Học ôn luyện theo cấu trúc đề thi môn toán vũ thế hựu

510.76
rS. VO THE HirU - NGUYEN VINH CAN

H419V

•

•

HOC a ON LUYIN
T H E O C A U T R U C D E THI
MON

I


TS. VU THE HlfU - NGUYEN VINH CAN

nioc & ON LUYEN
T H E O C A U T R U C D E THI

THi; VIEN TINH BINH THU*N

ON THI
DAI HOC
Ha

NQI

N H A X U A T B A N D A I HQC QUOC G I A H A N O I

H O C vA

NHA XUAT B A N DAI HQC QUOC GIA HA NQI

16 Hang Chuoi - Hai Ba TrUng - Ha Npi
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Che ban:

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Trinh

bay bia:

THAI HOC

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SACH LIEN KET
O NLUYEN THEO CAU TRUC D ET H I M O N TOAN THPT

IVla so: 1L - 65DH2013
In 2.000 cuon, I

S6'xu3it ban: 246 - 2013/CXB/8 - 33/DHQGHN, ngay 25/02/2013
Quyet dinh xua't ban so: 58LK-TI\i/QD - NXBOHQGHN ngay 01/03/2013.
In xong va nop li/u chieu quy II n§m 2013.


GlAl TICH

Hpc va on-luyen theo CTDT mon Toan THPT

3


M f

IJBISOTOHdPVflMSy'fiT
§1. HOAN VI, CHINH HdP, TO HOP
KIEN THlTC

1.

Quy t^c CQng v a quy t i c n h a n
a) Quy tdc cong :

Neu tap )igp A c6 m p h a n til, tap hgfp B c6 n p h a n tuf va giCTa A va B
k h o n g CO p h a n tijf chung t h i c6 m + n each chon m o t p h a n tuf thuoc A
hoac thuoc B .
b) Quy tdc nhdn :
Be hoan t h a n h m o t cong viec A p h a i thiic h i e n h a i cong doan. Cong
doan I CO m each thiic h i e n , 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 p h a i qua k cong doan. Cong
doan thijf i ( 1 < i < k ) c6 m i each t h i t h i c6 m i . m 2 . . . m k each de hoan
t h a n h cong viec A .
2.
Hoan vi
M o t tap hop A huTu h a n c6 n p h a n tuf ( n > 1). M 5 i each sap thuf til eac
phan tiif ciia t a p hop A duoc goi l a m o t hoan vi ciia n p h a n tuf eua A .
Dinh li : So hoan v i khac nhau ciia n p h a n til bang :
Pn = n ( n - l ) ( n - 2 ) . . . 2 . 1 = n !
3.
C h i n h hrfp
M o t tap hop A hOTu h a n gom n phan tuf (n > 1) va so nguyen k
(0 < k < n). M o i tap hop eon eua A gom k phan til sSp theo mot thuf tiT
nhat d i n h dLfgrc goi la mot chinh hap chap k cua n phan tuf.
Dinh li : So c h i n h hop chap k ciia n p h a n tuf bang :
A;; = n ( n - l ) ( n - 2)...(n - k + 1) =
(n - k ) !
4.

(Quy irde : 0! = 1).
T o hofp
Cho t a p h o p A hufu h a n c6 n p h a n tuf ( n > 1) va so nguyen k
(0 < k < n). M o i tap hop con gom k p h a n tuf ciia A (khong t i n h thuf tU

eac p h a n tuf) g o i l a m o t to hop chap k cua n p h a n tuf.
n'
A''
Dinh li : So to hop chap k cua n p h a n tuf l a : C'' =
= —(n-k)!k!
k!
He qua:

Cl=C:=l;

0^= Cr''; C ^ = C!; + C ^ \
HQC va on luy$n theo CTDT mon Toan THPT S 5


1.
a)

BAI TAP
C h o cac chuf so 2 , 3, 4 , 5, 6, 7.
C o b a o n h i e u so' t\i n h i e n c6 h a i chuf so ducfc t a o n e n til t a p h o p c a c
chuf so d a cho.
b)

C o bao n h i e u so t i i n h i e n c6 h a i chuf so k h a c n h a u difOc tao n e n tiT t a p
hcrp chuf so d a cho.

CHI
a)

DAN

D e t a o m o t so c6 h a i chuf so t a t h i f c h i e n h a i c o n g d o a n :

1. C h o n m o t chuf so l a m chuf so h a n g chuc : c6 6 k e t qua c6 t h e .
2.

C h o n m o t chuf so l a m chuf so h a n g d o n v i : c6 6 k e t qua c6 t h e .
T h e o q u y t ^ c n h a n so k e t q u a t a o t h a n h cac so c6 h a i chuf so tii t a p
hgfp 6 chCif so d a cho l a : n = 6 x 6 = 3 6 so'.

b)

L a p l u a n g i o n g n h i f c a u a ) n h u i i g liTu y sir k h a c b i e t so vdfi trifofng h o p
t r e n of cho so d u g c t a o t h a n h c6 h a i chuf so k h a c n h a u . D o do t a c6 k e t
q u a n h u sau :
1. C h o n m o t chuf so l a m chuf so' h a n g chuc : c6 6 k e t q u a c6 t h e .
2.

C h o n m o t chuf so l a m chuf so h a n g d o n v i : c6 5 k e t q u a c6 t h e ( v i chuf
so n a y p h a i k h a c chuf so h a n g chuc d a c h o n trifdrc do).
T h e o q u y t a c n h a n : so cac so c6 h a i chuf so k h a c n h a u difcfc t a o t h a n h
tCr t a p h o p 6 chuf so d a cho l a : n ' = 6 x 5 = 3 0 so.
Cdch

khac

: M o i so c6 h a i chuf so t a o t h a n h tiT 6 chuT so d a cho l a m o t

t a p hop c o n s^p thuf tiT g o m h a i p h a n tuf tiT 6 p h a n tuf d a cho. D o do so
cac so C O h a i chuf so k h a c n h a u t a o t h a n h tiT 6 chuf so d a cho l a so
c h i n h hop c h a p 2 cua t a p hop 6 p h a n tuf.

n = Ag = 6.5 = 30 so.

2.

C h o t a p h o p cac cha so 0, 1, 2, 3 , 4, 5, 6.
C o bao n h i e u so t U n h i e n c6 4 chuf so k h a c n h a u tCmg d o i tii t a p

b)

C o bao n h i e u so t u n h i e n c6 4 chuf so tii t a p h o p cac chuf so d a cho.

a)

hop

cac chuf so d a cho.

CHI
a)

DAN
Co 6 each c h o n chuf so h a n g n g h i n (chuf so d a u t i e n p h a i k h a c 0), 7 e a c h
c h o n chff so h a n g t r a m , 7 e a c h c h o n chuf so h a n g chuc v a 7 e a c h c h o n
chuf so h a n g

d o n v i . T h e o q u y tSe n h a n

: so each t a o t h a n h so t i f

n h i e n 4 chuf so tii t a p h o p 7 chiJ so d a cho l a :

N = 6 x 7 x 7 x 7
b)

= 2 0 5 8 so.

C o 6 e a c h c h o n chuf so h a n g n g h i n , k h i c h o n x o n g chuf so h a n g

nghin

c o n l a i 6 chuf so k h a c vdi chuf so h a n g n g h i n d a c h o n . V a y c6 6 e a c h
chon

chuf so h a n g

t r a m . K h i d a c h o n chuf so h a n g

nghin va hang

t r a m , e o n l a i 5 ehOf so k h a c v d i cac chuf so d a c h o n . D o do eo 5 e a c h

6 :S; IS. Vu The Hi/u - NguySn Vinh Cin


chpn chOf so h a n g chuc. Tifcfng tir, c6 4 each chon chOr so h a n g don v i .
Theo quy tac n h a n . So cac so txi n h i e n c6 4 chOf so khac nhau tCfng doi
difcfc tao t h a n h tix t a p hop 7 chuf so da cho l a :
N ' = 6 X 6 X 5 X 4 = 720 so.
Cdch lap luan khdc : M o i so tiT n h i e n c6 4 chOf so khac nhau tao
t h a n h tCr tap hop 7 chOf so da cho l a m o t c h i n h hgrp chap 4 ti^ t a p hgfp
7 chuf so m a cac c h i n h hgfp nay k h o n g c6 chuT so 0 or dau. Do do so cac

so CO 4 chijf so khac nhau tiT 7 chijf so l a :
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 n h i e u each de to xep h a n g vao l(Jp.
b) Co bao n h i e u each de to xep h a n g vao Idfp sao cho h a i b a n A v a B eua
to luon d i canh nhau va A dufng tri/dtc B .

CHI

D A N

a) So each xep h a n g bang so hoan v i ciia 10 p h a n tiif.
N i = 10! = 3628800 each.
b) Coi h a i b a n A va B n h i i m o t ngudi. Do do so each xep h a n g ciia to de
vao 16p t r o n g do h a i b a n A v a B d i l i e n nhau bang so hoan v i cua 9
phan tijf.
N2 = 9! = 362880 each.
4.
Co bao nhieu each xep 6 ngiicfi ngoi vao m o t ban a n 6 cho t r o n g cac
triiofng hcfp sau :
a) S^p 6 ngiTofi theo h a n g ngang ciia m o t ban a n d a i .
b) S4J) 6 ngiTori ngoi vong quanh m o t b a n a n t r o n .
CHI

D A N

a) M o i each ngoi theo h a n g ngang l a m o t hoan v i cua 6 p h a n tijf. So' each

sap xep l a : 6! = 720 each.
b) Gia suf 6 ngifofi a n diTOc d a n h so thijf t i f la : 1, 2, 3, 4, 5, 6 v a m o t each
sap xep theo b a n t r o n n h i i h i n h .
2 5 1 3 6 4

(1)

5

1 3 6 4 2

(2)

1 3 6 4 2 5

(3)

3 6 4 2 5 1

(4)

6 4 2 5 1 3

(5)

4 2 5 1 3 6

(6)
Neu t a eat b a n t r o n a v i t r i giCfa 2 va 4 r o i t r a i d a i theo b a n ngang
t h i t a CO hoan v i (1) tUofng ijfng m o t each xep ngiiofi ngoi theo ban a n

dai. TiTofng txi c a t of v i t r i giufa 5 va 2. N h u vay m o t each sap xep theo
ban t r o n tiiOng ufng vdri 6 each s a p xep theo b a n d a i . Do do so each
Hoc va on luyen theo CTDT mon Toan THPT

7


5.

x e p 6 ngiTofi n g o i q u a n h b a n a n t r o n l a :
N = — = 120 e a c h .
6
M o t t o CO 15 ngifofi g o m 9 n a m v a 6 nOf. C a n l a p n h o m cong t a e eo 4
ngUdri. H o i eo b a o n h i e u e a c h t h a n h l a p n h o m t r o n g m o i trifcfng h o p
sau d a y :
a) N h o m c6 3 n a m v a 1 nur.
b ) So n a m v a nCf t r o n g n h o m b a n g n h a u .
c) P h a i CO i t n h a t m o t n a m .

CHI

DAN

9 8.7
a) So e a c h c h o n 3 n a m t r o n g so 9 n a m l a : Cj! = "
= 84
1.2.3
So e a c h c h o n 1 niJ t r o n g so 6 nOr l a : Cg = 6
So each t h a n h l a p n h o m g o m 3 n a m v a 1 nCf (theo quy tSc n h a n ) l a :
N i = C^C^ = 5 0 4 e a c h .

b ) So e a c h l a p n h o m g o m 2 n a m v a 2 nuf l a :
N2=

C^C^ = —
'
'
1.2

1.2

= 540 each.

c) So e a c h t h a n h l a p n h o m 4 ngu'ofi t r o n g do c6 i t n h a t 1 n a m l a : 1
nam,

3 nO h o a c 2 n a m , 2 nuf hoae 3 n a m , 1 nur h o a c 4 n a m .
— Cg.Cg +

.Cg 4~ Cg.Cg "I" Cg

_ 6.5.4
9.8 6.5.
9.8.7 ^
9.8.7.6
= 9.
+
+
.6 +
= 1350 each.
1.2.3

1.2 1.2
1.2.3
1.2.3.4
Ghi chu : C u n g eo t h e l a p l u a n n h i f sau :
C a t o CO 15 ngiTcfi. So e a c h l a p n h o m 4 n g U d i t u y y l a :
^ 4 ^ ^ — 1^4 .x1 3o. 1^2 ^ ^ g g g ^ ^ ^ ^
_
1
5
.
~
1.2.3.4
1.2.3.4
So e a c h l a p n h o m 4 ngUofi t o a n nuf l a : C ^ = Cg = 6.5. = 15 e a c h .
1.2
So e a c h l a p n h o m 4 n g i / d i eo i t n h a t 1 n a m l a :
N = CJs - C^ = 1365 - 15 = 1 3 5 0 e a c h .
T r o n g m a t p h a n g eo n d i e m p h a n b i e t ( n > 3 ) t r o n g do eo d i i n g k

6.

d i e m n S m t r e n m o t d i i d n g t h S n g (3 < k < n ) . H o i c6 bao n h i e u t a m
g i a c n h a n cac d i e m d a cho l a d i n h .
CHIDAN
Cuf 3 d i e m k h o n g t h S n g h a n g t a o t h a n h m o t t a m g i a c . So cac t a p h o p
c o n 3 d i e m t r o n g n d i e m l a : C^. So cac t a p c o n 3 d i e m t r o n g k d i e m
t r e n diTcfng t h i n g l a : C^. So t a m g i a c c6 3 d i n h l a cac d i e m d a cho
l a : N = C^ - Cl t a m g i a c .

8 ;

TS. Vu The Hi/u - Nguyen VTnh Can


7. a) Co b a o nhieu so t i i n h i e n l a 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 n h i e n c6 6 chuf so doi mot khac nhau, trong do c6
dung 3 chuf so le, 3 chuf so chSn (chuf so dau t i e n phai khac 0).
CHI D A N

a) So can t i m c6 d a n g : x = a^agaga^agag t r o n g do a i ,

ae

l a y cac chOf

so 0, 1, 2,
8, 9 vdfi a i ?i 0, ai aj v d i 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 a i la chuT so le nen c6 5 each chon tiT cac chuf so 1, 3, 5, 7, 9.
Con l a i a2a3a4a5 l a m o t chinh hop chap 4 eiia 8 chuf so con l a i s a u k h i
da chon ae va a i . Theo q u y t^c n h a n , so cac so can xac d i n h 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 t a p 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 t a p X i la : Ci? = 10 each.
So each lay 3 p h a n tuf thuoc X2 l a : Cg = 10 each.
So' each ghep 3 p h a n tuf l a y txi X i v o i 3 p h a n tuf l a y tii X2 l a :
C^C^ = 10.10 = 100 each.

So' day so' eo thuf t i f eiia 6 p h a n tuf diioc ghep l a i l a :
100.6! = 72000 day.
Cac day so c6 chuf so 0 a dau g o m 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 b a i la :
N2 - C ^ C ^ 6 ! - C ^ C ^ 5 ! = 72000 - 7200 = 64800 so.
8.

M o t hop diing 4 v i e n b i do, 5 v i e n h i t r a n g va 6 v i e n b i vang. NgLfofi
ta chon r a 4 v i e n b i t i f hop do. H o i c6 bao nhieu each l a y de t r o n g so
b i j a y r a k h o n g dii ca 3 mau.

CHI D A N

Cdch 1 : So each chon 4 v i e n b i k h o n g d u 3 mau b a n g so' each chon 4
v i e n b a t k i trir d i so each chon 4 v i e n c6 ca 3 mau.
N = Cjg - (C^ .C^ .C^ + C^ .C\ + Cl .C\) = 645 each.
Cdch 2 : So each chon 4 v i e n b i k h o n g d u 3 m a u bang so each chon 4
v i e n m o t m a u (4 do, 4 t r a n g va 4 vang) cong v d i so each chon 4 v i e n
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 v a n g hoac 2 do, 2 v a n g hoae 3 do, 1 v a n g hoae 1 t r S n g , 3 v a n g
hoae 2 trSng, 2 v a n g hoac 3 t r a n g , 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

9.
Co 15 n a m va 15 nuT k h a c h du l i c h dijfng t h a n h vong t r o n quanh ngon
lijfa t r a i . H o i c6 bao n h i e u each xep de k h o n g eo triTcfng hop hai ngi/6i
eCing gidfi canh nhau.
CHI DAN
ThiTe h i e n sap xep bang each d a n h so 30 cho t r e n di/orng t r o n tii 1
den 30 va cho n a m dufng so le nuT dufng cho so chSn hoac ngi/gc l a i (2
each). Co 15! each sSp n a m dufng trong cae cho so' le (hoac chSn) va
15! each sdp nuf dufng t r o n g cae cho so ch^n (hoac le).
V i diidng t r o n 30 cho nen m o i each sSp xep nao do xoay tua 30 cho
theo dung t r a t tiT do ta cung chi eo mot each sap t r e n dirofng t r o n
(xem b a i so 4).
Do do so each sSp xep theo difcfng t r o n 30 k h a c h du l i c h theo yeu cau
2.(15!)(15!)
, ,
de la : N =
= 14!.15! each.
30
10. Chufng m i n h cae dang thufc :
a)
+
+ ... +
= ——- (1) t r o n g do A^ la c h i n h hop chap 2 eua n.
Ag
A3
A„
n
b)

CHI

C;; = C;;:; + Cl;}^ + ... + Cl:\) t r o n g do C; la to hop chap r ciia n.
DAN

a) V(Ji k e N, k > 2 ta c6 : A', = k ( k - 1)
^ =
=
- '
^
A^
k(k-l)
k-1
k
Thay k = 2, 3,
n vao (*) ta c6 ve t r a i ciia (1) la :
(1
(I
V
f 1
1^
— —
— — —
— —
+... +
[n-1
nj
u
2. l 2 3v
b) Theo t i n h chat eua to hop ta c6 :

=

Cn_3

(*)

+ C^^g

Cong ve vdi ve cae dang thufc t r e n ta

diTOe :

c:;-c::;+c-^3+c::u...+c:-uc:
Do C;; = C;::} = l n e n thay C;: d dang thufc cuoi bori C^:}

ta

dugfc

dang

thufc (2) can chufng m i n h .
11.

Chufng m i n h bat dang thile :
t r o n g do k e N, k < 2000,

C^ooi +

^ C\Z + CfZ

la to hop chap k eua n p h a n tuf.

1 0 t4l TS. Vu The Hi/u - Nguyin Vinh CSn


CHI DAN
V(Ji 0 < k < 1000 t h i
^2001
k+1
2001
k
'^2001 -

2001!

(k + l ) ! ( 2 0 0 0 - k ) !

k +1

k!(2001-k)!

2001!

2001-k

'-^2001 -

'^2001

-

, plOOO _ p l O O l
•• - '-^2001 ~ ^ 2 0 0 1

pk
*-^2001

<1

p k + 1 ^ plOOO
^2001 — ^2001

plOOl
^^2001
-111-k

M a t k h a c , v 6 i 1 0 0 0 < k < 2 0 0 0 t h e o t i n h c h a t ciia t o h o p C ' = C;;'^ t a

- CIZ'^

• ^ 2 0 0 1 ^ ^-^2001 ~" '^2001

' "^2001

< Cir, +CIZ\i 0 < 2000 - k < 1000,
-

'^2001

' ^2001

0 < 2 0 0 1 - k < 1 0 0 0 t h e o p h a n t r e n d a chufng m i n h .

C A C BAI TAP Tl/ GIAI
12.

TCr d i e m A d e n d i e m B n g i / d i t a c6 t h e d i q u a C h o a c d i q u a D v a
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 e a c h ,
t i r C d i t h a n g d e n B c6 3 e a c h . TCr A d i t h a n g d e n D c6 3 e a c h tix D d i
t h a n g d e n B eo 4 e a c h .

a) H o i txi A CO b a o n h i e u e a c h d i tdfi B ?
b)

H o i tCr A d e n B r o i til B trdf v e A

A /

\

CO b a o n h i e u e a c h ?
DS

:

3 X^/

a) 18 e a c h

b) (18)^ e a c h .

13.

4

D

TCr 7 chOf so 0, 1 , 2 , 3, 4, 5, 6 eo t h e g h i dirge b a o n h i e u so tiT n h i e n
m o i so' CO 5 chCT so k h a c n h a u tCrng d o i .

DS : 2 1 6 0 so.
14.
a)
b)

C h o t a p h o p cac chOf so X = |0; 1 ; 2 ; 3; 4 ; 5; 61.
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.
D u n g t a p h o p X c6 t h e g h i dirge b a o n h i e u so t i r n h i e n c6 5 chOf so'
k h a c n h a u tCrng d o i .

c) DCing t a p h g p X c6 t h e g h i dugc b a o n h i e u so tiT n h i e n c6 5 ehCt so
k h a c n h a u l a so e h S n .
DS
15.

:

a ) 6.7* so

b ) 6 l 5 . 4 . 3 so

e) A ^ + 15A^ so.

M o t t o h o c s i n h c6 5 n a m , 5 nOf x e p t h a n h m o t h a n g d o c .

a) Co b a o n h i e u e a c h x e p k h a c n h a u .
b) Co bao n h i e u each x e p h a n g sao cho h a i ngircfi dijfng k e n h a u k h a c g i d i .
DS : a ) 1 0 ! e a c h
16.

b ) 2(5!)^ e a c h .

M o t i g p CO 2 5 n a m h o c s i n h v a 2 0 nuf h o c s i n h . C a n c h o n m o t n h o m
c o n g 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 s a u

a) B a h o c 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)

B a hoc s i n h c6 i t n h a t m o t nuf.
DS

17.

:

a) C;;^ e a c h

b) 25.C^o e a c h

c) C'^^ -Cl,

each.

Co bao n h i e u e a c h p h a n p h o i 7 do v a t cho 3 n g U d i t r o n g cac trUcfng
h o p sau :
M o i n g u d i i t n h a t m o t do v a t v a k h o n g q u a 3 do v a t .

b)

M o t ngUofi n h a n 3 do v a t , eon 2 ngUcfi m o i ngUcfi h a i do v a t .

a)

DS
18.

: a) 3.C^C^ e a c h

M o t to

CO

b) S.CtCl

+ SCl.Cl

each.

9 n a m v a 3 nOf.

Co bao n h i e u e a c h c h i a t o t h a n h 3 n h o m m o i n h o m 4 ngUofi v a t r o n g

b)

Co bao n h i e u e a c h c h o n m o t n h o m 4 ngUcfi t r o n g do eo 1 nijf.

a)

m o i n h o m c6 1 nuf.
DS

: a) 3.C^ e a c h

b) 3.C;;.2C^ = 10080 each.

19. T i m cac so n g u y e n d u o n g x, y t h o a m a n cac d a n g thufe :
6
f)S

20.

:

X

^ ^ ^ " 5 ^

" 2 ^

•

= 8, y = 3.

Co bao n h i e u so t U n h i e n chain c6 4 ehuf so d o i m o t k h a c n h a u .
DS

21.

•.n=

Al+

4.8.8

= 7 6 0 so.

C h o d a g i a c d e u 2 n d i n h AiA2...A2n, n > 2 n o i t i e p t r o n g d u d n g t r o n .
B i e t r a n g so t a m g i a c c6 d i n h l a 3 t r o n g 2 n d i e m t r e n n h i e u g a p

20

I a n so h i n h ehuf n h a t eo d i n h l a 4 t r o n g 2 n d i n h t r e n . T i m so n .
£>S : n = 8.

22.

T i m so t U n h i e n n , b i e t r a n g C" + 2C;, + 4 C ' + ... + 2 " C " = 2 4 3 .
: n = 5.

T r o n g m o t m o n h o c , t h a y g i a o eo 3 0 c a u h o i k h a c n h a u , g o m 5 cau

24.

G i a i b a t p h u o n g t r i n h ( v d i h a i a n n , k G N)

23.

h o i k h o , 10 c a u h o i t r u n g b i n h v a 15 c a u h o i de. T i r 30 cau h o i do c6
t h e l a p dUcfc bao n h i e u de k i e m t r a g o m

5 cau k h a c n h a u sao

cho

t r o n g m 5 i de n h a t t h i e t p h a i eo d u b a l o a i cau h o i ( k h o , t r u n g b i n h ,
de) v a so c a u h o i de k h o n g i t h o n 2.
DS:n=
25.

Cl,ClCl+C',,C\,Cl+C%C\f

= 56785 d l .

C h o t a p hcfp A eo n p h a n tuf ( n > 4). B i e t r S n g so t a p h o p eon eo 4

p h a n tuf eua A g a p 2 0 I a n so t a p hofp c o n c6 2 p h a n tuf ciia A . T i m so'
t U n h i e n k sao cho so t a p h o p eon eo k p h a n tuf eua A l a I d n n h a t .
: n = 18, C^g > C\^'

flS

o

k = 9.

12 ;.'; TS. Vu Th§' Hyu - Nguygn VTnh Can


§2. NHI THlfC NIUTCfN
K I E N THLTC
1.

N h i thufc N i u t c f n
(a + b ) " = Cf,a"b° + Cla"-'h

+ ... + ClJa'^'^b'^ + ... + C > V = Xc;;a"-''b''
k=0

Ydi
2.

q u y vide a, b ^ 0, a" = b° = 1 , C° = 1 .

Tarn giac P a t c a n
Cac h e so' ciia n h i thufc N i u t o r n 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 g i a c dtfofi d a y g o i l a t a r n g i a c P a t c a n .
1

n =0
n =1

1

n =2

1

n =3

2

1

n =4

1

3

4

n =6

1
3

1;
4 : 1

6

5

n =5

1

10

10
20

15

6

5
15

1
6

1

T r o n g m 6 i k h u n g t h e h i e n t i n h c h a t t o n g h a i h e so h a n g t r e n

so h a n g or h a n g diTdfi h a y C^'^ + Cj^ =

bang

Cl;^i.

BAITAE^
26.

T i m cac so h a n g k h o n g chufa x t r o n g k h a i t r i e n n h i thufc N i u t O n ciia
vdfi X > 0.
/X

J
(Trich

de TSDH

kho'i D nam

2004)

CHI D A N
Vdfi

X >

0, t a

1

CO

: \/x = x ^ ;

I

_ i

= x "*

—j=r

%/x

1

f
/X +•

/x;

= (x'^ + x
7-k

_k

7-1
-

C ° x 3 + Cix

1

3 X 4 + c?x

7-2

2

^ X 4

7

+ ... + C)x^ x"-* + ... + C l x ' ^
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 c h o :

Hoc va on luyen theo CTOT mon Toan THPT ' 1 3


27.

C^'x

x"^ =C^x 3 ^ " * =C,'x*'

tufc la phai c6 :
- - = 0
3k = 4(7 - k) o k = 4
3

4
Vay so hang khong chijfa x trong khai trien la :
= 35.
Tim so hang chinh giijfa cua nhi thufc NiutOn : (x^ - xy)^*.

CHI DAN

KJiai trien nhi thufc (x^ - xy)^^ c6 15 so hang, so hang chinh giiJa la
so hang thuf 8 c6 dang :
C L ( x ^ r " ( - x y ) ^ = -CLx^^xV^ - -3432x^V^
28.

Tim so hang thuf tii cua khai trien nhi thufc

a
b- a

+

b2 ' - „a2

A"

. Biet

a

rang he so ciia so' hang thuf ba cua khai trien do bSng 21.
CHI DAN

Trong cong thufc nhi thufc NiutOn (A + B)" so hang thuf 3 ciia khai
trien c6 he so la : C? = 21 o
Vay so hang thuf tii cua khai trien

29.

~'^^ = 21

o n = 7

a
b^-a'^^'
la :
b-a • +

7-3
^b^-aM
a(b + a)
C?
= 35
[b-aj
b-a
^ a
J
Biet rSng tong tat ca cac he so cua khai trien nhi thufc (x^ + 1)" bang
1024. Hay t i m he so ciia so hang chufa x^^ trong khai trien do.

CHIDAN

(1 + x'r = ci +c\x' +cix' + ... + c y +... + c:y"

Cho X = 1 ta dirge : (1 + 1)" =

+ c;, + Cf, + ... +

+ ... + C;;

= 1024 = 2" = 2'*^ ^ n = 10
Do do he so cua x'^ la :

=

6!4!

= 210.
/

30.

2

Trong khai trien nhi thuTc NiutOn ' nx
14
chufa x\t rSng 5C;;-' = C'l

-I

1 ^ , X ^ 0, hay t i m so hang
x^

(Trich de TSDH khoi A -

2012)

CHI DAN

bCr

14

= Ct

n(n - l)(n - 2)
5n =
1.2.3
n(n'^ - 3n - 28) = 0

n = 7

TS. Vu The' Hifu - Nguygn VTnh Can


Thay n = 7 vao nhi thuTc Niutofn da cho t h i c6 :
1

-c*

12

.2,

I-]

X

X

2

+ ... +

V

f-1

So hang chufa x^ trong khai trien la so' hang thuf k + 1 sao cho :
7-k

12;

k
.Xy

X^
27-k

^ 2 ( 7 - k ) - k = 5=^k = 3

Vay so hang chufa x la : -C^
31.

CHI

1

7.6.5

1

^5^_35^3

1.2.3 2'
16
Tim he so cua so' hang chufa x^° trong khai trien nhi thufc NiutOn ciia
(2 + x)", biet rang
3"C° - 3""'C;, + 3"'C^ - 3"-''Cl + ... + (-1)"C,'; = 2048.
(Trich de TSDH khoi B - 2007)
vXy

D A N

Xet khai trien nhi thiifc Niutcfri :

(x - D" = c>" - c^x"-' + c'^x"-' - cf,x"-^ +... + (-1)"c;;

Cho X = 3 ta diroc :

2" = 3"c;; -3"-'c;, +3""'cf, -3"-'c^

+ ... + (-i)"c;;

= 2048

2" = 2048 = 2" => n = 11
Thay n = 11 vao khai trien (2 + x)" ta diioc :

(2 + x ) " = 2"c?, + 2^°c;jx +... + 2c;?x^°'+ c;;x"
32.

CHI

(*)

He so cua x^° trong khai trien (*) la : a^o = 2C\\ 22.
Khai trien bieu thufc P(x) = x ( l - 2 x f + x^(l + Sx)^** va viet P(x) diTdi
dang da thufc vdri luy thifa tang cua x. Hay t i m he so ciia x'' ciia da
thufc do.
D A N

Ta

CO :

x ( l - 2xy^ = x(C° - 2C^x + 2'C5'x' - 2''C^x' + 2'C5'x' - 2'C^x'^)
x ' ( l + Sx)"" = x^(C°o + 3Cj„x + 3'Cfnx' + 3'C;'nx' +
+ 3*C,'x" +3^C?„x^+... + 3 " ' C ; V ° )
=^ P(x) = C;|x + (C?o - 2C;)x' + ... + (3''C-^„ + 2''C^)x^ +... + 3^"c;°x
Vay he so ciia so hang chufa x'' la :
as =

1 f) q Q

+ 2'Ct = 2 7 . ^ ^ ^ ^ + 16.5 = 3320.
'
1.2.3
Hqc va on luyen theo CTDT mon Toan THPT

.'' 1 5


33.

CAC BAITAP lij GIAI
T i m so h a n g k h o n g chijfa x cQa k h a i t r i e n n h i thijfc N i u t o n .

if

X + —
X

.

j

DS : 924.

K h a i t r i e n va r u t gon P(x) = (x + 1)^ + (x - 2f t h a n h da thufc v d i luy

35.

K h a i t r i e n va r u t gon bieu thufc :
P(x) = ( 1 + x f + (1 + x)^ + (1 + xf + (1 + x)^ + (1 + x)^"
ta dirgfc : P(x) = aiox^° + agx® + asx** + ... + aix + ao
T i n h ag.
£>S : a« = 55.

34.

thtra giam dan ciia x. T i m he so cua cac so hang chufa x^ va x^.
DS : He so ciia x^ la : - 6 2 2 , ciia x^ la : 570.
36.

Chufng m i n h vdfi n nguyen diiOng t a c6 :

a) cL+cL+... + CL=cL+cL+... + c r .
b) c;, + 2Ci + 3Ci +... + nc;; - n2"-'.

CHI DAN
a) K h a i t r i e n P(x) = (x - 1)^" r o i cho x = 1.
b) K h a i t r i e n P(x) = ( 1 + x)". T i m P'(x) r o i t i n h P ' ( l ) .

T r o n g k h a i t r i e n n h i thufc

38.

Viet k h a i t r i e n Niutcfn, bieu thufc (3x - 1)^'', tU do chufng m i n h rSng :

37.

x

28 \
15
. H a y t i m so' hang

khong

phu thuQC X , biet rSng : C;; + C;;' + C" ' = 79.
DS :a = 792.
39.

T i m he so ciia so h a n g chufa x

t r o n g k h a i tri§n

+ X

Vx

biet

DS :a = 210.
40.

T i m he so ciia so h a n g chufa x^ t r o n g k h a i t r i e n

biet

C - j - C - = 7 ( n + 3).
DS :a = 495.

16 ;;'. TS. Vu The HUu - Nguygn Vinh Can


§3. XAC SUAT
K I E N THCTC
P h e p thijf n g S u n h i e n , k h o n g g i a n m a u
M o t phep thuf ( t h i n g h i e m ) c6 the lap l a i so I a n t u y y vdfi cac dieu
k i e n co ban giong nhau nhiftig k h o n g the xac d i n h chSc chSn, k e t qua
nao t r o n g m o i I a n thifc h i e n ma chi co the n o i k e t qua do thuoc m o t
tap hop xac d i n h t h i t a 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
M o t phep thuf ngau n h i e n T co k h o n g gian mau la E, m 6 i tap hop A c
E bieu t h i mot bien co ngdu nhien ( l i e n quan tdfi T ) . B i e n co ngau
n h i e n , chi gom m o t p h a n tuf ciia E dtfoc goi la bien co so cap. B i e n co
dac biet gom m o i p h a n tuf cua E la bien co chdc chdn. B i e n co k h o n g
chufa p h a n tuf nao ciia E la bien co khong the co, k i hieu 0. H a i b i e n
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 k h o n g gian mau E gom n b i e n co scf cap co
k h a n a n g xuat h i e n dong deu (dong k h a nang). B i e n co ngau n h i e n A
gom k b i e n co' so cap (ciia E) t h i xac sudt cua bien co ngdu nhien A,

1.

2.

3.

4.
a)
b)
c)

d)
5.

ki hieu P(A) la t i so: P(A) = - .
n
C a c tinh chat cua xac suat
B i e n co ngau n h i e n A bat k i ta deu co 0 < P(A) < 1.
P(0) = 0, P(E) = 1.
A va B la h a i b i e n co xung khSc (tufc A n B = 0) t h i
P(A u B) = P(A) + P(B)
Neu A va B la h a i b i e n co bat k i t h i
P(A L ^ B ) = P(A) + P(B) - P(A n B).
Neu A va A l a h a i b i e n co' ngau n h i e n ddi lap
(tufc la A u A = E, A n A = 0) t h i P(A) = 1 - P(A).
B i e n co d p c l a p v a q u y t ^ c n h a n x a c s u a t
H a i bien co ngau n h i e n A va B cCing l i e n quan vdfi m o t phep thuf ngau
n h i e n la doc lap uoi nhau neu viec xay r a hay k h o n g xay r a cua b i e n
CO nay k h o n g a n h hifdng t d i k h a n a n g xay ra cua b i e n co k i a .
Quy tdc nhan xac sudt
Neu hai b i e n co ngau n h i e n A va B doc lap vdfi nhau t h i
P(A n B) = P ( A ) ^ P ^

7/15 X23'

Ji

Hoc va on luyen theo CTDT mon Toan THPT

. 1 7


41. Tung mot dong tien dong chat va can do'i ba Ian.
a) Khong gian mau c6 bao nhieu phan tijf ?
b) Goi A la bien co, trong ba Ian tung c6 diing mot Ian xuat hien mat sap.
CHI DAN

a ) K i hieu S neu dong tien xuat hien mat sap va N neu dong tien xuat
hien mat ngiifa. Ket qua tung dong tien ba Ian bieu t h i bang day 3
chuf cai S hoac N . Nhu vay khong gian mau gom 8 phan t i i .
E = (NNN; NSN; SNN; NNS; NSS; SNS; SSN; SSS}.
b) Bien co A ba Ian tung dong tien co dung mot Ian xuat hien mat sap
bieu t h i boi tap hcfp
A = ISNN; NSN; NNSl.
Gia thiet dong tien la can doi va dong chat neu cac ket qua ciia phep
thijf la dong kha nang. Khong gian m l u co 8 phan tuf. Bien co A co 3
3
phan tuf, do do xac suat cua A la : P(A) = —.
8
4 2 . Trong mot hop co 4 vien bi mau do, 3 vien bi mau xanh (cac vien bi
chi khac nhau ve mau sic). Lay ngau nhien cung mot liic 3 vien bi.
Tinh xac suat de trong 3 vien bi lay ra co diing hai vien bi mau do.
CHI DAN

43.

Khong gian mau co
bien co sof cap (co
tap hcfp con 3 phan tuf
trong 7 phan tuf), moi each lay 3 vien bi la lay 1 tap hcfp do. So each
lay 2 vien bi do trong 4 vien bi do la C 4 each. So each lay 1 vien bi
xanh trong 3 vien bi xanh la C 3 . So each lay 3 vien bi co 2 vien bi
do, 1 vien bi xanh la C 4 . C 3 each.
Xac suat trong 3 vien bi lay ra co 2 vien bi do la : P(A) = — i - ^ = —.
C^^
35
Chon ngau nhien mot so tiT nhien co 3 chuf so. Tinh xac suat de so
duoc chon la mot so chan co 3 chuf so khac nhau.

CHI DAN

Goi A la bien co so ducfc chon co 3 chuf so khac nhau la so chSn.
Khong gian mau E la so cac so co 3 chU so (9 each chon chuf so' hang
tram, 10 each chon chuf so hang chuc, 10 e a c h chon ehuT so hang dofn
vi) la : 9 X 10 X 10 = 900 so.
So cac so CO 3 chuf so khac nhau la so tan cung la 0 la : 9.8 = 72 so
(9 each chon chuf so hang tram, 8 each chon chuf so hang chuc)
So cac so chfin co 3 chuf so khac nhau co chuf so hang dcfn vi khac 0 la
8.8.4 = 256 so.

18

,

TS. Vu The HUu - Nguyin VTnh Can

(4 each chon chuf so h a n g don v i , 8 each chon chiJ so h a n g 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 l a : P(A) = —

= 0,3644.

900

44.

Mot to hoc sinh co 10 ngiTcfi gom 6 nam va 4 nuf, chon ngau n h i e n mot
nhom 3 ngiiofi eiia to. T i n h xac suat xay ra mot tri/cfng hop diidi day :

a) Ca ba nguofi diioc chon deu l a n a m .
b) Co i t n h a t mot t r o n g ba ngUoti duoe chon l a nam.
CHI

DAN

10 9 8

a) K h o n g gian mau co : C^^ = —'—^

= 120 p h a n tijf.

1.2.3

Co C'l = ^'^'^

= 20 each chon 3 ngi/ofi deu l a nam. Xac suat b i e n co 3

1.2.3

ngiicfi dU'ge chon deu l a n a m l a : P(A) =

C'i

20 1

Cl

120

6-

b) Goi B la bien co 3 ngUdi dirge chon co i t n h a t 1 nam. B i e n co doi lap
eiia B l a 3 nguofi difgc chon deu l a niJ :
—
C'*
1
— 29
P(B) = — L = _ ^ P(B) = 1 - P(B) = — .
C^o
30
30
45.

CHI

Cho 8 qua can co k h o i lu'Ong I a n l i i g t l a 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 s u a t de t o n g k h o i
lu'Ong ba qua can dirge ehgn k h o n g virgt qua 9kg.
DAN

So each ehgn 3 qua can t r o n g 8 qua can (so p h a n tif eiia k h o n g g i a n
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)

46.

Xac suat ciia A : P(A) = — = 0,125.
56
Tung mot Ian h a i con sue sSc dong chat can doi.

a) T i n h xac suat b i e n co t o n g so cham t r e n hai con sue sSc b a n g 8.
b) T i n h xac suat b i e n co t o n g so cham t r e n hai con sue sac l a m o t so le
hoac mot so chia het eho 3.
CHI

DAN

K h o n g gian mau eo 36 p h a n tijf (6 x 6 = 36 cap so (i; j ) vdri i , j nguyen
dirgng 1 < i < 6; 1 < j < 6).

Hoc va on luyen theo CTDT mon Toan THPT

19


a) Cac b i e n co sof cap t h u a n Igfi b i e n co A ( t o n g so c h a m b a n g 8) l a : (2; 6),
(6; 2), (3; 5), (5; 3), (4; 4). X a c s u a t cua A l a : P(A) = — .

36
b ) B i e n co' B : t o n g so c h a m l a so le h o a c c h i a h e t cho 3.
G o i Bi l a b i e n co' t o n g so' c h a m b a n g so' l e , B2 l a b i e n co t o n g so c h a m
l a m o t so c h i a h e t cho 3, t h i
B = BiuB2
P(B) = P(B,)

+ P(B2)

- P(Bi n B2)

Bi x a y r a k h r m o t c o n sue sic n a y m a t c h i n , m o t con n a y m a t l e , co
18 b i e n co scf c a 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 a y r a vdfi 12 b i e n so so cap : (1; 2), (2; 1), (1; 5), (5; 1), (2; 4), (4;

2), (3; 3), (6; 6), (3; 6), (6; 3), (4; 5), (5; 4).
B i e n co Bi n
cap :

B2 t o n g so c h a m le v a c h i a h e t cho 3 g o m 6 b i e n co sof

(1; 2), (2; 1), (3; 6), (6; 3), (4; 5), (5; 4).
P(B) =

47.

P(Bi) + P(B2) - P(Bi n B2) = — + — - — = — = - .
36 36 36 36 3

H a i x a t h i i cCing b a n ( m o t e a c h doc l a p ) v a o m o t m u c t i e u m o i ngir&i
m o t v i e n d a n . X a c s u a t b S n t r i i n g d i c h t r o n g m o t I a n b S n ciia ngUdi
thuf n h a t l a 0,9;

cua ngiicfi thuf h a i l a 0,7. T i n h xac s u a t t r o n g m o i

t r i l o f n g hgfp sau :
a) Ca h a i v i e n d e u t r u n g d i c h .
b ) i t n h a t co m o t v i e n t r u n g d i c h .
c) C h i CO m o t v i e n t r i i n g .

CHI DAN
a) Ggi Ai l a b i e n co ngiTcfi thOf n h a t b a n t n i n g d i c h , A2 l a b i e n co ngiTcfi thijf h a i
b a n t n i n g d i c h . A l a b i e n co ca h a i v i e n deu t n i n g d i c h t h i A = Ai n A2. Do
h a i ngiicfi b a n doc l a p vcfi n h a u n e n Ai v a A2 l a doc l a p n e n

P(A) = P(Ai n A2) = P(Ai).P(A2) = 0,9.0,7 = 0,63.
b ) G o i B l a b i e n co co i t n h a t m o t v i e n d a n t n i n g d i c h :
B =

Ai u A2

P(B) = P(Ai u A2) = P(Ai) + P(A2) - P(Ai n A2)
= 0,9 + 0,7 - 0,63 = 0,97
c) G o i C l a b i e n co, h a i n g i i d i b a n m o i n g i i d i m o t v i e n c h i co m o t v i e n
t n i n g d i c h . B i e n co C x a y

r a k h i n g i i d i t h i i nha't t n i n g d o n g

thcfi

ngiiofi thuf h a i trUcft h o a c ngiTofi thijf nha't triigft, n g i f d i thijf h a i t n i n g .
P(C)

20

= P ( A i A 2 ^ A i A 2 ) = P ( A i A 2 ) + P ( A i A 2 ) - P ( A i A 2 n A1A2)

, . ' , TS. Vu The Huu - l\lguy§n Vinh Can


Do A i , A 2 doc l a p t h a n h thuf A i , A 2 doc l a p , A i v a A 2 doc l a p v a
CO A j A 2

P{C)
48.

n

A i A2

=0

bien

cho n e n

= P ( A i ) P ( A 2 ) + P ( A i ) P ( A 2 ) - 0 = 0,9.0,3 + 0,1.0,7 = 0,34.

M o t 16 h a n g c6 3 0 s a n p h a m , t r o n g do c6 3 p h e p h a m . Ngiic/i t a c h i a
n g a u n h i e n 16 h a n g t h a n h 3 p h a n , m o i p h a n 10 s a n p h a m .

a)

T i n h x a c s u a t de m o i p h a n c6 d i i n g m o t phe' p h a m .

b)

T i n h xac s u a t de c6 i t n h a t m o t p h a n c6 d u n g m o t p h e

CHI
a)

pham.

DAN
T a t h y c b i e n c h i a nhxi sau : L a y

n g a u n h i e n 10 s a n p h a m t r o n g

30

s a n p h a m t a c6 p h a n thuf n h a t . T r o n g 20 s a n p h a m c o n l a i l a y

ngau

n h i e n 10 s a n p h a m de c6 p h a n thuf h a i . C o n l a i l a 10 s a n p h a m

phan

thuf 3. So' e a c h c h i a n h i i v a y b S n g C^^.Cgo e a c h .
So e a c h c h i a de p h a n thuf n h a t eo m o t s a n p h a m x a u l a : C 3 C 2 7

(lay

m o t s a n p h a m x a u g h e p v 6 i 9 s a n p h a m t6't t r o n g 27 s a n p h a m to't).
So e a c h c h i a d e p h a n thijf h a i c6 m o t s a n p h a i m x a u l a : C g C i g . S6
each c h i a de m o i p h a n c6 m o t s a n p h a m x a u l a : C3C27.C2C18.
X a c s u a t de m o i p h a n eo d u n g m o t s a n p h a m x a u l a :
plp9

plp9

P ( A ) = ^^'^"f;

•'^ = 0 , 2 4 6 .

*^:i()^20

b)

G o i B l a b i e n c6 t r o n g 3 p h a n c6 i t n h a t m 6 t p h a n eo m o t p h e
Neu

pham.

g o i ( 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
(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

bo

P(B)

d o n g i a n hcfn.
So' e a c h c h i a de p h a n thuf n h a t c6 3 s a n p h a m x a u , cac p h a n c o n l a i
deu t o t l a : Ci^.C^^.C^i;.
So each c h i a de p h a n thuf h a i eo 3 s a n p h a m x a u , cac p h a n k h a e d e u
l a s a n p h a m t o t l a : Cl^^.Cl.C]-.
So each c h i a de h a i p h a n d a u cac s a n p h a m d e u t6't, p h a n thuf b a c6 3
s a n p h a m x a u l a : C27.C}".
_
^

pSpiT p l O , p l O p l i p V
' ~

plOplO

plOplO
'-';io'-^20

P ( B ) = 1 - P { B ) = 0,911.
Hoc va on luy§n theo CTOT mon Toan THPT .'. 21


BAi TAP Tl/GIAI
49. Tung dong t i e n can doi dong chat b o n I a n .
a) K h o n g gian m^u c6 bao nhieu p h a n tijf.
b) T i n h xac suat bien co dong t i e n n a y m a t sap diing 3 I a n .
0 T i n h xac suat bien co dong t i e n nay m a t sap i t n h a t 2 Ian.
a) 16 p h a n tuf
. ^ P(A)
. -= 1
c) P(B) = — .
b)
50.
a)
b)
0
d)

DS

16

M o t hop dung 7 qua cau do, 5 qua cau xanh (cCing k i c h thudc chi khac
mau s&c). C h o n ngau n h i e n 2 qua cau. T i n h xac suat :
H a i qua cau dixgc chon mau do.
Hai qua cau diicfc chon mau xanh.
Hai qua cau duofc chon m o t do, m o t xanh.
Hai qua cau dUOc chon co i t n h a t m o t qua do.
DS

: a) P(A) =

b) P(B)
pipi

c) P(C) =

d) P(D) = 1 C2
12

51.

Xep ngau n h i e n 5 chijr cai B, G, N , O, O t h a n h m o t day ngang. T i n h
xac suat de duoc chCf BOONG.
1
1
: P(A) = 1

DS

~ 5.4.3

52.

"

60'

M o t d o t x6 so p h a t h a n h 20.000 ve, t r o n g do co m o t giai dac biet,
100 g i a i n h a t , 200 g i a i n h i , 1000 g i a i ba va 5000 g i a i t u . T i n h xac
suat de m o t ngUcfi mua 3 ve t r i i n g m o t g i a i n h a t va 2 g i a i tU.
DS

53.

a)
b)
c)

54.
a)
b)

: P(A) =

^ 0,0009.
"^20000

Co h a i hop dung cac qua cau ciing k i c h thudc, t r o n g lUOng hop thuf

n h a t dung 7 cau trSng, 3 cau den. Hop thuf hai dUng 6 cau trSng 4 cau
den. R u t ngau n h i e n tU m o i hop m o t qua cau. T i n h xac suat :
H a i qua cau r u t r a deu mau t r d n g .
H a i qua cau r i i t r a cCing mau.
H a i qua cau r i i t r a khac mau.
DS : a) P(A) = 0,42
b) P(B) = 0,54
c) P(C) = 0,46.
Co h a i to hoc s i n h . To I co 9 nam, 6 nijf. To H co 7 nam, 8 nU. Chon
ngau n h i e n m o i to m o t ngUoti. T i n h xac suat :
H a i ngUcfi dugc chon co diing m o t nU.
H a i ngUcfi dugfc chon co i t n h a t m o t nam.
177
: a) P(A) = 114
b) P(B) =
DS

225'

225

22 • TS- Vu The Hifu - NguySn Vinh Can


CHOYEM € II. mm mm vn BAT PHifiG mm m s
§1. PHl/dNG TRINH,
BAT PHl/dNG TRINH BAC NHAT MQT AN
K I E N THCTC
Phifcfng t r i n h b ^ c n h a t m p t a n c6 the diia ve dang :
ax + b = 0;a, b e E, a ; t 0

1.

CO mot n g h i e m duy n h a t x = - —.
a
Phvfcfng t r i n h t i c h m p t a n c6 dang :
A(x).B(x) = 0
(1)
Tap nghiem ciia (1) l a tap n g h i e m ciia phiTcfng t r i n h A(x) = 0 hcfp v d i
tap nghiem ciia phiTOng t r i n h B(x) = 0.

2.

A(x).B(x) = 0 o " A ( x ) - 0
B(x)-0'
B a t phiicfng t r i n h b a c n h a t m p t a n c6 the difa ve dang :
ax + b > 0 hay ax + b > 0; a, b e M, a ?t O

3.

• Neu a > 0, ax + b > 0 CO tap n g h i e m

x>- —
a

• Neu a < 0, ax + b > 0 CO tap n g h i e m

x< — .
a
D a u c u a n h i thtfc b a c n h a t ax + b (a 0)

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

4.

Gia t r i x = - — t a i do f c6 gia t r i bang 0 g o i l a n g h i e m cua n h i thufc
a •
Ta CO d i n h l i ve dau ciia n h i thufc nhir sau :
Dinh li
+ Vdi cac gia t r i x < - — t h i f = ax + b c6 dau trai vdi dau ciia he so a.
a
+ Vori cac gia t r i x > — t h i f = ax + b c6 dau ciing dau vdfi he so a.
a

1.
,
a)

G i a i cac phiTofng t r i n h :
10-x 20-X 30-X _
+
+
=3
100
110
120

(1)

Hoc va on luyen theo CTOT mon Toan THPT ..'. 23

b)
- X + 1 = 0
c) x'*_+ 2x^ + 5x2 + 4x - 12 = 0
CHI DAN
a) (1) o

10-X

100

-1

-90-X
C5>

b) x'^ - x ' -

20-X

^ 110
-90-X

(2)
(3)
30-X

-1

120

-90-X

-1

=0

^

100 + 110 + 120 = 0
1
= Oc^ X = - 9 0 .
( 9 0 + x)
•
100 110 + 120
1 = 0 « x ' ( x - 1) - (x - 1) = 0
X +

o

( X - l)(x'^ - 1) = 0

»

x'' - 1 = 0
x - l =0

<=> X =

±1.

c) x S 2 x ^ + 5x2 + 4x - 12 = 0 » ( x ' - 1)+ 2 ( x ^ - l ) + 5(x^ - 1) + 4(x - 1) = 0
(X - DLx'* + 3x''^ + 8x + 12] = 0 o (x - l ) ( x + 2){x^ + x + 6) = 0
x-1 =0
1\
23
> 0 Vx)
(Bof v i x^ + x + 6 = X + x +2=0
Tap n g h i e m 1-2; 11.
2.
G i a i , b i e n l u a n theo t h a m so m phiiong t r i n h :
m^x - 3 m = 9x + m^
(1)
CHI DAN
(1) o (m^ - 9)x = m^ + 3 m
+ Neu m^ - 9 0, nghia la vdi m ?i ±3 t h i phiiong t r i n h (1) c6 nghiem :
m + 3m
m
X =
m^-9
m-3
+ N e u m = 3, phiTOng t r i n h (1) c6 dang : 9x - 9 = 9x + 9
Phiiong t r i n h nay v6 n g h i e m .
+ N e u m = - 3 t h i phifong t r i n h (1) c6 dang : 9x + 9 = 9x + 9
Phiforng t r i n h v6 d i n h (mpi so thiic deu la n g h i e m ciia phuong t r i n h ) .
3.
Cho phiiong t r i n h : a^x + b = ax + ab
(1)
t r o n g do a, b la cac t h a m so thiTc. G i a i , bien l u a n theo a, b phu'ong
trinh tren.

CHI DAN
( l ) o a ( a - l ) x = b(a - 1)
+ N e u a(a - 1) ;^ 0, tufc l a neu a 0 va a ;^ 1 t h i (1) c6 nghiem duy n h a t
^ ^ (a-l)b ^ b
a(a-l)
a
+ N e u a = 0, phuong t r i n h (1) c6 dang : Ox = - b
T r o n g t r u d n g hop nay c6 h a i k h a n a n g :
- N e u b ;^ 0 (tijfc l a a = 0, b ;^ 0) phiiong t r i n h v6 nghiem.
- N e u b = 0 (tufc l a a = 0, b = 0) m o i so thiic deu la nghiem.
24

TS Vu The Hi/u - Nguygn Vinh Can


+ N e u a = 1, phUofng t r i n h (1) c6 dang : Ox = 0, phifdng t r i n h c6 t a p
nghiem l a moi so thiTc.
4.
Giai cac phiiorng t r i n h :
a) ( X - 1)-^ + (x + 2f = (2x + 1)'^
(1)
b) 9a_x^ - ISx'^ - 4ax + 8 = 0
(2), a l a t h a m so.
CHI

D A N

a) ( 1 ) « (2x + l ) [ ( x - 1)^ - (x - l ) ( x + 2) + (x + 2f] = (2x + i f
o (2x + l ) [ ( x - 1)- - (x - I X x + 2) + (x + 2f - (2x + 1)^] - 0
o

(2x + l ) ( x + 2)(x - 1) = 0 <i:>

X

=

x = -2,

X =

1.

b) (2) o ax(9x-^ - 4) - 2(9x''^ - 4) = 0

(ax - 2)(3x + 2)(3x - 2) = 0
2
2
+ Neu a ^ 0, (2) c6 cac n g h i e m : x = —,
x =±—
a
3
2
+ Neu a = 0, (2) c6 cac n g h i e m : x = ± —.
3

G i a i cac b a t phLfomg t r i n h diidi day va bieu dien t a p n g h i e m ciia no
t r e n true so.

5.

a) 3(x + 2) - 1 > 2(x - 3) + 4

b) (x - 2)^ + 1 + x'"^ > 2x^ - 3x - 4.

C H I D A N

-

1

0

x -+ 2x
- 2- -25 o X > - 7
a) o 3 3x
5 >>2x
M % « M t
Tap n g h i e m dugc bieu d i e n d h i n h ben.
-^'"^ °
b) o x^ - 4x + 4 + 1 + x^ > 2x^ - 3x - 4
- 4 x + 3x > - 4 - 5

>

6.

X ^ — —9
^
\////////////////////

_X
• ^>>>>>>»m>»>>>>,
Bieu dien tap n g h i e m d h i n h ben.
Hinh h
G i a i va bien l u a n theo t h a m so m bat phifofng t r i n h :
x - l
x+1 ^
,,,
x+
>
(m + l ) x
(1)
m
m

C H I

D A N

c i > —X ^ —9

2
Dieu k i e n xac d i n h m =i'i 0. Chuyen ve, r u t gon ta di/dc : (m + 2)x > —
m
+ Neu m 0 va m > - 2 ( t h i m + 2 > 0) bat phifdng t r i n h c6 n g h i e m
2
x >
m ( m + 2)
2
+ Neu m < - 2 ( t a t n h i e n m 0) t h i (1) c6 tap n g h i e m x <

m ( m + 2)
+ Neu m = - 2 b a t phiiang t r i n h t r d t h a n h Ox > - 1
M o i so thuc deu l a n g h i e m ciia (1).
3X
b
7. G i a i , bien l u a n bat phi/cfng t r i n h vdi t h aHoc
m va
soonaluy?n
vatheo
b: x
+
1
>
1
—
1 )2 5
CTDT mon Toan THPT (•.'.
b
a