Kien Thuc Tong Hop

8/8/2019 Kien Thuc Tong Hop

1/55

L THUYT TON LTH Cao Hong Nam

Trang 1

KHO ST HM S

Vn 1: N TPCNG THCI. Tam thc bc hai:

x , 2ax bx c 0

a b 0

c 0

a 0

0

x , 2ax bx c 0

a b 0

c 0

a 0

0

Cho phng trnh : ax2 + bx + c = 0Gi sphng trnh c 2 nghim 1 2x ; x th:

1 2

bS x x ;

a

1 2

cP x .x

a

Pt c 2 nghim phn bit a 00

Pt c nghim kp a 00

Pt v nghima 0

a 0b 0

0c 0

Pt c 2 nghim tri du P 0 Pt c 2 nghim cng du 0

P 0

Pt c 2 nghim phn bit cng dng0

P 0

S 0

Pt c 2 nghim phn bit cng m0

P 0

S 0

II. a thc bc ba: Cho phng trnh : ax3 + bx2 + cx + d = 0Gi sphng trnh c 3 nghim 1 2 3x ; x ; x th:

1 2 3

bS x x x ;

a 1 2 2 3 3 1

cx .x x .x x .x ;

a

1 2 3

dP x .x .x

a

III.o hm:BNG O HM

(kx) ' k (ku)' k.u '

1(x ) ' .x

1(u ) ' .u '.u .

1( x ) '

2 x

u '( u ) '

2 u

'

2

1 1

x x

'

2

1 u '

u u

(sin x) ' cosx (sin u) ' u ' .cosu

(cosx) ' sin x (cosu) ' u' .sin u

2

1(tan x) '

cos x

2

u '(tanu)'

cos u

2

1(cot x)'

sin x

2

u '(cotu)'

sin u

x x

(e ) ' e u u

(e ) ' u ' .e 1

(ln x)'x

u '

(lnu)'u

a1

log x 'xlna

au '

log u 'ulna

x x(a ) ' a .lna u u(a )' u'.a .lna

Quy tc tnh o hm

(u v) = u v (uv) = uv + vu

2

u u v v u

v v

(v 0) x u xy y .u

o hm ca mt s hm thng dng

1.

2

ax b ad bcy y '

cx d cx d

2.

2 2

2

ax bx c adx 2aex be cdy y '

dx e dx e


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th c mt tim cn ng ld

xc

v mt

tim cn ngang la

yc

. Giao im ca hai tim

cn l tm i xng ca th hm s.

Cc dng th:adbc > 0 adbc < 0

5. Hm s hu t2

ax bx cy

a 'x b '

( a.a ' 0, t khng chia ht cho mu)

Tp xc nh D = b 'R \a '

.

th c mt tim cn ng lb '

xa '

v mt

tim cn xin. Giao im ca hai tim cn l tm

i xng ca th hm s. Cc dng th:

y = 0 c 2 nghim phn bit

a 0 a 0

y = 0 v nghim

a 0 a 0

CC BI TON LIN QUANKHO ST HM S

Vn 1. STIP XC GIA HAI

NG, TIP TUYN CANG CONG

ngha hnh hc ca o hm: o hm cahm s y = f(x) ti im x0 l h s gc ca tiptuyn vi th (C) ca hm s ti im

0 0 0M x ; f (x ) . Khi phng trnh tip tuyn

ca (C) ti im 0 0 0M x ; f (x ) l:

yy0 = f(x0).(xx0) (y0 = f(x0))

Dng 1: Lp phng trnh tip tuyn cang cong (C): y = f(x)

Bi ton 1: Vit phng trnh tip tuyn ca(C): y =f(x) ti im 0 0 0M x ; y

Nu cho x0 th tm y0 = f(x0).

Nu cho y0 th tm x0 l nghim ca phngtrnh f(x) = y0.

Tnh y = f (x). Suy ra y(x0) = f (x0).

Phng trnh tip tuyn l:

yy0 = f (x0).(xx0)

Bi ton 2: Vit phng trnh tip tuyn ca(C): y =f(x), bit c h sgc k cho trc.

Cch 1: Tm to tip im.

Gi M(x0; y0) l tip im. Tnh f (x0).

c h s gc k f (x0) = k (1)

Gii phng trnh (1), tm c x0 v tnh y0= f(x0). T vit phng trnh ca .

Cch 2:Dng iu kin tip xc.

Phng trnh ng thng c dng:

y = kx + m.

tip xc vi (C) khi v ch khi hphngtrnh sau c nghim:

f (x) kx m

f '(x) k

(*)

Gii h(*), tm c m. T vit phngtrnh ca .

0 x

y

0 x

y


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Ch : H s gc k ca tip tuyn c thc cho gin tip nh sau:

to vi chiu dng trc honh gc thk = tan song song vi ng thng

d: y = ax + b th k = a

vung gc vi ng thngd: y = ax + b (a 0) th k =

1

a

to vi ng thng d: y = ax + b mtgc th

k atan

1 ka

Bi ton 3: Vit phng trnh tip tuyn ca

(C): y = f(x), bit i quaim A AA(x ;y ) .Cch 1: Tm to tip im.

Gi M(x0; y0) l tip im. Khi :y0 = f(x0), y0 = f (x0).

Phng trnh tip tuyn ti M:yy0 = f (x0).(xx0)

i qua A AA(x ;y )nn:

yAy0 = f (x0).(xAx0) (1)

Gii phng trnh (1), tm c x0. Tvit phng trnh ca .Cch 2: Dng iu kin tip xc.Phng trnh ng thng i qua

A AA(x ;y )v c h s gc k: yyA = k(xxA)

tip xc vi (C) khi v ch khi hphngtrnh sau c nghim:

A Af (x) k(x x ) y

f '(x) k

(*)

Gii h(*), tm c x (suy ra k). T vitphng trnh tip tuyn .

Dng 2: Tm iu kin hai ng tip xc

iu kin cn v hai ng (C1): y = f(x)v (C2): y = g(x) tip xc nhau l hphngtrnh sau c nghim:

f(x) g(x)

f '(x) g '(x)

(*)

Nghim ca h (*) l honh ca tip imca hai ng .

Dng 3: Tm nhng im trn ng thng dm t c th vc 1, 2, 3, tip

tuyn vi th (C): y = f(x)

Gi s d: ax + by +c = 0. M(xM; yM) d.Phng trnh ng thng qua M c h s

gc k: y = k(xxM) + yM tip xc vi (C) khi h sau c nghim:

M Mf (x) k(x x ) y (1)

f '(x) k (2)

Th k t(2) vo (1) ta c:f(x) = (xxM).f (x) + yM (3)

S tip tuyn ca (C) v t M = S nghimx ca (3)

Dng 4: Tm nhng im m t c th vc 2 tip tuyn vi th (C): y = f(x)v 2 tip tuyn vung gc vi nhau

Gi M(xM; yM).Phng trnh ng thng qua M c h s

gc k: y = k(xxM) + yM tip xc vi (C) khi h sau c nghim:

M Mf (x) k(x x ) y (1)

f '(x) k (2)

Th k t (2) vo (1) ta c:f(x) = (xxM).f (x) + yM (3)

Qua M vc 2 tip tuyn vi (C) (3)c 2 nghim phn bit x1, x2. Hai tip tuyn vung gc vi nhau

f (x1).f (x2) =1T tm c M.

Ch : Qua M vc 2 tip tuyn vi (C) sao

cho 2 tip im nm v hai pha vi trc honh

th 1 2

(3)co2nghiem phan bietf(x ).f(x ) < 0

Vn 2. STNG GIAO CACC TH

1.Cho hai th (C1): y = f(x) v (C2): y = g(x).tm honh giao im ca (C1) v (C2)

ta gii phng trnh: f(x) = g(x) (*) (gi lphng trnh honh giao im).S nghim ca phng trnh (*) bng s giao


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im ca hai th.2. th hm s bc ba

3 2y ax bx cx d (a 0) ct trc honh ti 3im phn bit

Phng trnh 3 2ax bx cx d 0 c 3

nghim phn bit. Hm s 3 2y ax bx cx d c cc i, cc

tiu v C CTy .y 0 .

Vn 3. BIN LUN S NGHIMCA PHNG TRNH BNG

THC sca phng php:

Xt phng trnh: f(x) = g(x) (1) S nghim ca phng trnh (1) = S giao

im ca (C1): y = f(x) v (C2): y = g(x) Nghim ca phng trnh (1) l honh

giao im ca (C1): y = f(x) v (C2): y = g(x) bin lun s nghim ca phng trnhF(x, m) = 0 (*) bng th ta bin i (*) v mttrong cc dng sau:Dng 1: F(x, m) = 0 f(x) = m (1)

Khi (1) c thxem l phng trnh honh

giao im ca hai ng: (C): y = f(x) v d: y= md l ng thng cng phng vi Ox Da vo th (C) ta bin lun sgiao im

ca (C) v d. T suy ra s nghim ca (1)

Dng 2: F(x, m) = 0 f(x) = g(m) (2) Thc hin tng t, c tht g(m) = k. Bin lun theo k, sau bin lun theo m.

c bit: Bin lun s nghim ca phngtrnh bc ba bng th

C sca phng php:Xt phng trnh bc ba:

3 2

ax bx cx d 0 (a 0) (1) c th (C) S nghim ca (1) = Sgiao im ca (C)vi trc honh

Bi ton 1: Bin lun s nghim ca phngtrnh bc 3

Trng hp 1: (1) ch c 1 nghim (C) vOx c 1 im chung

C CT

f khong co cc tr (h.1a)f co 2 cc tr

(h.1b)y .y >0

Trng hp 2: (1) c ng 2 nghim (C)tip xc vi Ox

C CT

f co 2 cc tr(h.2)

y .y =0

Trng hp 3: (1) c 3 nghim phn bit (C) ct Ox ti 3 im phn bit

C CT

f co 2 cc tr(h.3)

y .y 0, x > 0

a.f(0) < 0 (hay ad < 0)

Trng hp 2: (1) c 3 nghim c m phn

y

x

m A(C)(d) : y = m

yC

yCTx


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bit (C) ct Ox ti 3 im phn bit c honh m

C CT

C CT

f co 2 cc try .y < 0

x < 0, x < 0

a.f(0) > 0 (hay ad > 0)

Vn 4. HM S C CHA DUGI TR TUYT I

1. th hm s y = f x (hm s chn)

Gi (C): y f(x) v 1(C ) : y f x ta thc hincc bc sau:

Bc 1. V th (C) v ch gi li phn th nm pha bn phi trc tung.

Bc 2. Ly i xng phn thbc 1

qua trc tung ta c th (C1).

2. th hm s y = f(x)

Gi (C): y f(x) v 2(C ) :y f (x) ta thc hin

cc bc sau:Bc 1. V th (C).Bc 2. Gi li phn th ca (C) nm pha

trn trc honh. Ly i xng phn th nmpha di trc honh ca (C) qua trc honh tac th (C

2).

3. th hm s y = f x

Gi 1(C ) : y f x , 2(C ) : y f (x) v

3(C ) : y f x . D thy v (C3) ta thc hincc bc v (C1) ri (C2) (hoc (C2) ri (C1)).

Vn 5. IM C BIT TRN TH CA HM S

Dng 1: Tm cp im trn th

(C): y = f(x) i xng qua ng thngd: y = ax + b

C sca phng php: A, B i xng nhauqua d d l trung trc ca on AB

Phng trnh ng thng vung gc

vi d: y = ax + b c dng: :1

y x ma

Phng trnh honh giao im ca v

(C): f(x) =1

x m

a

(1)

Tm iu kin ca m ct (C) ti 2im phn bit A, B. Khi xA, xB l ccnghim ca (1).

Tm totrung im I ca AB. Tiu kin: A, B i xng qua d I

d, ta tm c m xA, xB yA, yB A, B.

Ch :A, B i xng nhau qua trc honh

A B

A B

x x

y y

A, B i xng nhau qua trc tung A B

A B

x x

y y

A, B i xng nhau qua ng thng y = b

A B

A B

x x

y y 2b

A, B i xng nhau qua ng thng x = a

A B

A B

x x 2a

y y


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Dng 2: Tm cp im trn th

(C): y = f(x) i xng qua im I(a; b)

C sca phng php:A, B i xng nhauqua I I l trung im ca AB.

Phng trnh ng thng d qua I(a; b), ch s gc k c dng: y k(x a) b .

Phng trnh honh giao im ca (C)v d: f(x) = k(x a) b (1)

Tm iu kin d ct (C) ti 2 im phnbit

A, B. khi xA, xB l 2 nghim ca (1). Tiu kin: A, B i xng qua I I l

trung im ca AB, ta tm c k xA, xB.Ch :

A, B i xng qua gc to O A B

A B

x x

y y

Dng 3: Khong cch

Kin thc c bn:1. Khong cch gia hai im A, B:

AB = 2 2B A B A(x x ) (y y )

2.

Khong cch tim M(x0; y0) n ngthng : ax + by + c = 0:

d(M, ) = 0 02 2

ax by c

a b

3. Din tch tam gic ABC:S =

22 21 1

AB.AC.sin A AB .AC AB.AC2 2

Nhn xt: Ngoi nhng phng php nu, bi

tp phn ny thng kt hp vi phn hnh hcgii tch, nh l Vi-et nn cn ch xem li cctnh cht hnh hc, cc cng c gii ton tronghnh hc gii tch, p dng thnh tho nh lVi-et trong tam thc bc hai.

LNG GIC

Vn 1: N TPI. Gc v cung lng gic:1. Gi trlng gic ca mt sgc:

06

4

3

2

Sin 01

2 2

2

3

2 1

Cos 1 3

2

2

2

1

2 0

Tan 0 3

3

1 3

Cot 3 13

3 0

2. Cung lin kt: (cos i, sin b, ph cho)

x x2

x + x

2

+ x

Sin sinx sinx cosx sinx cosx

Cos cosx cosx sinx

cosxsinx

Tan tanx tanx cotx tanx cotx

Cot cotx cotx tanx cotx tanx

II. Cng thc lng gic:1. Cng thc c bn:

2 2sin a cos a 1 tana.cot a 1

2

2

11 tan a

cos a

2

211 cot a

sin a

2. Cng thc cng:

cos( ) cos .cos sin .sin

cos( ) cos .cos sin .sin

sin( ) sins .cos cos .sin

sin( ) sins .cos cos .sin

tan tantan( )

1 tan . tan

tan tantan( )

1 tan . tan


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3. Cng thc nhn i, nhn ba:2 2 2 2

cos2 cos sin 2cos 1 1 2sin

(cos sin )(cos sin )

sin2 2sin .cos

3

cos3 4cos 3cos 3sin 3 3sin 4sin 4. Cng thc h bc:

2 21 cos2xcos x 1 sin x2

(1 cosx)(1 cosx)

2 21 cos2xsin x 1 cos x2

(1 cos x)(1 sin x)

5. Cng thc bin i tng thnh tch:x y x y

cos x cos y 2 cos cos2 2

x y x ycos x cos y 2 sin sin

2 2

x y x ysin x sin y 2 sin cos

2 2

x y x ysin x sin y 2 cos sin

2 2

6. Cng thc bin i tch thnh tng:

1cos cos cos( ) cos( )

2

1sin sin cos( ) cos( )

2

1sin cos sin( ) sin( )

2

Mt sch cn thit:4 4 2 2sin x cos x 1 2.sin x.cos x 6 6 2 2sin x cos x 1 3.sin x.cos x 8 8 4 4 2 4 4

2 2 2 4 4

4 2

sin x cos x (sin x cos x) 2sin x.cos x

(1 2sin x.cos x) 2sin x.cos x

1sin 2x sin 2x 1

8

Trong mt sphng trnh lng gic, ikhi ta phi sdng cch t nh sau:

t t tanx

Khi :

2

2 22t 1 tsin 2x ; cos2x1 t 1 t

Vn 2: PHNG TRNH LNGGIC

I. Phng trnh c bn: x k2sinx sin k

x k2

x k2cos x cos k x k2

tan x tan x k k cot x cot x k k

Trng hp c bit:

sin x 0 x k , k sin x 1 x k2 k

2

sin x 1 x k2 k 2

cos x 0 x k k 2

cos x 1 x k2 k II. Phng trnh bc hai hay bc n ca mthm lng gic:

2asin x b sinx c 0 (1) 2acos x bcosx c 0 (2) 2a tan x b tan x c 0 (3) 2a cot x a cot x c 0 (4)

Cch gii:- t t l mt trong cc hm lng gic.

Gii phng trnh theo t v d dng tm cnghim ca phng trnh cho.III.Phng trnh a.sin x b.cos x c Cch gii:

- Nu 2 2 2a b c :phng trnh v nghim- Nu 2 2 2a b c : Ta chia hai v ca

phng trnh cho 2 2a b . Pt tr thnh:

2 2 2 2 2 2

a b csin x cos x

a b a b a b

2 2

ccos .sin x sin .cos x

a b

2 2

csin(x )

a b

Lu :2 2 2 2b asin ;cos

a b a b


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Bin th:a.sin x b.cosx csin y dcos y

Trong : 2 2 2 2a b c d a.sin x b.cos x csin y (c th c.cosy )

Trong :2 2 2

a b c IV.Phng trnh

2 2a.sin x b.sin x.cos x c.cos x d Cch gii:Cch 1:

- Xt cos x 0 x k2 ,k 2

Pt tr thnh: a = d.(kim tra ng sai v kt lunc nhn nghim cosx 0 hay khng?)

- Xt cos x 0 x k2 ,k 2

Chia hai v ca phng trnh cho 2cos x . Phngtrnh tr thnh:

2 2a.tan x b.tan x c d(1 tan x)

t t tanx ta d dng gii c phng trnh.Cch 2:

Dng cng thc hbc a v phng trnh III.Ch :i vi dng phng trnh thun

nhtbc 3 hay bc 4i vi sin v costa cngc cch gii hon ton tng t.V. Phng trnh

a(sin x cos x) b.sin x.cosx c 0 Cch gii:

t t sinx cosx

iu kin: t 2 Do t 2 sin x4

Ta c: 2 2 2t sin x cos x 2sin x.cosx 2t 1

sin x.cosx2

Pt tr thnh:2t 1

a.t b c 02

Ta d dng gii c.

Ch : i vi dng phng trnha(sin x cosx) b.sin x.cos x c 0

Bng cch t t sin x cos x 2 sin x 4

ta s gii c vi cch gii hon ton tng tnh trn.

VI.Phng trnh A.B 0 Cch gii:

- Dng cc cng thc bin i a vdng A.B 0

A 0

A.B 0 B 0

Vn 3: K THUT NHN BIT

Xut hin 3 ngh n phng trnh III. Xut hin 3 v gc lng gic ln ngh n

dng bin th ca phng trnh III. Xut hin gc ln th dng cng thc tng

thnh tch a v cc gc nh.

Xut hin cc gc c cng thmk ,k ,k

4 2

th c th dng cng thc tng thnh

tch, tch thnh tng hoc cung lin kt, hoc

cng thc cng lm mt cc k ,k ,k 4 2

Xut hin 2 th ngh n phng trnh IIIhoc cng c khnng l cc v cn li nhm

c (sin x cos x) trit 2 v

t sin x cos x 2 sin x4

Khi n gin cc gc, m cha a vcphng trnh quen thuc th ngh ngay nkhnng nhm nh, nhm ca. Lu , khnng tch phng trnh bc hai theo sin (hoccos) v tch hai phng trnh bc nht.

Ch : Gc ln l gc c so ln hn 2x.Ta ch s dng cng thc nhn ba khi a biton v sinx, 2sin x hoc cosx, 2cos x .

Vn 4: GII TAM GICI. Cng thc sin, cos trong tam gic:Do A B C nn:

a. sin(A B) sinC b. cos(A B) cosC

DoA B C

2 2 2 2

nn:

a. A B Csin( ) cos2 2 2


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b. A B Ccos( ) sin2 2 2

II. nh l hm s sin:a b c

2RSinA SinB SinC

III.nh l hm s cosin:2 2 2a b c 2bccosA

IV.Cng thc ng trung tuyn:2 2 2

2

a

2b 2c am

4

V. Cng thc ng phn gic:

a

A2bc.cos

2lb c

VI.Cc cng thc tnh din tch tam gic:a

1 1 abcS ah bcsin A pr

2 2 4R

p(p a)(p b)(p c)

I S

Vn 1: PHNG TRNH BCHAI

I. Phng trnh bc haiCho phng trnh bc hai 2ax bx c 0

(a 0) c 2b 4ac .

0 : phng trnh v nghim. 0 :phng trnh c nghim kp bx

2a .

0 : (3) c hai nghim phn bit2

1,2

b b b 4acx

2a 2a

II. nh l Viet (thun v o) Cho phng trnh 2ax bx c 0 c hainghim

1 2x , x th

1 2

1 2

bS x x

a

cP x .x

a

Nu bit S x yP x.y

th x, y l nghim ca

phng trnh2

X SX P 0 .III.Bng xt du ca tam thc bc hai

f(x) = ax2 + bx + c (a 0)

0 :

x y Cng du a

0 :

x 0

x

y Cng du a 0 Cng du a0 :

x 1

x 2

x

y Cng 0 tri 0 Cng

IV.Cch xt du mt a thc: Tm nghim ca a thc gm c nghim

t v nghim mu (nu a thc l phn thc) Lp bng xt du

Xt du theo quy tc Thng cng, li, chn khngCh : Khng nhn nhng im m hm s

khng xc nh.


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Vn 2: PHNG TRNH BCCAO

I. Phng trnh bc 3:3 2ax bx cx d 0(a 0)

Bc 1: nhm 1 nghim x Bc 2: chia 3 2ax bx cx d cho

( x ) (dng s Horner), a (4) vphngtrnh tch 2(x )(ax Bx C) 0 .

Ch : trng hp nghim phng trnh bc lnhn 3 ta cng c th gii tng t. Cch nhm nghim hu t: Nghim l

mt trong cc t s(c ca d vi c ca a)II. Phng trnh bc 4 c bit:

1. Phng trnh trng phng:ax4 + bx2 + c = 0 ( a 0 )

t t = x2, t 0 . (5) at2 + bt + c = 0.2. Phng trnh i xng:

ax4 + bx3 + cx2 bx + a = 0 ( a 0 )Bc 1: Chia 2 v cho x2,

2

2

1 1pt a x b x c 0

x x

.

Bc 2: t1

t xx

, a (8) vphng trnh

bc hai theo t.3. Phng trnh trng phng tnh tin:

(x + a)4 + (x + b)4 = c

ta b

t x2

, a (7) vphng trnh trng

phng theo t4. Phng trnh cn bng h s theo php

cng:(x + a)(x + b)(x + c)(x + d) = e vi a + c = b + d

t t = (x + a)(x + c), a (6) vphngtrnh bc 2 theo t5. Phng trnh cn bng h s theo php

nhn:

2x a x b x c x d mx vi ab=cd=p

tad

t x2

hoc t (x a)(x d)

6. Phng php h s bt nh:Gi sphng trnh bc 4:

x

4

+ ax

3

+ bx

2

+ cx + d = 0v c phn tch thnh

(x2 + a1x + b1) ( x2 + a2x + b2) = 0

Lc ta c:

1 2

1 2 1 2

1 2 2 1

1 2

a a a

a a b b b

a b a b c

b b d

Tip theo tin hnh nhm tm cc h s a1; b1;a2 ; b2 . Bt u t b1b2 = d v ch th vi cc gitr nguyn.

Ch : Phng php h s bt nh ny cnp dng rt nhiu cc dng ton i hi nhmt tha s chung hay phn chia phn s.

III.Phng php tham s, hng s bin thin:Phng php: Coi cc gi tr tham s, hng s lbin. Cn bin c coi lm hng s.

IV.Phng trnh

2 2a f (x) b.f (x).g(x) c g(x) 0

Trong bc f(x) v g(x) 2.

Xt g(x) = 0 tha phng trnh? Xt g(x) 0 chia hai v cho 2g(x) t

f(x)

t g(x) .

Vn 3: PHNG TRNH BTPHNG TRNH V T.

I. Cc cng thc:1. Cc hng ng thc ng nh: 2 A, A 0A A

A, A 0

2 22 2 B 3BA AB B A2 4

3 3 3(A B) A B 3AB A B 22 bax bx c a x

2a 4a

2. Phng trnh bt phng trnh chadu gi tr tuyt i:

2 2A B A B A B B 0A B

A B


12/55


Trang 12

A B B A B B 0A B

B A B

A B B 0

B 0

A B A B

3. Phng trnh bt phng trnh v t: A 0 B 0A B

A B

2A B B 0 A B A B 0 A B 0 B 0A B

A B

2

A 0 B 0A B

A B

2

B 0B 0A B

A 0 A B

3 3A B A B 2n 12n 1 A B A B 2n 2n A 0 B 0A B

A B

2n2n

B 0A B

A B

II. Cc dng ton thng gp:1. Phng trnh v t:

a. Dng c bn:

f x g x f x g(x) 0 f x g x

2

g x 0

f x g x

f x g x h x . t iu kinbnh phng hai v

Ch : y ta c th khng t iu kin,cbnh phng cc v mt cn, phng trnhmi l phng trnh h qu ca phng trnh cho. Do khi gii tm nghim ta phi th li.

f x g x h x k x Vi

f x h x g x k x

Ta bin i phng trnh v dng

f x h x k x g x

Bnhphng, gii phng trnh h qu. 3 3 3A B C

3 33A B 3 A.B A B C

S dng php th : 3 3A B C Ta c phng trnh:

3A B 3 A.B.C C

Th li nghim.b. t n ph:

Dng 1: t n pha vphng trnh 1 nmi:

2 2ax bx c px qx r trong a b

p q

Cch gii: t 2t px qx r iu kin t 0 Dng 2: Phng trnh dng:

2 2

P x Q x P x Q x

2 P x .Q x 0 0

Cch gii: t t P x Q x

2t P x Q x 2 P x .Q x Dng 3: Phng trnh dng:

P(x) Q(x) P(x).Q(x) 0 0

Cch gii:

* Nu P x 0

P x 0pt

Q x 0

* Nu P x 0 chia hai v cho P x sau t

Q xt

P x vi t 0

Dng 4: Phng trnh i xng vi hai cnthc:

a cx b cx d a cx b cx n

Cch gii: t t a cx b cx

a b t 2 a b

Dng 5: Phng trnh dng:2 2

x a b 2a x b x a b 2a x bcx m

Cch gii: t t x b iu kin: t 0


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Trang 13

a phng trnh v dng:2

t a t a c(t b) m

Dng 6: Phng php tham s, hng s binthin.

2 2

6x 10x 5 4x 1 6x 6x 5 0 c. Sdng n pha vhi xng, h

na i xng:Dng 1: Phng trnh dng

n nx a b bx a

Cch gii:t ny bx a khi ta c h:n

n

x by a 0

y bx a 0

Dng 2: Phng trnh dng:

2ax b r ux v dx e

trong a,u, r 0 v u ar d, v br e

Cch gii: t uy v ax b khi ta c h:

2

2

uy v r ux v dx e

ax b uy v


n ma f x b f x c

Cch gii: t n mu a f x , v b f x

Khi ta c h:

n m

u v c

u v a b

d. Nhn lng lin hip:Dng 1: Phng trnh c dng:

f x a f x b Cch gii:Nhn lng lin hp ca vtri khi

ta c h:

f x a f x b

af x a f x

b


f x g x a f x g x

Ch : Bi ton nhn lin hip thng dng nuta nhm c nghim ca bi ton v nghim l nghim duy nht.

Ta nn bin i nhn cho lng lin hiptng vic chng minh nghim duy nht cd dng.

e. Phng php hm s:Dng 1: Chng minh nghim duy nht

chng minh phng trnh f(x) = g(x) (*)c nghim duy nht, ta thc hin cc bc sau: Chn c nghim x0 ca phng trnh. Xt cc hm s y = f(x) (C1) v y = g(x)

(C2). Ta cn chng minh mt hm sng binv mt hm s nghch bin. Khi (C1) v (C2)giao nhau ti mt im duy nht c honh x0. chnh l nghim duy nht ca phng trnh.

Ch :Nu mt trong hai hm s l hmhng y = C th kt lun trn vn ng.

Dng 2: Bin lun tham s mt n phtheo cc phng php trn.Chuyn m theo n ph mDng cng co hm nh m tha bi

ton.f. Phng php nh gi:Phng php ny ch yu da vo cc bt

ng thc, o hm dnh gi so snh v tri vv phi. Nghim bi ton l khi ta i gii quytdu bng xy ra khi no ca cc ng thc tri v

phi.

2. Bt phng trnh v t:Phng php gii bt phng trnh cng

c chia thnh cc dng ging nh giiphngtrnh.

Ch :Lun t iu kin trc khi bnh phng.Mt s cng thc b sung:a. f (x) 0f(x) 0 g(x) 0g(x)

hoc

f (x) 0

g(x) 0

b. f (x) 0f(x) 0g(x) 0g(x)

hoc

f (x) 0

g(x) 0

c.2

B 0A1

B A B

d. B 0A 1A 0B

hoc

2

B 0

A 0

A B


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Trang 14

Vn 4: HPHNG TRNHI. Hphng trnh bc nht hai n:

1 1 1

2 2 2

a x b y c

a x b y c

Cch gii:

t 1 1

2 2

a bD

a b , 1 1

x

2 2

c bD

c b , 1 1

y

2 2

a cD

a c

1. D 0 : Hphng trnh c nghim duynht

x

y

x D / D

y D / D

.

2.x

D 0, D 0 hoc yD 0 : Hphng

trnh v nghim.

3. D = Dx = Dy = 0: H c v s nghim thaa1x + b1y = c1 hoc a2x + b2y = c2.II. H cha mt phng trnh bc nht:

1y c ax

ax by c b

1f (x,y) df x, c ax d

b

III.Hi xng loi 1:f (x,y) 0

g(x,y) 0

vi

f(x,y) f(y,x)

g(x, y) g(y, x)

Cch gii: tu x y

v xy

vi 2u 4v

IV.Hi xng loi 2:Dng 1:

f (x,y) 0

g(x,y) 0

vi

f(x,y) g(y, x)

g(x, y) f(y,x)

Cch gii:

f (x;y) g(x;y) 0 (x y)h(x;y) 0

f (x;y) 0 f (x; y) 0

x y 0

f(x;y) 0

h(x;y) 0

f(x;y) 0

Dng 2:f (x,y) 0

g(x,y) 0

trong ch c mt phng

trnh i xng.Cch gii:

Cch 1: a phng trnh i xng v dngtch gii y theo x ri thvo phng trnh cn li.

Cch 2: a phng trnh i xng v dngf (x) f (y) x y vi hm f niu.

V. Hng cp bc 2:2 2

1 1 1 1

2 2

2 2 2 2

a x b xy c y d

a x b xy c y d

Cch gii:

Xt y = 0. Xt y 0 khi t x ty v giiphng trnh bc hai n t

VI.H bc hai mrng:f (x, y) 0 f (x, y) 0

g(x, y) 0 .f (x, y) .g(x, y) 0

f (x,y) 0

(ax by c)(px qy r) 0

Ch : Mt s bi ton cn phi t n ph

chuyn v cc dng ton bit.Ngoi ra phngphp nh gi v phng php hm scng cthc dng gii.


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Trang 15

M - LOGARIT

Vn 1: CNG THCI. Hm sm y = ax (a > 0)

1. Tp xc nh: D 2. Tp gi tr: G (0; ) 3. Tnh n iu: 0 < a < 1: Hm nghch bin trn a > 1: Hm sng bin trn

4. Mt s cng thc c bn: 0a 1 (a 0) n

n

1a

a

m n m na .a a m n m na : a a nm m.na a m m m(ab) a .b m m

m

a a

b b

m mnna a II. Hm s logarit y = logax (0 a 1)

nh ngha: y = logax x = ay

1. Tp xc nh: D (0; ) 2. Tp gi tr: G 3. Tnh n iu: 0 < a < 1: Hm nghch bin trn D a > 1: Hm sng bin trn D

4. Mt s cng thc c bn: alog xa x lnxe x b blog c log aa c 2na alog x 2n log x

aalog b log b

ab

1log b

log a

cac

log blog blog a

a b alog b.log c log c a a alog (bc) log b log c

a a a

blog log b log c

c

III.Phng trnh v bt phng trnh m cbn:

1. f ( x )a

b 0a b

f(x) log b0 a 1

2. f (x) g(x)a a a 1

x : f (x), g(x)

0 a 1

f(x) g(x)

3. f ( x ) ab 0

f(x) log ba b

0 a 1 b 0

x : f (x)

4. f ( x ) ab 0

f(x) log ba b

a 1 b 0

x : f (x)

5. f (x) g(x)a a f(x) g(x)0 a 1

6. f (x) g(x)a a f(x) g(x)a 1

IV.Phng trnh v bt phng trnh logaritc bn:

1. a blog f(x) b f (x) a0 a 1

2. a alog f (x) log g(x) f (x) 0f(x) g(x)0 a 1

3. a blog f(x) b 0 f (x) a0 a 1

4. a blog f(x) b f (x) aa 1

5. a alog f (x) log g(x)0 a 1

0 < f(x) < g(x)

6. a alog f (x) log g(x)a 1

f(x) > g(x) > 0

V. Cc dng ton thng gp:1. Phng trnh m:a. a vcng c s:

Vi a > 0, a 1: f (x) g(x)a a f (x) g(x)

Ch :Trong trng hp c s c cha n s th:

M N

a a (a 1)(M N) 0 b. Logarit ho: f (x) g(x) aa b f (x) log b .g(x)


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Trang 16

c. t n ph:Dng 1:

f (x)P(a ) 0 f ( x )t a , t 0

P(t) 0

,

trong P(t) l a thc theo t.Dng 2:2f (x) f (x) 2f (x)a (ab) b 0

Cch gii:

Chia 2 v cho 2f(x)b , ri tf ( x )

at

b

Dng 3:f (x) f (x)a b m , vi ab 1 .

Cch gii: t

f (x) f (x) 1

t a b t d. Sdng tnh n iu ca hm s:

Xt phng trnh: f(x) = g(x) (1)on nhn x0 l mt nghim ca (1). Da vo tnh ng bin, nghch bin ca f(x)

v g(x) kt lun x0 l nghim duy nht. Nu f(x) ng bin (hoc nghch bin) th

f (u) f (v) u v

e. a v phng trnh cc phng trnhc bit:Phng trnh tch: A.B = 0

A 0

B 0

Phng trnh 2 2A 0

A B 0B 0

f. Phng php i lp:Xt phng trnh: f(x) = g(x) (1)

Nu ta chng minh c:f (x) M

g(x) M

th

(1)f (x) M

g(x) M

2. Bt phng trnh m:Cch gii: Tng tnh phng trnh m.

Ch : Trong trng hp c s a c cha ns th: M Na a (a 1)(M N) 0

3. Phng trnh logarit:a. a vcng c sVi a > 0, a 1:

a a

f(x) g(x)log f (x) log g(x)

f(x) 0 (g(x) 0)

b. M haVi a > 0, a 1:

alog f(x) b

alog f (x) b a a c. t n phd. Sdng tnh n iu ca hm se. a vphng trnh c bitf. Phng php i lpCh : Cc phng php lit k khng nu cch

gii c cch gii tng tphng trnh m.Khi gii phng trnh logarit cn ch iu

kin biu thc c ngha. Vi a, b, c > 0 v a, b, c 1 th:

b blog c log aa c 4. Bt phng trnh logarit:

Cch gii: Tng tnh phn phng trnh.Ch : Trong trng hp c s a c cha n

s th:

alog B 0 (a 1)(B 1) 0 ;

a

a

log A0 (A 1)(B 1) 0

log B

5. Hphng trnh m logarit:Cch gii: Kt hp cc cch gii ca phng trnhm logarit trn v phn gii phng trnh vhphng trnh i s.


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Trang 17

NGUYN HMTCH PHN

BNG NGUYN HM

Ham

so f(x)

Ho nguyen

ham F(x)

Ham so

f(x)

Ho nguyen ham

F(x)+Ca ax + C

x

+1x

+ C +1

(ax b) 1

a

1(ax b)

C1

1

x ln x C

1

ax b 1 ln ax b C

a

xa x

aC

lna

xe xe C ax be ax b1

e C

a

sinx -cosx + Csin(ax+b) 1

cos(ax b) Ca

cosx sinx + Ccos(ax+b) 1

sin(ax b) Ca

2

1

cos x tgx + C 2

1

cos (ax b) 1 tg(ax b) C

a

2

1

sin x

-cotgx + C 21

sin (ax b)

1 cotg(ax b) Ca

'u (x)

u(x) ln u(x) C

2 2

1

x a

1 x aln C

2a x a

tgx ln cosx C 2 21

x a 2 2ln x x a C

cotgx ln sinx C

Vn 1: NGUYN HM

I. nh ngha:Hm s F x gi l nguyn hm ca hm s

f x trn a, b nu F x f x , x a,b .

Ch : Nu F x l nguyn hm ca f x th

mi hm s c dng F x C ( C l hng s) cng

l nguyn hm ca f x v ch nhng hm s c

dng F x C mi l nguyn hm ca f x . Ta

gi F x C l h nguyn hm hay tch phn btnh ca hm s f x v k hiu l f x dx .

Nh vy: f x dx F x C II. Tnh cht:1. kf x dx k f x dx; k 0 2. f x g x dx f x dx g x dx

3. f x dx F x C th f u du F u C

Vn 2: TCH PHNI. nh ngha:

b

b

aa

f x dx F x F b F a

II. Tnh cht:1. b a

a b

f x dx f x dx

2. b ba a

kf x dx k f x dx (k 0)

3. b b ba a a

f x g x dx f x dx g x dx

4. b c ba a c

f x dx f x dx f x dx

5.Nu f x 0, x a;b th ba

f x dx 0

6.Nu f x g x th b ba a

f x dx g x dx ,

x a;b

7.Nu m f x M, x a;b th

b

a

m b a f x dx M b a

Ch :- Mun tnh tch phn bng nh ngha ta phi

bin i hm sdi du tch phn thnh tnghoc hiu ca nhng hm s bit nguyn hm.

- Nu hm sdi du tch phn l hm shu t c bc ca t ln hn hoc bng bc camu ta phi thc hin php chia t cho mu.


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Trang 18

Vn 3: TCH PHN I BIN SI. Cng thc:

.b

a

f x x dx f t dt

II. Nhng php i bin ph thng:Hm s c cha

n(x) t t (x)

Hm s c mu s t t l mu s

Hm s c cha (x) t t (x) hay

t (x)

Tch phn chadx

x t t ln x

Tch phn cha xe t xt e

Tch phn chadx

x t t x

Tch phn cha2

dx

x t

1t

x

Tch phn cha cos xdx t t sinx

Tch phn cha2

dx

cos x t t tgx

Tch phn cha2

dx

sin x t t cotgx .

Tch phn cha 2 2a x

t x = asint,

t ;2 2

Tch phn cha2 2

1

a x

t x = atant,

t ;2 2

Vn 4: TCH PHN TNG PHNI. Cng thc:

b bb

aa a

uv dx uv vu dx

hay b b

b

aa a

udv uv vdu

Cc bc thc hin:Bc 1:

u u(x) du u (x)dx (aoham)at dv v (x)dx v v(x) (nguyen ham)

Bc 2: Th vo cng thc (1).

Bc 3: Tnh ba

uv v suy ngh tm cch

tnh tipb

a

vdu

II. Nhng cch t thng thng:u dv

xP(x).e dx P(x) xe dx P(x).cosxdx P(x) cosxdx P(x).sinxdx P(x) sinxdx P(x).ln xdx lnx P(x)

Ch :Tch phn hm hu t:

- Nu mu l bc nht th ly t chia mu- Nu mu l bc hai c nghim kp th a vhng ng thc

- Nu mu l bc hai c hai nghim th ngnht thc

- Nu mu l bc hai v nghim th i bin s.Tch phn hm lng gic:

- Nu sinx,cosx c sm chn th h bc2 21 cos2x 1 cos2x

sin x ;cos x

2 2

- Nu sinx,cosx c sm l th tch ra ri t t- Nu c tan2x hoc cot2x th thm bt 1- Nu c tanx,cotx c tha v sinx,cosx rit t

- Nu c sina.cosb,sina.sinb,cosa.cosb th dngcng thc bin i tch thnh tng.

- Nhiu bi chng ta phi bin i cc hmlng gic a v cc dng c khnng tnhc.Ch : Tch phn trong cc thi i hc thngra di dng kt nhiu dng tnh tch phn. Vth, ttch phn ban u ta bin i v tng hochiu cc tch phn. Khi , tng tch phn ddng tch c bng cc phng php trn.(thng l mt tch phn i bin v mt tchphn tng phn).


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Trang 19

Vn 5: TCH PHN C CHADU TR TUYT I

Gi s cn tnh tch phnb

a

I f (x) dx .

Bc 1. Lp bng xt du (BXD) ca hm s f(x)trn on [a; b], gi s f(x) c BXD:

X a x1 x2 b

f(x) + 0 0 +

Bc 2. Tnh1 2

1 2

x xb b

a a x x

I f (x) dx f (x)dx f (x)dx f (x)dx .

Ch : Nu trong khong (a; b) phng trnhf(x) = 0 khng c nghim th:

b b

a a

f(x) dx f(x)dx

Vn 6: NG DNG CA TCHPHN

I. Tnh din tch hnh phng:1. Trng hp 1:

Din tch hnh phng S gii hn bi cc

ng y f (x), y g(x), x a, x b l:b

a

S f (x) g(x) dx

2. Trng hp 2:Din tch hnh phng S gii hn bi cc

ng y f(x), y g(x) l:

S f (x) g(x) dx

Trong , l nghim nh nht v lnnht ca f(x) = g(x).

Ch :

Nu trong khong ; phng trnhf(x) g(x) khng c nghim th:

f (x) g(x) dx f (x) g(x) dx

Nu tch S gii hn bi x = f(y) v x = g(y) thta i vai tr x cho y trong cng thc trn.

II. Tnh th tch khi trn xoay:1. Trng hp 1.

Th tch khi trn xoay V do hnh phng giihn bi cc ng

y f (x) 0 x a; b , y = 0, x = a v x = b

(a < b) quay quanh trc Ox l:b

2

a

V f (x)dx

2. Trng hp 2.Th tch khi trn xoay V do hnh phng gii

hn bi cc ng

x g(y) 0 y c; d , x = 0, y = c v y = d

(c < d) quay quanh trc Oy l:d

2

c

V g (y)dy

3. Trng hp 3. Th tch khi trn xoay Vdo hnh phng gii hn bi cc ngy = f(x), y g(x) , x = a v x = b

a b, f (x) 0, g(x) 0 x a; b quayquanh trc Ox l:

b

2 2

a

V f (x) g (x) dx

4. Trng hp 4. Th tch khi trn xoay Vdo hnh phng gii hn bi cc ng x = f(y),x g(y) , y = c v y = d

c d, f (y) 0, g(y) 0 y c; d quayquanh trc Oy l:

d

2 2

c

V f (y) g (y) dy

Ch : Cch gii tch phn c du gi trtuyt i nu trn.


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Trang 20

Chuyn : HNH HC KHNG GIAN

I. Kin thc c bn:1. Kin thc hnh hc 910:

1.1H thc lng trong tam gic vung:Cho tam gic ABC vung ti A c ng cao AH, ng trung tuyn AM. Ta c:

2 2 2AB AC BC 2AH BH.CH 2AB = BH.BC 2AC CH.BC

2 2 2

1 1 1

AH AB AC AH.BC AB.AC

b c b csin B , cosB , tan B , cot Ba a c b

M l trung im BC nn MA = MB = MC v M l tm ng trn ngoi tip tam gic ABC

1.2H thc lng trong tam gic thng:Cho tam gic ABC c cc cnh ln lt l a, b, c, ng trung tuyn AM.

nh l hm cos:a2 = b2 + c2 - 2bc.cosA

2 2 2b c a

cosA2bc

nh l hm sin:a b c

2RsinA sinB sinC

nh l ng trung tuyn:2 2 2

2 2

a

2(b c ) am AM

4

1.3Cc cng thc tnh din tch:Tam gic ABC:

ABC

1S BC.AH p.r

2

abc 1.AB.AC.SinA

4R 2.

p(p a)(p b)(p c)

Hnh thang ABCD(AB // CD), ng cao DH:

ABCD

1S (AB CD).DH2

Hnh vung ABCD cnh a:

ABCD

2

S AB.AC

1 AC.BD a2

Hnh chnht ABCD:

ABCDS AB.AD

Din tch hnh thoi ABCD:

ABCD

1S AC.BD

2

Din tch hnh trn:2

(O;R)S .R

Din tch hnh bnh hnh:S = cnh y x chiu cao

Din tch tam gic u:2

ABC

a 3

S 4

Tam gic vung ti A:

1

S AB.AC2


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Trang 21

1.4Tam gic - Cc trng hp bng nhau - ng dng ca tam gic:a. Trng hp bng nhau v ng dng ca tam gic thng:

Tam gic ABC c cc gc A;B;C cc cnh i din tng ng a;b;c. Chu vi 2p.Din tch S

Tnh cht:

Hai tam gic bng nhau th cc yu ttng ng bng nhau. Hai tam gic ng dng th :

T s gia cc yu t( khng k gc; v din tch) tng ng bng nhau v bng tsng dng.

T s din tch bng bnh phng t sng dng. Hai tam gic ng dng nu c 1 yu t vdi tng ng bng nhau th bng nhau.

b. Trng hp bng nhau v ng dng ca tam gic vung:Do 2 tam gic vung c gc vung tng ng bng nhau nn c sc bit so vi

tam gic thng:

Hai cnh gc vung bng nhau (t l ). Mt gc nhn tng ng bng nhau v 1 cnh gc vung bng nhau (t l). Mt cnh gc vung v cnh huyn bng nhau (t l).1.5nh l Thalet: Nhng ng thng song song nh ra trn 2 ct tuyn nhng on thng t l. Trong tam gic 1 ng thng song song vi cnh y khi v chkhi n nh ra trn 2cnh kia nhng on thng tng ng t l. Trong tam gic ng thng song song vi mt cnh th to vi 2 cnh kia 1 tam gicng dng vi tam gic cho ban u.1.6Cc yu tc bn trong tam gic: Ba ng trung tuyn ng quy ti 1 im: trng tm G cch nh bng 2

3mi ng.

Mi ng trung tuyn chia tam gic thnh hai phn c din tch bng nhau. Ba ng cao ng quy ti mt im: trc tm H. Ba ng trung trc ng quy ti mt im gi l tm ng trn ngoi tip, cn gi ltm ca tam gic. Ba ng phn gic trong ng quy ti mt im gi l tm ng trn ni tip.Mi ng phn gic chia cnh i din thnh hai phn t l vi hai cnh bn tng ng.1.7Cc tnh cht c bit:

Cho tam gic nhn ABC, ni tip ng trn tm O, ng knhAA, M trung im BC, H l trc tm, H i xng vi H qua BC.Ta c:- BHCA l hnh bnh hnh c tm l M nn A l im i xngca H qua M- H nm trn ng trn tm O.- 9 im gm trung im 3 cnh tam gic, trung im AH, BH, CH,v cc chn ng cao nm trn mt ng trn c tm l trung imOH c gi l ng trn Euler.


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2. Kin thc hnh hc 11:Quan h song song:

Bi 1: NG THNG SONG SONG VI MT PHNG

nh ngha:

Mt ng thng v mt mt phngc gi l song song nu chngkhng c im chung.

a / / (P) a (P)

a

(P)

nh l:

L1:Nu ng thng d khngnm trn mt phng (P) v songsong vi ng thng a nm trn

mt phng (P) th ng thng dsong song vi mt phng (P)

d (P)

d / /a d / /(P)

a (P)

d

a

(P)

L2: Nu mt ng thng songsong vi mt phng th n songsong vi giao tuyn ca mt phng v mt phng bt k cha n.

a / /(P)

a (Q) d / /a

(P) (Q) d

d

a(Q)

(P)

L3: Nu mt ng thng songsong vi 2 mt phng ct nhau thn song song vi giao tuyn ca haimt phng .

(P) (Q) d

(P) / /a d / /a(Q) / /a

a

d

QP

Bi 2: HAI MT PHNG SONG SONG

nh ngha:

Hai mt phng c gi l songsong nu chng khng c imchung.

(P) / /(Q) (P) (Q) Q

P

nh l:

L1:iu kin cn v 2 mtphng song song l trong mtphng ny cha 2 ng thng ctnhau cng song song vi mt

phng kia.

a,b (P)

a b I (P) / /(Q)

a / /(Q),b / /(Q)

Ib

a

Q

P

L2:Nu 2 mt phng song songvi nhau th mi ng thng nmtrong mt phng ny u song songvi mt phng kia.

(P) / /(Q)a / /(Q)

a (P)

a

Q

P


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L3: Cho 2 mt phng song song.Mt phng no ct mt phng nyth cng ct mt phng kia v 2giao tuyn song song vi nhau.

(P) / /(Q)

(R) (P) a a / /b

(R) (Q) b

b

a

R

Q

P

Quan h vung gc:Bi 1: NG THNG VUNG GC VI MT PHNG

nh ngha:

ng thng vung gc vi mtphng khi v ch khi n vung gcvi mi ng thng nm trongmt phng .

a (P) a c, c (P)

P c

a

nh l:

L1:Nu ng thng d vunggc vi hai ng thng ct nhau av b cng nm trong mp(P) thng thng d vung gc vimp(P).

d a , d b

a ,b (P) d (P)

a b A

d

ab

P

L2: (nh l 3 ng vunggc): Cho ng thng a c hnhchiu trn mt phng (P) l ngthng a. Khi mt ng thng bcha trong (P) vung gc vi a khiv chkhi n vung gc vi a.

a (P), b (P)

b a b a 'a ' a / (P)

a'

a

b

P

Bi 2: HAI MT PHNG VUNG GC

nh ngha:

Hai mt phng c gi l vunggc vi nhau nu gc gia chng

bng 900

.

0(P) (Q) ((P),(Q)) 90

nh l:

L1:Nu mt mt phng cha mtng thng vung gc vi mtmt phng khc th hai mt phng vung gc vi nhau.

a (P)(Q) (P)

a (Q)

Q

P

a

L2:Nu hai mt phng (P) v (Q)vung gc vi nhau th bt c

ng thng a no nm trong (P),vung gc vi giao tuyn ca (P)v (Q) u vung gc vi (Q).

(P) (Q)

(P) (Q) d a (Q)

a (P),a d

d

Q

P

a


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L3: Nu hai mt phng (P) v (Q)vung gc vi nhau v A l mtim trong (P) th ng thng a iqua im A v vung gc vi (Q)s nm trong (P)

(P) (Q)

A (P)a (P)

A a

a (Q)

A

Q

P

a

L4: Nu hai mt phng ct nhauv cng vung gc vi mt phngth ba th giao tuyn ca chngvung gc vi mt phng th ba.

(P) (Q) a

(P) (R) a (R)

(Q) (R)

a

R

QP

Bi 3: MI LIN H QUAN H SONG SONG V VUNG GC

1. a / /b

b Pa P

2.

a P

a / /bb P

3.

P / / Q

a Qa P

4.

a PP / / Q

a Q

5.

a ba / / P haya P

P b

Bi 4: KHONG CCH

1. Khong cch t1 im ti 1 ng thng, n 1mt phng:Khong cch tim O n ng thng a (hoc nmt phng (P)) l khong cch gia hai im O v H,trong H l hnh chiu ca im O trn ng thng a(hoc trn mt phng (P))

d(O; a) = OH; d(O; (P)) = OH

aH

O

H

O

P

2. Khong cch gia ng thng v mt phngsong song:Khong cch gia ng thng a v mt phng (P)song song vi ng thng a l khong cch tim O

bt k thuc ng thng a n mt phng (P)

a

H

O

P

3. Khong cch gia hai mt phng song song:Khong cch gia hai mt phng song song l khongcch tim thuc mt phng ny n mt phng kia.

H

O

Q

P

4. Khong cch gia hai ng thng cho nhau :Khong cch gia hai ng thng cho nhau l dion vung gc chung ca hai ng thng .

B

A

b

a


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Phng php: Dng on vung gc chung ca hai ng thng cho nhau a v b.

Cch 1: Gi s a b:

Dng mt phng (P) cha b v vung gc vi a ti

A. Dng AB b ti B

AB l on vung gc chung ca a v b.

a

bA

B

Cch 2: S dng mt phng song song.

Dng mt phng (P) cha b v song song vi a.

Dng hnh chiu vung gc a ca a trn (P).

T giao im B ca a v b, dng ng thng

vung gc vi (P) ri ly giao im A ca ng thngny vi a.

AB l on vung gc chung ca a v b.

b

a'

a

B

A

Cch 3: S dng mt phng vung gc.

Dng mt phng (P) a ti O.

Dng hnh chiu b ca b trn (P).

Dng OH b ti H.

T H, dng ng thng song song vi a, ct b tiB.

T B, dng ng thng song song vi OH, ct ati A.

AB l on vung gc chung ca a v b.Ch : d(a,b) = AB = OH.

a

b'

b

O

H

B

A

Bi 5: GC

1. Gc gia 2 ng thng trong khng gian:Gc gia 2 ng thng trong khng gian l gc hpbi hai ng thng cng phng vi chng, xut phtt cng mt im.

Lu : 0 00 a,b 90 b'

b

a'a

2. Gc gia ng thng v mt phng:ng thng khng vung gc vi mt phng: L

gc gia ng thng v hnh chiu ca n lnmt phng.

ng thng vung gc vi mt phng: gc giachng bng 900 P

a'

a

Phng php: Xc nh gc gia ng thng a v mt phng (P).

Tm giao im O ca a vi (P). Chn im A a v dng AH (P). Khi AOH (a,(P))


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3. Gc gia hai mt phng:Gc gia 2 mt phng l gc to bi 2 ng thng

ln lt vung gc vi 2 mt phngHoc l gc gia 2 ng thng nm trong 2 mt

phng cng vung gc vi giao tuyn ti 1 imba

QPP Q

ab

Phng php: Mun tm gc gia hai mt phng (P) v (Q) ta c th s dng mt trong cccch sau:

Tm hai ng thng a, b: a (P), b (Q). Khi : (P), (Q) a,b .Gi s (P) (Q) = c. T I c, dng a (P), a c

b (Q), b c

(P), (Q) a,b

[Tm mt phng (R) vung gc vi giao tuyn c = (P) (Q)(R) (P) = a; (R) (Q) = b (P), (Q) a,b ]

4. Din tch hnh chiu:Gi S l din tch ca a gic (H) trong mt phng (P)v S l din tch hnh chiu (H) ca (H) trn (P).

Khi ta c: S' S.cos P,P ' CB

A

S

Lu : Ngoi nhng vn nu thm phng php gii, hc sinh nn ch cc nh l c innghing cng chnh l phng php thng c s dng gii quyt cc vn .

MT SHNH THNG GP Hnh lng tr:l hnh a din c 2 y song song v cc cnh khng thuc hai y th song

song v bng nhau v gi l cc cnh bn. Hnh hp:l hnh lng trc y l hnh bnh hnh Hnh lng trng: l hnh lng tr c cnh bn vung gc vi y. Hnh lng tru:l lng trng v c y l a gic u. Hnh hp ng: l hnh hp c cnh bn vung gc vi y. Hnh hp chnht: l hnh hp ng c y l hnh ch nht . Ba di ca ba cnh xut

pht t mt nh gi l ba kch thc ca hnh hp ch nht. Hnh lp phng: l hnh hp ch nht c ba kch thc bng nhau. Hnh chp:l hnh a din c mt mt l mt a gic cn cc mt khc u l cc tam gic c

chung nh. Hnh tdin: l hnh chp c y l hnh tam gic. Hnh chp u: l hnh chp c y l a gic u v cc cnh bn u bng nhau. ng

thng ni tnh n tm a gic u gi l trc ca hnh chp. Trc ca hnh chp vung gc vi mtphng y. Hnh chp ct:l hnh a din to ra thnh chp c hai y l hai a gic ng dng nm

trong hai mt phng song song, cc mt bn l cc hnh thang.


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3. Kin thc hnh hc 12:Din tchth tch khi a din: Din tch xung quanh: bng tng din tch cc mt bn. Din tch ton phn: bng tng din tch xung quanh v din tch y.

1. TH TCH KHI LNG TR:

V = B.hvi B: l din tch y hnh lng tr

h: l ng cao hnh lng tr

TH TCH KHI HP CHNHT:

V = a.b.cVi a, b c l chiu di, chiu rng,chiu cao ca hnh hp ch nht.

TH TCH HNH LPPHNG:

V = a3Vi al di cnh hnh lp

phng

2. TH TCH KHI CHP:

V = 13

B.h

vi B: l din tch yh: l ng cao

3. T S TH TCH TDIN:Cho khi t din SABC v A, B,

C l cc im ty ln lt thucSA, SB, SC ta c:

SABC

SA'B'C'

V SA SB SC

V SA' SB' SC'

4. TH TCH KHI CHP CT:

h

V B B' BB'3

vi B, B: l din tch yh: l ng cao

BA

C

A'B'

C'


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IV.Mt cu ngoi tipni tip khi a din:Mt cu ngoi tip Mt cu ni tip

Hnh a din Tt c cc nh ca hnh a din unm trn mt cu

Tt c cc mt ca hnh a din u tipxc vi mt cu

Hnh tr Hai ng trn y ca hnh tr nmtrn mt cu Mt cu tip xc vi cc mt y v ming sinh ca hnh tr

Hnh nn Mt cu i qua nh vng trn yca hnh nn

Mt cu tip xc vi mt y v ming sinh ca hnh nn

V. Xc nh tm mt cu ngoi tip khi a din: Mt cu i qua mi nh ca hnh a din gi l mt cu ngoi tip hnh a din. Tm l

im cch u cc nh ca hnh a din, bn knh l khong cch t tm n mt trong ccnh .

Cch xc nh tm mt cu: Cch 1: Nu (n 2) nh ca a din nhn hai nh cn li di mt gc vung th tm camt cu l trung im ca on thng ni hai nh .

Cch 2: xc nh tm ca mt cu ngoi tip hnh chp.

o Xc nh trc ca y (l ng thng vung gc vi y ti tm ng trn ngoitip a gic y).

o Xc nh mt phng trung trc (P) ca mt cnh bn. o Giao im ca (P) v l tm ca mt cu ngoi tip hnh chp.

VI.Xc nh tm mt cu ni tip hnh chp:

Mt cu ni tip hnh chp l mt cu trong hnh chp v tip xc vi tt c mt bn vmt y ca hnh chp . Tm l im cch u tt c cc mt bn v y, bn knh lkhong cch ttm n mt trong cc mt y.

T din lun c mt cu ni tip, cc hnh chp khc c th khng c mt cu ni tip. Cch xc nh tm mt cu: Tm mt cu ni tip (nu c) l giao im cc mt phn gic

ca cc nh din hp bi cc mt bn v y.

Bn knh: 3tp

Vr

S

DIN TCHTH TCH

Cu Tr Nn

Din tch 2S 4 R xqS 2 Rh

tp xqS S 2S ay

xqS Rl

tp xqS S S ay

Th tch 34

V R3

2V R h 21

V R h3


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HNH HC TA OXY

Vn 1: TA PHNGI. nh l:Cho A A B BA(x , y ), B(x , y ) , 1 2a (a , a )

1. B A B AAB (x x ; y y ) 2. 2 2

B A B AAB AB (x x ) (y y )

.

3. 2 21 2

a a a

II. Tnh cht vect:Cho 1 2a (a , a )

, 1 2b (b ,b )

1. 1 12 2

a ba ba b

2. 1 2ka (ka ,ka ) 3.

1 1 2 2a b (a b ;a b )

4.1 1 2 2ma nb (ma nb ;ma nb )

5. 1 1 2 2a.b a b a b 6. a cng phng

1 2 2 1

a k.bb

a b a b 0

7.1 1 2 2a b a.b 0 a b a b 0

8. 1 1 2 22 2 2 2

1 2 1 2

a b a ba.bcos(a;b)

a b a a b b

9.1 2

AB (a ,a )

,1 2

AC (b ,b )

ABC 1 2 2 1

1S a b a b

2

III.Dng ton thng gp:1. A, B, C thng hng AB cng phng AC. 2. A, B, C lp thnh tam gic AB khngcng phng AC.

3. A,B,C,D l hnh bnh hnh AD BC. 4. M trung im AB: A B A Bx x y yM ;

2 2

5. M chia AB theo t s k 1:A B A B

x k.x y k.yM ;

1 k 1 k

6. Trng tmA B C

G

A B CG

x x xx

3G :

y y yy

3

7. Trc tm H: Gii h: AH.BC 0BH.AC 0

8. E chn phn gic trong: EB ABACEC

F chn phn gic ngoi:FB AB

ACFC

9. Tm ng trn ngoi tip ABCGii h:

2 2

2 2

IA IB

IA IC

Vn 2: NG THNGI. Phng trnh ng thng:1. Phng trnh tng qut :

0 0qua M(x ;y )

VTPT: n = (A;B)

: 0 0A(x-x )+B(y-y )=0

: Ax+By+C=0

2. Phng trnh tham s:2

0 0

1

qua M(x ;y )

VTCP : a = (a ;a )

:

0 1

0 2

x = x + a t(t R)

y = y + a t

3. Phng trnh chnh tc :2

0 0

1

qua M(x ;y )

VTCP : a = (a ;a )

: 0 0

1 2

x -x y -y=

a a

II. Vi tr tng i ca hai ng thng:1. ) 1 11 2

2 2

A B( ) (

A B

2. 1 1 11 2

2 2 2

A B C( ) / / ( )

A B C

3. 1 1 11 2

2 2 2

A B C( ) ( )

A B C

III.Vtr tng i ca hai im i vi mtng thng:


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Cho hai im1 1 1 2 2 2M (x ; y ), M (x ; y ) v

ng thng (d): Ax + By + C = 0, ta c:

1M hoc 2M nm trn (d)1 1 2 2(Ax By C)(Ax By C) 0 .

1 2M , M nm khc pha so vi(d)1 1 2 2(Ax By C)(Ax By C) 0 .

1 2

M , M nm cng pha so vi (d)

1 1 2 2(Ax By C)(Ax By C) 0 .

IV.Gc ca hai ng thng:1 2 1 2

2 2 2 2

1 1 2 2

A A B Bcos

A B A B

V. Khong cch tmt im n mt ngthng:

Cho () : Ax By C 0 v 0 0M(x ;y )

0 0

2 2

Ax By Cd (M, )

A B

VI.Ch : Trc Ox c pttq : y 0 Trc Oy c pttq : x 0 ng thng song song hoc trng vi Oy :

ax c 0 b 0

ng thng song song hoc trng vi Ox :by c 0 a 0

ng thng i qua gc ta :ax by 0 c 0

ng thng ct Ox ti A a;0 v Oyti B 0;b

x y1

a b a, b 0

ng thng qua im 0 0M x ;y v c h sgc k l : 0 0y y k x x

ng thng d qua im 0 0M x ;y v songsong vi ng thng : ax by c 0 c

phng trnh tng qut l: 0 0a x x b y y 0

ng thng d qua im 0 0M x ;y v vunggc vi ng thng : ax by c 0 c phng

trnh tng qut l : 0 0b x x a y y 0

Cho () : Ax By C 0 1. (d)/ / () (d) : Ax By m 0

2. (d) () (d) : Bx Ay m 0

VII.Dng ton thng gp:Dng 1 : Tm hnh chiu ca mt im M trnmt ng thng d :Cch 1:

Bc 1: Gi H l hnh chiu ca M trn d suyra ta ca H theo t

Bc 2: Tm ta vect MH theo t , tmVTCP u

ca d

Bc 3: Gii phng trnh MH . u = 0 c tsuy ra ta HCch 2:

Bc 1: Vit phng trnh ng thng quad qua M v vung gc vi d

Bc 2: Gii h : dd '

c ta im H

Dng 2 : Tm im i xng ca mt im Mqua mt ng thng d Bc 1: Tm hnh chiu H ca M trn d Bc 2: gi M l hnh im i xng ca Mqua d th H l trung im ca on MM , da

vo cng thc ta trung im suy ra ta M

Vn 3: NG TRNI. Phng trnh ng trn:1. Phng trnh chnh tc ng trn (C) tmI(a;b) , bn knh R:

(C): 2 2 2(x a) (y b) R

2. Phng trnh tng qut ng trn (C) tmI(a;b) , bn knh R:

(C): x2 + y2 2ax 2by + c = 0

(K:a2 + b2c > 0) v R = 2 2a b c

II. Phng trnh tip tuyn ca ng trn:1. Phng trnh tip tuyn TI 0 0M(x ;y ) :

0 0

0 0

quaM(x ;y ):

VTPT IM (x a;y b)

: 0 0 0 0(x a)(x x ) (y b)(y y ) 0

: 0 0 0 0x.x y.y a(x x ) b(y y ) c 0

2. iu kin tip xc: d(I, ) R


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III.Phng trnh ng thng i qua 2 tipim:

ChoM M

M(x ;y ) nm ngoi ng trn tm

I(a;b) bn knh R. T M dng 2 tip tuyn tipxc ng trn ti 2 im A, B. Phng trnh

ng thng AB c dng: 2M Mx a x a y b y b R

IV.Phng trnh tip tuyn chung ca haing trn:

Bc 1: Xt tip tuyn vung gc vi 0x :x a R v x a R . Kim tra tip tuyn thamn iu kin u bi?

Bc 2: Xt tip tuyn khng vung gc vi 0xc dng: y kx m . tm k v m: Ta gii h

lp c tiu kin tip xc. Nu (C1) v (C2) ngoi nhau: c 4 tip tuynchung.

Nu (C1) v (C2) tip xc ngoi: c 3 tiptuyn chung. Nu (C1) v (C2) ct nhau: c 2 tip tuynchung. Nu (C1) v (C2) tip xc trong: c 1 tiptuyn chung. Nu (C1) v (C2) lng nhau: khng c tiptuyn chung.

Vn 4: ELPI. nh ngha:

Cho 1 2 1 2F ,F co nh va FF = 2c (c > 0)

1 2M (E) MF MF 2a (a c 0)

II. Phng trnh chnh tc:2 2

2 2

x y(E) 1 (a,b 0)

a b

III.Cc tnh cht:1. Tiu im : 1 2F ( c;o), F (c;o) .2. Tiu c : 1 2FF 2c .3. nh trc ln: 1 2A ( a;0), A (a;0) .4. nh trc b : 1 2B (0; b), B (0;b) .5. di trc ln: 1 2A A 2a .6. di trc b : 1 2B B 2b .7. Tm sai : ce 1

a .

8. Bn knh qua tiu im : 1 M2 M

MF a e.x

MF a e.x

9. Phng trnh cnh hnh ch nht c s:x a

y b

10.Phng trnh ng chun 2axc

IV.Phng trnh tip tuyn ca Elip:1. Phng trnh tip tuyn TI

0 0M(x ;y ) :

0 0

2 2

x.x y.y: 1 (a,b 0)

a b

2. iu kin tip xc:Cho:

2 2

2 2

x y(E) 1 (a,b 0)a b v ng

thng () : Ax By C 0

() tip xc (E) 2 2 2 2 2A a B b C

Vn 5: Cc dng ton tam gicTrong mt phng Oxy cho tam gic ABC bit

im C(a;b) v hai ng thng ct nhau1 2d , d

khng i qua C ln lt c phng trnh tham s :

1 1 1

1

1 1 1

x x a td :y y b t

v 2 2 222 2 2

x x a td :y y b t

Hy tm ta cc nh A, B trong cc trnghp :Dng 1:

1 2d , d l hai ng cao.

Gi s d1l ng cao AM , d2l ng cao BN Vit phng trnh BC: (BC c VTCP l

VTPT ca d1i qua C)

Gii h 2

BC

d

ta im B

Tng t : Vit phng trnh AC (AC c VTCP l

VTPT ca d2v i qua C)

Gii h1

AC

d

c ta im A

Dng 2:1 2

d , d l hai ng trung tuyn.

Gi s d1: l trung tuyn AM ; d2 l trung tuyn

BN Md1 suy ra ta M theo t1 M l trung im CB suy ra ta B theo t1


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B d2 nn c h theo t1 v t2 . Gii h c t1suy ra ta im B

Tng t : Nd2 suy ra ta N theo t2 N l trung im CA suy ra ta A theo t2 A d1 nn c h theo t1 v t2 . Gii h c t2

suy ra ta im ACh : C th gii theo cch khc : Tm ta trng tm G ca tam gic ; Tm im i xng D ca C qua G Vit phng trnh ng thng qua d1 qua D

song song vi d2 Vit phng trnh ng thng qua d2 qua D

song song vi d1

Gii h 11

d 'd

ta A ;

Gii h 22

d '

d

ta B

Dng 3:1 2d , d l hai ng phn gic trong

ca gc A v gc B. Tm ta im C1l im i xng ca C

qua d1; 1C AB

Tm ta im C2l im i xng ca Cqua d2; 2C AB

Vit phng trnh tham s C1C2l phngtrnh ca AB

Ta ca A l nghim ca h : 1 21

C C

d

Ta ca B l nghim ca h : 1 22

C C

d

Dng 4:1

d l ng cao,2

d l trung tuyn.

Gi s d1: ng cao AM; d2: trung tuyn BN Vit phng trnh cnh CB (nh trn) Gii h

2

CB

d

tm ta im B

Dng tnh cht trung im N thuc BN ,N latrung im AC v A thuc AM suy ra ta im A

Dng 5: 1d l ng cao , 2d l phn gic

trong.Gi s d1: ng cao AM; d2: phn gic trong BN Vit phng trnh cnh CB

Gii h2

CB

d

ta im B

Tm ta im C2l im i xng ca Cqua d2 ( C2 thuc AB)

Vit phng trnh BC2 (BA) Gii h

1

BA

d

ta im A .

Dng 6:1

d l trung tuyn ,2

d l phn gic

trongGi s d1: ng trung tuyn AM; d2: phn gictrong BN

21

M d

MA MC

A d

ta im B.

Tm C2l im i xng ca C qua d2 Vit phng trnh tham s BC2 (BA) Gii h

1

BA

d

ta im A

Nhn xt: Hc sinh ch cn nm k cc dng 1, 2, 3 th

cc dng khc n gin hn.

Nu bi ton c lin quan n ng cao cnch n im hnh chiu ca nh bittrn ng cao hoc VTPT ca ng caohoc tm VTCP ca cnh v vit phng trnhtham s ca cnh tam gic

Nu bi ton c lin quan n trung tuyn cnlu n tnh cht trung im .

Nu bi ton c yu tng phn gic trongcn lu n im i xng ca nh bitqua ng phn gic trong .Ch : thi i hc thng s dng cc

tnh cht i xng tm (im), i xng trc(ng)lin quan n Php bin hnh 11. Ngoira s kt hp gia cc tnh cht ca ng trn vtam gic cng l dng ton rt thng gp.


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HNH HC TA OXYZ

Vn 1: TA IM VVECT

I. Ta ca vct:Trong khng gian vi h ta Oyz

1.1 2 3 1 2 3a (a ;a ;a ) a a i a j a k

2. i (1,0,0) ; j (0,1,0) ; k (0,0,1) 3. Cho 1 2 3a (a ; a ; a ) v 1 2 3b (b ;b ;b ) ta c :

1 12 2

3 3

a b

a b a b

a b

1 1 2 2 3 3a b (a b ;a b ;a b ) 1 2 3k.a (ka ;ka ;ka ) 2 2 21 2 3a a a a 1 1 2 2 3 3a.b a . b cos(a;b) a b a b a b

II.Ta im :Trong khng gian vi h ta Oxyz

1.

M M M M M MM x ; y ; z OM x i y j z k

2. Cho A A AA x ; y ; z v B B BB x ; y ; z ta c:

B A B A B AAB x x ;y y ;z z 2 2 2B A B A B AAB (x x ) (y y ) (z z )

3. Nu M chia on AB theo t s k MA kMB

th ta c :

A B A B A B

M M M

x kx y ky z kzx ; y ; z

1 k 1 k 1 k

(Vi k 1) c bit khi M l trung im AB (k =1 ) thta c:

A B A B A BM M M

x x y y z zx ; y ;z

2 2 2

III. Tch c hng ca hai vect v ngdng:

1. Nu 1 2 3a (a ; a ; a ) v 1 2 3b (b ;b ;b ) th:2 3 3 1 1 2

2 3 3 1 1 2

a a a a a aa,b ; ;

b b b b b b

2. Vect tch c hng c a,b

vung gc vi

hai vect a

v b

.

3. a,b a b sin(a,b)

.

4. ABC 1S [AB,AC]2

.

5. VHpABCDABCD= [AB,AD].AA' .6. VTdin ABCD = 1 [AB,AC].AD

6

.

IV.iu kin khc:1. a v b cng phng:

1 1

2 2

3 3

a kb

a,b 0 k R : a kb a kba kb

2. a v b vung gc:1 1 2 2 3 3a.b 0 a .b a .b a .b 0

3. Ba vect a, b, c ng phng a,b .c 0 4. A,B,C,D l bn nh ca t din AB, AC, AD

khng ng phng.5. G l trng tm ca tam gic ABC:

A B CG

A B CG

A B CG

x x xx3

y y yy

3

z z zz

3

6. G l trng tm t din ABCDGA GB GC GD 0

A B C D

G

A B C DG

A B C DG

x x x X

x 4

y y y yy

4

z z z zz

4

7. G l trng tm ca t din ABCD:GA GB GC GD 0

.

8. Chieu cao AH ke t nh A cua t dienABCD:

AH = ABCD

BCD

3V

S


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Vn 2: MT PHNGI. Phng trnh mt phng:1. Trong khng gian Oxyz phng trnh dng

Ax + By + Cz + D = 0 vi A2 + B2 + C2 0

l phng trnh tng qut ca mt phng, trong n (A;B;C) l mt vect php tuyn ca n.

2. Mt phng (P) i qua im M0(x0;y0;z0) vnhn vect n (A;B;C)

lm vect php tuyn c

dng :A(xx0) + B(yy0) + C(zz0) = 0

3. Mt phng (P) i qua M0(x0;y0;z0) v nhn1 2 3a (a ; a ; a )

v 1 2 3b (b ;b ;b )

lm cp vect

chphng th mt phng (P) c vect phptuyn:

2 3 3 1 1 2

2 3 3 1 1 2

a a a a a an a,b ; ;

b b b b b b

.

II. Vtr tng i ca hai mt phng:1. Cho hai mt phng (P): Ax + By + Cz + D = 0v (Q):Ax + By + Cz + D = 0 (P) ct (Q) A : B : C A: B: C (P) // (Q) A : A = B : B = C : C D :D

(P) (Q) A : B : C : D = A: B: C: D2. Cho hai mt phng ct nhau :

P : Ax By Cz D 0

Q :Ax By Cz D 0

.

Phng trnh chm mt phng xc nh bi(P) v (Q) l:m(Ax + By + Cz + D) + n(Ax + By + Cz + D)

= 0 (vi m2 + n2 0)III.Khong cch tmt im n mt phng:

Khong cch t M0(x0;y0;z0) n mt phng(): Ax + By + Cz + D = 0.

0 0 0

02 2 2

Ax By Cz Dd(M , )

A B C

IV.Gc ga hai mt phng:Gi l gc gia hai mt phng :

P : Ax By Cz D 0

Q :Ax By Cz D 0

. Ta c:

P QP Q

P Q

n .ncos cos(n ,n )

n . n

0 02 2 2 2 2 2

A.A' B.B' C.C'0 90

A B C . A ' B' C'

0

P Q90 n n

hai mt phng vung gc

nhau.

V. Cc dng bi tp:Dng 1: Vit phng trnh mt phng: Tm VTPT n A;B;C v im i

qua 0 0 0 0M x ; y ; z

Dng: 0 0 0A x x B y y C z z 0 Dng 2: Vit phng trnh mt phng qua baim A, B, C:

Tnh AB,AC Mp (ABC) c VTPT l n AB,AC

vqua A

Kt lun.Dng 3: Vit phng trnh mt phng i

qua im A v vung gc BC

Mt phng BC nn c VTPT l BC qua A

Ch :

Trc Ox cha i 1;0;0 Trc Oy cha j 0;1;0 Trc Oz cha k 0;0;1 Dng 4: Vit phng tnh mp l mt

phng trung trc ca AB.

Mt phng AB. Nn c VTPT l AB iqua I l trung im ca AB

Kt lun.Dng 5: Vit phng tnh mt phng i

qua im 0 0 0 0M x ;y ; z v song song vi mt

phng : Ax By Cz D 0

/ / . Nn phng trnh c dng:Ax + By + Cz + D= 0

0M D' Kt lun.Dng 6: Vit phng trnh mp (P) i qua haiim A, B v vung gc vi mp (Q)

Mt phng (P) c cp VTCP l: AB v VTPTca (Q) l Qn


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Mt phng (P) c VTPT lQ

n AB, n

v

qua A Kt lun.Dng 7: Vit phng trnh mp i qua cc

im l hnh chiu ca im 0 0 0M x ;y ;z trn cc trc to. Gi M1, M2, M3 ln lt l hnh chiu ca

im M trn Ox, Oy, Oz. Th M1(x0;0;0),M2(0;y0;0), M3(0;0;x0)

Phng trnh mt phng l:

0 0

x y z1

x y z

Dng 8: Vit phng trnh mp i qua

im M0 v vung gc vi hai mt phng (P)v (Q). (P) c VTPT l

Pn

(Q) c VTPT l Qn Mp c VTPT l P Qn ,n v qua Mo Kt lun.Dng 9: Ta im Mi xng ca M quamt phng

Gi M (x; y; z) l im i xng ca Mqua

Gi d l ng thng i qua M v d .Nn d c VTCP l n

Vit phng trnh tham s ca d Gi H d Ta im H l nghim ca hphng trnh

d :

:

Ta im H

V H l trung im ca MM Ta imM

Dng 10: Vit phng trnh mt phng tipdin ca mt cu (S) ti tip im A. Xc nh tm I ca mt cu (S) Mt phng : Mp tip din c VTPT : IA Vit phng trnh tng qut.

Vn 3: NG THNGI. Phng trnh ng thng:1. Phng trnh tham s ca ng thng:

Cho im0 0 0 0M (x ;y ;z ) l im thuc ng

thng v 1 2 3a (a ; a ; a )

l VTCP ca ngthng . Phng trnh tham s ca ng thng :

0 1

0 2

0 3

x x a t

y y a t (t R)

z z a t

2. Phng trnh chnh tc ca ung thng:Cho im

0 0 0 0M (x ;y ;z ) l im thuc ng

thng v1 2 3

a (a ; a ; a )

l VTCP ca ngthng . Phng trnh chnh tc ca ng thng :

0 0 0

1 2 3

x x y y z z

a a a

II. V tr tng i ca cc ng thng vcc mt phng:1. Vtr tng i ca hai ng thng:

Cho hai ng thng () i qua M c VTCPa

v () i qua M c VTCP a '

.

() cho () a,a ' .MM' 0 () ct () a,a ' .MM' 0 vi a,a ' 0

() // () [a,a']=0M '

hoc

'a;a = 0

a;MM' = 0

() ()

[a,a ']=0

M '

hoc

'a; a = 0

a;MM' = 0

2. V tr tng i ca ng thng v mtphng:

Cho ng thng () i qua

0 0 0 0M (x ;y ;z ) c VTCP 1 2 3a (a ; a ; a )

v mtphng ():

Ax By Cz D 0 c VTPT n (A;B;C)

.

() ct () a.n 0 () // () a.n 0

M ( )


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() nm trn () a.n 0M ( )

III.Khong cch:1. Khong cch tM n ung thng () iqua M0 c VTCP a

.0[M M,a]

d(M, )a

2. Khong cch gia hai ng cho nhau :0 0 0 0

qua M (x ;y ; z ):

VTCPa

;0 0 0 0

qua M' (x' ; y' ;z' )' :

VTCPa'

[a,a '].MM'd( , ')

[a,a']

Ch :* Nu () v () ct nhau hoc trng nhau th:

d((),()) = 0* Nu () v () song song th:

d((),()) = d(M , ()) = d(N , ())( trong M () v N ())

IV.Gc:1.

Gc gia hai ng thng:

0 0 0 0qua M (x ;y ; z )

:VTCPa

;

0 0 0 0qua M' (x' ; y' ;z' )

' :VTCPa'

1 1 2 2 3 3

2 2 2 2 2 21 2 3 1 2 3

a.a 'cos cos(a,a ')

a . a '

a .a ' a .a ' a .a '

a a a . a ' a ' a '

2. Gc gia ng thng v mt phng:() i qua M0 c VTCP 1 2 3a (a ; a ; a ) , mp() c

VTPT n (A;B;C)

.Gi l gc hp bi () vmp() , ta c:

1 2 3

2 2 2 2 2 21 2 3

Aa +Ba +Casin cos(a,n)

A B C . a a a

V. Dng ton thng gp:Dng 1: Vit phng trnh ng thng :

Cn bit VTCP 1 2 3a a ;a ;a vim 0 0 0 0M x ; y ; z

Vit phng trnh tham s theo cng thc.

Vit phng trnh chnh tc theo cng thc.Dng 2: Vit phng trnh ng thng khi:

: 1 1 1 1

2 2 2 2

A x B y C z D 0

A x B y C z D 0

c VTCP l : 1 1 1 1 1 12 2 2 2 1 2

B C C A A Ba ; ;B C C A A B

Cho z = 0 tm c im M0. Vit phng trnh ng thng.Dng 3: Vit phng trnhng thng iqua im 0 0 0 0M x ;y ; z v vung gc vi mt

phng : Ax By Cz D 0

Mp c VTPT l n A;B;C ng thng i qua im M0 v c VTCP

l n

Vit phng trnh ng thng.Dng 4: Vit phng trnh hnh chiu ca dtrn mt phng

Gi d l hnh chiu ca d trn mp Gi l mt phng cha d v Nn c cp VTCP l VTCP ca d l du

v n

l VTPT ca mt phng Mp c VTPT dn u ,n i qua im

M0d

Vit phng trnh tng qut ca Mp Phng trnh ng thng d:

:

:

Chuyn vphng trnh chnh tc (tham s).Dng 5: Vit phng trnh ng thng d quaim 0 0 0 0M x ; y ; z v vung gc vi hai

ng 1 v 2

1 c VTCP 1u 2 c VTCP 2u d vung gc vi 1 v 2 . Nn d c VTCP l

d 1 2u u ,u

Dng 6: Vit phng trnh ng thng d iqua im A v ct chai ng

1 v 2 .

Thay toA vo phng trnh 1 v 2 1 2A ,A


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Gi (P) l mt phng i qua im A v cha1

Gi (Q) l mt phng i qua im A v cha2

Phng trnhng thng d:

PQ

Chuyn vphng trnh chnh tc (tham s)Dng 7: Vit phng trnh ng thng d

P ct chai ng 1 v 2 .

Gi 1A P Gi 2B P ng thng chnh l ng thng ABDng 8: Vit phng trnh ng thng d // d1v ct chai ng 1 v 2 .

Gi (P) l mt phng cha 1 v(P) // d1 Gi (Q) l mt phng cha

2 v(Q) // d1

d P Q Phng trnh ng thng d

P :

Q :

Dng 9: Vit phng trnh ng vung gcchung ca hai ng thng cho nhau 1 v

2 .

Cch 1:

Gi1

u

v2

u

ln lt l VTCP ca1

v 2

Gi 1 2v u ,u Gi (P) l mt phng cha

1 v c mt

VTCP l v

. Nn c VTPT l P 1n u , v

phng trnh mt phng (P)

Gi (Q) l mt phng cha 2 v c mtVTCP l v

. Nn c VTPT l Q 2n u , v

phng trnh mt phng (Q)

Phng trnh ng vung gc chung ca 1 v 2 :

P

Q

Cch 2:

Chuyn phng trnh ng thng1

v2

v dng tham s.

Gi1M v 2N (M, N di dng tham

s). Tnh MN

.

Gii h: 12

MN.u 0

MN.u 0

. Tm c tham s

tm c ta im M, N vit phngtrnh MN.

Dng 10: Vit phng trnh ng thng dvung gc (P) v ct hai ng thng

1 v 2

Gi l mt phng cha 1 v c mtVTCP l

Pn

( VTPT ca (P) )

Gi l mt phng cha 2 v c mtVTCP l Pn

( VTPT ca (P) )

ng thng d Dng 11: Vit phng trnh ng thng d iqua im M0 vung gc vi ng thng 1 v

ct ng thng 2

Gi l mt phng i qua M0 v vunggc

1

Gi l mt phng i qua im M0 vcha 2

ng thng d Dng 12: Vit phng trnh ng thng d iqua giao im ca ng thng v mtphng v d ,d

Gi A Gi l mt phng i qua A v vung gc

vi . Nn c VTPT l VTCP ca

ng thng d

Dng 13: Tm ta im M' i xng ca M0qua ng thng d Gi M (x ; y ; z ) Gi (P) l mt phng i qua im M0 v

P d . Nn (P) nhn VTCP ca d lmVTPT

Gi H d P Ml im i xng ca M0qua ng thng

d. Nn H l trung im ca on M0M


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Ta c:

0H

0H

0H

x x 'x

2

y y 'y

2

z z '

z 2

M

Dng 14: Khong cch gia hai ng thngcho nhau v ' Gi u v u ' ln lt l VTCP ca v ' i qua im M0 , 0M' '

0 0

u,u ' .M M'd , '

u ,u '

Vn 3: MT CUI. Phng trnh mt cu:1. Phng trnh mt cu tm I(a;b;c) bnknh R

2 2 2 2x a x b x c R .

2. Phng trnh mt cu tm I(a;b;c) , bnknh 2 2 2R a b c d :

2 2 2x y z 2ax 2by 2cz d 0

vi a2 + b2 + c2d > 0II. Vtr tng i ca mt cu v mt phng:Cho mt cu

(S): 2 2 2 2

x a x b x c R

tm I(a;b;c)bn knh R v mt phng (P):

Ax + By + Cz + D = 0.

Nu d(I,(P)) > R th mt phng (P) v mt cu(S) khng c im chung. Nu d(I,(P)) = R th mt phng (P) v mt cu(S) tip xc nhau. Khi (P) gi l tip din camt cu (S) v im chung gi l tip im Nu d(I,(P)) < R th mt phng (P) v mt cu(S) ct nhau theo giao tuyn l ng trn c

phng trnh:

2 2 2 2x a y b z c R

Ax By Cz D 0

Trong bn knh ng trn2 2

r R d(I, (P)) v tm H ca ng trn lhnh chiu ca tm I mt cu (S) ln mt phng(P).

III.V tr tng i gia ng thng v mtcu:

Cho mt cu (S):(xa)2 +(yb)2+(zc)2 =

R2v ng thng (d) :0 1

0 2

0 3

x x a t

y y a t

z z a t

Mun tm giao im gia (d) v (S) , ta thay x,y, z trong phng trnh (d) vo phng trnh (S)ta c mt phng trnh bc hai theo t . Nu phng trnh theo t v nghim th (d) v(S) khng c im chung Nu phng trnh theo t c mt nghim t th(d) tip xc vi (S) . Khi (d) gi l tip tuynca mt cu (S) v im chung gi l tip im .Nu phng trnh theo t c hai nghim phn bitt1; t2 th (d) ct (S) ti hai im phn bit.IV.Dng ton thng gp:Dng 1: Vit phng trnh mt cu Xc nh tm I(a ; b ; c) ca mt cu Bn knh R Vit phng trnh mt cu

2 2 2 2

x a x b x c R

Dng 2: Vit phng trnh mt cu ngknh AB Gi I l trung im ca AB. Tnh to I

I l tm mt cu

Bn knh 1R AB2

Vit phng trnh mt cuDng 3: Vit phng trnh mt cu (S) c tmI (a; b; c) v tip xc vi :

Ax + By + Cz + D = 0

Mt cu (S) c tm I v tip xc vi . Nnc bn knh

R d I, I I I2 2 2

Ax By Cz D

A B C

Vit phng trnh mt cuDng 4: Vit phng trnh mt cu (S) ngoitip tdin ABCD Phng trnh mt cu (S) c dng

x2 + y2 + z2 + 2Ax + 2By +2Cz + D = 0

A, B, C, D thuc (S). Ta c hphng trnh Gii hphng trnh tm A, B, C, D Kt lun


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Dng 5: Lp phng trnh mt cu i qua baim A, B, C c tm nm trn mt phng Oxy

Gi I(xI ; yI ; 0) l tm ca mt cu, I Oxy Ta c AI2 = BI2 = CI2 Ta c hphng trnh

2 2

2 2AI BIAI CI

Gii hphng trnh tm I IA = R Kt lun

Vn 5: Cc dng ton tam gicTrong khng gian Oxyz cho tam gic ABC

bit im C(a;b;c) v hai ng thng ct nhau

1 2d , d khng i qua C ln lt c phng trnh

tham s :

1 1 1

1 1 1 1

1 1 1

x x a t

d : y y b t

z z c t

v2 2 2

2 2 2 2

2 2 2

x x a t

d : y y b t

z z c t

Hy tm ta cc nh A, B trong cc trnghp :

1 2d , d l hai ng cao ca tam gic .

1 2

d , d l hai ng trung tuyn ca tam gic.

1 2d , d l hai ng phn gic trong gc A , B 1d l ng cao, 2d l trung tuyn ca tamgic

1

d l ng cao,2d l phn gic trong ca

tam gic

1

d l trung tuyn, 2d l phn gic trong ca

tam gic Phng php: Tng tnh trong hnh

hc phng.

Ch : Hnh hc gii tch khng gian thii hc thng tp trung vo cc dng tonthng gp ca phng trnh ng thng, ccdng ton khong cch, im i xng nn hcsinh cn nm k(v hnh hc gii tch trong Oxythi khai thc yu t tam gic)

Vn 6: ng dng hnh hc gii tchgii cc bi hnh hc thun.

C sl lun:Nh ta bit trong vi cng c gii tch

ta c thtnh c din tch mt a gic, th tchmt khi a din, khong cch gia hai mt

phng, gia hai ng thng, gc gia hai mtphng, gia ng thng v mt phng, gia haing thng

V vy gii bi ton thun ty hnh hc ctha v mt bi ton hnh hc gii tch nu ta

xy dng mt h trc Oxyz hp l.Nhn xt:- u: Gii bi ton chn thun l tnh ton,khng suy ngh nhiu.- Khuyt: Khng thy c ci hay ca hnh hcthun ty, tnh ton phi ht sc cn thn.Mt s cch chn h trc Oxyz thng dng:

1. Vi hnh lp phng hoc hnh hp chnht ABCD.A'B'C'D'

2. Vi hnh hp y l hnh thoiABCD.A'B'C'D' 3. Vi hnh chp tgic u S.ABCD

4. Vi hnh chp tam gic u S.ABC5. Vi hnh chp S.ABCD c ABCD l hnh

ch nht v SA (ABCD)6. Vi hnh chp S.ABC c ABCD l hnh

thoi v SA (ABCD)7. Vi hnh chp S.ABC c SA (ABC) v

ABC vung ti A.8. Vi hnh chp S.ABC c SA (ABC) v

ABC vung ti B.9. Vi hnh chp S.ABC c (SAB) (ABC), SAB cn ti S v ABC vung ti C

10.Vi hnh chp S.ABC c (SAB) (ABC), SAB cn ti S v ABC vung ti A

11.Vi hnh chp S.ABC c (SAB) (ABC), SAB cn ti S v ABC vung cn ti C


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Mt cch tng qut: Chn trc h trc Oxynm trong mt phng y da trn cc tnh chtvung gc (O nm gc vung). Sau dng tiaOz vung gc vi Oxy ti O.

S PHC

Vn 1: CC NH NGHA TNHCHT.

I. Khi nim s phc Tp hp s phc: C S phc (dng i s) : z a bi

(a, b R , a l phn thc, b l phn o, i l n vo, i2 =1)

z l s thc phn o ca z bng 0(b = 0)

z l thun o phn thc ca z bng 0(a = 0)

S 0 va l s thc va l so. Hai s phc bng

nhau:a a '

a bi a ' b 'i (a,b,a ', b ' R)b b '

2. Biu din hnh hc: S phc z = a + bi (a,b R) c biu din bi im M(a; b) hay biu (a; b)

trong mp(Oxy) (mp phc)

3. Cng v trs phc:

a bi a ' b 'i a a' b b' i a bi a ' b'i a a' b b' i Si ca z = a + bi lz =abi u biu din z, u ' biu din z' th

u u '

biu din z + z v u u '

biu dinzz.

4. Nhn hai s phc:

a bi a ' b'i aa ' bb' ab' ba ' i k(a bi) ka kbi (k R)

5. S phc lin hp ca s phc z = a + bi lz a bi

z z ; z z ' z z ' ;


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1 12 2

z zz.z ' z.z ';

z z

2 2z.z a b z l s thc z z ;

z l so z z 6. Mun ca s phc: z = a + bi

2 2z a b zz OM z 0, z C , z 0 z 0 z.z' z . z ' z z

z ' z '

z z ' z z ' z z ' 7. Chia hai s phc:

12

1z zz

(z 0)

12

z ' z '.z z '.zz 'z

z z.zz

z ' w z ' wzz

8. Cn bc hai ca s phc:

z x yi l cn bc hai ca s phcw a bi

2

z w

2 2x y a

2xy b

w = 0 c ng 1 cn bc hai l z = 0 w 0 c ng hai cn bc hai i nhau Hai cn bc hai ca a > 0 l a Hai cn bc hai ca a < 0 l a.i

9. Phng trnh bc hai Az2 + Bz + C = 0 (*)(A, B, C l cc s phc cho trc, A 0 ).

2B 4AC 0 : (*) c hai nghim phn bit

1,2

Bz

2A , ( l 1 cn bc hai ca )

0 : (*) c 1 nghim kp:1 2

Bz z

2A

Ch : Nu z0 C l mt nghim ca (*)th

0z cng l mt nghim ca (*).

10. Dng lng gic ca s phc:

z r(cos i sin ) (r > 0) l dng lng

gic ca z = a + bi (z 0)

2 2r a b

acos

r

bsin

r

l mt acgumen ca z, (Ox,OM) z 1 z cos isin ( R)

11. Nhn, chia s phc di dng lng gic:z r(cos i sin ) , z ' r '(cos ' i sin ')

z.z ' rr '. cos( ') i sin( ') z r cos( ') i sin( ')

z ' r '

12. Cng thc Moavr:

n nr(cos i sin ) r (cosn i sinn ) ,( *n N )

ncos isin cos n isin n 13. Cn bc hai ca s phc di dng lng

gic: S phc z r(cos isin ) (r > 0) c hai

cn bc hai l:

r cos i sin2 2

hoc r cos i sin

2 2

r cos isin2 2

Mrng: S phc z r(cos isin ) (r > 0) c n cn bc n l:

n k2 k2r cos isin , k 0,1,...,n 1n n

Vn 2: CC DNG TONI. Thc hin cc php ton cng tr, nhnchia s phc.

p dng cc quy tc cng, tr, nhn, chiahai s phc, cn bc hai ca s phc.

Ch cc tnh cht giao hon, kt hp ivi cc php ton cng v nhn.II. Gii phng trnh - hphng trnh sphc:

- Gi s z = x + yi. Gii cc phng trnhn z l tm x, y thomn phng trnh.

- Gii phng trnh bc hai trong tp sphc, kt hp vi nh l Vi-et.


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- Ch : l ln ca mt s phc chkhng phi l tr tuyt i. (tr tuyt i l trnghp ring ca ln c nh ngha trn trc sthc).III.Tp hp im.

- Gi s s phc z = x + yi c biu dinim M(x; y). Tm tp hp cc im M l tm hthc gia x v y.

- Ch : Cc dng phng trnh ngthng, ng trn, conic.IV.Dng lng gic.

- p dng nh cc cng thc nu.Ch : Vic kt hp khai trin nh thc Newtontrong tp s phc chng minh cc ng thccng hay c s dng.

I S T HPXC SUT

Vn 1: HON VCHNH HPT HP

V. Quy tc m, cng v nhn:1. Quy tc m:

a. Quy tc:Vi iu kin l khong cch gia cc s bngnhau (cch u), ta c:

1

so ln nhat so nho nha

so cac sokhoang cach gia 2 so lien ke

t

.

b. Cc du hiu chia ht:Chia ht cho 2: s c ch s tn cng l 0, 2, 4,6, 8.

Chia ht cho 3: s c tng cc ch s chia htcho 3.

Chia ht cho 4: s c 2 ch s tn cng lpthnh s chia ht cho 4.

Chia ht cho 5: s c ch s tn cng l 0, 5.Chia ht cho 6: s chia ht cho 2 v 3.Chia ht cho 8: s c 3 ch s tn cng lpthnh s chia ht cho 8.

Chia ht cho 9: s c tng cc ch s chia htcho 9.

Chia ht cho 10: s c ch s tn cng l 0.Chia ht cho 11: s c hiu ca tng cc ch shng l v tng cc ch shng chn chia htcho 11 (VD: 1345729 v (1+4+7+9)(3+5+2) =11).

Chia ht cho 25: s c 2 ch s tn cng l 00,25, 50, 75.

2. Quy tc cng:1) Nu mt qu trnh (bi ton) c th thc hinc mt trong hai cch (trng hp) loi tr lnnhau: cch th nht cho m kt qu v cch th haicho n kt qu. Khi vic thc hin qu trnhtrn cho m + n kt qu.2) Nu mt qu trnh (bi ton) c th thc hinc k cch (trng hp) loi tr ln nhau: cchth nht cho m1 kt qu, cch th hai cho m2 ktqu, , cch th k cho mk kt qu. Khi vic

thc hin qu trnh trn cho m1 + m2+ + mkkt qu.

3. Quy tc nhn:


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1) Nu mt qu trnh (bi ton) c thc hintheo hai giai on (bc) lin tip nhau sao choc m cch thc hin giai on th nht, ng thing vi mi cch c n cch thc hin giaion thhai. Khi c mn cch thc hin qutrnh trn.

2) Nu mt qu trnh (bi ton) c thc hintheo k giai on (bc) lin tip nhau sao cho cm1 cch thc hin giai on th nht, vi micch c m2 cch thc hin giai on thhai, , c mk cch thc hin giai on th k. Khi, ton b qu trnh c m1.m2mk cch thchin.

VI.Hon vChnh hpT hp:1. Hon v:

nh ngha. Cho tp hp X gm n phn t phnbit n 0 . Mi cch sp xp n phn t ca Xtheo mt th tno c gi l mt hon vca n phn t. S cc hon v ca n phn tck hiu l Pn.

Pn= n! = 1.2n

2. Chnh hp:nh ngha. Cho tp hp X gm n phn t phnbit n 0 . Mi cch chn ra k 0 k n phn

t ca X v sp xp theo mt th tno cgi l mt chnh hp chp k ca n phn t. S ccchnh hp chp k ca n phn tc k hiu l

k

nA .

k

n

n!A

(n k)!

3. T hp:nh ngha. Cho tp hp X gm n phn t phnbit n 0 . Mi cch chn ra k 0 k n phn

t ca X c gi l mt t hp chp k ca nphn t. S cc t hp chp k ca n phn tck hiu l knC .

k

n

n!C

k!(n k)!

Nhn xt:1)iu kin xy ra hon v, chnh hp v thp l n phn t phi phn bit.

2) Chnh hp v t hp khc nhau ch l saukhi chn ra k trong n phn t th chnh hp c spth t cn t hp th khng.

VII. Phng php gii ton m:1. Phng php 1.

Bc 1.c k cc yu cu v s liu ca bi.Phn bi ton ra cc trng hp, trong mi trnghp li phn thnh cc giai on.

Bc 2. Ty tng giai on c th v gi thit biton s dng quy tc cng, nhn, hon v,chnh hp hay t hp.Bc 3. p n l tng kt qu ca cc trnghp trn.

2. Phng php 2.i vi nhiu bi ton, phng php 1 rt di. Do ta s dng phng php loi tr (phn b)

theo php ton A A X A X \ A .

Bc 1: Chia yu cu ca thnh 2 phn l yucu chung X (tng qut) gi l loi 1 v yu curing A. Xt A l phnh ca A, ngha l khngtha yu cu ring gi l loi 2.Bc 2: Tnh s cch chn loi 1 v loi 2.Bc 3:p n l s cch chn loi 1 tr s cchchn loi 2.

Ch :1) Cch phn loi 1 v loi 2 c tnh tng i,ph thuc vo ch quan ca ngi gii.

2) Gii bng phng php phn b c u im lngn tuy nhin nhc im l thng sai st khitnh slng tng loi.3*)Thng th ta xl cc iu kin trc, hocn gin cc iu kin ri gii quyt bi ton.

VIII. Phng php phng trnh, btphng trnh, hi s t hp:Bc 1: t iu kin cho bi ton.

-xP c iu kin l x

-k

nA ,k

nC c iu kin l k,n v 0 k n Bc 2: p dng cng thc tnh a bi tonvcc phng trnh, hphng trnh quen thuc.Bc 3: Gii phng trnh, bt phng trnh, h

phng trnh ri so iu kin chn nghim.Ch : Do tnh c bic ca nghim l s t

nhin nn i khi mt s bi ta phi nhmnghim, cn i vi nhng bi bt phng trnhi khi ta cng cn lit k cc nghim.


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Vn 2: NH THC NEWTON

I. nh ngha:Nh thc Newton l khai trin tng ly tha cdng:

n 0 n 1 n 1 2 n 2 2n n nn

k n k k n n k n k k

n n n

k 0

a b C a C a b C a b ...

C a b ... C b C a b

S hng th k+1 l k n k k k 1 nT C a b thng c gi l s hng tng qut.

Cc h s knC c tnh theo cng thc t

hp chp hoc da vo tam gic Pascal sau:

Tnh cht

1) k n kn nC C (0 k n)

2) k k 1 k n n n 1C C C (1 k n)

.

II. Phng php gii ton:1. Dng khai trin: Du hiu nhn bit: Cc h s ng

trc t hp v ly tha l 1 hoc 1 v1 xen knhau.

Khai trin na b hoc na b . Cng hoc tr hai v ca 2 khai trin trn.2. Dng o hm:

a. o hm cp 1: Du hiu nhn bit: Cc h s ng

trc t hp v ly tha tng dn t1 n n (hocgim dn tn n 1).

Xt khai trin (1):

n 0 1 2 2 k k n n

n n n n n1 x C C x C x ... C x ... C x

o hm 2 v ca (1). Thay s thch hp vo (1) sau khi o hm.

b. o hm cp 2: Du hiu nhn bit: Cc h s ng

trc t hp v ly tha tng (gim) dnt1.2 n (n1).n hoc tng (gim) dn t12n n2.

Xt khai trin (1):

n 0 1 2 2 n 1 n 1 n n

n n n n n1 x C C x C x ... C x C x

o hm 2 v ca (1) ta c (2):

n 11 2 3 2 n n 1

n n n nC 2C x 3C x ... nC x n 1 x

Tip tc o hm 2 v ca (2) ta c (3):2 3 4 2 n n 2

n n n n1.2C 2.3C x 3.4C x ... (n 1)nC x

n 2n(n 1)(1 x) .

Nhn x vo 2 v ca (2) ta c (4): n 11 2 2 3 3 n nn n n nC x 2C x 3C x ... nC x nx 1 x .

o hm 2 v ca (4) ta c (5):2 1 2 2 2 3 2 2 n n 1

n n n n

n 2

1 C 2 C x 3 C x ... n C x

n(1 nx)(1 x)

3. Dng tch phn: Du hiu nhn bit: Cc h sng

trc t hp (v ly tha) l phn s gim

dn t1 n1

n 1

hoc tng dn t1

n 1

n 1.

Xt khai trin (1):

n 0 1 2 2 n 1 n 1 n n

n n n n n1 x C C x C x ... C x C x

Ly tch phn 2 v ca (1) t a n b tac:

b b b b

n 0 1 n n

n n n

a a a a

1 x dx C dx C xdx ... C x dx

b b bn 1 b

2 n 10 1 n

n n n

a a aa

1 x x x xC C ... Cn 1 1 2 n 1

2 2 n 1 n 10 1 n

n n n

b a b a b aC C ... C

1 2 n 1

n 1 n 1

(1 b) (1 a)

n 1

.

Ch : Trong thc hnh, ta d dng nhn bitgi tr ca n. nhn bit 2 cn a v b ta nhn vo

s hng

n 1 n 1n

n

b aCn 1

.

4. Tm s hng trong khai trin nh thcNewtn:

a. Dng tm s hng thk: S hng th k trong khai trin n(a b) lk 1 n ( k 1) k 1

nC a b .

b. Dng tm s hng cha xm: S hng tng qut trong khai trin n(a b)

l k n k k f (k)nC a b M(k).x (a, b cha x).


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Gii phng trnh0

f (k) m k , s hng

cn tm l 0 0 0k n k k n

C a b v h s ca s hng cha

xm l M(k0).

Ch : S hng khng cha x th m = 0

c.

Dng tm s hng hu t: S hng tng qut trong khai trin n(a b) l

m r

k n k k k p q

n nC a b C . .

( , l hu t).

Gii h0

m

p(k ,0 k n) k

r

q

.

S hng cn tm l 0 0 0k n k k nC a b .d. Dng tm h s ln nht trong khai

trin Newton:

Xt khai trin n(a bx) c s hng tngqut l k n k k k nC a b x

.

t k n k k k nu C a b , 0 k n ta c dy hs l ku .

tm s hng ln nht ca dy ta gii hbt phng trnh

k k 1

0

k k 1

u u

ku u

.

H s ln nht l 0 0 0k n k k nC a b .

Vn 3: XC XUTI. Php thngu nhin v khng gian mu

1. Php thngu nhin:a. Khi nim: Php th ngu nhin (php

th ) l mt th nghim hay hnh ng m:

- Kt qu ca n khng on trc c .- C thxc nh c tp hp cc kt qu c thsy ra ca php th.

b. K hiu:Php th ngu nhin hay k hiu l : T

2. Khng gian mu ca php th:a. Khi nim : Tp hp tt c cc kt qu c

th xy ra ca php php th gi l khn

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