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

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

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.

Giii tich 12/1

193


Phan 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
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.

IV. TIEN TRINH DAY - HOC

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
cu the'.
Cau hdi 2
Thuc hien cac phep tfnh sau :

:(uf

a)

v^y

b) [yfnf
GV : Sau ddy, chung ta se nghien cdu ve luy thda vdi sd md thuc.

195


B. BAI Mdl
I . KHAI NIEM LtJY THlTA
HOATDQNGl
1. Luy thira vdi sd mu nguyen
• Thuc hien Qy I trong 5'

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

bam : (2 +3)^3 =.
Sau dd ta in not a
Cau hdi 3

ta dugc ke't qua.

Ggi y tra loi cau hdi 3

[^=^S.

Tfnh (Vs)^

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
196


a"' = a.a

a

n thijfa so

Vdi

a^O

aO=l.
a

- —.
a"

Trong bieu thiic a", ta ggi a la co sd, n la sdmu.
• GV neu chii y :
> Chu y. 0 va 0~" khdng cd nghTa.
HI. Neu mdt sd vi du ve luy thiia vdi so mii nguyen.
f n'\'^ / .3\5
H2. Tfnh
V "+ y

• GV neu vf du 1. GV ed the lay vf du tuong tu.
Hoat ddng cua GV
Cau hdi 1

Hoat ddng cua HS
Ggi y tra loi cau hdi 1

Hay doi so' hang thii nhat ra
ca so 3.
Tacd
Cau hdi 2

(1 v'° 27"^=3^°.3"^=3.

Ggi y tra loi cau hdi 2

Hay doi so hang thir hai ra ca
Tacd (0,2)"^.25"2=5^5~^ = 1
sd 5.
Cau hdi 3
Ggi y tra loi cau hdi 3
Hay doi so hang thii hai ra ca
so 2.
Tacd—.2^=2-12^=4
128

197


• GV neu vf du 2. GV cd the la'y vf du tuong tu.
Hoat dgng ciia GV

Hoat ddng ciia HS

Cau hoi 1

Ggi y tra loi cau hdi 1

Hay phan tich trong ngoac
[aV2(l + a^) - 2V2a]
thanh nhan tir.
= (aV2+ a ^ V 2 - 2 a V 2 )
= aV2(a^ - 1 )

Cau hdi 2

Ggi y tra loi cau hoi 2

Hay phan tich ngoai ngoac
thanh nhan tur.
Cau hdi 3
Tfnh B.

1

1
a^il-a"^)

a^-a

1
aia^--1)

Ggi y tra loi cau hoi 3
B= V2
HOAT DONG 2

2. Phirong trinh x" = b
• Thuc Men . ^ 2 trong 5'.
Hoat ddng cua GV
Cau hdi 1

Hoat dgng cua HS
Ggi y tra loi cau hdi 1

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 .

H4. Tim sd nghiem phuang trinh : x^°°^ = -2009, x^°°^ = 0 va x^^^^ = 2009
HOAT DONG 3
3. Can bac n
• GV neu van de :
Cho sd nguyen duang n, phuang trinh
a"" =b
dua de'n hai bdi todn ngugc nhau :
•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.
199


• 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

• 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 ^

3 trong 5

Hoat ddng cua GV
Can hdi 1
Tfnh

Hoat ddng cua HS
Ggi y tra Idi cau hdi 1

(^^j/b)"

Can hdi 2

Ta cd ( ^ ^ ^ b ) " = ab
Ggi y tra loi cau hdi 2

Tfnh ( ^ ) "
Cau hdi 3
Ke't luan.

Ta cd (^Vab)" = ab
Ggi y tra loi cau hdi 3
HS tu KL.

• GV neu va thuc hien vf du 3. GV cd the' thay bing vf du khac.
Cau a
Hoat ddng cua GV
Cau hdi 1
Tinh

Hoat ddng cua HS
Ggi y tra Idi cau hdi 1

^ . ^

Cau hdi 2

Tacd ^ . ^

= ^-32

Ggi y tra loi cau hdi 2

Bieu dien - 3 2 theo liiy thiia -32 = (-2)'
bac 5.
Ggi y tra Idi cau hdi 3
Cau hdi 3
Ke't luan.

^4 ^ - 8 = ^ - 3 2 = ^(-2)^ = - 2 .

Caub
Hoat ddng cua GV

Ggi y tra Idi cau hdi 1

Can hdi 1
Tfnh

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 ^

= 2^

_5

H9. Cd tdn tai 2 2 hay khdng ?
• GV neu va thuc hien vf du 4. GV cd the thay bing vf du khac.

cau a
Hoat ddng ciia GV
Cau hdi 1

Ggi y tra Idi cau hdi 1
/I

Tfnh

\

Ta cd
\oy

Can hdi 2
Tinh 4 2

202

Hoat ddng cua HS

8

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

Hoat ddng ciia HS

Cdu hoi 1

Ggi y tra Idi c&u hdi 1
5

Ta cd

5

Phan tfch x^y + xy^
nhan tir.
Cau hdi 2
So sanh
1 1
1 1
x 4 + y 4 va x 4 + 3 / 4

Cau hdi 3
Riit ggn bieu thiic tren.

thanh
5
5
1 1
X43/+ X3/4 =x3'(x4 + 3 / 4 )

Ggi y tra loi cau hdi 2
1

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

5
6
7
8
9
10

S'-n

1
1,4

3

1,41
1,414
1,4142
1,41421
1,414213
1,4142135
1,41421356
1,414213562

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.

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

1,4142135
1,41421356
1,414213562

• 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^"n—>+00

204

vdi

a = lim n

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\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

thiia ca sd a.
Cau hdi 3
Ggi y tra loi cau hdi 3
Hay lap bang bie'n thien cua Xem SGK.
ham so.
205


Ggi y tra loi cau hdi 4

Cau hdi 4
Riit ggn E.

E=a5

Thuc hien vf du 7 trong 5'

Hoat ddng ciia GV

Ggi y tra loi cau hoi 1

Cau hdi 1

52N/3 ^

Tfnh 5 ^ ^

^

Ggi y tra Idi cau hdi 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<3.

Can hdi 2

Ggi y tra Idi cau hoi 2

3
So sanh — va 1.
4
Cau hdi 3

4
Ggi y tra loi cau hdi 3

Hay so sanh hai so tren.

(if > (ir
206


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
a"' = a.a

a

n thira .so

aO = 1

Vdi a ^ O

a

=

1
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
sd b ne'u a" = b.

4. -^ -S = -4ab ; ^ = "£ ; ( ^ r = "V^ "V^ = f"'''''^'^
vfo

#^ = '^^;

'"

i^ i^ = i^-

[|a|, khi n chin

la^

nf p-q

1=^ = va
207


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"^^

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.

(\ ^
v^y

(c)(V2)'>(^/2)

208

•75

n
n


/ ,

(d)

/^ ^ \f&

A3

D

V3J ^tVs.

Trd Idi.
a

b

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)

•

Trd Idi.
a

b

S

D

c
S

d
D

3. Hay dien cac dau thich hgp trong cac dau : > , < , = vao chd trdng sau
(a)

^O

I

v^y

2;

/'jN^?

(c)

(b)

^iv ro
v^y

^,^747

v^y

.p7

]_

(d)

v2y

755

..2-^
K'^J

Trd Idi.
a
>

b
>

Giii tich 12/1

c
<

d
=

209


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

av^
(b)

/1 ^

(a)

(c)

v^y

2

a^

I

v2y

v2y

\^J
(d)
v-^y

Trd Idi.
b

a
<

<

c
>

d
=

5. Trong cac sd sau day, sd nao Idn hon 5.
(a) 253;

(b) 254

(c) 2 5 ^ ;

(d) 25'^

Trd Idi. (c).
6. Trong eac sd sau day, sd nao bing 5.
(a) 253;

(b) 252

(c)25^;

(d) 25'^

Trd Idi. (b).
7. Trong cac so sau day. so nao nhd hon 5.
fi
(a) 25 3 ;
1

(c) 2 5 ^ ;
Trd Idi. (d).
210

2

(b) 253
1

(d) 25"

/55

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

/

,x5

(b) (2~^) =-2^~

id)(2^f=2''

Trd Idi. (a).
11. Trong cac khing djnh sau khing djnh nao diing ?
(a) V2.V8 = 4;

(b) ^ . ^ = 4

(c) ^ . ^ = Vs ;

(d) ^ . ^ 4 = 8

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 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

Ggi y tra loi cau hdi 2

1

1

Tfnh 1444 : 94

-

-

Iri44^

1444 : 9 4 = 4 i r :

Cau hdi 3

=8

Tfnh
Ggi y tra Idi cau hdi 3

f 1 A-O-75
vl6y

+ 0,25 2

Cau hdi 4

(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

= (5"2) 2 - (2-3) 2 = 53 _ 22 = 121.
Bai 2. Hudng ddn. Sir dung tfnh chat cua luy thira.
GV cho HS len bang chira bai vdi nhiing ggi y sau day.

212


Hoat ddng cua HS

Hoat ddng cua GV
Can hdi 1
Hay giai cau a.

Ggi y tra loi can hdi 1
1

I

1

5

a3.Va =a3.a2 = a^
Cau hdi 2
Hay giai cau b.

Ggi y tra Idi can hdi 2
111

I I

b2 b3 Vb = b2^3*6 = 5
Cau hdi 3
Hay giai cau c.

Ggi y tra loi cau hdi 3
4

4 I

a 3 : V a = a3 ^ =a
Cau hdi 4
Hay giai cau d.

Ggi y tra Idi cau hdi 4
1

11

1

Vb :b6 =b3 6 =b6
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.
Cau a.
Hoat ddng cua GV
Cau hdi 1
Tfnh cac gia tri : 1^'^^ ; 2~^ ;

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-'
rn-3
v2;
213


caub.
Hoat ddng cua G V
Cau hdi 1
^o\-l
1
Tfnhcacgiatrj: 9 8 ° ; f - J

;325
Cau hdi 2

H o a t ddng ciia H S
Ggi y t r a loi cau hdi 1
98° = 1; -

= i ; 3 2 5 = ^ =2

Ggi y t r a loi cau hdi 2

Hay s i p xep theo thii tu tang
7
Ta cd 1 < 2 < — nen :
din.
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.
Cau a.
Hoat ddng ciia GV
Cau hdi 1
Phan tfch tii so thanh nhan tii.

Cau hdi 2
Riit ggn m i u so.

Cau hdi 3

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

Ggi y tra Idi cau hdi 1

Phan tfch tii sd thanh nhan tu.
T S = b5^5 _ b 5 5 = b - l .

Cau hdi 2
Riit ggn miu sd.
Cau hdi 3
Hay riit ggn bieu thiic.

Ggi y tra loi cau hdi 2
2 I
2_2
MS= b3^3_b3 3 = b - l

Ggi y tra Idi can hdi 3
1

65 ( ^ - ^ )

1

63(3/6-^)
Cau c.
Hoat ddng cua GV
Cau hdi 1
Phan tfch tir sd thanh nhan tir.

Hoat ddng ciia HS
Ggi y tra loi cau 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)

215


caud.
Hoat ddng cua GV
Cau hdi 1
Phan tfch tii so thanh nhantir.

Hoat ddng cua HS
Ggi y tra Idi cau hdi 1
i l l

1

TS= a3b3(b6+a6)
Cau hdi 2

Ggi y tra loi cau hoi 2

Riit ggn m i u so.

1

1

MS = a 6 + 66
Cau hdi 3
Hay riit ggn bieu thiic.

Ggi y tra Idi cau hoi 3

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