Ôn luyện theo cấu trúc đề thi môn Toán Phần 2 Ôn thi tốt nghiệp THPT TS.Vũ Thế Hựu, Nguyễn Vĩnh Cận

CHI

DAN

Xet dau tarn thuTc h(x) = 2x^ - 3x - 5, c6 h(x) = Oc^x = - l , x = -

2

2

5

5

h(x) > 0 vdi X < - 1 hoac x > — ; h(x) < 0 vdi - 1 < x < —
Xet tLfcmg tir vdi tam thufc g(x) = 4x^ - 19x + 12 ta lap diigfc bang sau :
X

+

^ _ h(x)
g(x)

+

g(x) = 4x2 - 19x + 12

+

h(x) = 2x' - 3x - 5

-1
0

0

-

J

r

5
2

-

-

+

+

-

0

0

^

+

-

0

+
0

-

+
+

TCf bang xet dau ta ket luan :
f(x) > 0 vdi X e

(-oo;

-1) u ' 3 .
v4'

5^
u (4;
2y

+oo)

rs ^

3^

f(x) < 0 vdi X e

f(n) = 0 vdi X = - 1 , X = 2
3
f(n) khong xac dinh vdi n = —, x = 4.
4
5x +phifong
2 > 0 t r i n h va bieu d ib)
4x + 3 < tren
0. true so.
18.a) 2x2
Giai- bat
l n x^
tap+ nghiem
HI

DAN

a) Xet dau tam thufc ve trai : fix) = 2x2 _ 5x + 2
A = 52 - 16 > 0, fix) = 0

Xj = -

= 2

, X2

fix) > 0 (f(x) cung dau vdi 2) vdi x < - hoac x > 2. Vay tap nghiem

2
1]
2; + c o ) . Bieu dien tap
cua bat phirong t r i n h la: T = {
-00;
—
I
2_
nghiem tren true so' (phan true so khong bi danh cheo).
1.

^ ] » « f ^

•

b) Lam nhiT cau a). Tap nghiem T = (-3; -1).
-3

-1

32 S TS. Vu The Hi^u - Nguygn VTnh CSn


19.

Giai he bat phifofng trinh va bieu dien tap nghiem tren true so :
(I)

x' + x - 6 > 0

(1)

x'-5x +4<0

(2)

CHI D A N

Tap nghiem cua (1) :
T i = (-oo; - 3 ] u [2; +oo).

Tap nghiem ciia (2) :
T2 = (1; 4).

-3

(D(2)i
(i)i

Tap nghiem cua he (I) : T = T i n T2 = [2; 4).
^
20. Trong moi trirdng hop, t i m cac gia t r i tham so m de :
a) PhtfOng t r i n h : (3m + l)x^ - (3m + l ) x + m + 4 = 0 (1) c6 hai nghiem
phan biet.
b) Phuong trinh : (m - 4)x^ + (m + l ) x + 2m - 1 = 0
(2) v6 nghiem.
CHI D A N

a) PhLfOng t r i n h (1) c6 hai nghiem neu 3m + 1 ;t 0 va biet thuTc
m ^ —

3
-.(3m + l)(m + 15) > 0
Tarn thufc theo bien m : A(m) = -3m^ - 46m - 15 c6 cac nghiem : m i = -15,
A = (2m +

- 4(m + 4)(3m + 1) > 0

<

m2 = - — CO gia t r i diiong (trai dau vdi - 3 ) vdi -15 < m < - —.
3
3
Ket luan : Phirong t r i n h (1) c6 hai nghiem neu -15 < m < - - .
b) Neu m = 4 phirong trinh (2) c6 dang : 5x + 7 = 0 c6 mot nghiem.
. fm ^ 4
PhUOng trinh (2) v6 nghiem neu \
^
'
^ •
[A = (m + l ) ' - 4 ( m - 4 ) ( 2 m - l ) < 0
A = -7m^ + 38m - 15 c6 gia t r i am (cung dau vdi - 7 ) neu m lay gia t r i
3
ngoai khoang hai nghiem la : m.^ ^ — va m2 = 5.

21.

DS : m < — hoac m > 5.
7
Tim cac gia t r i cua tham so m de bat phufong t r i n h
x^ - (3m - i)x + 3m - 2 > 0

(1)
dutJc nghiem dung vdi moi x thoa man I x I > 2.

CHI DAN

Tarn thufc fix) = x^ - (3m - l)x + 3m - 2 c6 biet thufc
A = (3m - 1) - 4(3m - 2) = 9(m - 1)'.
• Neu m = 1, A = 0=:> f(x) > 0 Vx e R, f(x) = 0 o x = 1, do do.
f(x) > 0 Vx:

> 2.
Hoc

6n luy?n theo CTDT mfln Todn THPT S 33


22.

Neu m vi 1 => f ( x ) = 0 <=> x i = 1, x = 3 m - 2.
Vdfi m < 1 t h i 3 m - 2 < 1 tap nghiem cua (1) la: ( - o o ; 3m - 2) u ( 1 ; +oo)
De X sao cho x | > 2 thuoc t a p n g h i e m t h i p h a i c6
- 2 < 3 m - 2 c^O < m < 1.
Vdfi m > 1 t h i 3m - 2 > 1 tap nghiem cua (1) la: ( - o o ; 1) u (3m - 2; +oo)
De t h o a m a n yeu cau b a i t o a n t h i p h a i c6
3m-2<2<=>lT o n g hap k e t qua : De

3
x

> 2 thuoc t a p n g h i e m cua (1) t h i m p h a i
4

lay gia t r i sao cho : 0 < m <
T r o n g m o i triTdng hop diidi day, h a y t i m cac gia t r i cua t h a m so m de

a) phirofng t r i n h : (m - l)x^ - 2mx + m + 5 = 0

(1)

CO m o t n g h i e m thuoc k h o a n g (0; 1).
b) phirong t r i n h : 4x^ - (3m + l ) x - m - 2 = 0

(2)

CO h a i n g h i e m thuoc k h o a n g ( - 1 ; 2).
CHI DAN
a) Neu m = 1, phtTong t r i n h (1) c6 n g h i e m duy n h a t x = 3 g (0; 1).
Neu m ^ 1, phirong t r i n h (1) la phiiong t r i n h bac h a i . A p dung d i n h l i
dao ve dau cua t a m thiJc, phiTong t r i n h
fix) = (m - l)x^ - 2mx + m + 5 = 0
CO m o t n g h i e m t r o n g khoang (0; 1) neu va chi neu
fIO).fll) < 0

o

(m + 5)4 < 0 o

m < -5.

b) PhiTcfng t r i n h (2) c6 h a i n g h i e m t r o n g khoang ( - 1 ; 2) neu m lay cdc
gia t r i thoa m a n cac dieu k i e n sau :
A = (3m +

+ 16(m + 2) > 0

af (-1) = 4[4 + (3m + 1) - m - 2] > 0
af (2) = 4[16 - 2(3m + 1) - m - 2] > 0
,
S
3m + l
-1 < —=
< 2
2
8

-3 < m < 5

- 8 < 3 m + 1 < 16

3
m > —
2
12
m < —

- 7 m + 12 > 0

9 m ' + 2 m + 33 > 0
2m + 3 > 0

23.

3
12
<=> — < m < — .
2
7

T i m cac gia t r i cua t h a m so m de phirong t r i n h
(m - 2)x^ - 2mx + 2 m - 3 = 0
(1)
CO h a i n g h i e m thoa m a n dieu k i e n - 6 < X i < 4 < X 2 .
£ 2 TS. Vy Th6' H^fu - NguySn Vinh CJn

34


CHI DAN
- 6 < x i < 4 < X2 o

af (4) = (m - 2)(10m - 35) < 0
af (-6) = (m - 2)(50m - 75) > 0

c>2
2
m-2
Tuy theo tham so m, hay so sanh 2 vdi nghiem cua phuang trinh
y - ( m + l ) x + m = 0(l)

CHI DAN
Ta CO : af(2) = lf(2) = 2 - m

24.

• Neu 2 - m < 0

o m > 2 t h i (1) c6 hai nghiem thoa man xi < 2 < X2.

• Neu 2 - m = 0 <=> m = 2 t h i (1) c6 cac nghiem 1 = x i < X2 = 2.
• Neu 2 - m > 0 o m < 2 t a xet them biet thufc
A = (m + 1)^ - 4m = (m - 1)^
S „ m+1 „ m-3
o\
2=
2=
< 0 (vi m < 2)
2
2
2
Vdi m = 1 ta CO nghiem kep x i = X2 = 1 < 2
Vdi m < 2 va m ^ 1 t h i X i < X 2 < 2.
25. Giai, bien luan theo tham so m he bat phiTOng t r i n h
(I)

x ' - (m + l)x + 2(m - 1) < 0 (1)
x' - (m + 2)x + 3(m - 1) > 0 (2)

CHI DAN
Tam thu-c ve trai (1) la : fix) = x^ - (m + l ) x + 2(m - 1) = 0

o X = m - 1 hoac x = 2
• Neu m - 1 < 2 o m < 3 t h i (1) c6 tap nghiem
• Neu m - 1 > 2

Tj = [m - 1; 2].

m > 3 t h i nghiem cua (1) la : Ti = [2; m - 1].

• Neu m = 3 t h i T i = 121.
Xet tam thufc ve trai cua bat phtTcfng trinh (2) :
g(x) = x^ - (m + 2)x + 3(m - 1) c6 cac nghiem x = m - l v a x = 3.
• Neu m < 4 tap nghiem cua (2) la : T2 = (-00; m - 1] u [3; +co).
• Neu m > 4 tap nghiem cua (2) la : T2 = ( - c o ; 3] u [m - 1; +00).
• Neu m = 4 : T2 = {3}.
Tap nghiem T = T i n T2 cua he (I) nhif sau :
+ Neu m < 3 t h i (I) c6 nghiem chung duy nhat x = m - 1.
+ Neu 3 < m < 4 t h i T = [2; m - 1].
+ Neu m = 4 t h i (I) c6 hai nghiem chung T = {2; 3}.
+ Neu m > 4 t h i T = [2; 3] u !m - 1}.
HQC va 6n luy?n theo CTDT m6n Toan THPT SS 35


1.

§4. PHlIC(NG T R I N H , B A T P H U C I N G T R I N H

CHlfA D A U

GIA TRI T U Y E T DOI

KIEN THLfC
Phx:ifofng t r i n h chtfa d a u g i a t r i t u y ^ t d o i
La phucfng t r i n h t r o n g bieu thufc cua no c6 chufa a n n a m t r o n g dau gia
t r i tuyet do'i.
Vt
: I 2x - 1 1 - 3x = 2.
Dudng l o i g i a i phirong t r i n h
a

neu

l a diTa

t r e n d i n h nghia

a > 0

-a
neu
a < 0
ta phan chia tap xac d i n h t h a n h nhufng tap nho de thay the bieu thufc c6
gia t r i tuyet doi bkng bieu thufc khong c6 gia t r i tuyet doi ti/cfng diTOng.
Dac b i e t : | A(x) | = | B(x) i o (A(x)f = (B(x)f'.
2.

B a t phu:cifng t r i n h chtfa d a u g i a t r i t u y $ t d o i
Cac phtrong t r i n h don g i a n chufa dau gia t r i tuyet doi :

M o t so t i n h chat cija gia t r i tuyet doi can n h d k h i g i a i bat phuong
t r i n h chufa dau gia t r i tuyet doi.

2. a.b

1.

I fix) I < g(x) <=> - g ( x ) < fix) < g(x).

c)

I f(x) I > g(x)

b)

I fix) I < I g(x) I «

a)

3.

o

f ' ( x ) < g'(x).

fix) < - g ( x ) hoSc fix) > g(x).

-a
a

4.

a + b

+

BAI TAP
26.

G i a i cac phuong t r i n h :
- 2x = 7
(1)

b)

X + 4
a
CHI DAN

X +

a) Theo d i n h nghia I x + 4

(A)
(1)

«
(B)
.

4

2x + 3

vdi

- ( x + 4) v d i

X >
X <

3x - 2

= 3

(2)

-4
-4

Jlx + 4 ) - 2 x = 7
X

> -4

- ( x + 4) - 2x = 7
-4

X <

63 TS. Vu Thg Hi^u - Nguyln VFnh CJn

36

-x + 4 = 7

_
(B)

o

<=>x = - 3 ;

r-3x-4 =7

= > v 6
nghiem
X < -4
-4
Vay phuong t r i n h (1) c6 nghiem duy nhat x = - 3 .
b) De CO bieu thufc khong chufa gia t r i tuyet doi ttrcrng duong ta lap bang sau
(A)

<

X >

3
"2

X

2

3

2x + 3

-2x - 3

0

2x + 3

2x + 3

3x- 2

-3x + 2

Bieu thufc (2)

-5x - 1 = 3

- X

Nghiem

X = — (loai)
5

X = 2 (loai)

-3x + 2

0

3x - 2

+5=3

5x + 1 = 3
X = — (loai)
5

Vay phiiong t r i n h (2) v6 nghiem (theo bang c6 nghia la vdi x < —
2
4
thi (2) CO bieu thufc : - 5 x - l = 3 = > x =
khong thuoc khoang
3 nen loai).
5
-2

27.

Tuong tir nhir vay, ta loai x = 2 va x = —.
5
Giaix cac
phiTOng
t
r

i
n
h
:
^-x
2x-4
= 3 ( 1 ) b ) 2 x ' + 6x + 8 + x^ - 1
a)

=30(2)

CHI DAN

a) x ^ - x = 0 « x = 0; x = l , 2 x - 4 = 0 o x = 2
x^ - X

x^-x

-x^ + x

vdi X < 0
vdi 0 <

2x - 4 vdi

2x- 4

-2x + 4

vdi

X <

X >

2

X <

2

hoac

x >1

1

Ta lap bang de d i theo doi.
X

x^-x
2x- 4

0

-00

x^ -

0

X

-2x + 4

1

-x^ +

-2x + 4

Bieu thufc (1) x^ - 3x + 4 = 3 -x^ Nghiem

3±V5
(loai)
2

X

X +

2

0

x^ -

x^-x

X

-2x + 4

+00

0

2x - 4

4 = 3 x^ - 3x + 4 = 3 x^ +

- 1 +Vs
2

3±V5
— - — (loai)

Hoc

X -

4=3

- 1 + V29
2

6n luyOn theo CTDT mSn Toan THPT S

37

Phifdng t r i n h (1) c6 cac nghiem : X j =

- i + Vs.
_

1

_ - 1 + V29
X 9

b) Ta lap bang sau :
X

-00

-4

x'^- 1

Ix^-ll

-2{x^ + 6x + 8)

2|x^ + ex + 8l

2(x2 + 6 x + 8)

0

B i e u thiifc (2) 3x^ + 12x + 15 = 30

-2

-1

1

0 2(x^ + Gx + 8
x^ - 1

-x2-12x-17=30

2(x^ + 6x + g
0

-x^ + 1

(c)

+CO

2(x^ + 6x + 8)
0

x^ - 1

(d)

(e).

Nghiem

(c) o
_^ ^2x + 15 = 30
(d)
x^ + 12x + 17 = 30
(e)
3x2 _^ -^2x + 15 = 30
TCr cac phirong t r i n h trong cac khoang ta t i m dtfoc cac nghiem cQa (2)
la : X = - 5 , X = 1.
Giai cac phirong t r i n h :
2x-l
-3x = 1
b) 2x - 3
X + 4 =6

28.
a)

c) x 2 - x = 3 x - 1
CHI DAN

2
b) X = 1 va X = —
c) X = 1, X = ±3.
a) X = 3
11
29. Giai cac bat phuong t r i n h va bieu dien tap nghiem tren true so.
a) | 3 x - l | < 2

(1)
b) i'2x+*l| > 3 ' (2)
CHI DAN

a) (1)

<;> - 2 < 3x - 1 < 2 <=> - 1 < 3x < 3
«

—

3

< X <

2x + l > 3

1

p3

1.

2x + 1 < - 3

b) (2)

•/W/MW/////////////,

<=:>

x < -2

-2

1

x>l

Giai bat phuong t r i n h va bieu dien tap nghiem tren true so'.
_|x2 - 2 x - 3 f < 3 x - 3
(1)

30.

CHI DAN

(1)

o-(3x-3)<(x2-2x-3)<3x-3
rx<-3

0

fx' + X - 6 > 0

< X <

5

x>2

x ' - 5x < 0

0

< X <

<=> 2

< X <

5.

5

3 8 C3 TS. Vu Th6' Huu - Nguyin VTnh CJn


§5. PHlIOfNG TRINH, BAT PHlJOfNG TRINH CHlfA CAN THtJC
KIEN THCfC
1.

PhvfoTng t r i n h chufa c a n thijfc
PhiTofng phap chung de g i a i phtTOng t r i n h c6 bieu thufc chufa a n nam
diTdi dau can l a n a n g l e n l u y thCra bieu thufc cua phifcfng t r i n h v d i cac
dieu k i e n d i k e m de c6 diroc phucfng t r i n h k h o n g con chufa a n t r o n g
dau can. Cung c6 the dSt a n p h u de d a n d e n cac phifcfng t r i n h , he
phuong t r i n h don g i a n va de g i a i h o n . M o t v a i dang phifOng t r i n h
chufa can thiJc dan g i a n l a :

Jg(x) > 0
a) VfOO=g(x)
^ |f (x) = (g(x))'
f (x) = g(x)

b) V f U ) = Vg(x)
2.

[f(x)>0,

g(x)>0'

B a t phu!ofng t r i n h chufa c a n thiJc
Dang CO b a n cua b a t phiTOng t r i n h chura can bac h a i .
f (x) > 0
a)

g(x) < 0

4Ux)>gix)

g(x) > 0

•

f(x) > [g(x)]^
f (x) > 0
b) V f O O < g ( x )

ojg(x)>0

f (x) < [ g ( x ) ] '
De g i a i cac b a t phiTOng t r i n h chufa cSn thufc, t a dtfa r a cac dieu k i e n
xac d i n h r o i luy thCra m o t each t h i c h hop cac ve cua b a t phiTOng t r i n h
de g i a m d a n cac dau can thufc, d a n d a n dtTa t d i b a t phifong t r i n h , he
bat phufong t r i n h k h o n g chuTa cSn thufc. Cung c6 t h e dat cac a n p h u
hoSc b i e n l u a n cac ve cua b a t phuong t r i n h de t i m n g h i e m .

BAITAP
31.
a)

G i a i cac phiTOng trinh :
x - V2x + 7 = 4

(1)

b) V3x + 4 - V x - 3 = 3.

CHI D A N
a) (1)

V2x + 7 = X - 4

x-4 >0
2x + 7 - ( x - 4 f
Hgc vJ On luy§n theo CTBT m6n To^n THPT S

39

<=>

<

X >

4

x ' - lOx + 9 = 0

b) ( 2 ) o V3x + 4 = 3 + V x - 3 c ^

o 9(x - 3) = (x - If
32.

X >

4

-irx =

1

C:>X =

9.

3Vx-3 = x - 1

3x + 4 = 6 + x + 6 V x - 3

x>3

x-3>0

o x^ - l l x + 28 = 0 <^

x =4
x = 7'

Giai cac phuang t r i n h :

a) 2x - x^ + V e x ' - 12x + 7 = 0

(1)

b) V x ' - 3x + 3 + V x ' - 3x + 6 = 3
(2)
CHI DAN
a) Nhan x e t : 6x^ - 12x + 7 = 6(x - 1)^ + 1 > 0 Vx e R.
Dat t = V e x ' - 1 2 x + 7 vdi t > 0, ta c6 :
x^ - 2x =
7-t'
6

va phirang t r i n h (1) dan den
t = -1

+t =0

(loai)

t =7

o

Vex' - 12x + 7 = t > 0

6x' - 12x + 7 = t '

<=> ex^ - 12x + 7 = 49 c:> xi,2 = 1 ± V s .
b) Nhan xet : x^ - 3x + 3 =
Dat t =

x^ -

(2)

(2)o
33.

3x + 3, ta

3

\

+ - > 0 Vx
4

X

2

CO :

G

R.

0
0
t ' + 3t = 9 + 1 ' - et

V t ' + 3t = 3 - t

t >0

t >0

2t + 3 + 2Vt(t + 3) = 9

Vt + V t T 3 = 3

ot = 1

Vx' - 3 x + 3 = l c : > x ^ - 3 x + 2 = 0 o x = l

hoac x = 2.

Giai phiTcfng t r i n h : Vx - 2 + V4 - x = x ' - 6x + 11

(11)

CHI DAN
Ve phai (1) la : x^ - 6x + 11 = (x - 3)^ + 2 > 2 Vx
Theo bat dang thiJc Bunhiacopxki ve t r a i
V x - 2 + V 4 - X = l . V x - 2 + 1.V4 - X
4 0 GH T S . VU Th§' Huu - Nguyen VTnh Can


x-2 >0

Do do : (1)

4-x> 0
N/X-2

+

V4-X

34. Giai cac phuong t r i n h :
a) ^ / ^ 7 3 4 - ^ / I ^ = l (1)
CHI

= 3.

ox

x' - 6 x + l l = 2
=2

b) ^/^r75+^x + 6- = fe + l l (2)

DAN

a) Cdc/i i : Lap phUdng hai ve (1) ta difoc :

( D o (x + 34) - (x - 3) - 3^x + 34.^/^r^[^x + 34 - ^ x - 3 ] = 1
(Ap dung hang dang thuTc : (a - b)^ = a^ - b^ - 3ab(a - b))
( D o ^(x + 34)(x - 3) = 12 o (x + 33)(x - 3) = 1728
"x = 30
o x ^ + 31x - 1830 = 0 o

x = -6l'

Cdch 2 : DSt an so phu u = ^x + 34, v = ^ x - 3 ta c6 :
(1) o u - V = 1, u^ = 37
1 = (u - v)^ = u^ - 3uv(u - v) o 1 = 37 - 3uv => uv = 12
u + (-v) = 1

Ta dilgfc he phirong t r i n h

u ( - v ) = -12
u va - V la nghiem cua phifcfng t r i n h

- X - 12 = 0
o Xi = - 3 , X2 = 4
^x + 34 = 4
'^x + 34 = -3
x = 30
hoac
o
X--61'
^/^r^ = 3
^/^r^ = -4
b) Dat u = ^x + 5,
V = ^x + 6 ta thay u^ + v^ = 2x + 11. Suy ra
(u + v f = + v^ + 3uv(u + v) => 2x + 11 = 2x + 11 + 3uv(u + v)
o uv(u + v) = 0 o ^x + 5.^x + 6.^2x + l l = 0
=0
o

^x + 6 = 0
^2x + l l = 0

35.

x = -5
o

x = -6
X

11

= --

Giai, bien luan theo tham so m cac phifOng t r i n h :

a) Vx + m + V x - m = V2m (1)
CHI

b) Vx^ - 2mx + 1 + 2 = m

(2)

DAN

a) DK : m > 0.
• Ne'u m = 0 t h i (1) c6 nghiem x = 0.
X > m >0
• Neu m > 0(1) o

X

2x + 2Vx^ - m ^ = 2m

o

HQC

> m >0

4^

m

=m-

X

6n luygn theo C T D T mOn Toan T H P T 0 41


X > m

<=>m-x>0

o x =m

x^-m'=(m-x)'
Ket luan : V d i m > 0, phuong t r i n h (1) luon c6 mot nghiem x = m .
b) TCr bieu thiJc cua phirong t r i n h t a t h a y dieu k i e n xac d i n h m > 0.
m-2>0
(2) o

Vx^ - 2mx + 1 = m - 2

x ' - 2mx + 1 = ( m - 2)2

m > 2
| ( x - m ) ' - 2m^-4m +3

(*)

V i 2m^ - 4 m + 3 = 2(m - 1)^ + 1 > 0 V m n e n phirong t r i n h (*) luon c6

36.

2 n g h i e m p h a n b i e t xi,2 = m ± V2m^ - 4 m
Ket luan : N e u m < 2 t h i (2) v6 n g h i e m ,
2 nghiem.
V d i cac gia t r i nao cua t h a m so m phifcfng
V3 + X + V6 - X - V(3 + x)(6 - x) = m

+3
n e u m > 2 phiTofng t r i n h c6
t r i n h c6 n g h i e m .
(1)

CHI DAN
Cdch 1 : D K X D : - 3 < x < 6. DSt u = VSHOC > 0,V = V6 - x > 0. K h i do
(1) <=> u + V - uv = m
(2)
(2)

=>

uv = u + V - m
(u + v ^2
) ' = ..2
u ' +. . 2 + 2uv = (3 + x ) + (6 - x ) + 2uv
= 9 + 2uv = 9 + 2(u + V - m )
(u + v)^ - 2(u + v ) + 2m - 9 = 0
u +V = 1+Vl0-2m

^

ju^ +

(loai u + v = 1 - V l 0 - 2 m )

=9

Ta t h a y u + v = Vu^ +
+ 2uv > Vu^ +
= 3 ( v i uv > 0)
Mat khac, theo b a t dSng thufc Co si - Svac
u + V = l . u + l . v < V ( l ' + l ' ) ( u ' + v ' ) - 3V2
V a y 3 < u + v < 3 V 2 ^ 3 < 1 + VlO - 2m < 3V2
~ ^< m < 3
2
6V2 — 9
Ket luan : Ydi
< m < 3 phiTcfng t r i n h (1) c6 n g h i e m .
2
Cdch 2 : D u n g phiTOng phap khao sat h a m so.
Dat t - V3 + X + V 6 - X vdri - 3 < x < 6.
T i m gia t r i \dn n h a t , gia t r i n h oV6
n h -a txcua
e n [^- 3 ; 6 ] . T a
- V3t +t rX
3 c6 :
t'(x) =
—,

=—,
,—=^ = 0 o X =—
2V31 + X 2 V 1
6-X
2V3 + X . V 6 - X
2
M i n t(x) = m i n - | t ( - 3 ) ; t - ; t(6)^ = min{3; 3^2; 3} = 3
v2y
42 ^

TS. VQ ThS' H^u - Nguygn VTnh C?n


Max t = 3V2
Vay t e [3; 3>/2]
Ta c6:

= (3 + x) + (6 - x) + 2^(3 + x)(6 - x) => V(3 + x)(6 - x) =

Ve trai cua phiTOng t r i n h (1) t r d thanh :
^t^-9^
= - - t ^ + t + - = m v d i t e [3; 3V2]
fit) = t 2
2
Khao sat SLT bien thien cua fit) tren [3; 3V2] ta thay :
f'(t) = - t + 1 < 0 Vt
t

[3; 3V2]

G

^ f t

3yf2

6\f2i — 9

TCf bang bien thien cho thay :
< m < 3 t h i phirong t r i n h (1)
C O nghiem.
3 7 . Giai cac phuong t r i n h :
b) V3x + 4 - V x - 3 = 3
a)
N / X - 1 = 13
d)>/5x-l = V 3 x - 2 - V 2 x - 3 .
c) ^Jl6-x + ^|9 + x= ^
X

CHI

+

D A N

a)
= 10
c) X = 0, x = 7

b)

4 va = 7
d) v6 nghiem.

X

38.
a)

X

=

X

Vdi gia t r i nao cua tham so m phiiOng trinh diTdi day c6 nghiem :
X

+ m = 2Vx + 3

b) V x - 1 + V 3 - x - V(x - 1)(3 - x) = m

C H I D A N

a) m < 4
3 9 . Giai cac phiiOng t r i n h :

b) 1 < m < V2.

a) 2 ^ 3 x - 2 + 3 V 6 - 5 x - 8 = 0

(1)

(Trich de thi tuyen sinh DH khoi A - 2009)
b)

3V2

+

X

-

6V2

-

X

+

4N/4

-

X '

= 10 - 3 X
(2)
(Trich de thi tuyen sinh DH khoi B - 2011)

C H I D A N

a) DKXD : 6 - 5x > 0 o
he phifcfng t r i n h :

x < - . DSt u = ^3x - 2, v = N / 6 - 5 X ,

f2u.3v = 8
5u^ + 3v^ = 8

^

V

V

> 0 ta

C6

=

15u' + 4u' - 32u + 40 = 0
Hoc va 6n luy?n theo CTBT m6n ToSn THPT S

43


b)

V =

8-2u
<=>

<=> i
(u + 2 ) ( 1 5 u ' - 2 6 u + 20) = 0

ru =
V =

-2
4

=^ X =

-2.

D K X D : - 2 < X < 2 . D a t u = V2 + x , v = V2 - x t a c6 h e :
(I)

+

= 4

3u - 6 v + 4 u v = 3v^ + 4

Tii (*) => 3 ( u - 2 v ) = ( u +

= 4

u - 2v = 0

( D o

+

= 4

u - 2v = 3

2vf

3 u - 6 v = 4v' - 4 u v + u'(*)
u - 2v = 0
u - 2v = 3

(A)

(B)

H e ( A ) CO n g h i e m v ^ = — o
5

40.

2-x

= — =>x

5

= —.
5

H e ( B ) v 6 n g h i e m . V a y phucfng t r i n h (2) c6 n g h i e m d u y n h a t x = - .
5
G i a i cac phtrcfng t r i n h :

a) 2Vx + 2 + 2>/^7T - Vx + 1 = 4
b)

(1)

(Trich
de thi tuyen
(2)

V2x - l + x ' - 3 x + l = 0

(Trich

de thi tuyen

sinh
sinh

DH khoi
DH khoi

D D -

2005)
2006)

CHI D A N
a) Cdch

1 : B K x > - 1 , b m h phiTofng h a i v e .
2Vx + 2 + 2Vx + 1 = 4 + Vx + l

(1)

4(x + 2 + 2Vx + l ) = 17 + x + 8Vx + l
x>-l
'4x + 8 = 17 + x

o

o

x> - 1

Cdch

3.

X =

2 : N h a n x e t x + 2 + 2Vx + 1 = (1 + yfx + lf

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

(1) <=>

<=>

x > - l

b)(2)<»

suy r a

[Vx + 1 - 2

X =

<

^x>-l

3.

- x ' + 3x - 1 > 0

= - x ^ +3x-l<=> .

V2x-1

2x - 1 =: (-x' + 3x - x ^ + 3x - 1 > 0
x' -6x=' + l l x ' - 8 x + 2 = 0

44 S5 TS. Vu Thg' Huu - Nguygn VTnh Can


3-V5
2

< X <

3 + V5
c:>x = l v x = 2 - V 2 .

2

( x - l ) ' ( x ' - 4 x + 2) = 0
41.

T i m m de phifofng t r i n h sau c6 h a i n g h i e m thiTc p h a n b i e t .
Vx^ + m x + 2 = 2x + 1

CHI

(1)
(Trich

de thi tuyen sink DH kiwi B -

2006)

D A N

(1)

1

2x + 1 > 0

o

X >

- -

2
f (x) = 3 x ' + (4 - m ) x - 1 = 0(*)

x ' + mx + 1 = (2x +

N h a n xet phu'cfng t r i n h (*) c6 he so cua x^ va he so tu do t r a i dau nen
( * ) CO h a i n g h i e m t r a i dau. TCr dieu k i e n x > - — va h a i n g h i e m ciia
(*) t r a i dau, suy r a (*) p h a i c6 m o t n g h i e m thuoc

nUa khoang

. Dieu nay xay ra k h i m thoa m a n cac dieu k i e n sau :
>0

3f

S

1

4-m

1

o

m>-.
2

42.

>—
2 ^ 2
G i a i b a t phiTOng t r i n h 6: V x ' - 32x - 1 0 > (x - 2) (1)

CHI

D A N

Bat phu'cfng t r i n h (1) l a dang cO b a n . Neu x - 2 < 0 t h i m o i x de can
bac h a i c6 n g h i a deu n g h i e m diing (1). Neu x - 2 > 0 t h i (1) c6 n g h l a
vdi x^ - 3x - 10 > 0. K h i do 2 ve cua (1) deu k h o n g a m , t a b i n h
phiiong h a i ve t h i dau b a t phu'cfng t r i n h v a n giuf chieu. TCr l a p l u a n
t r e n t a c6 the v i e t n h u sau :
x-2>0
x-2<0
hoSc ( B )

(1) o (A)
x' - 3 x - 1 0 > 0
- 3x - 10 > (x - 2 ) '
X

(A)«

< 2
x<-2

C:>X<-2

x>5
!x>2

(B) o ^

x>2
x>14

X

> 14

X - - 3 x - 1 0 > x^ - 4 x + 4
Tap n g h i e m cua b a t phu'cfng t r i n h (1) l a h o p cac t a p n g h i e m cua he
(A) va he ( B ) :
T = (-oo; - 2 ] u [14; +oo).
43.
G i a i cac bat phiJOng t r i n h :

ai) V l 1 - X - Vx - 1 < 2 (1)
b) Vx + 3 - V7 - X > V2x - 8 (2)
Hoc va 5n luyen theo CTDT m6n Toan THPT S

45


CHI

DAN

ll-x>0
a) (1) c = > V l l - x < 2 + V x - l « - x - 1 >

0

o

ll-x<[2 +Vx-lf

1 < x < 11
4- x< 2Vx-l

'1
4-x<0
l < x < l l

o

{41

4-x>0

o

< X <

4

{4 < x < l l
o

1

< X <

4

2
-12x + 2 0 < 0

(4 - x ) ' < 4(x - 1)
C:>2 < X < 11.
b) (2)c^ Vx + 3 > V7 - X + V2x - 8
DKXD : x + 3 > 0 , 7 - x > 0 , 2 x - 8 > 0
'4 < X < 7
(2)

o4
X + 3 > (7 - x) + (2x - 8) + 2 V 7 - x V 2 x - 8
4
[4
V(7 - x)(2x - 8) < 2
4

< X <

^

[(7 - x)(x - 4) < 2
4
7

x <5

<=> <

-x' + l l x - 3 0 < 0

X >

6

_Vay t a p n g h i e m cua (2) la : 4 < x < 5 u 6 < x < 7.
44.

Giai b a t phirong t r i n h : V3x^ + 5x + 7 - VSx^ + 5x + 2 > 1 (1)

CHIDAN

o
D K : 3x^ + 5x + 2 > 0
x < - 1 hoac x > — . Dat t = 3x^ + 5x + 2 thi
3
(1) t r d t h a n h : V t + 5 - >/t > 1 <=> V t + 5 > l + V t o t + 5 > l + t + 2Vt
x<-l

<=>Vt<2

t > 0

=>0
2
X >

t < 4

o (-2; - 1 ] u

3

TS. Vu The'

- -

3
3 x ' + 5x - 2 < 0

-1

X <

<=> <

-2

46 ^

< X <

r

2
3

—

3

Nguygn Vinh CSn

Hifii -

45.

G i a i bat phuong t r i n h :

^^^^^^^ + V^T^ > 4=^

(D
Vx - 3
Vx ^ 3
(Trich de thi tuyen sinh DH khoi A -

2004)

CHI DAN

D K X D : x ^ - 16 > 0, X - 3 > 0<=>x > 4
(1)

»

V2(x' - 1 6 ) + x - 3 > 7 - x
X

>4

(A)

<=>

fV2(x' - 1 6 ) > 10 - 2x
<

X >

4

10-2x<0
x>4
10-2x>0

(B)

46.

X

>4

2(x'-16) >(10-2x)'
He (A) CO n g h i e m : x < 5, he (B) c6 n g h i e m 10 - ^/34 < x < 5
Tap n g h i e m cua (1) l a : x > 10 - V34.
G i a i bat phifofng t r i n h :

a) V5x - 1 - Vx - 1 > V2x - 4
b)

(1)
(Trich de thi tuyen sinh DH khoi A -

x + l + Vx' - 4 x + l >3V^

2005)

(2)
(Trich di thi tuyen sinh DH khoi B - 2012)

CHI DAN

a) D K X D : 5x - 1 > 0, X - 1 > 0, 2x - 4 > 0 c:> X > 2
B i n h phiidng h a i ve (1), chuyen ve t h i c6 :
x> 2
X > 2
<=>
(1) «
X + 2 >V2(x-l)(x-2)
[(x + 2f > 2 ( x ' - 3x + 2)
X >

2

x ' - lOx < 0

o

2

< X <

10.

b) D K X D : x > 0 , x ^ - 4 x + l > 0 c : > 0 < x < 2-S
Ta CO X = 0 l a m o t n g h i e m cua (2)
X e t X > 0, chia h a i ve cho \fx t h i diroc : V x +

hoSc x > 2 + Vs

Vx

+Jx+—- 4 >3
V
X

Dat t = Vx + 4= t h i X + i - t^ - 6 t h i (2) c6 dang :
Vx
X
t +Vt^-6>3

Vt'-6>3-t

t > 2
G i a i bat phtfOng t r i n h : V x +
> — t a duoc 0 < V x < —, hoSc Vx > 2
Vx
2
2
Suy r a t a p n g h i e m cua (2) l a : 0 < x < — hoSc x > 4.
4

HQC V§ 6n luy§n theo CTDT mfln Jo&n THPT El 47


1.

§6. HE PHUCfNG TRINH NHIEU AN
KIEN THLfC
H $ phi^ofng t r i n h b a c n h a t h a i a n , b a a n
a) H e phuong t r i n h bac n h a t h a i a n (x va y) c6 dang:
(I)

ajX + b i y ^ C j
a 2 X + b 2 y = C2

K i hieu D =

(1)

(2)
=

aib2 -

a2bi

goi la d i n h thufc cila he ( I )

b2

-

C2

a2

C:

ai

b2

C2

b,

Ci

= Cib2 -

-

aiC2 -

C2bi

a2Ci

Quy tdc Crame. G i a i he phifcfng t r i n h bac n h a t .
Neu D 0 he (I) CO nghiem duy nhat (xo; yo) xac dinh bdi cong thijfc
D.,y

Xo =

D '

_
yo =

D

- N e u D = 0,
^ 0 hoac Dy ;^ 0 h e . ( I ) v6 n g h i e m .
- N e u D = Dx = Dy = 0 he ( I ) c6 v6 so n g h i e m l a t a p n g h i e m cua phiiong
t r i n h : aix + b i y = C i hoac ciia a 2 X + b 2 y = C2.
b) Gidi he phiiang trinh bac nhat hai an bdng phuang phdp do thi
+ b j y = c,
(1)
Cho he phiiong t r i n h : ( I )

-

aiX

[aaX +

\y

= C2

(2)

T r e n cung m o t m a t phSng t o a do Oxy, ve cac dirorng thSng (di) c6
phiiOng t r i n h (1) v a duTcfng t h a n g ( d 2 ) c6 phtfcfng t r i n h (2). K h i do t o a
do (xo; yo) ciaa giao d i e m (di) v a ( d 2 ) l a n g h i e m cua he ( I ) .
N e u (di) v a ( d 2 ) giao nhau he (I) c6 n g h i e m duy n h a t .
N e u d i // d2 he (I) v6 n g h i e m .
Neu d i v a d 2 trCing nhau, he (I) c6 v6 so n g h i e m .
Toa do m o i d i e m cua (di) (hay ( d 2 ) ) l a m o t n g h i e m .

BAI TAPi
47.

G i a i cac he phtfofng t r i n h sau:

a) ( I )

2x - 3y = - 4

x-2

b) ( I I )

3x + y = 5

2

I2-X

+ y=7
+ 5y - 3

48 S TS. Vu Thg' Hi;u - Nguyen Vinh CSn


CHI D A N

a) A p dung quy tac Crame cho he ( I ) .
2
-3
-4
D =
= 11 ^ 0,
=
3
1
5
2

-4

3

5

Dx

-3

= 11,

1

= 22

= ^ . 2 2 ^ 2
—i
D
11
11
N g h i e m duy n h a t cua he ( I ) l a (1; 2).
Xo =

D

—

b) Dieu k i e n xac d i n h

ox

^ 2. DSt X =

t h a y vao ( I I ) t a

X

dxsgc he phuong t r i n h bac n h a t v d i X va y.
[

(ir)

3X + y - 7
- 2 X + 5y = 3
X -

A p dung quy tSc Crame cho he ( I D t a difcJc
y =
48.

G i a i , b i e n l u a n theo t h a m so m he phiTcfng t r i n h
(I)

32
17
23
17

81
32
23'
y =
17
X =

6mx + (2 - m)y = 3
(m - l ) x - m y = 2

CHIDAN

Ta t i n h d i n h thufc cua he ( I ) .
D =

6m

2-m

m -1

- m

= -Gm^ - (m - 1X2 - m) = -5m^ - 3m + 2

D =0»

- 5 m ^ - 3 m + 2 = 0 « m = - l hoSc m = 5
+ Neu m = - 1 , D = 0, Dx = - 3 ;t 0 he ( I ) v6 n g h i e m .
2
22
+ N e u m = - , D = 0, Dx =
^0 he ( I ) v6 n g h i e m .
5
5
+ N e u m ^ -1, m ^ —, D = - 5 m ^ - 3 m + 2 ^ 0, he ( I ) c6 n g h i e m duy
5
n h a t (xo; yo) v d i ,
3
2-m
m +4
D.
2
- m

Xo =
D
D
5m^ + 3m - 2
6m
m-1

9m+ 3
D

-5m^ - 3m + 2 ã
Hoc

6n luyĐn theo CTDT mfln Jo&n THPT 0 49


49. Giai cac he phiTcrng trinh:
X
2y _ 29
x + 2 y + 1 15
a) (I) 2x
y
8
x + 2 y + 1 15

X + 2y + 3z = 10
b) (II) 2x + 3 y - z = l l
3x - 2y + z = 6

(1)

(2)
(3)

CHI DAN

a) DKXD: x ^ - 2 , y ^ - 1 . DM X = x + 2 Y = y + 1 thi (I) trd thanh:
X + 2Y = 29
15
(D
_8_
2 X - Y = 15
3
2
Ap dung quy tSc
3 ^Crame
, . 3 cho
. ^he =(I')2 ta^ tinh difdc X = —, Y = —.
Giai
y+1 3 ^
5
3
x +2 5
Vay (3; 2) la nghiem cua he (I).
X + 2y + 3z = 10
(1)
b) (II) J2x + 3 y - z = l l
(2)
3x - 2y + z = 6
(3)
Nhan phiicrng trinh (1) v6i - 2 dem cong vao phiTcfng trinh (2). Lai nhan

phifong trinh (1) vdi - 3 cong vao phifOng trinh (3) thi dufdc
x + 2y + 3z = 10 (1')
(II)i - y - 7 z = -9 (2')
- 8 y - 8 z = -24 (3')
Lai tiep tuc nhan phiTOng trinh (2') cua (II)i vdfi - 8 , cong vao phifong
trinh (3') thi diTcfc
x + 2y + 3z = 10 x = 3
(11)2 - y - 7 z = -9
y = 2.
48z - 48
z = l ta c6 the lap rieng bang cac he
Ghi chu: De thuc hien phep giai tren
so cua he (II) va thiTc hien nhtr sau:
'I 2 3 10^ ^1 2 3
10^
2 3 10^
2 3 -1 11 —> 0 -1 -7 -9
0 -1 -7 -9
6;
0 48 48j
10 -8 -8 -24;

13 -2 1

50

E3 TS. Vu The' HiAi - Nguyen Vinh C5n


2.

Mpt so
phxiofng t r i n h h a i a n d a n g d a c b i $ t
a) He gom mot phiiang trinh bdc hai, mot phiiong trinh
ax + by + c = 0
(1)
(I)
[ A x ' + Bxy + C y ' + D x + Ey + F = 0 (2)

bdc nhdt

G i a i he (I) b k n g phLrong phap the.
b) He phiiong trinh dot xiing loqi I
L a he phufdng t r m h co dang <
l g ( x , y ) = 0 (2)
Trong do f(x, y) va g(x, y) la cac bleu thufc doi xufng doi v 6 i cac bien x, y.
Cdch giai: D a t a n so phu S = x + y, P = xy.
Dieu k i e n can va du de he c6 n g h i e m la
- 4 P > 0.
rf(x,y) = 0 (1)
c) He phiiong trinh doi xiing loai H: ( I I )
fly,x) = 0
(2)
Cdch giai: Di/a viec g i a i he ( I I ) ve g i a i he: (F)
d) Phuang

trinh

(III)

'f(x,y)-f(y,x) =0
f(y,x) = 0

dang cap

fi(x,y) = gi(x,y)

(1)

f2(x,y) = g , ( x , y )

(2)

T r o n g do m 6 i phuong t r i n h cua he l a m o t dSng thiJc cua cac da thufc
dang cap cung bac.
Cdch giai: G i a i he ( I I I ) v d i x = 0 hoSc v d i y = 0
V d i x ;t 0 dat y = k x hoSc v6x y 0 dat x = k y r o i khuf a n de doi ve
g i a i phirong t r i n h m o t a n .

BAI TAP
50. G i a i , b i e n l u a n theo t h a m so m he phi/cfng t r i n h
3x + 5y = 13 (1)
(I)
x ' + 3 y ' = m (2)
CHI DAN
TCf (1)

X = l i z ^ y . The vao (2) t h i diroc
3
\

13-5y
+ 3 y ' - m = 0 o 52y^ - 130y + 169 - 9 m = 0 (2')

B i e t thufc cua (2'): A' = 65^ - 52(169 - 9m) = 4 6 8 m - 4563
4563 39
• Neu m <
468

= — , A' < 0, (2') v6 n g h i e m => (I) v6 n g h i e m .
4
HQC va 6n luy§n theo CTBT mOn Tcrin THPT S 5 1


39
5
9
Neu m = — , A' = 0, (2') c6 n g h i e m kep y = — => x = —
4
'
4
4
He (I) CO n g h i e m

[9

5^

4' 4

Neu m > — t h i (2') c6 h a i n g h i e m y i 2 =

4
•
'
n g h i e m cua he ( I ) .
51. G i a i he phiTOng t r i n h

xy +

X

52

tCr do suy r a h a i

+ y = 11

x V + xy^ = 30
(Trich

de thi vdo DHGTVT

-

2000)

CHI DAN
Dat

X

+ y = S, x y = P, t a c6:

fS + P = 11
S.P = 30

S v a P l a n g h i e m cua phtrong t r i n h :
X =5
X =6

x +y= 5
[xy = 6

hoac

X

- I I X + 30 = 0

+y= 6

xy = 5

G i a i cac he t r e n t a difoc cac n g h i e m (2; 3), (3; 2), ( 1 ; 5), (5; 1).
3y =
52. G i a i he phtfong t r i n h : (I)
3x =

y^+2
x^ + 2

(Trich

de thi tuyen sinh DH khoi B - 2003)

CHI DAN
TCr cac phifdng t r i n h cua (I) suy r a x > 0, y > 0.
3xV = y' + 2

'3xV = y ' + 2

(I)«

3xy(x - y) = y^ - x^

^ 3 y ' x = x^ + 2

3 x V = y^ + 2
x-y = 0

3xV = y ' + 2

C5>

(x - y)(3xy + x + y) = 0

3x'-x'-2 = 0

(A)

3x'y = y ' + 2

3xy

(A)c^

+ X

(B)

+y=0

<z>x = y = 1

y =X

TCr dieu k i e n x > 0, y > 0. Suy r a : 3xy + x + y > 0, do do he (B) v6
n g h i e m . V a y he (I) c6 n g h i e m duy n h a t x = y = 1.
52 a TS. Vij Thg' HUu - Nguyen Vinh CSn


X
y
- +- = a
y
X

53. G i a i v a b i e n l u a n theo t h a m so a h e phtfcfng t r i n h (I)

x +y = 8
(Trich
CHI

de thi tuyen

DHQG

Hd Nqi khoi B - 1997)

DAN
2

D K : x y ^ 0, t a c6: <

2

+y

X

<=> <

= axy

x +y =8

(x + y ) ' = (a + 2)xy
x +y = 8

fS = 8
- 4 P > 0 t a c6: (I) c^l
(a + 2 ) P = 64

Dat S = X + y , P = x y , d i e u k i e n
•

sinh

N e u a = - 2 he v6 n g h i e m .
8 = 8

• N e u a ^ - 2 , t a c6: <^

64

,
+

~ a+2

He CO n g h i e m k h i v a c h i k h i
- 4P = 64 - 4 . - ^ > 0 «
a+2

^—^ > 0
a+2

a < - 2 hoac a > 2

K h i do X , y l a h a i n g h i e m c u a p h i i O n g t r i n h :
z

64

- 8z +

X =

a +2

4+4

tufc l a
y =4 - 4

= 0 <=>zi,2 = 4 ± 4

a-2

x =4 - 4

a +2

hoSc

a-2

y =4+4

'a + 2

a-2

a +2
a-2
a +2
a-2
a +2

+ Vdi - 2 < a < 2 he v6 nghiem.
2x
54. G i a i h e p h i / o n g t r i n h :

+ l

.

y

(I)

(1)

^
X

2y + - = X
y
(Trich
CHI

de thi tuyen

sinh

(2)
vdo DHQG

Hd Noi khoi B - 1999)

DAN
C o n g v e vdfi v e , r o i trCr v e v d i v e t a dUOc:
^ x +y
3(x + y )
2(x + y ) +
=
—
xy
xy
o/

Y

2(x - y ) +

_

xy

II-

^(V

-

xy

f

(x + y ) 1 V

J

<

Y^

(X

1 +

-y)
V

1

^1

= 0

J

xy

2 ^

= 0

xyj

T i f do r u t r a diigc 4 h e p h i i o n g t r i n h sau:

HQC va 6n luyen theo CTBT m6n Toan THPT 0

53


x +y =0
x - y =0

X

+y =0

(loai)
xy

1- — = 0
xy

v6 n g h i e m

1-

1
xy

= 0

<=>

x - y =0

xy

x = ^/2; y = - V 2
X =

- V 2 ; y = V2

x =l ,

y - 1

x =-l, y = - l '

55. G i a i he phi/dng t r i n h :
a) ( I )

(2)

x ' + 2 x y + 3 y ' = 17

(1)

3 x ' + 2 x y + y^ = 1 1

b) ( I I )

x^-y^=7

(1)

x y ( x - y ) = 2 (2)

CHI D A N
a) V e t r a i c u a ( 1 ) v a ( 2 ) h e ( I ) l a cac d a thufc d a n g c a p b a c 2 d o i v d i x , y .
R o r a n g vdfi x = 0 h o a c y = 0 h e ( I ) v 6 n g h i e m .
G i a sijf X

x'(3 + 2k + k ' ) = l l

0, d a t y = k x t a dixac:

(1')

x ' ( l + 2 k + 3 k ' ) = 17 ( 2 ' )

C h i a v e vdfi v e (1") v a (2') t h i difotc
k' + 2k+ 3
11
4k2 - 3k - 10 = 0 ^ k = - - h o a c k = 2
3k' + 2k + 1
17

5
5
T h a y k = - — t a dtfoc y = — x , p h i f o n g t r i n h ( 1 ) t r d t h a n h
4
•
4
5

3x2 _^ 2 x — x
4

16

-X

=

11

<=> X =

±4^3

+5^/3

3

T h a y k = 2 t a se t i m dtTcfc x = ± 1 , y = ± 2

(4V3

N h i r v a y h e ( I ) c6 4 n g h i e m l a :

3

-5V3

-4V3

5V3

'

, ( 1 ; 2),

(-i;-2).
b)

N h a n x e t rSng x = 0 k h o n g n g h i e m diing he ( I I )
V d i x ;t 0, d a t y = k x t h i h e ( I I ) t r d t h a n h
x ' ( l - t = ' ) = 7 (1')

(ID

_f3

1

n

^ = - =:> 2 t ' - 5 t + 2 = 0
t(l - t)
2

x ' t ( l - t ) = 2 (2')

1
=> t = 2 h o a c t = 2
L a m t u o n g tii cau a) t a dtroc h e ( I I ) c6 h a i n g h i e m l a ( - 1 ; - 2 ) v a ( 2 ; 1).
5 6 . C h o h e phucfng t r i n h vdi t h a m so m .
(I)

(2)

x^ + y ' + x y = 3

(1)

x ' - y', + m ( x + y ) =

X

- y +m

V(Ji g i a t r i n a o c u a m t h i h e ( I ) c6 d u n g 2 n g h i e m .
54 E J TS. Vu The Huu - Nguyen Vinh C$n


CHI DAN
(Do

(x - y)(x + y) + m(x + y) = x - y + m

2

2

o

x ' +[xy+' +
y xy
- 1 =-30

o(A)

hoSc (B)

r ( x - y + m)(x + y - 1) = 0
'^^ x^ + y^ + xy = 3

X- y +m= 0

x^ + y^ + xy = 3
I x ' + y ' + xy = 3
He (A) CO hai nghiem (2; -1) va (-1; 2). De he (I) c6 dung 2 nghiem
thi he (B) phai c6 2 nghiem trung vdi cac nghiem cua he (A) hoSc v6
nghiem. Ro rang vdi moi m he (B) khong the c6 nghiem trung vdi cac
nghiem cua he (A). Vay can t i m gia t r i cua m de he (B) v6 nghiem.
(B)o

=
[3x' + 3ax + a ' - 3 = 0(2')

PhucJng trinh (2') v6 nghiem neu A = 9a^ - 12(a^ - 3) < 0
o a < -2V3 hoac a > 2>/3.

MOT SO HE PHl/dNG TRINH DAI SO KHAC
57. Giai he phirong t r i n h : (I)

CHI DAN
(Do

x^ + y + x^y + xy^ + xy = — (1)
4
x ' + y ' + x y ( l + 2x) = - (2)
4
(Trich de thi tuyin sink DH kiwi A - 2008)

x^ + y + xy(x^ + y) + xy = —(x^ +y)^ + xy = - -

Dat x^ + y = u, xy = V ta diroc he

u + uv + v = —
4
2
5
u'^ + V =

—

fu =

u^ - u - uv = 0
5

<=> i
u

+ V=

—

u

2

U^ + U +

2
u +

V=

1-

4
5

0

=0

+ V=

<=>

Hpc

(A)

5

2
u

4

u' + u + -

=

2
u^ +

5

V=

0

(B)

—

6n luy§n theo CTDT mSn Toan THPT El

55


(A)o

+ y =0
xy = - X

(B)<=>

J5
\

25

Vl6

+y =

3
xy = - -

He ( I ) CO h a i n g h i e m
58. G i a i he phirong t r i n h : ( I )

o

5.

X =

1, y = —
y
2

_J25
16

va

(2)

x^ + 2xy = 6x + 6

(1)

x" + 2 x V + xV^ = 2x + 9
(Trich

de thi tuyen sinh DH khoi B -

2008)

CHI DAN

\2

3x + 3

(x^ + xy)^ = 2x + 9
(Do

xy = 3x + 3 - - x '

= 2x + 9

(1')

xy = 3x + 3 -

(2')

x =0
(1')
- (x^ + 12x^ + 48x + 64) = 0 o x(x + 4)^ = 0
X = -4
4
V i x = 0 k h o n g n g h i e m dung (2') n e n k h o n g la n g h i e m cua he ( I )
17
V d i X = - 4 t a dirge y = — .
V a y he ( I ) c6 n g h i e m duy n h a t
59. G i a i he phtfong t r i n h :
xy + X + 1 = 7y
a) ( I )

xY

+ xy + 1 = 13y2

(x +

y)^-4 + l

-4;

'

17
4

j

(1)
(2)
(Trich de thi tuyen sinh DH khoi B (1)

x(x + y + 1) - 3 = 0
b) ( I I )

=0

2009)

(2)

(Trich de thi tuyen sinh DH khoi D - 2009)
CHI DAN
a) V d i y = 0 k h o n g n g h i e m dung he ( I ) , do do chia cac phtfong t r i n h cua
y

y

1

X

X + — + — =

he oho y ?t 0 t h i dugc

7

X

x ^ + -— h+ ^- = 13
y

56

y

Ea 15. Vu Thg' H\ju - Nguyin Vinh CSn