Học và ôn luyện theo cấu trúc đề thi môn toán

GlAl TICH Hpc va on-luyen theo CTDT mon Toan THPT 3... Moi tap hop eon eua A gom k phan til sSp theo mot thuf tiT nhat dinh dLfgrc goi la mot chinh hap chap k cua n phan tuf... Do do so

TS VU THE HlfU - NGUYEN VINH CAN

T H E O CAU T R U C D E THI

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n ta

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Che ban:

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Trinh bay bia:

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C DE T HI

ON T OA

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PT

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Q GH

N nga

y 01/03/2013

GlAl TICH

Hpc va on-luyen theo CTDT mon Toan THPT 3

Neu tap )igp A c6 m phan til, tap hgfp B c6 n phan tuf va giCTa A va B

khong CO phan tijf chung t h i c6 m + n each chon mot phan tuf thuoc A

hoac thuoc B

b) Quy tdc nhdn :

doan I CO m each thiic hien, cong doan I I c6 n each thiic h i e n t h i c6

m.n each d e hoan t h a n h cong viec A

Tong quat, de hoan t h a n h cong viec A phai qua k cong doan Cong

Mot tap hop A hOTu h a n gom n phan tuf (n > 1) va so nguyen k

(0 < k < n) Moi tap hop eon eua A gom k phan til sSp theo mot thuf tiT nhat dinh dLfgrc goi la mot chinh hap chap k cua n phan tuf

A;; = n(n - l ) ( n - 2) (n - k + 1) =

(n - k ) ! (Quy irde : 0! = 1)

BAI TA

P

1 Ch

o ca

c chu

f s

o 2 , 3 , 4 , 5 , 6 , 7

a) C

o b ao

hi eu

so' t\i

n hi en c6 h

ai chu

f s

o ducf

c ta

o ne

b) C

o ba

o nh ie

u s

o t

ii nh ie

c nh au difOc ta

o ne

o d

a cho

CH

I DA

N

a) De t ao m

ai co ng d oa

n :

1 Ch on m

ot chu

f s

o l am chuf s

o h an

g chu

c : c6

6 k

et qu

a c

6 t he

2 Ch on m

ot chu

f s

o l am chuf s

o h an

g do

n vi : c

6 6 k

et qu

a c

6 t he

Th eo q uy t

o th an

x 6 = 3

6 so'

b) La

p lu an g io ng n hif cau a ) n hu ii

g liT

u y sir kh ac b ie

o dug

c ta

o th an

c nh au Do do

1 Ch on m

ot chu

f s

o l am chuf so ' h an

g chu

c : c6

6 k

et qu

a c

6 t he

2 Ch on m

ot chu

f s

o l am chuf s

o h an

g do

n vi : c

6 5 k

y ph

ai kh ac chuf s

o h an

g chu

c da

ho

n trifdr

c do)

Th eo q uy ta

c nh an : s

c nh au difcfc ta

p ho

p 6 chuf s

o d

a ch

o la : n ' =

6 x

5 = 3

0 so

ai ph an tuf ti

T 6 p ha

n tu

f d

a cho

c nh au t ao

ha nh tiT 6 chuf s

h ho

p c

ha

p 2 cua t ap

hop

6 p ha

n tuf

n =

Ag = 6.5 = 3

0 so

2.

C ho t ap h op cac

cha

so

0, 1 , 2 , 3 , 4 , 5 , 6

a) Co bao nh ie

u s

o t

u n hi en c6 4 chuf s

o tii

t ap h op cac chu

f s

o da cho

b) Co bao n hi eu so t

U nh ie

n c

6 4 chuf s

o k ha

c nh au tCmg do

CH

I DA

N

a) C

o 6 each c ho

n chu

f s

o h an

g ng hi

n (chu

f s

o d au t ie

n ph

ai kh

ac 0) , 7

each

ch on chff s

o h an

g tr am , 7 each c ho

n chu

f s

o h an

g chu

c va

g do

n vi T he

o qu

y tS

e n ha

n : so each t ao

n 4 chuf s

o tii

ap h op

N =

x7 x7 x7 = 2 05

8 so

b) C

o 6 each c ho

n chu

f s

o h an

g ng hi

n, kh

i ch on x on

g chu

f s

o h an

i 6 chuf s

o k ha

c vdi

chuf s

o h an

g ng hi

n da c ho

o h an

g tr am Kh

i da c ho

n chu

f s

o h an

g ng hi

n la

i 5 ehOf s

o k ha

c vd

n

Do do eo

5 eac

h

6 :S;

IS V Cin

chpn chOf so hang chuc Tifcfng tir, c6 4 each chon chOr so hang don v i

Theo quy tac nhan So cac so txi nhien c6 4 chOf so khac nhau tCfng doi difcfc tao t h a n h tix tap hop 7 chuf so da cho la :

t h a n h tCr tap hop 7 chOf so da cho la mot chinh hgrp chap 4 ti^ tap hgfp

7 chuf so ma cac chinh hgfp nay khong c6 chuT so 0 or dau Do do so cac

so CO 4 chijf so khac nhau tiT 7 chijf so la :

N' = - Ag = 7 X 6 X 5 X 4 - 6 X 5 X 4 = 720 so

3 Mot to hoc sinh c6 10 ngUofi xep thijf tif thanh hang 1 de vao lorp Hoi

a) Co bao nhieu each de to xep hang vao l(Jp

b) Co bao nhieu each de to xep hang vao Idfp sao cho hai ban A va B eua

to luon d i canh nhau va A dufng tri/dtc B

a) S^p 6 ngiTofi theo hang ngang ciia mot ban a n dai

b) S4J) 6 ngiTori ngoi vong quanh mot ban an t r o n

Neu t a eat ban t r o n a v i t r i giCfa 2 va 4 r o i t r a i dai theo ban ngang

t h i t a CO hoan v i (1) tUofng ijfng mot each xep ngiiofi ngoi theo ban a n

dai TiTofng txi cat of v i t r i giufa 5 va 2 N h u vay mot each sap xep theo

ban t r o n tiiOng ufng vdri 6 each sap xep theo ban dai Do do so each

Hoc va on luyen theo CTDT mon Toan THPT 7

p 6

ngiTofi

n go

i qu an

h b an a

n t ro

n la :

N = — = 120

15 ngifofi

go

m 9 n am v

a 6 nOf

Ca

n l ap n ho

u eac

h th an

h la

p nh om t ro ng m

a) Nh om c6 3 n am v

a 1

nur

b) S

o na

m v

a nC

f t ro ng n ho

m ba ng n ha

u

c) Ph

ai C O

i

t n ha

t m

ot na

m

CH

I DA

p n ho

m g om

3 n am v

a 1 nCf (the

o qu

y tS

c n ha n)

la

:

Ni = C^C^

= 5 04 each

b) S

o eac

h la

p n ho

m go

m 2 n am v

a 2 nuf la :

N2

= C^C

^ = — = 540

c) S

o eac

h th an

h la

p nh om

4 ngu'of

i tr on

g d

o c

6 it n ha

t 1 n

e 3 n am , 1 nur hoa

c 4 n am

— Cg.C

g + Cg

7 ^ 9.8.7.6

= 9 + +

6 + = 135

0 each

g e

o th

e l ap lu an n hif sau :

Ca t

o C O

15 ngiTcfi

So eac

h l ap n ho

m 4

n gU di

t uy

y la :

_ 15.14.13.1

2

~ 1.2.3.

4 6.5

1.2

So eac

h l ap n ho

m 4 n gi/

di e

o it n ha

t 1 n am l

- 1

5 = 1350 each

6

Tr on

g ma

t ph an

g e

o n d ie

m ph an b ie

t (n > 3 ) tr on

n mo

t dii dn

g t hS ng (3 < k <

h

CH ID

AN

Cuf 3 d ie

m kh on

g th Sn

g h an

g ta

o t ha nh m

ot ta

m giac S

n di em l

a : C^ S

o ca

c ta

p co

n 3 d ie

m tr

th in

g la : C^

S

o ta

m gia

c c

6 3 d in

h la cac di

em

da

cho

la : N = C

^

Cl

t am giac

^4 ^

^

—

^ x o

^ ^ ^ gg

m 4 ngUofi

to an nuf la

: C

^ = Cg

-V Tn

h Ca

n

7 a) Co bao nhieu so t i i nhien la so chan c6 6 chiif so doi m o t khac nhau

va chuf so dau t i e n la chOf so le

b) Co bao nhieu so t i i nhien c6 6 chuf so doi mot khac nhau, trong do c6 dung 3 chuf so le, 3 chuf so chSn (chuf so dau tien phai khac 0)

C H I D A N

so 0, 1, 2, 8, 9 vdfi ai ?i 0, ai aj vdi 1 < i ?i j < 6

- V i X la so chSn nen ae c6 5 each chon tiT cac chuT so 0, 2, 4, 6, 8

- V i ai la chuT so le nen c6 5 each chon tiT cac chuf so 1, 3, 5, 7, 9

da chon ae va a i Theo quy t^c nhan, so cac so can xac dinh l a :

N i = S.S-Ag = 5.5.8.7.6.5 = 42000 so

b) M o t so theo yeu c a u de b a i gom 3 chuf so tii tap X i = |0; 2; 4; 6; 81 va

3 chuf so tCr tap hop X2 = I I ; 3; 5; 7; 91 ghep l a i va loai d i cac day 6

chuf so CO chuf so 0 dufng dau

So each lay 3 chuf so thuoc tap X i la : Ci? = 10 each

So each lay 3 phan tuf thuoc X2 l a : Cg = 10 each

So' each ghep 3 phan tuf lay txi X i v o i 3 phan tuf lay tii X2 l a :

C^C^ = 10.10 = 100 each

So' day so' eo thuf tif eiia 6 phan tuf diioc ghep l a i la :

100.6! = 72000 day

Cac day so c6 chuf so 0 a dau gom 2 chiJ so khac 0 ciia X i va 3 chuf so'

ciia X2 : So cac day so nhif t r e n la : C 4 C 5 5 ! = 7200 day

So cac so theo yeu cau de bai la :

N2 - C ^ C ^ 6 ! - C ^ C ^ 5 ! = 72000 - 7200 = 64800 so

ta chon ra 4 vien b i tif hop do H o i c6 bao nhieu each l a y de trong so

b i j a y ra khong dii ca 3 mau

C H I D A N

Cdch 1 : So each chon 4 vien b i khong du 3 mau bang so' each chon 4

vien bat k i trir d i so each chon 4 vien c6 ca 3 mau

N = Cjg - (C^ C^ C^ + C^ C\ + Cl C\) = 645 each

Cdch 2 : So each chon 4 vien b i khong du 3 mau bang so each chon 4

vien mot mau (4 do, 4 t r a n g va 4 vang) cong vdi so each chon 4 vien hai mau (1 do, 3 t r a n g hoae 2 do, 2 t r a n g hoac 3 do, 1 t r a n g hoae 1

do, 3 vang hoac 2 do, 2 vang hoae 3 do, 1 vang hoae 1 trSng, 3 vang hoae 2 trSng, 2 vang hoac 3 trang, 1 vang)

N = c : +Ct +CI+ ClCl + ClCl + C^C^ + + C^C^ = 645 caeh

Hoc va on luyen theo C T D T m o n loan T H P T SI 9

h von

g tro

n quan

h ngo

n

lijfa trai

CH

I DA

N

ThiTe hie

o s

o ch^

n (hoa

c le)

Vi diidn

g tron 30 cho ne

tr

on

(xem bai s

o 4)

+ = ——

- (1) tron

Cl;}^ + + Cl:\) tron

n

CH

I DA

N

a) V(J

i k

e N, k > 2 ta c6 : A', = k(k

- 1) ^ = =

(*)

-' ^ A^

k(

l) k-

(1 (I

vdi ve ca

e dan

g thuf

c tre

n ta diTOe :

c:;-c::;+c-^3+c ::u +c:-uc :

Do C;

; = C;::} =

l ne

n tha

y C;

: d dang thuf

c cuo

i bor

i C^:

} ta

dugfc

dang

thufc (2) ca

n chufn

g min

h

11 Chufn

g minh bat dan

g thil

e : C^ooi +

^ C\Z + CfZ

trong

do k

e N, k < 2000, l

a t

o ho

p cha

p k eua n phan

tuf

10 t4

l TS V

u Th

e Hi/

u Nguyin Vin

-h CS

n

C A C BAI TAP Tl/ GIAI

k h o n g CO diTcfng d i t h a n g tii C d e n D Til A d i t h a n g d e n C c6 2 each,

tir C d i t h a n g d e n B c6 3 each TCr A d i t h a n g d e n D c6 3 each tix D d i

a) D u n g t a p h o p X eo t h e g h i dufcfe bao n h i e u so tiT n h i e n eo 5 chiT so

b) D u n g t a p h o p X c6 t h e g h i dirge bao n h i e u so t i r n h i e n c6 5 chOf so'

cong t a c 3 ngiTdi H o i eo bao n h i e u each c h g n t r o n g m o i t r i r d n g h g p sau

a) B a hoc s i n h b a t k i eua Idp

b) H a i nijf s i n h v a m o t n a m s i n h

Hpc va on luyen theo CTDT mon Toan THPT I J : 1 1

c) Ba hoc si nh c6 i

t n ha

t mo

t nuf

DS :

a) C;;

^ ea ch b) 25.C^

o eac

h c )

u eac

h ph an p ho

i 7

do va

t ch

o 3 ngUdi

tr on

t ngUof

i nh an

3 d

o va

t, eo

n 2 ngUcfi

mo

i ngUcf

i ha

i d

o va

t

b) Mo

i ng ud

i

it nh

at mo

t d

o va

t va k ho ng qua

3 d

o va

CO

9 na

m va

3 nOf

a) C

o ba

o nh ie

u eac

h ch on m

ot nh om

4 ngUcf

i tr on

g d

o e

o 1 nijf

b) C

o ba

o nh ie

u eac

h ch ia to t ha nh

3 nh om m

oi nh om

1 nuf

DS :

a) 3.C

^ ea ch b) 3.C;;.2C

^ = 10080

x, y t ho

a ma

n ca

c da ng thufe :

20 C

o ba

o nh ie

u s

o tU n hi en chain c6

4 ehu

f s

o do

i mo

t kh ac n ha

u

DS •.n= Al+

4.8.8 = 76

0 so

21

Ch

o da g ia

c de

u 2n d in

h AiA2 A2n , n

> 2 n

oi tie

p tr on

g du

so ta

m gia

c c

6 di nh la

3 tr on

g 2n d ie

m tr en n hi

at e

o di nh la

4 tr on

g 2n d in

h tr en T im

Ti

m s

o tU n hi en n , b ie

t ra ng C" + 2C;, + 4 C' + + 2

"C

" =

243

: n = 5

23

Gi

ai ba

t ph uo ng t ri nh (v

di ha

i an n , k

G N )

24

Tr on

g mo

t mo

n hoc , th ay g ia

o e

o 3

0 ca

u ho

i kh ac n ha

u, go

m 5

cau

ho

i kh

o, 1

0 ca

u ho

i tr un

g b in

h va 15 cau h

oi de Tir 30 cau h

p dUcf

c ba

o nh ie

u d

e ki em t ra g om

5 ca

u kh ac n ha

u sa

o ch

o

tr on

g m5

i d

e nh

at th ie

t ph

ai e

o du b

a lo

ai ca

u ho

i (k ho , tr

so ca

u ho

i d

e kh on

g

it ho

n 2

DS:n= Cl,ClCl+C',,C\,Cl+C%C\f

= 56 78

5 dl

25

Ch

o ta

p hcf

p A

eo n p ha

so ta

p ho

p eo

n e

o 4

ph an tuf eu

a A g ap 2

0 Ia

n s

o ta

p hof

p co

n c

6 2 p ha

k sa

o ch

o s

o ta

p ho

p eo

n e

o k p ha

n tu

f eu

a A la I dn n

u Th§

' Hy

u Nguygn

-V Tn

h Ca

n

Cac h e so' ciia n h i thufc Niutorn ufng vdi n = 0, 1 , 2, 3, c6 t h e s a p x e p

diidfi d a n g t a r n giac dtfofi d a y g o i l a t a r n giac P a t c a n

So h a n g k h o n g chufa x l a so h a n g thuT k + 1 t r o n g k h a i t r i e n sao cho :

4(7 k) o

i trie

n l

a : = 35

-CH

I DA

-6 1

5 s

o hang, s

^r"

(-xy)^

CH

I DA

21 o ~'^^ =

a

a(b + a)

AN

(1 +

x'r =

ci +c\x' +cix' +

+ c

, +

Cf, + + +

+ C;;

= 102

4 = 2" = 2'*^ ^

5C;;-CH

I DA

N

/ 2 -I

0 n = 7

14 TS

V

u The ' Hif

u Nguygn VTn

-h Ca

n

Thay n = 7 vao nhi thuTc Niutofn da cho thi c6 :

He so cua x^° trong khai trien (*) la : a^o = 2C\\ 22

CAC BAITAP

lij GIA

I

33.

Tim so hang khon

g chijf

a x cQa kh

ai trie

n n

hi thijf

c Niuton

i trie

f + (1 + x)

^ + (1

+ xf

+ (

1 + x)^

+ (

1 + x)^"

ta dirgf

c : P(x) = aiox^° +

agx®

+ asx*

* + + aix + a

o

Tin

h ag

£>S : a« =

55

35.

Kha

i trie

n v

a r

ut go

n P(x) = (x + 1)

^ + (x

- 2f

thanh da thufc vd

h vdf

i n nguyen

diiOng t

a c

6 :

a) cL+cL+ + CL=cL+cL+

+

r

b) c; , +

2Ci + 3Ci +

+ nc;;

- n2"-'

CH

I D

AN

a) Kh

ai trie

n P(x) = (x

- 1)^

" r

oi ch

o x = 1

b) Kh

ai trie

n P(x) = (1 + x)"

Tim P'(x) ro

i tin

h P'(l)

37.

Viet kha

i trie

n Niutcfn, bie

n nh

m so' han

g khon

g

phu thuQC

X ,

biet rSn

g : C;; + C;;' + C" ' = 79

DS :a = 792

39.

Tim he so ciia s

o han

g chuf

a x trong

kh

ai tri§

n

DS :a = 210

40.

Tim he so ciia s

i trie

n

j-C- =7(

C-n + 3)

DS :a = 495

+ X

Vx biet

biet

16 ;;'

TS V

u Th

e H

U u Nguygn Vin

-h Ca

n

tap hop xac dinh t h i ta goi la phep thii ngdu nhien Tap hop t a t ca cac k e t qua co the co cua phep thijf ngau n h i e n goi la khong gian mdu

ciia phep thijf do

B i e n co n g a u n h i e n

Mot phep thuf ngau nhien T co khong gian mau la E, m6i tap hop A c

E bieu t h i mot bien co ngdu nhien (lien quan tdfi T) Bien co ngau nhien, chi gom mot phan tuf ciia E dtfoc goi la bien co so cap Bien co dac biet gom moi phan tuf cua E la bien co chdc chdn B i e n co khong chufa phan tuf nao ciia E la bien co khong the co, k i hieu 0 H a i bien

CO A, B ma A n B = 0 t h i A va B difofc goi la hai bien co xung khdc

X a c s u a t c i i a b i e n co' n g a u n h i e n

Phep thuf ngau n h i e n co khong gian mau E gom n bien co scf cap co kha nang xuat hien dong deu (dong kha nang) Bien co ngau n h i e n A

gom k bien co' so cap (ciia E) t h i xac sudt cua bien co ngdu nhien A,

Neu A va B la hai bien co bat k i t h i

d) Neu A va A la hai bien co' ngau nhien ddi lap

(tufc la A u A = E, A n A = 0) t h i P(A) = 1 - P(A)

5 B i e n co dpc l a p v a quy t ^ c n h a n x a c s u a t

Hai bien co ngau n h i e n A va B cCing lien quan vdfi mot phep thuf ngau

nhien la doc lap uoi nhau neu viec xay ra hay khong xay ra cua bien

CO nay khong anh hifdng t d i kha nang xay ra cua bien co kia

Quy tdc nhan xac sudt

Neu hai bien co ngau n h i e n A va B doc lap vdfi nhau t h i

P(A n B) = P ( A ) ^ P ^

g b

a Ia

n tun

g c6 diing mo

N

a)

Ki hie

u S neu don

CH

I DA

, mo

i eac

h la

y 3 vien b

i l

a la

y 1 tap hcf

p do S

o eac

h

lay 2 vien b

i d

o tron

g 4 vien b

i d

o l

a C4 each S

i xan

h la

C 4.

i d

o l

a : P(A) = —i-

^ =

— C ^^ 3

CH

I DA

u Th

e HU

u Nguyin VTnh Ca

-n

(4 each chon chuf so hang don v i , 8 each chon chiJ so hang t r a m , 8 each chon chuf so h a n g chuc)

So cae so co 3 chOf so khac nhau l a so chSn l a : n = 72 + 256 = 328 Xac suat ciia A la : P(A) = — = 0,3644

900

44 Mot to hoc sinh co 10 ngiTcfi gom 6 nam va 4 nuf, chon ngau nhien mot

a) Ca ba nguofi diioc chon deu la nam

b) Co i t n h a t mot trong ba ngUoti duoe chon l a nam

45 Cho 8 qua can co k h o i lu'Ong Ian liigt la 1kg, 2kg, 3kg, 4kg, 5kg, 6kg,

7kg, 8kg Chon n g a u n h i e n 3 qua can T i n h xac suat de t o n g k h o i lu'Ong ba qua can dirge ehgn k h o n g virgt qua 9kg

C H I D A N

So each ehgn 3 qua can t r o n g 8 qua can (so phan tif eiia k h o n g gian mau) l a : Co = = 56 each

^ 1.2.3

A la bien co tong khoi lirgng 3 qua can dirge ehgn khong qua 9kg Cae bien

CO so cap thuan Igi cho A (thuoe tap hgp A) co 7 bien co la :

(1; 2; 6), ( 1 ; 3; 5), (2; 3; 4), ( 1 ; 2; 3), ( 1 ; 2; 4), ( 1 ; 2; 5), ( 1 ; 3; 4)

Xac suat ciia A : P(A) = — = 0,125

56

46 Tung mot Ian hai con sue sSc dong chat can doi

a) T i n h xac suat bien co tong so cham t r e n hai con sue sSc bang 8

b) T i n h xac suat bien co tong so cham t r e n hai con sue sac l a mot so le hoac mot so chia het eho 3

a) Ca

c bie

n c

o so

f ca

p th ua

n Igf

i b ie

n c

o A ( to ng

so ch am b an

g 8 ) la

: (2

; 6)

,

(6; 2) , (3

; 5) , (5

; 3) , (4

; 4) X

ac su

at cu

a A la : P(A ) = —

36

b) Bi en co' B : t on

g s

o ch am la

so l

e hoa

c ch ia h

et ch

o 3

Go

i B i la b ie

n co ' t on

g so ' c ha

m ba ng so' le , B

2 l

a bi en

co to ng

so

ch

am

la m

ot s

o ch ia h

et ch

o 3 , t

hi

B = B iu B2

P (B) = P (B,) + P (B2)

- P(B

i n

B2)

Bi x ay r

a kh

r mo

t co

n su

e sic

n ay m

at ch in , mo

t co

n na

n c

o sc

f ca

p : (1;

2), (1;

4), (1;

6), (3; 2) , (3

; 4) , (3

; 6) , (5;

2),

(5; 4)

,

(5; 6), (2

; 1), (2

; 3), (2

; 5), (4

; 1), (4

; 3), (4

; 5), (6

; 1), (6

; 3), (6

; 5)

B2 x ay r

a vdf

i 1

2 b ie

n s

o s

o ca

p : (1;

2), (2; 1) , (1

; 5) , (5

; 1) , (2

; 6), (3

; 6), (6

; 3), (4

; 5), (5

; 4)

Bi en

co B

i n B2 to ng

so ch am

le v

a ch ia h

et ch

o 3 g om

6 b ie

n c

o so

f

cap :

(1; 2), (2

; 1), (3

; 6), (6

; 3), (4

; 5), (5

; 4)

P (B) =

P(Bi) +

P (B2)

- P(B

i n

B2) = — + — - — = — = -

n (m

ot eac

h do

c la p) va

o mo

t mu

c ti eu

c su

at bS

n tr ii ng d ic

h tr on

g mo

t Ia

n bS

n cii

a ngUd

i

thuf nh

at la 0,9;

cua ngiicfi thu

f ha

i la 0,7.

T in

i vi en deu t ru ng d ic

h

b) it n ha

t c

o mo

t vi en t ru ng d ic

h

c) Ch

i C

O m

ot vi en t ri in

g

CH

I DA

N

a) Ggi

Ai la b ie

n c

o ngiTcf

i thO

f n ha

t b an t ni ng dich, A

2 l

a b ie

g dich A la b ie

n c

o c

a ha

i v ie

n de

u t ni ng dich t

hi A = A

p vcf

i n ha

u ne

n A

i va A2 la doc la

p ne

n

P(A) = P(Ai

n A2) = P(Ai).P

(A2) = 0,

9 0, 7 = 0,63

b) Go

i B la b ie

n c

o c

o

it nh

at mo

t vi en d an t ni ng d ic

h :

B =

Ai u

A2

P (B) =

P(Ai

u A2

)

= P(Ai) +

P (A2)

0,97

c) Go

i C la b ie

n co , ha

i ng iid

i ba

n mo

i ng iid

i mo

t vi en c

g dic

h

Bi en

co C x ay r

a kh

i ng iid

i th

ii nha'

t tn in

i thij

f ha

Ai A2 ) = P (A iA 2) + P (A iA 2) -P (A iA

2 n A1A2)

20 ,.' , TS V

u Th

e Hu

u l\lguy§n Vin

-h Ca

n

Do A i , A2 doc l a p t h a n h thuf A i , A2 doc l a p , A i v a A 2 doc l a p v a b i e n

N e u goi ( i , j , k ) l a s6' s a n p h a m x a u t h e o thuf t i i ciia cac p h a n thuf

n h a t , thuf h a i , thuf b a vdfi i , j , k n g u y e n dUOng vdri 0 < i , j , k < 3 K h i do

B x a y r a ufng v6i cac bo ( 1 ; 1; 1), ( 1 ; 2; 0), ( 1 ; 0; 2), (2; 0; 1), (2; 1; 0),

(0; 1; 2), (0; 2; 1) B i e n e6' d o i l a p ciia B x a y r a tUOng ufng vdfi cac bo (3; 0; 0), (0; 3; 0), (0; 0; 3) Ro t i n h P ( B ) = 1 - P ( B ) t h o n g q u a t i n h P ( B )

0

50

a) b)

0 d)

51

52

53

a) b) c)

54

a) b)

Tung don

g tie

a) 1

6 pha

n tu

f ^

) = c) P(B

) = —

do, 5 qua cau xanh (cCing kic

n nga

u nhie

n 2 qua cau Tin

t xanh

DS

: a) P(A) =

pi

pi

c) P(C) = b) P(B

)

C 2

d) P(D) =

i B, G, N, O, O thanh mot da

y ngang

DS

: P(A) =

0 gia

i nh

DS

: P(A) = ^ 0,0009

cau den Ho

p thu

f ha

i dUn

g 6 cau trSng 4

Hai qu

a ca

u riit r

a cCin

g mau

DS

: a) P(A) = 0,42 b) P(B

) = 0,54 c) P(C

) =

0,46

Co ha

i t

o ho

c sinh T

o I

co 9 nam, 6

nijf T

o H

co 7 nam, 8

nU Cho

n

ngau nhien

m

oi t

o mo

t ngUoti T

t mo

t nam

DS :

a) P(A) =

114

22

5

b) P(B) =

177 225'

-h Ca

n

CHOYEM € II mm mm vn BAT PHifiG mm m s

§1 PHl/dNG TRINH, BAT PHl/dNG TRINH BAC NHAT MQT AN

N h i thufc bac nhat : f = ax + b (a ?t 0)

Gia t r i x = - — t a i do f c6 gia t r i bang 0 goi la nghiem cua n h i thufc

^ +

5x2 + 4x

N

(2) (3)

20

-X

^ 11

0 -1

30

-X

120 -1

-b) x'

^ x' -

X +

1 =

0 « x'(x

- 1) - (x

- 1) =

0 » x'' -

^ + 5x2 + 4x - 12 = 0

»(

x' 1)+

2(x

l) + 5(x

^^ 1) + 4(x

0 o (

x l)(x

+ 2){x^

+ x + 6) =

1-2; 11

2 Giai, bie

n lua

n the

o tha

m s

o m phiiong

trin

h :

m^x 3

-m = 9x + m

^ (1)

CH

I DA

N

(1)

o

(m^ 9)x = m

a l

a vd

i m ?i ±

3 t

hi phiion

g trin

h (1) c

6 nghie

m :

X =

m + 3m

m

m^

-9 m-3

+ Ne

u m = 3, phiTOn

g trin

h (1) c

6 dan

g : 9x

- 9 = 9

x +

9

Phiiong trin

h na

y v

6 nghiem

+ Ne

u m = -

3 t

hi phifon

g trin

h (1) c

6 dan

g : 9x + 9 = 9

x +

9

Phiforng trin

h v

6 din

h : a^x +

b = ax + a

b (1)

tron

g d

o a, b la cac tham so thiTc Giai, bie

h tren

CH

I DA

N

(l)oa(

a l)x = b(

a 1)

-+ Ne

u a(

a 1)

-; ^ 0, tuf

1 t

hi (1) c

b ^

b

a(

l) a

a-+ Ne

u a = 0, phuon

g trin

h (1) c

6 dan

g :

Ox = -

b

Trong trudng hop nay c6 h

ai kh

a nan

g :

- Ne

u b ;^

0 (tijf

c l

a a = 0, b ;^

0) phiion

g trin

h v

6 nghiem

- Ne

u b = 0 (tufc la

a =

0, b = 0) m

-h Ca

n

+ Neu a = 1, phUofng t r i n h (1) c6 dang : Ox = 0, phifdng t r i n h c6 t a p

nghiem la moi so thiTc

a 3

2 + Neu a = 0, (2) c6 cac nghiem : x = ± —

3

tren true so

Bieu dien tap nghiem d h i n h ben Hinh h

2

x >

m(m + 2)

2 + Neu m < - 2 (tat n h i e n m 0) t h i (1) c6 tap nghiem x <

m ( m + 2) + Neu m = - 2 bat phiiang t r i n h t r d t h a n h Ox > - 1

Moi so thuc deu la nghiem ciia (1)

I D

AN

(a'^ ab)x + b" -

ab a

b,.

N A /

A u

^

>

^

f 1) <

= > <

0 <

= > (a

x b) <

0 (a

hi (1) c

6 dan

g O

x - 0 < 0 v6 nghiem

< a < b th

h :

a) (2

x 3X5

- X)

> 0

b)2^

nghiem

x = —

<5va5-x<

0 vd

i x > 5

(2x 3X5 -

x)

- 0 + 0

X -G O

—5

- 0 +

X +

5

- 0 +

+

2x-l

X +

5

+

- 0 +

26 '

TS V

G Th

e Hii

u Nguygn Vin

-h Ca

n

Ta CO : (1) <=> < 0 <=> < 0 o x < —

2 3 6 7 10(2x + l ) 15(3x-2) 6(4x) 41x - 20 ^

Bat phiiong trinh (2) c6 tap nghiem la : x < - 2

Vay nghiem ciia bat phuofng trinh kep (*) la : -5 < x < - 2

Hoc va on luyen theo CTOT mon Toan THPT ', 2 7

§2 PHlfOiN

G TRINH , BA

a) Cong

thiic nghiem :

Tong qudt :

Biet thil

c A = b

^ 4ac

-+ Ne

u A < 0, (1) v

6 nghiem

+ Ne

u A = 0, (1

) C

O nghie

m ke

p x = 2a

+ Ne

u A > 0, (1

N/A

_ -h + ^fA

~ 2a ' "

t thuf

c th

u gp

n A' = b'^ -

ac

+ Ne

u A' <

0, (1) v

6 nghiem

b'

+ Ne

u A' =

a

+ Ne

u A' >

_ b' +

hi

S =

X

i + X2 =

a

P =

X1X2 = -

hi ca

c s

o

nay l

a nghie

m cii

a phu'On

g trin

h : x^ - Sx + P = 0

m s

o m phiiOng

trin

h :

Jm

- l)x

^ 2(m + l)x + m = 0 (1)

-CH

I DA

N

+ Ne

u m-

l =

0 c^

m=

l, phLfon

g trin

h (1) l

a ba

c nhat, c

6 nghie

m du

y

nhat x = —

4

+ Ne

u m

;tl=

>m

-l 9i O,

ap dung con

m + l)'-^ - (m

- Dm = 3

m +

1

+ Ne

u m = — t

hi A' =

0, c

6 nghie

m ke

p x, = x„ = ^ -

- —

Vdri m = 0, hai phi/ang t r i n h khong c6 nghiem chung

Gia sijf m ;t 0, biet thufc ciia (1) la : A i = m^ + 2 > 0, V m , ciia (2) la :

Biet thijfc A = (m - l)'^ - 4(5m - 6) = m'^ - 22m + 25 Ta can chon gia

t r i ciia m sao cho A > 0 de (1) c6 2 nghiem X i , X2 Theo dinh l i Vi-et :

Cac gia t r i m = 0 va m = 1 thoa m a n dieu k i e n A > 0 Vay m = 0 hoac

m = 1 la cac gia t r i can t i m

Cho phuan

g trin

h v6i

tha

m s

o m

- 2(

m + l)x + 2

m +

5 =

0 (1)

h (1) c

CH

I DA

N

Gia sij

f ( 1) C

i + X2)^

+ 8X 1X

^ +

8

A da

t min k

hi m = -

3 Gi

a t

ri m = -

3 tho

a man dieu kie

n :

A' = (m +

1)^

- (2

m + 5) >

0

Vay m = -

15.

Cho phuon

g trin

h : x^

- 8

x +

12 =

0 (1 )

Khon

g ca

n gia

i (1 ), ha

y la

p mo

t phuon

g trin

h c

6 ha

i nghie

m l

a

nghich dao cii

a ca

c nghie

m cii

a ( 1)

CH

I DA

la

X ], X

— Phifon

g trin

„

X + =

0

Thay

Xi + X2 =

h ca

n tim :

, 2

X + — =

16.

Tim cac gi

a t

ri cii

a tha

m s

o m trong

+ m = 0 (1) c

6 4 nghiem

phan biet

b) x'' - 2(m

6 h

ai nghie

m pha

n biet

CH

I DA

N

a) Phu'on

g trin

h (1) c

6 4 nghiem

neu phUcfng

trin

-S =

>0 <

::

>0

<m

<9

h (2) c

h :

X^ 2(m

S :

m <

3

30 ,.'

TS V

u The ' H

§3 BAT PHl/OiNG TRINH BAG HAI MQT AN

KIEN THCfC

a) Dinh li ve ddu cua tarn thiic bdc hai :

Goi A = b^ - 4ac la biet thijfc ciia t a m thufc (1)

• Neu A < 0, Vx e M, f(x) c6 cung dau vdfi a, nghia la : af(x) > 0, Vx e R

• Neu A = 0 t h i f(x) = 0 vdi x = - — va f(x) ciing dau vdi a, vdi m o i

2a

X ^ - — , nghia la : af{x) > 0 Vx e R, f(x) = 0 c:> x = - —

2a 2a

• Neu A > 0 t h i fix) c6 hai nghiem phan biet X i , X2 (xi < X2), fix) trai dau

vdfi a v6i x e (xi; X2) va ciing dau w6i he so a vdri x e (-co; xi) hoac x e

(X2; +ac), nghia la : af(x) < 0, Vx e ( x i ; X2)

af(x) > 0, Vx e ( - » ; xi) hoac Vx G (X2; +00)

b) Dinh li ddo ve ddu ciia tam thiic bdc hai

Cho t a m thufc f(x) = ax^ + bx + c (a ?t 0) va so a

Neu af(a) < 0 t h i f(x) c6 hai nghiem X i < X2 va X i < a < X2

f(x) = ax^ + bx + c = 0 (a ?t 0) CO hai nghiem phan biet

la CO SO a e Rlaf{a) < 0

fix) = ax^ + bx + c = 0

cac so cho triidfc, l a : f ( a ) f ( p ) < 0

c) So sdnh mot so vai cdc nghiem ciia tam thiic f{x) = ax^ + bx + c (a 0)

• Neu af(a) = 0 t h i a la nghiem

• Neu af(a) > 0 ta xet t h e m A = b^ - 4ac

i) Neu A < 0 t h i f(x) v6 nghiem, khong so sanh

I DA

- 5, c

6 h(x) =

o x = -1,

; h(x) <

0 v

di -1 < x < —

h(x) = 2x^ - 3x

- 5 + 0

0 +

+

g(x) = 4x^ - 19x +

12

+ 1

+ 0 -

- 0 +

h(x)

g(x) + 0

+ 0

-+

Tir ban

g xe

X

= 1,

X = -

2

, x = 4

h v

a bie

u die

n tap nghiem

tre

-^ + 4x + 3 < 0

CH ID

AN

a) Xe

t da

u tam

thuTc

ve tra

i : f(x) = 2x^ - 5x + 2

A = 5'-^

- 1

6 >

0, f(x) =

0 o

X

i = X2

= 2

f(x) >

0 (f(x) Clin

g da

u vdf

i 2) vdf

i x <—

hoac x > 2 Va

h la: T = {

Bieu

dien

tap

nghiem tre

n tru

e so' (pha

b) La

m nhii cau a) Ta

p nghie

m T

- (-3

; 1)

-1

'^MMM^C'

*

-3 -1

32

i: '".

TS V

O Th

e Hi^

u Nguyin Vln

-h Ca

n

19.Giai he bat phi/cfng t r i n h va bieu dien tap nghiem tren true so :

Tap nghiem ciia he (I) : T = T i n T2 = [2; 4)

20 Trong moi triTcfng hofp, t i m cac g i ^ t r i tham so m de :

3 ngoai khoang hai nghiem la : m , = - va m2 = 5

• Ne

u m

?t

1 f(x) =

0 o

X i

= 1, x = 3

p nghie

m cu

a (1) th

b) phiron

g trin

h : 4x^ - (3m + l)

x

m

2 =

0 (2)

; 2)

CH ID

AN

a) Ne

u m = 1, phifon

g trin

h (1) c

6 nghie

m du

y nha

t x = 3

a phifcfn

g trin

h ba

c hai A

g trin

h

f(x) = (m

0 m < -

5

b) Phiidn

g trin

h (2) c

; 2) neu m lay ca

c

gia tr

+ 2) >

0

af(

-l) = 4[

4 + (3m + 1) - m

- 2] >

0

af (2) = 4[16 - 2(3m + 1) - m

- 2] >

0

^ S 3m+

1 ^

-1 < — = <

9m^ + 2m + 3

3 >

0

2m + 3 > 0

-7m+

12 >

0

-8 < 3

1

-3 < m < 5

23

Tim cac gi

- 3 = 0

-h Ca

n

• Neu 2 - m < 0 <=> m > 2 t h i (1) c6 hai nghiem thoa man x i < 2 < X2

• Neu 2 - m = 0 o m = 2 t h i (1) c6 cac nghiem 1 = x i < X2 = 2

• Neu 2 - m > 0 < = > m < 2 t a xet t h e m biet thufc

• Neu m - 1 < 2 o m < 3 t h i (1) c6 tap nghiem Ti = [ m - 1; 2]

• Neu m - 1 > 2 o m > 3 t h i nghiem cua (1) la : Ti = [2; m - 1]

Xet tarn thtifc ve t r a i ciia bat phiiang t r i n h (2) :

g(x) = x^ - ( m + 2)x + 3(m - 1) c6 cac nghiem x = m - l v a x = 3

• Neu m < 4 tap nghiem ciia (2) la : T2 = (-00; m - 1] u [3; +GO)

• Neu m > 4 tap nghiem ciia (2) la : T2 =*{-oo; 3] u [ m - 1; +00)

Tap nghiem T = Ti n T2 cua he (I) n h u sau :

+ Neu m < 3 t h i (I) CO nghiem chung duy n h a t x = m - 1

§ 3.

PHl/OfNG

T RI NH , B AT P HU ON

G TR IN

H

CHlfA

G IA T RI T UY ET D

OI

KiEN THirC

1 Phulofn

g t ri nh chiJa

da

u g ia t

ri tu y^

t d

oi

La phiJcfng

trin

gia

tr

i tuye

t doi

W d« :

I

2x -

i gia

i phiTOn

g trin

h l

a diT

a tre

n din

=

I B(x

) | (A(x))^

= (B(x

))l

2 Ba

t phvfcfn

g t ri nh chufa

da

u gi

a t

ri tu y^

t d

oi

Cac phLrdn

g trin

g'(x)

f(x)

I >

g(x) f(x) <

-g(x) hoa

c f(x) >

g(x)

f(x)|

<

g(x) o-g(

h cha

t cii

a gi

a tr

h :

a) |x_+ 4

| 2x = 7 (1)

-CH

I DA

h nghi

a x + 4 | =

(A) (B)

X +

4 vdi

X >

-4

-(x + 4) vdr

i x < -

4

(1) « (x +

2x = 7

)-x >

-4

-(x + 4)-2x = 7

u Th

e Hy

u Nguy§n VTnh Ca

-n

Vay phiTcfng t r i n h (1) c6 nghiem duy nhat x = - 3

b) De CO bieu thiifc khong chufa gia t r i tuyet doi tiiOng duong ta lap bang sau

4

5

TiTOng t y nhif vay, ta loai x = 2 va x = —

PhUcfng trin

h (1) c

6 ca

c nghie

m : =

-2 -1

1 +^

2 6

x +

81

2(x-^ + 6x + 8 ) 0 -2 (x

- + 6x + 8

l 0 2(x^ +

6x + 8 ) 2(x^ + 6x + 8 ) 2(x^ +

6x

+ 8

)

x-^- 1 x'^

- 1 x^ -

1 0 -

x^

+ 1

0 x' -

1

Bie

u thijf

c (2 ) 3x^ + 12x + 1

5 =

30 -x

12 x- 17

^-=3

0 (c)

(d ) (e)

Ng hi

?m

(c) Sx

^ + 12x +

15 =

(e) o

3x2 ^ + 1

5, x = 1

x

3

a)

X = -

11 b)

X =

1 va

X = —

2 <

3x

1 <

2 -1 < 3

N

(1) -(3

x 3) <

(x^

- 2

x 3) <

x^

-5

x <

0 0

< X <

5

o 2

< X <

5

38

O TS V

n

§4 PHlJdNG TRINH, BAT PHl/dNG TRINH CHlfA CAN THlfC

KIEN THLfC

a) VfU) = g(x)

b) Vf(x)=>/gOO <=> i

1 Phiicfng trinh chiia can thi?c

PhiTcfng phap chung de giai phiicfng trinh c6 bieu thufc chufa an nam

dudi dau can la nang len luy thiia bieu thufc cua phiicfng trinh vdi cac dieu kien di kem de c6 diicfc phi/cfng trinh khong con chufa an trong dau can Cung c6 the dat an phu de dan den cac phiTofng trinh, he phifcfng trinh dcfn gian d§ giai hofn Mot vai dang phifong trinh chufa can thufc dofn gian la :

g(x) > 0 f(x) = (g(x))'

f (x) = g(x) f(x)>0, g(x)>0'

2 Bat phifofng trinh chufa can thufc

Dang ccf ban cua bat phiicfng trinh chufa can bac hai

f (x) > 0 g(x) < 0 g(x) > 0

f (x) > [g(x)]^

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

f (x) < [g(x)]^

De giai cac bat phifcfng trinh chufa can thufc, ta diia ra cac dieu kien xac dinh roi luy thiia mot each thich hop cac ve ciia bat phiTcfng trinh

de giam dan cac dau can thufc, dan dan difa t6i bat phiiong trinh, he

bat phiicfng trinh khong chufa can thufc Cung c6 the dat cac an phu hoac bien luan cac ve cua bat phifcfng trinh de tim nghiem

X >

4

.x' lOx +

x =

9

3>

x-0 x>

^ llx + 2

12x + 7 = 0 (1

)

CH

I DA

1 >

0 V

x G

Dat t -

h (1) da

n de

n

t^

7 =

49 o

Xi.2

= 1 ± x/S

t = -

1 (loai)

t =

7

6x' 12x +

7 t'

-b) Nhan xet : x^ - 3x +

3 =

2]

+ > 0

Vx e

Dat t = x

^ 3x +

3, ta

CO :

9 + t' - 6t

0 <

t <

3 ot

3x + 2 = 0ox

^-=l hoac x

11 (11)

CH

I DA

N

Ve pha

i (1) l

a : x^ - 6x +

11 = (x - 3)^ +

-e tr

ai

Vx

2 + V4

- X

- l.V

x

2 + 1.V4 -

x

< 7(1' +l')[(

2) + (4-x)] =

2

40 ';

T S^ Vu The Huu - Nguyin Vinh Can