[eBook] Phuong Phap Giai Nhanh de Thi Dai Hoc Mon Toan

8/9/2019 [eBook] Phuong Phap Giai Nhanh de Thi Dai Hoc Mon Toan

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Hong Vit Qunh

Toan hoc ph thngCc phng php gii nhanh thii hc


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1

Cc phng php gii ton i s vgii tchLi niu:

Sau 12 nm hc tp, gi y ch cn mt k thi duy nht ang ch i cc em l k thi ihc. y s l k thi kh khn nht trong sut 12 nm cc em ngi trn gh nh trng. K thii hc chnh l mt bc ngot ln trong cuc i ca mi hc sinh v th mi hc sinh cnphi chun b kin thc tht ton din v ni dung ca thi mang tnh lin tc. C l trong ccmn, mn ton vn lun chim v tr quan trng v l vt cn ln nht trn bc ng tin tging ng i hc. V th ti xin mo mui gp cht kin thc thu lm c trong qutrnh hc tp vit ln quyn sch ny. Hy vng y s l ti liu b ch cho cc em hc tp.

Quyn sch c chia thnh su n v bi hc v hai ph lc. Mi bi u l nhng phnquan trng, xut hin thng xuyn trong thi i hc. mi bi u c nhng c imsau:

Phn tm tt kin thc hc c trnh by ngn gn v tng qut nhm khi li phnkin thc qun ca cc em.

H thng cc bi lm c chn lc k lng, c tnh in hnh v khai thc ti a ccgc cnh ca vn nu ra, ng thi phng php gii ngn gn, trc quan cng nhiukinh nghm gii gip cc em c th hiu c ni dung bi gii v cch p dng cho ccdng thi s gp sau ny. ng thi, cc v d u c trnh by t c bn n nng cao.y l nhng bi trch ra t thi d tr ca cc nm trc v tham kho t nhng tliu ca cc thy c c nhiu nm kinh nghim trong qu trnh luyn thi nn m bo vmc v gii hn kin thc. Li gii trong cc v d ch l tng trng nhm mc ch nuln phng php gii, cc em v cc thy c khi tham kho cun ti liu ny c th tm ra vtrnh by cch gii v cch trnh by hp l hn. Cc em nn tp gii cc dng bi trn mtcch thun thc v c lp. sau khi gii xong mi xem phn li gii. l iu m tc gi kvng nhiu nht.

L gii cc phng php, a ra thut ton gii chung, a ra bn cht li gii, lphn li bnh, lu cui mi bi tp.Phn ph lc l 12 thi tiu biu theo cu trc thi mi nht do B GD&T cng b. Cc

thi c mc kh rt cao, i hi ngi lm phi t duy rt nhiu. Vi mc kh , timong rng khi cc em gii thun thc cc bi trong b thi ny cc em s c t tin v kinthc t im cao khi lm bi mn ton. Ph lc 2 l mt s mo dng my tnh onnghim c nh, phc v cho qu trnh gii cc bi tp v phng trnh tch nh lng gic, hphng trnh, phng trnh, cch gii nhanh bi ton hnh hc bng my tnh ng thi githiu thm phng php chia Horner gip cc em lm nhanh bi ton c chia a thc, phntch thnh tch

Vi d nh l s gii thiu quyn sch cho cc em trong thng cui cng trc khi thi ihc nn sch gin lc mt s phn khng cn thit v cc kin thc bn l, ch gii thiunhng trng tm ca thi nn bi tp c th cn t. Ti cng c li khuyn cho cc th sinh l

hy tm thm cc thi trn mng internet v y l kho kin thc v tn.Mc d rt c gng nhng cun sch rt c th cn nhiu thiu st do thi gain bin sonngn ng thi kinh nghim v s hiu bit cn hn ch. Rt mong c s gp ca bn c.Mi gp xin lin h vi tc gi qua a ch sau:

Hong Vit QunhKhu 6a Th trn Lc Thng Bo Lm Lm ng

Email: [email protected]: http://vn.myblog.yahoo.com/vquynh-qflower

Tel: 063-3960344 - 01676897717


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Bi I: ng dng phng trnh ng thng gii phng trnh cn thc.

VD1. Nhc li kin thc v ng thng.

1) Phng trnh tng qut:ng thng i qua M(x0;y0) v c vet php tuyn n

(A;B) th ng thng c phng trnh:

(d): A(x-x0)+B(y-y0)=0

(d): Ax+By+C=0

VD1. ng thng qua M(1;2) nhn n (2;1) lm vect php tuyn.(d): 2(x-1)+1(y-2)=0 (d): 2x+y-4=0

2) Phng trnh tham s:ng thng i qua M(x0;y0) v c vect ch phng a

(a1;a2)

(d):

+=

+=

tayy

taxx

20

10

VD2. ng thng qua M(3;4) nhn a (2;3) lm vtcp c phng trnh:(d):

+=

+=

ty

tx

34

23

VD3. Cho (d): x+y=4. Vit phng trnh tham s ca (d).Gii:Vect php tuyn : n

(1,1)

Vect ch phng : a(1,-1)

im i qua M(2;2)

(d) :

=

+=

ty

tx

2

2

VD2. ng dng

VD1. Gii phng trnh : 101238 33 =++ xx

Gii:

t: 83 +x =1+3t v 312 x =3-t k( -1/3 t1/3)

x3 +8=(1+3t)2 (*) v 12-x3 = (3-t)2 (**)

Ly (*)+(**) ta c 20=10t2+10 t2=1 t=1 hoc t=-1(loi) x3=8 x=2Tip:C phi bn ang t hi: thut ton no gip ta nhn thy c cch t n t ???


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Khng phi ngu nhin m ti li trnh by li vn ng thng, mt vn tng chng nhchng lin quan g n i s. Nhng gi y ta mi nhn ra c ng thng chnh l tuyt chiu gii phng trnh dng cn thc. Mu cht l:

B1: 101238 33 =++

YX

xx

T ta c phng trnh ng thng : X+3Y=10

B2: ta vit li phng trnh: X+3Y=10 theo tham s t

=

=

t-3Y

3t+1X

Lc ny phng trnh quy v 1 n t v vic gii phng trnh trn l khng kh. (V y l kin thlp nh) hiu r hn v phng php ny cc bn hy cng ti n vi VD2.

VD2. Gii phng trnh :

X

x 3+ +

Y

x3 2+ =1

Gii:

Gi (d): X=1+t v Y=0+t

(1) t

=+

=+

tx

tx

3 2

13(t1)

=+

+=+

3

2

2

213

tx

ttx

Ly phng trnh 2 tr pt1 ta c: -1=t3-t2 +2t-1 t3-t2 +2t=0 T=0 x=-2

Lu :Trong khi gii thi, cc bn nn trnh by t bc(1) tr i nhm m bo tnh ngn gn cho bi tonBc gi phng trnh ng thng ch nn lm ngoi giy nhp.

Trong bi trn ta c th t

=+

=+

vx

ux

3 2

3v quy v gii h phng trnh. Cc bn c th xem

cch ny nh mt bi tp. cc bn hy lm v so snh s u vit gia 2 phng php. Trong bi trn ta hn ch phng php ly tha v nu mun kh 2 cn thc khc bc trn, ta ph

^6 phng trnh. Ta s gp kh khn v s i mt vi 1 phng trnh kinh khng v ta phi gixt khi mi c th ra nghim.

VD3. Gii h phng trnh :

( )

( )

=+++

=+

2411

13

yx

xyyx( thi H nm 2005)

Gii:

t:

=+

+=+

ty

tx

21

21(-2t2)

+=+

++=+

441

441

2

2

tty

ttx

+=

++=

34

34

2

2

tty

ttx

Phng trnh(1) tr thnh: 2t2+6- )43)(43( 22 tttt +++ =3


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910 24 + tt =2t2+3

hoc

t=0 x=y=3

VD4. nh m phng trnh sau c nghim:

Gii: phng trnh c nghim:

mxf =)(Min f(x)mMax f(x)

t

=

+=+

txm

tmx

33

312(-1/3t3)

+=

++=+

2

2

693

9612

ttxm

ttmx

cng v vi v => 5m=10+10t2

2t2

+2=m f(t)=m

Vi f(t)= 2t2+2 min xc nh: D=[-1/3;3]F(t)=4t =>f(t)=0 t=0

t - -1/3 0 3 +F(t) - 0 +

F(t)20/9 20

2

M c nghim 2m20

VD3. Bi tp t luyn

1) Gii h phng trnh:

2) Gii h phng trnh:

3) Gii h phng trnh: 2 1 1 13 2 4

x y x

x y

+ + + =

+ =( thi d b1A 2005)

4) Gii phng trnh: 1 sin( ) 1 cos( ) 1x x + + = ( thi d b2A 2004)


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Bi II: Cc cch gii phng trnh v bt phng trnh

v t.

1)Ly ThaPhng php ly tha l phng php tng qut nht gii phng trnh c cn. Khi gp cc phngtrnh c dng cn phc tp nhng khi chng ta bit mo ly tha th c th gii bi ton mt cch ddng. y l mt phng php c bn, cc bn phi thc tp nhun nhuyn v phng trnh trong thi hc c lc rt d nhng ta li khng . cc bn hy theo di cc v d sau. Nhng trc ht hylu vn sau:

t iu kin Ly tha chn th hai v khng m Cc dng c bn:

BA =

=

2

0

BA

B

BA <

20

0

BA

B

BA >

>

+

1232

1

22 xxxx

x

>

1

1

x

x x=1


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VD3.

Gii:k: 2x+1>0 x>1/2Bpt (4x2-4x+1)(x2-x+2)36t t = (x2-x) bpt tr thnh:(4t+1)(t+2)364t2+9t-340t-17/4 hoc t2 x2-x-17/4 hoc x2-x2 x1 hoc x2

VD4. Gii bt phng trnh :Gii:

>

=+

02

0

0

2

2

2

xx

xx

xx

10 == xx

Lu : bt phng trnh trn cc bn khng nn ly tha tnh ton v qu trnh ly tha v nhn phn phrt mt thi gian. Hn na, khi quy v mt phng trnh h qu, chng ta gii rt d sai v khi giao cctp nghim s khng c gi tr no tha mn.Trong bi trn ti s dng cch nh gi theo kiu nh sau:

A B 0

>

=

0

0

0

A

B

B

chnh l mu cht ca bi ton

VD5. Gii phng trnh :Gii:

=

2

2

4

538

053

04

532

xx

x

x

x=3


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Lu :Trong phng trnh trn cc bn phi v nhanh mt cht v nu nh ta nguyn phng trnh cho ly tha th l mt iu khng cn g di bng ta s i mt vi chuyn ly tha 2 ln =>mt phng trnh bc 4. Phng trnh ny ta khng th bm my tnh. Nhng nu gii tay th phi gii xkhi mi ra trong khi thi gian khng ch i ai. ng thi chng ta khng cn gii iu kin vi v gimkho ch quan tm n bi lm v kt qu. Chng ta hy ch vit ci sn ca iu kin. sau khi gii ra

nghim ch vic th vo iu kin l xong.

2)Phng php t n ph:

CCH GII:( )( )( ) 0)();(

0)();(

0)();(

=

n

n

n

xuxuf

xuxuf

xuxuf

t= n xu )( Phng trnh hu t hoc h phng trnh

BI TP P DNG:VD1.

Gii:

t t= => t>0 ; t2+2= x

2+ x

3t=2(t2-1) t=-0.5 (loi) hoc t=2x2+x=6 x=2 hoc x=3

VD2.

Gii:

T= 1x

=+

xt

t

1

0

2

Phng trnh tr thnh:

t2+1-(t+1)=2 t2-t-2=0 t=2 hoc t=-1

x=5

VD3.

Gii:

=>


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pt tr thnh: t2+t+2=8 t=2 t=-3

TH1: t=2

TH2: t=-3

LOI II: ( )nn xvxuf )()( + { 0; 0; =0 }Phng php chung:

=

=

vxv

uxu

m

n

)(

)(=> a v h phng trnh.

VD1. 085632323 =+ xx ( tuyn sinh i hc 2009)Gii:

=

=

)0(56

233

vvx

ux

=+

=+

0832

3

8

3

5 23

vu

vu

=

=+

3

28

3

8

3

5 23

uv

vu

=

=

+

3

28

3

8

3

28

3

52

3

uv

uu

=

=++

3

28

0)202615)(2( 2

uv

uuu

=

=

4

2

v

u x=-2

LOI III: H PHNG TRNH A THCNhng h phng trnh ny ta rt thng hay gp trong thi i hc. lp 10, ta thng gp nhnphng trnh c tn l h i xng, ng cp Nhng h ny c cch gii n lin. nhng trong thi hc, ta khng h tm thy nhng dng . Nhng tt c cc h trn u quy v mt mi l Phntch thnh nhn t.


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VD1. Gii h phng trnh: ( )( )3

1 11

2 1 2

x yx y

y x

=

= +

(H A 2003)

Gii:K: xy0

Ta c ( ) ( ) 11 1 01

x yx yxyxy

= + = =

TH1:( ) ( )23 3

1

1 5

1 1 0 22 1 2 1

1 5

2

x y

x y x y x yx y

x x x y x x x

x y

= =

== = + = = + == + = +

= =

TH2:3

3 4

11

122 1

1 2 0

y

xy yx xy x

x x xx

=

= = = + = + + + =

M

2 2

4 2 1 1 32 0,2 2 2

x x x x x VN + + = + + + >

Vy nghim ca h l ( ) ( )1 5 1 5 1 5 1 5

; 1;1 , ; , ;1 1 1 1

x y + +

=

VD2. Gii h phng trnh: ( )( )

( )2

2

x 1 y(y x) 4y 1x, y R .

(x 1)(y x 2) y 2

+ + + =

+ + =

(D b A2006)

Gii:

( ) ( ) ( )21 1 4 0 * x y x y + + + =

t: 2 1 0; 4u x v x y= + > = +

H( )

( ) ( )

0 3

2 4

u yv

u v y

=

+ =Thay (4) vo (3) ta c: ( ) ( ) ( )3 2 . 0 1 2 0u u v v u v v + + = + + =

2 2 1 0v v + + = 2( 1) 0 1 3v v x y + = = + =

Vy (*) ( )2

21 21 0

1 3 02 53

x yx yx x

x yx y

= = + = + =

= ==

VD3. Gii h phng trnh( )

( )3 3

2 2

x 8x y 2yx, y R .

x 3 3(y 1) *

= +

= +

(D b 2A 2006)

Gii:

H( ) ( ) ( ) ( )

( )

3 33 3

2 2 2 2

3 6 4 2 12 4

3 6 3 6 2

x y x y x y x y

x y x y

= + = +

= =

Ly (2) thay vo (1) ta c

( ) ( )( )3 3 2 2 3 2 23 3 4 12 0x y x y x y x y x x y = + + = ( )2 212 0 x x xy y + = D thy x=0 th y=0. Th vo (*) ta thy khng tha mn. Vy y khng phi l nghim ca phngtrnh:


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10

( )( )2 2

2 2 2 2

3 4 012 0

3 6 3 6

x y x y x xy y

x y x y

+ = + =

= =

TH1:2 2 2

3 0 3 1 3

1 33 6 6 6

x y x y y x

y x x y y

= = = =

= = = =

TH2:2 2 2

78 4 784 4 13 13

3 6 13 6 78 4 78

13 13

y x x y x y

x y yy x

= == =

= =

= =

Vy nghim ca phng trnh l:

( ) ( ) ( )78 4 78 78 4 78

; 1;3 , 1; 3 , ; , ;13 13 13 13

x y

=

VD4. Gii h phng trnh ( )( ) ( )( ) ( ) ( )

2 2

2 2

13 1

25 2

x y x y

x y x y

+ =

+ =

(D b 2005)

Gii:Nhn c 2 v ca (1) cho 25. Nhn c 2 v ca (2) cho 13. Sau ly (1)-(2).

(1)-(2) ( ) ( ) ( ) ( ) ( ) ( )22 2 2 2 213( ) 25 0 13 25 0x y x y x y x y x y x y x y + + = + + =

( ) ( ) ( ) ( )2 2 2 212 26 12 0 2 12 26 12 0 x y x xy y x y x xy y + = + = D thy x=y khng tha mn h.

( ) ( )

( )( ) ( ) ( )

2

2 2 2

2

3 23

253 2 . 25 2

3 2 2 3 0 9 32 3

2 32525

325 1. 25

24 2

x yy

yx y y x

x y x yx y

x y x y x y x y x y

xy y

y

==

= = = = =

=+ = + = = = =

Li bnh:

Lm sao ta c th phn tch nhanh ( )2 212 26 12 x xy y + thnh nhn t( ) ( )3 2 2 3 x y x y ??Lc ny, cng c ca chng ta chnh l my tnh b ti! Cc bn hy lm nh sau:

Coi nh ta khng thy n y. vy nn ta c phng trnh bc 2 theo x:( )212 26 12 0x x + = Chchn cc bn u bit gii phng trnh bc 2 ny bng my CASIO. Ta bm c nghim

3 2

2 3x x= = . Lc ny ta gi li n y bng cch thm y vo sau cc nghim tm c.

3 2

2 3 x y x y= = . Quy ng b mu v mu l hng s. ta c nhn t cn phn tch. Lu l

( )2 212 26 12 0 x xy y + = ( ) ( )3 2 2 3 0 x y x y = . Nu gii bt phng trnh, bn nn ch n

du khi phn tch (Trng hp ny l du - : ( ) ( ) ( )2 212 26 12 2 3 2 2 3 0 x xy y x y x y + = = ) Khi gp dng phng trnh a thc c hng s pha v phi (hoc c th a c 2 phng trnh

v dng c hng s v phi), Ta nhn c 2 v ca phng trnh trn cho s v phi ca phngtrnh di v nhn c 2 v ca phng trnh di cho s phng trnh trn. Sau tr v theo


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v. Mc ch ca phng php ny l quy h v phng trnh tch sau tin hnh phn tch. Huht cc loi phng trnh a thc u gii c theo cch ny!

Bi tp t luyn

Bi 1.

4 3 2 2

3 2

1

1

x x y x y

x y x xy

+ =

+ =

Bi 2.( ) ( )

2 2 4

1 1 2

x y x y

x x y y y

+ + + =

+ + + + =

Bi 3.( )

( )

2 2

22 2

3

7

x xy y x y

x xy y x y

+ =

+ + =

Bi 4.( )

( )

3 2

3 2

log 2 3 5 3

log 2 3 5 3

x

y

x x x y

y y y x

+ =

+ =

Bi 5.

( )

( )2

2

1 3 0

51 0

x x y

x yx

+ + =

+ + =

Bi 6.

9 9

25 25 16 16

1x y

x y x y

+ =

+ = +

Bi 7.

4 3 2 2

2

2 2 9

2 6 6

x x y x y x

x xy x

+ + = +

+ = +

Bi 8.2 2 2

1 7

1 13

xy x y

x y xy y

+ + =

+ + =

Bi 9.

( )

3

4

1 8

1

x y x

x y

+ =

=

Bi 10.

2

2

2

2

23

2

3

yy

x

x

x y

+=

+=

Bi 11.

3

1 1

2 1

x yx y

y x

=

= +


13/63

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Bi III: Phng trnh lng gic.

Mt s cng thc lng gic cn nh:

1.2 2 2 2

2 2

1 1sin x cos x 1;1 tan ;1 cot .

cos sin

x x

x x

+ = + = + =

2.sin cos 1

tanx ;cot x ; tancos sin cot

x xx

x x x= = = .

3. Cng thc cng:sin( ) sin cos cos

cos( ) cos cos sin sin

a b a b asinb

a b a b a b

=

=

4. Cng thc nhn i: sin2x = 2sinxcosx

5. cos2x = cos2x sin2x = 2 cos2x 1 = 1 - 2 sin2x

6. Cng thc h bc: 2 21 cos 2 1 cos2

cos ;sin2 2

x xx x

+ = =

7. Cng thc nhn ba: Sin3x = 3sinx 4sin3x; cos3x = 4cos3x 3cosx.

8. Cng thc biu din theo tanx:2

2 2 2

2 tan 1 tan 2 tansin 2 ;cos2 ; tan 2

1 tan 1 tan 1 tan

x x x x x x

x x x

= = =

+ +

9. Cng thc bin i tch thnh tng

( )

( )

( )

1cos cos cos( ) cos( )

2

1sin sin cos( ) cos( )2

1sin cos sin( ) sin( )

2

a b a b a b

a b a b a b

a b a b a b

= + +

= +

= + +

10.Cng thc bin i tng thnh tch

sin sin 2sin cos2 2

sin sin 2cos sin2 2

cos cos 2cos cos2 2

cos cos 2sin sin2 2

x y x yx y

x y x yx y

x y x yx y

x y x yx y

+ + =

+ =

+ + =

+ =


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3

Cch gii cc phng trnh lng gic trong thi i hc:

Lu trc khi gii :Cc phng trnh lng gic trong thi i hc nhn qua mt hc sinh thng rt kh khn phc tpnhng chng u quy v nhng phng trnh n gin. thi i hc cc nm u xoay quanh bii v dng phng trnh tch, t n ph. Nm 2009, thi c bin i hn l phng trnh cubin i v dng cng thc cng. Nhn chung phng php gii dng ton ny l cc em hc thuc cc

cng thc trn y v rn luyn k nng phn tch a thc thnh nhn t

GII MT S THI TIU BIU:

1. Gii phng trnh:2sin 2 4sin 1 06

x x

+ + =

(1)

Gii:

(1) 3 sin 2 cos 2 4sin 1 0 x x x + + = ( ) 22sin 3 cos 2 2 2 sin 0 x x x+ =

( )2sin 3 cos sin 2 0 x x x + = sinx 0

1

3 cos sin 1 cos cos2 6

x k

x x x x

= =

= + =

5

26

72

6

x k

x k

x k

= = +

= +

2. Tm nghim trn khong (0; ) ca phng trnh :

Gii:Tm nghim ( )0,

Ta c 2 2x 3

4sin 3 cos2x 1 2cos x2 4

= +

(1)

(1) ( )3

2 1 cos x 3 cos 2x 1 1 cos 2x2

= + +

(1) 2 2 cos x 3 cos 2x 2 sin 2x =

(1) 2cosx 3 cos2x sin2x = . Chia hai v cho 2:

(1) = 3 1

cos x cos2x sin 2x2 2

( )cos 2x cos x6

+ =

( ) ( )

= + = +

5 2 7x k a hay x h2 b

18 3 6

2 2 34sin 3 cos 2 1 2cos ( )2 4

xx x = +


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Do ( )x 0, nn h nghim (a) ch chn k=0, k=1, h nghim (b) ch chn h = 1. Do ta c ba

nghim x thuc ( )0, l 1 2 35 17 5

x ,x ,x18 18 6

= = =

3. . Gii phng trnh : 32 2 cos ( ) 3cos sin 04

x x x

= (2)

Gii:

(2)3

2 cos x 3cosx sin x 04

=

( ) + =

+ + + =

3

3 3 2 2

cos x sin x 3 cos x sin x 0

cos x sin x 3cos xsinx 3cosxsin x 3cosx sinx 0

=

=3

cos x 0

sin x sin x 0

+ + + =2 3 2 3

cos x 0hay

1 3tgx 3tg x tg x 3 3tg x tgx tg x 0

=2sin x 1 =haytgx 1 x k

2

= + hay

= + x k

4

4. . Gii phng trnh : 22

cos 2 1( ) 3

2 cos

xtg x tg x

x

+ = ( d b khi B 2005)

Gii:

(2)2

2

2

2sin xcotgx 3tg x

cos x

=

= = = = +

2 31tg x 0 tg x 1 tgx 1 x k ,k Z

tgx 4

PHNG PHP T N PH TRONG PHNG TRNH LNG GIC:

A. t t=sinxCos2x= 1 sin2x = 1-t2 t[-1;1]

Tan2x =2

2

sin

cos

x

x=

2

21

t

t

Cos2x =2

1 2sin x = 1-2t2

Sin3x = 3 33sin 4sin 3 4 x x t t =

B. t t = cosx2 2 2

sin 1 cos 1 x x t = = 2cos 2 2 1x t= + 2 2

2

2 2

sin 1tan

cos

x tx

x t

= = 3 3cos3 4cos 3cos 4 3 x x x t t = =

C. t t= tanx


16/63

5

1cot x

t= 2

2

1cos

1x

t=

+

22

2sin

1

tx

t=

+

2

2

1cos2

1

tx

t

=

+

2

1sin2x=2t

1 t

+

2

2tan2

1

tx

t

=

+

sin cos tan

sin cos tan

a x b x a x b at b

c x d x c x d ct d

+ + += =

+ + +

D. t t=sinx cosx t 2; 2 sinxcosx

2 1

2

t =

sin2x= ( )2 1t +

( ) ( )2 3

3 3 2 2 1 3sin cos sin cos sin cos sin cos 12 2

t t x x x x x x x x t

+ = + + = =

NGUYN TC CHUNG GII PHNG TRNH LNG GIC

Bin i: t t Phn tch thnh tch

Nguyn tc : Ly tha H bc Tch Tng Tng Tch

Bin i khng c th i bin.

GII MT S THI TIU BIU:

Bi 1. 2cos 2 1

cot 1 sin sin 21 tan 2

x x x x

x = +

+

Gii:t t=tanx, pt tr thnh:

( )

2

2 2

2 2

1

11 1 21 0; 1

1 1 2 1

t

t t tt t

t t t t

+ = +

+ + +

3 22 3 2 1 0t t t + = 1t = tan 1 4 x x k

= = +

Bi 2. cos3 cos 2 cos 1 0 x x x+ = Gii:t t=cosx, pt tr thnh:

3 24 3 2 1 1 0t t t t + =


17/63

6

cos 11

21cos cos

32

xt

xt

= = ==

22

3

x k

x k

= = +

Bi 3. Gii phng trnh: 1 sin 1 cos 1x x + = ( thi d b2 A 2004) (1)

Gii:(1)1 sin cos 2 (1 sin )(1 cos ) 0 x x x x + = t t=sinx +cosx

2 1sin

2

txcosx

=

Pt tr thnh:2 1

1 2 1 02

tt t

+ + = 2 2 22 1 4 2 2 4 ( 1) 0 1t t t t t t + = + = =

Sinx+cosx =1 2 sin 14

x

+ =

sin sin4 4

x

+ =

x k=

Bi 4. ( )

2

2cossin 6 tan 1 sin 21 sin

x x x xx

+ + =+

Gii:t t=sinx [ 1;1]t pt tr thnh:

( )2 2

2

2

16 1 2 6 1 0

1 1

t tt t t t

t t

+ + = =

+

21 6

1sin 52

22

1 6sin sin3 1

arccos 23

x k

tx

x k

xt

x k

= + = = = +

== = +

Bi 5. 6 61

sin cos cos84

x x x+ = (1)

Gii:

(1) 23 1 3 1 cos 4 11 sin 2 cos8 1 cos84 4 4 2 4

x x x x

= =

t t=cos4x [ 1;1]t pt tr thnh:

( )22

43 1 1 16 42 41 2 1

3 34 2 4 2 44 16 42

k

xt x ktt

k x k xt

= += = + =

= + = +=


18/63

7

Bi tp t luyn

1 1sin 2x sin x 2cot g2x2sin x sin 2x

+ =

2

x3cos2

42

xcos

42

x5sin =

22cos x 2 3sinx cosx 1 3(sinx 3 cosx)+ + = +

gxcottgxxsin

x2cos

xcos

x2sin=+

( )( ) 12 cos 1 sin s in2 cos 22

x x x x + =

( )( )2sin 1 2cos 1 1x x+ = ( )3 3sin cos 2 1 sin cos x x x x+ = 2sin cos cos 1

2xx x =

4 4 3sin cos cos .sin 3 04 4 2

x x x x

+ + =

Cho phng trnh: 2sin cos 1sin 2cos 3

x xa

x x

+ +=

+(2) ( d b khi a 2002)

1. gii phng trnh khi a=13

2. tm a phng trnh (2) c nghim.

2tan cos cos sin 1 tan tan2x x x x x x + = +

( )2

4

4

2 sin 2 sin3tan 1

cos

x xx

x

+ =


19/63

8

Bi IV: Tch PhnLu trc khi gii thi:

Tch phn l bi ton rt thng xut hin trong thi i hc. K t nm 2002, khi bt u tin hnh th Ba chung cc dng ton tch phn v ng dng lun xut hin v l cu 1 im. Bi tp phn nkhng qu kh nhng vn phi i hi k nng phn on, phn tch , v nm r c cc cch lm b

ton tch phn c bn nh i bin s v tnh theo tch phn tng phn cc em cng theo di cc v ddi y.

NGUYN TC CHUNG GII BI TON TCH PHN:Gm c 2 phng php chnh:

A. I BIN: i bin loi 1:

( )( ) ( ). ' f u x u x dx t t=u(x)Ch : Cc biu thc c quan h o hm

GII CC V D:

VD 1. Tnh tch phn:2

2

0

sin 2

3 cos

xI

x

=+

Gii:t 23 cost x= + ( )2 cos sindt x x dx = 2sin2dt xdx =

X0

2

t 4 3

4

3

4 4

ln ln3 3

dt I t I

t

= = =

VD2. Tnh tch phn:6

2

dxI

2x 1 4x 1=

+ + + ( DB 1A 2006)

Gii:

t t= 21

4 1 4 12

x t x tdt dx+ = + =

X 2 6t 3 5

( )

( ) ( )

5 5 5

2 2

3 3 3

51 1 1 3 1ln 1 ln31 1 2 121 1

t dt dt dt tt tt t

+ = = + + = + + + +

VD3. Tnh tch phn:4

2

0 cos 1 tan

dxI

x x

=+

Gii:


20/63

9

t t= 22

1 tan 1 tan 2cos

dx x t x tdt

x+ = + =

X0

4

t 1 2

2 2

1 1

2 22 2 2 2 2

1

tdt I dt t

t= = = =

VD 4. Tnh tch phn:e

1

3 2 ln xI dx.

x 1 2 ln x

=

+

Gii:

t t= 21 2 ln 1 2 lndx

x t x tdt x

+ = + =

Xe

1t 2 1

( )( )

22 2

2

1 1

3 1 10 2 114

3

t I tdt t dt

t

= = =

1. i bin loi 2: Bc t ln hn bc mu: chia a thc Bc t nh hn bc mu:

Xt quan h o hm i bin Mu c nghim Tch phn thc Hm hu t (mu v nghim):

( )( )2 2

du

u x a+ t u(x)=atant

Hm cn thc:( )( )

22a u x+ t u(x)=atant

( )( )22 u xa t u(x)=asint (hoc u(x)=asint)

VD 5. Tnh tch phn: I=3

2

09

dx

x +

Gii:t x=3tan(t)

( )23 tan 1dx t dt = + X 0 3t

04


21/63

10

( )( )

24

2

0

3 tan 1 14

3 129 tan 10

t dtI t

t

+

= = =+

VD 6. Tnh tch phn:( )

5

2

21 9 1

dxI

x=

Gii:t x-1= 3sint

3cosdx tdt = X

15

2

t0

6

6 6 6

2 20 0 0

3cos cos cos6cos 69 9sin 1 sin 0

tdt tdt tdt I ttt t

= = = = =

VD 7. Tnh tch phn:3

2 21 3

dxI

x x=

+

Gii:

t x= 3 tan t ( )23 tan 1dx x dx = + X 1 3t

6

3

( )2 3 32222 2

2 26 6

13 tan 1 1 1 coscos

3 3 sinsin 13 tan 3 tan 3

cos cos

dtt tdtt I dxttt

t t

+ = = =

+

( )3

2

6

sin1 1 6 2 33

3 sin 3sin 9

6

d tI

t t

= = =


22/63

11

B. PHNG PHP TNH TCH PHN TNG PHN:

Cng thc:b b

a a

budv uv vdu

a= (1)

Cch ly phn cc tch phn:

K hiu P(x) l a thc. Khi gp hai dng nguyn hm sau y, ta thng dng phng php tch phtng phn:

Dng 1: ( ) lnP x xdx ta t u= ln x (Do lnx khng c nguyn hm)

Dng 2: ( ). sin( )cos( )

ax be

P x ax b dx

ax b

+

+ +

ta t u=P(x)

Vi cch y khi ly cng thc 1 ta s c bi ton dn ti nguyn hm ng dng vi bc ca P(x)thp hn

GII CC V D:

VD 1. Tnh tch phn:2

0

I (x 1)sin2xdx.

= + ( d b khi D 2005)

Gii:

t:( ) 2

0

11 1

cos 2 cos 2 1212 2 4s in2 cos 2

02

u x du dxx

I x xdxdv xdx v x

= + = +

= + = + = =

VD 2. Tnh tch phn:2

1

I (x 2)lnx dx.= ( d b khi D 2006)

Gii:

t:( ) 2

1ln

22

2

du dxu x x

dv x dx xv x

==

= =

22

1

2 52 ln 2 ln 4

12 2 4

x x I x x dx

= = +

VD 3. Tnh tch phn:

2

4

0

sin xdx

Gii:t t= 2 2 x t x tdt dx = =

X0

2

4

t0

2


23/63

12

2

0

2 sin B t tdt

=

Tnh2

0

sin I t tdt

=

t:sin cos

u t du dt

dv tdt v t

= =

= =

2

0

cos cos cos 0cos 0 sin 12 22 2

0 0

I t t tdt t

= + = + + =

B=2I=2

VD 4. Tnh tch phn: A=2

0

cosx

e xdx

Gii:

t:sin cos

x xu e du e dx

dv xdx v x

= =

= =

2 2 202

0 0 0

cos cos cos cos 0 cos 1 cos22

0

x x x x A e x e xdx e e e xdx e xdx

= + = + + = + (1)

Tnh2

0

cosx

K e xdx

=

t:cos sin

x xu e du e dx

dv xdx v x

= =

= =

2

2

0

sin sin2

0

x xK e x e xdx e A

= =

Thay vo (1):2

2 21

1 2 12

e A e A A e A

+

= + = + =

VD 5. Tnh tch phn: A=2

0

sin cos x x xdx

Gii:

t:22 sin cossin cos

du dxu x

v x xdxdv x xdx

==

==

Tnh: 2sin cosv x xdx= t : cos sint x dt xdx= =


24/63

13

V=3 3

2 cos

3 3

t xt dt C C

= + = +

Chn C=03cos

3

xv =

Vy3

3

0

cos 1 1cos

03 3 3 3

x A x xdx K

= + = +

(1)

Tnh ( )3 20 0

cos 1 sin cosK xdx x xdx

= =

t t=sin(x) cosdt xdx = X 0 t 0 0

( )0

2

0

1 0K t dt = =

Thay vo (1): 13 3 3A K = + =

VD 6. Tnh tch phn:2

3

sin

1 cos

x x D dx

x

+=

+

Gii:

2

2

3

sin

2cos

2

x xD

x

+= t:

( )

2

sin1 cos

1

tan2cos

22

u x xdu x dx

dv dx xx v

= + = +

==

Vy: ( ) ( )2

3

3 32sin tan 1 cos tan 1

2 2 2 3 2 3

3

x x D x x x dx K

= + + = + +

(3)

Vi: ( )2 2 2

2

3 3 3

1 cos tan 2cos tan sin2 2 2

x x xK x dx dx xdx

= + = =

12cos2

3

x

= =

Thay vo (3) ta c: D=( )9 2 3

18

+

Li bnh: tch phn tng phn ta c cch nh t u nh sau: nht log nh a (a thc) tamLng (Lng gic) T m. Trong php tnh tch phn tng phn, gp php no ng trc trong 4php trn, hy t u bng php !


25/63

14

Bi tp t luyn

Tnh tch phn: 3 20

sin . I x tgxdx

=

Tnh tch phn:7

3

0

2

1

xI dx

x

+=

+

Tnh tch phn: 20

ln

e

I x xdx=

Tnh tch phn: 4 sin0

( cos )x

I tgx e x dx

= +

Tnh tch phn:0

cos sin I x xdx

=

Tnh tch phn:3

2 2

6

tan cot 2 I x x dx

= +

Tnh tch phn: ( )2

2

2 1 cos2 I x dx

= +

Tnh tch phn: 3

6

sin 4 sin 3

tan cot 2

x x I dx

x x

=+

Tnh tch phn:10

5

dxIx 2 x 1

=

Tnh tch phn:e

1

3 2 ln xI dx.

x 1 2 ln x

=

+

Tnh tch phn:2

0

sin

1 sin

x xI

x

=+

Tnh tch phn: 360

sin sin

cos2

x xI

x

+=

Tnh din tch hnh phng gii hn bi parabol ( ) 2P : y x x 3= + v ng thngd : y 2x 1.= +

Tnh din tch hnh phng gii hn bi cc ng: ( ) ( ) ( )2

2 271 ; 2 ; 327

xC y x C y C y

x= = =


26/63

15

Bi V:Cc bi ton lin quan n ngdng ca o hm v th hm s.Lu trc khi gii thi:

Cc bi ton dng ny l cu chim 1 im, thng nm cu th 2 sau phn kho st hm s trong thi i hc. Mun gii c dng ton ny ta cn nm vng cc l thuyt v s tng, gim hm s, ccvn v cc tr, s tng giao gia hai th (iu kin tip xc ca hai ng cong) Cc v d diy s trnh by mt cch c h thng cc vn nu trn v cch gii n gin v d hiu nht. Ccbn tham kho cc v d sau y:

I: S TNG GIM CA HM S:Nhc li kin thc:Cho hm s ( ) y f x= c o hm trn min I

( ) 0; f x x I Hm s tng

( ) 0; f x x I Hm s gim

VD 1. Cho hm s: ( ) ( )3 2 21

23

y f x x mx m m x= = + +

Tm m hm s:a. Tng trn Rb. Gim trn (0;2)

c. Tng trn ( )4;+

d. Gim trn on c di bng 2

e. Tng trn 2 khong ( );4 v ( )2;+

Gii:

TX: D R= 2 2' 2 2 ' 2 y x mx m m m= + + = +

a. Ycbt ' 0 2 0 2m m +

b. Ycbt ( )

( )

2

2

' 0 0 2 01

' 2 0 3 2 0

y m mm

y m m

+

+

V

c. YcbtTH1: ' 0 2 0 2m m +

x - 0 2 +F(x) + - +

F(x)


27/63

16

TH2: ( ) 22' 0

' 4 0 9 14 0

44

2

m

y m m

mS

+ +

< + > + +

+ <


28/63

17

TH2: ( ) 20' 0

' 6 0 13 36 0

66

2

m

y m m

mS

> >

+


29/63

18

c. Hm s t C v CT ti im c honh >-1d. Hm s t C v CT ti im c honh nm trong [-2;3]e. Hm s t C v CT ti im c honh dngf. Hm s t C v CT ti im c honh tri du nhau

g. Hm s t C v CT ti x1;x2 sao cho ( )3 31 2x x+ nh nhtGii:

MX: D=R2 2' 2 2 1 y x mx m= + 2' 1m = +

' 0 > :

X X1 X2 +

Y + 0 - 0 +

YC

CT

a. Ycbt Hm s t cc i ti x=0

( ) 2' 0 0 2 1 0 2

2002

ym

mSm

= = =

><

b. Ycbt :

( )

2

2

11 0' 0

0' 1 0 2 2 0

11

1 12

mm

m y m m

mmS

m

< + > > > >

< < -1

( )

2

2

11' 0

0' 1 0 2 2 0

11

1 12

mm

m y m m

mmS

m

< >

> + > <

> > >

0 1m< <

d. Hm s t C v CT ti im c honh nm trong [-2;3]

Ycbt

( )

( )

( )

( )

2

2

' 01

' 2 02 4 3 0

1 1' 3 0 2 6 8 0

2 3 2 32

my

m m m

my m m mS

m

>


30/63

19

Ycbt ( ) 2

1 1

21 1' 022

' 0 0 2 1 0 122

00 2

2 0

m

m m

y m m

m mS

m

<

f. Hm s t C v CT ti im c honh tri du nhau

( )2

' 0 0 2 22 1 0

2 2' 0 1

ym m

m

< < < - V bng bin thin:


31/63

20

X -2 X1

2

SX2 3 +

Y + 0 - 0 +

YC

CT

T ta c( )

( )

' 2 0

' 3 0

y

y

. Vy l iu kin th 2 c biu hin rt r rng trn bng bin

thin. y thc ra l xt quan h v du ca h s a: ( )af nhng y khi ta bit r du

ca a th ch cn t du vo trc ( )f l c. y cng c th l bc rt gn th

gian m cc em nn lm, trnh khai trin mt thi gian.

-2

Sl tng hai nghim X1;X2 ca phng trnh y=0 hay bng

2

b

a

. R rng nu X1;X2 nm

trong [-2;3] th2

Scng phi nm trong on ny. V

2

b

a

l gi tr c th rt ra d dng t

phng trnh gc nn ta chn gi tr trung bnh ny lm iu kin.Nt tht th 3 c gb.

- Li khuyn l: khi gp nhng dng ton nh trn hc sinh hy v bng bin thin nhtrn ra giy nhp sau ty theo cu hi m in cc thng s thch hp vo bng. t mi hng gii u c phi by!

Ti c tham kho qua mt vi ti liu ca cc thy c gio th thy phn ln cc sch u trnh by lgii mt cch my mc, khng trc quan, nhiu lc c th coi l lun qun. . V d: tm m hm sy=f(x) tng trn (1;+ ), cc thy c trnh by trong sch cng nh trn lp theo phng php MinMax, xt nhiu trng hp Nhng cch gii khng phi l sai tuy nhin iu i khi lm kh cem hc sinh trong qu trnh t duy tm trng hp, nht l cc em hc sinh trung bnh. Phng php

xt du trnh by trn y va ngn gn r rng li khng b st trng hp. bi ton c n ginha.

Cch gii trn cng p dng c cho hm s2

2' ' '

ax bx cy

a x b x c

+ +=

+ +v dng o hm

( )

2

22

2' ' ' ' ' '

'' '

a b a c b cx x

a b a c b cy

a x b x c

+ +

=

+ +

. Trong trng hp ny, ty biu thc mu c nghim hay

khng ta t thm trng hp. V mu thc 0 nn khi xt du ta ch cn xt du t s tng tnh cc v d trnh by trn.

Dng hm s ny khng cn thng dng ( ch gii thiu s lc trong sch gio khoa) nn xu

hng ra ch xoay quanh 3 hm l: bc 3, trng phng v' '

ax by

a x b

+=

+.

Bi 2: Cho (Cm): ( )3 23 3 1 4 y x mx m x= + +

nh m :a. C(m) c hai im cc tr A;B sao cho AB thng hng vi C(1;-1)

b. C(m) c hai im cc tr A;B sao cho AB = 2 5 c. C(m) c hai im cc tr A;B sao cho AB cch u : 2y =

Gii:


32/63

21

MX: D=R

Ta 2 im cc tr tha h:' 0

( )

y

y f x

=

=

Vy: 2' 2 1 0 y x x m= + =

( )3 23 3 1 4 y x mx m x= + + ( ) ( )2

0

2 1 y x x m cx d ax b ax b = + + + + = +

( )2 2 1 ( 1) 2 5 y x x m x mx m = + +

( )

( )

2 2 1 0 1

2 5 2

x x m

y mx m

+ + =

= +

C(m) c hai cc tr (1) phi c 2 nghim phn bit ' 0 0m >

a. C(m) c hai im cc tr A;B sao cho AB thng hng vi C(1;-1)(2) phng trnh ng thng qua hai im cc tr l 2 5 y mx m= +

V AB thng hng vi C(1;-1) C AB nn: -1=-2m.1-m+5 2m =

Vy vi m=2 AB thng hng vi C(1;-1)

b. C(m) c hai im cc tr A;B sao cho AB = 2 5

( ) 2 12 '

1 2 x x ma

= =

( ) ( )2 1 2 12 2 4 y y m x x m m = = ( ) ( )2 2

2 1 2 1 2 5 AB x x y y = + =

2

1

16 4 20 5

4

m

m mm

= + = =

So snh k 1m =

c. C(m) c hai im cc tr A;B sao cho AB cch u : 2y = Ycbt ( ) ( ); ;d A d B = vi : 2y =

( )

1 2 1 2

1 2

1 2 1 2

2 22 2

2 2 4

y y y yy y

y y y y

= = =

= + =

( ) ( ) ( )1 2 1 22 5 2 5 4 2 2 10 4mx m mx m m x x m + + + = + + =

2 .2 2 10 4 1m m m + = =

Bi 3: Cho (Cm): ( )3 23 3 1 y x x m x= +

nh m :

a. C(m) c hai im cc tr A;B sao cho OAB vung ti Ob. C(m) c hai im cc tr A;B nm khc pha vi trc Oxc. C(m) c hai im cc tr A;B cng pha vi trc Oyd. C(m) c hai im cc tr A;B nm cch u ng thng y=5e. C ng thng i qua hai im cc tr cch gc ta mt khong bng 1

f. C ng thng i qua hai im cc tr tip xc vi ng trn ( ) ( )2 2

1 1 4x y + =

g. C ng thng i qua hai im cc tr to vi hai trc ta mt tam gic cnh. C ng thng i qua hai im cc tr to vi hai trc ta mt tam gic c din tch =8

Gii:


33/63

22

CT

CDy

1

x

1x

5y =

MX: D=R

Ta 2 im cc tr tha h:' 0

( )

y

y f x

=

=

( ) ( ) ( )

2

3 2 2

'2 1 0

3

3 3 1 2 1 1 2 1

y x x m

y x x m x x x m x mx m

= + + =

= + = + + + +

( )

( )

2 2 1 0 1

2 1

x x m

y mx m

+ + =

= +

C(m) c hai cc tr (1) phi c 2 nghim phn bit ' 0 0m > (*)a. C(m) c hai im cc tr A;B sao cho OAB vung ti O

Ycbt OA OB

.OAOB

vi( )

( )

;

;

A A

B B

OA x y

OB x y

=

=

( )( )1 2 1 2 1 2 1 20 2 1 2 1 0 x x y y x x mx m mx m + = + + + = ( ) ( ) ( )

22 2

1 2 1 2 1 24 2 2 1 0x x m x x m m x x m + + + + + =

( ) ( ) ( ) ( )22 21 4 1 2 2 . 2 1 0m m m m m m + + + + + + =

3 24 9 7 2 0m m m + + = ( )( )2

v 7

4 5 2 1 0

VN

m m m

=

+ =

1m = (tha iu kin(*))

b. C(m) c hai im cc tr A;B nm khc pha vi trc Ox

Ycbt 1 2. 0y y < ( )( )1 22 1 2 1 0mx m mx m + + <

( )( ) ( )22 2

1 2 1 24 2 2 1 0m x x m m x x m + + + + <

( ) ( ) ( )22 24 1 2 2 2 1 0m m m m m + + + <

( )( )23 2

0

4 9 6 1 0 4 1 1 0m m m m m

+ + < +

c. C(m) c hai im cc tr A;B cng pha vi trc OyYcbt 1 2 0x x > ( 1x cng du vi 2x ) 1 0 1m m + > <

d.

C(m) c hai im cc tr A;B nm cch u ng thng y=5Ycbt : y=5 ct (Cm) ti trung im AB. M l trung im AB c ta 1 2 ; 2 1

2

x xmx m

+ +

( )1;3 1M m 5 3 1 2Ycbt m m = =

So snh vi iu kin (*) ta thy m=2 l kt qu cn tm.

e. C ng thng i qua hai im cc tr cch gc ta mt khong bng 1: 2 1 : 2 1 0 y mx m mx y m = + + + =


34/63

23

Ycbt ( ); 1d O =( )

2 2

2 .0 0 11

2 1

m m

m

+ + =

+

( ) ( )2 2 2 21 2 1 3 2 0m m m m + = + + =

0

2

3

m

m

= =

So snh vi iu kin m>0 ta nhn thy khng c gi tr m tha mn yu cu bi ton.

f. C ng thng i qua hai im cc tr tip xc vi ng trn ( ) ( )2 2

1 1 4x y + =

Ycbt ( );d I R = vi tm I(1;1) v R=2

: 2 1 0mx y m + + =

( )2

2 .1 1 12

2 1

m m

m

+ + =

+

( )2 2 22 16 4 15 4 0m m m m + = + + =

0

4

15

m

m

= =

So snh vi (*) ta nhn4

15

m =

g. C ng thng i qua hai im cc tr to vi hai trc ta mt tam gic cn

Gi M l giao im ca v Ox:2 1 0 1

;00 2

mx m mM

y m

+ =

=

Gi N l giao im ca v Oy: ( )2 .0 1

0; 10

y m mN m

x

= +

=

Ycbt 1 11 1 . 1 02 2

M N m x y m mm m

= = =

1

12

1

2

m

m

m

=

=

=

D thy vi m=1, i qua gc ta , vi m=1

2

khng tha (*) nn loi. Vy ta chn

1

2m =

h. C ng thng i qua hai im cc tr to vi hai trc ta mt tam gic c din tch =8

Ycbt:1 1 1

.2 8 2

OMN M N S OM ON x y = =

( )211 1 1. 1

4 2 4 2

mmm

m m

= =

( )

2

2

2

2 1 122

2 12

mm

m m m

mm m VN

=

+ = =

+ =

So snh (*) vy c hai gi tr m tha mn: m=2 v m=0.5


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24

III: S TNG GIAO GIA HAI THNhc li kin thc:

Cho: ( ) ( )1 2: ; :C y f x C y g x= =

S giao im ca C1 v C2 l s nghim ca phng trnh honh giao im:

( ) ( ) f x g x=

c bit khi C1 tip xc C2:

( ) ( )

( ) ( )' '

f x g x

f x g x

=

=

Lu : Khng c s dng iu kin nghim kp lm dng ton tip xc ca hai th. hiu r hn, ta hy n vi cc v d sau:

Bi 1: Cho hm s ( ) ( )2 3 2

: 21

m

mx mC y m

x

=

v ( ) : 1d y x=

nh m (d) ct (Cm) ti hai im phn bit:

a)C honh ln hn -1b)C honh nh hn 2c)C honh nng trong khong [ ]2;3 d)C honh dnge)C honh tri du.

Gii:Phng trnh honh giao im gia (Cm) v d:

( ) ( )22 3 2

1 : 2 1 3 3 01

mx m x g x x m x m

x

= + + + =

x

1

x 2

S

2

x +

( )g x + 0 - 0 +

(d) ct (Cm) ti hai im phn bit g(x)=0 c hai nghim phn bit

( )

2' 0

1

1 0 2

m

m

g m

> >

<

(*)

a)C honh ln hn -1

Ycbt:

( )1 0

12

g

S

>

>

+ > > So snh vi (*) ta kt lun:

61

52

m

m

<

b)C honh nh hn 2

( )( )

2 04 4 1 3 3 0 3 0 3

1 11 222

gm m m m

Sm mm

> + + + > + <


36/63

25

So snh vi (*) ta kt lun:2

2 1

m

m

<

< <

c)C honh nng trong khong [ ]2;3

Ycbt:( )( )

( )( )

11

2 0 4 4 1 3 3 0 73 0 9 6 1 3 3 0 2

3 22 1 32 3

2

mg m mg m m m

mS m

+ + + + + + +

+

So snh iu kin (*) ta suy ra:11

17

m

d)C honh dng

Ycbt:

( ) 03 3 0 1

1 0 10

2

g om m

Sm m

>+ > >

+

So snh vi (*) ta suy ra: m>2

e)C honh tri du.

Ycbt: ( )0 0 3 3 0 1g m m< + < <

So snh iu kin (*) ( ) ( ); 2 2; 1m

Bi 2: Cho hm s ( )1

:1

xC y

x

+=

v ( ) : 1d y mx= +

Tm m d ct (C):a)Ti 2 im phn bit nm trn 2 nhnh ca th.b)Ti 2 im phn bit nm trn cng 1 nhnh ca th

Gii:Phng trnh honh giao im ca (C) v d:

( )1

1 11

xmx x

x

+= +

( ) ( )2 2 0 1g x mx mx = =

a)Ti 2 im phn bit nm trn 2 nhnh ca th. (Hnh 1)Ycbt: phng trnh (1) c hai nghim phn bit tha

1 21x x< <

x

1x

1

tiem can dung

2x +

( )g x Cng du m 0 Tri du m 0 Cng du m

( ) ( ). 1 0 2 0 2 0 0m g m m m m m < < < >


37/63

26

nh1H

nh2H

nh3H

Lu : Trng hp ny khng cn phi xt bit thc v khi d ct

C v 2 pha ca tim cn ng x=1 th mc nhin phng trnh

c 2 nghim, khng cn thit phi xt

b)Ti 2 im phn bit nm trn cng 1 nhnh ca th

(Hnh 2)

Phng trnh (1) c hai nghim phn bit tha:

1 2

1 2

1

1

x x

x x

<

0

0

8

m

m

m

<

8m <

Bi 3: Vit phng trnh ng thng ct th :( ) 3: 3 2C y x x= + ti 3 im phn bit A,B,C sao cho xA=2 v

BC= 2 2 .Gii: (hnh 3)

2 4A A

x y= =

Phng trnh ng thng qua A(2;4) l

( ): ( ) : 2 4A Ay k x x y y k x = + = +

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

( ) ( )3 33 2 2 4 3 2 2 x x k x x x k x + = + =

( )3 3 2 2 0 x k x k + + = ( ) ( )22 2 1 0 x x x k + + =

( ) 22

2 1

x

g x x x k

=

= + +

iu kin c BC:

Khi ta

( ) ( )1 1 2 2; ; ;B x y C x y tha

h:

( )

( )

2 2 1 0 1

2 4 2

x x k

y kx k

+ + =

= +

(1) 2 12 '

2 x x k a

= =

(2) ( )2 1 2 1 2y y k x x k k = =

( ) ( )2 2

2 1 2 1 2 2 BC x x y y= + =

3 34 4 2 2 4 4 8 0 1k k k k k + = + = =

y

y

2 2

x

y

( )

' 0 0 0

2 0 4 4 1 0 9

k k

g k k

> > >

+ +


38/63

27

Vy ( ): 1 2 4y x = +

Bi 3: Cho (C) ( ) 3 23 2 y f x x x= = + . Tm trn ng thng (d):y=-2 nhng im m t c th v

c n (C) :a. Ba tip tuyn phn bitb. Ba tip tuyn phn bit trong c 2 tip tuyn vung gc vi nhau

Gii:a. Ba tip tuyn phn bit

Xt ( ; 2) : 2 A a d y = .

Phng trnh ng thng qua ( ; 2 )A a v c h s gc :

( ) ( )2y k x a= .

tip xc vi (C) H phng trnh sau c nghim:

( ) ( )

( )

3 2

2

3 2 2 1

3 6 2

x x k x a

x x k

+ =

=

Thay k t (2) vo 1 ta c:

( )( ) ( )3 2 23 2 3 6 2 3 x x x x x a + =

( )3 22 3 1 6 4 0 x a x ax + + =

( ) ( )32 2 3 1 2 0 x x a x + =

( ) ( ) ( )22

2 3 1 2 0 4

x

g x x a x

=

= + =

T A k c ba tip tuyn phn bit n (C) phng trnh (3) c 3 nghim phn bit phng trnh (4) c 2 nghim phn bit khc 2

( )( )

( )( )

2

2

50 3 1 16 0 1*3

2 0 2.2 3 1 .2 2 0 2

g a a a

g a a

> > < >

+

b. Ba tip tuyn phn bit trong c 2 tip tuyn vung gc vi nhauKhi phng trnh (3) c 3 nghim phn bit:

0 1 22; ; x x x= ( vi x1;x2 l hai nghim ca phng trnh g(x)=0) v 3 tip tuyn ng vi h s gc l:

( ) ( ) ( )2 20 1 1 1 1 2 2 2 2' 2 0; ' 3 6 ; ' 3 6k f k f x x x k f x x x= = = = = =

V 0 0k = nn : Ycbt k1.k2=-1.

( ) ( ) ( ) ( )2 2 2 21 1 2 2 1 2 1 2 1 2 1 23 6 3 6 1 9 2 4 1 **x x x x x x x x x x x x = + + =

p dng nh l Viet cho phng trnh (4) ta c:1 2

3 1ax x

x

+ = v 1 2 1x x =

Do (**)3 1

9 1 2 4 12

a + =

55

27a = (tha iu kin (*)).

Vy im cn tm l55

; 227

A

.


39/63

28

DNG TON: H NG CONG TIP XC VI MT NG C NHPhng php:

Dng 1: Cho h ng cong ( )mC :y=f(x;m). chng minh ( )mC lun tip xc vi mt ng (C) c nh .

TH1:

( )mC :y=f(x;m). l hm a thc.

a : ( );y f x m=

v dng: ( ) ( ) ( ): 2n

y ax bm g x n nguy n= + +

.Xt ng cong ( ) ( ):C y g x= v chng minh h:

( ) ( ) ( )

( ) ( ) ( )1

' '

n

n

ax bm g x g x

na ax bm g x g x

+ + =

+ + =

C nghim m

TH2:

( )mC :y=f(x;m). l hm hu t: (Dng tng qut)

( ) tip xc vi (C) h sau c nghim

( ) ( )

( )( ) ( )

0 0

2

1

2

cax b k x x y

x d

ca k x a

x d

+ + = + +

= +

Gii h trn qua 3 bc:B1: nhn 2 v ca phng trnh (2) cho: x+d

( ) ( )3c

ax ad k x d x d

+ = ++

B2: (1)-(3):

( )0 02c

b ad k x d yx d

+ = ++

( ) ( )0 02

4c

k x d y ad bx d

= + + ++

B3: Thay (4) vo (2) s c 1 phng trnh theo k. gii phng trnh ny v tm m sao cho phng trnh

ng m .

Lu : cch gii trn c th p dng i vi hm sax b

cx d

+

+

Dng 2: Tm iu kin h ng cong tip xc vi 1 ng c nh:Dng iu kin tip xc.

II/ Mt s v d:

Bi 1: Cho ( ) ( )3 2 2: 2 2 1 2mC y x x m x m= + + + + + . Chng minh rng (Cm) lun tip xc vi mt ng

cong c nh.

Gii:

Ta c: ( ) ( )3 2 2: 2 2 1 2mC y x x m x m= + + + + + ( )2

3 2 2 x m x x x + + + + + Xt ng cong ( ) 3 2: 2C y x x x= + + +

( )mC lun tip xc vi (C): h sau c nghim:

( )

( )( )

2 3 2 3 2

2 2

2 21

2 3 2 1 3 2 1

x m x x x x x x

x m x x x x

+ + + + + = + + +

+ + + + = + +


40/63

29

Ta c: ( )( )

( )

20

12 0

x m

x m

+ =

+ =R rng vi mi m , h (1) lun c nghim x=-m

Vy m , (Cm) lun tip xc vi 1 ng cong c nh: ( )3 2: 2C y x x x= + + + .

Bi 2:

Cho ( ) ( ) ( )

2

2 2 4:m m x m mC yx m

+=

. Chng minh (Cm) lun tip xc vi hai ng thng c nh.

Gii:

( )( ) ( )22 2 4

:m

m x m mC y

x m

+=

( )

42y m

x m =

(Cm) lun tip xc vi ng thng ( ) : y ax b = +

H phng trnh sau c nghim m :

( ) ( )

( ) ( )

( )

2

42 1

4

2

m ax bx m

I

ax m

= +

=

Nhn 2 v ca phng trnh (2) cho: x-m

( ) ( )4

3a x mx m

=

Ly (1)-(3):

( ) ( ) ( )8 8

2 1 2 4m b am a m b x m x m

= + = + +

Thay (4) vo (2):

( ) ( )2

1 2 16a m b a + + =

( ) ( ) ( ) ( ) ( )2 22

1 2 1 2 2 16 0 *a m a b m b a + + + = H (1) c nghim m ( )* ng m :

( )

( ) ( )

( )

2

2

1 01

2 1 2 02 6

2 16 0

aa

a bb b

b a

= =

+ = = =

+ =

Vy (Cm) lun tip xc vi 2 ng thng c nh y=x+2 v y=x-6


41/63

30

Bi tp t luyn

1. Cho hm s ( ) ( )3 2 21 11 23 3

y x m x m m x= + + + . nh m hm s:

a) Tng trn Rb) Gim trn (0;1)c)

Tng trn (-;2)d) Gim trn on c di bng 3

e) Tng trn 2 khong (-;0) v (2; +)2. Cho hm s ( ) ( )3 2 2 3: 3 3 1 1mC y x mx m m x m= + + + + + + . Tm m :

a) (Cm) c im cc i nm trn x=5b) Hm s t cc i v cc tiu ti nhng im c honh >1c) Hm s t cc i v cc tiu ti x1 v x2 sao cho: 1 2

2 1

14

5

x x

x x

+ =

3. Cho hm s ( ) 3: 3 2mC y x x= + .a) Vit phng trnh tip tuyn c h s gc nh nhtb) Vit phng trnh tip tuyn i qua M(1;0)c) Tm trn Ox nhng im m t k c trn C ng:

mt tip tuyn hai tip tuyn Ba tip tuyn hai tip tuyn vung gc vi nhau

d) Tm trn ng thng x=1 nhng im m t k c trn C ng: mt tip tuyn hai tip tuyn Ba tip tuyn

e) Tm trn (C) nhng im m t k c trn C ng 1 tip tuyn.4. Cho hm s ( ) 4 2: 2 2 1mC y x mx m= + . Tm m (Cm) ct Ox ti bn dim phn bit c hon lp

thnh cp s cng.

5. Xc nh m phng trnh c nghim duy nht: 3 2 1 0 x mx+ = 6. Cho hm s ( ) ( )3 2 2 3: 3 3 1mC y x mx m x m= + . Tm m (Cm) ct Ox ti 3 im phn bit trong

c ng 2 im c honh m.7. Cho hm s ( ) ( )3: 1 1mC y x k x= + + + . Tm k (Ck) tip xc vi ng thng ( ) : 1y x = + 8. Cho hm s ( ) 3 2 3: 3 4mC y x mx m= + . Tm m (Cm) ct ng thng ( ) :d y x= ti A,B,C sao cho

AB=BC.

9. Cho hm s ( ) 2 1:2

m

xC y

x

+=

+. Chng t rng ng thng y=-x+m lun lun ct th ti hai im

phn bit AB. Tm m on AB ngn nht.

10.Cho hm s ( ) ( ) ( )2

3 1: 1m m x m mC yx m

+ +=

+. Trong m l tham s khc 0:

a) Tm nhng im m th khng i qua m .b) Chng minh rng th ca (1) lun tip xc vi 2 ng thng c nh.

11.Cho hm s ( ) ( ) ( ) ( ) ( )3 2: 3 3 1 6 1 1 1mC y m x m x m x m= + + + + + . Chng minh rng h th (Cm)lun lun i qua 3 im c nh thng hng.


42/63

31

Bi VI: Mt s dng ton khc cn lu .I/ Gii hn:Dng ton ny tng xut hin trong thi i hc t rt lu (nm 2002 2003) Tuy nhin rt lukhng thy xut hin trong thi i hc. Tuy nhin ta cng nn ch n dng ton ny. u ti xin trnh by phng php tng qut lm bi dng ny l Gi s hng vng bng h s btnh.

Bi 1. Tm33 2

21

5 7lim

1x

x x

x

+

Gii:

Ta c: ( )3 33 2 3 2

2 2 21 1

5 7 5 2 7 2lim lim 1

1 1 1x x

x x x x

x x x

+ + =

( )( )

3 3

21 1 2 3

5 2 1lim lim

1 1 5 2x x

x x

x x x

=

+

=( )

( )( )

( )2

1 3

1 3lim 2

81 5 2x

x x

x x

+ + =

+ +

( ) ( )

3 2 2

21 1 232 2 23

7 2 1lim lim

11 7 2 7 4

x x

x x

x x x x

+ =

+ + + +

=

( )( )

21 32 23

1 1lim 3

127 2 7 4

x

x x

=

+ + + +

Thay (2),(3) vo (1) c:3 1 11

8 12 24A

= =

Lu :Trong li gii ta thm s 2 vo t thc f(x). C l bn ang t hi: Ti sao phi thm s 2 ? Lm cch no nhn ra s 2 ?S 2 l hng t b xa! Mun lm dng bi ny, ta phi khi phc n. Mun khi phc s 2 ny talm nh sau:

B1: c R lun c: ( )33 2

2 2

5 7

1 1

x c x cf x

x x

+ =

B2: Trong cc s c . Ta tm s c sao cho x2-1 c cng nhn t chung vi ( ) 31 5 f x x c= v

( ) 3 22 7 f x x c= + . iu xy ra khi v ch khi c l nghim ca tuyn:

( )

( )

( )( )

1

2

1

2

1 02

1 026

1 0 21 0

fc

fcc

f cf

= =

= ==

= = =

chnh l l do ti sao 2 xut hin trong bi gii.

y l vic nn lm trong giy nhp. Khng nht thit trnh by trong bi lm.Qua v d trn ta nu ln thut ton sau:

Gi s ( )( )

( )

f xF x

g x= c gii hn

0

0


43/63

32

B1: Phn tch ( )( )

( )

( )

( )1 2f x c f x c

f xg x g x

+ = + .

B2: (Tm c): Gi ( )1;2;...i i = l nghim ca h g(x)=0

Khi c l nghim ca h:( )

( )

( )1

1

01;2;...

0

i

i

f ci

f c

+ ==

=

Vi c tm c th( )

( )1lim

ix

f x c

g x

+v

( )

( )2lim

ix

f x c

g x

s hoc l dng xc nh hoc l dng quen thuc.

Sau khi tm c, vic trnh by li gii nh lm.

BI TP P DNG:

A=3 2 2

0

3 1 2 1lim

1 cosx

x x

x

+ +

( d b 2002)

B=3

20

1 2 1 3limx

x x

x

+ +

II/Phng trnh v bt phng trnh m v logarit:y l dng ton cng rt thng xuyn xut hin trong thi. Nhn chung, dng ton ny khngkh. Tt c cc php bin i ch xoay quanh cc cng thc nu trong sch gio khoa. phnny, ti khng nu li cc cng thc trn. Xin trnh by cch gii ca 1 s thi gn y.Bi lm qua 2 bc:B1: t iu kin. (Nu iu kin qu phc tp th c th n bc 2 ri th nghim vo iu kin)B2: Bin i phng trnh hay bt phng trnh v dng n gin cng c s c 2 v:

M: Chia Logarit: loglog

log

ba

b

xx

a=

log lognn

aa

mx x

n

=

t n ph: ( )logat f x= phng trnh hu t hoc phng trnh m( )x

t a= phng trnh hu t. Phng php hm s

Bi 1. ( )22 2 2 3 1

2 3 1 2 3 181.4 78.6 16.9 0 1x x

x x x x +

+ + +

Gii:

( )

2 22 3 1 2 3 16 9

1 81 78 16 04 4

x x x x + +

+

( )2 22 3 1 2. 2 3 13 3

81 78 16 02 2

x x x x + +

+

t

22 3 1

3

2

x x

t

+

=

k: t>0

Phng trnh tr thnh: 216 78 81 0t t + 3 27

;2 8

t

22 3 1

23 3 27 1 2 3 1 32 2 8

x x

x x

+

+


44/63

33

2 2

2 2

3

2

02 3 1 1 2 3 0

12 3 1 3 2 3 2 0

2

2

x

x x x x x

x x x xx

x

+

+

2

1

2

x

x

Bi 2. Gii bt phng trnh: 1 1 1 1 x x xe e x+ +

Gii:

t:1

11 1

u x xu v x

v x

= + =

= +

Phng trnh tr thnh: u ve e u v =

( ) ( ) f u f v

Vi ( ) ; 1x f x e x x=

( )' 1 0x f x e = + > ( )f x tng.

Do u v 1 1 1 1 x x x x + +

Bi 3. Gii phng trnh: ( )2 3log 1 logx x+ = Gii:t

3log 3

t x t x= =

Do : ( ) ( )2log 1 1 2 1 3 2t

t t x t x+ = + = + =

221 3 1 3 1 3

1

2 2 2 2 2 2

t tt t + = + = +

( ) ( )2 f t f = 2t = (V ( )1 3

2 2

tt

f x

= + l hm gim)

2 9t x = =

Bi 4. Gii bt phng trnh: ( ) ( )12log log 4 8 1 1x

x

+

Gii:

K: ( ) ( )2 11 3 54 8 0 2 2 2 1 3

2

xxx x

> > > >

( ) ( )121 log log 4 8 logx

x x x+ ( ) ( )1 12 2 2log 4 8 log 4 8 log 2 x x xx

( )1 2 044 8 2 2 8 0 34 2 8

xx x x x

x

loaix

Bi 5. Gii h phng trnh:( )

( ) ( )2 39 3

1 2 1 1

3log 9 log 3 2

x y

x y

+ =

=

(H A 2005)


45/63

34

Gii:

k:1

0 2

x

y

<

( ) ( )3 3 3 32 3 1 log 3log 3 log log x y x y x y + = = =

Thay x=y vo (1) ta c:

( )( )1 2 1 1 2 2 1 2 1 x x x x x x + = + + =

( ) ( )1 2 0 1, 2 x x x x = = =

Vy h c hai nghim l (x;y)=(1;1) v (x;y)=(2;2)

Bi 6. Gii phng trnh: ( ) ( )4 22 1

1 1log 1 log 2 1

log 4 2x

x x+

+ = + + (D b 1A 2007)

Gii:K: x>1

( ) ( ) ( ) ( )4 4 41

1 log 1 log 2 1 log 2

2 x x x + + + =

( ) ( )4

1 2 1 1log 1

2 2

x xv x

x

+ = >

+

22 12 1

2

x xv x

x

= >

+

2 52 3 5 0 12

x x v x x = > =

BI TP P DNG:

1)( )22

24

log 6 72

11 log4

x x

x x

+

+ +

2)2 3 2 3

log log log log x x x x+

3) 2 2log 3 log 52 x x x+ =

4) ( ) ( )2 23 5log 15 log 45 2 x x x x+ = 5) ( ) ( )0.2 3 5log 2 log log 2 x x x + +

6) ( ) ( ) ( ) ( )22 33 log 2 4 2 log 2 16 x x x x+ + + + + =

7) ( ) ( )2 23

1log 3 1 2 log 1

log 2x

x x+

+ = + +

8) CMR: vi mi a>0, h phng trnh sau c nghim duy nht:

( ) ( )ln 1 ln 1x ye e x y

x y a

= + +

=

9) Gii h phng trnh:( ) ( )

( )2 2

2 2

2 2log 1 log,

3 81 x xy y

x y xy x y R

+

+ = +

=

(H A 2009)


46/63

2

10)Tm m phng trnh sau c ng 1nghim:

( ) ( )5 1 2 5 1 2x x

m x+ + =

11) 3 2 2 37 9.5 5 9.7 x x x x+ = +

12) ( ) ( )7 5

5 7

x x

=

13) ( ) ( )10

5 103 3 84 0x x

+ =

14) ( )3 316 6 4 8 2 0x xx x + + =

15)Tm m phng trnh sau c ng 1nghim:

2 2sin cos9 9x x m+ =

16) ( )23

log 3 1x x

x

>

17) ( )3 316 6 4 8 2 0x xx x + + =

18)Cho bt phng trnh:

( ) ( ) ( )22 2log 1 log 1 x ax a+ < + a) Gii bt phng trnh khi a=2b) Tm tt c gi tr ca a bt phng

trnh c nghim

19)2

23 9 6x x

x x x

= +

20) ( )2 23.25 3 10 .5 3 0x xx x + + =

21)Tm m phng trnh c 2 nghim tri

du:( ) ( )3 16 2 1 4 1 0x xm m m+ + + + =

22)Tm m phng trnh c nghim:

9 .3 2 1 0x x

m m + + =

23) ( )2 5 4 5 3 5 3 x x x+ + 24)Tm m h c nghim:

( ) ( )22 2

log log 1m x y x y

x y m

+ + =

=

25)Gii bt phng trnh:

2 0.5

15log log 2 2

16

x

26)Gii bt phng trnh

3 4 1 1

3 4

3 1 1log log log log

1 3 1

x x

x x

+

+


47/63

2

PH LC: MT S THI CN THAM KHO (Theo cu trc thi ca B GD& T 2010)

1:

A. PHN CHUNG:

Cu 1: Cho hm s (C) ( )( )2 21

14

y x m x= + , m l tham s.

1. Kho st v v th (C) khi m =32. nh m bit th hm s (C) ct Ox ti A v B sao cho 2 tip tuyn ti A v B vung gc.

Cu 2:

1. Gii phng trnh: 3 27cos 2 sin 2sin2

x x x+ =

2. Gii phng trnh: ( ) ( )4 4 4 2 x x x x x+ = Cu 3: Tnh gii hn:

( )2 2sin 20

log cos

lim 2 1x xx

x x

x

+

+

Cu 4: Cho hnh nn nh S c thit din qua trc SO=a l mt tam gic vung. Mt phng qua S v cng trn y ti A v B sao cho SAB u. Tnh th tch hnh cu ngoi tip hnh chp SOAB.

Cu 5: Cho x,y,z [ ]0;1 . Tm gi tr ln nht: ( ) ( ) ( )2 2 2

A x y y z z x= + +

B. PHN T CHN: (Th sinh ch c chn Cu 6 hoc Cu 7Cu 6: (Chng trnh chun)

a. Trong Oxy cho ABC c A(0;2), B(2;6), v : 3 1 0C d x y + = sao cho phn gic k t A song song

vi d. Tm ta C.

b. Trong Oxyz vit phng trnh ng thng qua A(0;1;2) ct1

1 1:

1 1 1

x y zd

= =

v hp vi

2

1 2 4

2 1 1

x y zd

+ = = =

mt gc 600

c. Cho ( ) ( ) ( )1

1 1 01 1 ... 1 ,

n n n

n na x a x a x a x x R

+ + + + = . Tm n bit

2 3 1231a a a+ + =

Cu 7: (Chng trnh nng cao)

a. Trong Oxy tm ( )2 2

: 16 3

x yM E + = bit khong cch t M n d: x+y=0 l ln nht

b. Trong Oxyz vit phng trnh mt phng qua M(1;2;2) v ct Ox, Oy, Oz ti A,B,C sao cho:

2 2 2 2

1 1 1 1

OA OB OC OM + + =

c. Bng cch khai trin: ( )2

1n

i+ hy chng minh: ( )0 2 4 22 2 2 2... 1 2 cos2

n n n

n n n n

nC C C C

+ + = ,

( ), 0n N n > .


48/63

3

2:

A. PHN CHUNG:

Cu 1: Cho hm s (C) 4 22

9 y x x= +

1. Kho st v v th (C)

2. Tm trn th (C) cc im A bit tip tuyn ti A ct (C) ti B v C sao cho AB=AC ( B,C khc A)Cu 2:

1.Gii phng trnh: ( ) ( )1 3 cos sin 3 cos cos 1 x x x x + =

2.Gii h phng trnh:2 23

2 2 2

3 4 5

x y x y

x x y

+ =

+ + =

Cu 3: Tnh tch phn:2

1 1 ln

edx

x x x+

Cu 4: Cho lng tr ng ABC.ABC c AB=a; BC=b v ABC vung cn ti A. Tnh th tch lng

tr.

( )2a b a< <

Cu 5: Cho [ ], 1;2 .x y Tnh gi tr ln nht v nh nht:

( ) ( )2 2 2 21 1 1 1

4 A x y x y x y x y

= + + +

B. PHN T CHN: (Th sinh ch c chn Cu 6 hoc Cu 7

Cu 6: (Chng trnh chun)

a. Trong Oxy tm ( )2 2

: 1

6 3

x yM E + = bit gc F1MF2 bng 60

0.

b. Trong Oxyz vit phng trnh tham s ng thng song song vi (P): 2x+2y-z-3=0 v ct hai

ng thng 12 1

:2 1 1

x y zd

= =

v 2

1 1:

1 2 1

x y zd

+= =

ti A v B sao cho AB=3

c. Gieo ng thi 3 con xc xc, tnh xc sut tch 3 s nt xut hin l 1 s chn.

Cu 7: (Chng trnh nng cao)a. Trong Oxy vit phng trnh chnh tc hypebol qua M(2;1) tha gc F1MF2 bng 60

0

b. Trong Oxyz vit phng trnh mt phng hp vi (Oxy) mt gc 450, song song vi Ox v cch Ox mt

khong bng 2

c. Cho z= 3 i+ . Tm s t nhin n>0 sao cho nz l s nguyn dng b nht.


49/63

4

3:

A. PHN CHUNG:

Cu 1: Cho hm s (C)2mx

yx m

+=

+

1. Kho st v v th (C) khi m =-1

2. Tm trn th (C) ct Ox ti A, Ct Oy ti B sao cho 2 tip tuyn ti A v B song songCu 2:

3.Gii phng trnh: 1cos 2 cos 3 sin2

x x x+ + =

4.Gii phng trnh: ( ) ( )2 22 3log 12 .log 12 2 x x x x+ =

Cu 3: Tnh tch phn:( )

2

4

0

sin3

1 cos

xdx

x

+

Cu 4: Tnh th tch hnh chp S.ABCD c y l hnh ch nht, chiu cao SA=a hp vi (SBC) v(SBD) cc gc 450 v 300

Cu 5: nh m h sau c nghim:

22

2

1

2 4

y x xy

x x y m

+ =

+ =

B. PHN T CHN: (Th sinh ch c chn Cu 6 hoc Cu 7)


a. Vit phng trnh ng trn i qua gc ta v ct Ox, Oy ti A,B sao cho AB=4 2 . Bit rngtm ng trn thuc d:x+y-4=0

b. Trong Oxyz vit phng trnh mt phng (P) qua M(1;1;0), song song vi

3: 4 5 3

x y z

d

= = v cchgc ta mt khong bng 1.

c. Tm ,a b R bit phng trnh 31 5

a b

z z+ =

+ c 1 nghim 1

5

1 2

iz

i=

+. Tm nghim cn li.

Cu 7: (Chng trnh nng cao)a. Tm ta 3 nh ABC vung cn ti A c trc i xng l x-2y+1=0; ; A Ox B Oy v

: 1 0C d x y + = .

b. Vit phng trnh tham s ca ng thng d qua M(1;2;0), song song vi (P):2x-y+z-1=0 v hp vi(Q): x+y+2z-1=0 mt gc 600

c. Trong hp ng 15 vin bi gm 4 bi , 5 bi xanh v 6 bi vng. Tnh xc sut chn c 4 vin bi c 3 mu.


50/63

5

4:

A. PHN CHUNG:

Cu 1: Cho hm s3

2

3

xy x= + c th (C)

1. Kho st v v th (C)2. Vit Phng trnh ng thng d qua gc ta O v ct (C) ti A v B (khc O) saocho 2 tiptuyn ca (C) ti A v B vung gc.Cu 2:

5.Gii phng trnh: tan tan sin 2 1 2sin 24 2 2 x x x x+ ++ = 6.Gii bt phng trnh: 2 2 3

2 2 5

x xx

x x

+

Cu 3: Tnh tch phn:44

4 4

0

sin

sin cos

xdx

x x

+

Cu 4: Tnh th tch hnh chp S.ABCD c y l hnh vung chiu cao SA. Bit SC=2a hp vi (SAB

mt gc 300.

Cu 5: Cho a,b,c>0 v a+b+c=1. Tm gi tr nh nht:2 2 2

3 3 3

3

a b c A a b c

+ += + +



I/ Trong Oxyz cho A(2;3;-1), B(5;-3;2) v (P): x+y+z-3=0:a.Vit phng trnh tham s ng thng d vung gc vi (P) v ct ng thng AB ti I sao cho

2 0 AI BI + =

b. Tm ( )M P sao cho AM2+2BM2 nh nht

II/ Hy phn phi 2010 im ln 2 ng thng song song sao cho tng s tam gic thu c l lnnht.

Cu 7: (Chng trnh nng cao)I/a Vit phng trnh ng trn trong Oxy i qua A(2;1), Tm thuc Oy v ct Ox ti B v C sao cho gcBAC bng 600b. Trong Oxyz cho A(0;1;2), B(1;-1;1), C(-1;3;0). Vit phng trnh tham s ng thng d vung gcvi (ABC) v ct (ABC) ti trc tm H ca ABC.

II/ nh m bit th hm s( )2 1 2 1 x m x m

yx m

+ + =

tip xc vi Ox.


51/63

6

5:

A. PHN CHUNG:

Cu 1: Cho hm s3

1

xy

x

=

+c th (C)

1. Kho st v v th (C)

2. Cho A(0;2). Tm trn (C) im M sao cho AM ngn nht.Cu 2:

1. Gii phng trnh: 2 2 3cos cos cos 3 cos 34

x x x x + =

2. Gii h phng trnh:2 2

2 2

1 13

1 11

x yx y

x y xy

+ + + =

+ = +

Cu 3: Tnh tch phn:

4

3

23

4

ln

1

x xdx

x+

Cu 4: Cho hnh chp S.ABCD c (SAB) (ABC), ABC u v ABC vung cn ti A. Tnh th tch

mt cu ngoi tip hnh chp Bit SC= 2a

Cu 5: Cho a,b,>0 v1 1

1a b

+ = . Tm gi tr nh nht:( )2 225

1 1 4

a b abA

a b a b= + +

+



I/ Trong Oxyz cho A(2;-1;2), B(3;-3;3); C(1;-2;4) v (P): 2x-3y+z+1=0:

a.Vit phng trnh tham s ng thng d i qua tm ng trn ngoi tip ABC v vung gcvi (P)

b. Tm ( )M P sao cho AM2+2BM2+CM2 nh nht

II/ Tm ,a b R bit 2 3 4 2009... Z i i i i i= + + + l nghim ca phng trnh 11 1

a b

z z+ =

+ . Tm

nghim cn li.


I/Trong Oxyz cho 1 : 1 2

2

x t

d y t

t

=

= +

+

; 21

:1 1 1

x y zd

= =

a Tm 1A d bit khong cch t A n d2 bng 6

b. Vit phng trnh mt phng (P) cha d2 v hp vi d1 mt gc 300

II/ Gii h phng trnh:

3 3

3

log log 22 6

log log 1

x

x y

x

y

y x

+ = + =


52/63

7

6:

A. PHN CHUNG:

Cu 1: Cho hm s (C)4

2 14

x y mx m= + + , m l tham s.

1. Kho st v v th (C) khi m =1

2. nh m bit th hm s (C) c 3 im cc tr to thnh tam gic c trc tm l gc ta Cu 2:

1. Gii phng trnh: sin 2 cos 2 tan6 3 4

x x x

+ + + = +

2. Gii h phng trnh:( ) ( )3 3 3 3

2 3

1 1 1 18

log log 12 3

x y x y x y x y

x y

+ + + + + =

=

Cu 3: Tnh tch phn:2

3

10 x

xdxI

e+

=

Cu 4: Tnh th tch hnh lng tr u ABCD.ABCD bit AC=a v gc gia BD v CD bng 600.

Cu 5: Cho a,b,c>0 v1 1 1

1a b c

+ + = . Tm gi tr ln nht:3 3 3 3 3 3

b c c a a bA

b c c a a b

+ + += + +

+ + +


a. Trong Oxy cho ABC vung cn ti A c din tch bng 2, bit 1 2 1 0 A d x y = + = v

2, : 2 0B C d x y + = . Tm ta A,B,C vi xA, xB>0.

b. Trong Oxyz vit phng mt phng (P) qua A(0;1;2), B(1;3;3) v hp vi ( ) : 2 0Q x y z = mt gcnh nht.

c. Tm s t nhin n tha: 3 2 31 1

1

7n n nC C A+ + =


a. Trong Oxy cho hai ng trn ( ) 2 2: 2 2 0mC x y mx my m+ + = v ( )2 2: 3 1 0C x y x+ + = . nh m

bit s tip tuyn chung ca hai ng trn l mt s l.

b. Trong Oxyz vit phng trnh ng thng d song song vi ( ) : 2 1 0P x y z+ + = v ct 2 ng thng

Ox v2 1

:

2 1 1

x y z + = =

ti 2 im A,B sao cho AB ngn nht.

c. Gii phng trnh: 4 2 1 0z z+ + = , z C .


53/63

8

7:

A. PHN CHUNG:

Cu 1: Cho hm s (C) 3 23 y x ax b= + , (1) ( ), 0a b >

1. Kho st v v th (C) khi a=1 b=42. nh a,b bit th hm s (C) c 2 im cc tr A v B sao cho OAB vung cn.

Cu 2:

3. Gii phng trnh: 2tan2 1 tan . tan2 sin 3

xx x

x

+ =

4. Gii h phng trnh:2 2 2 2

1 1 1

2

5 2 1

2

x y xy

x y x y

+ =

+ = +

Cu 3: Tnh gii hn:( )0

1lim

ln 1 sin

x

x

e x

x

+

+

Cu 4: Cho hnh chp S.ABCD chiu cao SA=2a, y l hnh thang vung ti A v B c AB=BC=aAD=2a. Mt phng qua trung im M ca SA cha CD, ct SB ti N. Tnh din tch t gic CDMN.

Cu 5: nh m bt phng trnh c nghim:

( )2

1ln 2 1

2 x x m x m

mx x+ + +

. Tm nghim tng ng


a. Trong Oxy cho ( ) ( ) ( )7;1 , 3; 4 , 1; 4 A B C . Vit phng trnh ng trn ni tip ABC.

b. Trong Oxyz vit phng trnh mt phng (P) qua gc ta , song song vi 1 1 2:1 2 1

x y zd + = =

v

hp vi1 2

:2 1 1

x y z+ = = mt gc 600

c. Tm h s ca 3x trong khai trin thnh a thc ca biu thc: ( )6

2 1x x+ .

Cu 7: (Chng trnh nng cao)a. Trong Oxy cho ng trn ( ) 2 2: 6 5 0C x y x+ + = . Tm M thuc trc tung sao cho qua M k c ha

tip tuyn ca (C) m gc gia hai tip tuyn bng 600

b. Trong Oxyz Cho ( )2;1;0M v ng thng d c phng trnh1 1

2 1 1

x y z += =

. Vit phng trnh

chnh tc ca ng thng i qua im M, ct v vung gc vi ng thng d.

c. Tm h s ca 3x trong khai trin thnh a thc ca biu thc: ( )5

2 1x x+ .


54/63

9

8:

A. PHN CHUNG:

Cu 1: Cho hm s (C)1

1

mxy

x

+=

+

1. Kho st v v th (C) khi m =-1

2. nh m bit tip tuyn ti im c nh ca h th (C) cch I(1;0) mt khong ln nhtCu 2:1. Gii phng trnh: 2 2sin sin 2 .sin 4 cos 2 x x x x+ = 2. Gii bt phng trnh : ( )2 3 2 32 2 7 2 2 15 x x x x+ + +

Cu 3: Tnh th tch vt th trn xoay sinh ra bi hnh phng to bi ( )1 1

: 1 1 ,C yx x

= + + trc Ox

v 2 ng thng x=1; x=2 quay quanh Ox.Cu 4: Cho hnh vung ABCD cnh a v hai ng thng 1 2;d d ln lt qua A v C v vung gc v

mt phng (ABCD). Ly1 2, N M d d sao cho , AM CN

cng chiu v c tng di bng 6a. Tnh

th tch t din MNBD

Cu 5: Gii h phng trnh:

2

2

1 1

1 ln

1 1

1 ln

xy x x y y

xy y y x x

+ = +

+ + = + +


Cu 6: (Chng trnh chun)a. Trong Oxy cho A,B l hai im trn ( ) 2:P y x= sao cho OAB vung ti A. Tm ta A,B

( )0Ay < bit OB ngn nht.

b. Trong Oxyz vit phng trnh mt phng (P) qua gc ta v song song vi1 1 2

:2 2 1

x y zd

= =

v cch d mt khong bng 1.

c. Cho a gic li n nh, bit rng s tam gic c nh v cnh chung vi a gic l 70. Tm s tamgic c nh chung v khng c cnh chung vi a gic.

Cu 7: (Chng trnh nng cao)a. Trong Oxy vit phng trnh chnh tc elip (E) qua M(2;1) sao cho

1 2. MF MF nh nht.

b. Trong Oxyz vit phng trnh mt phng (P) qua gc ta v ln lt hp vi 2 mt phng

( ) ( ): 1 0 v : 2 1 0Q x z R x y z+ = + + = cc gc 300 v 600

c. Tnh gi tr: ( )( )2 2008 2 3 20081 2 3 ... 2009 1 2 3 4 ... 2009 Z i i i i i i i= + + + + + + + .


55/63

10

9:

A. PHN CHUNG:

Cu 1: Cho hm s (C) ( ) ( )2 1 y x m x x= + 1. Kho st v v th (C) khi m =32. nh m bit (Cm) ct Ox ti A, ct Oy ti B sao cho hai tip tuyn ca (Cm) ti A v B vung gc.

Cu 2:

1. Gii phng trnh: 1 sin costan1 sin cos

x xx

x x

+=

+ +

2.Gii bt phng trnh : ( ) ( )2 2 22

log log log 0.25

7 5 2 3 2 2xx

x+

+ =

Cu 3: Tnh din tch hnh phng gii hn bi: ( ) 2: 2 3C y x x= v : 1d y x= +

Cu 4: Cho hnh chp S.ABCD chiu cao SA=a, y l hnh vung cnh a. chng minh AI (SBD) av2tnh th tch t din SIBD, bit I l trung im SC.

Cu 5: Tm gi tr nh nht tham s m h:2

2

1 13

2

x y

x y m

+ =

+ =c nghim x,y>0. Tm nghim tng

ng.


Cu 6: (Chng trnh chun)a. Trong Oxy cho ABC c ng cao v trung tuyn k t A l 2 4 0Ah x y= + + = , 2 0Am y= = v

ng trung tuyn k t B l : 3 11 21 0B

m x y+ + = . Tnh gc C

b. Trong Oxyz cho 1 22 1 2

: ,d : 21 2 11

x t x y z

d y t

z t

=

= = =

= +Chng minh rng c v s mt phng (P) cha

d2 v song song vi d1. Vit phng trnh (P) sao cho d2 l hnh chiu vung gc ca d1 ln (P)

c. Tm , x y R tha:

( ) ( )2

1 1 1

2 2 1 x y i y xi i =

+ + + +


a. Trong Oxy cho ( ) ( )

2 2

2 2: 1 , 0

x y

H a ba b = >

c hai tiu im l 1F 2; F . ng thng d qua 2; F vung gcOx v ct (H) ti M v N sao cho

1F MN u. Tm tm sai ca (H) v vit phng trnh (H) nu bit din

tch 1 4 3F MN =

b. Trong Oxyz cho A(-1;2;2), B(0;3;0). Hy tm trong (P) sao cho ABC u.

c. Mt ng thng tip xc vi th hm s3 3

4

xy

x= + v ct 2 ng tim cn ti A v B. Tnh din

tch OAB.


56/63

11

10:

A. PHN CHUNG:

Cu 1: Cho hm s ( ) ( )4

21 , 12

x y m x m

= + + c th (C) . m l tham s.

1. Kho st v v th (C) khi m=0

2. Chng minh rng th hm s (1) lun i qua 2 im A v B c nh. nh m bit 2 tip tuyn tA v B hp nhau gc 600Cu 2:

3. Gii phng trnh: 4sin 2 sin 1 3 sin 2 cos 23

x x x x

+ = +

4. Gii h phng trnh:2

2

4 8

3 12

x xy y

xy y x

+ =

+ + =

Cu 3: Tnh din tch hnh phng gii hn bi:lnx x

xy

e += , trc Ox v hai ng thng x=1;x=4.

Cu 4: Tnh th tch hnh chp S.ABC bit SA, SB, SC i mt hp vi nhau gc 600 v c di ln

lt l a, 2a, 3a.Cu 5: nh m phng trnh ( ) ( )( )2 3log 2 4 1 log 1 3 x m m x x + = + c nghim duy nht. Tm

nghim duy nht .



I/ Trong Oxyz cho2 1

:2 1 1

x y zd

+ = = v (P): x-y-1=0:

a.Vit phng trnh tham s ng thng d l hnh chiu vung gc ca d ln (P). Tnh gc gia

d v d.b. Gi A l giao im ca (P) v d. Vit phng trnh cc mt cu tip xc (P) ti A v ct d ti B

sao cho AB= 6

II/ Gii phng trnh:3

3 2 3 2

3 1log log log log

23

xx x

x

= +


I/Trong Oxyz cho A l giao im ca 1 : 1 2

2

x t

d y t

t

=

= +

+

v mt phng (P):x-2y+z=0

a Vit phng trnh chnh tc ng thng qua A vung gc vi d v hp vi (P) mt gc 300

b. Vit phng trnh mt cu c tm I thuc d, i qua A v ct P mt ng trn di 2 2

II/ Tm ( )0;2 bit th hm s( )2 2 cos 3 sin

1

x xy

x

+ + +=

c hai im cc tr l A v B

sao cho AB di nht, ngn nht.


57/63

12

11:

A. PHN CHUNG:

Cu 1: Cho hm s ( )2

, 11

xy

x=

c th (C) .

1. Kho st v v th (C) ca hm s (1)

2. Tm M trn (C) bit tip tuyn ti M to vi 2 tim cn ca (C) mt tam gic c chu vi b nht.Cu 2:

5. Gii phng trnh: 216sin 4cos 4 3 cos sin x x x x+ = + 6. Gii phng trnh: ( )35 5 2 x x x =

Cu 3: Tnh th tch vt th trn xoay sinh ra bi hnh trn ( ) ( ) ( )2 2

: 3 1 1C x y + = quay quanh trc

Oy.

Cu 4: Cho t din ABCD c AB=a, AC= 2a , AD=2a. ng thng AC hp vi AB,AD cc gc 450

AB hp vi AD gc 600. Tnh t s th tch ca t din v hnh cu ngoi tip t din.

Cu 5: Cho 2 2 2 1.a b c+ + = Chng minh rng: 3 3 3 3 1a b c abc+ + .



a. Trong Oxyz vit phng trnh mt phng (P) i qua H(1;2;3) v ct Ox, Oy, Oz ln lt ti A, B, C

sao cho H l trc tm ABC

b. Trong Oxyz vit phng trnh mt cu tm I Oz, i qua A(1;1;1) v ct (Oxy) mt ng trn di 2

c. Gii phng trnh : 0 1 2 22 3 4 .... 120 , xx

xC C C C N

+ + + + =

Cu 7: (Chng trnh nng cao)I/Trong Oxyz cho A(3;0;0) B(1;-2;8) v mt phng (P):x-2y+2z+6=0

a Tm M(P) sao cho AM BM +

nh nht.

b. Vit phng trnh mt phng (Q) qua A, B v ct (P) theo giao tuyn d hp vi AB gc 900

II/ Gii h phng trnh :

2 2

3 5 5 3

4 2 5.4

log log log .log

x x y x y

y xy xy

x y x y

+ = + =


58/63

13

12:

A. PHN CHUNG:

Cu 1: Cho hm s (C) ( ) ( )3

2 162 2 13 3

x y mx m x

= + +

1. Kho st v v th (C) khi m =02. Chng minh rng (Cm) lun tip xc vi 1 ng thng c nh ti 1 im c nh.Cu 2:

3. Gii phng trnh: ( )sin 3 sin 3 cos 1 x x x+ = 4.Gii bt phng trnh : 2

0.522 2

4log log 0, 25 log

xx

x+

Cu 3: Tnh tch phn:

1

4

01 2

x I dx

x=

Cu 4: Cho hnh tr c chiu cao bng bn knh y v bng a. Ly trn cc gn trn y (O) v(O) cc im A, B sao cho AB=2a. tnh gc gia hai ng thng OA, OB v th tch t din OOAB

Cu 5: Cho a,b>0 v1 1

1a b ab

+ =+

. Tm gi tr nh nht:2 2a b ab

Pab a b

+= +

+



a. Trong Oxy cho ABC c tm ng trn ngoi tip l I(2;1), AOy v ng thng

BC:3x y 10 0 = . Tm ta A,B,C bit gc BAC bng 450 v 0A By y> > b. Trong Oxyz cho A(0;1;0), B(1;-2;2). Hy vit phng trnh mt phng (P) qua O, B v cch A mt

khong bng2

2

c. Gii phng trnh : 44 1 0z + =

Cu 7: (Chng trnh nng cao)a. Trong Oxy cho ( ) 2: 2P y x= c hai tiu im l F . ng thng d quay quanh F ct (P) ti M,N.

Chng minh rng 1 1 MF NF

+ khng i.

b. Trong Oxyz vit phng trnh tham s ng thng qua M(1;-2;2). d OM v d hp vi Oy mt gc450

c. Tm h s ca 6x trong khai trin thnh a thc ca biu thc: ( ) ( )1 21 1

nnP x x x

+

= + + + . Bit h s ca

10x bng 10.


59/63

14

PH LC II: Cch gii nhanh bi ton bng my tnh b ti.Php chia theo s Horner.Trong cc k thi quan trng c mn ton, my tnh b ti c php s dng v tr thnh cng c khng

th thiu i vi th sinh. Tuy nhin t ai c th tn dng c ti a cc chc nng ca my tnh tronggii ton. Nay ti xin gii thiu mt s phng php tm nghim bng chc nng SOLVE ca my tnh. Bivit c vit vi my fx-570ES v ti cng khuyn cc em tp lm quen s dng my ny trong qu

trnh gii ton.

VD1. Tm nghim

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