HÀM SỐ LŨY THỪA HÀM SỐ MŨ VÀ HÀM SỐ LOGARIT HAY NHẤT

Luy thiia vdi sd mu nguyen; Phuang trinh x" = b ; Can bac n ciia mot sd duong; Liiy thira vdi sd mu hiin ti va luy thira vdi sd mu vd ti.. Cac tfnh chai ciia liiy thira vdi sd mu thuc; L

Ndi dung chfnh ciia chuong II :

' Luy thira la gi ? Luy thiia vdi sd mu nguyen; Phuang trinh x" = b ; Can bac n ciia mot sd duong; Liiy thira vdi sd mu hiin ti va luy thira vdi sd mu vd ti

Cac tfnh chai ciia liiy thira vdi sd mu thuc; Luy thira ciia mdt tfch va mot thuang, tfch hai liiy thira va thuang hai liiy thiia

Ham sd y = x" , dao ham ciia ham so y = x**, khao sat ham so y = x "

• Logarit la gi? Cac tfnh chat cua Idgarit Mot sd quy tic tfnh Idgarit, Idgarit ciia mot luy thiia, phuang phap ddi co so, Idgarit tu nhien va Idgarit thap phan

Ham sd mii va ham sd Idgarit :

Khai niem ham sd mu, khao sat ham so mu

Khai niem ham sd Idgarit, khao sat ham sd Idgarit

Phuong trinh mu va phuong trinh Idgarit : phuong phap giai va mdt so phuong trinh don gian

Bai phuang trinh mil va bai phuang trinh logarit : phuang phap giai va mdt sd phuong trinh don gian

IL MUC TIEU

1 Kien thiic

Nim dugc toan bd kie'n thiic co ban trong chuang da neu tren, cu the':

Khai niem luy thiia vdi sd mu thuc va mdt sd tfnh chat cua nd

Ham sd mu la gi; khao sat dugc ham sd mu

Khai niem ham sd logarit va khao sat chiing

• Giai dugc mdt sd phuang trinh va bai phuong trinh mu, Idgarit

Mdi quan he giiia ham sd mii va ham sd Idgarit

2 KT nang

Khao sat tdt ham sd mii va ham sd Idgarit

• Giai thanh thao phuong trinh, bit phuong trinh mQ va Idgarit

- Ve dugc dd thj cac ham sd mii va ham so Idgarit

Mdi quan he giiia hai ham sd tren

3 Thai do

Tu giac, tfch cue, dgc lap va chii ddng phat hien cung nhu ITnh hdi kie'n thiic

trong qua trinh boat ddng

- Cam nhan dugc su cin thie't cua ham so mii va ham sd Idgarit trong thuc te'

• Cam nhan dugc thuc te' ciia toan hgc, nha't la dd'i vdi mu va Idgarit

P h a n Z

C A c B A I SOAI!^

§1 Luy thufa (tieTt 1, 2, 3)

I M U C T I E U

1 Kien thirc

HS nim dugc :

Nhd lai liiy thira vdi sd mii nguyen

' xay dung dugc khai niem luy thira vdi so mu thuc

Hie'u va van dung dugc mdt so tfnh chat ciia liiy thiia vdi so mu thuc

2 KT nang

Sau khi hgc xong bai nay, HS phai bie't khai niem ciia liiy thira vdi so' mil thuc

Van dung dugc cac tfnh chat trong giai toan

Nim dugc mdi quan he giiia liiy thira vdi so mu thuc vdi phuang trinh

x " = b

" Lien he vdi mdt sd liiy thiia da hgc

3 Thai do

* Tu giac, tfch cue trong hgc tap

Biei phan biet rd cac khai niem co ban va van dung trong tirng trudng hgp

• Chuin bj cac hinh tir hinh 26 den hinh 27

Chuan bj pha'n mau, va mdt sd dd diing khac

2 Chuan bi cua HS

Can dn lai mdt sd kien thiic da hgc ve luy thira da hgc d Idp dudi

HI PHAN PHOI T H d l LUONG

Bai nay chia lam 3 tiet :

Tie't I : Td ddu din hit muc 3 phdn I

Tie't 2 : Tie'p theo din hit phdn I

Tii't 3 : Tiip theo din hit phdn II

IV TIEN TRINH DAY - H O C

A OAT VAIN DE

Cau hdi 1

Xet tfnh diing - sai cua cac cau sau day :

a) Vdi mgi a thi a.a = a^

b) Chi cd a > 0 thi mdi xay ra a.a = a^

GV : Khing djnh a) diing, cdn khang djnh b) sai Cd the din ra cac vf du

B BAI Mdl

I KHAI NIEM LtJY THlTA

HOATDQNGl

1 Luy thira vdi sd mu nguyen

• Thuc hien Qy I trong 5'

Sau dd ta in not a ta dugc ke't qua

Ggi y tra loi cau hdi 3

[^=^S

HS cd the sii dung may tfnh dien td

ba'm: (^^3)^5 =

• GV neu djnh nghia:

Cho n Id mgt sd nguyen duang

Vdi a la sdthuc tuy y, luy thica bdc n ciia a la tich cua n thda sda

a"' = a.a a

n thijfa so Vdi a^O

• GV neu vf du 1 GV ed the lay vf du tuong tu

Hoat ddng cua GV Hoat ddng cua HS

• GV neu vf du 2 GV cd the la'y vf du tuong tu

Hoat dgng ciia GV

Cau hoi 1

Hay phan tich trong ngoac

thanh nhan tir

Cau hdi 2

Hay phan tich ngoai ngoac

thanh nhan tur

Cau hdi 3

Tfnh B

Hoat ddng ciia HS Ggi y tra loi cau hdi 1

[aV2(l + a^) - 2V2a]

= (aV2+ a ^ V 2 - 2 a V 2 )

= aV2(a^ - 1 )

Ggi y tra loi cau hoi 2

a^il-a"^) a^-a

aia^-Ggi y tra loi cau hoi 3

Ggi y tra loi cau hdi 1

• Vdi mgi 6 e R, phuong trinh

o

X = b ludn cd mdt nghiem

Ggi y tra loi cau hoi 2

Vdi 6 < 0, phuang trinh x^ = b

khong cd nghiem

• Vdi 6 = 0, phuong trinh x = 0 cd

mdt nghiem x = 0

• Vdi 6 > 0, phuang trinh x = 6

cd hai nghiem trai da'u

• GV dua ra nhan xet:

Dd thi cua hdm sd y = x^*"*"' tuang tU dd thi hdm sd y = x^ vd dd thi hdm sd y = x'^* tuang tu do thi hdm sd y = x'^ Tu: dd ta cd ke't qud bien ludn so nghiem cua phuang trinh x" = b nhU sau

Trudng hap n le

Vdi mgi sdthUc b, phuang trinh cd nghiem duy nhdt

Trudng hap n chdn

Vdi b < 0, phuang trinh vd nghiem ;

Vdi b = 0, phuang trinh cd mgt nghiem x = 0 ;

Vdi b > 0 phuang trinh cd hai nghiem trdi dd'u

H3 Tim sd nghiem phuang trinh : \^ = 2008 , x^°°^ = -2008

H4 Tim sd nghiem phuang trinh : x^°°^ = -2009, x^°°^ = 0 va x^^^^ = 2009

•Bie't a tinh b

•Bie't b tinh a

Bdi todn thd nhdt la tinh luy thica cua mgt sd Bdi todn thic hai ddn de'n khdi niem lay cdn cua mdt sd

• GV neu djnh nghTa :

Cho sdthuc b vd sd nguyen duang n >2 Sda dugc ggi la cdn bdc n cua sdb ne'u a " = b

H5 Hay chiing minh 2 la can bac hai cua 4

H6 - 2 cd phai la can bac hai cua 4 hay khdng?

• GV neu bien luan:

Tddinh nghia vd kit qud bien lugn ve sd nghiem cua phuang trinh x'' — b, ta cd :

Vdi n le, Z? € R phuang trinh cd duy nhdt mgt cdn bdc n ciia b, ki hieu Id'ilb

^ b <0 : Khdng tdn tgi cdn bdc n cua b ; Vdi n chdn x"— b = 0 : Cd mot cdn bdc n cua b la sd'O;

b > 0 Co hai cdn trdi ddu, ki hieu gid tri duang

la y/b cdn gid tri dm la —yjb

• Thuc hien ^ 3 trong 5

Hoat ddng cua GV

Can hdi 1

Tfnh ^ 3 ^ 3

Hoat ddng ciia HS Ggi y tra Idi cau hdi 1

Tacd ^3V3 =^(V3)^

4 Luy thira vdi sd mu hufu ti

• GV neu dinh nghia :

HOATDQNG 4

m

Cho sdthuc a duang vd sdhifu ti r = — trong dd m e Z, n e N

Luy thita cua a vdi sd'md r Id sd af xdc dinh bdi

5 Luy thiira vdi sd mu vo ti

• GV neu van de va cho HS kiem tra bang sau bing may tfnh dien tir

S'-n

3

4,655536722 4,706965002 4,727695035 4,72873393 4,728785881 4,728801466 4,728804064 4,728804376 4,728804386

Sau dd cho HS thue hien tuong tu va dien vao bang sau :

GV cd the lay 2 nhdm HS va cho dien thi tren bang

2'-• GV neu dinh nghia :

Ta ggi gidi hgn ciia ddy sd I a'" I la Idy thita cua a vdi sdmii a

ki hieu Id a "

a " = lim a^"- vdi a = lim n

n->+oo

• GV neu chii y :

Tir djnh nghTa ta cd l " = l{a e R)

HIO Hay chiing minh chii y tren

HOATDQNG 6

II TINH CHAT CUA LUY THUA VCJl SO MU THUC

• Thuc hiin "pt 4 trong 5'

GV ggi 2 nhdm HS len bang thi vie't cac tfnh chat ciia luy thita vdi so mu nguyen duong Sa dd tong ket

• GV neu cac tfnh chat cua luy thira vdi sd mu nguyen duong

a" aP = a"'!" ; ^ = a"'^

a^

(a") a\0 _ ^ap iabf = a"b"; a a

bJ b"

Neu a> 1 thi a" > a" khi vd chi khi a> fi

Ni'u a< 1 thi a" < a" khi vd chi khi a> /3

• Thuc hien vf du 6 trong 5'

HOAT DONG 7

TOM TfiT Bfll HPC

L Cho n la mdt so nguyen duong

Vdi a la sd thuc tuy y, luy thilfa bac n dua a la tfch eiia n thiia sd a

Trong bie'u thirc a'" ta ggi a la co so, m la sd mu

2 sd nghiem ciia phuang trinh x" = b nhu sau :

Trudng hgp n le

Vdi mgi sd thue b, phuang trinh cd nghiem duy nhat

Trudng hgp n chin

Vdi b <0, phuong trinh vd nghiem ;

Vdi b = 0, phuong trinh cd mdt nghiem x = 0 ;

Vdi b>0 phuang trinh cd hai nghiem trai dau

3 Cho sd thuc b va sd nguyen duong n > 2 So a dugc ggi la can bac n cua

5 Cho sd thuc a duong va sd hiru ti r = — trong dd m e Z, n&N Luy thiia ciia a vdi sd mii / la so a^ xac dinh bdi

m

^L.m

a - a"- = yja

6 Ta ggi gidi ban ciia day sd j a'" j la luy thiira ciia a vdi sd mu a , kf hieu la a*^

a" = lim a'" vdi a = lim r^

n—>'+^ ;j_^+00

7 a" a^ = a"^^

iab)" = a"b";

a a'^

= a"-^ ; ia")^ = a"^ ;

," a

a a bJ h"

Neu a > 1 thi a " > a>^ khi va chi khi a> fi

Neu a < 1 thi a" < af^ khi va chi khi « > /?

HOAT DONG 8

MQT SO CfiU H 6 | TRfiC N G H I | M ON TQP Bfil 1

1 Hay dien diing sai trong cac khing djnh sau

4, Hay dien cac dau thich hgp trong cac dau : > , < , = vao chd trdng sau

8 Trong cac khing djnh sau khing djnh nao diing (a) 2^+2^ = 2 ^ (b) 2^-2^=2"^ (c) 2l2^ = 2^ ; (d) 2^ : 2^ = 25

Trd Idi ia)

12 Trong cac khing djnh sau khing djnh nao diing ? (a) V2 Vs = 2 ; (b) ^2.^4 = 4 (c) V2.V4 = Vs ; (d) V2.V4 = 2

Trd Idi id)

HOAT DONG

9-HaOfNG DfiN Bfil T6P SfiCH GIfiO KHOfi

Bai 1 Hudng ddn Dua ve luy thiia cua ciing mdt co sd rdi sir dung tfnh chat ciia

luy thiia vdi sd mu thuc

cau a) GV cho HS len bang chiia bai vdi nhimg ggi y sau day

Hoat ddng ciia GV Hoat ddng ciia HS

Cau hdi 1 Ggi y tra Idi cau hoi 1

Bai 2 Hudng ddn Sir dung tfnh chat cua luy thira

GV cho HS len bang chira bai vdi nhiing ggi y sau day

Bai 3 Hudng ddn Sir dung tfnh chit ciia liiy thira ciing ca so

GV cho HS len bang chiia bai vdi nhiing ggi y sau day

caub

H o a t ddng c u a G V

C a u hdi 1

^ o \ - l 1 Tfnhcacgiatrj: 9 8 ° ; f - J ;325

C a u hdi 2

Hay s i p xep theo thii tu tang

din

H o a t d d n g ciia H S Ggi y t r a loi c a u hdi 1 98° = 1; - = i ; 3 2 5 = ^ = 2

Ggi y t r a loi cau hdi 2

7

Ta cd 1 < 2 < — nen :

3 98° < 325 < f 3 j

Bai 4 Hudng ddn Su dung tuong tu vf du 6 trong SGK

GV cho HS len bang chira bai vdi nhiing ggi y sau day

4 1 4 2

T S = a 3 3 + a 3 3 = a ( a + l) Ggi y t r a loi c a u hdi 2

1 3 1 1

M S = a4 4 + a 4 4 = a + l Ggi y t r a loi c a n hdi 3

4 / _1 2 \

a 3 \ a 3 + a 3 /

1/ 2 _i) -"

a 4 \ a 4 + a 4 /

Hay riit ggn bieu thiic

Ggi y tra Idi cau hdi 1

1 1 2 2

T S = a 36 3 ( a 3 _ 6 3 ) Ggi y tra Idi can hdi 2

2 2

MS = a3 - 63 Ggi y tra loi cau hdi 3

1 1 2 2

a 36 3 ( a 3 - 6 3 ) 1

2 2 3 ^ a3 - 6 3

ia ^ 0, b ^ 0, a ^ b)

Bai 5 Hudng ddn Sd dung tfnh chat cua liiy thiira

,, 1 a) Ta cd 2V5 = V20, 3V2 = Vl8 nen 2V5 > 3V2 Vi ca sd - nhd hon 1 nen

o tir dd cd bit ding thiic cin chiing minh

b) Tuang tu, GVS = Vl08 > V54 = sVs ; 7 > 1 nen 7 ^ ^ > 7 ^ ^

" Khai niem ham so luy thu'a la gi ? Djnh nghia va mdt sd chii y

Tfnh dugc dao ham ciia ham so luy thira

Khao sat ham liiy thiia, chieu bien thien va dd thj

• Tu giac, tfch cue trong hgc tap

Bie't phan biet rd cac khai niem co ban va van dung trong tiing trudng hgp

cu the

- Tu duy cac van de ciia toan hgc mot each Idgic va he thdng

II C H U A N BI C U A G V VA HS

1 Chuan bi cua GV

Chuan bj cac cau hdi ggi md

Chuan bj pha'n mau va mdt sd dd diing khac

2 Chuan bi cua HS

• Can dn lai mgt sd kie'n thiic da hgc ve dao ham d Idp 11

III PHAN PHOI THCJI LUONG

Bai nay chia lam 2 tie't :

Tii't 1 : Tit ddu din hit phdn I vd II

Tii't 2 : Tii'p theo din hit phdn III; Phdn bdi tap

IV TIEN TRINH DAY - H O C

• GV neu khai niem ham so liiy thira

Hdm sd y - x"' vdi a eM., dugc ggi la hdm sd'luy thda

HI La'y mot sd vf du minh hga

• Thuc hien J^ 1 trong 5'

GV treo hoac chie'u hinh 7, hinh 8 lan bang va cho HS thuc hien

cau a)

Hoat dgng ciia GV

Cau hdi 1

Hay ve d6 thj cua ba ham sd tren

ciing mdt he true tga do

Cau hdi 2

Hay neu mdt vai nhan xet chung

Cau hdi 3

Dua vao dd thj hay neu tap xac

dinh cua cac ham so tren

Hoat ddng cua HS

Ggi y tra loi can hdi 1

GV ggi mdt HS ve va chinh sira lai

Ggi y tra Idi cau hdi 2

Cac dd thj ciing di qua diem (1 ; 1)

Ggi y tra loi cau hdi 3

2 ^

Tap xac djnh cua ham so y = x la

1

R, ciia ham so y = x^ la (0 ; + oo), cua ham so y = x~ la

R \ {0} Tii dd ggi y cho hgc sinh nhan xet tap xac dinh cua ham so luy thiia tuy thudc vao gia tri cua

so mii a nhung ludn chiia khoang

(0 ; + «))

• GV neu chii y :

Tap xdc dinh cua hdm sdluy thiia y = x" tuy thudc vdo gid tri cua

a Cu the,

Vdi a nguyen duang, tap xdc dinh la R ;

Vdi a nguyen dm hoac bang 0, tap xdc dinh /a R \ {O} ;

Vdi a khdng nguyen, tap xdc dinh Id (0;+oo)

HOAT DONG 2

II DAO HAM CUA HAM SO LUY THl/A

H2 Tfnh cac dao ham cua cac ham sd sau :

a ) y = x ^ + 2 x + l ; b) y = Vx

• GV neu dao ham cua ham luy thiia ;

Ham sd luy thira y = x" (« e R) cd dao ham vdi mgi x > 0 va

3

Q, — —

4 Ggi y tra Idi cau hoi 2

4 4 ^ Ggi y tra Idi cau hdi 3

a = V3

Thuc hien ^ , 3 trong 5'

III KHAO SAT HAM SO y = x"

• GV dat va'n de nhu sau:

H4 Trong viec khao sat ham sd y = x" vi sao ta chi xet tren tap (0 ; + oo) ? Sau khi HS tra Idi, GV neu If do va dat mdt sd cau hdi sau

H5 Hay tfnh dao ham ciia ham sd y = x"

H6 Trong trudng hgp a > 0 thi dau ciia dao ham la gi?

H7 Trong trudng hgp a < 0 thi da'u ciia dao ham la gi?

H8 Tim gidi han ciia ham sd khi x dan de'n 0 va oo

H9 Xac djnh tiem can ciia ham so

HIO Lap bang bie'n thien cua ham sd trong mdi trudng hgp

H l l Hay ve dd thj ciia ham sd trong mdi trudng hgp

H12 Dd thj ciia ham sd ludn di qua (1 ; 1), diing hay sai?

Sau dd GV tong ke't:

Gidi han dac biet

lim x " = 0, lim x " = +oo

X-^0^ X^+oo

Tiem can

Khdng cd

3 Bang bie'n thien (cho bang

thien va dd thj chia lam 2 trudng

a > l , a = l,0<c!:<l)

vao)

4 Dd thj

bie'n hgp

Ox la tiem can ngang,

Oy la tiem can diing ciia dd thj

3 Bang bie'n thien (cho bang bie'n thien va dd thj chia lam 2 trudng hgp ci;<0,a = 0 )

4 Dd thi

HI3 Hay hoan thien bang bien thien sau :

Bang bie'n thien

HI4 Dua vao dd thj sau, hay chi ra dd thj ciia ham sd trong mdi trudng hgp cu the

• GV neu chii y trong SKG:

Khi khdo sdt hdm luy thiia vdi sd'md cu the, ta phdi xet hdm sd dd tren todn bg tap xdc dinh ciia nd

• Thuc hien vf du 3 trong 5' GV cd the thay bdi vf du khac tuong tU

Tap xac djnh D = (0 ; +oo)

Ggi y tra Idi cau hdi 2

3 -I

1

Ggi y tra loi cau hdi 3

Ta cd y' < 0 tren khoang (0 ; +oo) ham so nghjch bien

Ggi y tra Idi cau hdi 4

Tiem can : lim y = +oo,

x^O^

nen

lim y = 0

a;->+co

Do thj cd tiem can ngang la true hoanh

va cd tiem can diing la true tung

H15 Hoan thanh bang bie'n thien sau

X

y

+00

y

H16 Hay ve do thj ciia ham so

• GV neu bang tdm tit sau: ,

Bang tdm tit cac tfnh cha't ciia ham sd luy thira y = x " tren khoang (0 ; +oo)

Ham sd luon nghjch bien

Tiem can ngang la Ox, tiem can dirng la Oy

Dd thi ludn di qua diem (1 ; 1)

HOATDQNG 4

TOM T ^ Bfil MPC

1 Ham sd y = x'^ vdi a e M, dugc ggi la ham sd luy thira

2 Tap xac dinh ciia ham sd luy thira y = x " tuy thudc vao gia trj ciia a Cu the,

Vdi a nguyen duong, tap xac djnh la R ;

Vdi a nguyen am hoac bing 0, tap xac djnh la E \ {O};

Vdi a khdng nguyen, tap xac djnh la (0; +00)

3 ham sd luy thira y = x" (« e R) cd dao ham vdi mgi x > 0 va

(x«)' = a x « - l

4 Cdng thiic tfnh dao ham ciia ham hgp dd'i vdi ham so luy thira cd dang

iu"y = au'^-^.u'

HOAT DONG 5

MQT SO Cfla Hdi TRflC NGHIEM KHflCH QOflN

Hay dien dung sai vdo d trdng sau

Cdu 1 Cho ham so y = x^

(a) Ham sd khdng cd dao ham

2

-ii-(b) Ham sd cd dao ham y' = — vx

(c) Ham sd ludn ludn ddng bie'n

(d) Ham sd khdng ed tiem can

Cdu 2 Cho ham sd y = x^

(a) Ham sd cd the' vie't lai la y = v x^

(b) Ham sd cd dao ham y' = — x ^

(e) Ham so luon luon nghjch bie'n

(d) Ham sd cd tiem can la hai true toa do

Cdu 3 Cho ham sd y = x ^

(a) Tap khao sat cua ham sd la R

2 3rT (b) Ham sd cd dao ham y' = — v x

(c) Ham sd ludn ludn nghjch bie'n

(d) Ham sd khdng ed tiem can

Cdu 4 Cho ham sd y = x ^

(a) Tap khao sat ciia ham sd la:

2 -I

(b) Ham sd cd dao ham y = - - x ^

(c) Ham sd ludn ludn ddng bie'n

D

D

D

(d) Ham sd cd tiem can la hai true toa do

Cdu 5 Cho ham sd y = (2x + 1)4

(a) Ham sd cd dao ham la y' = —(2x + 1 ) 4

(b) Ham sd ed dao ham la y' 3 1

2V2X + 1 (e) Ham sd ludn ludn ddng bie'n

(d) Ham sd khdng cd tiem can

Cdu 7 Cho ham sd y = (2x + 1 ) 4

(a) Ham sd cd dao ham la y • = — ( 2 x + l) 4

(b) Ham sd cd dao ham la y': 3 1

2t/(2x + l)'

(c) Ham sd ludn ludn nghjch bie'n

(d) Ham so khdng cd tiem can

Cdu 8 Cho ham sd y = (2x +1)

(a) Ham so cd dao ham la y' = (2x + 1 ) 2

3 1 (b) Ham sd cd dao ham la y' =

'p + lf

(c) Ham so ludn luon nghjch bie'n

(d) Ham sd khdng cd tiem can

Hdy chgn khdng dinh dung trong cdc cdu sau

Cdu 9 Cho ham sd y = x" cd dd thj sau

D

D

D

D

Cdu 10 Cho ham sd y = x* cd dd thi sau

y

(a) a > 1 ;

(c) a = 0 ;

Ti'd Idi ia)

Cdu 11 Cho ham sd y = x " cd dd thj sau

(b) 0 < a < 1 ; (d) a = 1

Cdu 14 Trong cac ham sd sau day, ham so nao ddng bie'n