Giải tích 12 Tác giả: Trần Sĩ Tùng, 2010

8/13/2019 Gii tch 12 Tc gi: Trn S Tng, 2010

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TRNG THPT TRNG VNG

GIAO AN

GIAI TCH 12CHUAN

HOC K I

Giao vien:Tran S Tung

Nam hoc: 2009 2010

www.daykemquynhon.ucoz.com


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NG DNG O HMKHO ST HM S

&

VTHCA HM S

Bi 1: Sng bin, nghch bin ca hm sBi 2: Cc trca hm sBi 3: Gi trln nht - gi trnhnht ca hm sBi 4: ng tim cn

Bi 5: Kho st sbin thin v vthca hm s



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Chng II:

HM SLUTHAHM SM

HM SLOGARIT

Bi 1: Lutha

Bi 2: Hm slutha

Bi 3: Logarit

Bi 4: Phng trnh mv phng trnh logarit

Bi 5: Bt phng trnh m- logarit



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Chng III:

NGUYN HMTCH PHN V NG DNG

Bi 1: Nguyn hm

Bi 2: Tch phn

Bi 3: ng dng ca tch phn trong hnh hc



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Chng IV:

SPHC

Bi 1: SphcBi 2: Cng, trv nhn sphcBi 3: Php chia sphcBi 4: Phng trnh bc hai vi hsthc



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Gii tch 12 Trn STng

2

GV hng dn HS nu nhnxt vthca hm s.

Nhn xt:thca hm sng bintrn K l mt ng i ln ttri sang phi.

th ca hm s nghchbin trn K l mt ng i

xung ttri sang phi.

7' Hot ng 2: Tm hiu mi lin hgia tnh n iu ca hm sv du ca o hm Da vo nhn xt trn, GVnu nh l v gii thch.

2. Tnh n iu v du cao hm:nh l: Cho hm s y = f(x)c o hm trn K. Nu f '(x) > 0, x K

th y = f(x) ng bin trn K.Nu f '(x) < 0, x K th y = f(x) nghch bin trn K.

Ch :Nu f (x) = 0, x K th f(x) khng i trn K.

15' Hot ng 3: p dng xt tnh n iu ca hm sHng dn HS thc hin.

H1.Tnh yv xt du y?

HS thc hin theo s hngdn ca GV.1.a) y= 2 > 0, x

x +

y'

y

+

b) y= 2x 2x +

y'

1

0

y

VD1: Tm cc khong niu ca hm s:a) 2 1y x=

b)2

2y x x=

5' Hot ng 4: Cng cNhn mnh: Mi lin quan gia o hm

v tnh n iu ca hm s.

4. BI TP VNH: Bi 1, 2 SGK. c tip bi "Sng bin, nghch bin ca hm s".

IV. RT KINH NGHIM, BSUNG:.........................................................................................................................................................................................................................................................................................................................................................................................................................................................

xO

y

xO

y



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Trn STng Gii tch 12

1

Ngy son: 10/08/2009 Chng I: NG DNG O HM KHO STV VTHHM S

Tit dy: 02 Bi 1: SNG BIN, NGHCH BIN CA HM S(tt)

I. MC TIU:Kin thc:

Hiu nh ngha ca sng bin, nghch bin ca hm sv mi lin hgia khi nimny vi o hm.

Nm c qui tc xt tnh n iu ca hm s.Knng:

Bit vn dng qui tc xt tnh n iu ca mt hm sv du o hm ca n.Thi :

Rn luyn tnh cn thn, chnh xc. Tduy cc vn ton hc mt cch lgic v hthng.II. CHUN B:

Gio vin: Gio n. Hnh vminh ho.Hc sinh:SGK, vghi. n tp cc kin thc hc vo hm lp 11.

III. HOT NG DY HC:1. n nh tchc: Kim tra sslp.2. Kim tra bi c:(5')

H.Tm cc khong n iu ca hm s 42 1y x= + ?.Hm sng bin trong khong (0; +), nghch bin trong khong (; 0).

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung10' Hot ng 1: Tm hiu thm vmi lin hgia o hm v tnh n iu ca hm s

GV nu nh l m rng vgii thch thng qua VD.

x

y

y

+0

0+ +

0

+

I. Tnh n iu ca hm s2. Tnh n iu v du cao hm

Ch :Gi s y = f(x) c o hm

trn K. Nu f (x) 0 (f(x) 0),x K v f(x) = 0 chti mts hu hn im th hm sng bin (nghch bin) trn K.

VD2: Tm cc khong niu ca hm sy = x3.

7' Hot ng 2: Tm hiu qui tc xt tnh n iu ca hm s

GV hng dn rt ra qui tc

xt tnh

n

iu c

a hm s.

II. Qui tc xt tnh n iuca hm s1. Qui tc1) Tm tp xc nh.2) Tnh f(x). Tm cc im xi(i= 1, 2, , n) m ti ohm bng 0 hoc khng xcnh.3) Spx xp cc im xi theoth t tng dn v lp bngbin thin.4) Nu kt lun vcc khong



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Gii tch 12 Trn STng

2

ng bin, nghch bin cahm s.

15' Hot ng 3: p dng xt tnh n iu ca hm s

Chia nhm thc hin v giHS ln bng.

GV hng dn xt hm s:

trn 02

;

.

H1.Tnh f(x) ?

Cc nhm thc hin yu cu.a) ng bin (; 1), (2; +)

nghch bin (1; 2)b) ng bin (; 1), (1; +)

1.f(x) = 1 cosx0(f(x) = 0 x = 0)

f(x) ng bin trn 0 2;

vi 02

x

< < ta c:

f x x x( ) sin= >f(0) = 0

2. p dng

VD3: Tm cc khong n

iu ca cc hm ssau:a) 3 2

1 12 2

3 2y x x x= +

b)1

1

xy

x

=

+

VD4:Chng minh:sin>x x

trn khong 0; 2

.

5' Hot ng 4: Cng cNhn mnh: Mi lin quan gia o hmv tnh n iu ca hm s. Qui tc xt tnh n iu cahm s.

ng dng vic xt tnh niu chng minh bt ngthc.

4. BI TP VNH: Bi 3, 4, 5 SGK.




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Trn STng Gii tch 12

1


Tit dy: 03 Bi 1: BI TP SNG BIN, NGHCH BIN CA HM S

I. MC TIU:Kin thc:

Hiu nh ngha ca sng bin, nghch bin ca hm sv mi lin hgia khi nimny vi o hm.

Nm c qui tc xt tnh n iu ca hm s.Knng:

Bit vn dng qui tc xt tnh n iu ca mt hm sv du o hm ca n.Thi :


Gio vin: Gio n. Hthng bi tp.Hc sinh:SGK, vghi. n tp cc kin thc hc vtnh n iu ca hm s.

III. HOT NG DY HC:1. n nh tchc: Kim tra sslp.2. Kim tra bi c:(Lng vo qu trnh luyn tp)

H..

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Xt tnh n iu ca hm s

H1.Nu cc bc xt tnh niu ca hm s?

H2.Nhc li mt squi tc xtdu bit?

1.

a) B:3

2;

, NB:

3

2;

+

b) B:2

03

;

,

NB: ( )0; , 23;

+

c) B: ( )1 0; , ( )1; +

NB: ( )1; , ( )0 1;

d) B: ( ) ( )1 1; , ; +

e) NB: ( ) ( )1 1; , ; + f) B: 5( ; )+ , NB: 4( ; )

1. Xt s ng bin, nghchbin ca hm s:

a) 24 3y x x= +

b) 3 2 5y x x= +

c) 4 22 3y x x= +

d) 3 11

xy

x

+=

e)2 2

1

x xy

x

=

f) 2 20y x x=

7' Hot ng 2: Xt tnh n iu ca hm strn mt khongH1.Nu cc bc xt tnh niu ca hm s?

1.a) D = R

( )

2

22

1

1

xy

x

'

=

+

y= 0 x = 1b) D = [0; 2]

2

1

2

xy

x x

'

=

y= 0 x = 1

2. Chng minh hm s ngbin, nghch bin trn khong

c chra:a)

2 1

xy

x

=

+

, B: 11( ; ) ,

NB: 1 1( ; ),( ; ) +

b) 22y x x= , B: 0 1( ; ) ,NB: 1 2( ; )



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Gii tch 12 Trn STng

2

15' Hot ng 3: Vn dng tnh n iu ca hm s GV hng dn cch vndng tnh n iu chngminh bt ng thc. Xc lp hm s. Xt tnh n iu ca hm strn min thch hp.

a) tan , 0;2

=

y x x x .

2' tan 0, 0;2

=

y x x

y= 0 x = 0y ng bin trn 0;

2

y(x) > y(0) vi 02

<


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Trn STng Gii tch 12

1


Tit dy: 04 Bi 2: CC TRCA HM S

I. MC TIU:Kin thc:

M tc cc khi nim im cc i, im cc tiu, im cc trca hm s. M tc cc iu kin hm sc im cc tr.Knng: Sdng thnh tho cc iu kin tm cc tr.

Thi : Rn luyn tnh cn thn, chnh xc. Tduy cc vn ton hc mt cch lgic v hthng.

II. CHUN B:Gio vin: Gio n. Hnh vminh ho.

Hc sinh:SGK, vghi. n tp cc kin thc hc vtnh n iu ca hm s.III. HOT NG DY HC:

1. n nh tchc: Kim tra sslp.2. Kim tra bi c:(3')

H.Xt tnh n iu ca hm s:2

( 3)3= x

y x ?

.B:4

; , (3; )3

+

, NB:

4;3

3

.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung10' Hot ng 1: Tm hiu khi nim cc trca hm s

Da vo KTBC, GV giithiu khi nim C, CT cahm s.

Nhn mnh: khi nim cc tr

mang tnh cht "a phng".

H1.Xt tnh n iu ca hms trn cc khong bn tri,bn phi im C?

1.Bn tri: hm sB f(x)0Bn phi: h.sNB f(x) 0.

I. KHI NIM CC I,CC TIUnh ngha:Cho hm s y = f(x) xc nhv lin tc trn khong (a; b)

v im x0(a; b).a) f(x) t Cti x0h > 0,

f(x) < f(x0), x S(x0, h)\ {x0}.

b) f(x) t CT ti x0h > 0,f(x) > f(x0), x S(x0, h)\ {x0}.

Ch :a) im cc tr ca hm s;Gi tr cc tr ca hm s;im cc trca thhm s.b) Nu y = f(x) c o hm

trn (a; b) v t cc trti x0(a; b) th f(x0) = 0.

10' Hot ng 2: Tm hiu iu kin hm sc cc trGV phc ho th ca cchm s:a) 2 1= +y x

b) 2( 3)3

= x

y x

a) khng c cc tr.b) c C, CT.

II. IU KIN HMSC CC TRnh l 1: Gi s hm s y =

f(x) lin tc trn khong K =

0 0( ; ) +x h x h v c o hm

trn K hoc K \ {x0} (h > 0).



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Gii tch 12 Trn STng

2

T cho HS nhn xt milin h gia du ca o hmv s tn ti cc tr ca hms.

GV hng dn thng quavic xt hm s =y x .

a)f(x) > 0 trn 0 0( ; )x h x ,

f(x) < 0 trn 0 0( ; )+x x h th x0

l mt im Cca f(x).b) f(x) < 0 trn 0 0( ; )x h x ,

f(x) > 0 trn 0 0( ; )+x x h th x0

l mt im CT ca f(x).

Nhn xt:Hm s c th tcc tr ti nhng im m ti o hm khng xc nh.

15' Hot ng 3: p dng tm im cc trca hm sGV hng dn cc bc thchin.H1. Tm tp xc nh. Tmy. Tm im m y = 0 hockhng tn ti. Lp bng bin thin. Da vo bng bin thin kt lun.

1.a) D = Ry= 2x; y= 0 x = 0im C: (0; 1)b) D = Ry= 23 2 1 x x ;

y= 0 1

1

3

= =

x

x

im C:1 86

;3 27

,

im CT: (1;2)

c) D = R \ {1}

2

2' 0, 1

( 1)= >

+y x

x

Hm skhng c cc tr.

VD1:Tm cc im cc trcahm s:

a)

2

( ) 1= = +

y f x x b) 3 2( ) 3= = +y f x x x x

c)3 1

( )1

+= =

+

xy f x

x

5' Hot ng 4: Cng cNhn mnh: Khi nim cc tr ca hms. iu kin cn v iu kinhm sc cc tr.

4. BI TP VNH: Lm bi tp 1, 3 SGK. c tip bi "Cc trca hm s".




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Trn STng Gii tch 12

1


Tit dy: 05 Bi 2: CC TRCA HM S(tt)

I. MC TIU:Kin thc:




Hc sinh:SGK, vghi. n tp cc kin thc hc vtnh n iu v cc trca hm s.III. HOT NG DY HC:


H.Tm im cc trca hm s: 3 3 1= +y x x ?.im C: (1; 3); im CT: (1; 1).

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung5' Hot ng 1: Tm hiu Qui tc tm cc trca hm s

Da vo KTBC, GV cho HSnhn xt, nu ln qui tc tmcc trca hm s.

HS nu qui tc. III. QUI TC TM CC TRQui tc 1:1) Tm tp xc nh.2) Tnh f(x). Tm cc im ti f(x) = 0 hoc f(x) khng

xc nh.3) Lp bng bin thin.

4) T bng bin thin suy racc im cc tr.

15' Hot ng 2: p dng qui tc 1 tm cc trca hm sCho cc nhm thc hin. Cc nhm tho lun v trnh

by.a) C: (1; 3); CT: (1; 1).b) C: (0; 2);

CT:3 1

;2 4

,

3 1;

2 4

c) Khng c cc tr

d) C: (2; 3); CT: (0; 1)

VD1:Tm cc im cc trcahm s:a) 2( 3)= y x x

b) 4 23 2= +y x x

c)1

1

=

+

xy

x

d)

2 1

1

+ +

=+

x x

y x

5' Hot ng 3: Tm hiu qui tc 2 tm cc trca hm s GV nu nh l 2 v giithch.

nh l 2:Gisy = f(x) c o hm cp2 trong 0 0( ; ) +x h x h (h > 0).

a) Nu f(x0) = 0, f(x0) > 0th x0l im cc tiu.b) Nu f(x0) = 0, f(x0) < 0



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Gii tch 12 Trn STng

2

H1.Da vo nh l 2, hy nuqui tc 2 tm cc trca hms?

1.HS pht biu.th x0l im cc i.

Qui tc 2:1) Tm tp xc nh.2) Tnh f(x). Gii phng trnh

f(x) = 0 v k hiu xil nghim3) Tm f(x) v tnh f(xi).

4) Da vo du ca f(xi) suyra tnh cht cc trca xi.

10' Hot ng 4: p dng qui tc 2 tm cc trca hm sCho cc nhm thc hin. Cc nhm tho lun v trnh

by.a) C: (0; 6)

CT: (2; 2), (2; 2)

b) C:4

= +x k

CT:3

4

= +x k

VD2:Tm cc trca hm s:

a)4

22 64

= +x

y x

b) sin2=y x

5' Hot ng 5: Cng cNhn mnh: Cc qui tc tm cc trcahm s. Nhn xt qui tc nn dngng vi tng loi hm s.

Cu hi: i vi cc hm ssau hy chn phng n ng:

1) Chc C.2) Chc CT.3) Khng c cc tr.4) C Cv CT.

a)3 2 5 3= + +y x x x

b)3 2 5 3= + +y x x x

c)2 4

2

+=

x xy

x

d)4

2

=

xy

x

a) C Cv CTb) Khng c Cv CTc) C Cv CTd) Khng c Cv CT

i vi cc hm a thc bccao, hm lng gic, nndng qui tc 2. i vi cc hm khng co hm khng thsdng quitc 2.

4. BI TP VNH: Lm bi tp 2, 4, 5, 6 SGK.




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Trn STng Gii tch 12

1


Tit dy: 06 Bi 2: BI TP CC TRCA HM S

I. MC TIU:Kin thc:



II. CHUN B:Gio vin: Gio n. Hthng bi tp.


1. n nh tchc: Kim tra sslp.2. Kim tra bi c:(Lng vo qu trnh luyn tp)

H.

.3. Ging bi mi:

TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Sdng qui tc 1 tm cc trca hm s

Cho cc nhm thc hin.

H1. Nu cc bc tm imcc trca hm stheo qui tc1?

Cc nhm tho lun v trnhby.1.a) C: (3; 71); CT: (2; 54)b) CT: (0; 3)c) C: (1; 2); CT: (1; 2)

d) CT:1 3

;2 2

1. Tm cc im cc tr cahm s:a) 3 22 3 36 10= + y x x x

b) 4 22 3= + y x x

c)1

= +y xx

d)2

1= +y x x

15' Hot ng 2: Sdng qui tc 2 tm cc trca hm sCho cc nhm thc hin.

H1. Nu cc bc tm imcc trca hm stheo qui tc2?

Cc nhm tho lun v trnhby.1.a) C: (0; 1); CT: (1; 0)

b) C:6

= +x k

CT:6

= +x l

c) C: 24 = +x k

CT: (2 1)4

= + +x l

d) C: x = 1; CT: x = 1

2. Tm cc im cc tr cahm s:a) 4 22 1= +y x x b) sin2= y x x c) sin cos= +y x x

d) 5 3 2 1= +y x x x

10' Hot ng 3: Vn dng cc trca hm sgii tonH1.Nu iu kin hm slun c mt Cv mt CT?

1. Phng trnh y = 0 c 2nghim phn bit.

3.Chng minh rng vi mi m,hm s 3 2 2 1= +y x mx x



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Gii tch 12 Trn STng

2

Hng dn HS phn tch yucu bi ton.

H2.Nu x = 2 l im C thy(2) phi tho mn iu king?

H3.Kim tra vi cc gi trmva tm c?

2' 3 2 2= y x mx = 0 lunc 2 nghim phn bit.= m2+ 6 > 0, m

2.

y(2) = 0 1

3

=

=

m

m

3.m = 1: khng thomnm = 3: thomn

lun c mt im C v mtim CT.

4. Xc nh gi tr ca m

hm s

2 1+ +=

+

x mx

y x m t Cti x = 2.

3' Hot ng 4: Cng cNhn mnh:

iu kin cn, iu kin hm sc cc tr. Cc qui tc tm cc tr cahm s.

4. BI TP VNH: Lm cc bi tp cn li trong SGK v bi tp thm. c trc bi "Gi trln nht v gi trnhnht ca hm s".




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Trn STng Gii tch 12

1


Tit dy: 07 Bi 3: GI TRLN NHT V GI TRNHNHTCA HM S

I. MC TIU:Kin thc:

Bit cc khi nim GTLN, GTNN ca hm strn mt tp hp s. Nm c qui tc tm GTLN, GTNN ca hm s.

Knng: Bit cch tm GTLN, GTNN ca hm strn mt on, mt khong. Phn bit vic tm GTLN, GTNN vi tm cc trca hm s.




1. n nh tchc: Kim tra sslp.

2. Kim tra bi c:(5')H.Cho hm s 3 2 1y x x x= + . Hy tm cc trca hm s. So snh gi trcc trvi

2 1y y( ), ( ) ?

.1 32

3 27Cy y

= =

, 1 0

CTy y( )= = ; 2 9y( ) = , 1 0y( )= .

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu khi nim GTLN, GTNN ca hm s

T KTBC, GV dn dt n

khi nim GTLN, GTNN cahm s. GV cho HS nhc li nhngha GTLN, GTNN ca hms.

GV hng dn HS thc hin.

H1. Lp bng bin thin cahm s?

Cc nhm tho lun v trnhby.

1.+

++

03 1f x f

( ; )min ( ) ( )

+

= =

f(x) khng c GTLN trn(0;+)

I. NH NGHA

Cho hm s y = f(x) xc nhtrn D.

a)

0 0

Df x M

f x M x D

x D f x M

max ( )

( ) ,

: ( )

=

=

b)

0 0

Df x m

f x m x D

x D f x m

min ( )

( ) ,

: ( )

=

=

VD1: Tm GTLN, GTNN cahm ssau trn khong (0; +)



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Gii tch 12 Trn STng

2

10' Hot ng 2: Tm hiu cch tm GTLN, GTNN ca hm strn mt khong GV hng dn cch tmGTLN, GTNN ca hm slintc trn mt khong.

H1. Lp bng bin thin cahm s?

1. +

++

1 6

Ry ymin ( )= =

khng c GTLN.

II. CCH TNH GTLN,GTNN CA HM SLINTC TRN MT KHONG

Da vo bng bin thin xc nh GTLN, GTNN cahm s lin tc trn mt

khong.VD2:Tnh GTLN, GTNN ca

hm s 2 2 5y x x= + .

10' Hot ng 3: Vn dng cch tm GTLN, GTNN ca hm sgii ton GV hng dn cch giiquyt bi ton.

H1.Tnh thtch khi hp ?

H2.Nu yu cu bi ton ?

H3.Lp bng bin thin ?

1.

22 02

aV x x a x x( ) ( )

= <


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Trn STng Gii tch 12

1


Tit dy: 08 Bi 3: GI TRLN NHT V GI TRNHNHTCA HM S (tt)

I. MC TIU:Kin thc:

Bit cc khi nim GTLN, GTNN ca hm strn mt tp hp s. Nm c qui tc tm GTLN, GTNN ca hm s.

Knng: Bit cch tm GTLN, GTNN ca hm strn mt on, mt khong. Phn bit vic tm GTLN, GTNN vi tm cc trca hm s.



Hc sinh:SGK, vghi. n tp cc kin thc hc vcc trv GTLN, GTNN ca hm s.III. HOT NG DY HC:


2. Kim tra bi c:(5')H.Tm GTLN, GTNN ca hm s 2 3 2y x x= + ?

.3 1

2 4Rmax y y

= =

; khng c GTNN.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung12' Hot ng 1: Tm hiu cch tm GTLN, GTNN ca hm slin tc trn mt on

TKTBC, GV t vn ivi hm s lin tc trn mt

on.GV gii thiu nh l.

GV cho HS xt mt s VD.T dn dt n qui tc tmGTLN, GTNN.VD: Tm GTLN, GTNN ca

hm s 2y x= trn on cchra:

a) [1; 3] b) [1; 2]

-1 1 2 3

-8

-6

-4

-2

2

4

6

8

x

y

a)

[ ]1 31 1y y

;min ( )= =

[ ]1 3

3 9max y y;

( )= =

b)[ ]1 2

0 0y y;

min ( )

= =

[ ]1 22 4max y y

;( )

= =

II. CCH TNH GTLN,GTNN CA HM STRN

MT ON1. nh lMi hm s lin tc trn mton u c GTLN v GTNNtrn on .

2. Qui tc tm GTLN, GTNNca hm slin tc trn on[a; b] Tm cc im x1, x2, , xntrn khong (a; b), ti f(x)bng 0 hoc khng xc nh.

Tnh f(a), f(x1), , f(xn), f(b).Tm sln nht M v snhnht m trong cc strn.

[a b][a b]M max f x m f x

;;( ), min ( )= =

25' Hot ng 2: Vn dng cch tm GTLN, GTNN ca hm sgii tonCho cc nhm thc hin. Cc nhm tho lun v trnh

by.VD1: Tm GTLN, GTNN ca

hm s 3 2 2y x x x= + trn



21/155

Gii tch 12 Trn STng

2

Ch cc trng hp khcnhau.

23 2 1y x x' =

10 3

1

xy

x

'

= =

=

1 59

3 27y

=

; 1 1y( ) =

a)y(1) = 1; y(2) = 4

[ ]1 21 1 1y y y

;min ( ) ( )

= = =

[ ]1 22 4max y y

;( )

= =

b)y(1) = 1; y(0) = 2

[ ]1 01 1y y

;min ( )

= =

[ ]1 0

1 59

3 27max y y

;

= =

c)y(0) = 2; y(2) = 4

[ ]0 2 1 1y y;min ( )= =

[ ]( )

0 22 4max y y

;= =

d)y(2) = 4; y(3) = 17

[ ]2 32 4y y

;min ( )= =

[ ]( )

2 33 17max y y

;= =

on:a) [1; 2] b) [1; 0]c) [0; 2] d) [2; 3]


Cch tm GTLN, GTNN cahm slin tc trn mt on. So snh vi cch tm GTLN,GTNN ca hm slin tc trnmt khong.

4. BI TP VNH: Lm bi tp 1, 2, 3 SGK.




22/155

Trn STng Gii tch 12

1


Tit dy: 09 Bi 3: BI TP GI TRLN NHT V GI TRNHNHTCA HM S

I. MC TIU:Kin thc:Cng c:

Cc khi nim GTLN, GTNN ca hm strn mt tp hp s. Cc qui tc tm GTLN, GTNN ca hm s.

Knng: Tm c GTLN, GTNN ca hm strn mt on, mt khong. Phn bit vic tm GTLN, GTNN vi tm cc trca hm s.



Hc sinh:SGK, vghi. n tp cc kin thc hc vcc trv GTLN, GTNN ca hm s.III. HOT NG DY HC:


2. Kim tra bi c:(Lng vo qu trnh luyn tp)H.

.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Luyn tp tm GTLN, GTNN ca hm slin tc trn mt on

H1.Nu cc bc thc hin ? 1.

a) [ ]

[ ]

4 44 4

0 50 5

41 40

8 40

y y

y y

[ ; ];

[ ; ];

min ; max

min ; max

= =

= =

b) [ ]

[ ]

0 30 3

2 52 5

1 564

6 552

y y

y y

[ ; ];

[ ; ];

min ; max

min ; max

= =

= =

c) [ ]

[ ]

2 42 4

1111

20

31 3

y y

y y

[ ; ];

[ ; ];

min ; max

min ; max

= =

= =

d)11 11

1 3y y[ ; ] [ ; ]min ; max

= =

1. Tnh GTLN, GTNN cahm s:

a) 3 23 9 35y x x x= + trn cc on [4; 4], [0; 5].

b) 4 23 2y x x= + trn cc on [0; 3], [2; 5]

c)2

1

xy

x

=

trn cc on [2; 4], [3; 2].

d) 5 4y x= trn [1; 1].

15' Hot ng 2: Luyn tp tm GTLN, GTNN ca hm slin tc trn mt khong

H1.Nu cc bc thc hin ? 1.a) 4R

ymax = ; khng c GTNN

b) 1R

ymax = ; khng c GTNN

c) 0R

ymin = ; khng c GTLN

d)0

4y( ; )min

+

= ;khng c GTLN

2.Tm GTLN, GTNN ca cchm ssau:

a)2

4

1y

x=

+

b) 3 44 3y x x=

c) y x=

d)4

0y x xx

( )= + >



23/155

Gii tch 12 Trn STng

2

10' Hot ng 3: Vn dng GTLN, GTNN gii ton Hng dn HS cch phntch bi ton.H1. Xc nh hm s ? TmGTLN, GTNN ca hm s?

1.3) S = x (8 x), (0 < x < 8)S ln nht th x = 4.maxS = 16

4) P =48

xx

+ ( )0 4 3x<

P nhnht th x = 4 3

minP = 16 3

3. Trong s cc hnh ch nhtc cng chu vi 16 cm, hy tmhnh ch nht c din tch lnnht.

4. Trong s cc hnh ch nhtcng c din tch 48 cm2, hytm hnh ch nht c chu vinhnht.

5' Hot ng 4: Cng cNhn mnh: Cc cch tm GTLN, GTNN

ca hm s. So snh vi cch tm GTLN,GTNN ca hm slin tc trnmt khong. Cch vn dng GTLN,GTNN gii ton.

4. BI TP VNH: c trc bi "ng tim cn".

IV. RT KINH NGHIM, BSUNG:...................................................................................................................................................

......................................................................................................................................................................................................................................................................................................



24/155

Trn STng Gii tch 12

1


Tit dy: 10 Bi 4: NG TIM CN

I. MC TIU:Kin thc:

Bit khi nim ng tim cn ng, tim cn ngang ca thhm s.Knng:

Tm c ng tim cn ng, tim cn ngang ca thhm s. Cng ccch tm gii hn, gii hn mt bn ca hm s.



Hc sinh:SGK, vghi. n tp cch tnh gii hn ca hm s.III. HOT NG DY HC:


H.Cho hm s2

1

xy x

= . Tnh cc gii hn: x xy ylim , lim + ?

. 1x

ylim

= , 1x

ylim+

= .

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu khi nim ng tim cn ngang ca thhm s

Dn dt tVD hnh thnhkhi nim ng tim cnngang.

VD: Cho hm s

2

1

x

y x

=

(C). Nhn xt khong cch tim M(x; y) (C) n ngthng : y = 1 khi x .H1. Tnh khong cch t Mn ng thng ?

H2. Nhn xt khong cch khi x +?

GV gii thiu khi nim

ng tim cn ngang.

1.d(M, ) = 1y +

2.dn ti 0 khi x +.

I. NG TIM CNNGANG1. nh nghaCho hm s y = f(x) xc nh

trn mt khong v hn.ng thng y = y0 l timcn ngangca thhm sy= f(x) nu t nht mt trong cciu kin sau c thomn:

0x

f x ylim ( )+

= ,

0x

f x ylim ( )

=

Ch :Nu

0x x

f x f x ylim ( ) lim ( )+

= =

th ta vit chung

0x

f x ylim ( )

=

20' Hot ng 2: Tm hiu cch tm tim cn ngang ca thhm s Cho HS nhn xt cch tmTCN .


2. Cch tm tim cn ngangNu tnh c 0

xf x ylim ( )

+

=

hoc 0x

f x ylim ( )

= th ng



25/155

Gii tch 12 Trn STng

2

H1.Tm tim cn ngang ?

H2.Tm tim cn ngang ?

1.a) TCN: y = 2b) TCN: y = 0c) TCN: y = 1d) TCN: y = 0

2.a) TCN: y = 0

b) TCN: y =

1

2 c) TCN: y = 1d) TCN: y = 1

thng y = y0l TCN ca thhm sy = f(x).VD1:Tm tim cn ngang cuthhm s:

a)2 1

1

xy

x

=

+

b) 21

1

xy

x

=+

c)2

2

3 2

1

x xy

x x

+=

+ +

d)1

7y

x=

+

VD2:Tm tim cn ngang cuthhm s:

a)2

1

3

xy

x x

=

b)3

2 1

xy

x

+=

c)2

2

3 2

3 5

x xy

x x

+=

+

d)7

xy

x=

+

3' Hot ng 3: Cng cNhn mnh: Cch tm tim cn ngang cathhm s.

4. BI TP VNH: Bi 1, 2 SGK. c tip bi "ng tim cn".




26/155

Trn STng Gii tch 12

1


Tit dy: 11 Bi 4: NG TIM CN (tt)

I. MC TIU:Kin thc:

Bit khi nim ng tim cn ng, tim cn ngang ca thhm s.Knng:

Tm c ng tim cn ng, tim cn ngang ca thhm s. Cng ccch tm gii hn, gii hn mt bn ca hm s.



Hc sinh:SGK, vghi. n tp cch tnh gii hn ca hm s.III. HOT NG DY HC:


H.Cho hm s2 3

1

xy x

+

= (C). Tm tim cn ngang ca (C) ? Tnh 1x ylim , 1x ylim+ ?

.1x

ylim

= ,1x

ylim+

= + .

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu khi nim ng tim cn ng ca thhm s

Dn dt tVD hnh thnhkhi nim tim cn ng.

VD:Cho hm s

2

1

x

y x

= cth(C). Nhn xt vkhongcch t im M(x; y) (C)n ng thng : x = 0 khi x1+?

H1. Tnh khong cch t Mn ?H2. Nhn xt khong cch khi x 1+?GV gii thiu khi nim tim

cn ng.

1.d(M, ) = 1x .

2.dn ti 0.

II. NG TIM CNNG1. nh nghang thng x = x0 gl tim

cn ng ca thhm sy= f(x) nu t nht mt trong cciu kin sau c thomn:

0x xf xlim ( )

+

= +

0x x

f xlim ( )+

=

0x xf xlim ( )

= +

0x x

f xlim ( )

=

20' Hot ng 2: Tm hiu cch tm tim cn ng ca thhm s GV cho HS nhn xt cchtm TC.


2. Cch tm tim cn ngca thhm s

Nu tm c0x x

f xlim ( )+

= +

hoc0x x

f xlim ( )+

= ,



27/155

Gii tch 12 Trn STng

2

H1.Tm tim cn ng ?

H2.Tm tim cn ng v timcn ngang ?

1.a) TC: x = 3b) TC: x = 1c) TC: x = 0; x = 3d) TC: x = 7

2.a) TC: x = 1; x = 2

TCN: y = 0b) TC: x = 1; x = 2

TCN: y = 0

c) TC: x =1

2

TCN: y =1

2

d) TC: khng c

TCN: y = 1

hoc0x x

f xlim ( )

= +,

hoc0x x

f xlim ( )

=

th ng thng x = x0l TCca thhm sy = f(x).

VD1: Tm tim cn ng cathhm s:

a)2 1

3

xy

x

+=

b)2 1

1

x xy

x

+=

c)2

1

3

xy

x x

=

d)1

7y

x=

+

VD2: Tm TC v TCN cathhm s:

a)2

1

3 2

xy

x x

=

+

b)2

3

2

xy

x x

=

+

c)3

2 1

xy

x

+=

d)

2

2

3

2

x x

y x x

+

= + +

3' Hot ng 3: Cng cNhn mnh: Cch tm tim cn ng vtim cn ngang ca thhms. Nhc li cch tnh gii hnca hm s.

4. BI TP VNH:

Bi 1, 2 SGK.IV. RT KINH NGHIM, BSUNG:.........................................................................................................................................................................................................................................................................................................................................................................................................................................................



28/155


29/155

Gii tch 12 Trn STng

2

15' Hot ng 2: Luyn tp tm iu kin thc tim cnH1. Nu iu kin thhm sc ng hai TC?

1. mu c 2 nghim phn bit. nghim ca mu khng lnghim ca t.a) vi m, th lun c 2TC.

b) 2 3 12 3 1

m

m

<

>

c)9

44

m

m

<

3.Tm m thhm s cng hai TC:

a)2

3

2 2 1y

x mx m

=

+ +

b)2

22

3 2( 1) 4

xy

x m x

+=

+ + +

c)2

3

2

xy

x x m

+=

+ +

5' Hot ng 3: Cng cNhn mnh: Cch tm tim cn ng vtim cn ngang ca thhm

s. Nhc li cch tnh gii hnca hm s.

4. BI TP VNH: Bi tp thm. c trc bi "Kho st sbin thin v vthca hm s".




30/155

Trn STng Gii tch 12

1


Tit dy: 13 Bi 4: KHO ST SBIN THIN V VTHCA HM S

I. MC TIU:Kin thc:

Bit stng qut kho st hm s. Bit cc dng th ca cc hm s bc ba, bc bn trng phng, hm phn thcax b

ya x b' '

+=

+.

Knng: Bit cch kho st v vthca cc hm strong chng trnh. Bit cch tm giao im ca hai th. Bit cch dng thca hm sbin lun snghim ca mt phng trnh.


II. CHUN B:

Gio vin: Gio n. Hnh vminh ho.Hc sinh:SGK, vghi. n tp cc kin thc hc vkho st hm s.III. HOT NG DY HC:


H.Nhc li nh l vtnh n iu, cc trca hm s?.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung10' Hot ng 1: Tm hiu skho st hm s

GV cho HS nhc li cchthc hin tng bc trong s

.H1. Nu mt s cch tm tpxc nh ca hm s?

H2. Nhc li nh l v tnhn iu v cc tr ca hms?

H3.Nhc li cch tm tim cnca thhm s?

H4. Nu cch tm giao imca thvi cc trc to?

1. Mu # 0. Biu thc trong cn bc haikhng m.2.HS nhc li.

3.HS nhc li.

4. Tm giao im vi trc tung:

Cho x = 0, tm y. Tm giao im vi trchonh:

Gii pt: y = 0, tm x.

I. SKHO ST HMS

1. Tp xc nh2. Sbin thin

Tnh y.

Tm cc im ti y = 0hoc ykhng xc nh.

Tm cc gii hn c bit vtim cn (nu c).

Lp bng bin thin. Ghi kt qu v khong niu v cc trca hm s.3. th

Tm togiao im ca thvi cc trc to.

Xc nh tnh i xng cath(nu c).

Xc nh tnh tun hon (nuc) ca hm s.

Da vo bng bin thin vcc yu t xc nh trn v.



31/155

Gii tch 12 Trn STng

2

5' Hot ng 2: p dng kho st v vthhm sbc nhtCho HS nhc li cc iu bit vhm s y ax b= + , sau cho thc hin kho st theos.

Cc nhm tho lun, thchin v trnh by.+ D = R+ y= a

+ a > 0: hs ng bin+ a < 0: hs nghch bin

+ a = 0: hs khng i

VD1:Kho st sbin thin vvthhm sy ax b= +

10' Hot ng 3: p dng kho st v vthhm sbc haiCho HS nhc li cc iu

bit vhm s 2y ax bx c= + + , sau cho thc hin kho sttheo s.

Cc nhm tho lun, thchin v trnh by.+ D = R+ y= 2ax + ba > 0

2

b

a +

+ +

4a

a < 0

2

b

a +

4a

VD2:Kho st sbin thin vvthhm s:

2y ax bx c= + + (a 0)

12' Hot ng 4: Cng cNhn mnh: Skho st hm s.

Cc tnh cht hm s hc.Cu hi: Kho st s binthin v vthhm s:

a) 2 4 3y x x= +

b) 2 2 3y x x+= +

4. BI TP VNH: c tip bi "Kho st sbin thin v vthca hm s".

IV. RT KINH NGHIM, BSUNG:

......................................................................................................................................................................................................................................................................................................

...................................................................................................................................................



32/155

Trn STng Gii tch 12

1


Tit dy: 14 Bi 4: KHO ST SBIN THIN V VTHCA HM S(tt)

I. MC TIU:Kin thc:


ya x b' '

+=

+.



II. CHUN B:



H.Nhc li skho st hm s?.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung25' Hot ng 1: Tm hiu kho st hm sbc ba

Cho HS thc hin ln ltcc bc theo s.

Cc nhm thc hin v trnhby.

+ D = R+ y= 23 6x x+

y= 0 20

x

x

= =

+x

ylim

= ;x

ylim+

= +

+ BBT

+ x = 0 y = 4y = 0 2

1x

x

= =

+ th

II. KHO ST MT SHM A THC V HM

PHN THC1. Hm s

3 2y ax bx cx d= + + + (a 0)


3 23 4y x x= +



33/155

Gii tch 12 Trn STng

2


Cc nhm thc hin v trnhby.+ D = R

+ y= 23 1 1x( ) < 0, x

+x

ylim

= + ;x

ylim+

=

+ BBT

+ x = 0 y = 2y = 0 x = 1

+ th


3 23 4 2y x x x= + +

10' Hot ng 2: Tm hiu cc dng thca hm sbc ba

5' Hot ng 3: Cng cNhn mnh: Skho st hm s. Cc dng th ca hm sbc ba.Cu hi: Cc hm ssau thucdng no?

a) 3y x x= b) 3y x x= +

c) 3y x x= d) 3y x x= +

Cc nhm tho lun v trlia) a > 0, > 0 b) a > 0, < 0c) a < 0, < 0 d) a < 0, > 0

4. BI TP VNH: Bi 1 SGK. c tip bi "Kho st sbin thin v vthca hm s".




34/155

Trn STng Gii tch 12

1



I. MC TIU:Kin thc:


ya x b' '

+=

+.



II. CHUN B:




3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung25' Hot ng 1: Tm hiu kho st hm sbc ba



+ D = R+ y= 24 1x x( )

y= 0 1

10

x

x

x

= =

=

+x

ylim

= + ;x

ylim+

= +

+ BBT

+ thx = 0 y = 3

y = 0 3

3

x

x

=

=

Hm s cho l hm schn th nhn trc tunglm trc i xng.


PHN THC2. Hm s

4 2y ax bx c= + + (a 0)


4 22 3y x x=



35/155

Gii tch 12 Trn STng

2


Cc nhm thc hin v trnhby.+ D = R

+ y= 22 1x x( ) +

y= 0 x = 0+

xylim

= ;x

ylim+

=

+ BBT

+ th

x = 0 y =3

2

y = 0 x = 1 thnhn trc tung lm trci xng.


42 3

2 2

xy x= +

10' Hot ng 2: Tm hiu cc dng thca hm strng phng

5' Hot ng 3: Cng cNhn mnh: Skho st hm s. Cc dng th ca hm sbc bn trng phng.Cu hi: Cc hm ssau thucdng no?

a)4 2

y x x= b)4 2

y x x= +

c) 4 2y x x= d) 4 2y x x= +

Cc nhm tho lun v trli





36/155

Trn STng Gii tch 12

1



I. MC TIU:Kin thc:


ya x b' '

+=

+.



II. CHUN B:




3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung25' Hot ng 1: Tm hiu kho st hm snht bin



+ D = R \ {1}

+ y=2

3

1x( )

+

< 0, x 1

+ TC: x = 1TCN: y = 1

+ BBT+

+

+ thx = 0 y = 2y = 0 x = 2Giao im ca hai tim cn

l tm i xng ca th.


PHN THC3. Hm s

ax by

cx d

+=

+

(c 0, ad bc 0)


2

1

xy

x

+=

+



37/155

Gii tch 12 Trn STng

2



+ D = R \1

2

+ y=2

5

2 1x( )+> 0, x

1

2

+ TC: x = 12

TCN: y =1

2

+ BBT1

2

1

2

+

+

1

2 + th

x = 0 y = 2y = 0 x = 2

th nhn giao im ca 2tim cn lm tm i xng.


2

2 1

xy

x

=

+

10' Hot ng 2: Tm hiu cc dng thca hm snht bin

5' Hot ng 3: Cng cNhn mnh: Skho st hm s. Cc dng th ca hm snht bin.Cu hi: Cc hm ssau thucdng no? Tm cc tim cnca chng:

a)2 1

1

xy

x

+=

b)

2 1

1

xy

x

+=

+

Cc nhm tho lun v trli



0

ad bc > 0

x

y

0

ad bc < 0

x

y



38/155

Trn STng Gii tch 12

1



I. MC TIU:Kin thc:

Bit stng qut kho st hm s. Bit cc dng th ca cc hm s bc ba, bc bn trng phng, hm phn thc

ax by

a x b' '

+=

+

.



II. CHUN B:



H.Tm togiao im ca thhai hm s: 2 22 3 2y x x y x x,= + = + ?

. ( )5 7

1 02 4

; , ;

.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung10' Hot ng 1: Tm hiu cch xt stng giao ca cc th

T KTBC, GV cho HS nucch tm giao im ca hai th.

(1) gl phng trnh honhgiao im ca hai th.


III. STNG GIAO CACC THCho hai hm s:y = f(x) (C1) v y = g(x) (C2).

tm honh giao im ca(C1) v (C2), ta gii phngtrnh: f(x) = g(x) (1)

Gis(1) c cc nghim l x0,x1, Khi , cc giao im l

( ) ( )0 0 0 1 1 1M x f x M x f x; ( ) , ; ( ) ,

Nhn xt: S nghim ca (1)bng s giao im ca (C1),(C2).

25' Hot ng 2: p dng xt stng giao ca hai thCho HS thc hin.

H1. Lp pt honh giaoim?

Hng dn HS gii pt bc ba.

Cc nhm thc hin v trnhby.1.

a) 3 2 3 23 5 2 2 3x x x x + = +

3 23 5 8 0x x + = x = 1

VD1: Tm to giao imca thhai hm s:

a) 3 23 5y x x= + (C1)3 22 2 3y x x= + (C2)



39/155

Gii tch 12 Trn STng

2

Ch iu kin mu khc 0.

H2.Lp pt honh giao imca thv trc honh?

H3.Nu iu kin thcttrc honh ti 3 im phn bit

b) 22 4

2 41

xx x

x

= + +

3 23 0

1x x

x

=

0

3x

x

=

=

c)

2

3 11

xxx = +

22 1 0x( ) =

1

2x =

2.2 21 3 0x x mx m( )( ) + =

3.Pt c 3 nghim phn bit

2 2 3 0x mx m + = c 2nghim phn bit, khc 1

20

1 3 0m m

>

+

2 21

m

m

<

: (1) c 1 nghim

2

2

m

m

= =

: (1) c 2 nghim

2 < m < 2: (1) c 3 nghim

Da vo th, bin lun theom snghim ca phng trnh:

3 23 2x x m+ = (1)

15' Hot ng 3: n tp bi ton tip tuynH1.Nhc li ngha hnh hcca o hm ?

GV hng dn HS cch giibi ton 2. (Bi ton 3 dnhcho HS kh gii).

H2. Nu dng phng trnhng thng i qua (x0; y0) vc hsgc k?

H2.Tm togiao im ca(C) v trc honh ?

1.Hsgc ca tip tuynk = f(x0).

2. 0 0y y k x x( ) =

3. 32 3 0x x+ = 12

x

x

= =

+ Pttt ca (C) ti (1; 0):y = 0

+ Pttt ca (C) ti (2; 0):y = 9(x 2)

V. TIP TUYNBi ton 1:Vit phng trnhtip tuyn ca (C): y = f(x) ti

im ( )0 0 0M x f x; ( ) (C).

0 0 0y y f x x x'( ).( ) =

(y0= f(x0))

Bi ton 2:Vit phng trnhtip tuyn ca (C): y = f(x),bit tip tuyn c hsgc k. Gi (x0; y0) l to catip im. f(x0) = k (*)

Gii pt (*), tm c x0.T vit pttt.

Bi ton 3:Vit phng trnhtip tuyn ca (C): y = f(x),

bit tip tuyn i qua imA(x1; y1).

VD2: Vit phng trnh tiptuyn ca th (C) ca hms sau ti cc giao im ca(C) vi trc honh:

32 3y x x= +


Cch gii cc dng ton.

4. BI TP VNH: Bi 5, 6, 7, 8, 9 SGK.




42/155

Trn STng Gii tch 12

1


Tit dy: 19 Bi 4: BI TP KHO ST SBIN THINV VTHCA HM S


Skho st hm s. Bit cc dng th ca cc hm s bc ba, bc bn trng phng, hm phn thc

ax by

a x b' '

+=

+

.

Knng: Bit cch kho st v vthca cc hm strong chng trnh. Bit cch tm giao im ca hai th. Bit cch dng thca hm sbin lun snghim ca mt phng trnh. Bit vit phng trnh tip tuyn ca thhm s.


II. CHUN B:Gio vin: Gio n. Hthng bi tp.Hc sinh:SGK, vghi. n tp cc kin thc hc vkho st hm s.


H..

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Luyn tp kho st sbin thin v vthhm sbc ba

H1.Nhc li cc bc kho stv vthhm sbc ba?

Cc nhm thc hin v trnh

by.1.a)

+

+

b)

+

+

1. Kho st s bin thin v

vthhm s:a) 32 3y x x= +

b) 3 2 9y x x x= + +

x

y

O 11

4

2

22



43/155

Gii tch 12 Trn STng

2

15' Hot ng 2: Luyn tp kho st sbin thin v vthhm sbc bn trng phng

H1.Nhc li cc bc kho stv v th hm s bc bntrng phng?


1.a)

+

++

b)

+

2. Kho st s bin thin vvthhm s:

a) 4 22 2y x x= +

b) 2 42 3y x x= +

-3 -2 -1 1 2 3

-1

1

2

3

4

5

6

7

8

9

x

y

-2 -1 1 2

-1

1

2

3

x

y

10' Hot ng 3: Luyn tp kho st sbin thin v vthhm snht bin

H1.Nhc li cc bc kho stv vthhm snht bin?


1.a)

+

+

b)

+

+12

1

2

1

2

3. Kho st s bin thin vvthhm s:

a)1 2

2 4

xy

x

=

b)2

2 1

xy

x

+=

+

-4 -3 -2 -1 1 2 3 4 5 6 7

-4

-3

-2

-1

1

2

3

4

x

y

O

-3 -2 -1 1 2 3 4 5

-3

-2

-1

1

2

3

x

y

O

3' Hot ng 4: Cng c

Nhn mnh: Cc bc kho st hm s. Cc dng thca cc hms.

4. BI TP VNH: Bi 5, 6, 7, 8, 9 SGK.




44/155

Trn STng Gii tch 12

1


Tit dy: 20 Bi 4: BI TP KHO ST SBIN THINV VTHCA HM S(tt)


Skho st hm s. Bit cc dng th ca cc hm s bc ba, bc bn trng phng, hm phn thcax b

ya x b' '

+=

+.

Knng: Bit cch kho st v vthca cc hm strong chng trnh. Bit cch tm giao im ca hai th. Bit cch dng thca hm sbin lun snghim ca mt phng trnh. Bit vit phng trnh tip tuyn ca thhm s.


II. CHUN B:Gio vin: Gio n. Hthng bi tp.Hc sinh:SGK, vghi. n tp cc kin thc hc vkho st hm s.


H..

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Luyn tp xt stng giao gia cc th

H1. Nu k th hm s

ct trc honh ti 3 im phnbit ?

H2.Nu k thcc hms ct nhau ti 2 im phnbit ?

1.Pt honh giao im c 3

nghim phn bit:3 23 1 2 1 0mx mx m x( )+ =

21 2 1 0x mx mx( )( )+ + =

21

2 1 0 2

x

mx mx ( )

=

+ =

(2) c 2 nghim pb, khc 1

00

2 2 0

m

m

'

>

10mm

< >

2.Pt honh giao im c 2nghim phn bit:

22 32

1

x x mx m

x

+= +

21

x m

x

=

21

2

x m

m

=

1.Tm m thhm ssau

ct trc honh ti ba im phnbit:

3 23 1 2 1y mx mx m x( )= +

2.Tm m thcc hm ssau ct nhau ti hai im phnbit:

22 32

1

x x my y x m

x;

+= = +



45/155

Gii tch 12 Trn STng

2

15' Hot ng 2: Luyn tp bin lun snghim ca phng trnh bng thH1.Kho st v vthhms?

H2.Bin i phng trnh?

H3.Bin lun sgiao im ca(C) v (d)?

1. Cc nhm kho st v vnhanh thhm s.

-3 -2 -1 1 2 3

-2

2

x

y

m+1

O

2. 3 3 0x x m + =

3 3 1 1x x m + + = + 3.

22

m

m

< >

: pt c 1 nghim

2

2

m

m

=

=

: pt c 2 nghim

2 < m < 2: pt c 3 nghim

3. Kho st v v th (C)

ca hm s: 3 3 1y x x= + + .Da vo th (C), bin luns nghim ca phng trnhsau theo m:

3

3 0x x m + =

10' Hot ng 3: Luyn tp vit phng trnh tip tuyn ca thhm sH1. vit pttt, cn tm ccgi trno ?

1.x0, y(x0).

4 20 0

1 1 71

4 2 4x x+ + =

0 1x =

Ti7

14

;

, pttt l:

7 2 14

y x( ) = 124

y x=

Ti7

14

;

, pttt l:

72 1

4y x( ) = +

1

24

y x=

4.Vit phng trnh tip tuyn

ca (C): 4 21 1

14 2

y x x= + +

ti im c tung bng7

4.


Cch gii cc dng ton.

4. BI TP VNH: Bi tp n chng.




46/155

Trn STng Gii tch 12

1


Tit dy: 21 Bi 4: BI TP N CHNG I


Tnh n iu ca hm s. Cc trca hm s, GTLN, GTNN ca hm s. ng tim cn. Kho st hm s.

Knng: Xc nh thnh tho cc khong n iu ca hm s. Tnh c cc i, cc tiu ca hm s(nu c). Xc nh c cc ng tim cn ca thhm s(nu c). Lp bng bin thin v vthhm smt cch thnh tho. Tnh c GTLN, GTNN ca hm s. Gii c mt sbi ton lin quan n kho st hm s.



Hc sinh:SGK, vghi. n tp cc kin thc hc vkho st hm s.III. HOT NG DY HC:

1. n nh tchc: Kim tra sslp.2. Kim tra bi c:(Lng vo qu trnh luyn tp)

H..

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Luyn tp kho st hm s

H1. Nu k hm s ngbin trn D ?

H2.Nu k hm sc 1 Cv 1 CT ?

H3. Phn tch yu cu biton?

1.f(x) 0, x D

23 2 2 1 0x mx m( ) + ,x

2 2 1 0m m' = + m = 1

2.f(x) = 0c 2 nghim phnbit.

2 2 1 0m m' = + > m 1

3.Gii bt phng trnh:f(x) > 6x

6x 6m > 6x m < 0

1.Cho hm s:3 23 3 2 1 1f x x mx m x( ) ( )= + + a) Xc nh m hm sngbin trn tp xc nh.b) Vi gi tr no ca m, hmsc mt Cv mt CT.c) Xc nh m f(x) > 6x.

25' Hot ng 2: Luyn tp gii cc bi ton lin quan n kho st hm sCho HS lm nhanh cu a).H1. Nu k ng thnglun ct (C) ti 2 im phnbit ?

1. Pt honh giao imlun c 2 nghim phn bit.

32

1

xx m

x

+= +

+

2.a) Kho st v vth(C)

ca hm s3

1

xy

x

+=

+

b) Chng minh rng vi mi



47/155

Gii tch 12 Trn STng

2

H2. Nhn xt tnh cht ca

honh cc giao im M, N ?

H3.Tnh MN ?

H4.Tnhf(x), f(sinx) ?

H5. Gii pt f(x) = 0? Suy ranghim ca pt:f(sinx) = 0 ?

H6. Tnh f(x) v gii pt0f x''( )= ?

22 1 3 0

1x m x m

x

( ) + + + =

23 16

2 0m' ( ) = +

2.l cc nghim ca pt:2

2 1 3 0x m x m( )+ + + =

1

23

2

M N

M N

mx x

mx x.

++ =

=

3.2 2 2

M N M NMN x x y y( ) ( )= +

= 25

3 164

m( ) +

5

16 204

. =

minMN = 2 5 khi m = 3

4.f(x) = 2 4x x 2 4f inx x inx'(s ) sin s=

5. 20 4 0f x x x'( ) = =

1 17

2x

= [1; 1]

Pt:f(sinx) = 0 v nghim.

6.1

2 1 02

f x x x' '( ) = = =

Pttt ti1 47

2 12;

:

17 1 47

4 2 12y x

= +

m, ng thng 2y x m= + lun ct (C) ti hai im phnbit M, N. Xc nh m sao chodi MN l nhnht.

3.Cho hm s

3 21 1 4 63 2

f x x x x( )= +

a) Gii pt: 0f inx'(s ) = .b) Vit pttt ca th hm sti im c honh l nghim

ca phng trnh 0f x' '( )= .

3' Hot ng 3: Cng cNhn mnh: Cch gii cc dng ton.

4. BI TP VNH: Chun bkim tra 1 tit chng I.




48/155


49/155

Gii tch 12 Trn STng

2

Cu 7: thhm s2

1

3

xy

x x

=

c bao nhiu tim cn:

A. 0 B. 1 C. 2 D. 3

Cu 8: thhm s2

3

2

xy

x x

=

+

c bao nhiu tim cn ng:

A. 0 B. 1 C. 2 D. 3

B. Phn tlun:(6 im) Cho hm s: 3 23 3y x x= + .a) Kho st sbin thin v vth(C) ca hm s.

b) Da vo th(C), bin lun theo m snghim ca phng trnh: 3 23x x m+ = .

V. P N V BIU IM:A. Phn trc nghim:Mi cu ng 0,5 im

Cu 1 Cu 2 Cu 3 Cu 4 Cu 5 Cu 6 Cu 7 Cu 8B A D B C A D C

B. Phn tlun:Mi cu 3 im

a) 3 23 3y x x= + D = R 23 6y x x' = + y= 0 x = 0, x = 2

x x

y ylim ; lim +

= = +

+

+

x = 0 y = 3;

x = 1 y = 1; x = 3 y = 3

b) 3 23x x m+ = 3 23 3 3x x m+ = (*)

04

m

m

< >

: (*) c 1 nghim 04

m

m

= =

: (*) c 2 nghim 0 < m < 4: (*) c 3 nghim

VI. KT QUKIM TRA:0 3,4 3,5 4,9 5,0 6,4 6,5 7,9 8,0 10

Lp SsSL % SL % SL % SL % SL %

12S1 5312S2 5412S3 54

VII. RT KINH NGHIM, BSUNG:.........................................................................................................................................................................................................................................................................................................................................................................................................................................................

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

x

y



50/155

Trn STng Gii tch 12

1

Ngy son: 10/09/2009 Chng II: HM SLUTHA HM SMHM SLOGARIT

Tit dy: 23 Bi 1: LUTHA

I. MC TIU:Kin thc:

Bit cc khi nim v tnh cht ca lu tha vi smnguyn, lu tha vi smhu tkhngnguyn v lutha vi smthc. Bit khi nim v tnh cht ca cn bc n.

Knng: Bit dng cc tnh cht ca lutha rt gn biu thc, so snh nhng biu thc c cha lutha.


II. CHUN B:Gio vin: Gio n. Hnh vminh ho.Hc sinh:SGK, vghi. n tp cc kin thc hc vlutha.


H.Nhc li mt squi tc lutha vi smnguyn dng?

.3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu lutha vi smnguyn

H1. Nhc li nh ngha v tnhcht ca lu tha vi s mnguyn dng ?

H2.Bin i cc shng theo csthch hp ?

H3.Phn tch cc biu thc thnhnhn t?

1.

( )

mm n m n m n

n

nm mn n n n

n n

n

aa .a a ; a

a

a a ; (ab) a .b

a a

b b

+ = =

= =

=

2.10

3 10 91 .27 3 .3 33

= =

4 2 4 4(0,2) .25 5 .5 1 = = 9

1 7 91

128 . 2 .2 42

= =

A = 8.3.

2

2 1 1

a 2 2 2a 2(a 1)

(1 a ) a =

+

3

2 2

a 1

1 a a(a 1)

=

B = 2

I. KHI NIM LUTHA1. Lutha vi smnguynCho n l mt snguyn dng.

Vi a tu: n

n tha so

a a.a....a=123

Vi a 0: 0 nn

1a 1; a

a

= =

(a: cs, n: sm)Ch :

0 n0 , 0 khng c ngha.

Lu tha vi s m nguyn ccc tnh cht tng t nh lutha vi smnguyn dng.

VD1:Tnh gi trca biu thc10

3

94 2 1

1A .27

3

1(0, 2) .25 128 .

2

= +

+ +

VD2:Rt gn biu thc:3

2 1 1 2

a 2 2 2 aB .

(1 a ) a 1 a

=

+ (a 0, a 1)



51/155

Gii tch 12 Trn STng

2

8' Hot ng 2: Bin lun snghim ca phng trnh nx b= H1.Da vo th, bin lun snghim ca cc phng trnh:

3 4x b, x b= = ?

GV hng dn HS bin lun. T nu nhn xt.

2. Phng trnh nx b= (*)a) n l:(*) lun c nghim duy nht.b) n chn:+ b < 0: (*) v nghim.+ b = 0: (*) c 1 nghim x = 0

+ b > 0: (*) c 2 nghim inhau.

15' Hot ng 3: Tm hiu khi nim v tnh cht cn bc n

Da vo vic gii phng trnhnx b= , GV gii thiu khi nim

cn bc n.

H1.Tm cc cn bc hai ca 4?

Lu HS phn bit k hiu 2 gitrcn bc n ca mt sdng.

GV hng dn HS nhn xt mtstnh cht ca cn bc n.

H2.Thc hin php tnh ?

1.2 v 2.

2.

A = 5 32 2 =

B = ( )33

3 3=

3. Cn bc na) Khi nimCho b R, n N

* (n 2). S a

glcn bc nca b nu na b= .Nhn xt:

n l, b tu : c duy nht mtcn bc n ca b, k hiu n b n chn:+ b < 0: khng c cn bc n cab.

+ b = 0: cn bc n ca 0 l 0.+ b > 0: c hai cn tri du, k

hiu gi trdng l n b, cn gi

trm l n b .b) Tnh cht ca cn bc n

n n na. b ab= ;n

n

n

a a

bb=

( )m nn ma a= ; nk nka a=

n n a khi n leaa khi n chan

=

VD3:Rt gn biu thc:

A = 5 54 8. ; B = 3 3 3


nh ngha v tnh cht ca lutha vi smnguyn. nh ngha v tnh cht ca cnbc n.

4. BI TP VNH: Bi 1 SGK.

IV. RT KINH NGHIM, BSUNG:......................................... .................................................. .................................................. .................................................................. .................................................. .................................................. .................................................................. .................................................. .................................................. .........................



52/155

Trn STng Gii tch 12

1


Tit dy: 24 Bi 1: LUTHA (tt)

I. MC TIU:Kin thc:

Bit cc khi nim v tnh cht ca lutha vi smnguyn, lutha vi smhu tkhng nguyn v lutha vi smthc.

Bit khi nim v tnh cht ca cn bc n.Knng:

Bit dng cc tnh cht ca lutha rt gn biu thc, so snh nhng biu thc c chalutha.



Hc sinh:SGK, vghi. n tp cc kin thc hc vlutha.III. HOT NG DY HC:


2. Kim tra bi c:(3')H.Nu mt stnh cht ca cn bc n?

.3. Ging bi mi:

TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu lutha vi smhu t

GV nu nh ngha.

H1.Vit di dng cn thc?

H2. Phn tch t thc thnhnhn t?

1.

A = 31 1

8 2=

B = 33

1 14

84

= =

2.5 5 1 1

4 4 4 4x y xy xy x y

+ = + C = xy.

4. Lutha vi smhu t

Cho a R, a > 0 vm

rn

= ,

trong m Z, n N, n 2.m

nr mna a a= =

c bit:1

nna a=

VD1:Tnh gi trcc biu thc

A =

131

8

; B =324

VD2:Rt gn biu thc:

C =

5 5

4 44 4

x y xy

x y

+

+

(x, y > 0)

8' Hot ng 2: Tm hiu khi nim lutha vi smv tGV cho HS nhn xt kt qu

bng tnh 3 nr . T GV nunh ngha.

HS tnh v nu nhn xt. 5. Lutha vi smv tCho a R, a > 0, l sv t.Ta gi gii hn ca dy s

( )nra l lu tha ca a vi s



53/155

Gii tch 12 Trn STng

2

m, k hiu a .

nr

a alim = vin

rlim =

Ch : 1 1 = (R)15' Hot ng 3: Tm hiu tnh cht ca lutha vi smthc

H1.Nhc li cc tnh cht calu tha vi s m nguyndng ?

H2.Nu tnh cht tng tcholutha vi smthc ?

H3.Bin i tv mu vlutha vi csa ?

H4.Ta cn so snh cc sno?

1.HS nhc li.

2.Cc nhm ln lt nu tnhcht.

3.7 1 2 7 3

a a a.+ =

( )2 2

2 2 2a a

+

=

D = 5a

( )3 1

3 1 2a a

+

=

5 3 4 5a a a. = E = a

4.V cng csnn chcnso snh cc sm.

2 3 12 18 3 2= < = 2A < B

II. TNH CHT CA LUTHA VI SMTHCCho a, b R, a, b > 0; ,

R. Ta c:

a a a. +

= ;a

aa

=

( )a a

= ; ab a b( ) . =

a a

b b

=

a > 1: a a > >

a < 1: a a > <

VD3.Rt gn biu thc:

D =

( )

7 1 2 7

2 22 2

a a

a

.+

+

(a > 0)

E =( )

3 13 1

5 3 4 5

a

a a.

+

VD4:So snh cc s:

A = 2 35 v B = 3 25

3' Hot ng 4: Cng cNhn mnh: nh ngha v tnh cht calu tha vi s m hu t, smthc.

4. BI TP VNH: Bi 2, 3, 4, 5 SGK.




54/155

Trn STng Gii tch 12

1


Tit dy: 25 Bi 1: BI TP LUTHA


Khi nim v tnh cht ca lu tha vi smnguyn, lu tha vi smhu tkhngnguyn v lutha vi smthc.

Khi nim v tnh cht ca cn bc n.Knng:

Bit dng cc tnh cht ca lutha rt gn biu thc, so snh nhng biu thc c chalutha.





2. Kim tra bi c:(Lng vo qu trnh luyn tp)H..

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Luyn tp php tnh lutha

Cho cc nhm thc hin ccphp tnh.H1. Bin i a v lu thavi csthch hp ?

H2. Phn tch cc biu thcthnh nhn t?

Ch sdng cc hng ngthc.

A =2 2

5 59 27. = 23 9=

B = 32 8=

C =

3 5

2 2 40+ =

D = 3 25 2 121 =

2.A = a

B = 2a b C = a b

1.Tnh

A =2 25 59 27. B =

3 34 4144 9:

C =50 7521 0 25

16

,

,

+

D =1

3 6b b:

2.n gin cc biu thc:

A =( )( )3 14 4

4 1 23 3 3

14

a a

a a a

a

+

+

B= ( ) ( )1 2 2 1 2 4

3 3 3 3 3 3. .a b a a b b + +

C= ( ) ( ) ( )1 1 1 1 1 14 4 4 4 2 2. .a b a b a b + +

15' Hot ng 2: Luyn tp php tnh cn thcH1. Nhc li nh ngha lutha vi smhu t?

1.

A =56a

B = bC = a

3.Cho a, b R, a, b > 0. Vitcc biu thc sau di dnglutha vi smhu t:

A =1

3a a. B =11

632b b b. .



55/155

Gii tch 12 Trn STng

2

H2.Phn tch tv mu thnhnhn t?

D =1

6b

2.

A =1

11

b

b

=

(b 1)

B = ( )1 1 2 2

3 3 3 3

2 2 33 3

1a b a b

ab

a b

=

C =( )1 1 1 13 3 6 6

31 1

6 6

a b a bab

a b

+=

+

C =4

33a a: D =1

3 6b b:

4.Cho a, b R, a, b > 0. Rtgn cc biu thc sau:

A =

( )

( )

15 54 15

233 23

b b b

b b b

B =

1 1 1 1

3 3 3 3

3 32 2

a b a b

a b

C =

1 13 3

6 6

a b b a

a b

+

+

10' Hot ng 3: Luyn tp vn dng tnh cht lutha

H1. Bin i a v cng cs?

H2.Sdng tnh cht no?

1.a) x = 1

b) x =8

9

c) x =5

4

d) x =3

2

1.a) x < 3 (a < 1)b) x < 2 (a < 1)

c) x 1)

d) x < 1

5.Gii phng trnh:a) 54 1024x =

b) 1 31

832

x=

c) ( )2

2 13 3

9

xx

=

d) 0 2 0 008x, ,=

6.Gii bt phng trnh:

a) 0 1 100x

, >

b)100

0 39

x, >

c)1

39 3

x<

d) 11

27 33

x x.

<

3' Hot ng 4: Cng cNhn mnh: Cch vn dng nh ngha v

tnh cht ca lu tha giiton.

4. BI TP VNH: Bi tp thm. c trc bi "Hm slutha".




56/155

Trn STng Gii tch 12

1


Tit dy: 26 Bi 2: HM SLUTHA

I. MC TIU:Kin thc:

Bit khi nim v tnh cht ca hm slutha. Bit cng thc tnh o hm ca hm slutha. Bit dng thca hm slutha.

Knng: Bit kho st hm slutha. Tnh c o hm ca hm slutha.





H.Cho VD mt shm slutha hc?

. 21

y x y y xx

; ;= = = ,

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung18' Hot ng 1: Tm hiu khi nim hm slutha

H1. Cho VD mt s hm lutha v vthca chng ?

H2.Nhn xt tp xc nh cacc hm s ?

GV nu ch .

H3. Da vo yu tno xcnh tp xc nh ca hm slu tha ? T ch ra iukin xc nh ca hm s?

1. Cc nhm tho lun vtrnh by.

12 1 2y x y x y x y x; ; ;

= = = =

-3 -2 -1 1 2 3

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

x

y

y = x

y = x2

y = x-1

y = x1/2

3.Da vo sm.

a) 1 x > 0 D = (; 1)b) 22 0x >

D = 2 2( ; )

c) 2 1 0x D = R \ {1; 1}

d) 2 2 0x x > D = (; 1) (2; +)

I. KHI NIM

Hm s y x= vi R glhm slutha.

Ch : Tp xc nh ca hm

sy x= tuthuc vo gi trca :nguyn dng: D = R

0

nguyenam

=

: D = R \ {0}

khng nguyn: D = (0;+)

VD1: Tm tp xc nh ca cc

hm s:

a)1

31y x( )

=

b)3

2 52y x( )=

c) 2 21y x( )=

d) 2 22y x x( )=



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Tit dy: 27 Bi 2: HM SLUTHA (tt)

I. MC TIU:Kin thc:

Bit khi nim v tnh cht ca hm slutha. Bit cng thc tnh o hm ca hm slutha. Bit dng thca hm slutha.

Knng: Bit kho st hm slutha. Tnh c o hm ca hm slutha.





H.Nu tp xc nh v cng thc tnh o hm ca hm slutha?.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu kho st hm slutha

GV hng dn HS kho st

v v th hm s y x= theo tng bc ca skhost.

Cc nhm tho lun v trli.

III. KHO ST HM S

LUTHA y x=

Tp kho st

Sbin thin

Gii hn c bit

Tim cn

Bng bin thin

th

y x

=

(

> 0)(0; +)

1 0y x = > , x > 0

0

0xx

x xlim ; lim

+ +

= = +

Khng c

y x

=

(

< 0)(0; +)

1 0y x = < , x > 0

0

0xx

x xlim ; lim

+ +

= + =

TCN: trc OxTC: trc Oy

Ch : Khi kho st hm slu tha vi s m c th, ta

phi xt hm s trn tonbtp xc nh ca n.



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20' Hot ng 2: p dng kho st v vthca hm sluthaH1. Thc hin cc bc khost v vth?

H2. Thc hin cc bc khost v vth?

1. Cc nhm tho lun vtrnh by.D = (0; +)

743

4y x'

= < 0, x D

TC: x = 0; TCN: y = 0BBT:

th

2. Cc nhm tho lun vtrnh by.D = R \ {0}

4

3y

x

' = < 0, x D

TC: x = 0; TCN: y = 0BBT:

+

+

thHm s 3y x = l hm s lnn thnhn gc tolmtm i xng.

VD1:Kho st sbin thin v

vthhm s34y x

= .

VD2:Kho st sbin thin v

vthhm s 3y x

=

3y x

=

5' Hot ng 3: Cng cNhn mnh: Tnh cht v th ca hmslutha.

Bng tm tt> 0 < 0

o hm 1y x'

= 1y x' =

Chiu bin thin Lun ng bin Lun nghch binTim cn Khng c TCN: trc Ox

TC: trc Oyth Lun i qua im (1; 1)

4. BI TP VNH: Bi tp thm. c trc bi "Logarit".




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Tit dy: 28 Bi 2: LOGARIT

I. MC TIU:Kin thc:

Bit khi nim v tnh cht ca logarit. Bit cc qui tc tnh logarit v cng thc i cs. Bit cc khi nim logarit thp phn, logarit tnhin.

Knng: Bit vn dng nh ngha tnh mt sbiu thc cha logarit n gin. Bit vn dng cc tnh cht ca logarit vo cc bi ton bin i, tnh ton cc biu thc

cha logarit.Thi :


Gio vin: Gio n.Hc sinh:SGK, vghi. n tp cc kin thc hc vlutha.


H.Gii cc phng trnh: 2 8 3 81 2 3x x x; ;= = = ?.

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu khi nim logarit

Dn dt t KTBC, GV nunh ngha logarit.

H1.Nhn xt gi trbiu thc

a?

H2. Thc hin php tnh vgii thch ?

1. a> 0, b > 0

2.

a) 2 8log = 3 v32 8=

b) 13

9log = 2 v2

19

3

=

c) 12

4log = 2 v21 4

2

=

d) 31

27log = 3 v 3

13

27

=

I. KHI NIM LOGARIT1. nh nghaCho a, b > 0, a 1.

ab a blog

= =

Ch :khng c logarit ca sm v s0.

VD1:Tnh:a) 2 8log b) 1

3

9log

c) 12

4log d) 31

27log

10' Hot ng 2: Tm hiu tnh cht ca logaritGV hng dn HD nhn xtcc tnh cht.

a0= 1 1 0

alog =

2. Tnh chtCho a, b > 0, a 1.



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a1= a 1a

alog =

1.

a) 32 53 log = ( )32

5 23 5log =

b) 12

8log =3

1

2

13

2log

=

c)2

1

74log

= ( )22

1 27 12

7

log =

d)5

131

25

log

= ( )52

1 23 15

3

log

=

1 0 1a

a a

b

a

a

a b alog

log ; log

; log ( )

= =

= =

VD2:Tnh:

a) 32 53 log b) 12

8log

c)2

174

log

d)5

131

25

log

10' Hot ng 3: Tm hiu qui tc tnh logarit

H1.Cho 3 51 2

2 2b b,= = . Tnh

2 1 2 2 2 1 2b b b blog log ; log+ .

So snh kt qu?

GV nu nh l.


1.

2 1 2 2

2 1 2

3 5 88

b b

b b

log log

log

+ = + =

=

2 1 2 2 2 1 2b b b blog log ; log+ =

2.

a) = 6 36 2log =

b) 1 1 1

2 2 2

1 1 12

3 3 3log log log= +

c) = 13

27 3log =

d) = 5 125 3log =

II. QUI TC TNH

LOGARIT1. Logarit ca 1 tchCho a, b1, b2> 0, a 1.

1 2 1 2a a ab b b blog ( ) log log= +

Ch :nh l trn c thmrng cho tch ca n sdng:

1 1a n a a nb b b blog ( ... ) log ... log= + +

VD3:Tnh:

a) 6 69 4log log+

b) 1 1 12 2 2

1 32 2

3 8log log log+ +

c) 1 1 13 3 3

95 3

5log log log+ +

d) 5 55

753

log log+

3' Hot ng 4: Cng cNhn mnh: nh ngha logarit. Qui tc tnh logarit.

4. BI TP VNH: Bi 1, 2 SGK. c tip bi "Logarit".




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Tit dy: 29 Bi 2: LOGARIT (tt)

I. MC TIU:Kin thc:

Bit khi nim v tnh cht ca logarit. Bit cc qui tc tnh logarit v cng thc i cs. Bit cc khi nim logarit thp phn, logarit tnhin.


cha logarit.Thi :


Gio vin: Gio n.Hc sinh:SGK, vghi. n tp cc kin thc hc vlogarit.


H.Nu nh ngha logarit v tnh:2 1

4

1log ; log 2

8?

.3. Ging bi mi:

TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung15' Hot ng 1: Tm hiu cc qui tc tnh logarit

Tng t nh logarit ca 1tch, GV cho HS nhn xt.


GV hng dn HS chngminh.

1.a) =

2log 8 3=

b) = 31

log 29=

c) =

1

5

log 25 2=

d)7

1log 1

7=

ta

b b alog = =

II. QUI TC TNHLOGARIT2. Logarit ca 1 thngCho a, b1, b2> 0, a 1.

a a a

bb b

b

11 2

2

log log log=

c bit:a a

bb

1log log=

VD1:Tnh:a)

2 2log 120 log 15

b) 3 3log 16 log 144

c)1 1

5 5

log 16 log 400

d)7 7

log 30 log 210

3. Logarit ca 1 luthaCho a, b > 0; a 1; tu:

a ab blog log =

c bit:

n

a ab b

n

1log log=



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H2.Thc hin php tnh ? 2.

a) =2

72

2log 2

7=

b) =1

25

1log 5

2

=

VD2:Tnh:

a)1

72

log 4

b)5 5

1log 3 log 15

5

10' Hot ng 2: Tm hiu cng thc i csH1.Cho a = 4, b = 64, c = 2.Tnh

a c cb a blog , log , log . T

rt ra nhn xt?

GV hng dn HS chngminh.


1.

c a ca b blog . log log=

( )a bc c

b alog

log log=

=a c

b alog . log

2.

a)8 2

1log 9 log 9

3=

b)4 2 2

1log 15 log 15 log 15

2= =

c)1

31 3

27

log 2 log 2

=

III. I CSCho a, b, c > 0; a, c 1.

c

a

c

bb

a

loglog

log=

c bit:

a

b

ba

1log

log= (b 1)

aa b b

1log log

= (0)

VD3:Tnh:a)

3 8 6log 6.log 9.log 2

b) 4log 152 c)1

27

log 2

3

10' Hot ng 3: Tm hiu khi nim logarit thp phn, logarit tnhin GV gii thiu khi nimlogarit thp phn v logarit t

nhin.

GV hng dn HS s dngMTBT tnh.

HS theo di v thc hnhtrn MTBT.

2

log3log 3 1,5850

log2=

3

ln0,8log 0,8 0,2031

ln 3=

IV. LOGARIT THP PHN,LOGARIT TNHIN

1. Logarit thp phnb b b

10lg log log= =

2.Logarit tnhin

eb bln log=

Ch :Mun tnha

blog vi a

10 v a e, bng MTBT, tac th s dng cng thc ics.

3' Hot ng 4: Cng cNhn mnh: Qui tc tnh logarit.

Cng thc i cs.

4. BI TP VNH: Bi 3, 4, 5 SGK.




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Tit dy: 30 Bi 2: BI TP LOGARIT


Khi nim v tnh cht ca logarit. Cc qui tc tnh logarit v cng thc i cs. Cc khi nim logarit thp phn, logarit tnhin.


cha logarit.Thi :


Gio vin: Gio n. Hthng bi tp.Hc sinh:SGK, vghi. n tp cc kin thc hc vlogarit.

III. HOT NG DY HC:1. n nh tchc: Kim tra sslp.2. Kim tra bi c:()

H..

3. Ging bi mi:TL Hot ng ca Gio vin Hot ng ca Hc sinh Ni dung25' Hot ng 1: Luyn tp cc qui tc tnh logarit

H1.Nu qui tc cn sdng ?

H2.Nu qui tc cn sdng ?

H3.Nu cch so snh ?

1.A = 1

B =4

3

C = 9 + 16 = 25D = 16.25 = 400

2.

A = 4 3 25 6 7+ +

B = 2 26 8+ C =

Giải tích 12 Tác giả: Trần Sĩ Tùng, 2010

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Transcript of Giải tích 12 Tác giả: Trần Sĩ Tùng, 2010