ChiroNq II

HAM SO LUY THUA - HAM SO MU
VA HAM SO LOGARIT
Phan

1

OTTOlVG VAN D E C U A C H C W ^ G
I. NOI DUNG
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.

192


IL MUC TIEU
1. Kien thiic

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
cu the.
Tu duy eac va'n de ciia toan hgc mdt each Idgic va he thdng.
II. CHUAN BI CUA G V VA HS
1. Chuan hi cua GV
Chuan bj cac cau hdi ggi md.
194


• 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.

Hoat ddng ciia GV
Cau hdi 1

Hoat dgng cua HS
Ggi y tra loi cau hoi 1

Tfnh 1,5"^

1,5^ =5,0625.

G\: ggi HS thue hien.

HS cd the sii dung may tfnh dien tii
bam: 1.5M = 5.0625.

Cau hdi 2

Ggi y tra Idi cau hoi 2

3
r
2^
Tfnh :
. 3j

I 3j

21'

HS cd the sir dung may tinh dien tir


Hay bien luan so nghiem • Vdi mgi 6 e R, phuong trinh
o
phuang trinh x = b.
X = b ludn cd mdt nghiem.
Can hdi 2
Ggi y tra loi cau hoi 2
Hay bien luan sd nghiem
Vdi 6 < 0, phuang trinh x^ = b
phuang trinh x"^ = b .
khong cd nghiem.

198


• 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 .
0 Co hai cdn trdi ddu, ki hieu gid tri duang
la y/b cdn gid tri dm la —yjb

• GV neu mdt sd tfnh cha't:

^

^fa
' ^

'ilb^'if^

' % ) m" =

^

. ^ ^

fa
V6

a, k h i ^ l e
lai, khi n chin

#^ = "^; i^ i^ = i^; ^ = i^.
200


• Thuc hien ^

Hoat ddng ciia HS

^3^3

Tacd ^3V3 =^(V3)^

201


Cau hdi 2

Ggi y tra loi cau hdi 2

Ket luan.

^/3V3 = 3 | ( V ^ = V3
H7. Tfnh

^ . ^
HOATDQNG 4

4. Luy thira vdi sd mu hufu ti
• GV neu dinh nghia :
m . trong dd m e Z, n e N
Cho sdthuc a duang vd sdhifu ti r = —
Luy thita cua a vdi sd'md r Id sd af xdc dinh bdi
m
,
a^ = a n =^a'^
H8. Phai chang ^


3_3i - 1
V8 2

Ggi y tra loi cau hdi 2
- 2 . ^

4-3 =

1^
8


Cau hdi 3

Ggi y tra loi cau hdi 3
1

1

Tinh : a "

a^

='^

• GV neu va thuc hien vf du 5. GV cd the thay bing vf du khac.
Cau a
Hoat ddng cua GV


1
1
1
x 4 + y 4 =;j;4 + ^ 4

Ggi y tra loi cau hdi 3
5
5
^ ^ x 4 3 ; + xy4 ^
^

+

^

HOATDQNGL
5. Luy thiira vdi sd mu vo ti
• GV neu van de va cho HS kiem tra bang sau bing may tfnh dien tir.

203


n

^n

1
2
3
4

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.

n

1
2
3
4
5
6
7
8
9
10

2'-

^n

1
1,4
1,41
1,414
1,4142
1,41421
1,414213

a\0 _ ^ap
(a")

iabf = a"b";

a
bJ

a
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 ddng cua GV
Cau hoi 1
Hay dua tii sd ve ciing luy
thiia ca so a.
Cau hoi 2

Hoat ddng ciia HS
Ggi y tra loi cau hdi 1
^N/7+1

^2-N/7^^N/7+1+2-V7^^3

Ggi y tra Idi cau hdi 2

Hay dua miu sd ve ciing luy
(a^-2)^^2^a-2


Cau hdi 2

53^=718

Tfnh 5 ^ ^

Ggi y tra loi cau hdi 3

Cau hdi 3
Hay so sanh hai so' tren.
Thuc hien ^ ,

Hoat ddng ciia HS

HS tu ke't luan.

4 trong 5'

Hoat ddng cua GV
Can hdi 1
So sanh v8 va 3.

Hoat ddng cua HS
Ggi y tra Idi cau hoi 1

V8
a = a"-^ ; ia")^ = a"^ ;
a'^

iab)" = a"b";

a
.bJ

,"

a

a
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

D

(a) 2^ > 2 ^
(b)

'^ 1V
v2.


D

c

d

S ' D

S

2. Hay dien diing sai trong cac khing djnh sau
(a) 2 ^ ^ > 2 ^

D

(b) 2 ^ ^ =

D

2^

(c)(2V2)%(V8)

n

(d)(2^/2)^=(^/8)

•


(b)

^iv ro
v^y

^,^747

v^y

.p7

]_

(d)

v2y

755

..2-^
K'^J

Trd Idi.
a
>

b
>

Giii tich 12/1


v2y

\^J
(d)
v-^y

Trd Idi.
b

a


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. (c).
9. Trong cac khang djnh sau khing djnh nao diing ?
(a) 2^+2^ = 2 ^

(b) 2^-2^ =2"^

(c) 2^2^ = 2^^ ;

(d) 2^ : 2^ = 4

Trd Idi. (d).
10. Trong cac khing djnh sau khing djnh nao diing ?
\5

,c

(a) (2') = 2 " ;
(0(2=)"'=.

>15



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 HS

Hoat ddng ciia GV

Ggi y tra Idi cau hoi 1

Cau hdi 1
2

4

2

6

4 6

95.275 =35.36 =35 +—
5 =32=9_

Tinh 95.275
Can hdi 2


(1 ^-°'^^
~ \

U6J

--

+0,25 2 = 2

+5

2^ = 40

^4

Ggi y tra loi cau hdi 3

Tfnh
(0,04)-^'^-(0,125) 3

A--

2

(0,04)-!'^-(0,125) 3 =

' 1 \2 ri^
25 J

8j

fir'

Hoat ddng cua HS
Ggi y tra Idi can hoi 1

,3.".,;2-.4;(if=8

bJ

Cau hdi 2

Ggi y tra loi cau hdi 2

Hay s i p xep theo thii tu tang
Ta cd — < 1 < 8 nen :
2
dan.
2-'
H o a t ddng cua H S
Ggi y t r a loi cau hdi 1
4 1 42
T S = a 3 3 + a 3 3 = a ( a + l)
Ggi y t r a loi cau hdi 2
13
11
MS= a4 4 + a 4 4 = a + l
Ggi y t r a loi can hdi 3

Hay riit ggn bieu thiic.
4 / _1
2\
a3 \a 3 + a 3 /

1/ 2
a4\a4 +a

214

_i) -"
4/


caub.
Hoat ddng ciia HS

Hoat ddng cua GV
Can hdi 1


1 1 2

2

T S = a 36 3(a3 _ 6 3 )
Cau hdi 2
Riit ggn m i u so.

Ggi y tra Idi can hdi 2
2

2

MS = a3 - 63
Cau hdi 3
Hay riit ggn bieu thiic.

Ggi y tra loi cau hdi 3
1 1 2

2

a 36 3(a3 - 6 3 )
1
3 ^
2
2
a3 - 6 3
ia ^ 0, b ^ 0, a ^ b)


1 1 1

1

a363(66 + a 6 ) g^^—
1
1
~ ^"^
a 6 + 66
(a > 0, 6 > 0, a2 + 6^ > 0).

Bai 5. Hudng ddn. Sd dung tfnh chat cua liiy thiira.
a) Ta cd 2V5 = V20, 3V2 = Vl8 nen 2V5 > 3V2. Vi ca sd,, -1 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 ^ ^

216