toan 12

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TMTTLTHUYTVPHNGPHPGIITON12

MC LC

Chng I: NG DNG O HM1. S ng bin - nghch bin ca hm s .............................................42. Cc tr ca hm s.....................................................................................................63. GTNN - GTLN ca hm s............................................................................ 124.Timcn.............................................................................................................................13

5. Kho st hm s........................................................................................................14

6.Mt s bi ton lin quan n hm s, th....................... 17

Chng II: HM S M, LY THA,LGARIT1. M, ly tha vlgarit ......................................................................................292. Phng trnhm.......................................................................................................333. Phng trnhlgarit.............................................................................................354.Btphng trnh m, lgarit....................................................................36

Chng III: NGUYN HM -TCHPHN1. Nguyn hm....................................................................................................................37

2.Tchphn ...........................................................................................................................41

3.ng dng hnh hc ca tchphn .......................................................45Chng IV: S PHC..............................................................................................................47Chng V: TH TCH KHI A DIN, KHI TRN

XOAY..........................................................................................................................................49

Chng VI: PHNG PHP TA TRONG KHNGGIAN

1. H ta trong khng gian......................................................................... 512. Phng trnhmtcu .........................................................................................553. Phng trnhmtphng .................................................................................604. Phng trnhng thng ..........................................................................665. V

trt

ngi

...........................................................................................................73

6. Khong cch v gc................................................................................................757. Tm mt s im c bit..............................................................................77

2

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TMTTLTHUYTVPHNGPHPGIITON12Chng VII: MT S KIN THC B SUNG

1. Tam thc bc hai,PT, BPT bc hai ...................................................792. Xt du biu thc...................................................................................................84

3.Gii hn v cc v ti v cc ca hm s..................................89

4. o hm..............................................................................................................................925. Cngthc lng gic v phng trnhlng gic ...........95

PH LC: Kinh nghim lm bi thi mn Ton....................................102

Trn con ng thnh cng khng c du chn

ca k li bing.

3

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NG DNG O HM

Bi 1: S NG BIN NGHCH BIN CA HM S

* nh ngha:- y = f(x)ng bin trn Kx1, x2 K : x1

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f(x)>g(x) ,x(a; b)ta qua cc bc sau:f(x)>g(x),x(a,b)

f(x)g(x)> 0,x(a,b)

2. t h

(x

)=f

(x

)g

(x

)

4

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y' y'


3. Tnh h'(x)v lp bng bin thinca

h (x). T suy ra kt qu.

Bi ton 3: Tm iu kin hm sy =lun lun gim) trn min xc nh

f(x)lun lun tng (hoc

- Cc hm s

2

y =ax3 +bx2 +cx +d(a 0)vy =

ax +bx +c(a 0)lun lun tng (hoc lun lun

gim)Ax

+B

trn min xc nh ca n khi v ch khi y' 0 (hoc y' 0 )xD . Nu a c cha tham s th xt thm trng hp a=0

(i vi hm bc 3)

a > 0

0

(hoc

a < 0)

0

- Hm s y=

ax +b

cx+dlun lun tng (hoc lun lun gim) trn

min xc nh ca n khi v ch khixD

y'> 0 (hocy'< 0 )

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5

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TMTTLTHUYTVPHNGPHPGIITON12Bi 2: CC TR CA HM S

Bi ton 1: p dng quy tc 1 tm cc tr ca hm s1. Tm min xc nh

2. Tm f'(x)3. Tm cc im ti f'(x)= 0

hoc

f'(x)khng xc nh(gi

chung l im ti hn).4. Sp xp cc im theo th t tng dn v lp bng xt du

o hm.5. Nu kt lun v cc tr.Bng tm tt:

x a

f'(x) +

f(x)

xo b

-

C

x a

f'(x) -

xo b

+

f(x)

CT

Bi ton 2: p dng quy tc 2 tm cc tr ca hm s

1. Tnh f'(x). Gii phngtrnh

f'(x)= 0 .

Gi xi (i= 1,2,...)l cc nghim ca phng trnh.

2. Tnh f"(x)v

f"(xi )

3. Da vo du ca xi theo nh l sau:

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nh l:f"(xi )suy ra kt lun v cc tr caim

6

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Gi s hm s y =f(x)c o hm cp hai trn khong (a; b )cha im x

ov f'(xo )= 0 . Khi :

a) Nub) Nu

f"(xo )> 0th

f"(xo )< 0th

xo l im cc tiu.x

ol im cc i.

Bi ton 3: Tm iu kin ca m hm s t cc tr ti mt im

cho trc.p dng nh l Fec-ma:

Gi s y =f(x)c o hm tiim

x =xo

.

Khi nu y =f(x)t cc tr ti

im

x =xo

th f'(xo )= 0 .

Ch : Nu f'(xo )= 0 th cha chc hm s t cc tr tiim

x =xo

. Do khi tm c m th p h i t h l i.

Bi ton 4: Tm iu kin hm s c cc i v cc tiu2

Cc hm s y =ax3 +bx2 +cx +dva y =ax +bx +cc mt cc iAx +B

v mt cc tiu khi v ch khi phng trnhy'= 0 c hai nghim phnbit (khi hin nhin y i du hai ln khi qua cc nghim). Nuhm hu t th phi khc nghim mu.Bi ton 5: Vit phng trnh ng thng i qua hai im cc tr

ax

2

+bx +c1. Cho hm s y =Ax +B

(C)

- Nu (C) c hai im cc tr- Th phng trnh ng thng qua hai im cc tr l

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(ax2 +bx +c)'

2a b

y =(Ax +B )'

hay y =A

x +A

2. Cho hm s y =ax3 +bx2 +cx +d(C)- Nu (C) c hai im cc tr v chia y cho y ta c

y =y'.A(x)+x +

7

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(

'

TMTTLTHUYTVPHNGPHPGIITON12- Th phng trnh ng thng qua hai im cc tr l

y =x +

y'(x0 )= 0Bi ton 6: iu kin hm s t cc tr tix0

:y"x

0

(hoc) 0

y'(x0 )= 0 )y'oidaukhi quax0

Bi ton 7: iu kin hm s t cc i tix0

:

y'(x0 )=0

y"(x0 )< 0

(hoc

y'(x0 )= 0 )y'oida ut+sangkhi quax

0

Bi ton 8: iu kin hm s t cc tiu tix0 :

y'(x0 )=0

y"(x0 )

> 0

(hoc

y'(x0 )= 0 )y'oida utsang+khi quax

0

Bi ton 9: iu kin hm s t C,CT ti

> 0

x1, x

2tha

Ax +Bx =C1 2Ax

1+Bx

2=C: x +

x

=b

vix1, x

2l nghim ca y'= 0

1 2 a cx x = 1 2 a

Bi ton 10: iu kin hm bc 3 c C,CT v hai gi tr cc tr

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y'

cng du:

iu kin hm bc 3 c C,CT l > 0

a 0

8 NGUYN TUN KIT

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y

TMTTLTHUYTVPHNGPHPGIITON12 Gi A(x1 ; y1 ),B (x2 ; y2 )l hai im cc tr. Ta c

y(x1 ).y(x2 )> 0 (trng hp tri du th ngc li)

Ch : Hm s vit thnh:y =P(x).y'+mx +n (ly hm schia

cho o hm)

y(x1 )=mx1 +ny(x2 )=mx2 +n

Bi ton 11: iu kin hm s bc 3 c C,CT nm v hai pha i

vi trc tung: iu kin ycbt c tha mn ly'= 0 c hai nghim

tri du. Khi P=c

< 0a

Bi ton 12: Cch tnh nhanh gi tr cc tr ca hm hu t

ax2 +bx +cy =mx +n

Tm cc im cc tr ca hm s (nghim ca phng trnhy=0)

cctr=aoha m

cuaTS

aoha mcuaMS

=2ax +b

ri thay x cc tr vo phnm

s ny ta chm hu t

ycctrtng ng, v cch tnh trn ch p dng cho

Bi ton 13: Tm m hm trng phngcc tr lp thnh mt tam gic u:

TX: D=R

y =ax4 +bx2 +cc 3im

Tnh y'= 4ax3 + 2bx =2x(2ax2 +b ),

x = 0y'= 0 2

2ax +b = 0

x = 0x2 =

b(a 0)(1)

2a

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9

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TMTTLTHUYTVPHNGPHPGIITON12 Ycbt tng ng phng trnh (1) c hai nghim phn bit

khc 0. Khi b

> 02a

Bi ton 14: iu kin hm s(;)(C

)

x =ly'()= 0

y''

(

) 0

y =f(x)(C)t cc tr bngti

Bi ton 15: Hm trng phng c 3 im cc tr lp thnh mttam gic. Tnh din tch tam gic :

Tnh y', tm 3 im ti hn, suy ra 3 im cc tr A, B, C. Tnh din tch tam giac ABC theo cng thc:

S=1 |xy'x' y |vi

AB= (

x;y)

2 AC=(x';y')Bi ton 16: Tm m hm trng phng c 3 im cc tr lpthnh mt tam gic u:

TX: D=R

Tnh x = 0y'= 4ax3 + 2bx; y'= 0

2ax2 +b = 0

x = 0 bx2 = 2a (a 0)(1)

iu kin ycbt c tha l phng trnh (1) c hai nghim

phn bit khc 0. Khi : b

> 0 (* )2a

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TMTTLTHUYTVPHNGPHPGIITON12 Vi iu kin (*), gii phng trnh

x = 0 y =c(A)

y'= 0 x = b

2ay =? (B

)

. Tm c 3 im cc tr

bx = 2a

y =? (C

) AB2

=AC2

A, B, C. Do tam gic ABC u nn AB

2 =BC2

, t tm

c m v ch nhn nhng m tha iu kin (*).

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11 NGUYN TUN KIT

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TMTTLTHUYTVPHNGPHPGIITON12Bi 3: GI TR NH NHT GI TR LN NHT

CA HM S* nh ngha:

f(x

)m,

x

K

- min y =m K x0 K : m =f(x0 )

- max y =Mf(x )M ,x K

K

*Dng ton:x0 K : M=f(x0 )

Bi ton 1: Tm GTNN, GTLN ca hm s trn mt khong

tm GTNN v GTLN ca hm s y =f(x)trn khng (a; b )talp

bng bin thin ca hm s trn khong (a; b )ri da vo mkt

lun.Bi ton 2: Tm GTNN, GTLN ca hm s lin tc trn mt on

a; b

Cch 1: C th lp bng bin thin ri da vo m kt lun.Cch 2: Qua 3 bc:

1. Tm cc im x1, x

2,..., x

n

trn

a;b

m ti f'(x)= 0

hocf'(x)khng xc nh.

2. Tnh f(a ), f(b ), f(x1 ), f(x2 ),..., f(xn ).3. Tm s ln nht M v nh nht m trong cc s trn. Khi :

M=max f(x),m =min f(x)a ;b a ;b

Bi ton 3: Tm m phng trnh f(x)=m c nghim trn D: Xt hm s y =f(x)trn D, tm maxy, miny hoc tm tp

gi

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tr ca y t kt lun c m.

12 NGUYN TUN KIT

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0

lim

TMTTLTHUYTVPHNGPHPGIITON12Bi 4: TIM CN

1. Cch tm tim cn:

Nu lim y =()th ng

thngx x

x =x0

l tim cn ng.

Nu lim y =y0

th ng thngx

y =y0

l tim cn ngang.

Nu hm s vit thnh y =thngax +b+

Sod

Mauso(chia a thc)

m lim

Sod

=0th ngthng y =ax +b l tim cn xin.x Ma uso

* ng thng

y =ax +b gi l TCX ca hm s

f(x)y =f(x)

a =x x

b =

lim(f(x )

ax)x ax +b

2. Cc ng tim cn ca th hm s

TC : x =

d

y =cx +d

l :

c

TCN : y =a

c3. Cho M thuc (C). Tnh tch cc khong cch t 1 im trn (C) n2 tim cn:

Gi M(x0 ; f(x0 ))(C). Tm TC, TCX (hoc TCN) d=d(M,TC).d(M,TCN) l mt hng s.

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13 NGUYN TUN KIT

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TMTTLTHUYTVPHNGPHPGIITON12Bi 5: KHO ST HM S

1. S kho st:1. Tp xc nh:

2. S bin thin:

D =

a) Xt chiu bin thin ca hm s:- Tm o hm- Tm cc im m ti o hm bng 0 hoc khng xc nh.- Xt du o hm, suy ra chiu bin thin ca hm s.

b) Tm cc tr.c) Tm cc gii hn v tm tim cn (nu c)

d) Lp bng bin thin.* Ch : Kt lun v tnh ng bin, nghch bin phi trc BBT3. Da vo bng bin thin v cc yu t xc nh trn v

th. Ch :- v th chnh xc nn tnh thm ta ca mt s im,

c bit cn tm ta cc giao im ca th vi cc trc ta.

- Cn lu cc tnh cht i xng trc, i xng tm.2. Cc dng th:

1. Hm s bc ba: y =ax3 +bx2 +cx +d(a 0)

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th nhn im un lm tm i xng.

2. Hm s trng phng: y =ax4 +bx2 +c (a 0)

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TMTTLTHUYTVPHNGPHPGIITON12 th nhn trc Oy lm trc i xng.

3. th hm s y =ax +b

(c 0); adbc 0cx +d

th nhn giao im hai tim cn lm tm i xng.* Ch : M(x0 ; y0 )(C): y=

f(x )y0=

f(x0 )

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16 NGUYN TUN KIT

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TMTTLTHUYTVPHNGPHPGIITON12Bi 6: MT S BI TON

LIN QUAN N HM S V THBi ton 1: S tng giao ca cc th (bng phng trnh honh giao im)

Cho hai ng cong (C1 ): y=

f(x),(C2 ): y =g(x).

xt s tng giao gia (C1 ),(C2 )ta lp phng trnh honh giao im f(x)=g(

x

)

(1)

1. (C1 )khng c im chung vi (C2 )pt (1) v nghim.2. (C1 )ct (C2 )ti n im phn bit pt (1) c n nghimphn

bit. ng thi nghim ca pt (1) l honh giao im ca

(C

1 )v

(C

2 ).

Ch : Nu phng trnh honh giao im c dng

Ax2 +Bx +C= 0 .Ta bin lun theo A v . Tc l:- Nu A=0. Ta c kt lun c th v giao im ca (C1) v (C2).- Nu A 0. Tnh

+ 0: c hai giao im. Nu phng trnh honh giao im c dng

ax3 +bx2 +cx +d= 0 . a phng trnh ny v dng:

(x )(Ax2 +Bx +C)= 0 (Chia Horner, a 0)x =

Ax2 +Bx +C=0(1)Bin lun theo phng trnh (1) ta suy ra c s giao

im.Bi ton 2: Da vo th bin lun s nghim ca phng trnh

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F(x,m )=0

(1)

1. Bin iF(x,m )= 0 vdng

f(x)=g(m ).

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TMTTLTHUYTVPHNGPHPGIITON122. S nghim ca phng trnh (1) l s giao im ca th hm

s y = f(x)v ngthng

y =g(m )

3. Da vo th bin lun cc trng hp.

Ch : y =g(m )l ng thng song song vi trc Ox vct

trc Oy ti im c tung bngy

g(m )

y=f(x)

O 1

g(m)

x

y=g(m)

Bi ton 3: Phng trnh tip tuyn iu kin tip xc Dng 1: Phng trnh tip tuyn ti im thuc th:

Phng trnh tip tuyn ca (C):

M(xo ; yo )(C)l:y =f(x)ti im

y y0

= f'(x0 )(x x0 )Trong : + M(x0 ; y0 )gi l tip im.

+ k= f'(x0 )l h s gc ca tip tuyn. Dng 2: Phng trnh tip tuyn bit h s gc k:- Nu tip tuyn song song vi ng thngy =ax +b th k=a

- Nu tip tuyn vung gc ng thngy=

ax+

b th k=

1

a- Tip tuyn hp vi chiu dng ca trc honh mt gc th

k=tan1. Gii phng trnh f'(x)=k tm

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x0

l honh tip im.

2. Tnh y0

=f(x0 ).

18 NGUYN TUN KIT

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c

TMTTLTHUYTVPHNGPHPGIITON123. Phng trnh tip tuyn l y =k(x x0 )+y0 Dng 3: Tip tuyn i qua im A(xA ; yA )

1. Gi k l h s gc ca tip tuyn (d). Khi phng trnh ca(d) c dng y =k(x xA )+y

A. f'(x)=k

2. (d) tip xc vi (C) thi v ch khi h f(x)=k(x xA )+yA

nghim (h c n nghim th c n phng trnh tip tuyn)

3. Gii h tm c honh tip im lx0 v h s gc k.4. Thay vo phng trnh ca (d) ta c tip tuyn cn tm. Dng 4: Vit phng trnh tip tuyn (d) bit tip tuyn to vi

ng thng (): y=ax+b mt gc bng (0 0

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g()0

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TMTTLTHUYTVPHNGPHPGIITON12Bi ton 5: iu kin hm trng phngti 4 im phn bit:

y =ax4 +bx2 +cctOx

* Lp phng trnh honh giao im ca (C) v Ox:

ax4 +bx2 +c = 0

t=x2 0

at2 +bt+c =0(1)

* iu kin ycbt c tha l (1) phi c hai nghim dng phn

> 0

bit. Khi

P> 0S> 0

Bi ton 6: iu kin hm trng phng ct Ox ti 4 im phnbit lp thnh CSC:* Lp phng trnh honh giao im ca (C) v Ox:

ax4 +bx2 +c = 0

t=x2 0

at2 +bt+c =0(1)

* iu kin ycbt c tha l (1) phi c hai nghim dng phn

> 0bit. Khi

P> 0

S> 0

(*)

* Vi iu kin (*) c tha ta c 4 im c honh lp thnh CSC

nn (1) phi c hai nghim dng phn bit tha t2

= 9t1

(2).

b

Theo nh l Vit

t1 +t2 =a(3)

t .t

=1 2c

(4)a* T (2), (3), (4) ta gii ra tham s, ch nhn tham s khi m tha iukin (*).Bi ton 7: Tm m d:cho AB=l:

y =m ct (C) ti hai im phn bit A, B sao

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> 0

> 0

TMTTLTHUYTVPHNGPHPGIITON12* Lp phng trnh honh giao im ca (C) v d. Bin i phng

trnh ny v dng Ax2 +Bx +C= 0 (1)

* iu kin d ct (C) ti hai im phn bit l

A 0

(1)(* )

* Gi A(x1 ; m ),B (x2 ; m )l hai giao im ca (C)v d;

x1, x

2l

nghim ca (1). Ta c:

AB = (x x )2

=|x x |=|x x |=

=2 '

=l. T tm2 1 1 2 2 1

| a | | a |c m, ch nhn nhng m tha iu kin (*).Bi ton 8: Tm m d: y =m ct (C) ti hai im phn bit A, B saochoAB c di ngn nht:* Lp phng trnh honh giao im ca (C) v d. Bin i phng

trnh ny v dngAx

2

+Bx +C= 0(1) A 0* iu kin d ct (C) ti hai im phn bit l

(1)(*)

* Gi A(x1 ; m ),B (x2 ; m )l hai giao im ca (C)v d;

x1, x

2l

nghim ca (1). Ta c

AB = (x x )2

=|x x |=|x

x |=

=2 '

. T tm2 1 1 2 2 1

| a | | a |iu kin ca m AB nh nht, ch nhn m tha (*).Bi ton 9: Tm m d: y =m ct (C) ti hai im phn bit A, B saocho OA OB vi O l gc ta :* Lp phng trnh honh giao im ca (C) v d. Bin i phngtrnh ny v dng Ax2 +Bx +C= 0

(1) A 0* iu kin d ct (C) ti hai im phn bit l

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> 0(1) (*)

21 NGUYN TUN KIT

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TMTTLTHUYTVPHNGPHPGIITON12* Gi A(x1 ; m ),B (x2 ; m )l hai giao im ca (C)

v d;

x1, x

2l

nghim ca (1). Ta c OA OB nn ta c OA.OB = 0 . T y tmc m, ch nhn nhng m tha (*).Bi ton 10: Tm m d: y =ax +b ct (C) ti hai im phn bittrn cng mt nhnh ca (C):* Lp phng trnh honh giao im ca (C) v d. Bin i phng

trnh ny v dng Ax2 +Bx +C= 0 (1).

A 0* iu kin ycbt c tha l

s.

(1)> 0A.g()> 0

vi l nghim ca mu

Bi ton 11: Tm m d: y =ax +b ct (C) ti hai im phn bit

trn cng hai nhnh khc nhau ca (C)* Lp phng trnh honh giao im ca (C) v d. Bin i phng

trnh ny v dng Ax2 +Bx +C= 0 (1).A 0

* iu kin ycbt c tha l

s.

(1)> 0A.g()< 0

vi l nghim ca mu

Bi ton 12: Tm nhng im trn (C):y =vung gc vi ng thngy =ax +b.

f(x)m ti tip tuyn

* Gi M0 (x0 ; y0 )(C). H s gc ca tip tuyn

ti

M0

l f'(x0 ).

Gii phng trnh f'(x0 ).a =1. T y tmc

x0

v c c M0

.

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Bi ton 13: CMR mi tip tuyn ca (C):y =giao im hai tim cn:

f(x)u khng qua

22 NGUY

N TUN KIT

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TMTTLTHUYTVPHNGPHPGIITON12* Ta giao im I hai tim cn l nghim ca h phng trnh:

TiemcanngTiemcanxien (hay TCN)

* Lp phng trnh tip tuyn qua I, kt qu l khng c tip tuyn.T ta c iu phi chng minh.

Bi ton 14: Cho M(C), tip tuyn ti M ct hai tim cn ca(C)

ti A, B, gi I l giao im hai tim cn. CMR M l trung im caAB. Tnh din tch tam gic IAB:

* Gi M(x0 ; f(x0 ))(C). Phng trnh tip tuyn ti M ly y

0=f'(x0 )(x x0 )y

=

f'(x0 )(x x0 )+y0 .

* Tm giao im ca tip tuyn vi TC l A* Tm giao im ca tip tuyn vi TCX l B.* Tm giao im I ca hai tim cn.* Kim tra cng thc M l trung im AB, t ta c iu phi chngminh.

* Tnh vectIA,IB . T tnh din tch tam gic IAB (kt qu l mthng s.Bi ton 15: CMR tip tuyn ti im un l tip tuyn c h s gcnh nht (hoc ln nht):

* Tm h s gc ca tip tuyn ti imun I(x0 ; y0 )

lf'(x0 ).

* Gi h s gc ca tip tuyn bt kl

f'(x). Ta chng minh

f'(x)

f'(x0 )(trong trng hp ln nht ta lm ngc li).

Bi ton 16:Tm nhng im trn ng thng :

c th k c 2, 3 tip tuyn n (C):y =y

0m t

* Gi M(a; y0 )(). Vit phng trnh d qua M v c h s gc k

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l:

y y0

=k(x a )y =k(x a)+y0 .

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* iu kin d l tip tuyn ca (C) f(x)=k(x a )

+y0

f'(x)=k

(1).

Mun t M v c 2,3 tip tuyn th (1) c 2,3 nghim.Bi ton 17: CMR mi tip tuyn ca (C) to vi hai tim cn 1 tamgic c din tch khng i:

* Gi M(x0 ; f(x0 ))(C). Phng trnh tip tuyn ti M ly y

0=f'(x0 )(x x0 )y

=

f'(x0 )(x x0 )+y0 .

* Tm giao im ca tip tuyn vi TC l A* Tm giao im ca tip tuyn vi TCX l B.* Tm giao im I ca hai tim cn.* Kim tra cng thc M l trung im AB, t ta c iu phi chngminh.

* Tnh vectIA,IB . T tnh din tch tam gic IAB (kt qu l mthng s.Bi ton 18:Tm trn (C) nhng im c ta l cc s nguyn:

Sod* Hm s vit thnhy =Thng+

Mauso(chia a thc)

* Do x, y nguyn nn Mu s = c ca S d.Bi ton 19: Tm nhng im trn (C) cch u hai trc ta :

* Nhng im trn (C) cch u hai trc ta l nghim ca h

phng trnhy=

f(x)

y=

hoc

f(x)

y =x y =xBi ton 20: Tm nhng im trn (C) i xng nhau qua gc ta :

* Gi

.

A(x0 ; y0 ),B (x0 ;y0 )l hai im i xng nhau qua gcta

* Thay ta A, B vo phng trnh ca hm s ta c h phngtrnh. Gii h ny ta c ta im cn tm.Bi ton 21: Tm nhng im trn th hm nht bin sao cho tng

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khong cch t n hai tim cn t GTNN:

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F(X)lBi ton 25: CMR th (C) nhn ng thng

xng:x =x

0lm trc i

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* Bng php tnh tin theo vectOIviI(x0 ;0), h trc Oxy thnhh

X=x xtrc IXY. Ta c cng thc i trc: Y=y 0

0 x =X+x0

y =Y(1)

* Thay (1) vo hm cho ta c Y=F(X). Kimchng

hm chn.

Bi ton 26: Tm tp hp im (qu tch)

F(X)l

* Tm ta imM(x;y)theo mt thams

x =g(m )

y =h (m )

* Kh m t h trn ta c phng trnhF(x;y)= 0 .* Gii hn: da vo iu kin tn ti im M hay iu kin khi kh

m tm iu kin ca x hoc y.Kt lun: tp hp im M l ng (L) c phng trnh

F(x;y)= 0 tha iu kin bc 3.Bi ton 27: Tm im c nh m h (Cm)lun i qua:* Bin i phng trnhy =f(x,m )v

dng

Am +B = 0 (hay

Am2 +Bm +C= 0 (n m)).* Ta im c nh l nghim ca h phng trnh

A = 0A = 0B = 0

(hayB = 0)

C= 0

Bi ton 28: S tng giao gia 2 th m trong tham s m cbc 1 (tc l trong biu thc khng cha m2, m3)Gi s bi ton tm giao im ca ng cong qui v tm nghim ca

phng trnh f(x)=g( x)

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TMTTLTHUYTVPHNGPHPGIITON12* Bin i (1) v dng

thc.

F(x)=m (2), y F(x) c th l hm phn

* Lp bng bin thin ca hm sy =F(x

)* Da vo bng bin thin ta bin lun s nghim ca (2), v t suyra kt lun i vi (1).Nhn xt: Phng php ny cng c bit c ch cho bi ton tm m nghim ca phng trnh, h phng trnh,... tha iu kin chotrc no v mt s bi ton khc v tm m.

Bi ton 29: Cc php bin i th:

* T th hm s

1. V (C)y =f(x)(C)suy ra th hm sy=

f(x ) (C')

2. Gi nguyn phn th (C) nm pha trn trc honh; ly i xngca phn th (C) nm pha di trc honh qua trc honh.3. Xa phn th nm pha di trc honh, th cn li chnh l(C)

y

1 x

th hm s y = f(x

)

(phn nt lin, nt t l phn c xa)

* T th hm s

1. V (C)

y =f(x)(C)suy ra th hms

y =f(x )

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TMTTLTHUYTVPHNGPHPGIITON122. Xa phn th (C) nm pha bn tri trc Oy v cha li phn thnm bn phi.3. Ly i xng phn th ca (C) bn phi trc Oy qua Oy, ta cc th (C).

y

1x

th hm s y =f(x ) (phn nt lin, nt t l phn c xa)

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b

m


HM S M, LY THA,LGARIT

M, LY THA V LGARIT1. Ly tha, cn bc n:a) nh ngha:

* an =a.a.......a (a

,n*)n thaso

b) Tnh cht:Vi a,b *; m, n ta c:

* am an =am +n *

* a0 = 1;

a=amn

an =1

an

an

n n

* (ab )n

=anbn *

a =

a

* (am)

n=amn

* Nu: 0 1v m >nth:

am >an

* Nu 0

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c

TMTTLTHUYTVPHNGPHPGIITON122. Lgarit:

a)nh ngha: logab) Tnh cht:

b =c b =ac (0 0)

Cho a,b>0,ngha: a 1. Cc tnh cht sau c suy trc tip t nh* loga 1 = 0 * loga a = 1

* aloga b =bc) So snh logarit:

Cho a,b,c>0, c 1. Ta

c:

* loga ak =k(k)

*logc a = logc b a =b

* Neuc > 1th: logc a < logc b a 0, a 1. Ta c: loga (x1x2 )= loga x1 +loga x2

Logarit ca mt thng:x1Cho a,x1 ,x2 > 0, a 1. Ta c:

loga = logax2

x1 loga x2

Logarit ca mt ly tha:

Cho a,b > 0, a 1 . Ta c:log

bk =klog b (k)a a

log bi c s: log b= ca

log a

*loga b =1

logb a(b 1)

c bit: *log ka b =1

. logk

a b (k 0)

*loga b = loga c.logc b (0

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Hm c bn Hm hp

1/ (x

)/

=.x1

/1 12/ x

=x2

3/ ( x )/

=1

2 x

(u

)/

=.u 'u1

/1 u '

u =

u 2

( u )/

=u '

2 u

4/ (ex

)/

=ex

5/ (ax)/

=a x .ln a(e

u

)/

=u '.eu

(au)/

=u ' au ln a

6/ (lnx )/

=1

x

7/ (ln x )/

=1

x8/ (log x )

/ =1

9/ (log x )/

=1

(ln u )/

=u '

u

(ln u )/

=u '

u(log u )

/ =u '

(log u )/

=u '

TMTTLTHUYTVPHNGPHPGIITON12 Logarit thp phn:

- Logarit c s 10 gi l logarit thp phn- log10 a thng c vit l lg a hoc log a

Logarit t nhin:- Logarit c s e gi l logarit t nhin. (e 2,71828...)- loge a thng c vit l lna

Bng o hm ca hm s ly tha, hm s m v hm s logarit:

a x lna

au ln a

ax ln

a

au ln a

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TMTTLTHUYTVPHNGPHPGIITON12PHNG TRNH M

1. Phng php a v cng c s:

Vi a > 0,a 1. Ta c: af(x )

=ag(x )

2. Phng php t n ph:Dng 1:

A.a2x +B.a x +C= 0

A.a3x +B.a2x +C.ax +D = 0

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

f(x)=g(x)

t ax =tDng 2:

(t> 0)

A.a2x +B (ab )x

+C.b2x = 02x x

A

a

+B

a

+C= 0 b b x

a t: =t

b (t> 0)

Dng 3: A.a x +B.b x +C= 0 viax .bx = 1

t:a

x

=t(t> 0). Khi :

b

x

=

1

t3. Phng php logarit ha: Vi M> 0,0

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f(x)=g(x)nu c nghim th nghim l duy nht.

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Cho y =f(x)l hm tng (hoc gim). Khi phng trnhf(x)=knu c nghim th nghim l duy nht.

y =ax

tng nu a > 1 v gim nu 0

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a a

a a

TMTTLTHUYTVPHNGPHPGIITON12PHNG TRNH LOGARIT

1. Phng php a v cng c s:Vi 0 0 hoacg(x)> 0

Ch : loga

f(x)=M

f(x)=

aM

(khng cn t iu kin ca

f(x))2. Phng php t n ph:

Dng 1: A.log2 x +B.log x +C= 0 (a > 0,a 1)t: log

ax =t

Dng 2: A.loga

x +B.logx

a +C= 0(

a > 0,a 1)1

t: logax =t.Khi

3. Phng php m ha:

logx

a =t(x > 0, x 1)

( )= ( )= Mloga f x M f x a4. Phng php dng tnh n iu:D on nghim v chng minh nghim l duy nht.

Vi 0

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/


NGUYN HM TCHPHN

NGUYN HM

1. nh ngha:

Hm s F(x)c gi l nguyn hm ca hms

f(x)trn khong

(a; b )nu vi mi x thuc (a; b ), ta c: F'(x)=2. nh l:

f(x)

Nu F(x)l mt nguyn hm ca hms

f(x)trn khong (a; b )th:

a) Vi mi hng s C, F(x)+Ccng l mt nguyn hm cahm

s f(x)trn khong .b) Ngc li, mi nguyn hm ca hm s f(x)trn khong

(a; b )

u c th vit didng

s.

F(x)+Cvi C l mt hng

Ngi ta k hiu h tt c cc nguyn hm ca hms

f(x)dx . Nh vy:

f(x)l

f(x)dx =F(x)+CF'(x)=

3. Cc tnh cht ca nguyn hm:

f(x)

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Nguyn hm cc hm s s cpthng gp

Nguyn hm ca cc hm s h

(di y t=

t(x

)* dx =x +C+1

* xdx =

x+C( 1)

* dx

=ln x +C(x 0)

dx 1x2 x* = +C

* ex dx =ex +C

x

* a x dx =a

+C(0

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Ta thng chn C= 0 v =G (x) Cc dng c bn: Cho P(x)l mt a thc.

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P(x)

P(x)

A B C

+ = = + + . QuyQ (x) (

ax+

b)(

mx+

n

)2

ax +b mx +n(

mx+

n)

2

ng mu v cui cng, ng nht h s vi P(x) ta tm cA,B,C. T bin i c bi ton cho v dng n gin hn tnh.* Ch : Trong qu trnh gii ton cn ch n cng thc

f(x )+g(x )

f(x)

g(x )= + a bi ton v dng n gin hn.

h (x)

h (x)

h (x )

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b

a

TMTTLTHUYTVPHNGPHPGIITON12TCH PHN

b

1. nh ngha: f(x)dx =F(x)

a

2. Cc tnh cht ca tch phn:a

=F(b )F(a )

1. f(x)dx = 0a

b a

2.

f(

x)

dx =

f(

x)

dx

a b

b b

3. kf(x)dx =kf(x)dx (k)a a

b b b

4. f(x)g(x)dx =

f(x)dx

g(x)dx

a a a

b c b

5. f(x)dx =f(x)dx +f(x)dxa a c

6. f(x) 0 trnon

a;b

b

f(x)dx 0a

7. f(x)g(x)trnon

a;b

b b

f(x)dx g(x)dxa a

8. m f(x)Mtrnon

b

a; b

m (b a )f(x)dxM(b a)a

3. Cc phng php tnh tch phn Phng php tch phn tng phn:

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M rng: k2 x2 dx .

t:x =ksin t,t ;

a 2 2

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TMTTLTHUYTVPHNGPHPGIITON12NG DNG HNH HC CA TCHPHN

1. Tnh din tch hnh phng gii hn bi 1 ng cong vtrc honh:

Cho hm s y =f(x)(C)lin tc trn on a;

b. Din

tch hnh phng gii hn bi (C),trc honh v hai ng thng

x =a, x =b c tnh bi cngthc:

b

S=a

f(x)dx

2. Din tch hnh phng gii hn bi hai ng cong:

Cho hai hm s y =f(x)(C) v y =g(x

)

(C) lin tc trn

on a;b

. Din tch hnh

phng gii hn bi (C), (C) vhai ng thng x =a, x =b,c tnh bi cng thc:

b

S=a

f(x)g(x)dx

Ch :

- Trong trng hp cha cho cn a,b th phi gii phng trnhhonh giao im tm cn. Nghim nh nht l cn dia, nghim ln nht l cn trn b.

- tch tch phn c cha du gi tr tuyt i c 2 cch:

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dyc

d

Hay: V=g2 (y)dyc

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1. S i: i2 =12. nh ngha:

S PHC

- S phc z l biu thc c dng:

a gi l phn thc. b gi l phn o. i gi l n v o.

z =a +bi, a,b ,i2 =1

- Tp hp s phc k hiu l . Vy 3. S phc bng nhau:

a =a'Cho hai s phcz =a +bi,z'=a'+b'i ,z =z'

b =b'4. Biu din hnh hc ca s phc:

Cho s phc z =a +bi , imM(a; b )trong mt phng ta

Oxygi l im biu din cho s phc z

Gi s s phc z =a +bi c biu din biim

M(a; b ).

di ca vectOMgi l mun ca s phc z, k hiu: z . Vy:

5. S phc lin hp:

z =OM = a2 +b2

- S phc z =a bigi l s phc lin hp ca s phc z =a +bi

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(


TH TCH KHI A DIN, KHI

TRN XOAY

I. Th tch khi a din:

1. Th tch khi lp phng cnh a: V=a3 (vtt)2. Th tch khi hp ch nht c ba kch thc a,b,c l V=a.b.c

(vtt)3. Th tch khi lng tr c din tch y l B, chiu cao l h l;

V=

B.h (vtt)

4. Th tch ca khi chp c din tch y B, chiu cao h l: V=1

Bh3

(vtt)5. Th tch khi chp ct c din tch hai y l B v B, chiu cao

h l:

V=1

B +B'+36. Mt s tnh cht:

BB'

)h (vtt)

T s th tch ca hai khi a din ng dng bng lp phng ts ng dng

Cho khi chp S.ABC. Trn cc on thng SA, SB, SC lnlt ly 3 im A, B, C khc vi S. Khi :

VS .A' B'C'

VS.ABCII. Th tch khi trn xoay:

1. Mt nn trn xoay:

=SA'

.

SB'.SC' SA SB

SC

Cho hnh nn N c chiu cao l h, ng sinh l, bn knh y R- Din tch xung quanh ca hnh nn: S

xq=Rl(vdt)

- Din tch ton phn: S =S +S=Rl+R

2

tp xq ay

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I,(

I,(

I;(

I,(


PHNG TRNH MTCU

1. Phng trnh chnh tc:

Phng trnh mt cu tm I(a; b;c)bn knh R:(x a )2 +(yb)2 +(zc)2 =R2

2. Phng trnh tng qut:Trong khng gian Oxyz, phng trnh :

x2 +y2 +z2 2ax 2by 2cz +d= 0

vi a2 +b2 +c2 d> 0

l phng trnh mt cu tmI(a; b;c), bn knhR = a2 +b2 +c2 d

Ch : Nu phng trnh cho di dng

x2 +y2 +z2 + 2ax + 2by + 2cz +d= 0

vi a2 +b2 +c2 d> 0

th mt cu c tmI(a;b;c), bn knh R=a

2

+b2

+c2

d

3. V tr tng i gia mt cu (S) v mt phng ():* Nu d >R : mt phng v mt cu khng c im chung

* Nu d =R : mt phng ()tip xc mt cu (S), khi ()gi

l tip din ca mt cu (S).

iu kin mt phng ()tip xc mt cul

d =R

* Nu d

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- DoMNu1. T y tm c t1,t2 v c M,NMNu2

- ng vung gc chung qua M v nhn MN lm VTCP.Dng 10: Vit phng trnh ng thng (d) qua A v ct haing thng d1, d2 cho trc:

A dd1 M

N d2

C1:* Chuyn d1,d2 v phng trnh tham s* GiMd1, Nd2 (ta M,N cha

t1,t2 ). Tnh AM, AN.

* Do AMcng phng AN nn t kcng phng tm c t1,t2 v c c M,N.* ng thng cn tm qua A v c VTCP

AMCch khc:* Vit phng trnh mt phng (P) qua A v

cha (d1 )(xem dng 3 ca phng trnhmt phng)* Tm giao im M ca mt phng (P) v

(d2 )

* (d)l ng thng qua 2 im A,M(dng 1)

Dng 11: Vit phng trnh ng hng (d) qua A, vung gc vct ng thng :

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V TR TNG I

1. CM()ct(): Ta chng minh A : B : CA' : B':C'

2. CM()(): Ta chng minhA

=B

=C

=D

A' B' C ' D'

3. CM()//(): Ta chng minhA

=B

=C

D

A' B' C '

D'

4. CMd ,d 'ng phng: Ta chngminh

u,u'.MM'=0

vi

Md ,M 'd'

5. CMd ,d ' ct nhau: u,u'.MM'=0

6. CM d // d: Ta chng minh

v a : b : c a' : b' : c'

a : b : c =a' : b' : c'(x'

0

x0 )

:(y'

0

y0 )

:(z'

0

z0 )7. CM dd: Ta chng minh

a : b : c =a' : b' : c'=(x'0 x0 ): (y'0 y0 ): (z'0 z0 )

8. CM d v d cho nhau: ta chng minh

Md ,M 'd '

u,u'.MM'0

vi

9. CM d ct(): Ta chng

minh:

aA+

bB+

cC

0

aA +bB +cC= 0

10. CM d//(): Ta chngminh

M

0 dM0

()

aA +bB +cC= 011. CM d (): Ta chng minh

Ch :

M

0 dM0

()

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D

1 1 1 2 2 2

= ((1),(D

2))

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a

a

TMTTLTHUYTVPHNGPHPGIITON12- Nu phng trnh bc hai ax2 +bx +c = 0 c h s a,b,c tha

ca+b +c =0

th phng trnh c hai nghim l:x1 = 1,x2 =a

- Nu phng trnh bc hai ax2 +bx +c = 0 c h s a,b,c thac

ab +c = 0 th phng trnh c hai nghiml:

x1 =1,x2 =a

4*. Xc nh du cc nghim s ca phng trnh bc 2:ax2 +bx +c = 0 :- Phng trnh c hai nghim tri du ac < 0

0- Phng trnh c hai nghim phn bit cngdu

c> 0

a

0

- Phng trnh c hai nghim cng dng

b

> 0

c> 0

a 0

- Phng trnh c hai nghim cng m

b

< 0

III. Bt phng trnh bc hai:

c > 0a

1. N: Bt phng trnh bc hai l mnh cha bin thuc 1 trong 4dng sau:ax2 +bx +c >0;

ax2 +bx +c

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- Tm cc nghim ca P(x), gi s cc nghim lx

1,x

2,...,

xn

v x1

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P(x) (tnh lun c nghim ca h(x))Bng xt du:an .bm .ck > 0 , gi s c nghim bi chnl

xn1

, P(x)khng xc

nh ti xn (tc xn l nghim ca mu)

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x -2 1/2 1 2f(x) + || - 0 + 0 - || - || +

x3 2x2 + 3x


c. f(x )=(2 x )(x3 2x2 + 3x 2)= 0 2 x = 0

x = 2

x = 2(x 1)(x2 x + 2)= 0 x = 1

Bng xt du: (Tch cc h s ca x c m cao nht l -1.10)

3 +

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- Nu m >n : th limf(x)

= hoc limf(x)

= ty

x g(x ) x g(x )theo bi ton (xc nh du da vo qui tc mc 1).

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v

v

v2

v2


5) Gi s u =u(x),v =v(x) l cc hm s c o hm tiim

x thuc khong xc nh, ta c:

(u +v )' =u '+v '(u v )' =u 'v '

(u.v )' =u ' v +uv '

=u ' =

u ' v uv '

(k.u

)' =

k.u '

1 ' =

v'

(v =v (x ) 0)

6) o hm ca hm hp: Nu hm s u =g(x)c o hm ti

x l u 'x v hm s y =f(u)c o hm ti ul y '

u th hm hp

y =f(g(x))c o hm ti xl:

y 'x =y 'u .u 'x

7) o hm ca hm s lng gic:Bng tm tt:

(sinx )' = cosx(cosx )' = sinx

1

(sin u )' =u '.cos u(cos u )' =u '.sinu u '

(tanx )'= cos2 x

(tan u )'= cos2 u

(cotx )' =

1

sin 2

x

(cot u )' =

u '

sin 2

u8) Mt s cng thc khc:

y =ax +b

y ' =cx +d ax2 +bx +c

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TMTTLTHUYTVPHNGPHPGIITON12CH 5:CNG THC LNG GIC V

PHNG TRNH LNG GICI. Cng thc lc gic:1. T s lng gic ca mt s gc cn nh:

Gc00 300 450 600 900 1200 1350 1500 1800

0

6

4

3

2

2

3

3

4

5

6

sin 01

2

2

2

3

21

3

2

2

2

1

20

cos 1 32 22

12

0 12

22

32

1

tan 01

3 1 3 || 3 1 1

30

cot||

3 11

30

1

3 1 3 ||* Cng thc lng gic c bn:

sin 2 x + cos2 x =1

tanx.cotx = 1

1

cos2 x= 1 + tan 2

x

1sin 2

x

= 1 + cot 2 x

tanx =sinx

cosxcotx =

cosx

sinx2. Cng thc bin i tch thnh tng:

cosa.cosb =1

[cos(a b)+ cos(a +b)]2

sin a.sinb =1

[cos(a b)cos(a +b)]2

sin a.cosb =1

[sin(a b)+ sin(a +b)]2

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cos2 a =cos 2a +1

2

sin2 a =1cos2a

2

tan2 a =1cos2a1+cos2a

sin3 a =3 sin a sin3a

4

cos3 a =3cosa +cos3a

4

7. Cng thc cng:sin(a +b) = sin acosb +cosasinb

sin(a b) = sin acosb cosasin

b cos(a +b)=cosacosb sin a

sin b cos(a b) =cosacosb + sinasin b

Ngoi ra ta cng c cng thc sau vi mt s iu kin:

tan(a b) =

tan(a +b) =

tan a tanb

1+ tan a.tanb

tan a +tanb

1 tan a.tanb

(*)

(**)

(*) c iu kin: a +k,b +k,a b +k

(**) c iu kin:

2 2 2

a

+k,b

+k,a +b

+

k

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sin a =

cos a =

tan a =

2t

1 +t2

1 t2

1 +t2

2t

1 t2

, a

2+k

9. Cng thc lin h gia 2 gc b nhau, ph nhau, i nhau v hn

km nhau 1 gc hoc

:2

9.1. Hai gc b nhau:sin(a) = sin a

cos(a) =cosatan(a) = tan a

cot(a) = cot a

9.2. Hai gc ph nhau:

sin(

a) = cos a2

cos(a) = sin a2

tan(

a) = cot a2

cot(

a) = tan a2

9.3. Hai gc i nhau:sin(a) = sin acos(a) = cos atan(a) = tan

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TMTTLTHUYTVPHNGPHPGIITON129.4 Hai gc hn km nhau

:

2

sin(a +

) = cos

a2

cos(a +

) = sin

a2

tan(a +

) = tan

a2

cot(a +

) = cot

a2

9.5 Hai gc hn km nhau :sin(a +) = sin a

cos(a +) =cos

a tan(a +) = tan

a cot(a +) = cota

9.6. Mt s cng thc c bit:sinx + cosx= 2 sin(x +

)

=2 cos

x

4

4

sinx cosx= 2 sin(x

) =

2 cosx +

=

4 4

cosx sinx=

2 cosx +

4

III. Phng trnh lng gic:1. Phng trnh c bn:

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b


tanx

=t2

2t 1t2(1) a +b +c = 01+t

2

1+t2

Gii phng trnh bc hai i vi t, d dng gii c phng trnh (1).

Cch 3: Do a2 +b2 0 , chia hai v ca phng trnhcho

a2 +b2 :

(1)

t :

a

a2

+b2

sinx +b

a2 +b2

cosx =c

a2

+b2

a

a2 +b2 a2 +b2

= sin

= cos

(1) sin(x +) =

Ch : Ta lun c :

| a sinx +b sinx |

c

a2 +b2

a2 +b2

(y l phng trnh c bn).

Du "=" xy ra khi v ch khi sin(x + a) = 1.4. Phng trnh i xng i vi sinx v cosx:

a(sinx + cosx) + bsinxcosx = c (1) (a, b, c l hng s)Gii phng trnh (1) bng cch t :

sinx + cosx = t , |t| 2a (1) v phng trnh

bt

2

+ 2at (b + 2c) =0. Gii phng trnh (2) vi |t| 2 .5. Phng trnh lng gic s dng nhiu n php bin i lnggic:

y l dng ton ch yu trong cc k thi H-C, cch gii chyu l s dng cc php bin i lng gic thng dng a

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phng trnh v mt trong cc dng trn hoc dng phng trnh tchm mi tha s l mt phng trnh c bn gii gii.

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TMTTLTHUYTVPHNGPHPGIITON12Trong qu trnh bin i cn ch trnh s dng hai php

bin i tri ngc nhau, v ch quan st rt ngn thi gian v ccbc gii, mun th cn nm tht vng cc cng thc lng gic,nht l nhng cng thc c bit thng hay dng.

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TMTTLTHUYTVPHNGPHPGIITON12thc m ta nh khng chc chn. Cn m bo c sc kho tt nhttrc khi d thi. Cn tp thc dy sm vo bui sng (t thc dys sng khoi v c trng thi tm l tt hn b gi dy).

Khi nhn c thi cn c tht k phn nh u l cccu hi quen thuc v d thc hin (u tin gii trc), cn cc cuhi kh s gii quyt sau. Th t cc cu hi c gii l theo khnng gii quyt ca th sinh, khng nn b l thuc vo th t trong bi. C th nh gi mt cu hi no l d v lm vo giy thinhng khi lm mi thy kh th nn dt khot chuyn qua cu khcgii c d dng, sau cn thi gian th quay li gii tip cu kh

y. Trong khi thi khng nn lm qu vi v cu d ( ri c sai stng tic) v ng sm chu thua cu kh. Hy tn dng thi gian thid li cc cu lm mt cch cn thn v tp trung cao tm racch gii cc cu kh cn li.

(TS Nguyn Cam, khoa Ton - Tin H S phm TP.HCM)

lm bi thi H t im cao

Thc hin nguyn l 3 Nguyn l ny c c ng v theo th t: "ng - -p".

ng chin lc lm bi: Thc hin theo chin thut: "Htnc vc n xng", tc l cu quen thuc hoc d lm trc, cu khlm sau. Nu cu kh th b qua, khng lm ra hoc lm sai thnguy c trt H khng ln (bn ch thua rt t ngi lm c cukh), nhng nu cu d m khng gii c, lm sai, lm khng nni n chn th bn rt d trt (v bn s thua hng vn ngi lmc cu d). ng p s: Nu bi lm c p s ng, b cc n thgio vin chm ln 1 c th cho im ti a v nh k hiu dthng nht im vi gio vin chm ln 2. Nu p s sai th thnggio vin s tm im sai gn nht chm cho nhanh. V vy ngp s l rt quan trng, thm ch c nhiu ngi lp lun cha chnh

xc nhng vn c im ti a. ng chng trnh SGK: Lm ngp s nhng bn phi dng kin thc hc trong chng trnh SGK.ng thi gian: C nhiu TS khng bit phn b thi gian, trnh byqu cn thn dn n c cu

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TMTTLTHUYTVPHNGPHPGIITON12gii xong trn giy nhp nhng ht thi gian vit vo bi thi.Cng c nhiu TS lm bi nhanh nhng khng xem li bi k nn bmt im ng tic.

cc cu hi: TS cn iu tit thi gian lm ht cc cuhi theo trnh t t d n kh, trnh tn qu nhiu thi gian chomt cu hi khng cn gi suy ngh cu khc. Trnh by y :Do thang im chi tit n 0,25 nn nhng bi c lp lun y sd t im ti a.

Tm li gii p: Khi gp mt bi ton, bn cn u tin cch

gii c bn x l nhanh m khng nn loay hoay mt thi gian tmcch gii p. Tuy nhin mt s bi ton ng cp li cn n li giithng minh, ngn gn. Trnh by p: Mc d trong mn Ton yu tp b xem nh hn rt nhiu so vi yu t ng, nhng nu 2 bi thi

toan 12

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