ĐẠI SỐ LỚP 10 NÂNG CAO TIẾP THEO

Khdi niem "nghiem" ciing dugc dinh nghia tuong tu cho cdc bdt phuong trinh dang fix > gix, fix < gix va fix > gix.. Bien ddi tuong duong cac bat phuong trinh Ciing nhu vdi phuang trinh,

Chuang U fifll f HUDTIG TRlnH

Bat ding thuc va bat phuong trinh la nhung khai niem ma

chung ta da lam quen 6 lop duoi Chuong nay se hoan thien

hon cac khai niem do, dong thoi cung cap cho chiing ta nhung

ki^n thuc mdi nhu van de xet dau cua nhi thiic bac nhat va dau

cCia tam thuc bac hai Chung co nhieu ung dung quan trong trong

viec giiii va bien lu^n cac phuong trinh va bat phuong trinh

Chung ta can nam vung cac kien thuc do, dong thdi ren luyen

kl nang ap dung chiing de giai cac bai toan trong khuon kho

ciHa chuong tnnh

103

t /^_

B A T D A N G T H l / G

V A C H U N G M I N H B X T D A N G T H t T C

1 On tap va bo sung tmh chat cua biit dang thurc

Gia sir a vab la hai sd thuc Cac menh dd "a > b", "a < b", "a > b'\ "a < b"

dugc ggi la nhung bdt dang thdc

Ciing nhu cac menh dd Idgic khac, mdt bdt dang thiic cd thd ddng hoac sai

Chiing minh mdt bdt ddng thiic la chiing minh bdt dang thiic dd diing

Dudi day la mdt sd tinh chdt da bidt ciia bdt dang thtfc

a>b a>b

Vi du 1 Khdng dung bang sd hoac may tinh, hay so sanh hai sd >/2 + >/3 vd 3

Gidi Gia suf V2 + \/3 < 3 Do hai vd ciia bdt dang thiic dd ddu dugng ndn

V2 + >^ <3 <^ (V2 + Sf<9 <^ 5 + 2V6<9

<^ 2yj6 < 4 <» V6 < 2 O 6 < 4, vd Ii

vay V2 + V3 > 3 •

104

Ndu A, B la nhiing bidu thiic chiia bidn thi "A > B" la mdt menh dd chiia bidn Chiing minh bdt dang thiic A>B (vdi dieu kien nao dd ciia cdc bidn), nghia la chiing minh menh dd chiia bidn A>B diing vdi tdt ca cac gia tri cua cac bidn

(thoa man didu kien dd)

Tvi nay, ta quy udc : Khi ndi ta cd bdt dang thiic A > B (trong dd A va B la

nhiing bidu thute chiia bidn) ma khdng neu didu kien ddi vdi cdc bidn thi ta bidu rang bdt dang thiic dd xay ra vdi mgi gia tri cua bidn thudc R

Vi du 2 Chiing minh rang x^ > 2(x -1)

Gidi x ^ > 2 ( x - l ) o x ^ > 2 x - 2 <» x ^ - 2 x + 2 > 0

o x ^ - 2 x + l + l>0<=> (x-l)^ + l > 0

Hidn nhidn (x - 1) + 1 > 0 vdi mgi x nen ta cd bdt dang thiic cdn chiing minh n

Vi du 3 Chiing minh rang ndu a, b, c la dd dai ba canh cua mdt tam giac thi

ib + c - a)ic + a- b)ia + b - c) < abc

Gidi Ta cd cac bdt dang thiic hidn nhidn sau :

2 2 2

a > a -ib-c) =ia-b + c)ia + b-c) b^ >b^-ic-af = ib-c + a)ib + c-a) c^ >c^-ia-bf = ic-a + b)ic + a- b)

Do a, b, c la dd ddi ba canh cua mdt tam giac ndn tdt ca cdc vd ciia cac bdt

dang thiic tren ddu duong Nhan cac vd tuong ling cua ba bdt dang thiic tren,

ta duoc

a V c ^ >ib + c- afic + a- hfia + b-cf '

Ldy can bdc hai cua hai vd, ta dugc bdt dang thiic cdn chiing minh D

2 Bat d^ng thurc ve gia tri tuyet doi

Tur dinh nghia gid tti ttiyet ddi, ta suy ra cac tinh chdt sau day

Sau day la hai bdt ddng thiic quan ttgng khdc vd gia tri tuydt ddi (vidt dudi

dang bdt dang thiic kep)

a\ -\b\<\a+b\<\a\ + \b\ (vdi mgi a, ft e M)

Ta chiing minh bdt dang thiic | a + 61 < | a | + 161 Thdt vdy

\a + b\<\a\+\b\ <^ ia + b)^<a +2\ab\+b^

<:>a+2jab + l3'<a+2\ab\+^<^ab<\a\\

Bdt dang thiic cudi ciing ludn dung ndn ta cd bdt ddng thiic cdn chiing minh

m l Sd dung bdt ding Ihdc vCfa ehCtng minh vd ding thdc \a\ = \a + b + i-b) \ 64 chdng minh bdt ding there \a\-\b\<\ a+b\

3 Bl(t Akng thurc giura trung binh cdng \k trung binh nh^n^^^

a) Ddi vdi hai sd' khong dm

a + b ,

Ta da bidt la trung binh cdng cua hai sd a va b Khi avab khdng am

thi 4ab ggi Id trung binh nhdn cua chiing Ta cd dinh li sau ddy

DINHU

Vdi mgi a > 0, ft > 0 ta cd

a + b

> yfab

Dang thiic xay ra khi va chi khi a = b

Ndi cdch khdc, trung binh cdng cua hai sd khdng dm ldn han hodc bdng trung

binh nhdn cua chiing Trung binh cdng cua hai sd'khdng dm bdng trung binh nhdn cua chiing khi vd chi khi hai sddd bdng nhau

(1) Ngudi ta con goi IJI bit ding tiiiic C6-si (Augustin-Louis Cauchy, 1789 - 1857)

Chitng minh Vdi a > 0, ft > 0, ta cd

^^-4ab =]-ia + b-2y[ab) = -i4a -Sf >0

Dodd

a + b

> yfab

Dang thiic xay ra khi vd chi khi (Va - >/ft)^ = 0, hie la a = ft

H2J Trong hinh 4.1, cho AH = a, BH = b Hay tinh

cdc doan OD vd HC theo a vd b Tir dd suy ra bat

ding thdc glOa trung binh edng vd trung binh nhdn

Chicng minh Gia sir hai sd duong x vd y cd tdng x + y = S khdng ddi Khi dd,

— = — > yfxy nen xy < — Dang thiic xay ra khi vd chi khi x = y

Do dd, tich xy dat gid tri ldn nhdt bang — khi va chi klii x = y

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Gia sir hai sd duang x va y cd tich xy = P khdng ddi Khi dd

^ ^ ^ > 4xy = yfP nen x + y > 2>/P,

Dang thiic xay ra khi va chi khi x = y Do dd, tdng x + y dat gid tri nhd nhdt

bdiig 2 v P Idii va chi khi x = y •

vay gia tri nhd nhdt cua ham s6fix) = x + — v d i x > 0 la/(>M) = 2v3

b) Doi vdi ba sd Ichdng dm

flf + ft + C

Ta da bidt la trung binh cdng cua ba sd a, ft, c Ta ggi Vflftc la

trung binh nhan cua ba sd dd Ngudi ta ciing chiing minh dugc kdt qua tuong

tu dinh Ii tren cho trudng hgp ba sd khdng am

Vdi mgi a > 0, ft > 0, c > 0, ta cd

3

Dang thiic xay ra khi va chi khi a = b = c

Ndi each khdc, trung binh cdng cOa ba sd khdng dm ldn han hodc bdng trung

binh nhdn cua chiing Trung binh cdng ciia ba sd'khdng dm bdng trung binh

nhdn cua chiing khi vd chikhi ba sddd bdng nhau

Vi du 6 Chiing minh rang ndu a, ft, c la ba sd duong thi

Khi ndo xay ra dang thiic ?

Gidi Vi a,b,c la ba sd duong nen

a + b + c > 3 ¥abc (dang thiic xay ra khi va chi khi a = ft = c) va

i,i4>33-i-'

a b c V abc dang thiic xay ra khi va chi khi — = — = — a b c

Do dd (a + ft + c) —+ - + -I I I

a b c >3Vaftc.3 3 / — =9 \ abc

Dang thiic xay ra khi va chi khi

Vdy dang thiic xay ra Idii va chi khi a = b = c

a = b = c

1-1-1

a b c

a

H3| Phat biSu kd't qua tuong tuhd qud d phin a) cho trudng hgp ba so duang

Cau hoi va bdi tap

1 Oiling minh rang, ndu a > ft va aft > 0 thi — < —

a ft

2 Chting minh rdng nira chu vi cua mdt tam gidc ldn hon dd ddi mdi canh cua tam gidc dd

3 Chiing minh rang a^ + b^ + c^ > ab + be + ca vdi mgi sd thuc a, ft, c

Ding thiic xay ra khi va chi khi a = ft = c

4 Hay so sdnh cdc kdt qua sau day :

a) V2000 + V2005 vd V2002 + V2003 (khdng diing bang sd hoac may tinh);

b) yla + 2 + yJa + A va yfa + yfa + 6 ia > 0)

109

1 1 4

5 Chiing minh rdng, ndu a > 0 v a f t > 0 t h i — + — >

a ft a + b

6 Chiing minh rdng, neu a > 0 va ft > 0 thi a^ + ft^ > abia + ft) Dang thiic

xay ra khi nao ?

7 a) Chiing minh rdng a^ + aft + ft^ > 0 vdi mgi sd thuc a, ft

b) Chiing minh rang vdi hai sd thuc a, ft tuy y, ta cd a^ +b^ > a^b + ab^

8 Chiing minh rdng, ndu a, ft, c la dd dai cdc canh cua mdt tam gidc thi

11 Chiing minh rang :

a) Ndu a, ft la hai sd ciing ddu thi —+— > 2 ;

ft a b) Ndu a, ft la hai sd trdi ddu thi - + - < - 2

Bdi doc them

BAT DANG THQC BU-NHI-A-C6P-XKI(^)

1 Bat dang thiic Bu-nhi-a-cop-xki doi vdi hai cap sdthi/c

Vdi hai cap sd thUc (a,b) va (x,y) ta cd

iax + byf < ia' + b\x- + y\

Ding thiic xdy ra khi va chi khi ay = bx

Chdna minh

D l dang chdng minh ding thdc sau :

iax + by-f + iay - bxf = ia^ + b\x^ + y \

Udt Ichac, do iay - bxf > 0 nen

iax + byf + iay - bxf > iax + byf

lit do suy ra bat ding thdc can chdng minh

Ding thdc x^y ra l<hi va chi l<hi ay-bx = 0, tdc la ay = bx

ChOv Khi xy ?t 0, dieu l<ien ay = bx cdn dUOc viet dUdi dang - = - •

X y

2 Bat ding thiic Bu-nhi-a-cop-xki ddi vdi hai bg ba sdth^c

C6 the chdng minh ket qud sau :

Vdi hai bd ba sd thuc (ai, a2, a;}),(bi, b2, b^, ta cd

(ai&l + ^2^2 + «3''3)^ ^ ('^i^ + ^ + ^3)(bi + bl + bl)

Neu fc, bob'ii^O thi ding thdc xdy ra khi v^ chi khi — = ^ = ^ •

15 Mdt khach hang ddn mdt cxta hang ban hoa qua mua 2 kg cam da ydu cdu

cdn hai ldn Ldn ddu, ngudi ban hang dat qua cdn 1 kg Idn dia cdn bdn phai

va dat cam len dia can ben trai cho ddn khi can thdng bdng vd ldn sau, dat qua cdn 1 kg len dia can ben trai va dat cam len dia can ben phai cho ddn khi can thang bdng Ndu cdi cdn dia dd khdng chinh xdc (do hai cdnh tay don dai, ngdn khdc nhau) nhung qua can la dung 1kg thi khach hang cd mua dugc diing 2 kg cam hay khdng ? Vi sao ?

16 Chiing minh rang vdi mgi sd nguyen duong n, ta cd :

19 Chiing minh rang ndu a, ft, c, d la bdn sd khdng dm thi a+b+c+d A4 > abed

20 Chiing minh riyng :

a) Neu x^ + y^ = 1 thi I x + y I < >/2 ; b) Ndu 4x - 3y = 15 thi x^ + y^ > 9

DAI CLfONG v £ B X T PHLfONG TRINH

1 Khai niem bat phuong trinh mot an

trinh mgt dn ; x ggi la dn sd (hay dn) vd 3) ggi la tap xdc dinh

ciia bdt phuang trinh dd

Sd XQ e 2) ggi la mdt nghiem cua bdt phuang trinh fix) < g(x) niufixf)) < g(xQ) la menh di diing

Khdi niem "nghiem" ciing dugc dinh nghia tuong tu cho cdc bdt phuong trinh dang

fix) > gix), fix) < gix) va fix) > gix)

Gidi mdt bdt phuong trinh la tim tdt ca cdc nghiem (hay tim /op nghiem) cha

bdt phuang trinh dd

CHU Y Trong thuc hanh, ta khdng cdn vidt rd tap xac dinh 2) cua bdt

phuang ttinh ma chi cdn neu didu kidn dd x e 9) Dieu kien dd ggi

la diiu kiin xdc dinh cua bdt phuang trinh, ggi tdt la dieu kien cua

Dudi ddy, chiing ta chi ndi tdi bdt phuong ttinh dang fix) < gix) Ddi vdi cac bdt

phuong ttinh dang fix) > gix), fix) < gix) vafix) > gix), ta ciing cd cdc kdt qua

Niufiix) < giix) tuang duang vdif2ix) < g2ix) thi tdviit

fiix) < giix) o /2(x) < g2ix)

H2| Cdc khang djnh sau ddy ddng hay sai ? Vi sao ?

a) x+ Vx-2 > Vx-2 <=>X> 0 ;

b) i4x^f <\ <^x-\<\

CHUY Khi mudn nhdn manh hai bdt phuang trinh cd cung tdp xac dinh 3) (hay cd cung didu kidn xac djnh ma ta ciing ki hieu la 2)) va tuong duong vdi nhau, ta ndi:

- Hai bdt phuong trinh tuong duang tren 3), hoac

- Vdi dieu Iden 3), hai bdt phuang trinh la tuofng duang vdi nhau

Vl du 1 Vdi didu kien x > 2, ta cd > l < : > l > x - 2 D

3 Bien ddi tuong duong cac bat phuong trinh

Ciing nhu vdi phuang trinh, d ddy chiing ta quan tdm ddn cac phep bidp ddi

khdng lam thay ddi tdp jighidm cua bdt phuofng trinh Ta ggi chiing la cac jdiep

bien doi tuang duang Phep bie'n doi tuang duang biin mdt bdt phuang trinh

thdnh mdt bdi phuang trinh tuang duang vdi nd Chang ban, vide thuc hidn cac

phep bidn ddi ddng nhdt d mdi vd cua mdt bdt phuang trinh va giii nguydn tdp

xac dinh cua nd Id mdt phep bidn ddi tuofng duong

Dudi ddy la dinh Ii vd mdt sd phep bidn ddi tuong duang thudng diing Cac ham

sd ndi ttong dinh Ii nay ddu dugc cho bdi bidu thiic

DINH U

Cho bdt phuang trinh fix) <gix) cd tap xdc dinh S), y = hix)

la mdt hdm so xdc dinh trin 2)

Khi dd, trin 3), bdt phucmg trinh fix) < gix) tuang duang vdi

mdi bdt phuang trinh :

I) fix) + hix) < gix) + hix);

2)/(x)/i(x) < gix)hix) niu hix) > 0 vdi mgi x e 2);

3)fix)hix) > gix)hix) niu hix) < 0 vdi mgi x s9)

Chicng minh Sau day, ta chi chiing minh ket luan 3) Cac kdt luan khdc ciing

dugc chiing minh tuong tu

Ndu XQ thudc 2) thi /(XQ), ^(XQ) va /i(xo) la cdc gia tri xdc dinh bang sd,

hon niia, vi ft(x) ludn dm ndn /i(xo) < 0 Do dd, dp dung tinh chdt cua bdt

ddng thiic sd, ta cd

fiXo) < giXo) <:> fiXo)hiXQ) > g(Xo)/l(Xo)

Tii dd suy ra rdng hai bdt phuang ttinh cd ciing tap nghidm, nghia la chiing

tuong duong vdi nhau D

H3J ChOng minh cdc khing dinh trong vi du 2

H4| Cdc khing djnh sau ddy dOng hay sai ?Visao?

a)x+ — <1 + — <:>x<l;

115

HE QUA

Cho bdt phuang trinh fix) < gix) cd tap xdc dinh 2)

1) Quy tdc ndng lin luy thiCa bdc ba

fix)<gix)^\fix)f<[gix)f

2) Quy tdc ndng lin luy thuca bdc hai Niu fix) vd gix) khdng dm vdi mgi x thudc 2) thi

fix)<gix)<:>\fix)f<[gix)f

Tuofng.tir, ta ciing cd quy tdc nang len luy thiia bac le va nang len luy thiia bac chan

H5| Giai bat phuang trinh sau ddy (bang each binh phuong hai vd), giai thich rd cac phep bien ddi tuong dUOng da thuc hien :

| x + l | < l x |

Cau hoi va bai tap

21 Mdt ban lap luan nhu sau : Do hai vd cua bdt phuong trinh v x - 1 < |x|

ludn khdng am ndn binh phuong hai vd, ta dugc bdt phuong trinh tuang duong

X - 1 < X Theo em, lap luan tten cd dung khdng ? Vi sao ?

22 Tim didu kidn xdc dinh rdi suy ra tap nghiem cua mdi bdt phuofng trinh sau :

24 Trong bdn cap bdt phuang trinh sau ddy, hay chgn ra cdc cap bdt phuong trinh

tuong duong (ndu cd):

a) X - 2 > 0 va x^(x - 2) < 0 ; b) x - 2 < 0 vd x^(x - 2) > 0 ;

c) X - 2 < 0 vax^(x- 2) < 0 ; d) x - 2 > 0 vax^(x- 2) > 0

BXT PHLfONG TRINH vA Hfi BXT PHLTONG

TRINH B Ac N H X T M 6 T X N

Trudc day, chiing ta da lam quen vdi bdt phuang trinh bdc nhdt mot dn Dd la

bdt phuang trinh cd mdt ttong cdc dang ax + ft<0, ax + ft<0, ax + ft>0,

ox + ft > 0, trong dd a va ft la hai sd cho trudc vdi a ^t 0, x la dn

HI] Cho bit phuang trinh mx < mim + 1)

a) Giii bdt phuong trinh vdi m = 2

b) Giai bdt phuong trinh vdi m = —\/2

Nhu vdy, ndu a va ft la nhiing bidu thiic chiia tham sd thi tap nghiem ciia bdt

phuong trinh phu thudc vao tham sd dd Vide tim tap nghiem ciia mdt bdt

phuong trinh tu^ theo cdc gid tri cua tham sd ggi Id gidi vd biin ludn bdt

phuang trinh dd

Dudi ddy, chiing ta chu ydu ndi vd each giai va bien luan bdt phuong trinh

dang ax + ft < 0 Ddi vdi cac bdt phuong trinh dang cdn lai, each giai cung

tuong tu

1 Giai va bien lu^n bat phuong trinh dang ax + b<0

Kit qua giai vd bidn luan bdt phuong trinh

ax + ft<0 (1) dugc ndu ttong bang sau day

1) Ndu a > 0 thi (1) <» x < — Vay tap nghidm ciia (1) la 5 =

a 2) Ndu a < 0 thi (1) <=> X > — Vdy tdp nghiem cua (1) la S =

a

3) Ndu a = 0 thi (1) o Ox <-ft Do dd :

- Bdt phuang ttinh (1) vd nghiem (S = 0 ) ndu ft > 0 ;

- Bdt phuang ttinh (1) nghiem dung vdi mgi x (S = R) ndu ft < 0

ft,, , r b]

V aJ ( b \

— ;+oo

\ a J

117

CHUY Viec bidu didn cdc tap nghidm tten true sd se rdt cd ich sau ndy, Chang ban, phdn khdng bi gach d tten hinh 4.2 bidu didn tdj nghiem cua (1) vdi a > 0

Hinh 4.2

Vi du 1 Giai va bien ludn bdt phuofng trinh

mx + l>x + m (2) Gidi Bdt phuofng trinh (2) tuong duong vdi

im-l)x>m^-l (3)

Ta cd

1) Ndu m > l t h i / n - l > 0 nen (3) «> x > ——- <» x > m + 1 m^

m^

m 1 -1

- 1 2) Ndu m < l t h i m - l < 0 nen (3) <=> x < <» x < m + 1

m-l 3) Ndu OT = 1 thi bdt phuong trinh ttd thanh Ox > 0 nen nd vd nghidm

Kit ludn : -Ndu w > 1 thi tap nghiem cua (2) la 5 = (m + 1 ; +oo)

- Ndu m<l thi td.p nghidm ciia (2) la 5 = (-QO ; m + 1)

- Ndu m = 1 thi tap nghiem cua (2) la ,S = 0 D

H2| TCrke't qua tren, hay suy ra tdp nghiem cQa bdt phuang tnnh

mx+i >x + m :

V i d u 2 Giai va bien ludn bdt phuang trinh

• 2mx >x + Am-3 (4) Gidi Bdt phuang ttinh (4) ttiong duang vdi

( 2 m - l ) x > 4 m - 3 (5)

1 Am-3

l ) N d u / n > - thi2m-l>0nen(5)<i>x>-^^^—^•

2 2 m - l 2) Ndu m < - thi 2m - 1 < 0 nen (5) o x < •

2 2 m - l

3) Ndu m= — thi (5) ttd thanh Ox > - 1 , bdi vay nd nghidm dung vdi mgi x

Kit ludn : - Ndu m > — thi tdp nghidm ciia (4) laS =

- Ndu m< — thi tap nghiem cua (4) la 5 =

- Ndu m = — thi tdp nghiem cua (4) la 5 = ^

2 Giai he bat phuong trinh b$c nhsi't mot an

Tuong tu nhu he phuang trinh, tap nghiem cua mdt hd bdt phuong trinh la giao ciia tdt ca cac tap nghiem ciia cac bdt phuong trinh ttong he

X + 1 > 0

Gidi Giai ldn lugt ttoig bdt phuong trinh cua he, ta dugc :

(6) (7) (8)

Tap nghiem cua (6) la S^ = — 0 0

x > - l Tap nghiem cua he bdt phuong trinh da cho la

S = /

Dd de xac dinh tap nghiem S, ta bidu didn cdc tdp nghiem tten true

sd bdng each gach di cac didm (phdn) Ichdng thudc tdp nghidm cua tiing bdt phuang trinh ttong hd, phdn cdn lai se bidu didn tdp nghiem cdn tim (h.4.3)

-1 -1

Hinh 4.3

H 3 | 77m cac gid tri cOa x di ddng thdi xay ra hai ding thdc:

|3x + 2| = 3x + 2 v a | 2 x - 5 | = 5 - 2 x

Hu&ng ddn.\A\ = A<:^A>Ova\B\ = -B<:>B<Q

Vi du 4 Vdi gid tri nao ciia m thi he bdt phuang trinh sau cd nghidm ?

rx + m < 0 (9)

\ - x + 3 < 0 (10)

Gidi Ta cd (9) o x < -m Tdp nghiem ciia (9) la (-00 ; -m\

(10) <:> X > 3 Tdp nghidm ciia (10) la (3 ; +00)

Vdy tap nghiem ciia he Id S = (-00 ; -m] n"(3 ; +00) Hd cd nghifim khi vd

chi khi S ?!^ 0 , hie la 3 < - m hay m < - 3 •

Cau lioi va bai tap

25 Giai cdc bdt phuofng trinh :

121

DAU CUA NHI THLfC BAC N H A T

1 Nhi thurc b$c nhat va d^u cua no

Nhidu bai toan ddn den viec xet xem mdt bidu thvcc fix) da cho nhan gid tri am (hoac duang) vdi nhiing gia tri nao cha x Ta ggi vide lani dd la xet ddu cda biiu thdc fix) Dudi day, ta se tim bidu vd nhi thdc bac nhdt va ddu cua nd

a) Nhi thurc b^c nhat

DINH NGHiA

Nhi thdc bdc nhdt (dd'i vdi x) la biiu thdc dang ax + b, trong ddava

ft la hai so cho trudc vdi a ^ 0

Ta da bidt, phuofng trinh ax + ft = 0 (a 9^ 0) cd mdt nghidm duy nhdt XQ= —

a Nghidm dd cung dugc ggi la nghiem ciia nhi thicc bdc nhdt fix) = ax + b No

cd vai trd rdt quan trgng trong viec xet ddu ciia nhi thiic bac nh^t fix)

b) Dau cua nhi thurc bac nhdt

Dat XQ = — , ta vidt nhi thdc bdc nh^t fix) = ax + b nhu sau

Khi X > XQ thi X - XQ > 0 nen ddu cua a(x - XQ) trung vdi ddu cua a

Khi X < XQ thi X - Xo < 0 nen ddu cua a(x - XQ) ttdi vdi ddu cua a

Ttt dd ta cd

DINH U (vd dau cua nhj thdc bac nhat)

Nhi thicc bdc nhdt fix) = ax + b cUng ddu vdi hi so a khi x ldn han nghiim vd trdi ddu vdi hi sd'a khi x nhd hem nghiim ciia nd

Kdt qua cua dinh li nay dugc tdm tdt trong bang sau

y

-\

0 a<0 '^

Nhi thiic da cho duang khi x < 1,5

vaamkhix> 1,5

Hinh 4.4

H1| Hdy giai thich bhng dd thi (h.4.4) cdc ket qud cOa djnh II trdn

2 Mot so umg dung

a) Giai bdt phuong trinh tich

Ta xdt cdc bdt phuong ttinh cd thd dua vd mdt ttong cdc dang P(x) > 0, P(x) > 0,

P(x) < 0, Pix) < 0 vdi Pix) Id tich cua nhiing nhi thiic bac nhdt

Vi du 1 Giai bdt phuong ttinh

- Sap xdp cac gia tri tim dugc cua x theo thii tu tang : - 1 , — 3 Ba sd nay chia

true sd thanh bdn khoang Ta xac dinh ddu cua Pix) ttdn tiing khoang bang cdch

Idp bang sau ddy ggi la bdng xet ddu cua Pix)

- 0 + +

+ 0

+

-+ 0 - 0 -+ 0

Trong bang xet ddu, hang ttdn ciing ghi lai bdn khoang dugc xet cua true sd,

ba hang tidp theo ghi ddu cua cdc nhan td bac nhdt trdn mdi khoang (dua vao

dinh li vd ddu cua nhi thiic bac nhdt) ; hang cudi ghi ddu cua Pix) ttdn mdi

khoang bang cdch ldy "tich" cua cac ddu cung cdt d ba hang ttdn

Dua vdo bang xet ddu, ta cd tdp nghiem cua bdt phuong trinh (1) la

5 = ( - o o ; - l ) u

b) Giai bat phuong trinh chura an d mdu

O day, ta chi xet cac bdt phuang trinh cd thd dua ve mdt ttong cdc dang

S ^ < ^' 5 4 - ^ ' 5 4 - 0 ' ^ > 0 , ttong dd ^(x) vd Qix) Id tich cua

Qix) Qix) Qix) Qix)

nhiing nhi thiic bdc nhdt Di giai cac bdt phuofng tririh nhu vay, ta lap bang xet

ddu cua phan thdc Pjx)

Qix) Khi lap bang xet ddu, nhd rang phai ghi tdt ca cac nghiem cua hai da thiic Pix) va Qix) len ttiic sd Trong hang cudi, tai nhiing

didm ma Qix) = 0, ta diing ki bidu || dd chi tai dd bdt phuang trinh da cho

Gidi Ta cd

x - 2 2x - 1 3(2x - 1) - 5(x - 2)

Tir dd suy ra tap nghidm cua (2) la 5 = (-oo ; -7] u | - ; 2 D

c) Giai phuong trinh, bat phuong trinh chura ^n trong ddu gia tri tuyet ddi

Mdt ttong nhiing each giai phuang trinh hay bdt phuong trinh chiia dn ttong

ddu gia tri tuydt dd'i la sir dung dinh nghia dd khd ddu gid tti tuydt ddi Ta

thudng phai xet phuofng trinh hay bdt phuong trinh ttong nhidu khoang (doan,

nira khoang) khdc nhau, tten do mdi bidu thiic nam ttong ddu gid tri tuyet ddi

ddu cd mdt ddu xdc dinh Sau day la mdt vi du don gian

Vi du 3 Giai bdt phuang ttinh

12x - 11 < 3x + 5 (4)

Gidi

1) Vdi X < - , ta cd

4 (4) <» 1 - 2x < 3x + 5 o 5x > ^ o X > —-•

1 4 1 Kdt hgp vdi didu kidn x < - , ta dugc —- < x < - Vdy tdp cdc nghiem thoa

-^ 3 z man didu kien dang xet la khoang

/ 4 1

I 5 2J

125

2) Vdi X > - , ta cd

2 (4) <» 2x - 1 < 3x + 5 <:> X > - 6 Kdt hgp vdi didu kien x > —, ta dugc x > — Vdy tdp cdc nghidm thoa man didu kien dang xdt la niia khoang !;+»

Tdm lai, tap nghiem cua bdt phuong trinh (4) Id

S = A j _

" 5 ' 2 u 1 ; + - | ; + o o |

Cau lioi va bai tap

32 Ldp bang xdt ddu ciia cdc bidu thdc :

1 - X 2x + 1 d)|(V2-V3)x+l|<V3+V2

b)

2 < 1

2x - 1 3 - X Ixl < 1

b) 3 - 2 x (3x - l)(x - 4) < 0 ;

/ ^ BXT PHLTONG TRINH vA

1 B^'t phirong trinh b$c nhat hai an

a) Bat phuong trinh bac nhdt hai ^n va mien nghiem cua nd

Mdi cap sdixQ ; yo) sao cho OXQ + ftyo + c < 0 ggi la mdt nghiem

cua bdt phuang trinh ax + by + c<0

Nghiem cua cac bdt phuang ttinh dang ax + fty + c > 0, ax + fty + c < 0 vd

ox + fty + c > 0 dugc dinh nghia tuong tu

Nhu vay trong mat phdng toa dd, mdi nghidm cua bdt phuang trinh bac nhdt hai dn dugc bidu didn bdi mdt didm va tap nghidm cua nd dugc bidu didn

bdi mdt tap hgp didm Ta ggi tap hgp didm dy la mien nghiim cua bdt

phuong trinh

Dudi ddy, chiing ta se thdy midn nghidm cua bdt phuofng ttinh bdc nhdt hai dn

la mdt nda mat phang

b) Gach xac dinh mien nghidm ciia bdt phuong trinh b^c nhl(t hai ^n

Vide xdc dinh midn nghiem cua mdt bdt phuang ttinh bdc nhdt hai dn (hay bidu didn hinh hgc tdp nghidm cua nd) ttong mat phdng toa dd dua ttdn dinh li dugc thda nhdn sau ddy

128

DINHU

Trong mat phdng tog do, dudng thdng id) : ax + by + c = 0 chia mat phdng thdnh hai nica mat phdng Mdt trong hai nica mat phdng dy ikhdng ki bd id)) gdm cdc diim cd tog do thoa mdn bdt phuang trinh ax + fty + c > 0, nica mat phdng cdn lai ikhdng

ki bd id)) gdm cdc diem cd tog do thoa mdn bdt phuang trinh

ax + fty + c < 0

Tur dinh 11, ta suy ra

Niu (XQ ; yo) Id mdt nghiim cua bdt phuang trinh ax + by + oQ ihay ax + fty + c < 0) thi nica mat phdng (khdng ki bd (d)) chiia diim M(xo ; yo) chinh Id miin nghiem ciia bdt phuang trinh dy

vay dd xdc dinh midn nghiem cua bat phuofng trinh ax + fty + c < 0, ta lam

nhu sau:

- Ve dudng thdng id): ax + by + c = 0 ;

- Xet mdt diim M(xo ; yo) khdng ndm trin id)

Ni'u axQ + ftyo + c < 0 thi nica mat phdng ikhdng ki bd id)) chda diim M la mien nghiem cua bdt phuang trinh ax + fty + c < 0

Ni'u axQ + byo + c> 0 thi nica mat phdng ikhdng ki bd id)) khdng chiia diim M la miin nghiim cua bdt phuang trinh ax + by+ c < 0

C H U Y Ddi vdi cac bdt phuong ttinh dang ox + fty + c < 0 hoac ax + fty + c > 0 thi niidn nghiem la niia mat phang ke ca bd

Vi du 1 Xdc dinh midn nghiem cua bdt phuong trinh 3x + y < 0

Gidi Tren mat phang toa dd, dudng thang ( ^ : 3x + y = 0 chia mat phdng

thdnh hai nda mat phang

9 afus6io(NC)-sT-A 129

Chgn mdt didm bdt ki khdng thudc dudng thang

dd, chang ban didm M(0 ; 1) Ta thdy (0 ; 1)

Ididng phai la nghidm cua bdt phuong trinh da

cho vay mien nghiem cdn dm la nda mat phang

bd id) khdng chda didm M(0 ; 1) (Trdn hinh 4.5,

midn nghiem la nda mat phdng khdng bi gach) n

H1| Xdc dinh miin nghiem cOa bat phuang trinh x + y>0 Hinh 4.5

He bat phurong trinh bac nhat hai an

Dudi day la mdt vi du vd hi bdt phuang trinh bdc nhdt hai dn

' 3 x - y + 3 > 0 (I) • - 2 x + 3 y - 6 < 0 2x + y + 4 > 0 Trong mat phang toa dd, ta ggi tap hgp cac didm cd toa dd thoa man mgi bdt

phuong trinh ttong bd la mien nghiem ciia he Vay midn nghiem cua he la

giao cac midn nghiem cua cac bdt phuofng trinh ttong he

Dd xdc dinh midn nghiem cua he, ta diing phuong phap bidu didn hinh hgc nhu sau :

- Vdi mdi bdt phuang trinh trong hi, ta xdc dinh miin nghiim cua

nd vd gach bd miin cdn lai

- Sau khi lam nhu trin ldn lu0 dd'i vdi tdt cd cdc bdt phuang trinh trong hi trin ciing mdt mat phdng tog do, miin cdn lai khdng bi gach chinh la miin nghiim ciia hi bdt phuang trinh da cho

Vi du 2 Xac dinh midn nghiem cua he bdt phuofng trinh (I)

G/a7.'Trudc hdt, ta ve ba dudng thang :

( J i ) : 3 x - y + 3 = 0 ;

id2): -2x + 3y - 6 = 0 ;

(fifg): 2x + y + 4 = 0

Thd ttrrc tidp ta thdy (0 ; 0) Id nghiem cua ca ba

bdt phuong trinh Didu dd cd nghia la gdc toa dd

thudc ca ba mien nghidm cua ba bdt phuang

trinh ciia he (I) Sau khi gach bd cac midn khdng

thich hgp, midn khdng bi gach trdn hinh 4.6

(khdng kd bien) Id mien nghiem cua he (I) n

H2| Xdc djnh miSn nghiem eOa he bdt phuang trinh

y-3x>0

x - 2 y + 5 < 0 5x + 2y + 10>0

Mot vi du ap dung vao bai toan liinh te

V& dd tim midn nghiem ciia he bdt phuang trinh bac nhdt cd lien quan chat

che ddn Quy hogch tuyin tinh Do la mdt nganh toan hgc cd nhidu dng dung

ttong ddi sdng va kinh td Sau day la mdt vi du don gian

Bai toan

Ngudi ta du dinh diing hai loai nguydn lieu dd chidt xudt it nhdt 140 kg chdt A

va 9 kg chdt B Td mdi tdn nguydn lieu loai I gia 4 trieu ddng, cd thd chidt xudt dugc 20 kg chdt A va 0,6 kg chdt B Td mdi tdn nguydn lieu loai II gia 3 trieu ddng, cd thd chidt xudt dugc 10 kg chdt A va 1,5 kg chdt B Hdi phai

diing bao nhieu tdn nguyen lieu mdi loai dd chi phi mua nguydn lieu la it nhdt, bidt rdng co sd cung cdp nguyen lieu chi cd thd cung cdp khdng qud 10 tdn nguydn lieu loai I va khdng qua 9 tdn nguydn lieu loai II ?

Phdn tich bdi todn Ndu sir dung x tdn nguyen lidu loai I va y tdii nguyen lieu loai II thi theo gia thidt, cd the chidt xudt dugc (20x + lOy) kg chdt A va (0,6x + l,5y) kg chdt B Theo gia thidt, x va y phai thoa man cac didu kidn :

0 < x < 1 0 v a 0 < y < 9 ;

20x + lOy > 140, hay 2x + y > 14 ;

0,6x + l,5y > 9, hay 2x + 5y > 30

Tdng sd tidn mua nguyen lieu la r(x ; y) = 4x + 3y

Bai todn da cho ttd thanh : Tim cac sd x va y thoa man he bdt phuang trinh

0 < X < 10

6 < y <9

2x + y > 14 2x + 5y > 30, (H)

sao cho r ( x ; y) = 4x + 3y cd gia tri nhd nhdt

131

Bai toan nay ddn ddn hai bdi todn nhd sau :

Bdi todn 1 Xac dinh tdp hgp (5) cac didm cd toa dd (x; y) thoa man hd (U) Bdi todn 2 Trong tdt ca cdc didm thudc (5), tim didm (x; y) sao cho Tix',y)

cd gid tri nhd nhdt

• Viec giai bai toan 1 chinh Id viec xdc dinh midn nghiem cua hd bdt phuong

trinh (II) ma ta da lap dugc

H3| Ki6m tra lai rang miin nghidm

iS) cOa he (II) Id miin tdgidc ABCD

tren hinh 4.7 (ki ca bidn)

• Di giai bai todn 2, ta thiia nhan

rdng bidu thdc Tix ; y) cd gid tri

nhd nhdt va gid tri dy dat dugc tai

mdt ttong cac dinh cua td giac

ABCD (xem bdi dgc them ttang

133) Bdng cdch tim toa dd cdc

dinh A, B, C, D rdi so sdnh cac gia

tti tuofng dng cua Tix ; y), ta dugc

r(5 ; 4) = 32 Id gid tti nhd nhdt

vay dd chi phi nguyen lieu it nhdt, cdn sd dung 5 tdn nguyen lieu load I va

4 tdn nguyen lieu loai II (khi dd, chi phi tdng cdng la 32 tridu ddng)

Hinh 4.7

Cdu iioi va liai tap

42 Xdc dinh midn nghidm cua mdi bdt pKuong trinh hai dn :

44 Mdt gia dinh cdn it nhdt 900 don vi prdtdin va 400 dan vi lipit trong thdc an mdi ngdy Mdi kildgam thit bd chda 800 dan vi prdtdin va 200 don vi lipit Mdi kildgam thit Ign (beo) chda 600 don vi prdtdin va 400 don vi lipit Bidt rdng gia dinh nay chi mua nhidu nhdt la 1,6 kg thit bd va 1,1 kg thit Ign ; gia tidn 1 kg thit bd la 45 nghin ddng, 1 kg thit Ign la 35 nghin ddng Gia sd gia dinh dd mua x kildgam thit bd va y kildgam thit Ign

a) Vidt cac bdt phuang trinh bidu thi cdc didu kidn cua bai toan thanh mdt hd bdt phuong trinh rdi xac dinh midn nghiem (5) cua he dd

b) Ggi T (nghin ddng) la sd tidn phai tta cho x kildgam thit bd va y kildgam thit Ign Hay bidu didn T theo x va y

c) 6 cau a), ta thdy (5) la mdt midn da gidc Bidt rdng T cd gia tri nhd nhdt tai

(xo; yo) vdi (XQ ; yo) la toa dd cua mdt trong cdc dinh cua (S) Hdi gia dinh

dd phdi mua bao nhidu kildgam thit mdi loai dd chi phi la it nhdt ?

Bdi doc them

M O T P H U O N G P H A P TIM CL/C TRj CUA BIEU T H a c

Pix ;y) = ax + by TREN M O T M I I N DA GlAC L 6 |

BAI T O A N : 77m gid tri, nhd nhdt hay Idn nhdt cQa biiu thdc P(x;y) = ax + by (b * 0)

trdn mdt miin da giac phing Idi (ki ea bidn)

Bdi todn dd c6 nghTa \d :

Cho bilu thdc Pix;y) = ax + by ib * 0) va mdt mien da giac Idi (5), l<e ca bidn, trong mat phlng toa dd Oxy Hay tim gia tri nho nhat (hay Idn nhat) ciia P{x ; y) vdi (x ; y) \d

tea d^ cCia cac diem thudc (S)

Cdch gidi Ta ludn c6 thd gl^ thidt rang b>0, bdi vi ndu ft < 0 thi ta cd the nh§n d i

hai vd vdi - 1 v^ b^i toan tim gia trj nho nhdt (hay Idn nhat) ciia Pix ; y) se trd th^nh b^i to^h tim gid trj Idn nhdt (hay nhd nhdt) cCia -Pix •,y) = -ax + b'y, trong do b' = -b> 0

133

Tap cac diem (x ; y) de Pix ; y) nh§n gia tri p la

dudng thing ax + by=p, hay y = — x + — fiudng

b b

thing nay cd he sd gdc bang — va cat true tung

b tai didm M(0 ; m) vdi m = ^ (h.4.8) Ki hieu dudng

b thing nay la ( d ^ ) V\ b > 0 nen viec tim gia trj

nho nhat (hay Idn nhat) cCia Pix ; y) = p vdi

(x ; y) e (5) quy ve viec tim gia trj nho nhat

(hay Idn nhdt) cCia m = —, tdc la tim diem M 6

b

vj tri thap nhat (hay cao nhat) tren true tung sao

cho dudng thing (J^) c6 ft nhat mdt diem chung

• Khi tim gia tri nho nhat ciia Pix ; y), ta cho dudng thing id^) chuyen ddng song

song vdi chinh nd td mdt vj tri nao dd 6 phia dudi mien (5) va di len cho den khi (d„)

Ian dau tien di qua mdt diem (XQ ;yo) nao dd cOa (5) Khi do, m dat gia tri nho nhdt

va tuong dng vdi nd la gia trj nho nhat ciia P{x ; y) Dd la

• Khi tim gia trj Idn nhat cCia Pix ; y), Xa cho dudng thing (cf^) vdi he sd gdc

-b chuyen ddng song song vdi chinh nd td mdt vj tri nao dd 6 phia trdn mien (S) vd di xudng cho den khi id^) Ian dau tien di qua mdt diem (XQ ;yo) riao dd ciia (S) Khi

dd, m dat gia trj Idn nhat va tuong dng vdi nd la gia trj Idn nhdt ciia Pix ; y) Dd la

PixQ;yo) = 0X0 +^3^0

cHoy

Qua each lam tren, ta thay rang7'(x fy) dat gia trj nho nhat (hay Idn nhdt)'

tai mdt dinh nao dd cOa da giac (S)

Ap dung each lam tren vao bai toan 2 neu trong §5, ta thdy khi (d„) di qua dinh

A(5 ; 4) thi m nhd nhat Dieu dd cd nghTa la Tix ; y) dat gia trj nho nhdt khi x = 5 v&

y = 4 Khi dd, 7(5 ; 4) = 32

Luyfn tap

45 Xac dinh midn nghidm ciia cac bdt phuang trinh hai dn :

a)x + 3 + 2(2y + 5 ) < 2 ( l - x ) ; ' b) (1 + V3)x-(1-V3)y>2

46 Xdc dinh midn nghiem cua cac he bdt phuang trinh hai dn :

' 3 x - 2 y - 6 > 0 b)

b) Trong (S), hay tim didm cd toa dd (x; y) lam cho bidu thvtcfix; y) = y - x c d gia tri nhd nhdt, bidt rkigfix; y) cd gia tri nhd nhdt tai mdt ttong cac dinh cua (5)

48 Bdi todn vitamin

Mdt nha khoa hgc nghidn ciiu vd tac ddng phdi hgp cua vitamin A va vitamin B

ddi vdi CO thd con ngudi Kdt qua nhu sau :

i) Mdt ngudi cd thd tidp nhan dugc mdi ngay khdng qua 600 don vi vitamin A

va khdng qua 500 dofn vi vitamin B

ii) Mdt ngudi mdi ngay cdn td 400 ddn 1000 don vi vitamin ca A ldn B

iii) Do tdc ddng phdi hgp cua hai loai vitamin, mdi ngay, sd don vi vitamin B khdng It ban ^ sd don vi vit^nin A nhung khdng nhidu hon ba ldn sd don vi vitamin A

Gia sii X vd y ldn lugt Id sd don vi vitamin A va fi ma ban dung mdi ngay a) Ggi c (ddng) la sd tidn vitamin ma ban phai tta mdi ngay Hay vidt phuang ttinh bidu didn c dudi dang mdt bidu thdc cua x vd y, ndu gia mdt don vi

vitamin A Id 9 ddng vd gid mdt don vi vitamin B la 7,5 ddng

135

b) Viet cac bat phuong trinh bidu thi cac didu kien i), ii), vd iii) thanh mdt hd bdt phuang trinh rdi xac dinh midn nghidm (5) cua he bdt phuofng trinh dd c) Hm phuang an dung hai loai vitamin A va 5 thoa man cac didu kidn tten dd

sd tidn phai tra la ft nhdt, bidt rdng c dat gid tri nhd nhdt tai mdt trong cac dinh

cua mien nghiem (5)

VAI NET VE LICH sCf QUY HOACH TUYEN TINH

Kup-man, Dan-dich vd Kan-to-rd-vich d Luc-xdm-bua ndm 1976

Td thdi cd dai, khi thuc hien cac cdng viec ciia minh,

loai ngudi da ludn hudng tdi each lam tdt nhat trong

cac each lam c6 the dugc (tim phuong an tdi uu trong

cac phuong an) Khi toan hgc phat trien, ngudi ta da

md hinh hoa toan hgc cac viec can lam, nghTa la bieu

thj cac muc tieu can dat dugc, cac yeu cau hay cac

dieu kien can thoa man bang ngdn ngd toan hgc de

tim Idi giai tdi LTU cho nd Td dd, hinh thanh nen cac bai

toan tdi uu

diuy hoach tuyen tinh la ITnh vUc toan hgc nghien cdu

cac bai toan tdi Uu vdi hdu han bien (an), trong dd,

muc tieu va cac dieu kien rang budc dugc bieu thj

bang cac ham sd, cac phuong trinh hay bat phuong

trinh tuyen tfnh (bac nhat)

Cd the ndi, ngudi dau tien quan tSm ddn Quy hoach tuyen tinh 1^ L V Kan-tofd-vich

(Leonid Vitalyevich Kantorovich, 1912 - 1986) Trong cudn "Cac phuong phdp toan hgc trong td chdc va ke hoach hoa san xuat" (NXB Dai hgc Qudc gia Le-nin-grat,

1939), dng d§ neu bat vai trd ciia mdt Idp bai toan Quy hoach tuyen tinh va de xudt

thuat toan sd bd de giai chiing Tuy nhien, Quy hoach tuyen tinh chi dugc nhidu ngudi biet den vao nam 1947, khi G.B Dan-dich (George Bernard Dantzig, 1914 -

2005) cdng bd thuat toan don hinh de giai cac bai toan Quy hoach tuySn tinh Cung

nam dd, T C Kup-man (Tjalling Charles Koopmans, 1910 - 1985) da chi ra rdng

Quy hoach tuyen tinh la cdng cu tuyet vdi dd phan tfch li thuyet kinh te cd dien

N&m 1975, Kan-to-rd-vich va Kup-man da dugc Vien Han lam Hoang gia Thuy Diln trao giai thudng Nd-ben ve khoa hgc kinh te

Ngay nay, trong thdi dai may tinh dien td, Quy hoach tuyen tinh van dugc tidp tue

nghien cdu nham tim ra cac thuat toan tdt hon

DXU CUA TAM THirC BAc HAI

1 Tam thurc b$c hai

DINH NGHIA

Tam ihdc bac hai (ddi vdi x) Id biiu thdc dang ax + bx + c, trong do a, b, c la nhiing so cho trudc vdi a ^ 0

Theo dinh nghia trdn, cac bidu thdc

fix) = -V2x^ + 3x + 1, gix) = x^ -5 va hix) = -o?

la nhiing tam thdc bdc hai

Nghidm cua phudng ttinh bdc hai ax^ + 6x + c = 0 ciing dugc ggi Id nghiim cua tam thicc bdc hai fix) = ax +bx + c

Cdc bidu thdc A = b^ - Aac vd A' = b'^ -acv6ib = 2b' theo thd ttJ ciing dugc ggi Id biet thdc vd biit thdc thu ggn cua tam thdc bdc hai fix) = ax +bx + c

Trong §4, ta da xet ddu cua nhi thdc bdc nhdt vd dp dung dd giai mdt sd bdt phuang ttinh Trong bdi nay va cac bdi tidp theo ciia chuong, ta se xet ddu cua tam thdc bdc hai va dp dung nd dd giai cdc bdt phuong trinh vd phuong trinh bdc hai ciing nhu mdt sd phuang trinh va bdt phuong ttinh khdc

2 Ddu cua tam thurc b^c hai

Ta se quan sdt dd thi cua ham sd bac hai di suy ra dinh li vd ddu cua tam thdc bdc hai/(x) = ax^ + bx + c

Ddu cha fix) phu thudc vdo ddu cua bidt tiidc A va bd sd a

137

Trong tiing trudng hgp, ddu cua^x) dugc ndu ttong cac bang sau :

1) A < 0 (tam thdc bdc hai vd nghidm)

Cung d^u vdi a

iaAx) > 0 vdi moi x e K)

2) A = 0 tam thdc bac hai cd nghiem kep XQ = —T—

Cung dSai | Khdc diful Ciing

vdi a Ovdia Od&vdia

Cac kdt qua tren dugc phat bidu ttong dinh li sau day

DINH U (vd dau cua tam thdC bac hai)

Cho tam thdc bdc hai fix) = ax^ + bx + cia^ 0)

Ni'u A < 0 thi fix) cUng ddu vdi hi sda vdi mgi x e

Ni'u A = 0 thi fix) cung ddu vdi hi sda vdi mgi x ^ —

2a Ni'u A > 0 thi fix) cd hai nghiim Xj vd Xj (x^ < X2) Khi dd, fix)

trdi ddu vdi hi sda vdi mgi x ndm trong khodng (xj ; X2) (tdc la

vdi Xj < X < X2), vd fix) cUng ddu vdi hi sd a vai mgi x ndm

ngodi doan [xj ; X2] (ticc la vdi x < Xj hodc x > X2)

CHUY

Cung nhu khi giai phuang ttinh bac hai, khi xet ddu tam thdc bac

hai, ta cd thd dung biet thdc thu ggn A' thay cho A vd ciing dugc cac

kdt qua tuong tu

Vf du 1 /(x) = 2x^ - X + 1 > 0 vdi mgi x e R vi tam tMc fix) cd A = - 7 < 0

va a = 2 > 0 •

Vf du 2 Xdt ddu ciia tam thdc bac hai fix) = 3x^ - 8x + 2

G/a7 Via = 3 >(yva fix) cd hai nghiem x, = i^Lll^.^ ^2 = i ± ^

(de thdy Xj < X2) ntn fix) > 0 (cung ddu vdi a) khi x e (-00 ; x{) u (x2 ; +00), va

fix) < 0 (ttdi ddu vdi a) khi x e (xj; X2) D

Cung cd thd ghi kdt qua ttdn ttong bang xet ddu cha fix) nhu sau :

X

fix) = 3x^-Sx + 2

- 0 0 Xj X2 +00

+ 0 - 0 +

HI Xdt ddu cda cdc tam thdc bdc hai sau:

SI) fix) = -2x^ + 5x + 7 ; b) gix) = -2x^ + XN/S - 7 ; e) hix) = 9x^ - 12x+4

139

Nhdn xet

Td dinh li vd ddu cua tam thdc bdc hai, ta thdy chi cd mdt trudng hgp duy nhdt

ttong dd ddu cua tam thdc khdng thay ddi (ludn am hodc ludn duong), dd Id

khi A < 0 Luc dd, ddu cua tam thdc trung vdi ddu cua hd sd a Do do, ta cd

Gidi Vdi m = 2 thi fix) = -2x + I ldy ca nhiing gid tri am (chang han

/(I) = -1) Do dd, gid trim = 2 khdng thoa man didu kidn doi hoi

Vdi m^2, fix) Id tam thdc bac hai vdi bidt thdc thu ggn A' = m - 1 Do dd

cau hoi vd bdi tap

49 Xet ddu cac tam thdc bdc hai sau :

a) (m^ + 2)x^ - 2(m + l)x + 1; b) im + 2)x^ + 2im + 2)x + m + 3

51 Tim cac gid tti cua m di mdi bidu thdc sau ludn dm

a)-x^ + 2 / n V 2 x - 2 m ^ - l ; b) (/n - 2)x^ - 2(/M - 3)x +/n - 1

52 Chdng minh dinh Ii vd ddu cua tam thdc bac hai

Hudng ddn Vdi cdc trudfng hgp A < 0 va A = 0, sd dung hd thdc da bidt

fix) = aix -xi)(x-X2) hay afix) = a ix-Xi)(x-X2),

ttong dd Xj va X2 la hai nghiem cua tam thdc bac hai fix)

B A T PHLTONG TRINH B A C HAI

1 Dinh nghTa va each giai

Bdt phuang trinh bdc hai^un x) la bdt phuang trinh cd mdt trong cdc dang fix) > 0,fix) < 0,fix) > 0,fix) < 0, trong do fix) Id mdt tam thdc bdc hai

Cdch gidi Di giai bdt phuong trinh bdc hai, ta dp dung dinh li vd ddu cua tam

thdc bdc hai

Vf du 1 Giai bdt phuang ttinh

141

Gidi Tam thdc bdc hai 2x^ - 3x + 1 cd hai nghiem Xj = - vd X2 = 1 va cd

he sd a = 2 > 0 nen

1 2x^ - 3x + 1 > 0 <=> X < — hoac x > 1

2 Vdy tap nghiem cua (1) la j -00 ; - j u (1 ; + 00)

Ta bidu didn tap nghiem cua (1) tten ttnic sd (h.4.9)

2 B^t phuofng trinh tich va bat phuong trinh chura an of m&u thurc

Vi du 2 Giai bdt phuong trinh

2 x ^ + 3 x - 2

x ^ - 5 x + 6 G/a7 Ta xdt ddu cua bidu thdc

Ddu cha fix) dugc cho ttong bang sau

-1

2

0 +

0 — 0

2 +

+

^ • 1

3 +

0 -

-0

1

+ + +

+00

• *