

a + b = x„ <=> x„ = - y -

=> Gia tri tham so.

Giiii

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

Dieu kien 0 < x < 2.

Bui/c 2:

Dieu kien can: Gia su (1) c6 nghiem x„ , khi do:
^x^ + V2-X0
<»

= m <=> ^ 2 - ( 2 - x „ ) + ^ 2 - x „ =

72-Xi, + ^ 2 - ( 2 - x „ )

2.

m

Dien kien dir Thuc hien viec thu lai.

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

cu the:
Ccicli I: Phuang phap ddr an pint:

=

m
x +a

tiic la 2 - X|, cung la nghiem ciia (1).
Vay ( 1 ) CO nghiem duy nhat khi
2 - x „ = x„<=>x„= 1.

Khi do phuang trinh dugc chuyen thanh he:
u +V= c

K h i do:
(1) <=> 1 + 1 = in «

, dieu kien u, v > 0.

V = V b - x

m = 2.

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

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:

Chu y-

X c t h a m so y = V x + a + V b - x l i e n tap D = [ - a, b ] , tCr d o xac d i n h :
•

D a o h a m l o i g i a i phuang t r l n h y ' = 0.

•

Bang bien thien.

1

T r o n g phan xac d i n h dieu k i e n can ta c6 the su d u n g :
•

Bat dang thilc Bunhiacopski n h u sau:
V T ^ +V ^

K h i do phuang t r l n h c6 n g h i e m d u y nhat k h i va c h i k h i d u a n g 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.

<=> l o g ^ ^ ( V 4 - x + Vx + 5 ) < 1.
V a y p h u a n g t r l n h c 6 n g h i e m k h i va c h i k h i :

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

D e nghj ban doc m a rong c h o phuang t r l n h :

V4-X
1

' ! y a - t ' ( x ) + 7b + f ( A ) = C.
V i dii 20:

< V ( l + ' H 4 - x + x + 5) = 3>/2

_

Vx+5

~

1

X =

la n g h i e m d u y nhat.

Bat dang thuc Cosi n h u sau:

T u n a de phuang t r l n h sau c 6 n g h i e m d u y nhat:
l o g ^ ^ ( V 4 - x + V x + 5 ) = a.

(V4^ +V x ^ ) ' =

(1)

4 - X + X + 5 + 2 ^ ( 4 - x ) ( x + 5)
< 9 + ( 4 - x ) + (x + 5 ) = 18

Gicii
Dii'ii

kicn ccin: G i a sir (1) c6 n g h i e m x,,, k h i do:

< » V 4 - x + V x + 5 £ 3 V 2 <X> l o g ^ ^ ( v ' 4 - x + V x + 5 ) < 1.
V a y p h u a n g t r l n h c 6 n g h i e m k h i va c h i k h i :

«

^

log3^(^(-l-X|,)+5+74-(-l-X|,)) = a
V4-X

Tuc la k h i do - 1 - x,, cung la n g h i e m cua (1).
2.

V a y ( 1 ) CO n g h i e m d u y nhat k h i

= \/x + 5 <=> X = - -

la n g h i e m d u y nhat.

Bai toan tren c o n c 6 the dirac g i a i bang phuang phap h a m so, n h u

sau:
1

X(| —

X(|

X|| —

—.

V i e t l a i (1) duofi dang:

V 4 - x + V x + 5 - log^y^a

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

vdi a > 0

'

(2)

Phuang t r l n h (1) c6 n g h i e m d u y nhat

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

«

<=> d u d n g thiing y = l o f i ^ ^ a cdt d 6 t h i h a m so y = V 4 - x 4 / x + S

V a y a = 11a dieu k i e n can de phuang t r l n h c6 n g h i e m d u y nhat.
Dicii

(2) CO n g h i e m d u y nhat

tai d i i n g m o t d i e m .

kicn dir. V d i a = 1, phirong t r l n h (1) c6 dang :

X e t h a m so y = V 4 - x + V x + 5 .
4-x >0
X

+5 > 0

<=>

4 - x + x + 5 + 2 V ( 4 - x ) ( x + 5) = 18
- 5
[ 4 ( 4 - x ) ( x + 5) = 81

< X <

-5

< X <

^

4

2V(4-x)(x+5) =9

4

,

4x-^+4x + l = 0

V a y , a = 1 phuang t r l n h c6 n g h i e m d u y nhat.

1

«

X =

2

.

•

M i e n xac d i n h D = [ - 5, 4 ] .

•

Dao ham:
y =

-

1

«

.

2V4-X

y' = 0 <=>

,\

1
2Vx+5'

T1 = 2V4^

Vx + 5 = V4 -

+

^

0
,=
2Vx + 5

X <=> X =

-

^ .


Neu a = 1, k h i do:

Being bien thien:

I

X

-GO

-

- 1/2

5

1

X =

X € | 0,271)

sinx = -

o

^~

T i m a, b de phirong t r i n h :

6

6

Vay. v o i b = 0 va a = 1 co d u n g 2 nghiem phan biet thuoc [0,

I2.sinx - 1 1 + I2sinx - al = b

CO d i i n g 2 n g h i e m phan biet thuoc [0,

7t
—

_ 57t

Ttr do dieu k i e n la a = I .
V i (III 21:

1 <=> 2sinx - I

(3) <=> (t - 1)' < 0 <=> t

(1)

2n).

Gicii

III.BAITAPD^
Biii tap 1:

NGHI

T u n m de cac phuong trinh sau c6 nghiem d u y nhat:

Dat I = 2 s i n x . dieu k i e n Itl < 2.

a.

mx"* + ( m + 3)x'^ + m " - m = 0.

K h i d o p h u o n g t r i n h c6 dang:

b.

( m - 2 ) x ' - 2 m x ' - ( m - 5)x- + 2 m x + m - 2 = 0.

c.

(x + 2 ) ' + (x + 6)-' = n r - 2 .

cl.

x ( x - 2 ) ( x + 2)(x + 4) = 2 m .

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

(2)

K h i do (1) CO d i i n g 2 n g h i e m phan biet thuoc [0, 2n)

Bai tap 2:

<=> (2) CO n g h i e m duy nhat thuoc [ - 2, 2 ] .
Dicti kien

can

T i m m de cac bat phuong t r i n h sau c6 nghiem d u y nhat:
^

G i a sir (1) c6 n g h i e m la t = t,, suy ra
lt„ - 11 -i- ll„ - al
b

b.

=> 1 + a - t|, cung la n g h i e m ciia (2).
V a y (2) CO n g h i e m d u y nhat k h i

> 1 + vn\

2'^-"< 1 + n r .

Bai tap 3:

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

T u n m de cac bat phuong t r i n h sau c6 n g h i e m d u y nhat:

a.

log,„(x' + 2) < l o g , m .

b.

log,,,,,(1x1+ l ) + l o g , ( l + m ' ) < 0 .

Bai tap 4: T u n m de cac phuong t r i n h sau c 6 nghiem duy nhat:
t„ = 1 + a - t„
T h a y t,, =
1 ^

'2

t„ = ^

.

vao (2), ta duoc:

% ' • 2

r

D o c h i n h la dieu k i e n can de phuong t r i n h c6 n g h i e m duy nha't.
Dicii

kien di'i:

It-11 + I t - a l = l a - I I
o l t - l l + lt-al =

l(t-l)-(t-a)l

c^(t-l)(t-a)<0
N e u a?tl (ta gia sir k h i do a < 1), k h i do:
(3) o

x ( 4 - x ) + m = Ix-21.

b.

Ix + n r l + Ix + I I = Im + I I .

c.

m x - - 2 ( m - 1 )x + 2 = Imx - 21.

a.

Vl-x" +

b.

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

•

a.

Bai tap 5: T u n m de cac phuong trinh sau c6 nghiem d u y nhat:

- l l + | i l l - a l = b o la - I I = b

a < t < 1, tuc la (3) k h o n g c6 n g h i e m d u y nhat.

(3)

= m.

+

+ 2 ^ f l ^

c.

Vl-x-

d.

V4-X

e.

HK

f.

ifZ'-l

g.

In).

+

=m.

+ v/s-x^ = m .

+ Vx + 2

= m.

= m.

+ ^-^"x + v x ~ +

+ VxTT

= m(

=m. .
+

ifx^).


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

X- - 2insin'x + 4 = 0.

b.

X - - m.sin(cosx) + m - = 0.

c.

x ' - ni.tg(cosx) 4 n-.~ ~- 0.

BAI TOA N 2
GiAi B A IT O A N D U Y N H A T

Bid tap 7: Tun a, b, de phuong trinh sau c6 nghiem duy nhat:
\/(ax + b)2

+ ^(ax-br

+ ^(a-x2 - b ^ =

NGHIEM

C H O H E PHUaNG TRINH, H E B A T PHl/O'NG TRINH

.

1.1>HIT()N(J F H A P
Voi yeu cau:
" Tim dicn klcn ciia tham so(i;id
hat phiioiiii trinh:

sir la m) dc he pliuang trinh, / t f

f(x,y,m)<0

(1)

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:
(pi(x„,y„) = x„
9 2 ( x „ , y ( , ) = y„

Gia tri ciia (x,„ y,,).

(2)
c.

Thay (2) vao (1) ta xac dinh dugc dieu kien can cho
tham so m de (1) c6 nghiem duy nhat, gia su meD,„ C) budc nay can su dung kien thiic ve dieu kien c6
nghiem duy nhat cho phuong trinh hoac bat phuang
trinh.

Biioc3:

Dieu kien dir. V o i meD,,,, ta di kiem tra lai tinh duy nhat
nghiem cho (1).

Biioc 4:

3a

Thong thuong trong buac nay, ta chi^ phai xet cac he
phuong trinh, he ba'l phuang trinh cu the (thuang la khong
CO tham so hoac neu c6 thi da dugc dan gian di nhieu). Ket
qua ciia buac nay cho phep ta loai di khoi tap D„, cac gia tri
khong thich hgp ciia m.
Ket hop ba buac giai tren ta tim dugc dap so.
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.

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.

II. V i DU MINH HOA
Vi du 1:

Bien doi he phuang trinh v6 dang:

Cho he phuang trinh:
+y

X
X

U +yr-2xy =m

=111

X +

+y =6

xy =

. 36-m

2

G'uii
Dicii kicii cciii:

36 - m

= 0.

(1)

He CO nghiem duy nhat

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:
= y„.

<r> (1) CO nghiem duy nhat o

2xf| = m
2x„ = 6

A',,, = 0 < = > m - 1 8 = 0 < = > m = 1 8 .

Khi do he c6 nghiem x = y = 3.

(*)

Khi do, he c6 dang:

Vay, vai m = 18 he phuang trinh c6 nghiem duy nhat.
Cdcli 2: Siiditng plutang plidp the.
Bien doi he ve dang:

m = 18.

x~ + ( 6 - x ) " = m

<=>

y = 6-x

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

2 x ^ - 1 2 x + 36

i.i-0

(2)

y = 6-x

(I)

He CO nghiem duy nhat

Voi m = 18, ta duac:
+ y- = 18
x +y =6

<=> phuang trinh (2) c6 nghiem duy nhat
fx + y = 6
XV = 9

«
<=> X = y = 3 la nghiem duy nhat.

A',|, = 0 »

m - 18 = 0 «

m = 18.

Khi do he c6 nghiem x = y = 3.
Vay, vai m = 18 he phuang trinh c6 nghiem duy nhat.

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

Cdch 3: Sit dung phuang phap do thi.

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
khi:
.
B
.
y
x„ = y,,

•
Bitoc

y =6

+y =6 .

khi do, X , y la nghiem ciia phuang trinh :

Xac dinh m de he c6 nghiem duy nhat, xac dinh nghiem do.

X(,

X

^

2:

(**)

Thay (**) vao he ta duac gia tri ciia tham so. Do chinh
la dieu kien can de he c6 nghiem duy nhat.

Dieii kien dii.

.

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)

= Vi^

<=> d ( 0 , (d)) = R »

«

m = 18.

Vi + i

Khi do, he c6 nghiem x = y = 3.
Vay, vai m = 18 he phuang trinh c6 nghiem duy nhat.

.14

35


Ccuh

4: Si'cclungphucfngplidp

luang gidc

hoa.

Dieu kien dir. Vcfi m = 1, ta duac:

N h a n xet r i n g v a i m < 0, he v6 n g h i e m , do do ta xet v 6 i m > 0.

(log2(x + y) = l o g j l x y ) ^ \^ + y "^^y ^
|log2(xy) = 6 - x y
[xy = 4

T u phuang t r l n h t h i i nhat ciia he:
v2

}

X

y

-I-

V i d u 3:

Cho he phuang t r l n h :
X

= Vm

y =

sin t

'
,tG[0,

2.

X = y =

V a y , v 6 i m = 1 he c6 n g h i e m duy nhat (2, 2).

=1

dat:
X

|x + y = 4
o
[xy = 4

271).

(3)

V m cos t

«

42^\t + - ) = 6

in + 2

(I)

[x"y+ xy'^ = m + 1

Xac d i n h m de he c6 n g h i e m duy nhat.
Gidi

T h a y (3) vao phucrng t r l n h t h u hai ciia he, ta duoc:
Vim sint + 4m cost = 6

+ xy + y =

Dieu kien edn: N h a n xet rang neu he c6 n g h i e m (x,„ y,,) t h i (y,,, x„) c u n g
la n g h i e m cua he, do do he c6 n g h i e m duy nhat k h i :

4

x„ = y„.
« . s i n ( t + ^ ) = ^ .

(4)

(*)

K h i do:

He CO n g h i e m duy nhat
phifong t r l n h (4) c6 n g h i e m duy nhat tren tap [0, 2n)
^

3V^

2xo

= m + 1

m =

—

4

T u n m de he p h u a n g t r l n h sau c6 n g h i e m d u y nhat:

ieu kien di'i
•

V o i m = 1, ta duac:
^ j ^ ^ | x y + (x + y) = 3

log2(x + y) = m logjCxy)

[xy(x + y) =

log2(xy) = 6 - x y

2

k h i do X + y va x y la n g h i e m ciia phuang t r l n h :

Gidi
Dieu kien:
X

kien

X

+ y>

xy >

0

0

<=> X ,

y > 0.

t ' - 3t + 2 = 0

o

t = 1
1=2

(dir.

ixy =
X

(vn)

2

+y =

o
2

la n g h i e m d u y nhat ciia he.
"

K h i do he c6 dang:
log2(X||+Xo) = !r.log2 x?| " 0 ^ " llog2 4 = mlcg2
log2 Xo = 6 - Xo

+ y = 1

[xy = l

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

do do he c6 n g h i e m d u y nhat t h i x„ = y„.

X,|

=z

V o i m = - 1, ta dugc:
^ j ^ ^

<=> m = 1.

D o c h i n h la dieu k i e n can de he cc n g h i e m d u y nhat.
36

.

D o chfnh la dieu k i e n can de he c6 n g h i e m d u y nhat.

V a y , v o i m = 18 he phuang t r l n h c6 nghiem duy nhat.

Dial

m = -3

2xo-xr,-2x,| + l = 0

= 1 c > m = 18.

K h i do he c6 n g h i e m x = y = 3.

V i d u 2:

(I)»

m = l

in = 2 x o - 1

X() + 2 X ( ) = m + 2

jxy + (x + y) = l
[xy(x + y) =

0

N h a n thay he l u o n c6 hai cap n g h i e m (0, 1) va ( 1 , 0).

X

= y = 1


•

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

V d i m = - - , ta duac:
4

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

<=>

xy + (x + y) = ^
4

(I)«

(2) v6 nghiem & (1) c6 nghiSiii kep
( i ) & (2) CO nghiem kep u „

xy(x + y) = ^
4

A,

k h i do X + y va x y la n g h i e m ciia p h u o n g t r i n h :
X

+y = I

<=>

1

t = l

- 4m - 3 < 0

<0

\o

m^ + 2ni - 3 = 0

Aj. < 0

ni^ + 2 m - 3 < 0

A,

<=>

=0

Aj, = A , = 0

" 4

+y =—
4
xy = 1
X

2 "

3 •
m = --

-4m-3=0
m^+2m-3 = -4m-3 = 0

1 ^ m+1

(vn)

m =1

ni = 0

2

V a v , v d i m = 1 hoac m = - • - he d i i cho c6 n g h i e m d u y nhat.
<=>x = y =

•

n d u 4:

Tim

la n g h i e m d u y nhat ciia he.

•

m de he sau c6 n g h i e m d u y nhat:
x ^ + y ^ = 17

V a y , v d i m = 1 hoac m = -

he da cho c6 n g h i e m d u y nhat.

log2 x + log2 y =

m

iidi

da tha'y t r o n g v i d u tren, c h i i n g ta c6 the thuc h i e n bang

Chii y: Nhu
ciich sau:

4

Dieu kien :
x>0

Dat:

y>0'

""^^

Dicii

^,dieukienS'-4P>0.

xy = P

kien can:

N h a n xet rang neu he co n g h i e m (x,,, y,,) t h i cung

n g h i e m (y,„ x„), do do he co n g h i e m d u y nhat t h i x,, = y,,.

V i e t l a i he d u d i dang:

K h i do he co dang:

(x + y) + x y = m + 2

S+ P = m+ 2

(x + y).xy = m + l

S.P = m + l

17
x,-)+x?)=17

k h i do S, P la n g h i e m ciia p h u o n g t r i n h :

1

17

m = log, y

l o g 2 X() + l o g 2

X()

.

=ni
21og2 , | y

=

"1

t- - ( m + 2)t + m + 1 = 0
Dieu

t = l
t =

111

17

kien dii : V d i m = logj y , he co dang:

+1

(x + y)

x^ + y - = 1 7
)x + y = I
[xy =

111

x +y =
xy = 1

+1

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

111 + r

»

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

(1)

log2 x + log2 y = log2

<=> X

(2)

= y =

1.

y

«

17
xy =

y

17
V2

•
17

IS

17

Vay, voi m = l o g j y

x + y = V34

- 2 x y = 17

he co n g h i e m d u y nhat.

«

17
xy = —

y: V i d u tren, c h i i n g ta c6 the thuc hien bang each sau:

^hii

Tniv-ng

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

V i e t lai he phuong t r i n h d u d i dang:
= 17

X -

log2(x.y) = m
x,y>(l

X+ Y

x2+y2 = i7

(x + y ) - - 2 x y = 17

Vx + -y/y =

x.y = 2"

x.y = 2'

x+y

1

V7

r

0
1

4t^-4t + l < 0

^2

1

,in+l

= VlV + 2'

= i - i

<=>t=^«>/x=Vy

= 2

X.y = 2'-'

ci> X = y =

K h i do X , y la n g h i e m cua phuong t r i n h :
17 +

+ 2 ' " = 0.

2 m+l

(1)

la n g h i e m d u y nhat cua he

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

He CO n g h i e m d u y nhat
»

A„)=0

<=>

<

,m+l

<=> m = log;

X

17+ 2""-" - 4 . 2 " " = 0

2a

+y< m

<=>T'' = 17
•>0

t^+(l-t)^ 0
0
<=>

2t--2t + l - m < 0

<=>

i 1-V2m-1

^ 1+ V 2 m - I

17

17

V a y , v o i m = logj y
V I ( U I 5:

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

( 1 ) CO n g h i e m kep duong

1-V2m-1 , 0}
{ Jmaxj^^^P^.O) < y < l
•V
2

o

max {

^

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

2

thoa m a n dieu k i c n dau bai.

• •

Tim m de he bat phuong t r i n h sau c6 n g h i e m d u y nhat:
_ yfx+yjy

V a v , m = - la dieu k i e n can va d u de he c6 n g h i e m d u y nhat.

=1

X+ y < m

(I)

Gicii

2

Chu

•

y: V i d u tren, c h u n g ta c6 the thuc hien bang cdc each sau:
T r u o c het, dat:

D i e u k i e n x , y > 0.
DieII kien can:
G i a s u he c6 n g h i e m (x,„ y,,) =:> (y,„ x,,) cQng la n g h i e m cua he . V a y
he CO n g h i e m d u y nhat t h i dieu k i e n can la: x„ = y„.
K h i do he ( I ) c6 dang:

^

, u, v>0.

K h i d o he c6 dang:

(1)

fu + v = l
u^ + v - < m

2 ^ = ^ = ^ m > i .
2x < m
2

Ccicli 1: Siidiing
V =

V a y m > ^ la dieu k i e n can de he c6 n g h i e m d u y nhat.
Dien kien dir.
Vdfi m > ^ , ta xet hai t i u o n g h o p :

An

•

[v = Vy

(II)

(2)

pinrang phdp bien doi tifc/ng diiWig:

1 -11

ir+(l-u)

0<

u <

1

(III)

[ f ( u ) = 2u - 2 u + l - m < 0 (3)

V a y ( I ) CO n g h i e m d u y nhat <=> ( I I I ) c6 n g h i e m d u y nhat.
Xet (3), ta c6:

A=

2 m - 1.
41


Trudc het can c6

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

A>0»2m-l>0<=>m>-.
2

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

Khi do tap nghiem ciia (1) la:
1 - V2in-1

^

x+y < 1

^ - l + -v/2iTi-l

2

2

Khi do (II) CO nghiem duy nhat
^

1-V2m-1

.

^

1

2

Vay, vdi

(II)

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

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

2

= ^ h? c6 nghiem duy nhat.

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

Cc/c// 2; Si'fdiing phifc/ngphdp do thi
taidiem M ( ^ , ^ )

Triayng lu/p 1: V 6 i m<0 thi (III) c6 v6 so nghiem.
Tnayng lu/p 2: V o i m > 0

Triayng lu/p 2: V 6 i m > -

h? (I) c6 v6 so nghiem, that vay vdi

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

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

"

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

X = - , ta duoc:
2

''2

TamO
2
3
(y-1)^ -"''"'"4

B k i n h R = Vm

l-,|m + ^ < y < l +jni + ^

Vay he c6 nghiem duy nhat khi (d) tiep xiic voi (C) tai diem l e A B
«

d(I, (d) = R «

4-

=

AA^

<=> m = - .

V2

« l - | i i

| < y < ^

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

Vay, voi m = ^ he c6 nghiem duy nhat.
Vi du 6:

+

2

Tun m de he bat phuong trinh sau c6 nghiem duy nhat:
X + y + ^2xy + m > 1

Chu y: Co the sir dung phuong phap do thi de giai vi du tren bang viec
bie'n doi tuong duong he ve dang:

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,,.
Khi do he (I) c6 dang:
2x„ < 1

m>2(x„-1)^-1

2(x„ -1)^ < m +1

x „ - l < - -

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

111 > 2 ( x „ -

ir - 1

(Xo-l)^4

=> m>
2

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

7 2 x y + 111 > l - ( x + y)
X+ y < 1

x+ y <1
2xy + iii > [ l - ( x + y ) f
x + y(x-1)' +(y-l)^
0)
(2)

.+y-l=0

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-?

iTa can ( * ) phai c6 n g h i e m d u y nhat

X 2 l a tap c a c d i e m t i o n g h i n h tron ( C ) c 6 :

<=> A = 0 <=> m = 2.

T a i n 1(1.1)

/ a y , dieu k i e n can de he c6 n g h i e m d u y nhat la m = 2.

Bkinh R = Vm + 1

5K'/' kiei^ 'J''- V<^'^'

V a y he c6 n g h i e m d u y nhat k h i ( d ) tiep xuc vcfi (C)

« d(l, (d) =

R c=> -J= = slm + \> m = -

V a y , vtfi m = - V i till 7:

V2

(x-ir+(y + l)'^2

- .

2

=^(x-

y =

- X+ m

kien

can:

V a y , he c6 n g h i e m d u y nhat k h i m = 2.
Vi d u 9:

Tim m de he sau c6 n g h i e m d u y nhat:
x"+(y +1)" < 111

N h a i i xet rang, neu he c6 n g h i e m (x,,, y,,) t h i c u n g c6

(x + l)^+y^ < m

n g h i e m (y,,, x„), d o d o he c6 n g h i e m d u y nhat i h i
Gicii

x„ = y„.

Dieu kien

Khi do:
(*) <=> x,| = xo - x,| + m <=> xfi - 2x„ + m = 0.

(3)

D o x„ d u y nhiit nen p h u o n g t r i n h (3) c6 n g h i e m d u y nhat
«

A',,, = 0 < : i > l - m = 0 < » m = l .

D o c h i n h la dieu k i e n can de he c6 n g h i e m d u y nhat.
Dieu kien dir. Vdfi m = 1, he c6 dang:
X = y^ - y + 1

y = X" - X + 1

» ( x -

1)'<4

N h a n xet l i n g x = y = 0 thoa m a n he ( I I ) .

(*)

Gicii
Dieu

l)- + (y+ l ) '+ (x+ l ) '+ ( y -

<=>x- + y " < 0 < = > x = y = 0.

T u n m de he sau c6 n g h i e m d u y nhat:

- y +1)1

^jY)

.

(x + l ) ' ^ + ( y - l ) ^ ^ -

he c6 n g h i e m d u y nhat.

X =

= 2, he c6 dang:

can:

G i a sir he c6 n g h i e m (x,„ y„), suy ra (y,„ x„) c u n g la n g h i e m cua he.
V a y de he c6 n g h i e m d u y nhat t h i dieu k i e n can la x„ - y„.
Khi do:
xr, + ( x „ +

l)-
+2x„-m+

1<0

(1)

Ta can ( 1 ) phai c6 n g h i e m d u y nhat

<=>A = 0 < = > m = - .

-l->

X + y = y^ - y-+ 1 +

- x + 1
V a y d i e u k i e n can de he c6 n g h i e m d u y nhat la ' i i = ^ •

l ) ' + (y - i r = 0 0

[x = l
y =

r

Dieu kien du:

V o i m = ^ , he c6 dang:

N g h i e m thoa m a n he va la n g h i e m d u y nhat.
V a y , vcfi m - 1 he c6 n g h i e m d u y nhat.
V i d H 8:

T u n m de; cac he sau c6 n g h i e m d u y nhat:
(x-i)^+(y+i)-
(X M ) ^ + y ^

x ' + (y +

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

«(xV^

Giiii
Dieu kien can: G i a sir he c6 nghiem (x,„ y„), -suy ra (y,„ x„) cQng la nghiem
cua he. V a y de he c6 nghiem duy nhat t h l dieu kien can la x„ - y„.
Khi do:
(x„ AA

x - + ( y + 1)- <

+ (x„ +

m «

2x?, + 2 - m < 0

-

(ID

< ^

+ (X +

+ ^ ) ^ + (yV^

+

y'
2 x ' + 2x + 2 y ' + 2 y + 1 < 0

+ ^ ) ^ < 0 « x

= y= - i .

N h a n xet rang x = y = - ^ thoa m a n he ( I I ) .
V a y , he c6 n g h i e m d u y nhat k h i m = ^ .

45


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

V a i m = 8, he c6 dang:
X

xy + x^ = 8 ( y - l )

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

xy + x^ = 8 ( y - l )

•

x y s y^ = 8 ( x - l )

X | la tap cac diem trong hinh tron:
(C,,:^""'<"-''

y = -8

Vi (111 11:

Tain I j l - 1 , 0 )

T i ' i i m de he sau c6 nghiem duy nhat:

B k i n h R j = Vm '

x-^ = y 2 + 7 x ^ - m x

(1)

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

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

<=> m = - .

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

Xii =

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

yd-

Khi do:

xy + x^ = m ( y - l )

( 1 ) « xo = X?) +1x1 - m x „ c : >

\

xy + y = m(x - 1 )

Gicii

=0
x?)-8xo+m = 0

(3)

Do x„ la nghiem duy nhat, nen
A'(3)<0

Die It kien can:
(3) v6 nghiem

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

A ' = 0

<=> 16 - m < 0

m > 16.

(3) nghiem kep bang 0
4 =0

Dieu kien dir.
Vai m > 16, ta c6:

Khi do:
(!)<=> 2x,^, - m X | , + m = 0.

x"* = y^ + 7x^ - mx

(3)

(X

Do x,| duy nhat nen phuang trinh (3) c6 nghiem duy nhat
A',,, = 0 <=> m^ - 8m = 0 »

- y)[x^ + (y - 6)x + (y^ - 6y + m)] = 0

111 = 0

X ' = y^ -I- 7 x ' - mx

111 = 8 •

x =y

(II)

=y^ +7x^ - mx

Do chinh la dieu kien can de he c6 nghiem duy nha't.

(III)

x ^ + ( y - 6 ) x + ( y - - 6 y + in)] = 0

Dieii kien du
•

72 = 0

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

(Q):

I.L = 2

y = -8-x

- X

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

{—-~T\

X2 la tap cac diem trong hinh tron:

<=> I , ! . = R, + R,

2x^ - 8 x + 8 = 0

x =y

.

[BkinhR|=NA^

•

<=>

= y

V d i m = 0, he c6 dang:

(*)

Giai (II)
x =y=0

x = y
xy +

X

=0

xy + y - = 0

Ta thay he c6 v6 so nghiem thoa man y = - x => loai.

X

=y

x =0

x(x^ - 8 x + m) = 0
<=> X

=y

x^ - 8 x + m = 0 (**)

<=>

x =y

x2 - 8 x + m = 0 (**)

0
47


Vi du 13:

bai phirang trinh (**) v6 nghiem do:

Tnn m de he sau c6 nghiem duy nhat:

A = 16 - m < 0, vai V m > 16
•

(X, -

Coi (*) la phuong trinh bac 2 theo x, ta c6:

= - 3 ( y - 2 ) ' - 4 ( m - 1 2 ) < 0 . V m > 16

Khi do:

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

(3)

( 1 ) « ( x „ - 2 ) ^ + xr, = m « 2 x r , - 4x„ + 4 - m = 0.

Tun m de he sau c6 nghiem duy nhat:

Do x„ duy nhat, nen phuong trinh (3) co nghiem duy nhat

x(4y' - 3 ) = in

<=> A',,, = 0 <=> 2m - 4 = 0 o

(I)
111

m = 2.

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

Gidi
Dieu kien can:

Dieu kien clii : V o l in - 2. he co dang;
(x-2)'+y-- =2

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„. Khi do:
(1) <» x„(4xr, - 3) = m o 4xf) - 3x„ = m.
Do x„ duy nhat nen (1) phai c6 nghiem duy nhat «
Dieu kien du:

hnl > 1.

(II)

x-+(y-2)2=2

(1)

^

(X

<=> X

- 2)- + y ' + x ' + (y - 2)- = 4 <»

(X

-

+ (y -

=0

= y = 1.

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

Voi Iml > 1, ta c6:

Vay, he co nghiem duy nhat khi m = 2.
x(4y- - 3 ) =

x ( 4 y ' - 3 ) = in

»

(2)

x„ = y„.

Do do (*) v6 nghiem « • (II) v6 nghiem.

y(4x^ - 3 ) =

(1)

^Gicii
Dieu kien can : Gia su he c6 nghiem (x„, y„), suy ra (y,„ x„) cung la
nghiem ciia he. Vay de he c6 nghiem duy nhat thl dieu kien can la

A = (y - 6)- - 4(y' - 6y + m)

(I)

= 111

x^+(y-2)^=m

Giai (III)

Vi du 12:

D^+y^

^

( x - y K 4 x y + 3) = 0

X

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

111

(II)

= y

tren, cu the:
Goi X | va X j Ian luot la tap nghiem ciia (1) va (2).

x(4y" - 3 ) = m
(HI)

Ta co:

4xy + 3 = 0

•

Giai (II)

X | la tap cac diSm lien duoiig tron (C,) co:
T a m i l (2,0)

(II) <=> ' "I /
4x

-3x =

<=>x = y = ^-(Vm + V111"
'
2

111

- 1

+ V m - Vnr -

1 ).

•

Uiai (III)

«

4xy

- 3x =

4xy - - 3

111

^

[- 3y - 3x = m

l^\y

= -3

x+y=--^
xy = - l

.

X , la tap cac diem tren dudng tron (Cj) co:
T i m I2 (0.2)

111

(III)

BkinJiR, =^/lrl '

(IV,

BkiiihRj =\/m

Vay he co nghiem duy nhat khi va chi
kiti (C|) tiep xiic ngoai voi (Cj)

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.
A

O

2 I2
V 1
0
2.

. Vay, he co nghiem duy nhat khi m = 2.

hr
40


Do x,i duy nhat <=> phuong trinh (3) c6 nghiem duy nhat

Vi du 14: T i m a de he bat phucfng trinh sau c6 nghiem duy nhat:
Vx + 1 + ^
4 I +

<-d

c=> A',„ = 0 <=> m^ - 8m = 0 <=>

Vx < a

Till = 0

m

= 8 '

Dieu kien du

Gidi

-

Dieu kien x, y > 0.

V o l m = 0, tij' (2) ta thay

Dieu kien can:

Igx.lgy + Ig" X = 0

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

IgxClgy + igx) = 0

Igx.lgy+ lg-y = 0

•

=> neu a < 1 thi he vo nghiem.

he CO v6 so nghiem =^ m - 0 loai.

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

Voi m = 8, nhan thay

Dieu kien dir. Xet hai trudng hop:
V o i a = 1, he CO dang:

Igx.lgy+ lg"x = 8 ( l g y - l )
I g x . l g y + Ig^ y = 8 ( l g x - l )

<z> X = y = 0 la nghiem duy nhat cua he.
+1 +

TrCr tCing ve he phuong trinh, ta duoc :

<1

Ig-x - Ig-y = - 8(igx - Igy)

Voi a > 1, xet cac cap nghiem ciia he c6 x = 0 he co dang:
Bien doi bii't phuong trinh (1) ve dang:
y <(a-ir

yjy + l

7

(Igx - lgy)(lgx + Igy + 8) = 0
^

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

rigx = lgy
^iav = - 8 - l g x "

y
Khi do he phuong trinh tuong duong voi:

=> 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.

Igx = I g y

Chu y: V o i each lap luan tuong t i i nhu tren ta c6 the giai duoc bai toan
voi yeu cdu:

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

X

Igx = - 8 - l g y

Igx = 2

= y

c ^ x = y = 100

72 = 0 (vn)

Khi do dieu kien la a > l .

Vay, voi m = 8 he c6 nghiem duy nhat.

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 )

21g^x-81gx + 8 = 0

(I)

Vi du 16:

Tim m de he sau c6 nghiem duy nhat:

(D

|l

Igx.lgy + lg-y = i n ( l g x - l )

X +

11 + I y - 3 1= 111

| l y + 11 + I

Gidi

X-

3 1= m •

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;

Dic'u kien can: Nhan xet rang, neu he c6 nghiem (x,„ y,,) thi:
|l

X()

+ I I + I y,| - 3 1= m

| l yo + 11 + 1

(1) o 2 1 g ' x o - m l g x „ + m = 0.
Dat t =

lgX||,

(2) «

3 1= m

[ l (2 - y,,) - 3 I + I (2 - x,,) + 11= m

(2)
tii'c la he cung c6 nghiem (2 - x,,, 2 - y,,)

ta duoc:

2t' - mt + m = 0.

X,) -

Jl (2 - x , | ) - 3 I + I (2 - y,)) + 11= m

(3)

Mat khac he cung c6 nghiem (y,,, x„).