Ôn luyện theo cấu trúc đề thi môn Toán Phần 1 Ôn thi tốt nghiệp THPT TS.Vũ Thế Hựu, Nguyễn Vĩnh Cận

510.76
6-454L

d

TS. VO THE HlTU - NGUYEN VINH CAN

TIICO C n U TRUC DC Till
MON


TS. VU THE HlTU - NGUYEN VINH CAN

ON

LUVEN
m

THEO cAu TDUC DEI THI

M6]\N
ON THI TOTNGHIEP THPT QUOC GIA
(2 trong 1)

lUHA XUAT BAN DAI HOC QUOC GIA HA NOI


GIJII

TICH

Hoc va On luy§n theo CTDT mfln Toan THPT E I 3


IK

tX

yYENflEl.fllllSOTOHIfPVflXIJCSU''

§1. HOAN VI, CHINH HdP, TO HOP
KIEN THLTC
1.

Quy tac C Q n g va quy t ^ c n h a n

a) Quy tdc cong :
Neu tap hc(p A c6 m phan tuT, tap hop B c6 n phan tuf va giufa A va B
khong CO phan tuf chung t h i c6 m + n each chon mot phan tuf thuoc A
hoac thuoc B .
b) Quy tdc nhdn :
De hoan thanh mot cong viec A phai thuc hien hai cong doan. Cong
doan I CO m each thifc hien, cong doan I I c6 n each thiTc hien t h i c6
m . n each de hoan thanh cong viec A .
Tong quat, de hoan thanh cong viec A phai qua k cong doan. Cong
doan thuf i (1 < i < k) c6 m i each t h i t h i c6 m i . m 2 . . . m k each de hoan
thanh cong viec A .
2.
Hoan vi
Mot tap hop A hufu han c6 n phan tuf (n > 1). Moi each sap thuf tii cac
phan tuf cua tap hop A dage goi la mot hoan vi cua n phan tuf cua A .

Dinh li : So hoan v i khac nhau cua n phan tuf bang :
P„ = n(n - l)(n - 2)...2.1 = n!
3.
C h i n h hofp
Mot tap hop A hufu han gom n phan tijf (n > 1) va so nguyen k
(0 < k < n). Moi tap hop con cua A gom k phan tuf sap theo mot thuf t i i
nhat dinh dLfOe goi la mot chinh hop chap k cua n phan tuf.
Dinh li : So chinh hop chap k cua n phan tuf bang :
A;; = n ( n - l ) ( n - 2 ) . . . ( n - k + l ) = -

4.

^,
(n-k)!

(Quy \x6c : 0! = 1).
Tohtfp
Cho tap hop A hufu han c6 n phan tuf (n > 1) va so' nguyen k
(0 < k < n). Moi tap hop eon gom k phan tuf eiia A (khong tinh thuf tiJ
cac phan tuf) goi la mot to hap chap k cua n phan tuf.
n!
- A''
Binh li : So to hop chap k cua n phan tuf la : Cj^ =( n - k ) ! k ! k !
Hequd:

Cl^Cl=l;

ik-i

C); = Cr" ; C);,, = C^ + C^

HQC fln luy?n theo CTDT mOn ToSn THPT 0 5


BAI

TAP

1.
Cho cac chff so 2, 3, 4, 5, 6, 7.
a) Co bao n h i e u so tir n h i e n c6 h a i chiJ so dirge tao n e n tCr t a p hgp eae
chCr so da eho.
b) Co bao n h i e u so tir n h i e n c6 h a i ehuf so khac nhau dugc tao n e n ttf tap
hgj) cac chiJ so da cho.

CHI DAN
a)
1.
2.

b)

1.
2.

De tao m o t so c6 h a i chuf so t a thiTe h i e n h a i eong doan :
Chgn m o t chuf so l a m ehuf so h a n g ehuc : eo 6 k e t qua c6 the.
Chgn m o t chuf so l a m ehuf so h a n g don v i : eo 6 k e t qua c6 the.
Theo quy t&c n h a n so' k e t qua tao t h a n h cac so' c6 h a i ehuf so tCr tap
hgp 6 chuf so da eho la : n = 6 x 6 = 36 so.

L a p luan giong n h u eau a) n h i m g liTu y sir khac biet so vdi trirdng hgp
t r e n d cho so dirge tao t h a n h eo h a i chCT so khac nhau. Do do t a c6 k e t
qua nhif sau :
Chgn m o t chuf so l a m ehOf so h a n g ehuc : eo 6 k e t qua eo t h e .
Chgn m o t ehuf so l a m chuf so hang don v i : eo 5 k e t qua eo the (vi chQ
so nay p h a i khac chij; so h a n g ehue da ehgn triTdrc do).
Theo quy tie n h a n : so eae so eo h a i chuf so khac nhau dirge tao t h a n h
tCr t a p hgp 6 eha so da cho la : n' = 6 x 5 = 3 0 so.
Cdch khac : M 6 i so eo h a i ehiJ so tao t h a n h tCr 6 chuf so da eho la mot
tap hgp con sap thuf t i i gom h a i p h a n tuf tu" 6 p h a n tuf da cho. Do do so
cac so C O h a i chuf so khac nhau tao t h a n h tCr 6 ehuf so da eho la so
e h i n h hgp chap 2 cua tap hgp 6 p h a n tuf.
n =
= 6.5 = 3 0 so'.

2.
Cho t a p hgp cac chuf so 0, 1, 2, 3, 4, 5, 6.
a) Co bao nhieu so tir n h i e n c6 4 chuf so tir tap hgp cac ehuf so da eho.
b) Co bao n h i e u so tiT n h i e n eo 4 ehuf so khac nhau tiTng doi tiT tap hgp
eae chuf so da eho.

C H I DAN
a) Co 6 each ehgn chuf so hang n g h i n (chuf so dau t i e n phai khac 0), 7 each
ehgn chiT so' h a n g t r a m , 7 each chgn chuf so' h a n g ehuc va 7 each ehgn
ehuf so h a n g don v i . Theo quy t^c n h a n : so each tao t h a n h so tiT
n h i e n 4 chuf so tCr t a p hgp 7 chij" so da eho l a :
N = 6 x 7 x 7 x

7 = 2058

so.

b) Co 6 each chgn chuf so' h a n g n g h i n , k h i ehgn xong ehC? so' h a n g n g h i n
con l a i 6 chOr so khae v d i ehuf so h a n g n g h i n da chgn. Vay c6 6 each
chgn ehuf so h a n g t r a m . K h i da chgn chiT so h a n g n g h i n va hang
t r a m , eon l a i 5 ehu: so khac v d i cac chuT so da chgn. Do do eo 5 each

6 a

TS. Vu Thg' Huu - Nguyen Wnh

C?n


chon chOf so h a n g chuc. Ttfang t i i , c6 4 each chon chOf so h a n g dcfn v i .
Theo quy tac n h a n . So cac so t u n h i e n c6 4 chijr so khac nhau tCrng doi
ducfc tao t h a n h til t a p hop 7 chOf so da cho l a :
N ' = 6 X 6 X 5 X 4 = 7 2 0 so.
Cdch lap luan khdc : M o i so t i i n h i e n c6 4 chOr so khac nhau tao
t h a n h tCr t a p hop 7 chCf so da cho l a m o t c h i n h h o p chap 4 tCr t a p h o p
7 chuf so ma cac c h i n h hop n a y k h o n g c6 chOf so' 0 or dau. Do do so' cac
so C O 4 chOr so khac nhau til 7 chuf so l a :
N' =
= 7 X 6 X 5 X 4 - 6 X 5 X 4 = 7 2 0 so.
3.

Mot to hoc sinh c6 10 ngufofi xep thCf td thanh hang 1 de vao Icfp. H o i
a) Co bao n h i e u each de to xep h a n g vao Idfp.
b) Co bao nhieu each de to xep h a n g vao Idp sao eho h a i b a n A va B cua
to luon d i canh nhau va A dufng triTdfc B.

CHI

DAN

a) So' each xep h a n g bang so' hoan v i cua 10 p h a n tuf.
N i = 10! = 3628800 each.
b) Coi h a i b a n A va B n h i i m o t ngifdi. Do do so each xep h a n g cua t o de
vao Idp t r o n g do h a i b a n A va B d i l i e n n h a u bang so h o a n v i cua 9
p h a n tuf.
N2 = 9! = 362880 each.
4.
Co bao nhieu each xep 6 ngUc/i n g o i vao m o t b a n a n 6 cho t r o n g cac
trirdng hop sau :
a) Sip 6 ngiidi theo h a n g ngang cua m o t b a n a n d a i .
b) Sip 6 ngLfcfi ngoi vong quanh m o t b a n a n t r o n .
CHI

DAN

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 tuf. So each
sap xep la : 6! = 720 each.
b) Gia sijf 6 ngiicfi a n dirge d a n h so thuf t u l a : 1, 2, 3, 4, 5, 6 v a m o t each
sap xep theo b a n t r o n nhiT h i n h .
2

/^^^T^K^
^

\ ^ _ 3 _ ^

j

(1)

5 1 3 6 4 2 (2)

^\ 3 6 4 2 5

J.

Neu
thi
dai.
ban

5 1 3 6 4

(3)

3 6 4 2 5 1

(4)

6 4 2 5 1 3

(5)

4

(6)

2 5 1 3 6

t a cat b a n t r o n d v i t r i giufa 2 v a 4 r o i t r a i d a i theo b a n ngang
t a C O hoan v i (1) tiTcfng lirng m o t each xep ngudi n g o i theo b a n a n
Tirong t i i cat d v i t r i giufa 5 va 2. N h i i vay m o t each sSp xep theo
t r o n tifOng ufng v d i 6 each sap xep theo b a n d a i . Do do so each
Hpc v4 On luy$n theo CTDT mfln Toan THPT £3 7


xep 6 ngiicri ngoi quanh b a n a n t r o n l a :
N = — = 120 each.
6
5.
M o t to CO 15 ngiJcfi gom 9 n a m va 6 niJ. C a n lap n h o m cong tac c6 4
ngiTcfi. H o i c6 bao n h i e u each t h a n h l a p n h o m t r o n g m o i triTofng hop
sau day :
a) N h o m c6 3 n a m va 1 nvC.
b) So n a m va nOf t r o n g n h o m bSng nhau.
c) P h a i CO i t n h a t m o t n a m .
CHI D A N
9 87
a) So each chon 3 n a m t r o n g so 9 n a m l a : Cg = "
= 84
1.2.3
So each chon 1 nuf t r o n g so 6 nuf la : Cg = 6
So each thanh lap nhom gom 3 nam va 1 nuf (theo quy tac nhan) la :
Ni = C^C^ = 504 each.
b) So each l a p n h o m gom 2 n a m va 2 nuf la :

N 2 = C ^ C ^ = — — = 540 each.
' '
1.2 1.2
c) So each t h a n h l a p n h o m 4 ngudi t r o n g do c6 i t n h a t 1 n a m la : 1
n a m , 3 nuf hoSe 2 n a m , 2 nOr hoSc 3 n a m , 1 nuf hoSc 4 n a m .
N = C^.C^+C^C^+C^C^+C^
^ 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 : Cung c6 the l a p l u a n nhiJ sau :
Ca to CO 15 ngUdi. So each l a p n h o m 4 ngUcJi t u y y l a :
^4
15.14.13.12
, ,
Ci5 = —
— = 1365 each.
1.2.3.4
6 5
So each l a p n h o m 4 ngudi t o a n nuf la : Cg = Cg = — ^ = 15 each.
So each l a p n h o m 4 ngiTdi c6 i t n h a t 1 n a m la :
N = C^5 - C^ = 1365 - 15 = 1350 each.
6.

T r o n g mSt p h a n g c6 n d i e m p h a n b i e t ( n > 3) t r o n g do c6 dung k
d i e m n k m t r e n m o t d i T d n g t h a n g (3 < k < n ) . H o i c6 bao n h i e u t a m
giae n h a n eac d i e m da cho la d i n h .
CHI D A N
C i i 3 d i e m k h o n g t h a n g h a n g tao t h a n h m o t t a m giae. So eac t a p hdp
con 3 d i e m t r o n g n d i e m la : C^. So eac t a p con 3 d i e m t r o n g k diem
t r e n dirdng t h a n g la : C^. So t a m giae c6 3 d i n h la eac d i e m da cho
l a : N = C ^ - C ^ t a m giae.

8

£ 3 TS. Vu ThS' Hi/u - Nguyen V i n h C?n


7. a) Co bao n h i e u so tu n h i e n l a so chSn c6 6 chOr so doi m o t khac nhau
v a chiJ so dau t i e n l a chui so le.
b) Co bao nhieu so tuT n h i e n c6 6 chuf so doi m o t khac nhau, trong do c6
dung 3 chCif 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 dang :
so 0, 1, 2,

8, 9

vdi

x = ajagaga^Hgag

aj ;t 0,

ai

aj

vdi

t r o n g do
1<

i

ai,

ae

l a y cac

chijr

;t j < 6.

- V i X la so chSn nen ag c6 5 each chon tii cac chuf so 0, 2, 4, 6, 8.
- V i a i la chuf so le nen c6 5 each chon tif cac chuf so 1, 3, 5, 7, 9.
Con l a i a 2 a 3 a 4 a 5 l a m o t c h i n h hop chap 4 cua 8 chuf so con l a i sau k h i
da chon ae va a i . Theo quy tSc n h a n , so cac so can xac d i n h l a :
N i = 5.5.Ag = 5.5.8.7.6.5 = 42000 so.
b) M o t so theo yeu cau de b a i gom 3 chuf so tCr t a p X i = {0; 2; 4; 6; 81 va
3 chuf so tii t a p hop X2 = { 1 ; 3; 5; 7; 9) ghep l a i va l o a i d i cac day 6

chuf so CO chuf so 0 dufng dau.
So each l a y 3 chuf so thuoc t a p X i l a : C 5 = 10 each.
So each l a y 3 p h a n tijf thuoc X2 l a : C 5 = 10 each.
So each ghep 3 p h a n tuf l a y tii X i v d 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 tir cua 6 p h a n tuf diicfc ghep l a i l a :
100.6! = 72000 day.
Cac day so c6 chuf so 0 of dau gom 2 chuf so khac 0 cua X i v a 3 chijf so
cua X2 : So cac day so n h u t r e n l a : C ' . C g . S ! = 7200 day.
So cac so theo yeu cau de b a i l a :
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 b i t r S n g va 6 v i e n b i v a n g . NgUcfi
ta chon r a 4 v i e n b i tii hop do. H o i c6 bao n h i e u each l a y de t r o n g so
b i l a y r a k h o n g du ca 3 m a u .

CHI D A N

Cdch 1 : So each chon 4 v i e n b i k h o n g du 3 m a u b a n g so each chon 4
v i e n b a t k i trCr d i so each chon 4 v i e n eo ca 3 m a u .
N = C\, -{ClC\.C\+Cl.C\.C\+Cl.C\.C\)

= 645

each.

Cdch 2 : So each chon 4 v i e n b i k h o n g du 3 m a u b a n g 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) eong v d i so each chon 4 v i e n
h a i m a u (1 do, 3 t r a n g hoac 2 do, 2 t r a n g hoac 3 do, 1 t r S n g hoac 1

do, 3 v a n g hoSe 2 do, 2 v a n g hoac 3 do, 1 v a n g hoSc 1 t r a n g , 3 v a n g
hoac 2 t r a n g , 2 v a n g hoac 3 t r a n g , 1 vang).
N=C^ +

+

+ C^C^ + ClCl + C^C^ +... + C^C^ = 645 each.
Hoc va On luyen theo CTBT mfln ToSn THPT S

9


DAN

CHI

Co 15 n a m va 15 nuf khach du l i c h duTng t h a n h vong t r o n quanh ngon
lufa t r a i . H o i c6 bao nhieu each xep de k h o n g c6 triTdng hop h a i ngiTdi
cung g i d i canh nhau.

9.

ThiTc h i e n sAp xep b k n g each danh so" 30 cho t r e n diTdng t r o n tii 1
den 30 va cho n a m duTng so le nuf dufng cho so chkn hoSc ngiioc l a i (2
each). Co 15! each sSp n a m dufng t r o n g eae ch6 so le (hoac ehSn) va
15! each sap nuf dufng t r o n g cac eho so ehSn (hoftc le).
V i dudng t r o n 30 cho n e n m o i each sap xep nao do xoay tua 30 eho
theo diing t r a t tiT do t a cung chi c6 m o t each sap t r e n ducfng t r o n
(xem b a i so 4).
Do do so each sSp xep theo diiofng 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 cac dSng thuTc :
1
1
= ——- (1) t r o n g do A^ la c h i n h hop chap 2 cua n .
a) -ir + —7r + -. + A„
n

b) c;; = c;;:'i + c;;:'^ +... + c;::; (2) trong do c;; la to hop chap r cua n .
CHI

DAN

a) V d i k e N , k > 2 t a eo : A^ = k ( k - 1)

J_
k(k -1)

k-1

- ~

(*)

T h a y k = 2, 3,
n vao (*) ta c6 ve t r a i cua (1) l a :
(1
1^ (1
1^
_^
1 _ n-1
1
^
+ —
+... + ( 1
—
n
n
2j
^n-l
nj
U
3j
b) Theo t i n h chat cua to hop ta c6 :

u

lr-l

Cn-2 = Cn.3 + C;;_3

Cong ve v d i ve eae dang thufc t r e n t a dirgfc :

c : ; = c - + C - U C - 3 + ... + C - + C :

Do C[ = C[:\ 1 n e n thay C; d dSng thufc euoi boi C^:} t a dUdc dang
thiJc (2) can chufng m i n h .
11.

Chufng m i n h b a t dSng thufc : C^QOI + C'OM ^ C^^i +
t r o n g do k e N , k < 2000,

^Zl

la to hop chap k cua n phan tuf.

£ 2 TS. Vu ThS' Huu - Nguyen Vinh C$n

10


CHI

DAN

V d i 0 < k < 1000 t h i

CLi
C
^

2001!
k!(2001-k)!

pk

^ pk+1
^ pk+2
'-^2001 - '-^2001 - ^ 2 0 0 1

(k + l ) ! ( 2 0 0 0 - k ) !
2001!

^
^ plOOO _ p l O O l
- ••• - '-'2001 " '-^2001

k+ 1
2001-k

^

p k
*-^2001

<1

p k +1
^2001

pi

^ p 11000
i
4- -ilOOl
^2001

'-^2001
- ^ 2

M a t khac, v d i 1000 < k < 2000 theo t i n h chat cua to hop
c6 : C^oo, 4- C^-. =

+ C---

= C"""^ t a

< C - ° + C^ol v i 0 < 2000 - k < 1000,

0 < 2001 - k < 1000 theo p h a n t r e n da ehufng m i n h .

OAQJiMJiMUmk
12.

TCr diem A den d i e m B ngtrdi t a c6 the d i qua C hoac d i qua D v a
k h o n g C O difdng di t h a n g tCr C den D. TCr A d i t h S n g den C c6 2 each,
tCr C d i t h a n g den B c6 3 each. TCr A d i t h ^ n g den D c6 3 each tCr D d i
t h a n g den B c6 4 each.

a) H o i tCr A c6 bao n h i e u each di t d i B ?

2

b) H o i tCr A den B r o i tCr B t r d ve A

A /

\

C O bao n h i e u each ?
DS :

a) 18 each

3 \ ^ / 4

b) (18)2
13.

D

TCr 7 chOf so 0, 1, 2, 3, 4, 5, 6 c6 the ghi dirge bao n h i e u so tu n h i e n
m o i so' C O 5 chuT so' khac nhau tCrng doi.
£)S : 2160 so.

Cho tap hop cac chuf so X = {0; 1; 2; 3; 4; 5; 6}.
a) Dung tap hop X c6 the ghi dufOc bao nhieu so tU n h i e n c6 5 chuf so.
b) Dung tap hop X c6 the ghi dirge bao n h i e u so tU n h i e n c6 5 chuf so
khac nhau tCrng doi.
c) D u n g tap hgp X c6 the g h i difgc bao n h i e u so t i f n h i e n c6 5 chur so
khac nhau la so chSn.
c) A^ + 15AI so.
b) 6 l 5 . 4 . 3 so
BS :
a) 6.7^ so

14.

15.

M o t to hoc s i n h c6 5 n a m , 5 nuf xep t h a n h m o t h a n g doc.

a) Co bao n h i e u each xep khac nhau.
b) Co bao nhieu each xep hang sao cho hai ngircri dufng ke nhau khac gidi.
DS : a) 10! each
b) 2(5!)' each.
16. M o t Idp C O 25 n a m hoc s i n h va 20 nff hoc s i n h . C a n chon m o t n h o m
cong tac 3 ngiTdi. H o i c6 bao nhieu each chon t r o n g m o i triTdng hop sau
a) Ba hoc sinh bat k i cua Idp.
b) H a i nu" sinh va m o t n a m s i n h .
HQC

6n luy$n theo CTDT mOn Toan THPT S

11


c) Ba hoc s i n h c6 i t n h a t m o t nuf.
DS :

a) C% each

b) 25.C^o each

c) C^^ - C^^ each.

17. Co bao n h i e u each p h a n phoi 7 do v a t cho 3 ngirdi t r o n g cac t r u 6 n g

hop sau :
a) M o t ngLfcfi n h a n 3 do v a t , con 2 ngiicfi m o i nguf6i h a i do v a t .
b) M o i ngirdi i t n h a t m o t do v a t va k h o n g qua 3 do v a t .
DS : a) S^.Cl
each
b) 3.C^C,' + 3C^.C^each.

18.

M o t to

CO

9 n a m va 3 nOr.

a) Co bao nhieu each chon m o t n h o m 4 ngUofi t r o n g do c6 1 nuf.
b) Co bao nhieu each chia to t h a n h 3 n h o m m o i n h o m 4 ngiroti va trong
m o i n h o m c6 1 nuf.
DS : a) 3.C^ each
b) 3.C^.2C^ = 10080 each.
19. T i m cac so nguyen dirong x, y thoa m a n eac dang thufc :

DS :
20.

X

= 8, y = 3.

Co bao n h i e u so t i i n h i e n ehSn c6 4 chuf so doi m o t khac nhau.

DS

21.

-.11=

Al+ 4.8.8 = 760 so.

Cho da giac deu 2 n d i n h A i A 2 . . . A 2 n , n > 2 n o i t i e p t r o n g difc/ng t r o n .
B i e t r a n g so t a m giac c6 d i n h l a 3 t r o n g 2 n d i e m t r e n nhieu gap 20
I a n so h i n h chuf n h a t c6 d i n h la 4 t r o n g 2 n d i n h t r e n . T i m so n .
DS :n = 8.

22.

T i m so t\i n h i e n n , b i e t r a n g C° + 2C;, + 4C^ + ... + 2"C;; = 243.
5 S : n = 5.

23.

Giai bat

phiTdng

t r i n h (vdi h a i a n n , k

^"^^

24.

G

N)

< eoA"::'

(n-k)!
T r o n g m o t m o n hoc, t h a y giao c6 30 eau h o i khac nhau, gom 5 cau
h o i k h o , 10 cau h o i t r u n g b i n h va 15 eau h o i de. Tii 30 cau h o i do eo
the l a p difcfe bao nhieu de k i e m t r a gom 5 cau khac nhau sao cho
t r o n g m o i de n h a t t h i e t p h a i eo du ba loai cau h o i (kh o, t r u n g b i n h ,
d i ) va so cau h o i de k h o n g i t h o n 2.
DS:n^

25.

+ C?,C;„C^ + C%C\f = 56785 d l

ClClCl

Cho t a p h d p A eo n p h a n tijf ( n > 4). B i e t r k n g so t a p hdp con c6 4
p h a n tuf cua A gap 20 I a n so t a p h a p con c6 2 p h a n tuf cua A . T i m so
tiT n h i e n k sao cho so t a p hap con eo k p h a n tuf cua A l a lorn n h a t .
£)S : n = 18, Cf^ > C\^^ o k = 9.

12 El TS. VO Thg' Hi;u - NguySn VTnh C?n


§2. N H I T H l f C NIUTOfN
KIEN T H l f C i

1.

N h i thuTc N i u t t f n
(a + b ) " = C ° a " b ° + C\a"-'h

+ ... + C;:a"-''b'' + ... + C > " b " = ^ C ' a " - ' ' b ' '

V d i q u y Lfdc a, b ^ 0, a° = b° = 1 ,

2.

= 1.

Tarn giac P a t c a n
Cac h e so c u a n h i thufc N i u t o n ufng v d i n = 0, 1 , 2 , 3, ... c6 t h e sSp x e p
dudi d a n g t a r n g i a c d u d i d a y g o i l a tarn gidc
1

n = 0
1

n =1

1
2

1

n = 2

3

1

n = 3
n =

Patcan.

4
1

n = 6

10;
15

6

l i

1 4 ;

6

|5

1

n = 5

;3

4

1

1

10

120

1

5
6

15

T r o n g m o 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 b a n g
so h a n g d h a n g d u d i h a y C^-' +0^=

iBAIIAP
26.

C^,.

A

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 c i i a
\

vdi
(Trich

X >

de tuyen

0.
sink

DH kiwi

D nam

2004)

CHI D A N
Vdi

1
-0, t a C O : ^/x = x^ ; - p . = x

X >

7-1

^

7-k

+ ... +

7-2

2

= ( x 3 + x " M ' = C ? X 3 + C ^ X 3 x"^ + C , ' X 3 X '

+ - ± =

C^X 3

k

x " * + ... +

C?X 4

So h a n g k h o n g c h i i a x l a so h a n g thuf k + 1 t r o n g k h a i t r i e n sao c h o :

'

Hoc va 6n luygn theo CTDT m6n Toan THPT £3 13


C^X

x"" =

3

k

7-k

C^X

7-k

k

3 "4

tufc la p h a i c6 :

= C^x"

- ^ = 0

3k = 4(7 - k ) o

Vay so h a n g k h o n g chura x t r o n g k h a i t r i e n la :

k = 4.
=35.

D A N

CHI

T i m so h a n g c h i n h giijfa cua n h i thufc Niutcfn : (x^ - xy)^".

27.

K h a i t r i e n n h i thufc (x^ - xy)^'* c6 15 so hang, so h a n g c h i n h giufa la
so h a n g thuf 8 c6 dang :
Cl,(x^)"-^(-xy)^ = -Clx'\xy
28.

= -3432x^V^

T i m so h a n g thuf tir ciaa k h a i t r i e n n h i thiJc

a
b-a

+

b^-a^

. Biet

r a n g he so cua so h a n g t h i i ba cua k h a i t r i e n do b k n g 2 1 .
CHI DAN
T r o n g cong thufc n h i thufc N i u t o n (A + B)° so h a n g thuf 3 cua k h a i
t r i e n c6 he so la : C!; = 21 <^ n ( n _ - l ) ^
V a y so h a n g thuf tiX cua k h a i t r i e n

- 35

b-a

n = 7
b^-a^^'

h-a

+•

la :

a(b + a)
b-a

29. B i e t rSng tong t a t ca cac he so cua k h a i t r i e n n h i thufc (x^ + 1)" bkng
1024. Hay t i m he so cua so hang chufa x^^ t r o n g k h a i t r i e n do.
CHI DAN
(1 + x^)" = C° + C^x^ + C^x* + ... + C^x^^ + ... + C > ^ "
Cho X = 1 t a dircfc : (1 + 1)" = C°n + C",n + C' + ... + C^ + ... + C"
= 1024 = 2" = 2^° ^ n = 10
10!
Do do he so cua x'^ la : Cf„ =
= 210.
'°
6!4!
30.

^nx^

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

1

, X ^ 0, hay t i m so' hang

14

chufa x ^ b i e t rSng SC;;-^ =
(Trich de tuyen sink DH khoi A CHI DAN
n ( n - l ) ( n - 2)
5Cr' = Cl o 5n =
1.2.3
<:> n(n^ - 3n - 28) = 0 <=> n = 7

2012)

14 rS TS. Vu The' Huu - Nguy?n Vinh CJn


Thay n = 7 vao nhi thufc Niuton da cho t h i c6 :
(,,1

2

\

1
X

v2.

+ ... +

v2.

v2.

+ . . . - C7 1
So hang chufa x^ trong khai trien la so hang thuf k + 1 sao cho :
f 2 S}-^ r 1 A''
^ 2(7 - k) - k = 5 =^ k = 3.
X
27-k
v2.
4

: -C^ rx^^ r r

5 l i . r<3

Vay so hang chufa x la
31.

3

7.6.5
1.2.3

1
2'

16

Tim he so' cua so' hang chufa x^° trong khai trien nhi thufc Niuton cua
(2 + xf, biet rang

3"c° - 3"-'c;, + 3"-'c^ - 3"-'c^ +... + (-irc;; = 2048.

(Trich de tuyen sink DH khdi B - 2007)
CHI D A N
Xet khai trien nhi thufc Niuton :
(x - ir = c°x" - c;,x"-* + c^x"-' - c^x"-3 +... + (-D^C;;
Cho X = 3 ta di/oc :
2" = 3"C"„ - 3"-^C;, + 3"-'C^ - 3"-'C^ + ... + (-1)"C;; = 2048
2" = 2048 = 2 " n = 11
Thay n = 11 vao khai trien (2 + x)" ta diTgfc :

(2 + x)^^ = 2''c'i, + 2'°cjix+... + 2c;;x'°'+ c;;x"

(*)

He so cua x^° trong khai trien (*) la : a^^ = 2CJi = 22.
32. Khai trien bieu thufc P(x) = x(l - 2x)^ + x^(l + 3x)^° va viet P(x) dudi dang
da thufc v6i luy thtra tang cua x. Hay tim he so cua x^ cua da thufc do.
CHI D A N
Ta

CO :

x(l - 2x)' = x(C° - 2C^x +

2'C5'x'

- 2'C^x^ + 2*C^x'' - 2'C^x^)

x^Cl + 3x)^° = x^(C?o + 3C;oX + 3^C?oX^ + 3^C?oX^ +

+ 3'C^oX^ + 3'C%x' + ... + 3^''C;°x^°)
^ P ( x ) = C°x + ( C ° o - 2 C ^ ) x 2 + . . . + (3^C?o+2^C5*)x^+... + 3^'^C;°x^^
Vay he so cua so hang chufa x^ la :
as = 3'C^„ + 2^Ct = 2 7 . ^ ^ ^ + 16.5 = 3320.
1.2.3
HQC V4 On luygn theo CTDT mOn Toan THPT E3 1 5

I


33.

T i m so h a n g k h o n g chiJa x cua k h a i t r i e n n h i thufc N i u t o n .

1
X + —

34.

x_
DS : 924.

K h a i t r i e n va r i i t gon bieu thufc :
P(x) = (1 + x)^ + (1 + x)^ + (1 + x)^ + (1 + x)'' + (1 + x)^"
ta diioc : P(x) = aiox^° + agx^ + agx^ + ... + aix + ao
T i n h as.
DS : as = 55.

35.

K h a i t r i e n va r u t gon P(x) = t x + if + (x - 2)^ t h a n h da thufc v d i luy
thiia giam dan cua x. T i m he so cua cac so hang chufa x^ va x^.
DS : He so cua x^ la : - 6 2 2 , cua x^ la : 570.

36.

Chufng m i n h v d i n nguyen ducfng t a c6 :

a) Cl+Cl+...
b)

+ Cl^Cl+Cl+...

+ Cl:-\

C > 2C^ + 3C^ + ... + 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 ) .
37.

Viet k h a i t r i e n Niutdn, bieu thufc (3x - 1)^^, tiT do chufng m i n h r&ng :

31 6 p 0

olSpl
^16

ql4p2

'16

Ql3p3

'16
28

38.

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

X\/x + X

. Pl6

'16

_

Ol6

A"

. H a y t i m so h a n g k h o n g

1^

phu thuoc X , b i e t rSng : C;; + C""' + C""' - 79.
£>S : a = 792.
39.

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

f 1
— + x^

biet

u
c L i + c L i + ... + C L . , = 2 - - i .
DS :a = 210.
40.

T i m he so cua so' h a n g chufa x^ t r o n g k h a i t r i e n

f 1
I—
— + Vx^

biet

c:;:i-c::,3 = 7(n+3).
D S : a = 495.
1 6 £2 TS. Vu ThS' H^u - Nguygn Vinh CSn


§3. X A C S U A T

1.

P h e p thijf n g a 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 nghiem) c6 the lap l a i so I a n t u y y vdti cac dieu
k i e n co b a n gio'ng nhau nhiTng k h o n g the xac d i n h chSc chfin, k e t qua
nao t r o n g m o i I a n thifc h i e n m a chi c6 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 c6 the c6 cua phep thuf ngSu n h i e n goi la khong gian mdu
cua phep thijf do.
B i e n co n g a u n h i e n
M o t phep thijr ngau n h i e n T co k h o n g gian mSu la E, m o i t a p hop A e
E bieu t h i mot bien co ngdu nhien ( l i e n quan t d i T ) . B i e n co ngfiu
n h i e n , chi gom m o t p h a n tijf cua E duoc goi la bien co so cap. B i e n co
dac b i e t 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
chiJa p h a n tuf nao cua E la bien co khong the co, k i h i e u 0 . H a i b i e n
C O A , B ma A n B = 0 t h i A va B dirge 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 c6 k h o n g gian m a u E gom n bie'n co so cap c6
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 sc( cap (cua E) t h i xac suat cua bien co ngau nhien A,

2.

3.

ki hieu P(A) la t i so:

4.
a)
b)
c)

d)
5.

P(A) = n
C a c tinh chat ciia xac suat
B i e n co' ngau n h i e n A bat k i t a 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 khac (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 ^ B ) = P(A) + P(B) - P(A n B).
Neu A va A la 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 a c n h a n x a c s u a t
H a i b i e n co ngau n h i e n A va B cCing l i e n quan v d i m o t phep thuf ngau
n h i e n la doc lap vai 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 htfdng t d i k h a n a n g xay r a cua b i e n co k i a .
Quy tdc nhdn xac suat
Neu h a i b i e n co ngau n h i e n A va B doc lap v d i nhau t h i
P(A n B) = P(A).P(B).

41.
a)
b)
CHI
a)

b)

42.

CHI

43.
CHI

18 ^

BAI

TAP

Tung mot dong tien dong chat va can doi ba Ian.
Khong gian mau co bao nhieu phan tuf ?
Goi A la bien co, trong ba Ian tung co dung mot Ian xuat hien mat sap.
DAN
K i hieu S neu dong tien xuat hien mSt sap va N neu dong tien xuat
hien mat ngufa. Ket qua tung dong tien ba Ian bieu t h i bang day 3
chuf cai S hoSc N . Nhii vay khong gian mau gom 8 phan tuf.
E = INNN; NSN; SNN; NNS; NSS; SNS; SSN; SSSl.

Bien co A ba Ian tung dong tien co dung mot Ian xuat hien mSt sap
bieu t h i bdi tap hop
A = {SNN; NSN; NNS|.
Gia thiet dong tien la can doi va dong chat neu cac ket qua cua phep
thuf la dong kha nang. Khong gian mSu co 8 phan tuf. Bien co A co 3
3
phan tuf, do do xac suat cua A la : P(A) = —.
8
Trong mot hop co 4 vien bi mau do, 3 vien bi mau xanh (cac vien bi
chi khac nhau ve mau sSc). Lay ngau nhien curig mot liic 3 vien bi.
Tinh xac suat de trong 3 vien bi lay ra co dung hai vien bi mau do.
DAN
Khong gian mSu co
bien co scr cap (co C 7 tap hop con 3 phan tuf
trong 7 phan tuf), moi each lay 3 vien bi la lay 1 tap hop 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) =
= —.
35
Chon ngau nhien mot so tir nhien co 3 chuf so. Tinh xac suat de so
dtfcfc chon la mot so c h i n co 3 chuf so khae nhau.
DAN
Goi A la bien co so dtroc chon co 3 chuf so khae nhau la so chSn.
Khong gian mau E la so cac so co 3 chOr so (9 each chon chuf so hang
trSm, 10 each chon cha so hang chuc, 10 each chon chuf so hang don
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 ebon chuf so hang ehuc)

So cac so chSn co 3 chur so khdc nhau co cha so hang dcfn vi khac 0 la
8.8.4 = 256 so.
TS. Vu The' Huu - Nguy§n VTnh C?n

;

'

~


(4 each chon chuf so h a n g ddn v i , 8 each chon chuT so h a n g t r a m , 8
each chon chuf so h a n g chuc).
So cac so C O 3 chOr so khac nhau l a so chkn l a : n = 72 + 256 = 328.
Xac suat cua A l a : P(A) = —

« 0,3644.

900

44.

M o t to hoc sinh co 10 ngtrdi gom 6 n a m v a 4 nuT, chon ngSu n h i e n m o t
nhom 3 ngiiofi cua to. T i n h xac suat xay r a m o t trLfcfng hop dudfi day :
a) Ca ba ngiTdi diioc chon deu l a n a m .

b) Co i t n h a t m o t t r o n g ba ngiidi diTOc chon l a n a m .
CHI D A N
a) K h o n g gian m a u co :

C^Q

=

^^'^'^

=

120 p h a n tuf.

1.2.3

Co Cg = ^"^'^ = 20 each chon 3 ngiTdi deu l a n a m . Xac suat b i e n co 3
1.2.3

.-,3

ngi/cfi dtfOc chon deu l a n a m l a : P(A) =

C?o

20

1

120 6

b) Goi B l a b i e n co 3 ngiTofi diTOc chon co i t n h a t 1 n a m . B i e n co d o i l a p
cua B l a 3 ngiTdi dtfoe chon deu l a nuf :
—

1
— 2Q
P(B) = ^
= — =^ P ( B ) = 1 - P ( B ) = — .
C?o
30
30

45.

Cho 8 qua can co k h o i liTOng I a n iMt l a 1kg, 2 k g , 3 k g , 4 k g , 5 k g , 6 k g ,
7kg, 8kg. Chon ngau n h i e n 3 qua can. T i n h xac suat de t o n g k h o i
luong ba qua can di/crc chon k h o n g virot qua 9kg.

CHI D A N
So each chon 3 qua can t r o n g 8 qua can (so p h a n tuf cua k h o n g gian
mau) l a : C o = ^"'^'^ = 56 each.
^ 1.2.3

A la bien co tong khoi luong 3 qua can diTcfc chon khong qua 9kg. Cac bien
CO sq cap thuan loi cho A (thuoc tap hop 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 cua A : P(A) = — = 0,125.
56

46. Tung m o t I a n h a i con siic sac 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 h a i con sue s^c 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 h a i con sue sac l a m o t so le
hoSe m o t so chia h e t cho 3.
CHI DAN

K h o n g gian m a u co 36 p h a n tuT (6 x 6 = 36 eSp so ( i ; j ) v d i i , j nguyen
dirong 1 < i < 6; 1 < j < 6).
HQC

fa.

On luy?n theo CTDT mCn Toan THPT S

19


a) Cac bien co so cap thuan Igi bien co A (tong so cham bang 8) la : (2; 6),
(6; 2), (3; 5), (5; 3), (4; 4). Xac suat cua A l a :P(A) = — .
36
b) B i e n co B : t o n g so cham l a so le hoac chia h e t cho 3.
Goi B i l a b i e n co t o n g so cham b k n g so le, B2 l a b i e n co t o n g so cham
la m o t so chia h e t cho 3, t h i
B = B i u B2
P(B) = P ( B i ) + P(B2) - P ( B i n B2)
B i xay r a k h i m o t con sue sac n a y m S t chSn, m o t con n a y m a t le, co
18 b i e n CO so cap : ( 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 vdi 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 B i n B2 tong so' cham le va chia h e t cho 3 gom 6 bien co' so
cap : ( 1 ; 2), (2; 1), (3; 6), (6; 3), (4; 5), (5; 4).

47.

P(B) = P ( B i ) + P(B2) - P ( B i n B2) = — + — - — = — =

36 36 36 36 3
H a i x a t h u cung b ^ n ( m o t each doc lap) vao m o t muc t i e u m 6 i ngiTdi
m o t v i e n dan. Xac suat b a n t r i i n g dich t r o n g m o t I a n bSn cua nguTdi
thur n h a t l a 0,9; cua ngudi thur h a i l a 0,7. T i n h xac suat t r o n g m o i
t r i r d n g h o p sau :

a) Ca h a i v i e n deu t r i i n g dich.
b) i t n h a t co m o t v i e n t r i i n g dich.
c) C h i CO m o t v i e n t r i i n g .
C H I

D A N

a) Goi A i la bien co ngLTcri thijf nhat ban t n i n g dich, A2 la bien co ngUcfi thtf hai
ban t n i n g dich. A la bien co ca hai vien deu t n i n g dich t h i A = A i n A2. Do
hai ngu6i ban doc lap v d i nhau nen A i va A2 la doc lap n e n
P(A) = P ( A i n A2) = P(Ai).P(A2) = 0,9.0,7 = 0,63.
b) Goi 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 r i i n g dich :
B = A i u A2
P(B) = P ( A i u A2) = P(Ai) + P(A2) - P ( A i n A2)
= 0,9 + 0,7 - 0,63 = 0,97
c) Goi C l a b i e n co, h a i ngLfdi bSn m 6 i ngudi m o t v i e n c h i co m o t v i e n
t r i i n g dich. B i e n co C xay r a k h i ngiidi thiir n h a t t r i i n g dong t h d i
ngtrdi t h i i h a i t r u g t hoSc ngudi t h i i n h a t triiOt, ngtfcfi t h i i h a i t r i i n g .
P ( C ) = P ( A i A 2 u A i A 2 ) = P(AiA2) + P ( A i A 2 ) - P ( A i A 2 n A i A g )
2 0 53 TS. Vu ThS' Hi^u - Nguygn Vinh CSn


Do A i , A2 doc lap t h a n h thuf A^, A2 doc l a p , A i va A2 doc l a p va b i e n
CO A1A2

n A1A2

= 0

cho n e n

P(C) = P(Ai)P(A2) + P(Ai)P(A2) - 0 = 0,9.0,3 + 0,1.0,7 = 0,34.
48.

M o t 16 h a n g c6 30 san p h a m , t r o n g do c6 3 phe p h a m . Ngirdi t a chia
n g i 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 san p h a m .

a) T i n h xac suat de m o i p h a n c6 diing m o t phe p h a m .
b) T i n h xac suat de c6 i t n h a t m o t p h a n c6 dung m o t phe p h a m .
CHi DAN
a) Ta thirc h i e n chia n h u sau : L a y ngSu n h i e n 10 san p h a m t r o n g 30
san p h a m t a c6 p h a n thuf n h a t . T r o n g 20 san p h a m con l a i lay ngau
n h i e n 10 san p h a m de c6 p h a n thuf h a i . Con l a i la 10 san p h a m p h a n
thuf 3. S6' each chia n h i i vay b k n g CH.cH

each.

So' each chia de p h a n thuf n h a t eo m o t san p h a m xau la : C3C27 (lay
m o t san p h a m xau ghep v d i 9 san p h a m to't t r o n g 27 san p h a m t6't).
So' each chia de p h a n thuf h a i c6 m o t san p h a m xau la : C2C18. S6'
each chia de m o i p h a n c6 m o t san p h a m xau la : CgCgy.CgCig.
Xac suat de m 6 i p h a n eo dung m o t san p h a m xau la :
P(A) = ^^^f-^,f^« = 0,246.
plOplO

'-'30^20

b) Goi B l a b i e n eo' t r o n g 3 p h a n cd i t n h a t m o t p h a n c6 m o t phe p h a m .
Neu goi ( i , j , k ) la so' san p h a m xau theo thuf t u cua cac p h a n thuf
n h a t , thiJ h a i , thuf ba v d i i , j , k nguyen dtfctng v d i 0 < i , j , k < 3. K h i do
B xay r a ufng v d i 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 c6' doi lap cua B xay r a tucfng lifng v d i cac bo
(3; 0; 0), (0; 3; 0), (0; 0; 3). Ro r a n g t i n h P(B) = 1 - P(B) t h o n g qua t i n h
P(B) dan g i a n han.
So each chia de p h a n thuf n h a t eo 3 san p h a m x a u , cac p h a n con l a i
deu t o t la : C^.C^^.C^g.

So' each chia de p h a n thuf h a i eo 3 san p h a m x a u , cac p h a n khae deu
la san p h a m t o t la : C^^.C^.Cjy.
S6' each chia de h a i p h a n dau cac san p h a m deu to't, p h a n thuf ba c6 3
san p h a m xau la : C27.Cj°.
*-^3'^27*^20 + '-^27'-'3'-^17 +

^^^> =

^27^17

T^w^o
'-'30*-^20

P(B) = 1 - P ( B ) « 0,911.
HQC

6n luy§n theo CTDT men To^n THPT E

21


BAl TAP Tir GIAI
49.
T u n g dong t i e n can doi dong chat bon I a n .
a) K h o n g gian mau c6 bao n h i e u p h a n tuf.
b) T i n h xac suat b i e n co dong t i e n xuat h i e n mSt sap diing 3 Ian.
• c) T i n h xac suat b i e n co dong t i e n xuat h i e n m a t sap i t n h a t 2 Ian.

DS : a) 16 p h a n tuf
50.

a)
b)
0
d)

b) P(A) = 4

c) P(B) = —.

M o t hop dufng 7 qua cau do, 5 qua cau x a n h (cung k i c h
m a u s^c). Chon ngau n h i e n 2 qua cau. T i n h xac suat :
H a i qua cau diTOc chon m a u do.
H a i qua cau duoc chon m a u x a n h .
H a i qua cau diicfc chon m o t do, m o t xanh.
H a i qua cau du'Oc chon co i t n h a t m o t qua do.

DS : a) P(A) =

c) P(C) =
51.

c]cl
p2
"-"12

b) P(B) =

thudc

16

chi khac

p2
*^12

d) P(D) = 1 -

p2
^12

Xep ngau n h i e n 5 chCf cai B, G, N , O, O t h a n h m o t day ngang. T i n h
xdc suat de dufoc chuf BOONG.
1 J _

DS : P(A) = 1

5.4.3

•^5

52.

60'

M o t dot 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 giai n h i , 1000 g i a i ba v a 5000 g i a i tii. T i n h xac
suat de m o t ngLfcri 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 t\X.

DS : P(A) =

C i100
nn -•'^5000
C

0,0009.

"^20000

53.
Co h a i hop diing cac qua cau cung k i c h thudc, t r o n g luang hop thuT
n h a t diTng 7 cau t r a n g , 3 cau den. Hop t h i i h a i difng 6 cau t r a n g 4 cau
den. Rut ngau n h i e n tCr m o i hop m o t qua cau. T i n h xac suat :
a) H a i qua cau r u t r a deu mau trSng.
b) H a i qua cau r u t r a cung mau.
c) 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.
54.
Co h a i to hoc s i n h . To I co 9 n a m , 6 nOf. To H co 7 nam, 8 nuf. Chon
ngau n h i e n m o i to m o t ngUdi. T i n h xac suat :
a) H a i ngirdi duoc chon co dung m o t nuf.
b) H a i ngiidi duoc chon co i t n h a t m o t nam.

DS : a) P(A) = —

b) P(B) = ^'^'^

225

225

22 £3 TS. Vu Th§' Hyu - Nguyen Vinh C5n


CHOYEW i II. PHIlOfNG TRJHH VH BflT PHIfJHG TBJNHflHIS
§1. PHlJOfNG TRINH,
BAT PHlJCfNG TRINH BAG NHAT MQT AN
1.

Phxidng t r i n h b a c n h a t m p t a n c6 t h e dtfa ve dang :
ax + b = 0;a, b e S , a 0
CO m o t n g h i e m duy n h a t x = - —.
a
Phvtofng t r i n h t i c h m p t a n c6 dang :
A(x).B(x) = 0
(1)

Tap n g h i e m cua (1) l a t a p n g h i e m cua phiTOng t r i n h A(x) = 0 horp v d i
tap n g h i e m cua phu'ong t r i n h B(x) = 0.

2.

A(x).B(x) = 0

"A(x) = 0
B(x) = 0 •

3.

B a t phtfofng t r i n h h^c n h a t m p t a n c6 t h e dtfa ve d a n g :
ax + b > 0 hay ax + b > 0; a, b e R, a ?i 0
• N e u a > 0, ax + b > 0 CO t a p n g h i e m

x>- —
a

• N e u a < 0, ax + b > 0 CO t a p n g h i e m

x< — .
a
D a u c u a n h i thufc b a c n h a t ax + b (a ;t 0)
Nhi thufc bac n h a t : f = ax + b (a ^ 0)

4.

Gia

t r i x = - — t a i do f c6 gia t r i hkng 0 goi l a n g h i e m cua n h i thufc.
a
Ta CO d i n h 11 ve dau cua n h i thufc n h u sau :
Dinh U
+

V6i cac gia tri x < - — t h i f = a x + b c 6 dau t r a i v6i dau cua he so a.
a

+

V d i cac gia t r i

,
a)

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

1.

x > - — t h i f = ax + b c6 dau cung dau vcfi he so a.
a

(1)

Hoc va Sn luy?n theo CTDT mOn ToSn THPT S 2 3


b) X - X - X + 1 = 0
c) x^ + 2 x ' + 5x^2 + 4 x - 12 = 0

CHI DAN
a)

(1)

o

10-X

100

-1

-90-X

100

X +

20-X

110
-90-X

+
100

1 = 0 «

(2)
(3)
30-X

-1

120

-90-X

+

110
1

(90 + x )
b ) x^ - x^ -

o

1

+ 110

120

1 ^
120

-1

-0

^

= 0
= Oci> X =

-90.

x^Cx - 1) - ( x - 1) = 0
(x - l ) ( x ^ - 1) = 0

x ' - l = 0

o

x - l = 0

o

X =

±1.

c) x^ + 2x^ + 5x2 + 4 x - 12 = 0 o ( x ^ - 1 ) + 2(x^ - 1 ) + SCx^ - 1) + 4 ( x - 1) = 0

o ( x - l ) [ x ^ + Sx^ + 8 x + 12] = 0 <^ ( x - l ) ( x + 2)(x2 + x + 6) = 0
,
(Bai

x - l = 0
x + 2 = 0

T a p n g h i e m {-2;
2.

o
v i x + x

f

+ 6=

1
2

x +

I

\

+ —>0

Vx)

1).

G i a i , b i e n l u a n t h e o t h a m s o ' m phufong t r i n h :

CHI DAN

m \ 3 m = 9x + m^

(1)

( I ) c : > ( m 2 - 9 ) x = m2 + 3 m
+

N e u m ^ - 9 ;t 0, n g h i a l a v 6 i m
X —

+
+

m^ + 3 m

±3 t h i phuTcfng t r i n h (1) c6 n g h i e m :

m

—

m^ - 9
m - 3
N e u m = 3, p h u o n g t r i n h (1) c6 d a n g : 9 x - 9 = 9 x + 9

Phiicfng t r i n h n a y v6 n g h i e m .
N e u m = - 3 t h i p h t f o n g t r i n h (1) c6 d a n g : 9 x + 9 = 9 x + 9
Phucfng t r i n h v 6 so n g h i e m ( m o i so t h u c deu l a n g h i e m ciia p h i f o n g t r i n h ) .

3.

C h o p h u o n g t r i n h : a^x + b = a x + ab

(1)

t r o n g do a, b l a cac t h a m so t h i f c . G i a i , b i e n l u a n t h e o a, b p h i / o n g
trinh tren.

CHI DAN
+

( l ) c : > a ( a - l ) x = b ( a - 1)
N e u a(a - 1)
0, tufc l a n e u a ?t o v a a ;^ 1 t h i (1) c6 n g h i e m d u y n h a t
(a-l)b
b
X =

+

=

a(a-l)

—.

a

N e u a = 0, p h u o n g t r i n h (1) c6 d a n g : Ox = - b
T r o n g t r i r d n g h o p n a y c6 h a i k h a n a n g :

24

a

N e u b = 0 (tufc l a a = 0, b = 0) m o i so t h t f c d e u l a n g h i e m .

-

N e u h ^ 0 ( t i i c l a a = 0, b ?i 0) p h i i d n g t r i n h v 6 n g h i e m .

-

TS, Vu The Hi/u - Nguyen Vinh CSn


+ Neu a = 1, phtrcfng t r i n h (1) c6 dang : Ox = 0, phiicfng t r i n h c6 t a p
n g h i e m la m o i so thuc.
4,
G i a i cac phifong t r i n h :
a) (x - if + (x + 2f = (2x + 1)^
(1)
b) 9ax^ - 18x^ - 4ax + 8 = 0
(2), a la t h a m so.
CHi DAN

a) ( 1 ) « (2x + l ) [ ( x - (x - I X x + 2) + (x + 2f] = (2x + 1)^
o (2x + l ) [ ( x - if - (x - l ) ( x + 2) + (x + 2f - (2x + 1)^] = 0
<=> (2x + l ) ( x + 2)(x - 1) = 0 o
b) (2)
+
+
5.
a)
CHI
a)

b)

6.

x =

X

= - 2 , x = 1.

ax(9x^ - 4) - 2(9x^ - 4) = 0

(ax - 2)(3x + 2)(3x - 2) = 0
2
2
Neu a ?t 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 bat phUcfng t r i n h dufdi day va bieu d i e n t a p n g h i e m cua no
t r e n true so.
3(x + 2) - 1 > 2(x - 3) + 4
b) (x - 2)^ + 1 + x^ > 2^ - 3x - 4.
DAN
-7
0
X
c^3x + 5 > 2 x - 2
<z> 3x - 2x > - 2 - 5 <=> X > - 7
^
Tap n g h i e m diTOc bieu d i e n 0 h i n h ben.
inx a
o x^ - 4x + 4 + 1 + x^ > 2x^ - 3x - 4 o - 4 x + 3x > - 4 - 5
o - x > - 9 « x < ^ = 9
Q
-1
Bieu dien t a p n g h i e m b h i n h ben.
Hinh
G i a i va b i e n l u a n theo t h a m so m b a t phirong t r i n h :

x +
_
CHI DAN

— - > '^^"*"-(m + l ) x
m

m

g

^

b

(1).

2
Dieu k i e n xac d i n h m ; t 0. Chuyen ve, r u t gon t a diTOc : ( m + 2)x > —
m
+ Neu m 5t 0 va m > - 2 ( t h i m + 2 > 0) b a t phtfong t r i n h c6 n g h i e m
2
X >

m ( m + 2)
+ Neu m < - 2 ( t a t n h i e n m ;^ 0) t h i (1) c6 t a p n g h i e m x <
+ Neu m = - 2 bat phiJOng t r i n h t r d t h a n h Ox > - 1
M o i so thuc deu la n g h i e m cua (1).

m ( m9 + 2)

7. G i a i , b i e n l u a n b a t phiTcfng t r i n h v d i t h a m so a va b: x + 1 > — + — (1)
b
a
HQC fln luy§n theo CTOT m5n Toan THPT S

25

CHI DAN
(a^ - ab)x +
- ab .
a-b,
,^ ^ ,
(1) <=>
;
— < 0 <=> — — ( a x - b) < 0 (a 0, b 0)
ab
ab
+ N e u a = b ? i 0 t h i (1) c6 dang Ox - 0 < 0 v6 n g h i e m .
+ N e u a < b < 0 hoSc 0 < b < a t h i (1) c6 n g h i e m x < —.
a
+ N e u a < 0 < b hoSc b < 0 < a hoac b < a < 0 hoSc 0 < a < b t h i (1) c6

8.

tap n g h i e m l a : x > —.
a
G i a i cac b a t philcfng t r i n h :

a) (2x - 3)(5 - x) > 0
X + 5
CHI DAN
a) D u n g d i n h l i ve dau cua n h i thufc : ax + b (a
3

0)

2x - 3 CO n g h i e m x = —
3
Neu X < — t h i 2x - 3 c6 dau a m ( t r a i dau v<5i 2).
3
Neu X > — t h i 2x - 3 CO dau dtrong (cung dau vdi 2).
2
Tuong tir 5 - X > 0 v d i X < 5 va 5 - X < 0 v d i X > 5.
3
L a p bang xX e t dau cua t i c h (2x - 3)(5 - x) nhir sau :
-co
—
5
2x- 3
5 -

X

+

+

+

-

0

+00

+

0

-

(2x - 3X5 - x)
0
+
0
Can cur bang x e t dau t a t h a y t a p n g h i e m cua b a t phildng t r i n h l a
— < X <

2
b) Dieu k i e n xac d i n h : x ^ - 5 .
L a p bang x e t dau nhir cau a) :
X

5
5

5.

-5

-00

—

—

2x- 1
X +

2x-l
X +

26 S TS. Vu Thg' Huu -

+CC

-

+

+

+

-

+

0

0

0

+

Nguyin VTnh C§n