luyện giải đề thi đại học 3 miền môn tóan theo từng chuyên đề

- Xac dinh mot so diem dac biet cua do thi, chang han tim diem uon, giao diem cua do thi vdti cac true toa dp trong triTdng hdp do thi khong c^t cac true toa dp hoSc vice tim tpa dp gi

Nh^ gi^o uu tij - TTi^c siToAn hpc NGUYEN VAN THONG (Chu bifen) Gl^ng vi«n chfnh - Thac sITo^n hpc NGUYEN VAN MINH

Ph6 gi^o su - mn si: NGUYEN VAN HIEU

Left noi ddu

Chiing toi la cac anh em ruot thjl, muo'n vie't quyen sach nay cho the he sau on

tap de chuan bi thi v^o dai hoc, vi bU'dc v^o triTdng dai hoc, ngtfcfi hoc sinh b^l

dau mot ddi song mdi vk c6 tiTdng lai tiTdi sang khi chpn dtfdc mot triTcJng dai hoc

tot, ch^c ch^n r^ng triT^ng nay se chon diem cao NhiT vay trong qua trinh on

luyen cac em can mot tai lieu tUcfng thich Dieu nay se diTdc thoa man neu cAc em

chiu kho on tap theo cac chuyen de mh chung toi bien soan, ch^c ch^n thi se dai

diTpc diem cao

Quyen sdch nay gom 20 chu de trong diem theo cau true de thi cua Bo giao

due hang nam, cac vi du duTa ra ttfcfng doi kh6 va c6 hiTdtng dan giai Tie'p theo

la 28 bp de thi cho cac khoi A, B, D cua tri/cfng chuyen Le Quy Don - Da N^ng

dilng de thi thijr tiTng nSm qua 66 danh gid diTdc hpc sinh, nit ra kinh nghiem

giang day TriTdng Le Quy Don - Da NSng hang nam deu c6 ti 16 dau vao cac

trirdng dai hpc la 100%

Rieng nam 2010 c6 so hpc sinh dat diem cao diTcfc xep thu" 4 trong bang xep

hang cua Bp giao due va Dao tao

Cic chuyen de chiing toi deu soan tilf can ban den nang cao, dp kho dii de

cho cac hpc sinh khd va gioi tU' luyen, neu thong hieu taft ca, ch^c ch^n ring cac

em se giai du'pc de thi Dai hpc mpt cdch de dang, nh\i chiing toi da tilfng luyen

rat nhieu em do thu khoa cic tru'dng Dai hpc danh tieng Dieu quan trong hdn

nifa vdi each vie't ciia mOt gido vien day chuyen Todn lau nim, nen each gpi

md, d i n d^t mang dam n6t t\i duy nang cao, hpc sinh se diTcJc phat trien kha

nSng Toan hpc khi hpc quyen sdch n^y

Mong muon ciia chiing toi l^m the nko de c^c em tif hpc to't mon Toan, va

kha nang thi dau v^o cdc tri/dng Dai hpc Idn phai nh5 vko sir kien tri cua cic

em "c6 cong mai s^t c6 ng^y nen kim" cd nhan day that diing vay

Quyen sich nay cung 1^ ky nipm 3 nim cic em hpc Toan vdi thay: Bich

Lien, PhiTdng Thao, Anh ThiT, Mai HiTdng, Thiiy Hien, Th^o Uyen Cic em da

giiip thay chinh sufa ban thao, mpt cich nhipt tinh trdch nhipm

Chiic c^c em se th^nh cong my man trong ky thi s^p idi

Chu bien: Nguyen VSn Thong (To trirdng To Todn trirdng chuyen Le Quy Don - Da NJng)

Nhd sdch Khang Vi$t xin tran trgng giai thi^ tai Quy dgc gia va xin

Idng nghe tnoi y kien dong gop, de cuan sdch ngdy cdng hay han, bo ich hart

Thu xin giH ve: „,.^- _

Cty T N H H Mpt Thanh Vien - Djch Van Hoa Khang Vi?t

71, Dinh Tien Hoang, P Dakao Qu^n 1, TP H C M

Tel: (08) 39115694 - 39111969 - 39111968 - 39105797 - Fax: (08) 39110880

Hoac Email: khangvietbookstore@yahoo.com.vn

'K^mdt m MJOl C«0 TRONG DI^M

GI(3l HAN vA TfNH UfeN TUC C6A MOT H A M S6

X ^ X ( ) _ ^ 1 lim g(x) L2 vdi

X-*X()

Neu 3s > 0 sao cho f(x) < g(x) Vx € (xo - e; x,) + s) va ton tai 1 , ,

lim f ( x ), lim g(x) thi lim f(x) < lim g(x)

Ne'u Pi(x„) = Qi(x,)) = 0 thi phan tich tie'p •

0

P,(x) = ( x - X o ) P 2 ( x )

Q , ( x ) = ( x - X o ) Q 2 ( x )

Q u ^ trinh khuT dang v6 dinh ^ la qua trinh k h u r cac nhan tuT chung (x-Xo)'' s e

difng l a i k h i nhan du'dc gidi han xac dinh ttfc la Qk 5^ 0

PhiTdng phap: SuT dung cdc hKng d^ng thtfc de nhan lien hdp 6 tuT va mau

nh^m true cac nhan tijr (x - x,,) ra khoi can thtfc

D e n day cac gidi han diTdc tinh theo dang 2

3 Gi6i hain v6 djnh —

00

PhiTdng ph^p: X 6 t I = l i m ^ v d i P(x), Q(x) la cac da thi?c hoSc cic ham

x-»xo g(x)

daj s6 G p i bac P(x) = p; bac Q(x) = q v^ m = min(p, q), k h i d6 chia ca va

mau cho x " ta c6 ket luan sau:

C6ngiyTNi:ii : ; W / / A / U ; ^ ,UI

• Ng'u p < q thi ton tai gidi han

• N e u p > q thi khong ton tai gidi han

4 Gi6i han dang v6 dinh « - oo

Phifdng phap: B i e n d d i diTa ve dang gidi han — '

1 0>t

T i m gidi han sau

Vx + V x - V x

l i m x-++<»L

(x + V x j - x

= l i m = l i m

J x^+«>^x + ^/^+^/x "

5 Gidi han dang v6 djnh oo 0 ,

Phifdng phap: Du'a ve dang v6 djnh —

6 Gidi h^n dang vd djnh ham lUging giac

PhtTdng phap: Suf dung cac ket qua gidi han cd ban sau dSy:

V

= l i m x->0 X x-»0

sin ax

I

ax sin ax

, sinax , sinax

= a h m a => h m = a

x ^ O X „ r x->() ax

bx

Sinax

h m sinax h m c o s b x = h m — c o s b x , sinax , ax a x ^ x-*() tanbx x->() sinbx x->obx sin Dx b

bx

Luyfn gidi di truOc kp ihi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyin Van ThOng

7 Gidi han dang v6 djnh 1°

8 Gidi han d?ng v6 djnh cua ham mu va Idgarit: ^ ^^'^ *

Phifdng phap: S^rdung l i m ^ ^ = 1; lim ^ = 1

Luy(n gUU d6 trade thi DH 3 miin Bdc Trung Nam Todn hoc - Nguyen Van ThOng

2^ 2 2"

= 3 - 1 = 2

e-^" - 1 Icos^x + c o s ^ x - l

•= Hm x->()

e" - V c o s x + l n ( l + x^) ' •!

x^ 2(l + V ^ ) Y x 2 ' l

HrfdngdSngiai

^^"'^ l n ( l + x^) + -x , 2 ,

4 ,

Luyfn giil di trudc thi DH 3 miin Bdc, Trung Nam Todn hpc - NgiQ^n Van Tlidng

C A C B A I T O A N K H A O S A T H A M s6

I T 6 M T A T L I T H U Y E T :

1 Cac bu6c khao sat si/ bien thien va ve do thi cua ham so

Bifdc 1 Tim tap xac djnh cua ham so

Btfdc 2 Xet sU' bien thien cua ham so

- Tim gidi han tai v6 cifc va gidi han v6 cifc (neu c6) ciia ham so Tim cic

dircJng tiem can ciia do thi (neu c6) l> "

- Lap bang bien thien cua ham so, bao gom: Tinh dao h^m cua ham so, xet

dau dao ham, xet chieu bie'n thien va tim ciTc tri cua ham so (neu c6), dien

cac ket qua vao bang ^.^ ^

BMcJ Ve d6 thi cua ham so ^ ^ j

- Ve cac du'dng tiem can cua do thi (neu c6)

- Xac dinh mot so diem dac biet cua do thi, chang han tim diem uon, giao

diem cua do thi vdti cac true toa dp (trong triTdng hdp do thi khong c^t cac

true toa dp hoSc vice tim tpa dp giao diem phtJfc tap thi bo qua phan nay),

- Nhan xet ve do thj: Chi ra true va tam doi xuTng cua do thi (neu c6, khong

yeu cau chufng minh)

2 Diem uon cua do thj: la diem U(X(); yo) ciia do thj sao cho lie'p tuye'n tai dd

di xuyen qua do thi, tiJc la ton tai mot khoang (a; b) chuTa diem Xo sao cho

tren mot trong hai khoang (a; x,,) hoSc (X(i; b) tiep tuye'n cua do thi tai diem

U nam phia tren do thi, con tren khoang kia tiep tuyen n^m phia du'di do thi

Ta thU'dng suT dung ket qua sau day de tim diem uon cua do thi:

Neu ham so y = r(x) cd dao ham cap hai tren mot khoang chuTa diem Xo,

f"(xo) = 0 va r ' ( x ) doi dau khi x di qua diem Xo thi U(X(); f(X())) la mot diem

uon cua do thi ham so'y = f(x)

3 Giao diem cua hai do thi

a Phu"dng trinh (xac djnh) hoanh dp giao diem ciia hai do thj y = l(x) va

y = g(x) (ciing ve tren mot mat phing tpa dp) la f(x) = g(x) (1)

b So nghiem cua (1) cung chinh la so giao diem cua hai do thi y = f(x) va

y = g(x) Dac biet: PhiTdng trinh (1) cd nghiem (v6 nghiem) khi va chi khi hai

do thj citt nhau (khong cat nhau) i

c Ta thudng gap u-irdng hpp phiTdng trinh (1) cd dang f(x) = m, trong dd m la

tham so va ham so f(x) khong chtfa tham so m TriTdng hdp nay ta cd bai

C6ng ty TNHH MTVDWH Khang VH

toan xdl giao diem cua do thi y = f(x) vdi drfdng thing y = m (vuong gdc vd true tung va c^t true tung tai diem cd tung dp m)

4 SI/ tiep xuc cua hai dudng cong •

a Dinh nghla: Hai ham so' f(x) va g(x) tiep xiic nhau tai diem M(xo; y,,) neu M

la mot diem chung cua chung va hai di/dng cong cd chung tie'p tuye'n vdi nhau tai M

b Dinh It: Hai diTdng cong y = f(x), y = g(x) tiep xuc vdi nhau khi va chi khi he

ff(x) = g(x) phiTdng trinh an x sau day cd nghiem: \

[f'(x) = g'(x)

(He phiTdng trinh nay la h? phiTdng trinh \ic dinh hoanh dp tiep diem cua

haidirdng) ' - •

I I C A C B A I T O A N M I N H H Q A

Bai 1. Cho ham so y = f(x) = - x ' + 2x^ + | x "^^

a Khao sat h^m so

b Tiep tuyen cua do thi (C) cua ham so tai goc tpa dp, cat (C) tai diem M Tinh tpa dp diem M

c Bien luan theo k so giao diem cua (C) va du'dng thing (d): y = kx

Luy?n gidi dS IruOc kp thi DH 3 miin Bdc, Trung, Nam Todn hgc - NguySn Vdn Th6n^

H a m so" nghich bien tren cac khoang — ; + o o

+ NhSn x6t: D 6 thi nhan diem I '2 46^

3 ' 2 7 lam tam doi xtfng

b PhiTdng trinh tiep tuyen (A) cua (C) tai O (0; 0) la:

D a t g ( x ) = x ^ - 2 x + k - - la tam thiJc bac hai CO A' = - - k

B a i 2 Cho ham so y = f(x) = - x ^ + 3x^ + 3(m^ - l)x - 3m^ - 1 (1), m la tham so

a Khao sat s i f b i e n thien va ve do thi ham s 6 ' ( l ) k h i m = 1

b T i m m de ham so' (1) c6 cifc dai, ciTc tieu va cac diem cUc t r i cua do thi ham

so (1) each deu goc toa do O

a Khao sdt s\i bien thien v^ ve do thi tfng vdi m = 1

b Chtfng minh r^ng (C^) luon di qua hai diem co dinh A, B v d i m o i m

c T i m m de cdc tiep tuyen v d i (C J tai A, B vuong g6c v d i nhau

Hifdng d i n giai

a Kh3o sdt hkm so i?ng v d i m = 1 (Hoc sinh tif giSi)

b G o i A(x,); yo) la diem c o d i n h cua (C), k h i d6: ' '

yo = - x ' * + 2 m x o - 2 m + 1; V m , ^; ; ;); ^ /o,^

0 2m(x2 - l ) - ( y „ + x ^ - l ) = 0 ; V m

c T i e p tuyen tai A ( l ; 0) c6 he so g6c k, = r ( l ) = - 4 + 4m Vrr'Jid i f ) S. iiiS

T i e p tuyen tai B ( - l ; 0) c6 he so goc k j = f ' C - l ) = 4 - 4 m ;,,,; oiJd:-! ?

T i e p tuye'n tai A, B vuong goc v d i nhau <=> k|.k2 = - 1 IT

B a i 4 Cho ham so y = f(x) = x " - 2 m V + 1 (1) vdi m la tham so

a Khao sat ham so (1) khi m = 1

b T i m m de do thi h^m so (1) c6 ba diem cifc t r i 1^ ba dinh ciia m o t tam giac

b T i m c a c d i e m thuoc (C) c6 loa do la nhiTng so nguyen

c T i m d i e m M e (C) de long khoang each liT M den hai du'dng tiem c a n cua

=> y = 2 la tiem can ngang cua do thi ham so

b Bang bien thien: y ' = , < 0 ; V x e D

H a m so nghich bie'n tren cac khoang (-oo; 1); ( 1 ; +oo)

3 D o thi (Hoc sinh M ve hinh) ^J'^ ' '

U^ngua ae miOc Ic m uii t mien oat;, irung, num ivunrm- lyguyen vun i

tivng-Goi M (X(,; y„) e (C) Khi d6:

d ( M ; d , ) = |xo-l|

' 2 x 0 - 1 1

d ( M ;d2)= y o - 2

Xo - 1 - 2 X o - 1 Khi d6, tong khoang c^ch \.\S diem M(X(); y o ) den hai di/cfng tiem can Ik

Aihly. Bai 6. Cho hamsoy = f(x) = x + 1

a Khdo sdt sif bie'n thien va ve do thi (C) ciJa ham so' da cho

b Tim toa do diem M e (C), bie't tie'p tuyen cua (C) tai M c^t hai true toa dp

Ox, Oy tai A, B va AOAB c6 di?n tich bkng - _

4

Hi/dng d i n giai

a Khao sdt hkm so (Hoc sinh tiT giai)

b Gpi M (xo; yo) 6 (C); y„ = f(xo)

PhiTdng trinh tie'p tuyen cua (C) tai M(xo; yo) la (d): y = f (xu)(x - XQ) + f(xo)

y = (Xo+1)' (Xo -x + - 2x^ +1)'

Toa dp giao diem A cua (d) vk Ox Ik: A( -Xo ; 0)

( 2x2

Tpa dp giao diem B ciia (d) va Oy la: B 0; ^

De thay AOAB vuong tai O S^QAB = ^ 0 A 0 B

X n

4 2 "(xo+1)^ 4x^-(Xo+1)2=0: 2xo + X() + 1 = 0

2 x 2 - X o - l = o' Ket luan: M, ;M2 (1;1)

X o =

-X o = l

if

III C A C BAI TOAN TI; L U Y f N c 6 H U 6 N G DAN

Bai 1 Khao s^t sir bie'n thien vk ve do thi (C) cua hkm so y = f(x) = x^ - 3x^ + 1

1\i do, bien luan theo m so nghiem cua phiTdng trinh sau:

b Vdi gia tri nko cua m do thi cija hkm so (1) tiep xuc vdi Ox Khao skt vk ve

do thi hkm so (1) gik tri tim diTpc cua m

c Xkc dinh m do thi hkm s6'(l) c^t true hoknh tai 3 diem phan bi^t

a Khio sat si/ bie'n thien vk ve do thi (C): y = f(x) = x' + x - 1

b Gpi Xo Ik nghipm cua phiTPng trinh f(x) = 0 Chtfng minh r^ng:

Luyfn giii di trade kp thi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyln Van ThOng

a Tim m dc ham so' co mot diem cifc tri

b Viet phiTcfng trinh tiep tuycn cua f C , ^ di qua g6'c toa do >f{ £ > tu »

a • a < 10 : Phi/cfng trinh v6 nghiem

• a = 10 : Phu'dng trinh c6 1 nghiem _ ^ r

• 10 < a < 11 : Phu'dng trinh c6 4 nghiem j ^'g;

• a = 11 : PhiTcfng trinh c6 3 nghiem

• a > 11 : Phifdng trinh c6 2 nghiem

b (P) tiep xuc vdi (C) khi va chi khi hoanh do tiep diem la nghiem cua he

phiTcfng Irinh

x ^ - 2 x ^ + l = m x 2 - 3 4

• m = 4x - 4 x = 2 m x 7^ - 2 ft)

B a i 6 C h o ( C ) : y = f ( x ) = - x ' * - 3 x ^ + -

19^'-a Viet phiTdng trinh tiep tuyen cua (C) tai diem x = Xo

b Chufng minh rang hoanh do giao diem cua (C) va (d) la nghiem ciia phiTdng

Bai 7 Cho ham so y = r(x) = p

a Khao sat sir bien thicn va ve do thi (C) cua ham so

b Tim m de di/c^ng thang (d): y = 2x + m c^t (C) tai 2 diem phan bi^t A va B

tim tap hdp trung diem I cua AS t m + A« j 8- ^ (£)^j '"" '

Hrfdng dSn giai '(.m.utll

b —=- = 2x + m o <^ ,

x + 1 [ 2 x ^ + ( m + 4)x + m + 4 = 0 -f + - f i : ^ r : Dieu kien de (d) c^t (C) tai 2 diem phan biet la m < - 4 hoSc m > 4

Tap hdp trung diem I cua AB la phan diTdng thang 2x + y - 4 = 0, gom

trinh = 2m - 1 c6 dung 2 nghiem x x; sin X + 2

HiAJngdSngiai al =.(x)t mil

* Khao sat va ve do thj (Hoc sinh tiT lam)

* Dat t = sin x; x e [0; 71J ^ t e [0; IJ n « (,., •^rmd^ fa; ;

PhiTdng trinh trd thanh: = 2m - 1 (1) " '

Dieu kien de phiTdng trinh c6 dung 2 nghiem x e [0; n] la phtfdng trinh (1) co

dung 1 nghiem t e [0; 1)

Khao sat ham so g(t) = = ^ iren [0; 1) => - < g(t) <3z:>-<m<2

KHAO SAT MpT SO HAM SO DA THllfC N A N G C A O

Bai 1 a) Bie't r^ng d6 thi ciaa hhm so'y = (3a' - l ) x ' - (b' + l ) x ' + 3c'x + 4d co

hai diem cifc tri la: (1; - 7 ) , (2; -8) Hay xac dinh tong M = a" + b' + c' + d' b) Chu-ng minh rang do thi ham so y = x ' + 2 m V + 1 luon citt difdng thang

y = X + 1 tai diing hai diem phan biet vdi moi gia trj m

Hifdng d i n giai a) De ddn giSn, ta dat A = 3a^ - 1, B = -(b^ + 1), C = 3c^ D = 4d, h^m so da

cho chinh m y = Ax^ + Bx^ + Cx + D

TCf cdc dieu kien n^y, ta tim diTdc A = 2, B = - 9 , C = 12, D = -12 ta tinh

diTcIc cdc gia tri tiTcfng tfng la a = ±1, b = 2, c = ±2, d = - 3 - ^

V a y M = a^ + b^ + c^ + d^= 1^ + 2.2^ + 3^= 18 x ^ i

b) Phi/dng trinh hoanh dp giao diem cua hai do thj Ik

x" + 2 m V + 1 = X + 1 o x(x' + 2m^x - 1) = 0 -j.,,

De thay phifdng trinh nay co nghiem x = 0, ta can chifng minh phifdng trinh

c6n lai c6, nghipm duy nha't khac 0

That vay, xet ham so f(x) = x ' + 2mx^ - 1

Ta c6 f (x) = 3x^ + 2m^ > 0 nen ham so nay dong bien tren tap so thi/c

Hdnnffa lim f ( x ) = lim (x"^ + 2m^x - 1 ) =-oo

Va lim f ( x ) = lim ( x ' ' + 2 m ^ x - l ) = +oo

Nen phi/dng trinh f(x) = 0 luon c6 nghi$m

Suy ra phi/dng trinh f(x) = 0 c6 dung mot nghiem nay cung khdc 0 do

f(0) = - l T a c 6 d p c m i _

Bai 2 Cho ham so y = -x^ - 3x^ + mx + 4 vdi m 1^ tham so thifc

a) Khdo sat sir bien thien va ve do thi do thi hhm so khi m = 0

b) Tim tat ca cac gid tri cua tham so m de h^m so da cho nghich bien tren

COngtyTNHHMl ^ L v vii KHung vict

H^m so nghjch bien tren moi khoang (-oo; - 2 ) , (0; +oo) va dong bien tren khoang (-2; 0)

Ham so dat ciTc dai tai x = 0 va dat ciTc tieu lai x = - 2

• Bang bie'n thien ' • I i ' •

Do thi cii true lung tai diem

(0; 4) va cat true hoanh tai

d i e m ( l ; 0 ) , ( - 2 ; ( ) )

Do thi cua ham so: t

b) Ham so da cho nghich bien tren khoang (0; +co) khi va chikhi ^ , X y' = -3x^ - 6x + m < 0 Vx > 0

» 3x^ + 6x > m, Vx > 0

- 0 0

Hinh 7 Do thi ham soy = - J C ' - 3 X ^ - ^ 4

Ta CO bang bien thien cua ham so g(x) = 3x^ + 6x tren (0; +oo)

Li^n giii di tniOc Aj> thi DH 3 miin Bdc, Trung, Nam Todn hQc - NguySn Van Thdng

3

— ; + t x 3

4

H a m so dong bicn tren

Va nghich bi6'n tren

H a m so dat ciTc dai tai (0; 1) va dat ciTc tieu tai

( 3^ - c o ; 0 ; —

I 4 ; 4 j

( 3 49 ^ r 3 4 9 ^ 4 ' 32 J 4 ' " 32 J

t e [ - l ; 1] va ro rang moi gia trj

cua t (t e [ - l ; 1 ] ) cung cho ta duy

nha't mot gia tri cua x phiTdng

trinh da cho ti/dng diTdng vdi

f(t) = St' - 91^ + 1 = 1 - m

Day chinh la phifdng trinh hoanh

do giao d i e m cua hai do thi

32 Htnh 8 Do thi ham soy = 8x^-9x^ + 1^

o m < 0 V ni > — ihi phiTdng trinh vo nghiem

Neu 1 - m = 1 o m = 0 ihi phiTdng trinh co diing mot n g h i e m

COng ty TNHH MTV DWH Khang Vi?t

N c u 0 < 1 - m < 1 o 0 < m < 1 Ihi phiTdng Irinh c6 dung hai nghiem

N e u A ' , , < 1 - m < 0 o 1 < m < ^ Ihi phifdng U-inh c6 dung bon nghiem

N e u I - m = <=> m = |i ihi phi/dng trinh c6 dung hai nghiem

32 32

B a i 4 Cho ham s 6 > = f(x) = m x ' + 3mx' - (m - l ) x - 1 vdi m la tham so' '

a) Khao sat sir bien thien va ve do thi cua ham so khi m = 1 b) Xac djnh taft ca cac gia tri m dcf ham scTy = l(x) khong c6 cifc tri

H i m so nay khon g c6 cifc trj khi v4

chl khi phiTdng Irinh y' = 0 khong c6

n g h i e m hoSc c6 n g h i e m kep, tiirc la

A ' < 0 « 9m^ + 3m(m - 1 ) = 12m^ - 3m < 0 <=> 0 < m < ^

Luy(n giii di trade kj> thi DH 3 miSn Bdc, Trung, Nam ToOn hoc - NguySn Van ThOng

VSy dieu kien can nm la 0 < m < -

a) Hoc sinh tiT khao sat ,v.,

b) Ta CO y' = - 6 x ' + 12x Goi M(x„; y„) \k tiep diem cua (C) vdi tiep tuy^n can tim

Khi do y,i = -2xi'', + 6xf, - 5 _ ^ ,,, „,,

Phifdng trinh tiep tuycn cua (C) tai M y - y,, = f (X())(x - x„) hay

Ta CO cdc tung do tiTcJng iJng la y ( l ) = - 1 , y(-2) = 35

- V d i M ( l ; - l ) thi phiTdng trinh tiep tuyen can t i m i a :

y + l = 6 ( x - l ) c ^ y = 6 x - 7

- V d i M ( - 2 ; 35) thi phu'dng trinh tiep tuyen can tim Ik '

y - 35 = - 4 8 ( x + 2) o y = - 4 8 x - 61

Bai 6 Cho ham so y = x ' - 3x^ - 9x + m vdi tham so m

a) Khao sat sif bic'n thien va ve do thj hkm so da cho khi m = 0

b) Tim tat ca cac gid tri m de do thi ham so c^t true hoanh tai 3 diem phan biet

c6 hoanh dp lap thanh cap so cpng

Hifdng dSn giai

a) Hoc sinh tiT khao sit

b) Do thi h i m so da cho c^t true hoanh tai ba diem phan biet c6 hoknh do lap

thinh cap so'cpng khi va chi khi phiTdng trinh sau c6 ba nghi^m phan biet lap

thanh cap so cpng: x"^ - 3x^ - 9x + m = 0 (*)

Theo djnh l i Vi-et thi tong ba nghi^m cua phU'dng trinh (*) l i 3 Do d6, theo

tinh chat cua tong cua cap so cpng thi 1 chinh la nghi$m cua phu'dng trinh (*)

hay l ' - 3 1 ' - 9.1 + m = O o m = l l

V d i m = 11, phiTdng trinh (•) trd thanh

x ' - 3x^ - 9x + 11 = 0 o ( X - l)(x^ - 2x - 11) = 0

Ceng ty TNHH MTV DWH Khang Vi(t

Phildng tiinh nay c6 ba nghi^m phan bi^t Ik 1 - 2>/3 , 1 , 1 + 2%/3 thoa man de bai

Vay gia tri can tim cua m la m = 11

Bai 7 Cho hkm so y = f(x) = x ' - 2x^ c6 do thj (C) f"" ' ' <• '

a) Khao sat va ve do thj (C). 'i ,i,.-r t , j t-f v j ! ' " ^

b) Tren do thi (C) lay hai diem phan bipt l i A v i B c6 hoinh dp Ian li/dt \h a, b

Tim dieu kipn cija a, b de tiep tuye'n ciia (C) tai cAc diem A va B song song

vdi nhau

Htf^ngdSngiai

mm so y = f(x) x" - 2x^ c6 tap xac djnh D = R

Gidi han cua ham so: lim y = +oo ; Um y = +oo i

S\ibien thien cua ham so': N d - u y r « i x

T a c 6 y ' = 4 x ' - 4 x = 4 x ( x ^ - l ) , y ' = O o x = 0,x = + l " ^ Ham so dong bie'n tren ( - 1 ; 0), ( 1 ; +oo) va nghjch bien tren (-oo; - 1 ) , (0; 1)

Diem cifc dai ciia do thi Ik (0; 0), diem cifc tieu cua do thi ham so' la ( - 1 ; - 1 ) ,

( 1 ; - 1 ) Bang bie'n thien iftj' '*

Ta c6 y" = 12x^ - 4 = 4(3x^ - 1), y" = 0 « x = ± Cac diem uon cua do thi la

f S 5^ (J3 5]

I 3 ' ^ [3- 9J

* Do thi cua ham so' b) Ta CO f ( X ) = 4x^ - 4x

Gpi a, b Ian li/dt la hoanh dp cua A vk B

Hp so gdc cua tiep tuye'n cua -3 (C) tai A va B Ian li/dt la

kA = f (a) = 4a^ - 4a, ke = f (b) = 4b' - 4b

Tiep tuye'n tai A va B Ian lUdt c6 phiTdng trinh la

y = f (a)(x - a) + f(a) = f (a)x + f(a) - af (a)

y = f(b)(x - b) + f(b) = r(b)x + f(b) - bf(b)

Hai tieo tuve'n n^v song song hokc trung nhau khi \k chi khi

Luy^n gidi di truOc thi DH 3 mijn Bdc Trung Nam Todn hoc - NguySn Van Thdne

ICA = ke « 4 a ' - 4a = 4 b ' - 4b c:>(a - b)(a' + ab + b ' - 1) = 0

Do A va B phan bict n c n a * b, suy ra a ' + ab + b ' = 1

M 3 l k h i c , hai tiep tuyen cija (C) tai A va B irung nhau k h i va chi k h i ^ j

a- + ab + b^ = l,a ^ b ^ a^ + ab + b^ = l,a ;t b ( h% ; ^j, ^)|,;f>{

f(a) - a f ' ( a ) = l"(b) - bl"'(b) [-3a^ + 2a^ = -Bb"* + 2b^ \ •<!] hh i t ; n T ,

Giai he nay, ta diTdc hai nghiem la (a; b) = ( - 1 ; 1), (a; b) = ( 1 ; - 1 )

V a y dieu k i e n can va du dc hai tiep tuyen ciia (C) t a i A va B song song v d i

nhau la a"^ + ab + b^ = 1, a 9t+1, a b

KHAO SAT MQT SO HAM SO PHAN THaC NANG CAO

B a i 1 BicTt rang del thi ham so' y = ^ , ac ^ 0 ad - be ^ 0 c6 tam doi xtfng

a) ChiJng minh rang v d i m o i m ^ 1 thi do thi cua h a m so y = ^?^^^—^——?^

luon tiep xiic v d i diTcfng phan giiic cua goc phan tu" thi? nha'l ^

K\ - ' 1* + ( m + 2)x + 2m + 2

0) l i m m de t i c m can xien cua do thi ham so y = tiep

x + 2 xuc v d i dirdng cong (C): y = x ' - 3 x ' - 8x i '

Hifdng dSn gial

a) Ta can churng minh r^ng y = — ^ ^ ^ " ^ luon tiep xuc v d i y = x v d i m o i

X - 1

m ?t 1

D i e u k i e n de hai diTdng nay tiep xuc nhau la he phiTPng trinh sau c6 nghicm

Ceng ty TNHH MTVDWH Khang Vi?,

X + m = x - ^ - 3 x ^ - 8 x

l = 3 x ^ - 6 x - 8

m = x-^ - 3 x ^ - 9 x x^ - 2 x - 3 = 0

o m ^ x " * - 3 x ^ - 9 x

X = - 1 V X = 3

V d i X = - 1 , ta CO m = 5 va v d i X = 3 Ihi m = - 2 7 ^ ' Vay CO hai gia tri m can tim la m = 5, m = - 2 7 j , - i)Oi ;

Bai 3 a) Cho ham so y = ^ '^^^ ^ ^ trong do p ^ 0 p^ + q^ = 1 (1)

Tim la't ca cac gia t n p, q sao cho khoang each giiya hai diem cifc tri la VlO

b) ChiJng minh rang v d i m o i m thi do thj cua ham so y = X + ( m + l)x + m + l

Do do, dieu k i e n dc do thj ham so nay c6 hai diem ciTc tri \k phi/dng trinh sau

CO hai nghiem phan biet p x ' + 2(q - l ) x - p = 0 (*)

Dieu nay liTPng difdng v d i A' > 0, p 0 o (q - 1 )^ + p^ > 0

V i p ^ nen do thj luon cd hai diem ciTc t r i

G p i X,, X2 Ian liTdl la hai nghiem cua phiTdng trinh (•), day cung chinh la cac cifc tri cua ham so da cho

Luy(n gidi dS trudc thi DH 3 miin Bdc Trung, Nam Todn hoc - Nguyin Van Thdng

Giai phifdng trinh nay, ta thu diTdc nghiem thoa man de bai la q = 0 Thuf lai

ta tha'y thoa Vay cac gia tri can tim la (p; q) = ( 1 ; 0), (p; q) = ( - 1 ; 0)

b) Ta CO y = x + m +

x + 1

y ' = l - - 1

(x + i y , y ' = 0 o (X + 1 ) - = 1 o X = 0 , X = 2

Do do, hai diem cifc trj cua do thi chinh la (0; m + 1), (-2; m - 3) nen khoang

each giffa hai diem nay la d = V(0 +2)^ + (m + 1 - m + 3)^ - 2 y / 5 khong doi

Ta CO diem phai chiJng minh | p + »:} +^x>yC. (l ^ ^xVo + xS)

c) Hoc sinh tirkhao sat

Bai 4 Cho hamso y = c6 do thi (C)

4 ( x - 3 ) a) Khao sdt va vc do thi cua ham so da cho „ , , , v

b) Tim tpa do diem M thupc (C) sao cho tic'p tuycn cua (C) tai M c^t hai true

tpa do Ox, Oy Ian liftJt tai hai diem A, B v& dien tich tam giac OAB \h -

8 c) Tim nhurng diem u-en (C) co khoting each den true hoknh gap 3 Ian den true tung

Nen ham so da cho nghich bien tren tiifng khoang xdc dinh

Ham so da cho khong c6 cifc tn , -v^ » v> < - i ' - a i j

, a c dir&ng tiem can: Ta c6 pB t ar 6 r > u v M fr < ; T Urn y = hm

b) V d i moi diem M ba't ki thuoc do thi (C), tim gia tri nho nha't cua tong khoang

each tCr M den hai true toa do. , >, j , n ;, ^ t,tij HU (.,• t ,\m-'

c) T i m nhOrng diem tren (C) C O toa do la cap so nguycn ,•>',,, ; ' ] , , ' }

HMng dSn giai \

a) Hoc sinh tU" khao sat ' ' * (*,,**'

b) Theo de bai, ta can tim gia tri nho nha't cua bieu thiJc sau \

x - 1

til

A = + 2x + 3 3 vf;"'!'?'- f':'~; mil \

5 " Ta tha'y rang vdi x = 1 thi A = 1 nen ta chi can x6t cac gia trj cua x khac i

sao cho A < 1, turc la chi can xet |x| < 1 o - 1 < x < 1 Khi do, do 2x + 3 > 0

nen ta chi can xet 2 tru'dng hdp sau:

1

B a i 6, Cho ham so y = 2x - 1 + •

^ x - 1 a) Khao sat va ve do thi ham so' ,

b) T i m toa do diem M ba't ki thuoc do thi sao cho tong khoang each tit M den

hai du'dng tiem can nho nha't

Hifdng d§n giai ,(iOttteriq;iliji3!i%

a) Hoc sinh tuf giai '

• " YiT

b) Ta tha'y tiem can duTng cua do thi la x = 1 va ti?m can xien la x = 2x - 1.'

X6t mot diem M nkm tren do thi c6 toa dp la M X ();2X(,-1 +

Xn - 1 , X„ I

o o n g ly iisnti M I V uvVH'Khang Vii

Tong khoang each tiT M tdi hai diTcJng tiem can ciia (C) la

HiTdng d§n giai

a) Ta cd y = x^ 3x + 3 +

-• y' = 2x - 3 2 x ^ - 3 x ^ - a

x^

Ta cd do thi cua ham ^ g " " " " ""^"^ ^ ^ f " ' '

y = 2x^ - 3x^ nhi/ dirdi day: ;^ ^ [ ~ 7 DiTa theo do thj cua ham so nay thi ta tha'y i /, rhng cac gid tri a can tim la - 1 < a < 0 /

V d i dieu kien tren, theo cong thuTc ve toa '

dp edc diem ciTc trj thi de tha'y cac ciTc tri ciia do thi ham so da cho se nlim tren

-3

parabol y = 3x^ 6x + 3 c^ dinh , W^*t, at>n m f o ^tuMA x

thdi h^m so nay lien tuc tren tap so thifc nen phiTdng trinh f(x) = 0 c6 ba

nghiem phan biet thuoc cac khoang ( - o o ; - 1 ) , ( - 1 ; 0), (0; +oo)

Do do, phi/dng trinh (•) c6 ba nghi?m phSn biet hay do thi da cho c6 ba diem

uon

G'd sur ho^nh do cua mot trong cac diem uon cua do thi h^m so da cho 1^ x„

va day cung la nghiem cua phiTcfng trinh (*) hay X Q + 3axo + 3(a - 1)X(, - 1 = 0 ;

khi d6, tung do trfdng uTng cua diem n ^ y chinh 1^ yo = —5-^ Ta se tim

Xo + Xo + 1

mot quan he tuyen tinh giffa Xo, y o

Tir dieu k i ^ n cua Xo, ta thafy r^ng

X Q + 3axf, + 3 a x „ + 3a - 1 = 3 x „ + 3a o (x,) + 3a - l)(xo + x,, +1) = 3 ( X Q + a ) '

S „ y ra y„ - + x„ +1) ^ X Q + 3a - 1 ^

3(Xo + x„ +1) 3

Do d6, cdc die m uon cua do thi cung thoa man quan h$ tuyen tinh nhif tren,

ttfc Ik chung t h i n g hang Du'dng t h i n g di qua cdc diem uon tifdng tfng chinh

x + 3 a - l

MQT S6 DANG TONG H0P TOAN LIEN QUAN DEN ^

DO TH| HAM S6 NANG CAP '

Bk\ Cho ( C „ ) c6 phifdng trinh y = x ' + ( m - l ) x ' - ( m + 3)x - I

a) Khao sdt ve do thi (C) ciia hkm so k h i m = 1

b) Chtfng minh r^ng v d i m p i m, hkm so' c6 ci/c d^ai, ci/c ti€u Vie't phifdng trinh

dirdng t h i n g d i qua c i c d i e m cifc dai ciJc tieu cua do thi

c) T i m nhffng cap d i l m nguyen tren (C) do'i xtfng v d i nhau qua diTdng t h i n g

y = X v£l khong nKm tren di/dng t h i n g d6

J) T i m tham so Ihifc m dc tpa dp cifc dai va ciTc lieu cua ( C m ) nam ve hai phia

nan luon cd 2 nghiem phan biet Suy ra v d i m p i m, ham so c6 ciTc dai, ci/c ticu Thi;c hicn phep chia y cho y', ta cd

va do chinh la phUdng trinh diTdng t h i n g di qua hai die m ciTc tri

c) N c u diem A cd tpa dp lii (x; y) thi diem doi xtfug cua A qua diTdng t h i n g

y = X cd tpa dp la (x; y) V i the ycu cau cua bai toan tiTdng diTdng vdi viec

tim nghiem nguyen (x; y) v d i x^y cua he phi/dng trinh I

Thur lai vko he, ta nhan bp nghiem ( 2 ; - 1 ) ( - 1 ; 2 ) ' ' '

V a y la ti m dirpc cap die m nguyen duy nha't d o i xuTng v d i nhau qua di/dng

t h i n g y = X va khong nam tren di/dng t h i n g do la (2; - 1 ) , ( - 1 ; 2) ,

Lu^n gUu di trUOC thi DH 3 m,.'n / U / ' \.,:m / >i h \ , • 7 , ',,;/( Thdng

B a i 2 Cho ho diTcfng cong (C J : y = x ^ + mx

-x - m a) Khao sat ve do thj ( C ) cua ham so khi m = 1

b) Xac dinh m de ham so c6 ciTc dai, ci/c tieu Vie't phiTctng trinh diTdng thing di

qua cac diem ci/c dai va cifc tieu cua do thj ham so'

c) Tim cac diem trong mat phlng sao cho c6 dung hai di/cfng cua ho ( C J di qua

H i f d n g dSn giai

—x^ + X — 1 • 1

a) Khi m = 1, ham so trd thanh y = x

-x - 1 ' -x - 1 Mien xac d j n h D = R \ { 1 }

1 x ( 2 - x ) ' ^ ^ t e U Taco y = - 1 + = —^

y' = 0 k h i x = 0 h o a c x = 2 ^, j

H^m so tang tren (0; 1) va (1; 2), giam tren (-00; 0) va (2; +00)

Ham so dat ciTc tieu b i n g 1 khi x = 0 v^ dat ciTc dai bllng -3 khi x = 2

• Gidi han: Um y = lim

H^m so CO tiem can xien y = - x va ti^m can dtfng x = 1

Bang bie'n thien

c6 ciTc dai v^ ciTc tieu khi vS chi khi m^O

Khi 66 hai diem cifc tri c6 Ipa dp ti/dng ufng la (0; m) va (2m, -3m) Suy ra

dudng thing di qua hai diem cifc tri c6 phiTdng trinh l i y = m - 2x

c) Gia sur (x,), y„) la mot diem trong mSt phing ma c6 dung hai diTdng cong ( C J

di qua Khi do phifdng trinh y„ = '"^o ~ "^^

X o - m (1)

CdngtyTNHH MTVl)\> / / , Vi(t_

c6 diing hai nghiem phan bipt (an so la m)

Viet lai (1) diTdi dang m^ - (x,, + y„)m + X(,(x ,i + y„) = 0, ta suy ra tat ca nhuTng

diem (xo; yo) thoa man yeu cau bai toan la nhCTng diem c6 toa do thoa man

(x,) + y())(yo - 3x()) > 0

Bai 3 Cho ham so y = x ' - 3(m + l)x^ + 2(m^ + 4m + l ) x - 4m(m + 1) ( C J

a) ChiJng minh rang ( C ^ ) luon di qua mot diem co djnh khi m thay doi

b) Tim m sao cho ( C m ) c i t true hoanh tai 3 diem phan biet

c) Khao sat ve ve do thj ham so' khi m = 1

H i M n g dSn giai * ^ ^ x (>: nM si ui il / f

a) Gia suTdiem co djnh la (x„; y„) Khi do ta c6 f"^' tJ'' f^

yo = - 3(m + l)xf) + 2(m^ + 4m + l)x„ - 4m(m + 1) diing vdi moi m

Vict dang thuTc tren nhu'da thiJc theo m, ta difdc (2X() - 4)m^ + (-3 xf, + 8x0 - 4)m + xf, - 3XQ + 2xo - y,, = 0 vdi mpi m

' 2 x o - 4 = 0 ' Dieu nay xay ra khi va chi khi -3xf) + 8x„ - 4 = 0

x f , - 3 x f , + 2 x o - y o = 0

Tur day giai ra dU'pc Xo = 2, yo = 0

t») ( C m ) c i t true hoanh tai 3 diem phan biet khi va chi khi phtfdng trinh

x ' - 3 ( m + l ) x ^ + 2(m^ + 4 m + l ) x - 4 m ( m + 1 ) = 0 (1)

CO 3 nghiem phan biet TiT cau a) ta thay ring x = 2 la nghiem cua phiTdng

trinh tren Nhd do, bie'n doi da thufc ve trai, ta diTdc phiTdng trinh tiTdng diTdng (x - 2)(x^ - (3m + l)x + 2(m^ + m)) = 0

Tilf day ta thay r i n g (1) co 3 nghiem phan biet khi va chi khi phi/dng trinh (2)

- (3m + l ) x + 2(m^ + m) = 0 (2)

Dieu nay xay ra khi va chi khi | A = (3m + l ) ^ - 8 ( m ^ + m ) > 0

l 2^ - ( 3 m + l).2 + 2 ( m 2 + m ) ^ 0 Giaira ta diTdc m 1

Vay vdi m ;t 1 thi ( C n , ) c i t true hoSnh tai 3 d i l m phan bi$t

B ^ i 4 Cho ham so y =— ( C )

x - 1 3) Khdo sdt sir bie'n thien v^ ve do thi (C)

^) Tim hai diem A, B thupc (C) v^ doi xtfng nhau qua difdng t h i n g y = x -

1-Luyfn giat ae truac Kytnttitijmi g

Hifdng dSn giai

a) Hoc sinh tiT khao sal ! if, i f \ ^.r t>i|V

b) N e u M ( x , y) M ' ( x ' , y') 1^ hai d i e m d o i xtfng v d i nhau qua du"dng thSng

I > ^ • ^ / ^ l 1 : 2 x ^ + ( 6 - m ) x + 4

Bai 5 Cho h ^ m so y = ^

mx + 2 a) Chtfng m i n h r ^ n g v d i m o i gi4 trj cua m , do thi h a m so luon d i qua m o t d i e m

C O dinh duy nha't Xic dinh tpa dp cua diem do

b) K h a o sat va ve do thi cua ham s6' k h i m = 5

Hifctng dan giai

Gia sur (x,,; y,,) la d i e m co djnh ciia do thj ham so

L'u- 1 ' ' 2 x , ^ ) + ( 6 - m ) x „ + 4 ^

K h i do ta C O y;, = — — ^ '—il v d i m o i m

mx,, + 2

Suy ra mXd.yi, + 2y„ = 2 xf, + (6 - m ) X ( , + 4 v d i m p i m

V i e t phirong Irinh tren diTdi dang nhj thiJc iheo m , ta diTdc

(xoyo + x,i)m + 2y() - 2 xf, - 6x1, - 4 = 0 vdi mpi x

Suy ra Xoy„ + x„ = 0 vii 2y„ - 2 x,^, - fix,, - 4 = 0 h

G i a i he nay ta diTdc n g h i c m duy nhal x„ = 0, y„ = 2 Tit do suy ra do thi ham

so luon di qua d i c n i co djnh duy nha't c6 tpa dp (0; 2)

( x - 1 ) ^

x + 2

pai 6. Cho h a m so y = a) Khao sat sir b i c n thien va ve do thi h a m so da cho

b) B i e n luan theo m so nghiem cua phi/dng trinh

Hifdng d§n giai

a) Hoc sinh t i f g i a i b) + m < 0: phiTdng trinh v6 nghipm + m = 0: phiTdng trinh co 1 nghiem duy nhat;

+ 0 < m < 12: phi/dng trinh co 2 nghiem;

+ m = 12: phiTdng trinh c6 3 nghiem;

+ m > 12: phiTdng Irinh co 4 nghiem

= m

','•'•1: 'l:'

Sd liTdc each g i a i : TiT do thi G d cau a), ta suy ra do thj G' cua ham so

v = ^ ^ ~ ' ^ nhir sau: V d i x > - 2 , G' chinh la G V d i x < - 2 , G' la anh doi

x + 2 xiJng cua G qua true hoanh

TCr do thi G' ta cd ket qua b i e n luan neu tren

Bai 7 T i m m dc phi/dng trinh sau cd 4 nghiem phan biet:

Cdch 2 Del thj h a m so' y = 4|x|-^ - 3 | x | - 1 g 6 m hai nhanh, nhanh 1 la d6 thi ham

so' y = - 4 x ' + 3x - 1 v d i X < 0 va nhanh 2 la d6 thj h a m so' y = 4x- - 3x - 1 vdi X > 0 T r o n g k h i do y = mx - m la diTdng thdng quay quanh d i e m A ( l ; 0) Hay vie'l phi/dng trinh lie'p tuye'n ke tiT A den nhanh 1 cua d6 thj h a m so'

J o

BAI TAP T O N G H0P L U Y g N THI DAI H Q C V A L U Y g N H Q C SINH GIOI

+ 3 x ^ - 2

Bai 1. Cho ham so y = -x^ + 3x^ - 2 (C) ' ' ^

a) Khao sat sir bic'n thien va ve do thj (C) Tif do vc do thj (C) y =

b) Tim tren (C) nhffng diem ma qua do chi ke diTdc mot ticp tuyen vdi (C)

H\i6ng dSn giai

a) Hoc sinh tiT khao sat

b) Gia sur M(x„; y,,) la mot diem tren (C) Ta giai bai loan vict phiTOng trinh ticp

tuyen cua (C) qua M Gia su" ticp tuyen (t) kc lit M den (C) tiep xiic vdi (Cj

tai N(x,; y,) Khi do phiTdng trinh cua (t) co dang

y - y , = ( - 3 x | + 6 x i ) ( x - x i )

y o - y i = ( - 3 x f + 6 x , ) ( X o - x , ) (1) Ngoai ra, do N thuoc (C) nen ta CO ^

y, = - x ) ' + 3x^ - 2 • (2)

Nhir vay toa do tiep diem la nghiem cua he (1), (2) Ycu cau bai toan o he

(1), (2) v('Ji an la (x,; y,) co nghiem duy nhat

Thay y, tif phiTdng trinh (2) vao phi/dng trinh (1), ta di/dc

2x-| - 3(x,) + l)xf + 6x„x, -1 -y,) = 0

Lai thay 2 x i ' - 3 x „ x ? + x i l - 3 x f + 6 x „ x , - 3 x i ^ - 0 ^ •

o ( x , - x „ ) 2 ( 2 x , + x , ) - 3 ) = 0 (3)

Ta thay he (1), (2) co nghiem duy nhat khi va chi khi ~ ^ = x,, o x,, - 1

Ttrdo tinh difdc y„ = 0

Vay M(l; 0) la diem duy nha't tren (C) ma qua 66 co the kc dung mpt Uep

tuyen vdi (C)

Bai 2 Cho ham so y = - 3x + 2 (C)

a) Khao sat su" bien thien va ve do thj (C)

b) Xet 3 diem A, B, C ihdng hang va thuoc (C) Gpi A', B', C' la giao diem cua

(C) vdi tiep tuyen cua (C) lai A, B, C ChiJng minh rang A', B', C th^ng hang

c) Tim tren do thj (C) cac diem doi xuTng nhau qua 1(0; 2)

Nhgn xet: X6t 3 diem A, B, C thuoc (C) co hoanh dp liTdng uTng la XA, XR, XC

Khi do A, B, C thang hang khi va chi khi XA + XB + Xc = 0

ChuTng minh Gi a suT A, B, C nam tren difdng thang co phU'dng trinh y = ax +b

Khi do XA, XB, XC la nghiem cua phU'dng trinh

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

Ap dung djnh li Vi-ct, ta suy ra XA + XB + Xc = 0

NgiTdc lai, gia suT XA + XB + Xc = 0 Viet phifdng trinh difdng thang di qua A, B

c^t (C) tai C thi theo phan thuan ta co XA + XB + Xc = 0 suy ra Xc- = Xc suy ra C' trilng C va co nghla la A, B, C thang hang Nhan xet diTdc chuTng minh Quay trd lai bai toan, do A, B, C th^ng hang ncn theo nhan xet, ta co

XA + XB + Xc = 0 Theo (1), ta co XA- + XB- + Xc = -2(XA + XB + XC ) = 0

Tiep tuc ap dung nhan xet ta suy ra A', B', C th^ng hang (dpcm)

Bai 3. Cho ham so' y - — - 3x - - CO d6 thj (C)

Dat f(x) = x' - 3x' + 1 thi f(-l) = -3, f(0) = 1, f(l) = -I, f(3) = 1 nen theo tinh

chat ham lien tuc, phiTdng Irinh y' = 0 c6 3 nghiem XA, XB, XC thoa man dieu ki^n -1 < XA < 0 < XB < 1 < Xc < 3 Tir do suy ra dpcm

LuySn gUU dS trade thi DH 3 miin Bdc, Trung Nam ToOn hoc - NguySn Van ThOng

b) DiOn tich tarn gidc ABC c6 the tinh theo cong thtfc

Thay vao ta tinh di/ac X Y = 3 + 3 - 24 = -18

Tir day suy ra (X - Y ) ' = (X + Y ) ' - 4XY = 9 - 4(-18) = 81 * V

-jj^i 4 Xet parabol (P) c6 phiTdng trinh y = x l M6t diTdng tr6n c^t (P) tai cac

diem A, B, C, D c6 hoanh do tUdng u'ng la a, b, c, d Chtfng minh r^ng a + b +

c + d = 0

Hrf(?ngdSn giai o - J < i v , ,.,,5

Gia suT phu-dng trinh du^dng tron c6 dang (x - A)^ + (y - B)^ = R I

Thay y = x' vao va bien ddi, ta du"dc phu"ctng trinh hoanh do giao diem cua (P) va dirdng tron la x* - (2B - l)x^ - 2Ax + A^ + - R^ = 0 (1)

Theo gia thiet a, b, c, d la 4 nghiem ciia phu'dng trinh (1)

Ap dung dinh l i Vi-et cho phu'dng trinh bac 4, ta c6 a + b + c + d = 0

Bai 5 Trong mat phang vdi he toa do De-cac vuong goc Oxy, cho di/dng cong

(C): y = 2x'' - 3x^ + 2x + 1 va difdng thing (d): y = 2x - 1

a) Chilng minh rang diTdng cong (C) va diTdng thang (d) khong cat nhau u

b) Tim tren (C) diem A c6 khoang each den (d) la nho nhat

HUiing, dSn giai

a) PhiTdng trinh hoanh do giao diem cua (C) va (d) co dang

2x^ - 3x^ + 2x + 1 = 2x - 1

o 2 x ' ' - 3x^ + 2 = 0 (1)

Dat t = x^ thi ta duTdc 2t^ - 3t + 2 = 0 Phu'dng trinh cuoi cung nay c6:

A = 9 - 16 < 0, do do CO nghiem Tif do (1) v6 nghiem va nhu" vay ta co dpcm

b) Xet diem A(X(); yo) thuoc (C) Ap dung cong thtfc tinh khoang each, ta tim

duTdc khoang each tiT A den (d) b^ng d = yo • 2x0 + 1

Tur day ta tim diTdc hai diem A Ih A,

Chu y: Co the kicm tra lai r^ng tiep tuyen ciia (C) tai A , , A2 trung nhau va

song song vdi (d) Sir kien nay khong ngau nhien Co the chiJng minh diTdc r^ng neu A(x„; y„) thuoc (C) la diem co khoang each den (d) la nho nhat thi

•Jircfng thing qua A va song song vdi (d) se tiep xuc vdi (C) Day cung la mot ttnh cha't co the diing de giai bai loan khoang each. (Ou!, 11 'ii! n ':i> I v

Luy?n glai d? trwfr i-f tWi lui ' misn ttac. imng, nam wan niM- - lyguyen vurrrmmg

B a i 6 Cho ham so y = x ' + 3 X + 3

x + 1 a) K h a o sat va ve do Ihi ( C ) cua ham so

b) Xac dinh hai d i e m A , B Ian liTdl d tren hai nhanh cua (C) sao cho A B ngrii,

nhat

Hifdng d§n giai 0 ^ b +

a) Hoc sinh tiT khao sat

b) X e t d i e m A ( X A ; Y A ) ihtipc nhanh phai (tfng v d i X A > - 1 ) va B ( X B ; ya) thuot

iii nhanh trai (tfng v d i XQ < - 1 ) r oia a % tnii i

a) Khao sat sir b i e n thien va ve do thi (C) ' ' *

b) Dirdng thang (d): y = - x + m c^t do thj (C) tai hai d i e m phan bipt M , N Tim

phu'dng trinh quy tich trung d i e m I cua doan M N

HUdng dSn giai

b) Phu'dng trinh hoanh dp giao d i e m cija (d) va (C) c6 dang

x^ + 2 x + 5

x + 1

= - X + m o 2x + (3 - m ) x + 5 - m =: 0 Theo cong thiJc tinh tpa dp trung d i e m va dinh l i V i - e t , ta co

B a i 8 Cho hp (C J : y m^x^ + 1 T i m tren dU'dng thang y = 1 d i e m ma khong

V a y tat ca nhCTng d i e m can l i m la nhffng d i e m cd tpa dp (a; 1) v d i a < 1

B a i 9 T i m tat ca cac gia tri m sao cho (x + l ) ( x + 3 ) ( x ' + 4x + 6) > m v d i m p i x

HiiTdng d§n giai

X6t hhm so y = f ( x ) = (x + l ) ( x + 3 ) ( x ' + 4x + 6) thi ba't dang thifc da cho

dung v d i m p i x k h i va chi k h i m < m i n f ( x ) Dat t = x ' + 4x + 4 = (x + 2 ) ' > 0 thi f(x) = ( X - + 4x + 3 ) ( x ' + 4x + 6) = (t - l ) ( t + 2) = t ' + t - 2 = g(t)

Khao sat ham so' g(t) v d i t > 0, dc dang t i m di/dc m i n g ( t ) = - 2

t>()

Tiir do suy ra m i n I"(x) = - 2 va nhU' vay dap so cua bai toan la m < - 2

B a i 10 B i e n luan theo m so nghiem cua phu'dng trinh x ' - (4 + m ) | x | + 5 + I m = 0

Hrfdng d§n giai > i

+ m < - ^ : phi/dng trinh cd 2 nghiem;

+ m = - — : phu'dng trinh c6 3 nghiem;

+ —^ < m < - 2 : phu'dng trinh cd 4 nghiem;

+ m = - 2 : phu'dng trinh cd 2 nghiem;

+ - 2 < m < 2: phu'dng trinh vd nghiem;

+ m = 2 : phu'dng trinh cd 2 n g h i e m ; + m > 2: phu'dng trinh cd 4 nghiem

Htfdng dan: V i c t phUdng trinh l a i du'di dang

, a L - < y •

x 2 - 4 X + 5

X - 2

• = m

4^ + 5

Do thi h a m so'd vc' trai difOc suy ra tiT do thi hiim so' y = b a n g

X - 2

* c d c h gii? nguyen n h a n h p h a i (u'ng vdi x > 0) hOp \d\h doi xu'ng c u a nhanh

nay true tung

Bai 12 Cho f(x) = x ' - 3x ChiJng minh rSng neu x > y thi f(x) - f(y) > - 4 Dan

bang xay ra khi va chi khi nao?

' Hi/(?ng d§n gial *») " ^ n ( P ) ^ l i ft!) OD

Khao sat do thj ham so f(x) = x ' - 3x, ta thay f(x) dat cUc dai bkng 2 khi x =

- 1 va dat ciTc ticu b^ng - 2 khi x = 1 -»> r ' M * * " i ' « • \ r«J

//i/i/i 15 Doth f ham soy = x' - 3x

• Ncu y < - 1 thi i"(y) < 2 Ta c6 2 triTimg hdp:

+ Neu X < - 1 thi do l"(x) tang tren -oo; -1) nen la c6 l(x) - f(y) > 0 > - 4

• + N e ' u x > - 1 t h i l ( x ) > - 2 d o d 6 f ( x ) - r ( y ) > - 4 ^

• Ncu y > - 1 ihi X > - 1 va ta c6 f(x) > - 2 Ta lai xct 2 triTtJng hdp:

+ Ncu y < 1 thi f(y) < 2 va ta c6 r(x) - l(y) > - 4

+ Ncu y > 1 ihi do h i m so f(x) tang len (1; +oo), la c6 f(x) - l"(y) > 0 > - 4

Vay trong moi irifdng hdp ta dcu co f(x) - f(y) > - 4 (dpcm)

PHlJdNG T I ^ B X T PHl/dNG TRiNH M C vA L 6GARIT

Hfe PHl/dNG TRiNH, Hfe B X T PHIJCJNG TRiNH VA NfU L 6GAWT

I T 6 M T A T L I T H U Y E T jjuw-'u.] u,u,

1 phi/dng trinh cd ban

a phuang tiinh mu CO ban a" = m ( 0 < a ^ 1)

Neu m < 0 thi phifdng trinh v6 nghiem

, • 7 !

Neu m > 0 thi phu'dng trinh co nghiem duy nhat x = logam

b Phuang trinh Idgarit ca ban i

log„x = m (0 < a ^ 1) o X = a'" 3"^^^^ ^''^ '

2. Mpt so phu'dng phap giai phu'dng trinh mu va logarit 8'>Mi>

FhiTdng phap du'a vc cung cd so, phi/dng phap dat an phu, phifdng phap

logarit hoa, phu'dng phap sijf dung tinh dong bicn, nghjch bicn cua ham so

C h i i y : " • i ,

• Cho 0 < a ?t 1 Khi d6 ^ ' ^ \ + a"^'= a ' ' * ^ ' f ( x ) = g(x)

+, loga f(x) = log, g(x) « f(x) = g(x) > 0 < •'

• Cho ham so y = f(x) dong bie'n tren D va x,, e D

Khi do, phu'dng trinh l"(x) = f(X()) co nghiem duy nhat x = Xo tren D

3 phu'dng trinh mu va Idgarit

Phrfdng phap chung

• Khi giai cac he phi/dng trinh mu va logarit, ta difa vao cdc phep bicn doi ve

luy thiTa, mu v^ logarit va dilng cac phiTdng phap the, phiTdng phap cong dai

so, phifdng phap dat an phu,

• Can chu y dat dieu kien dc he phifdng trinh c6 nghiem, dac biet la cac bieu

thiJc n^m trong logarit ,

4 Bat phifdng trinh mu va Idgarit

a- Bat phuang trinh mu ca ban

• Ba't phifdng trinh a" > m ( 0 < a 7i 1) ' Neu m < 0 thi ba't phifdng trinh nghiem dung vdi moi gia tri cua x

Neu m > 0 va a > 1 Ihi nghiem cua bat phifdng trinh la x > log„m

Neu m > 0 va a < I l h i nghiem cua bat phifdng trinh la x < log^m ,

• Bat phifdng trinh a" < m (0 < a 1) •/ ,

Ne'u m < 0 thi ba't phifdng Irinh v6 nghiem '

45

Liiy?n gidi di trade thi DH 3 miSn Bdc Trung, Nam Todn hoc - NguySn Van ThOng

Ne'u m > 0 va a > 1 Ihi nghicm cua bat phtfdng trinh la x < logam

Neu m > 0 va a < 1 thi nghicm ciia bat phUdng trinh la x > iogam

b Bdt phuang trinh Idgarit ca ban

• Bat phiWng trinh logaX > m (0 < a 1) !•;

Neu a > 1 thi nghiem cua ba't phU'dng trinh la x > a'" j Ji-, j- fj

Neu 0 < a < 1 thi nghiem cua bat phU'dng trinh la a < x < a"" ,5^,;.,^

• Bat phU'dng trinh logaX < m (0 < a ^ 1)

Ne'u a > 1 thi nghiem cua bat phU'dng trinh la 0 < x < a'"

Ne'u 0 < a < 1 thi nghicm cua phU'dng bat phU'dng trinh la x > a'"

c Phuang phdp giai bdt phuang trinh mu vd logarit

Cung nhU'phi/(Jng trinh mu va logarit, dc giai bat phi/dng trinh mu va logarit ta

cung siif dung cac phifdng phap: di/a ve cung cd so, dat an phu, phiTdng phap

logarit hoa, phiTdng phap sOc dung tinh dong bicn, nghjch bie'n cua ham so,

Chii y : Khi giai cac bat phU'dng trinh logarit, ta phai dac biet chu y dieu kien

xac dinh cua bat phifdng trinh

ChuTng minh rkng ne'u x, y, z lap nen cap so'nhan thi

log^ M - logy M _ log^ M logy M - log^ M logy M

o x = l (Thoa dieu kien)

Vay phiTdng innh c6 nghi$m x = 1

b PhiTdng trinh tiTdng diTdng vdi:

l0g2 [(x^ + X + l)(x2 - X + 1)] = log2 [(x^ + X^ + l)(x^ - X^ + l )

o ( x ^ ^ l f - x ^ = ( x ^ + l f - x ^ o x « - x ^ = = O o ^ ^ : J '

c- PhiTdng trinh tiTdng diTdng vdi: 3" + 5* - 6x - 2 = 0 (1)

De thay x = 0; x = 1 l i hai nghi?m ciia phiTdng irinh (1)

X6t ham so f(x) = 3' + 5' - 6x - 2 tren R-

f (x) = 3Mn3 + S'lnS - 6 J ' f"(x) = 3"(ln3)' + 5x(ln5)' > 0; Vx e R ' ~ - i

=> f ( x ) dong bien tren R .: id.'tt 3" ^ ' '

A Matkhdc:

LuySn mi d? trade ky thiPH 5 mmBUc Trung, Nam loan / tySn van 'IhO

f(()) = ln3 + ln5 - 6 = lnl5 - 6 < 0

r ( l ) = 31n3 + 5 1 n 5 - 6 > 0

Suy ra ton tai duy nhat a e (0; 1) sao cho f (a) = 0

Bang bien thien

f(x) +00~.^ _

Vay phiTdng trinh c6 hai nghiSm x = 0, x = 1

Bai 3 Giai phu'dng trinh: logjoos = x'' - 3x^ -1

Dat t = x ^ t > 0 Phi/cfng trinh trd thanh: t^ - 3t - 1 = 0

Xet ham so f(t) = t^ - 3t - 1 tren [ 0 ; + 0 0 )

f (t) = 3t^ - 3

f'(t) = 0 o t = l

t = - l X6t bang bie'n thien

2(4cos'(p - 3cos(p) = 1 <=> cos3(() = ^ <=> cp = ^

Vay phiTdng trinh c6 dung hai nghiem x = ^2cos-^ \x= 2cos —

^ a l 5 Gi^i phircjng trinh 2'" + 3'" = 2" + 3"^' + x + 1

HUdng dSn giai

Phifdng trinh liTcfng difdng vdi

2 ^ ' + 3 ' " + 2 ' ' = 2 ' ' + ' + 3 ' ' ^ ' + x + l (1) ' •

AO

Khi do, phufdng trinh g(x) = 0 c6 nhieu nha't 2 nghi^m tren R

De thay x = 0, x = 1 la hai nghi^m cua g(x)

Trirdc he't, ta chufng minh bat d^ng thiJc sau:

Va > 0; 0 < a < 1, ta luon c6: a" < 1 + a(a + 1) (*)

That vay,

• Neu a = 1 thi ba't ding thrfc luon dung

• Neu 0 < a < 1 thi xet h^m so f(a) = 1 + a(a + 1) - a" tren (0; +oo)

Dang thiJc xay ra khi va chi khi a = 1

* Trd lai b^i toan:

Vdi mpi x e 4^2 , ta luon c6:

= 2''"^" l + sin^x(sinx-cosx) =2'""^''(cos^x + sin^x(sinx-cosx))

o x = —

sinx-cosx = l 2

:-2

• 2 -2

VT (1) < cos^ X + sin^ x(sin x + cos x) + 2''" \cos^ x + 2''" " sin^x(sinx - cosx)

<, cos^ x(l + 2'''"^'') + sin^x 2sinx + ^ 2 n

x + 1

a In - I n ( x ' - x + 1 ) > 0 b iog,(log3(9''-72))<l

Hifdng d§n giai

J B f | J I I ^ I I / r Tif iruiic K.f I \,ti mil., J I m i g , I i n A , • J>guKe.ii > u » J

iiung-X + 1

>ln(x^ - X + 1) Baft phiTdng trinh ti/cfng diTcJng vdi: In

Bat phiTdng trinh tiTdng difdng vdi: log3(9'' - 72) < x

o 9* - 72 < 3' (3' - 9)(3' + 8 ) < 0 < : > 3 " < 9 < : : > x < 2

Vay nghiem cua ba't phuTdng trinh la logy73 < x < 2

Bai 8 Giai cac ba't phifdng trinh

Li^n gidi M truOc kj/ thi DH 3 miin Bdc Trung, Nam Todn hgc - Nguyin Van Thdng

Bai 10. Tim tat ca gia trj cua tham so' a sao cho bat phtfOng trinh sau diTdc

nghiem diing vdi moi x ^0

Do do, dieu kien de phiTdng trinh (1) c6 nghiem la m > >/3 - ^fi;/;.,!,!

Bai 12 Tim m de moi x e [0; 2J dcu thoa man bat phU'dng trinh log2 Vx^ - 2 x + m + 4^1og4(x^ - 2 x + m) < 5 (1)

Trong do f(x) = x^ - 2x + m

Tac6f(x) = x ^ - 2 x + m l i e n t u c t r e n [ 0 ; 2 ] ; i

f'(x) = 2 x - 2

f (x) = 0 < ^ X = 1 Bang bien thicn

Luy^n giii dg truOc thi DH i '

Ke't luan: khi m = 3 ihi f(x) = 0 c6 nghiem x = 0, x = 1

Bai 14 Tim tat ca cac gia Iri cua m de l n ( l + x) > x - mx^ Vx > 0

Hiidng dSn giai

Xet ham so f(x) = l n ( l + x) - x + m.x^ Hen tuc tren [0; +QO )

Ta CO f (x) = — 1 + 2mx = — ( 2 m x + 2m -1)

1 + x x + 1 + N e u m < O t h i f ' ( x ) < 0 ; V x > 0

=> l(x) giam tren [0; +oo) => f(x) < f(0) = 0; V x > 0

1 /ff! 1 ^

2mx + 2 m 1

~ 2 j

I + x > 0 ; V x > 0 + Neu m > - ihi r(x) =

Vay m > - la ket qua can tim. ^ I j , | V Bai 15 Giai ba't phi/dng trinh: x'* - S.e""'> x ( x ^ e ' " ' - 8)

Hvldng d§n giai

Ta CO bat phi/dng trinh tifdng diTdng vdi: (x - e''"')(x^ + 8) > 0

Xet ham so r(x) = x - e*"' lien tuc tren R

Vay f(x) < 0 Vx e R hay X - e'"' < 0; Vx e R

Khi do, ba't phiTOng trinh ti/dng diTdng vdi | ^ <^ I'^ ' o <! o X < -2

X >

Ta CO phiTdng trinh (2) tifdng diTdng vdi (x y)(x +y) = a <=> x y

-x + y Thay vaophiWng trinh (l),tadirdc

Luyen gidi dS trade thi DH 3 mu'n H,h InuiiiJ^ajn / ; • • VgMyg/i Van TMng

log2(x +y) + log, = 1 <=> log2(x + y) = log,(x + y)

.x + y ;

Tru<yng h<fp / : a = 2, he phi/dng trinh trd thanh: log2(x + y) = log2(x + y)

x 2 - y 2 = 2

De thay (2;V2); (2; -^2 ) m cac nghiem cua he phv/dng trinh ' ' •

Suy ra he phifdng trinh khong c6 nghi?m duy nhat a = 2 (loai)

Trudn}' hop 2:&^2, he phiTcfng trinh trc( thanh

b Tim gia tri Idn nhat ciia m sao cho h? phUdng trinh c6 nghiem (x; y) thoa

man dieu kien 3x + 2y < 5

Ket luan: V d i m = 5 thi h? phiTdng trinh c6 nghiem (x; y) = ( 1 ; 1)

b H? phtfdng trinh (1) tiTdng diTcing vdi:

b=i

(l + log3 5).log3m r-* 1 + log3 m (l + loga5)logim ,

Do3x + 2 y < 5 n e n a < 5 — - 2 ^ - ^ — £ ^ = logj a < logj 5

1 + log3 m

^ I o g 3 n L l l 2 g 3 £ < o o - l < l o g 3 m < l o g 3 5 o i < m < 5 ^

V§y gia tri Idn nha't cua m la 5

Bai 18 Tim m de he phi/cfng trinh sau c6 nghiem:

Hifdng din giai

phifctng trinh tU'cfng diTcJng vdi

qua can tim

(2)

(3)

Luy^n gidi di irudc ky thi DH 3 ; ^ / , • iQC - Nguyln Van ThOng

X<3t ham so f(t) = t + logit lien luc tren (0; +00)

f ( I ) = 1 + 1 > 0 , V t > 0 I - V Y;:; ri \

t •••• rn

'S'i'i-t i n 3

=> f(l) la ham tang tren (0; +00)

Khi do f(u) = f(v) <=>u = v < = > x - a = y - 2 a o x - y = - a

> 0 PhiTdng trinh trd thanh 3t^ + t 2 = 0 <=> t = 1 hoac t =

->/2T^] > 0. PhiTdng trinh trd thanh - + t = 2 < = > t = l = > x = 0

d PhiTdng trinh tuTdng di/dng vdi +

H^m so f(x) =

X = 2 nghipm duy nha't

PhiTdng trinh tiTdng diTdng v d i

61

Bai 2 Giai cac phifdng trinh sau

a log2(x^ - 1 ) = log I (X - 1 ) b logjx + S.logsx = 5 + logj x l o g a

4

-X •

1 - 7 5

b Phi/dng Irtnh tiTdng difdng v d i (log2X - 5 ) ( 1 - logsx) = 0 o x = 5 hoSc x = 3'

c D i e u kipn: \ Phifdng trinh ti/dng difdng v d i

, X > - l

l 0 g 2 ( X + 1) = <=>log2(X + l) = 4 C : >

l0g2(X + l )

l o g 2 ( x + l ) = 2 l0g2(X + l ) = - 2 <=>

x = 3

3

x = —

4

d Phifdng trinh tifdng difdng v d i logy x + ^ + 9 ' ' = 9 ' ' o x = 9 2 = 1

e Phifdng trinh tifdng difdng v d i

Bat t = 2\g trinh trd thanh t^ - t = - 4 m ' '

X 6 t ham so f(t) = t^ - t lien tuc tren (0; +co)

• 0 < m < —

2 ^

B a i 5 T i m m dc phi/dng Irinh sau c6 nghiem duy nha't

ln(mx^ + (2m - 3)x + m - 3) = ln(2 - x)

, Hvtiln^ dan giai

PhiTcfng trinh tifdng diTJng vdi

3^ = 3 >a-2x Hi/dng d S n giai

Khi do de he phufdng trlnh c6 nghiem thl (1) c6 nghiem thoa 0 < v < 3

Khao sat ham so f(v) = 3 - v + - tren (0; 3), ta diTdc a > - 1

V

B a i 8 Giai bat phiTdng trlnh: '

5x + yj6\^+ - x"* log2 X > (x^ - X) log2 X + 5^6 + X - x^ (1)

TiJc la neu co nghiem sc co nghiem duy nha't p "* i p i ""^' * ' "

2) Neu y = r(x) ddn dieu U-en D thl f(u) = f(v) khi va chi khi u = v vcifi moi u, v e D

3) Neu y = IXx) co dao ham den cap k va lien tuc tren D, dong th(li f"^*(x) co dung m nghiem phan biet thl phu'dng trlnh f " ' " " ( x ) = 0 se co khong qua

X e i ham so f(x) = logjx + log3(2x - 1) + log5(7x - 9) v d i x > '''' '

" e u CO nghiem sc co nghiem duy nha't

M i f(2) = 2 nen phi/dng trlnh da cho c6 nghifim la x = 2

E>i^u kien x > 0 V i c t phiTdng trlnh da cho di/di dang log3 x - ^ = 0 ;

X 6 t h a m so f(x) = logj X — J v d i X > 0 ' 27

1 81

T a CO r'(x) = — — + —- > 0 Vx > 0 ncn ham so y = l(x) ddng bicn (0; +c

x l n 3 X"*

ma f(3) = 0 ncn phiTOng trinh = 0 c6 nghicm duy nha'l Ircn (0; Vj

nghicm ciia phiTtJng trinh da cho la x = 3

c) Chia ca hai vccua phiTdng trinh cho 9 \a difdc 5 + 2

Ncn ham so f(x) nghich bicn ircn R , ma 1(2) = 1 ncn phifdng trinh

Vay nghicm can tim cua phifttng trinh da cho la x = 2

Bai 2 Giai phiTdng trinh rn g i

a) 3* + 5" = 6x + 2

c) 7'-' = 1 + 2 i o g 7 ( 6 x - 5 ) \

Hif<}ng dSn giai a) Xct ham so f(x) = 3* + 5" - 6x - 2 veil x e R

Ta CO r(x) = 3Mn3 + 5Mn5 - 6 la ham so lien tiic va

r(0) = In3 + ln5 - 6 < 0, f d ) = 31n2 + 51n5 - 6 > 0

Ncn phiTdng trinh r(x) = 0 c6 nghiC^m duy nhat x = x„

Bang bicn thicn

f(x)

A ^ - — *

Tir bang bicn thicn ta lhay phi/cJng trinh r(x) = 0 c6 khong qui hai nghici'

phan bicl Ma 1(0) = f ( l ) = 0 ncn moi nghicm cua phiTcJng trinh da cho '

Vay l"(t) d"ng hicn ircn R do do 1 — j —

i l i i - = i — ^ , hay X = 2 Vay phiftJng irinh da cho c6 nghiym la x = 2

c) Dicu kicMi x > ^ Dat y - 1 = log7(6x - 5) thi 7^"' = 6x - 5 o i i (1) PhiTdng trinh da cho IrcF lhanh

7""' = l + 21og7(6x-5)-^ = l + 61og7(6x-5) = 6 y - 5 Tir (1) va (2) suy ra '

7"-' - 7 ^ - ' = 6 y - 6 x , h a y 7""' + 6 ( x - l ) - 7 y - ' + 6 ( y - l ) (3)

Dc lhay f(l) = 7' = 7' + 6t la ham so dong bicn ircn R , ma (3) c6 dang

f(x - 1) = f(y - 1) ncn (3) ti/dng diTdng vdi x = y

Khi do phi/cJng trinh da cho c6 dang 7""' - 6x + 5 = 0

Tir bang bicn ihicn la lhay phi/dng trinh f(x) = 0 chi c6 khong qua hai

nghicm phan biCn Ro rang g ( l ) = g(2) = 0 nen x = 1 x = 2 la cic nghicm

Vay phiTifng trinh da cho c6 hai nghipm 1^ x = 1 hoac x =: 2

•^ai 4 Giai bat phUdng Innh

a)7''<5^+2.3V2N/3^ ^

^) •og2JVx2-5x + 7 + l ) + l o g 3 (x2+9-5x) ^5

6

LuySngididg trudc ky thi DH 3 miSn Bdc, Truny, Sun: !ndn ho, S^u)Cn ian Ihdng

7

I n — < 0 V x e R

7 Ham so f(x) nghjch bicn Iren R nen nghicm cua ba't phi/dng trinh (1) la

nhiTng gia tri cua x thoa man x < 2

Tap nghiem cua bat phiTdng trinh da cho 1^ (-oo; 2) H ) V

5 V

-b) Ta CO X - 5x + 7 = X — + - > 0 Vx 6 R nen bat phi/dng trinh xac djnh

vdi mpi X e R Vict ba't phi/dng Irinh da cho difdi dang

+ 31og3(x^ + 9 - 5 x ) < 5

21og2 V x ^ - 5 x + 7 + 1

Dai t = Vx^ - 5 x + 7 , t > ^ T a d i r d c 21og2(l + l ) + 31og3(t^+2)<5

X6t f(t) = 21og2(l + 1) + 31og3(t^ + 2) ihi f ( l ) = 5 nen bat phiTdng irinh dMc

— ; + c o , nen bat phiTdng trinh (2)

liTdng diTdng vdi t < 1 Do do Vx^ - 5 x + 7 < 1 hay x^ - 5x + 6 < 0, nen 2 < x

< 3 Vay tap nghiem ciJa ba't phiTdng trinh da cho la [2; 3]

jvjen ham so y = f(x) dong bicn tren khoang (0; +co)

y«, f(2) = 0 ncn bal phifdng trinh f(x) > 0 c6 nghiem ia x > 2

Vay tap nghiem can lim \k [2; +00)

P H l / O N G PHAP M O H O A L O G A R I T H O A

Nh§n xet: PhiTdng phap giiip la chuyen mpl phifdng trinh hay ba't phiTdng

trinh mu (logaril) ve mot phifdng trinh hay bat phu-dng logaril (mu) ma la da

• bic'tcachgiai l y j ^ v " ' l y j " " " ^ " '

B a i l G i a i phifdng trtnh _^ - M H.irf,., V

l o g , l o g x = l o g , l o g , X ^ " ^ > b) 5\ =500 c) 31og, l + V x + N / X =21og2>yx , ^ v j "

Ta cung c6 iog2logsX = l nen log-iX = 2" (2)

Vi log.x = log.S.logsX nen lif (1) va (2) la c6 5' = 2'.log25, hay

Vay phifdng trinh c6 nghiem la x = 2

^) Vic'i phif(Jng trinh da cho difc^i dang 5\2 ^ ' ^ =5\2~, hay 2

Lay logaril cd so 2 hai ve la c6 log, 2 = log,5^-" hay 3 x - 1 •2 = ( 3 - x ) l o g 2 5 (3)

-2

V X J ^3 •3-x

Phudng irinh (3) lifdng difcJng vdi

- 3)(l + xIog25) = 0, suy ra X = 3 hoAc X =-log52 ' ' '"^^

^ i i y phifdng trinh dii cho cd hai nghiC'm la x = 3 hoac x = -log52 •• '

kien x > 0 Dai 6l = 31og,,(l + N/X + ^ ) = 2log2 \fx , ^ , • ^

c) Di6

irinh (5) c6 nghicm duy nhal I = 2, do do x = 2 ' ' = 4096

Vay nghicm cua phifitng trinh da cho la x = 4096

Bai 2 Giai ba't phifitng Irinh

a) i " - - * " <2^-'» " b) x'"«-*^ > 8 x ^ c) log7X<log3(s/^ + 2 )

Hif(}ng dSn giai a) Lay logaril cit so 3 hai vc, la diTitc *

log3 3''^ < log, 2""'' hay x-^ - 4 x < ( x - 4 ) l o g , 2 , X - 4 (1)

Phifdng irinh (1) vict biTdc di/i^i dang (x - 4)(x - Iog32) < 0

V i logi2 < log,3 =1

Ncn lap nghicm ciia ba'l phifiJng Irinh da cho Ih [logil; 4]

b) Dicu kicn x > 0 Lay logaril c(* so 2 hai vc cua bat phi/tlng trinh, ta diTdc

logjx'"'^'^''" >iog2(8x-),hay log24x.log2x>logjCSx^) (2)

Dai 1 = log.x ihi

- log2 4x = log, 4 + logj X = 2 + 1 , logj(8x^) = logj 8 + iogj x^ = 3 + 2l "

Ncn phirdng trinh (2) In'* thanh 1(2 + t) S 3 + 2t, hay t^ ^ 3

V i the t > 73 hoilc t < S

Ttf do la CO nghicm cua ba'l phifdng irinh da cho la | - o o ; 2 ^

c) Dicu kicn x > 0 Dai t = logvx ihi x = 7' (3)

(Dc Ihi luycn sinh Dai hoc khoi A nam 2(K)8)

(Dc thi luycn sinh Dai hoc khoi D nam 2010)

lif }/ HiAJnR dSn jjiai ^

f x > 0 a) Dieu ki<;n

V d i x > 1 ihi | x - l | = x - 1 ncn(l)tri)thanh f '

<x + 3)(x - 1) = 4x hay x^ - 2x - 3 = 0 .suy ra x = -1, x = 3 ;

So sanh dicu kiOn la CO X = 3

Vi^i 0 < X < I thi | x - l | = 1 - X ncn (1) inH lhanh ^ r y

(x + 3)(l - x) = 4 x h a y x - + 6 x - 3 = 0,suy ra x = - 3 ± 2 > / 3

71

So sanh dicu kicn, ta nhan x = -3 + 2>/3

Vay cac nghiem can lim cua phiTctng trinh la x = 3 hoSc x = -3 + 2^3

Rut gon (2), difdc x^ x 4 = 0 ncn x

-So sanh vdi dicu kipn ta c6 nghiem can tim la x =

Gia silr X = coscp vdi (p e (0; n) PhiTdng trinh (3) trd thanh

4cos^cp-3cos(p = — hay cos3(p = — nen (p = ± —+ - ! ^ ^ , k e Z

2 2 9 3 j j

-Do (p 6 (0; 71) nen (p- ^ suy ra (3) c6 ba nghiem la

cos—; cos—; cos — >

9 9 9 J Ma phiTdng trinh bac ba co khong qua ba nghicMii

nen do la moi nghiem cua (3)

V i cos — < 0 ; cos — < 0 ncn vdi t > 0 thi t = cos—

9 9 9

Vay nghiOm can tim cua phiTdng trinh da cho la x = log,^^ '2cos^^

9 J 'ir>v

+ X)

72

V i 2x^ + X - 1 = (2x - l)(x + 1) n e n di^u ki^n \ic d i n h ciia phiTdng trinh \k

i < X ^ 1 Khi do phiTdng Irinh da cho tiTdng diTdng vdi

10g2x-l(2x-l)(X + l)+10g,^,(2x-l)^ =4 £ ''^1 r)J - ' ' '

Hay log2x-i(X +1) + 2 log.^i (2x -1) - 3 = 0

Dat t = log2x- ,(x + 1) t h i i o g , „ ( 2 x - l ) = ^ nen la c6 I "^'^ ' ^

r 1 \

1 + 1-3 = 0 h a y l ' - 3 l + 2 = 0 n e n l = 1,1 = 2 ^ " ' ^ ^ V ^ xj ^ij' •

I

+) Vdi t = 1 ihi log2x- i(x + 1) = 1 nen x + 1 = 2x - 1 hay x = 2

+) Vdi I = 2 thi log2x- i(x + 1) = 2 nen X + 1 = (2x - 1)^ hay x = 0 (loai), x =

+) 22^=2""^-'' thi x ' - 4 = 2V5rr2 nentaco: x " " * - 8 - 2 ( N / ^ - 2 ) = 0

2 Di»ng bieu thiirc lien hdp, ta co (x - 2)

Vay cac nghiem can tim cua phiTdng trinh 1^ x = 1, x = 2

Bai 4 Giai ba't phiTdng trinh

x^ + 3 x - 4 > 0

V a y nghiem can t i m cua ba't phiTdng trinh 1^ x < - 4 hoSc x > 2

Chu y: Qua b a i l a p t r c n , ta tha'y khong phai luc nao cung n c n dat d i e u kien

+) G i a i bat phiTcJng trinh (2) ta c6

+) G i a i ba't phiTdng trinh (3) ta c6

" 0 < x < l _ x > 2 '*

' x < 0

2-sf2<x<2 + sf2

K e t hdp ta diTdc tap nghipm can t i m la [ 2 - N/2;l) U (2:2 + N/2

c) V i e t ba't phiTdng trinh da cho v c dang

log.,(4' + 144) - log., 16 < logsS + log5(2'" ^ + 1)

Yi the 2"' ' < 2 hay x" - 2x - 2 < 1, nen - 1 < x < 3

Vay nghiem can l i m la [-1;3 Bai 5 G i i i i phifdng trinh

a) x ' " " ' = x - 3 ' " « ' ' - x ' " « 2 '

b) l o g : ( l + cosx) = 2co.sx c) 5' + 2 ' = 3 ' + 4 '

Ta CO a'"^'''' = c'"^''" n e n phifdng trinh da cho c6 dang ^^^ '^^ ^^^ ,

9i"i:2^ ^ x^ 3 ' " " ' - 3 ' " " " hay 3'""' (3'"'52 x - x ^ + l ) = 0

D o d o 3 ' " " ' - x ^ + 1 = 0 Dat I = logiX ta difdc 3' - 4' + 1 = 0 hay 4 ; + — = 1 (1)

De lhay f ( t ) =

1 r3Y

— + — \h h a m so nghich b i e n f ( l ) = 1 n e n phifdng

irtnh (1) CO n g h i p m duy nha'l t = 1 Suy ra x = 2

V a y phifdng i r i n h da cho CO nghiOm duy nhat X = 2

E>ai t = cosx thi l i f phifdng trinh c6 dieu k i p n cua t la - 1 < t < Phifdng trinh da cho trci thanh log2(l +1) = 2t hay 4' - t - 1 = 0 (2)

1-Xet ham so l(t) = 4' - t - 1 vdi - I < t < 1

{1-Tit bang bien thien ta thay diTc^ng th^ng y = 0 c^t do ihi ham so y = f(t)

khong qua hai diem, nen phU'dng trinh f(t) = 0 c6 khong qua hai nghiem phari

biet Ma f(0) = f I]

2, = 0 nen (2) c6 diing hai nghiem la t = 0, t = —

Tir do ta tim diTctc cac nghiem cua phifdng trinh da cho la

X = — + kn hoSc x = ± — + k27i vc'Ji k e Z

3 3

c) Viet phifdng trinh da cho ve dang 5" - 4' = 3' - 2\

Gia sur x = a la nghiem cua phU'dng trinh V

Ham so' f(t) = (t + 1)" - t" lien tuc va c6 dao ham trcn (0; +oo), ddng ihdi

f ( x ) =

2 + 4" 3 4''ln4.(2 + 4 ' ' ) - 4 ^ M n 4 l _ 2 4 M n 4 1

(2 + 4")^ 3 (2 + 4") 3

V i the r(x) = 0 khi va chi khi ^'""^"^ - - - 0, hay (2 + 4") - 6 In 4.4" = 0 (3)

(2 + 4") 3

PhiTdng trinh (3) Ih phi/dng trinh bac hai theo an la 4" nen phi/dng Irinh nay

CO nhieu nhat la 2 nghiem, suy ra phU'dng trinh f(x) = 0 co nhieu nhat 3

f 1 ^ nghiem Ta thay f(0) = 0, 1" - = 0 , 1(1) = 0 nen cac nghiem c^n tim cua

phifdng trinh da cho la x = 0, x ^ , x = 1

COng ty TNHH MTVDWH Khang Vi$t

e) Dieu kipn x > 0

pat t = 10g7X thi x'"S7 " = 11'"87 " = 1 1' , X = 7'"«7 * = 7' Phi/dng trinh da cho trd thanh 11' + 3' = 2.7' hay = 2 (4)

.7 l n ^ - > 0 nen r(t) la ham so dong bie'n nen

f (t) = 0 CO khong qua mot nghiem, do do phuTdng trinh f(t) = 0 co khong qua

hai nghiem phan biet „ ^ ^ ,

Ma f(0) = f( 1) = 0 nen cac nghiem can lim cua (4) la t = 0, t = 1

Vay nghiem cua phU'dng trinh da cho la x = 1; x = 7 , ^ ^ j ^ , , ,

0 Dieu kien x > 0

Vic't phiTdng trinh da cho ve dang log2(2''-1)-logj x = I + x - 2 " hay

2 " - l + log2(2"-l) = x + log2X

Vi ham so 1(1) = t + logjt ddng bie'n tren (0; +oo) nen

i{T - 1) = f(x) khi va chi khi 2^ - 1 = x (5) <

Ttfdng tir cau a, ta tim dufdc cac nghiem ciia (5) la x = 0, x = 1 ''

So sanh vdi dieu kien ta co nghiem can tim 1^ x = 1

Bai 6 Giai phi/dng trinh a) 5" + 3' + 2' = 28x - 18 ' b) (4* + 2)(2 - x) = 6 !} hi

13Ix c) 5"+2" - 2 - - + 441og2 2 - 5 " +

d) 4 " + 2 " = l l x - ^ - 9 x 2 + ^ x + 2

3 3

Hi^dng d i n giai a) Xet ham so f(x) = 5" + 3" + 2" - 28x + 18, x e

Ta CO f ( X ) = 5'ln5 + 3Mn3 + 2"ln2 - 28

r(x) = 5"ln'5 + 3'ln'3 + 2'\n^2 > 0 Vx e R

Vi r(x) > 0 Vx G R nen f (x) la ham so dong bie'n tren R

Ma f ( X ) lien tuc tren R va f (0), f (2) > 0 nen f (x) co duy nhat mot nghiem x,,

Bang bie'n thien * , ; • •

Luy^n gidi di iruOc Aj* thi DH 3 miin Bdc Trung, Nam Todn hoc - Nguyin Van Thdne

Tii bang bicn ihicn suy ra phiTdng irinh f(x) = 0 c6 khong qua hai nghicni

phan bict M a f( 1) = 1(2) = 0 ncn moi nghicm can l i m la x = 1, x = 2

Vay cac nghiC'Hi can lini la X = 1, X = 2

TiTdng luf cau a, la l i m diTdc nghicm cua (3) 1^ x = 0, x = 3

Vay cac nghicm can l i m ciia phi/dng irinh la x = 0, x = 3

d) X < 5 t h a m s o l ( x ) = 4 ' + 2 ' - j x U 9 x ^ - y X - 2 x e R

•«-f ( x ) = 4Mn4 + 2Mn2 - 14x^ + 18x - —

Ta CO r ( x ) = 4 ' i n ' 4 + 2Mn'2 - 2Xx + 18

r"(x) = 4Mn'4 + 2Mn'2 - 28, l'-"(x) = 4Mn''4 + 2Mn'*2

V i t^^'(x) > 0 V x e R nen phifdng irinh f(x) = 0 c6 khong qud bon nghicm

phan biC'l M i l f(0) = f ( l ) = f(2) = f(3) = 0 nen cac nghicm can tim ciia

phi/dng irinh da cho x = 0, x = 1 x = 2, x = 3

B a i 7 G i a i hi; phi/dng trinh

y > 0

1 '"^ „v.^

•

Tir phiTiJng Irinh ihi? nhal ciia he la c6 !og2(x - 3y) = 2, hay x - 3y = 4 (1)

Lay logaril c^.^ so' x hai ve phu'itng irinh ihiJ hai, la diTdc

log, X + l o g , y'"*^^ y = l o g , , hay 1 + log; y = ^ l o g , y (2)

Giai phifctng Irinh (2) di den log,y = 2; log.y = ^

2 0 < v , 0 < 3 c : +) V(^i log^y = 2 Ihi y = x^ nen k e l hdp vdi (1)

Suy ra 3 x ' - x + 4 = 0 (v6 nghicm) +) V d i l o g , y = — i h i X = y ' nen k e l hdp vdi (1)

So sanh vdi dieu k i c n ta c6 S = 4; P = • "v.;

Do do u , v la nghiem cua phifdng trinh l^ - 4t + - = 0 (3)

Giai (3) ta dUWc t = ' ^ - ^ T I T d 6 suy ra ^5, ; ' ; ^