<span class='text_page_counter'>(1)</span><div class='page_container' data-page=1>

xuflr

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2015

s6

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nn

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pxd

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vA

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xoc co s6

Tru s6: 187B Gi6ng V6, Ha Ndi.

DT Bi6n tdp: (04) 35121607; DT - Fax Ph6t hdnh, Tri su: (04) 35121606

</div>
<span class='text_page_counter'>(2)</span><div class='page_container' data-page=2>

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;iti:ir:r.. .l

i'i.',so

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c

vE

ctnt

rHU'oNG

LE VAN THIEM

Gido su LO Vdn Thiemld Ch0 tich ddu ti0n

.

c0a Hoi Toan hoc Vi0t Nam. Ong ld nhd todn
'hqc..n6i ti5ng, co nhung dong gop lon trong

nghiOn cuu vd ung dung Todn hoc. Ong c0ng
ld mot trong nh0ng nguoi ddt ndn mong cho

'r gido dqc dai hoc

0

nuoc ta, nguoi thAy c0a

nhi6u thd he cdc nhd to6n hoc Vi6t Nam. GidLo

su LO Vdn Thiemluon ddnh sU quan tAm ddc
bi6t ddn viec gidng dqy todn hoc d cdc truong
pnd tnOng. Ong

la

mot trong nhting nguoi
sdng lAp h6 th6ng phd thOng chuy6n todn vd
Tap chi Toan hoc va Tudi tr6.

Gidi thudng

L€

Vdn Thi6m do Hdi Toan hoc
Viet Nam

ddt ra

nhdm g6p phdLn

ghi

nhAn

nh0ng thdnh tich xudt sdc c0a nhung thdy c0

gido vd hoc sinh phd thOng da khdc phuc kho
khan dd dqy vd hoc to6n gr6i, dong vien hoc
sinh di sAu vdo mon hoc co vai tro dac bi6t
quan trong trong su phdt tridn lAu ddi c0a n6n

khoa hoc nuoc nhd. Gidi thuilng LO Van Thi€m
cring ld su ghi nhfn cong lao crla Gido su

ld

Vdn Thi6m, mOt nhd todn hoc l6n, m6t nguoi
thdy da h6t long vi su nghiOp giAo duc.

HFgAI#'ruGLE

YAN

TTITEM

ruA&€

E#E€

HA

HUY

KHOAI (Vi6n

Todn

hqc

Viat Nam)

Tu khi ra dai, Giei thuhng Le Van Thi€m dd nhAn dugc su 0ng h0 to lon vd
tinh thdn vdr vAt chdt c0a cQng ddng toAn hoc vd xA hQi. Dac biQt, sau dip kjt

ni6m 40 ndm Vi6t Nam tham gia Olympic To6n hoc Qu6c t5, mot cuu hQc sinh
chuyen toAn

(di

dd nghi khOng n6u t6n) da 0ng ho Quy gi6i thuo'ng s5 ti6n

1 tyi ddng.

II.

GIAITHU'ONG

LE

VANITI{IEIVI 2014

Hoi Todn hqc Vi6t Nam quydi dinh trao Gidi thudng

L€

Van Thi6m ndm
2014 cho c6c nhd gi6o vd hoc sinh sau dAy:

'1.

_{Co gido}

_{Nguyen Ngoc}

_Xu&n,

_THFT

_{chuyen Hodng}

_Vdm_T'hu,
hloa tsinh

*

Sinh nam 1981.

*

Tham gia dEy ToAn ho'n 1 1 ndm, trong d6 dqy chuy6n Todn 10 ndm.

*

C6ng tdc trong mot truo'ng

gip

nhi6u kho khdn,

a Trong 3 ldn phu tr6ch chinh Doi tuydn dA c6 11 hoc sinh doat gidi Qudc
gia, 2 hoc sinh doat Huy chuong Bac Olympic To6n Singapore mo r6ng.
o Nhidu bdi vi6t, chuyen d6 cho ciic hOi thiio.

o 9 ndm ld gido vi6n dqy gi6i, chi6n si thi dua cdp Tinh nam hoc2013-2014,

giAo vi6n ti6u bidu kh6i THPT chuy6n tinh Hoa Binh.

a Khi ld hoc sinh dd tung doat gidi trong ky thi hgc sinh gi6i Qu6c gia,
2, Vu'ong Nguy0n Thuy Duong,

hoc sinh

THPT

chuydn

LO

_Qu!'00n,

Dd

Nang

*

Huy chucrng Vdng Olympic 30/4 todn midn Nam.

*

GiAi Ba hQc sinh gi6iToan Qudc gia nam 2013.

*

Gitii Nhdt hQc sinh gi6iTo6n Qudc gia

nim2014.

a Huy chuong Bac Olympic Todn Qu6c td ndm 2014.

3. NguyOn The

Hodn, hoc sinh

THPT

chuy6n

KHTN-DHQG Ha NOi

a Gidi Nhl hQc sinh

gi6iTodn

Qudc gia nam 2014.

*

Huy chuong

Viing

Olympic Todn Qu6c td ndm 2014.
o Guong mat tr6 tieu bidu DHQG Ha NOi.

o Guong

mit

tr6 tieu bidu th0 do Ha N6i ndm 2014.

4.

Trdn

H0ng Qudn,

hoc

sinh

THPT

chuy€n Thdi

Binh

*

GiAi Ba hoc sinh gi6i To6n Qu6c gia ndm 2013.
o GiAi Nhl hQc sinh gi6iToiin Qudc gia ndm 2014.

*

Huy chuong Vdrng Olympic Toan Qu6c t6 ndm 20'14.

5. Vo Quang Hung,

hoc

sinh

THPT

chuyOn ltlguy6n

Binh

Khi0m,

Quang

Nam

o Gidi Nhi Olympic Todn Hd Noi mo rong (2013).

o Huy chuong Bac Otympic Todn Duy6n hdi dOng

bing

Bdc BQ.

r

Giai Nhdt ky thi hoc sinh gioi Qu6c Gia ndm hgc 2013

_-

2014.

L6 trao gi6i da ctugc td chuc tai cu6c Gdp

m{t

ddu xudn

cta-H6i

Todn
hoc Vi6t Nam

tai

He N0i.

ngiy

71312015.

na.

</div>
<span class='text_page_counter'>(3)</span><div class='page_container' data-page=3>

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que

tr6n:

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phu<rng trinh:

x2

_{-(3+2a)x+40-a=0}

c6

nghiQm

nguy6n.

H6y tim

c6c nghiQm nguYdn <16.

Cflu 3

(1,5

di€m).1)

Cho

hQ phucrng

trinh

(

)x+mY

=3Y11

l*-Y

=m2

-2

v\i

x,y

lir

dn,

m

lir

tham s6.

tim

m

aC

tQ

phucrng

trinh

c6

nghiQm

duy

nh6t

(x;y)

thoa

mdn

x2

-2x-y>0.

2)

Cho

o,

b,

c

la.

d0

d}ri

ei6c thoa mdn di0u kiqn

iri

nho nh6t

cria

bi6u thuc

" b+c-a c+a-b

a+b-c

Ciu

4

(3

diim).

Cho

tam

gigc AB-C

c6

_!a

g6c

nhen, n6i ti6p

cluong

trdn

(O)

_{.(AB <}

AO.

Cic

ti6p tuy6n

vO

(O) t4i B

vi

C cdt nhau

t4i N. Vc

aaj,

dU

song song

vbi

BC. Dudng thdng

MN

cdt

dudng trdn (O)

tqi

Mvd

P.

1)

cho

ai6t

_fir+ftr=fi,

tintr

<10

ddi doan

BC.

2) chung minh

ring

_{# =#}

3) Chrmg

minh

ring

ile

,

ounelP

<l6ng

quy.

Ciu

5

(1,5

di€m).1)

Cho cluong

trdn

t6m

O _Q6n

kinh

1, tam

gi6c

ABC c6

c6c

<linh

A, B, C ndm

trong

ducrng

trdn vd

c6.diQn

tich lon

hon ho{c

bAng

1..

Chung

minh

rdng

tli6m

O

ndm trong

ho{c

ndm

trdn c4nh

cira

tam

gi6c

ABC.

2)

Cho

tfp

6={t;2;3;...;t6\.

HaY

tim

s6

nguydn ducrng

k

_{.nhd nh6t sao.cho trong m5i}

tfll

lon

gOm E

phin

tu

c;ua

A

dAut6n tai

hai.s6

phdn

biQt

a,b

md

a2

+bz

li

mdt

sd

nguy6n t6.

NGUYEN VAN XA

(GV THPT Yan Phong s6 2,

Bdc

Ninh)

Sru

tdm

ba canh cria

mQt

tam

2c+b=abc.

Tim

gi5

7

_xe,

_LECM

vir

a,EMA c(:

Ciil

_{"",, ";:}

chuns, EM

=E-.

EA

EM.

Do

d6 A,ECM

a

LEMA

(c.g.c)

+EMC=EAM.

Mit

iAil=6a,

ne"

ifri=6i

+AD

llBM=5h=614=fiA.

Vpy

tam

gi6c

ABD

c6n tPi.B.

c)

Gqi N=BJ\AD

thl

BJL

AD

tai

N

₌

N

h

trung

di6m

cin

AD

_=AN=DN=S-.

y61

6fgy

2

c6

NKttBM

_BM

_-N{=4.

Ta

c6

ffu=B=+ffy.

JB.

2

^e

frN=!frE

2^

ncn

fifi=frds.

X6t

AA,IN

vd

LMBO

c6

AJN=MOB,

suy ra
AAJN

A

LMOB

(g.g)

./N

AN

.,TN

AN

_-.JN-

AN

.=

oB=

MB-

2oB=

zMB'

JB-

2MB

laco

NK

_-

AN

(-l{\=rvr=4{

_+ex=NK=A!

_.

MB

2MB\

JBl

2

)

Ti

AK:

NKvd AN

:

DN

₌

KD

:

3KA,

rnd6

_#=+.

Bii

6.

Gie

sir

n=ab

(a, b ldc6c

cht

s6, a

I

0)
Theo

diu

bdi, ta

c6:

U+O.aUiaU+(lOa+b)ia+bt

a.

Ddt

b=ma(meN.,mcl0).

Tt

d6

;=10

a+b=l}a+ma

chia h6t

cho zaz.

N€n
1O at ma

=10

i

m>

m

e{t;

Z;

5\.

o

Nliu rr

:

1

thi

b:

a.Ta c6:llat

az*lli

a*a=7.

Dodoa:b:1,tac6

ab=17.

o N6u

z

:

2

rhl b

:

2a.Ta

c6 c6c s6

tZ

zq;

36. Cdc
s6

n;zq;36

th6a man dA bdi.

o Ntiu

z

:

5 ta c6

b:

5a

+bi.5.

NCn b

:

5, suY ra

a:

1.

56

oU=tS

(th6a mdn Ad Ua1l.

Vay c6

nlm

s6 thoa mdn d6u bdi, d6 ld:

ll;

12;

15;24;36.

NGUYEN

OTIC

TAN

gP.

HA

chi

Minh)

.

TONN

HOC

</div>
<span class='text_page_counter'>(7)</span><div class='page_container' data-page=7>

Chui'n

[i

cho lrilhi

tdt ns[i0p illPT

vi

thi uio

Oai hoc

trinlr

Q

hu.ng ta thdy rang trong

di

thi

Dai

hpc cdc ndm

v

_{gdn ddy, cac bdi}

_phuong

_trinh

_(P7),

_bdt

phtrong

trinh

(BPT),

h€

phuong trinh

(HPT)

duoc

gidi

bdng cdch nhdn

luctng hAn _,hW,

_.ddt,nhdn

ta

chung

dua

,i

pr,

BPT

tich ld riit phii

bi€n.

Nhiiu

bqn dqc thudng ddt cdu

hdi:

Co

sd

dO c6 cdch

gidi

nhu vQy ld

gi?

De

giilp

cdc em hoc sinh c6 co sd

dii

tim duqc

ldi gidi

bdi

todn bdng cach nhdn luqng

li€n

hqp, c[ing nhu

d6n

dau

cdu,v€

PT, BPT,,HPT

trong d€

thi

THPT

_Qu6c

gia

sdp

tbi

bdi

viAt. ndy

trinh

bdy

cct

sd dd

c6 ldi giai

cho cac cdu

vi

PT,
BPT,

HPT

trong.di

thi

Dai

hoc cdc ndm

trudc

ddy
th6ng quc mQt s6

thi

dq sau

Thi

dy

1.

(EH

kh6i

A

_-

2014)

Gidi he

phaong

Cdch 2. Ta

c6:

(1)e

_{li(r:fl=tz-x,lrfi}

)

y(12-

*')=Aa-2ax.tD1

+

x'?(tz-y)

o

lTy

_-144

+24x$2J

_-12x2

= 0

e

llxz -

zax

_"lnj

+

D(12

_-

y)

=

o

<+

l2(x

-JO))'z

_=o

<+

*=,12-t

[x>o

o<

_'

|.Y

=

12-

x2'

Thay y

_=12-

x2

vio

PT

(2)

ta

iluoc

x3

_{-8x -l=2JT6-x'z}

<> x3

_-8x

_-z

=z(Jto-V

-t)

<+(.r-3)(,r2+3x+1=p\

'

Jlo-

xz

+l

*

(,,-3)[(

x2

+3x

+l

y*Sf.l

=

o

' 'L'

'

Jto-xz

+tl

<)

x=3>

_J

=3

.

Thu lai

ta

dugc

nghigm

cira

HPT

U

(:;:).

Cach 3. DAt

;

₌

(x;

Jn:F),i,

=

(,ln

t

;

_Ji).

ra

c6

l;l

=lil=

Jo

Pr

(1)

ez(x.{rz-y

+$@-q)=z.tz

-, -, -r2

e2.a.b=a

+b'

_{e\a-b) =0e}

a-

b

*x-JO).rl'>o

rhay

y

_{=12- xz"uY#bjf,}

u-n"

xj_gx_l=2,1T0_x,

<)

x3

-8x-l

₌

z(Jio-7-t)

<+(x-3)(x:*:r*l;=ffi

. t

2(x$\

1

e(x-s)l

_{' 'L'}

(x:+3x+

l)+$!

_'

l=6

Jl0-xz+l-l

ta nrn,n-roru,

T?EI#8E

₅

Lrri

gi,rti

Cach

t.sr,

[-2$

t

x

<2Ji

-vuLtt

t'-tt'

l, <y<12'

laco:

.[*=*?,a,p@:fl=':+t-Do

d6

x,!tz-

y

+r5(u:*t)

. x2

+12-

y

_*y

+12-

_{x2 _}

_r,

22--'

vdv

i'J--\

PT

(1)

e

"

l'=

o

ly=12_x2'

Thay y

_=12-

x2

vdo

PT

(2)

ta clugc:
x3

_-8x

_

t=2,[19-*z

<=) x3

_-8.r-

z+z(t-JiO-r,

₎₌

o

(

_z(x+3)

₎

<+

(-r-:)l

\ /(

x2

+3x

+;

a-

-1]

_l=

0

(3).

1+J1o-x2

₎

Do

x >

o

suy

ra xz

+3x

+11-2('I1)-

,

g.
1+

J10-

x2
NCn

PT

(3)

<+

x:3

.YOi

x

=3

ta

_{dugc y =}

3.

Vfly

h0

phuong trinh

c6 m6t nghiCm

ta

(:;

:).

</div>
<span class='text_page_counter'>(8)</span><div class='page_container' data-page=8>

<>J=3-y=3.

Thir lai

ta

dugc

nghiQm cira

HPT

U

(:;3)

.

t

Binh

luQn: Qua ba"c6ch

gi6i

tr6n

ta

th6y

ring,

phuong

trinh

(l)

c6 th6

tlugc

su

lf

bang nhi6u crich

kh6c nhau, nhung sau

khi

thay y=12-xz

viro
phucrng

trinh

(2) ta dugc PT

p _gy_1=2[g_yz

(3).

Thi

PT (3) ttuqc gi6i

bing

c6ch nhdn

luqng

li6n hqp,
d4t nhAn tu chung dua v6

tich

ld tlon gian nhat.

Tuy

nhi6n

cin cir

niro

d6

ta

bii5n

d6i

PT

p

_gy-1=2$$*yz

(3)

thdnhPT

xr-8x4:2([6A-t)t

Xin

dugc

trinh

bdy

c[n

cf

d6 nhu sau:

-

Ddu

ti6n ta

dirng

M6y tinh b6

flii

(MTBT)

dC

tht

nghiQm, ta

th6y

PT (3)

c6 hai

nghiQm

ld x=-1

vd

x=3.

Tuy

nhi6n

ta chi

quan

tdm

t6i

nghiQm

x:3

md

kh6ng

quan t6m

t6i

nghiQm

x=-1

vi di6u

kiQn

c6

nghi6m

li

x>0

.

Nhu

vfy

ta

phii

ldm xudt

hiQn

d4i

lucrng

x-3

<16

dat

nh6n

tu chung,

nghia

ld

ta

phii

dua

PT (3)

.r,4

dang

(x-a)7(x)=o.

Mu6n vfly

ta phAi

tim s6

a

sao cho bi6u

thtc

ZJTO

_-

x'z

_-

a

sau

khi

nhdn

lugng li6n hqp xu6t

hiQn

dai

luqng

x=3.

- Do

PT (3)

chi

c6 rnQt nghiQm

x=3 n6n

cdch dcrn

gi6n nh6t dC

tim

a

ld ta

thay

x=3

vdo PT

2[g-x2

-a=0

ta

tim

dugc

a=2.

(Luu

y

ld

do PT

(3) c6 duy nh6t m6t nghiQm nguy6n n6n ta

mdi

ding

c6ch

tr6n,

nt5u

PT (3)

c6

hai

nghiQm nguy6n

ho{c

kh6ng

nguy6n

ta

sE

ding c6ch

kh5c

sE

dugc trinh

bdy 6 phdn sau).

- Do tl6 ta c6

PT

x3

_{-8x-1 =2JT6:P}

<+x3-8x-3=zJtO-x'-2.

Thi

dqt

2.

(DH

kh6i

8-2014)

Gidi

h€

phactng

trinh

f(t-.v,1/,.-,

+x:z+(x-y-t).F

(1)

fzy,

-:r+6)+

t

=ZJi2

-

J4x

-srt

Q)

Ldi

gidi.

DK:

(*).

Ta

c6

:

o

V6i

y = L, thay vdo

PT (2)

ta

tlugc

:

9-3x=0<>x=3.

o

_V6i

y

₌

x-1,

DK

(*)e

1 <

x

a2.PT

(2)

trd

thdnh:2x2-x-3=Jr-e

2(x2

_-

x

_-

1) +

(x

-r-

J2

_-

x)

=

o

r

₁

_)^

o(.r2

_.

_x_l)l

_,\

2*-.

_

t=n

x-l

+,12-x

₎

<> -r2

--r-1 -

0

:--,

'-)^- 2

_"

=

I

tJ5

'

Tt

d6

ta

tim

dugc

nghiOm

ctra hQ

phuong

trinh

td:

(3rr;,I

l*-6,-tI6)

\-

2

)

l

Binh lufln:

o

Khi

dgc

ldi

gi6i

HPT h6n,

chlc chin

nhi6u ban sE

d[t

c6u

h6i:

"Co

sd ndo AC Uii5n

d6i

PT (1) thenh PT

(3)?".

Sau il6y ld mQt cSch tt6

tri

ldi

c6u

h6i

tr6n.
-

Thay

x=3

vdo PT

(l)

tadugc:

(t-y)J:-y

+3=2+(2-y)Jy

(a).

Dirng

MrBr

tim

dugc nghiQm ctra

PT

(a)

ld

y=t

vit

y=/

.

-

Thay

x=4

vito PT (1) ta dugc:

(r-r)!4_)

+4=2+(z-y)J,

(b).

Dune

MrBr

tim

dugc nghiQm

cta

PT

(b)

ld

y=I

vd

y=3

.

-

Nhu v4y

h

rhdy

ring

v6i

x=3 hoic

x=4

thi

PT

(1)

1u6n 1u6n c6 nghiQm

y=l

. Ta dq do6n PT

(1)

c6

th6 dua dusc v6 apng

_{(y-1)/(r;r)=o}

.

-

Mat

kh6c

ta thiy ring

khi x=3

thi

PT (1)

c6
nghiQm

y=2 .Y-hi

x=4

thl

PT

(1)

c6 nghiOm

y=3.

Nhu

vfy

m6i

quan hQ

gita

x

vit

y

ld y=l"-1.

7u

dqr do6n

PT

(1)

c6

ttr6 dua

dugc

vd

dang

(x-y-t)s(x;r)=o

.

-

Tri

c6c nhfln x6t tr€n, ta dU do6n

rlng

PT (1) c6 th6
dua dusc vc aang

(r

-

t)(x

_-,

_-r)n(x;r)

= o .

Tri

d6

cho

ta dinh hu6ng

eC Uirin

d6i

PT (1)

thanh

Pr

(3).

-

Tuy

nhi6n, ntiu bdi niro cflng

phii

lQp lufln nhu h6n

thi

sE r6t m6t

thdi

gian, c6 16

t5t

hcrn h6t

le

c6c ban

phii

chiu

kh6 gi6i

nhiAu

bdi

tQp, d6

bi6n

k!

ndng
thdnh

ky

x6o,

sao

cho

"nhin

vdo

bdi

todn

ta

thiy"

ngay c6ch gi6i.

r Ngodi

ra, trong

khi

gi6i HPT tr6n

thi

d6n d6n PT:

2*z-*4=$-i

19.

Ta c6: PT (4) tuong duong

v6i

[v=o

lx -2y

>0

lax

-5y

-3

>

0

pr

(1)

o(1-y)(,*E-y-r)+("r-y-t)(t-fr)=o

1a;

(

,---L)=o

t-v)(x-v-t)t51;I

t+ly)

l=l

!:x-I'

<+(

*[

,.

TONN

H9C

</div>
<span class='text_page_counter'>(9)</span><div class='page_container' data-page=9>

z(x2

_{-x-t)+(x:-Jf-)=o}

(5)
C6c ban sE d6t c6u

h6i

tl6u

ld co

sd AC UlCn

d6i

PT

(4)

thenh

PT (5).

Xin

dusc

hinh

bdy co

sd

d6

nhu

sau:

- DAu ti6n ta dirng

MTBT

dC

thir

nghiQm, ta th6y PT

(4)

c6 nghiQm (gAn dung)

ld

x=1,618033989...Ni5u

nhAm nhanh ta th6y

ring

x=1,6180:agSg...=l+J5

.

2

Me

x=1+6

ld nshidm cria

PT

fl-x-l=0

ho[c

ld

2"

nghiQm

PT

_-xz+x+1=0.

Nhu

vQy

chring

ta

ph6i
ldm xu6t hiQn dai

lugng

fl-x-l

ho[c

_-*+x+l

d6

d[t

nhan tu chung.

- Do

PT

(4) chi

chria mQt cdn

thirc

n6n

ta

lQp lufln

ngin

ggn

nhu

sau:

Trong

PT

2*z-v-3=$4

ta
cAn

hm

xu6t hiQn bi6u

thirc

x2

-x-l

OC e6t

nlan tu

chung n6n ta bir5n AOi

pnan

2P-x-3

trudc vd

lim

xu6t hiQn

2(xz-x-l),

nghia

ld

ta

bi6n

d6i

PT
2az

_-y-3=Q-a

thinh

2(*-x-t)+(x-t-,0-x)=o

.

Nhmg

ntiu PT

chria nhi6u

hon

mQt

cin

thric

thi

ta
kh6ng thd ldm nhu tr6n dugc.

Xin

trinh

bdy phuong
ph6p t6ng qu6t nhu sau:

-

Trong PT

(4)

chfta

.84,

khi

d6 ta phdi th6m bort

mQt i14i

luqng

a

, nghia

ld

ta

biiln

OOi

_1D-x

tnanfr

Q-x-a

sao

cho

sau

khi

nhdn

lugng

1i6n

hqp

thi

xu6t hi6n bi6u

thric

xz

-x-l

ho(c

_-fl

+x+1

.

Luu

f

ld biilu thric

cAn ,,rr6t

hi6n ld

bQc

hai

n6n

a

khdng
ph6i ld mQt si5

md

a

phbic6 dpng

d=ax+b

.

-

Khi

d6

Jz-r

_-a=Jz-,

-(o,r+b)=

=9+\

-'-'"/

"lr-x+(ax+b)

Sau

khi

nh6n 1i6n

hqp

xong xudt hiQn

_-axz

ndnta

sE

cho

2-x-(ax+bf

_=-vra*a1

e(ax+b)z

= xz

_-2x

+1

>

ax+b= x

*l

.

- Nhu

vfy

ta phii th6m

vot

m6t dai luqng

ld x+l

vdo

pT 2xz_y_3=E_x

.

- Voi

phuonC ph6p ti5ng qu6t

tr6n,

c5c ban

c6

th€
gi6i ttuqc c6c PT, BPT chila

cln

thric bdng c6ch
nhAn luqng li6n hgrp mQt c5ch dE ddng.

- D6n d6y c6 16 c5c b4n sE th6y

ring

d6

giei

mQt PT
chria cdn

thirc

bing

cdch nhdn

luqng

li6n

hqp

don

giin

nhu th6 ndo.

Thl

dyt

3.

(DH

kh6i

D

_-

2014)

Giai

bdt phutrng

trinh

;

(.r +

t)'[i

+2

+(x

+6),[i +7

> rz +7 -r + t

2

(l).

Ldi gidi.

EK:

x

>

_-2.

V6i

di6u

kiQn

tr6n,

BPT (1) tuong

duong

v6i

(x

+

r)(J

x.+

z

_-

2) +

(x

+

6)(,En

_-

3)

-(x'z+2x-B)>0

<> ( _-r₊

l\-E-

₊( _{.r +}

o\-E-\

/Jx+2+2

\

/Jx+1+3

-(x-z)(x+4)

> o

<+(*-zt[#

_

+4

_--(..++)l

>oe)

'

'l,lx+2+2

,lx+1

+3

'

'-l

(^

Do

x)

_-2

n€n

)x+2-0.

suuru

[x+6>0

"

x+l

x+6 / ,\

x+2

_r r,-I4r=_

Jx+2+2'

Jx+1

+3

\^

'

'/

Jx+2+2

x+2 x+6 x+6

I

^

--r---/ll

2

'

Jx+1

+3 2

Jx+2+2

Do

cl6

BPT

(2)

a

x-2

< 0

<>

x<2.

So

s6nh

v6i

tli6u

kiQn

ta

dugc

nghiQm cira

BPT

ld

_-2<x<2.

t

Binh

luQn: Ta th6y

ring

bu6c

bitin d6i

BPT

(1)
thdnh BPT

(x +

r)(,{i

+Z _-z) + (x +

e)(8

+f

4)

_-(xz + 2 x _-8

) >

0

(3 )

h

m6u ch6t cria bdi gi6i.

Vfly

co s& 6 tl6u d6 c6 dugc
bu6c biiSn dOi trenZ

Xin

trinh

bdy

nhu

sau:

Diu

ti6n

ta

thay d5u "

)

"

bdi

d6u

":".

Nghia ld

ta thay BPT

bing

PT. Dung

MTBT

d6

tim

nghiQm ta th6y PT

(x+t).{i+T+(x+6)Jii=xz

+7

x+12

c6

mQt
nghiQm

duy nh6t

x=2.

Nhu vfly

ta

ph6i

ldm

xu6t
hiQn

dai lugng

x-2

A6 Aat nnan

trl

chung.

Khi

tl6 ta
can

tim hai sri

a

vd

_B

sao cho

HPT

_{E-"=o

llx+l

-B=0

c6 nghiQm

x=2

.

Thay

x=2

viro hg trCn ta

tim

dugc

1"^=:

Ddy chinh ld co so o,i ui6n aoi

ser

1t;

lF=3

thenh BPT (3).

Thi

d1r

4. (DH

khii

B

_-

2013)

Giai

hQ phao'ng

trinh

l2rt

+

yt

_-3ry

+3x

_-2y

+l=0

(1)
1

\+r'

-r'

+x+4=,{Tx+y

+,tx;$

Q)'

</div>
<span class='text_page_counter'>(10)</span><div class='page_container' data-page=10>

Liri

sidlr".

_"'-'

{2x+r

> o

l.x+4y

>

0'

Ta c6:

PT (1)

tucrng <luong

vdi

[--.-o-.,

r

(x+r-y)(zx+

I

_-y)

=

o

<)

l'

-

"^

:'

.

LY='+1

o

Vdi

_Y

_=2x*

1 _,thaY

vio

PT

(2)

ta

clugc

:

.[4x

+

|

+

_Jrx

+

4

=

3-3x

(3).

Ta

c6

_/(x)

=

J4x

+r

+

_Jgx

+

4

tl6ng

bir5n.

s(r)=3-3x

nghfchbi6n'

NOn

PT (3)

ntiu c6

nghiQm

thi

ld

nghiQm duy

nrr6t.

Ua /(o)=s(O)=3

nen PT (3) c6

nghiQm

duynh6t

x=0=)=1.

o

V6i

y

=

x+1,

thay vdo

PT

(2)

ta <lugc

PT

$i+l+J5x+4

=3x2

-x+3

(4)

<=

[#;;f

_-(.r

+

r)]

*

_[Js,

+a

_-(x

+

z)]

=3(x-1)x

(s)

^

-x2+x -

-xz+x

_

-al-:_,\

-.l3rTT+(x+l)

'

,[ix+++(x+2)

-\"

o(

.c

_-."\(

J-*..:l--

*:) =

o

'

'IJ3r*l

+(x+l)

_{^l5x+4+(x+2) )}

lx2-x=0

el

" l

II

- _I - rI-A

LJ3*+t

+tr+l)

'

..6Lx+4

+(x+2\

T

e

x2

_-x=o<+l

'[x=l'

*=o

Tt

d6

ta

tim

cluqc nghiQm cira

hQ

phucmg

trinh

u

(o;t),(t;z).

0 Binh lu$n:

Dring

MTCT

di5

tim

nghiQm, ta th6y
PT (4) c6 hai nghiQm

ld

r=0

vi

x=1

.

Md

x=0 vd

x=l h

nghiQm

ctra

PT

fl-x:O

ho{c

PT

-x2+x:0.

Tt

d6 ta c6 co

sO <tC Uitin

d6i PT

(4)
thAnh PT (5)

bing

mQt trong hai c6ch nhu sau

o

Cdch 1.

Ta

cAn

tim

hai

sii

a

vd

_B

sao cho PT

.$x+t-(ax+f)=O 61

c6 hai

nghiQm

ld x=0

vd

x=l

.

Thay

x=0

vd .r=1

vlro PT

(6)

ta dugc hQ PT

theo

a

vd

_B.Gi6i

HPTtheo

a,

_B

ta

dugc

a=1

vd p=1.

Tuong

t.u

ta

cdn

tim hai s5

a

vd

b

sao

cho

PT

$*aa-(ax+b)=g

0)

c6 hai nghiQm ld

r=0

vd

r=1.

Thay

x=0 vli

x=l

viro

PT (7)

ta

tluqc

hQ

theo

a

vd

b.

Gi6i

h0

HPT

theo

a

,

b

ta

dugc a=1 vir b=2

.

o

Cdch2.Tac6

$r

ar

_-(a

x+

A=-:(?!:

P):,

-'

$x+t+(a**9)

Cho

3;+1-(

ax+

p)2 =-1s2

+.r

ta dugc

(ax+

_B)'

=x2

+2x+l+qx+

B

=x+1

.

(Luu

y

ld

cho 3x+7-(ax+Bf

=-*'**

chir

kh6ng
ph6i

cho

3x+l-(ax+Pf

=x'-*1.

Tucrng t.u ta c6

5

* +4

_-(,x

+

d=Y!,

't

(?'.bl,

.7sx+++(ax+b)'

Cho

5x+4-(

ax+b)' =-vz

'"x

ta du-o. c
(ax+b)2

=7'

a4x+4>ax+b=x*2

.

Chirng

tdi

hy

vgng

ring

qua nhirng

thi

du

vd

nhimg

binh lufln tr6n

phAn

nio

sE

girip

c6c

em

hgc

sinh

t.u

tin khi g{p

c5c

bdi

to6n v€,PT,

BPT,

HPT trong

ki

thi

THPT Qu6c Gia

sdp

t6i.

DC

thdnh thao

phucrng

ph6p nh6n

lugng

li6n

hqp,

c6c em

hdy

thir ldm

c6c

bdi

tflp

sau

BAI

TAP

Gi6i

c6c

PT,

BPT, HPT

sau

t.

31,1V +

JV

+8

-z

=

Jx\r5.

2.

J,

a2',

J5*

a

+zJBx

+9

=

4x2.

n-1

-2

:.

Vzx-S *

rl-f

=

*.

4.

6x3

_-5xz

-l}x-tO+JZx-z+Jlx+z

<0.

s.

z.lr'**ll

+_r2

_4<-L.

1l x+4

,l*r+l

6.

_JTx+4-zJ2-x

rH.

J9x2 +16

(_

,

.

),tiltn

+ t=

4(x

+

t)'

+

Ji

^[*

+,

(fi

+a

+,tS

+5 =

J2

x, + J-2 xz +

4JIy

4A

.

N

TO6N

HOC

I

icruaEA

_s"tttt-"'o

-

|

Jix,

+txylry

+,{T*TIxyTff

=3(x+y)

.i.<_

</div>

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