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n§m 2013.
GlAl
TICH
Hpc
va
on-luyen theo
CTDT
mon

Toan
THPT 3
Mf IJBISOTOHdPVflMSy'fiT
§1.
HOAN
VI,
CHINH
HdP, TO HOP
KIEN
THlTC
1. Quy t^c
CQng
va quy tic nhan
a)
Quy tdc
cong
:
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 thi
c6 m + n
each
chon
mot phan tuf thuoc
A
hoac
thuoc
B.
b)
Quy tdc
nhdn
:
Be
hoan thanh mot
cong
viec
A
phai thiic hien hai
cong
doan.
Cong
doan
I
CO
m
each
thiic hien,
cong

doan
II
c6 n
each
thiic hien thi
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
thijf
i (1 < i < k) c6 mi
each
thi thi
c6
mi.m2 mk

each
de
hoan
thanh
cong
viec
A.
2. Hoan
vi
Mot tap
hop A
huTu han
c6 n
phan tuf
(n > 1).
M5i
each
sap
thuf
til eac
phan
tiif
ciia tap hop
A
duoc
goi la
mot hoan
vi
ciia
n

phan tuf eua
A.
Dinh
li : So
hoan
vi
khac nhau ciia
n
phan
til
bang
:
Pn
=
n(n- l)(n-2) 2.1
= n!
3.
Chinh
hrfp
Mot
tap hop A
hOTu
han 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.
Dinh
li : So
chinh hop
chap
k
ciia
n
phan tuf bang
:
A;;
=
n(n

-
l)(n
-
2) (n
- k +
1)
=
(n
-
k)!
(Quy irde
: 0! = 1).
4.
To
hofp
Cho
tap hop A
hufu
han c6 n
phan
tuf (n > 1) va so
nguyen
k
(0
< k <
n). Moi tap
hop con
gom
k
phan tuf ciia

A
(khong
tinh
thuf
tU
eac phan tuf)
goi la
mot
to hop
chap
k cua n
phan tuf.
n'
A''
Dinh
li : So to
hop
chap
k
cua
n
phan tuf
la : 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
BAI TAP
1.
Cho cac chuf
so 2, 3, 4, 5, 6, 7.
a)
Co bao
nhieu so' t\i nhien
c6
hai chuf
so
ducfc
tao nen til tap hop
cac
chuf
so da
cho.
b)
Co
bao nhieu
so
tii nhien
c6
hai chuf
so
khac nhau
difOc
tao nen tiT tap
hcrp chuf

so da
cho.
CHI DAN
a)
De
tao mot
so c6
hai chuf
so ta
thifc hien hai
cong
doan
:
1.
Chon
mot chuf
so
lam chuf
so
hang chuc
: c6 6
ket qua
c6
the.
2.
Chon
mot chuf
so
lam chuf
so

hang don
vi : c6 6
ket qua
c6
the.
Theo
quy
t^c
nhan
so
ket qua tao thanh
cac so c6
hai chuf
so tii
tap
hgfp
6
chCif
so da
cho
la : n = 6 x 6 = 36
so'.
b) Lap luan
giong
nhif
cau
a)
nhuiig liTu
y
sir khac biet

so
vdfi trifofng hop
tren
of
cho
so
dugc
tao thanh
c6
hai chuf
so
khac nhau.
Do do ta
c6 ket
qua nhu sau
:
1.
Chon
mot chuf
so
lam chuf so' hang chuc
: c6 6
ket qua
c6
the.
2.
Chon
mot chuf
so
lam chuf

so
hang don
vi : c6 5
ket qua
c6
the
(vi
chuf
so nay phai khac chuf
so
hang chuc
da
chon trifdrc do).
Theo
quy tac nhan
: so
cac
so c6
hai chuf
so
khac nhau
difcfc
tao thanh
tCr tap hop
6
chuf
so da
cho
la : n' = 6 x 5 = 30 so.
Cdch khac

:
Moi
so c6
hai chuf
so
tao thanh tiT
6
chuT
so da
cho
la
mot
tap
hop
con s^p
thuf
tiT gom hai phan tuf tiT
6
phan tuf da cho.
Do do so
cac
so
CO hai chuf
so
khac nhau
tao
thanh tiT
6
chuf
so da

cho
la so
chinh
hop
chap
2
cua tap
hop 6
phan tuf.
n
=
Ag
=
6.5
=
30
so.
2. Cho tap hop cac cha
so 0,
1, 2,
3,
4,
5, 6.
a)
Co
bao nhieu
so
tu nhien c6
4
chuf so tii tap hop cac chuf so

da
cho.
b)
Co
bao nhieu
so
tU nhien
c6 4
chuf
so
khac nhau tCmg doi
tii
tap hop
cac chuf
so da
cho.
CHI DAN
a)
Co 6
each
chon chuf
so
hang nghin (chuf
so
dau tien phai khac 0),
7
each
chon chff
so
hang tram,

7
each
chon chuf
so
hang chuc
va 7
each
chon
chuf
so
hang don
vi.
Theo
quy tSe
nhan
: so
each
tao
thanh
so tif
nhien
4
chuf
so tii
tap hop
7
chiJ
so da
cho
la :

N
=
6x7x7x7
=
2058
so.
b)
Co 6
each
chon chuf
so
hang nghin, khi chon
xong
chuf
so
hang nghin
con lai
6
chuf
so
khac
vdi
chuf
so
hang nghin
da
chon.
Vay c6 6
each
chon chuf

so
hang tram.
Khi da
chon chuf
so
hang nghin
va
hang
tram,
eon lai
5
ehOf
so
khac vdi
cac
chuf
so da
chon.
Do do eo 5
each
6 :S; IS. Vu The Hi/u - NguySn Vinh Cin
chpn chOf
so
hang
chuc.
Tifcfng
tir,
c6
4
each

chon chOr
so
hang don vi.
Theo
quy tac nhan.
So cac so
txi nhien
c6
4
chOf
so
khac nhau tCfng doi
difcfc tao thanh tix tap hop
7
chuf
so da cho la
:
N'
=
6 X 6 X 5 X 4
=
720
so.
Cdch
lap
luan khdc
:
Moi
so
tiT nhien

c6 4
chOf
so
khac nhau
tao
thanh
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 di canh nhau
va A
dufng
tri/dtc
B.
CHI
DAN
a)
So
each
xep hang bang
so
hoan vi ciia
10
phan
tiif.
Ni
=
10!

=
3628800
each.
b) Coi hai ban
A
va B
nhii
mot ngudi.
Do do
so
each
xep
hang ciia
to
de
vao 16p trong
do
hai ban
A
va
B
di
lien
nhau bang
so
hoan
vi cua
9
phan
tijf.

N2
=
9!
=
362880
each.
4.
Co bao
nhieu
each
xep
6
ngiicfi
ngoi
vao
mot ban
an
6
cho
trong
cac
triiofng
hcfp sau
:
a) S^p
6
ngiTofi theo hang ngang ciia mot ban
an
dai.
b) S4J)

6
ngiTori ngoi vong quanh mot ban an
tron.
CHI
DAN
a) Moi
each
ngoi theo hang ngang
la
mot hoan vi cua
6
phan
tijf.
So'
each
sap xep
la
:
6!
=
720
each.
b) Gia suf
6
ngifofi
an
diTOc danh
so
thijf
tif la

:
1, 2, 3, 4, 5,
6
va
mot
each
sap xep theo ban
tron
nhii
hinh.
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
ta
eat ban
tron
a
vi
tri giCfa
2
va
4
roi
trai
dai theo ban ngang
thi
ta
CO hoan
vi (1)
tUofng
ijfng
mot
each
xep
ngiiofi ngoi theo ban
an
dai.

TiTofng txi cat
of vi
tri giufa
5
va 2.
Nhu vay mot
each
sap xep
theo
ban
tron
tiiOng
ufng
vdri
6
each
sap
xep
theo ban dai.
Do do
so
each
Hoc
va
on luyen theo
CTDT
mon
Toan
THPT 7
xep 6

ngiTofi
ngoi quanh
ban an
tron
la :
N
= — = 120 each.
6
5. Mot to
CO
15
ngifofi
gom 9 nam va 6 nOf. Can lap
nhom
cong tae eo 4
ngUdri.
Hoi eo bao
nhieu
each
thanh
lap
nhom trong
moi
trifcfng
hop
sau
day :
a)
Nhom
c6 3 nam va 1

nur.
b) So nam va nCf
trong nhom bang nhau.
c)
Phai
CO
it
nhat
mot nam.
CHI
DAN
9 8.7
a) So each chon 3 nam
trong
so 9 nam la : Cj! = " = 84
1.2.3
So each chon 1 niJ
trong
so 6 nOr la : Cg = 6
So each
thanh
lap
nhom
gom 3 nam va 1 nCf (theo quy tSc
nhan)
la :
Ni
= C^C^ = 504 each.
b) So each lap
nhom

gom 2 nam va 2 nuf la :
N2= C^C^ = — = 540 each.
' ' 1.2 1.2
c) So each
thanh
lap
nhom
4
ngu'ofi trong
do c6 it
nhat
1 nam la : 1
nam,
3 nO hoac 2 nam, 2 nuf hoae 3 nam, 1 nur hoac 4 nam.
— 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 :
Cung
eo the lap
luan
nhif
sau :

Ca
to
CO
15
ngiTcfi.
So each lap
nhom
4
ngUdi
tuy y la :
_ 15.14.13.12
~ 1.2.3.4
6.5.
1.2
So each lap
nhom
4
ngi/di
eo it
nhat
1 nam la :
N
=
CJs
- C^ = 1365 - 15 = 1350 each.
6.
Trong
mat
phang
eo n

diem phan biet
(n > 3)
trong
do eo
diing
k
diem
nSm
tren
mot
diidng thSng
(3 < k < n). Hoi c6 bao
nhieu
tam
giac
nhan
cac
diem
da cho la
dinh.
CHIDAN
Cuf
3
diem khong thSng hang
tao
thanh
mot tam giac. So cac tap hop
con 3
diem trong
n

diem
la : C^. So cac tap con 3
diem trong
k
diem
tren
diTcfng
thing
la : C^. So tam giac c6 3
dinh
la cac
diem
da cho
la:
N = C^ - Cl tam giac.
^4
^
^—^xo^
^ ^ggg ^^^^
1.2.3.4
So each lap
nhom
4
ngUofi toan
nuf la : C^ = Cg = = 15 each.
8 ; TS. Vu The Hi/u -
Nguyen
VTnh
Can
7.

a)
Co
bao
nhieu
so
tii
nhien
la so
chan
c6 6
chiif
so doi mot
khac
nhau
va
chuf
so
dau
tien
la
chOf
so le.
b)
Co bao
nhieu
so
tii
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).
CHI
DAN
a)
So can
tim
c6
dang
: x =
a^agaga^agag

trong
do ai, ae lay cac
chOf
so
0, 1, 2, 8, 9
vdfi
ai ?i 0, ai
aj
vdi
1 < i
?i
j < 6.
- Vi X
la so
chSn nen
ae
c6 5
each chon
tiT
cac
chuT
so 0, 2, 4, 6, 8.
- Vi
ai la
chuT
so le
nen
c6 5
each chon
tiT

cac
chuf
so 1, 3, 5, 7, 9.
Con
lai
a2a3a4a5
la
mot
chinh
hop
chap
4
eiia
8
chuf
so con
lai sau khi
da
chon
ae
va
ai. Theo
quy
t^c
nhan,
so cac so can xac
dinh
la :
Ni
=

S.S-Ag
= 5.5.8.7.6.5 =
42000
so.
b) Mot
so
theo
yeu
cau
de bai gom 3
chuf
so
tii tap
Xi =
|0;
2; 4; 6;
81
va
3
chuf
so
tCr tap
hop X2 =
II;
3; 5; 7; 91
ghep lai
va
loai
di cac day 6
chuf

so
CO
chuf
so 0
dufng
dau.
So each lay
3
chuf
so
thuoc tap
Xi la : Ci? = 10
each.
So each lay
3
phan
tuf thuoc
X2 la : Cg = 10
each.
So' each ghep
3
phan
tuf lay
txi
Xi
voi 3
phan
tuf lay tii
X2 la :
C^C^

=
10.10
= 100
each.
So' day
so' eo
thuf
tif
eiia
6
phan
tuf diioc ghep lai
la :
100.6!
=
72000
day.
Cac
day so c6
chuf
so 0 a
dau
gom 2
chiJ
so
khac
0
ciia
Xi va 3
chuf

so'
ciia
X2 : So cac day so
nhif
tren
la :
C4.C5.5!
=
7200
day.
So
cac so
theo yeu
cau de
bai
la :
N2 -
C^C^6!-C^C^5!
=
72000
-
7200
=
64800
so.
8. Mot
hop
diing
4
vien

bi do, 5
vien
hi
trang
va 6
vien
bi
vang.
NgLfofi
ta
chon
ra 4
vien
bi
tif
hop do.
Hoi
c6 bao
nhieu
each
lay de
trong
so
bijay
ra
khong
dii
ca 3
mau.
CHI

DAN
Cdch
1 : So
each chon
4
vien
bi
khong
du
3
mau bang
so'
each chon
4
vien
bat
ki trir di
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
bi
khong
du
3
mau bang
so
each chon
4
vien
mot mau
(4 do, 4
trang
va 4
vang) cong vdi
so
each chon
4
vien
hai
mau

(1 do, 3
trang
hoae
2 do, 2
trang
hoac
3 do, 1
trang
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 CTDT
mon
loan THPT
SI 9
9. Co 15 nam va 15 nuT khach du lich dijfng thanh
vong
tron quanh ngon
lijfa
trai.
Hoi c6 bao nhieu
each
xep de khong eo triTcfng hop hai ngi/6i
eCing
gidfi canh nhau.

CHI DAN
ThiTe
hien sap xep bang
each
danh so 30 cho tren di/orng tron tii 1
den 30 va cho nam dufng so le nuT dufng cho so chSn
hoac
ngi/gc
lai (2
each).
Co 15!
each
sSp nam dufng trong cae cho so' le
(hoac
chSn)
va
15!
each
sdp nuf dufng trong cae cho so ch^n
(hoac
le).
Vi
diidng tron 30 cho nen moi
each
sSp xep nao do
xoay
tua 30 cho
theo
dung trat tiT do ta cung chi eo mot
each

sap tren dirofng tron
(xem bai so 4).
Do do so
each
sSp xep
theo
difcfng tron 30 khach du lich
theo
yeu cau
2.(15!)(15!)
, ,
de la : N = = 14!.15!
each.
30
10. Chufng minh cae dang thufc :
a) + + + = ——- (1) trong do A^ la chinh hop
chap
2 eua n.
Ag A3 A„ n
b) C;; = C;;:; + Cl;}^ + + Cl:\) trong do C; la to hop
chap
r ciia n.
CHI 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
trai
ciia (1) la :
(1

(I
V
f
1 1^
— — — — —
+ +
— —
u
2. l2
3v
[n-1 nj
b)
Theo
tinh
chat eua to hop ta c6 :
= Cn_3 + C^^g
Cong
ve vdi ve cae dang thufc tren ta
diTOe
:
c:;-c::;+c-^3+c::u +c:-uc:
Do C;; = C;::} = l nen thay C;: d dang thufc cuoi bori C^:} ta
dugfc
dang
thufc (2) can chufng minh.
11.
Chufng minh bat dang thile :
C^ooi
+ ^ C\Z + CfZ
trong do k e N, k < 2000, la to hop

chap
k eua n phan tuf.
10 t4l TS. Vu The Hi/u -
Nguyin
Vinh
CSn
CHI
DAN
V(Ji
0 < k <
1000 thi
2001!
^2001
(k
+
l)!(2000-k)!
k + 1
k
+
1
2001
k
k!(2001-k)!
'^2001
-
'-^2001
-
'^2001
-
2001!

, plOOO
_
plOOl
••
-
'-^2001
~ ^2001
<1
2001-k
pk pk +
1
^
plOOO plOOl
*-^2001
^2001 — ^2001
^^2001
-111-k
Mat
khac,
v6i 1000 < k <
2000
theo
tinh
chat ciia
to hop C' =
C;;'^
ta
•
^2001 ^
^-^2001

~"
'^2001
'
"^2001
-
'^2001
' ^2001
0
<
2001
- k < 1000
theo
phan
tren
da
chufng minh.
- CIZ'^
<
Cir,
+CIZ\i 0 <
2000
- k < 1000,
CAC BAI TAP Tl/
GIAI
12.
TCr
diem
A den
diem
B

ngi/di
ta c6 the di qua C
hoac
di qua D va
khong
CO
diTcfng
di
thang
tii C den D. Til A di
thang
den C c6 2
each,
tir
C di
thang
den B c6 3
each.
TCr
A di
thang
den D c6 3
each
tix D di
thang
den B eo 4
each.
a) Hoi txi
A
CO

bao
nhieu
each
di
tdfi
B ?
b) Hoi tCr
A den B
roi
til B
trdf
ve A A / \
CO
bao
nhieu
each
?
DS
: a) 18
each
3 X^/ 4
b) (18)^
each.
D
13. TCr
7
chOf
so 0, 1, 2, 3, 4, 5, 6 eo the ghi
dirge
bao

nhieu
so
tiT nhien
moi
so'
CO
5
chCT
so
khac
nhau tCrng doi.
DS
: 2160 so.
14.
Cho
tap
hop cac
chOf
so X = |0; 1; 2; 3; 4; 5;
61.
a) Dung tap
hop X eo
the ghi
dufcfe
bao
nhieu
so
tiT nhien
eo 5
chiT

so.
b) Dung
tap hop X c6 the ghi
dirge
bao
nhieu
so
tir nhien
c6 5
chOf
so'
khac
nhau tCrng doi.
c) DCing
tap hgp X c6 the ghi
dugc
bao
nhieu
so
tiT nhien
c6 5
ehCt
so
khac
nhau
la so
ehSn.
DS
: a) 6.7* so b)
6l5.4.3

so e) A^ +
15A^
so.
15. Mot
to hoc
sinh
c6 5
nam,
5
nOf
xep
thanh mot hang
doc.
a)
Co bao
nhieu
each
xep
khac
nhau.
b)
Co bao
nhieu
each
xep
hang
sao cho
hai ngircfi dijfng
ke
nhau

khac
gidi.
DS
: a) 10!
each
b)
2(5!)^
each.
16.
Mot igp
CO
25 nam hoc
sinh
va 20
nuf
hoc
sinh.
Can
chon
mot
nhom
cong
tac 3
ngiTdi. Hoi
eo bao
nhieu
each
chgn
trong moi
trirdng

hgp sau
a)
Ba hoc
sinh
bat
ki
eua Idp.
b) Hai
nijf
sinh
va mot
nam sinh.
Hpc va on luyen theo CTDT mon Toan THPT IJ: 11
c) Ba hoc sinh c6 it nhat mot nuf.
DS : a) C;;^
each
b)
25.C^o
each
c) C'^^ -Cl,
each.
17.
Co bao nhieu
each
phan phoi 7 do vat cho 3 ngUdi trong cac trUcfng
hop sau :
a) Mot ngUofi nhan 3 do vat, eon 2 ngUcfi moi ngUcfi hai do vat.
b) Moi ngudi it nhat mot do vat va khong qua 3 do vat.
DS : a)
3.C^C^

each
b) S.CtCl + SCl.Cl
each.
18.
Mot to CO 9 nam va 3 nOf.
a) Co bao nhieu
each
chon mot nhom 4 ngUcfi trong do eo 1 nijf.
b) Co bao nhieu
each
chia to thanh 3 nhom moi nhom 4 ngUofi va trong
moi nhom c6 1 nuf.
DS : a) 3.C^
each
b)
3.C;;.2C^
= 10080
each.
19.
Tim cac so nguyen duong x, y thoa man cac dang thufe :
6 ^^^"5^ "2^ •
f)S : X = 8, y = 3.
20.
Co bao nhieu so tU nhien chain c6 4 ehuf so doi mot khac nhau.
DS •.n= Al+ 4.8.8 = 760 so.
21.
Cho da
giac
deu 2n dinh AiA2 A2n, n > 2 noi tiep trong dudng tron.
Biet rang so tam

giac
c6 dinh la 3 trong 2n diem tren nhieu gap 20
Ian
so hinh ehuf nhat eo dinh la 4 trong 2n dinh tren. Tim so n.
£>S : n = 8.
22.
Tim so tU nhien n, biet rang C" + 2C;, + 4C' + + 2"C" = 243.
: n = 5.
23.
Giai bat phuong
trinh
(vdi hai an n, k G N)
24.
Trong mot mon hoc, thay
giao
eo 30 cau hoi khac nhau, gom 5 cau
hoi
kho, 10 cau hoi trung binh va 15 cau hoi de. Tir 30 cau hoi do c6
the lap
dUcfc
bao nhieu de kiem tra gom 5 cau khac nhau sao cho
trong m5i de nhat thiet phai eo du ba loai cau hoi (kho, trung binh,
de) va so cau hoi de khong it hon 2.
DS:n=
Cl,ClCl+C',,C\,Cl+C%C\f
= 56785 dl.
25.
Cho tap hcfp A eo n phan tuf (n > 4). Biet rSng so tap hop eon eo 4
phan tuf eua A gap 20 Ian so tap hofp con c6 2 phan tuf ciia A. Tim so'
tU

nhien k sao cho so tap hop eon eo k phan tuf eua A la Idn nhat.
flS : n = 18, C^g > C\^' o k = 9.
12 ;.'; TS. Vu Th§' Hyu -
Nguygn
VTnh
Can
§2.
NHI
THlfC NIUTCfN
KIEN
THLTC
1.
Nhi
thufc
Niutcfn
(a
+
b)"
=
Cf,a"b°
+
Cla"-'h
+ +
ClJa'^'^b'^
+

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

Ydi
quy
vide
a, b ^ 0, a" = b° = 1, C°
=1.
2.
Tarn
giac Patcan
Cac
he
so'
ciia
nhi
thufc
Niutorn
ufng
vdi n = 0, 1, 2, 3, c6 the sap
xep
diidfi
dang tarn
giac
dtfofi
day goi
la
tarn
giac Patcan.
n
= 0
n
= 1

n
= 2
n
= 3
n
= 4
n
= 5
n
= 6
1
1
1
1
2 1
1
3
14
6
3
1;
4:1
10
5 1
20
15 6 1
Trong
m6i
khung
the

hien
tinh
chat
tong
hai
he so
hang tren bang
so
hang
or
hang diTdfi
hay
C^'^
+
Cj^
=
Cl;^i.
6
5
10
15
BAITAE^
26.
Tim
cac so
hang khong chufa
x
trong khai trien
nhi
thufc NiutOn ciia

CHI
DAN
/X
J
vdfi
X > 0.
(Trich
de
TSDH
kho'i
D nam
2004)
1
I _i
Vdfi
X > 0, ta
CO
:
\/x
=
x^;
—j=r
= x "*
%/x
f
/X
+•
/x;
1 7-1
1

7-2
2
= (x'^ + x -
C°x3
+ Cix 3 X 4 + c?x ^ X 4
7-k
_k 7
+
+ C)x^
x"-*
+

+
Clx'^
So
hang khong chufa
x la so
hang thuT
k + 1
trong khai trien
sao
cho
:
Hoc va
on
luyen theo CTOT mon Toan THPT
'13
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.
27.
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 :
CL(x^r"(-xy)^ = -CLx^^xV^ -
-3432x^V^
28.
Tim so
hang
thuf tii cua khai trien nhi thufc
2 „2 A"
a b'-a
+
.

Biet
b
- a 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 ~'^^ = 21 o n = 7
Vay so
hang
thuf tii cua khai trien
a
b^-a'^^'
•
+
C?
7-3
^b^-aM
[b-aj
^ a J
= 35

b-a
a(b + a)
b-a
la :
29.
Biet
rSng
tong tat ca cac he so cua khai trien nhi thufc (x^ + 1)"
bang
1024. Hay tim he so ciia so
hang
chufa
x^^ trong khai trien do.
CHIDAN
(1
+ x'r = ci +c\x'
+cix'
+

+ cy + + c:y"
Cho X = 1 ta dirge : (1 + 1)" = + c;, + Cf, + + + + C;;
= 1024 = 2" = 2'*^ ^ n = 10
Do do he so cua x'^ la : = = 210.
6!4!
30.
Trong khai trien nhi
thuTc
NiutOn
chufa
x\t

rSng
5C;;-'
= C'l
CHI
DAN
/ 2 -I
' nx 1 ^
, X ^ 0, hay tim so
hang
14 x^
(Trich de TSDH
khoi
A - 2012)
bCr = Ct 5n =
n(n - l)(n - 2)
1.2.3
n(n'^ - 3n - 28) = 0 n = 7
14 TS. Vu The'
Hifu
-
Nguygn
VTnh
Can
Thay
n = 7
vao nhi thuTc
Niutofn
da
cho thi
c6 :

1
12
-c*
.2,
X
I-]
X
2
+
+
V
f-1
So hang chufa x^ trong khai
trien
la
so' hang thuf
k + 1 sao
cho
:
^2(7-k)-k
= 5=^k = 3
7-k
k
X^
12;
.Xy
27-k
Vay
so
hang chufa

x la
: -C^
1
vXy
7.6.5 1 ^5^_35^3
1.2.3
2'
16
31.
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)
CHI
DAN
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" (*)
He
so

cua x^° trong khai
trien
(*) la :
a^o
=
2C\\ 22.
32.
Khai
trien
bieu thufc
P(x) = x(l - 2xf +
x^(l
+
Sx)^**
va
viet
P(x)
diTdi
dang
da
thufc vdri luy thifa tang
cua x.
Hay tim
he so
ciia
x''
ciia
da
thufc
do.

CHI
DAN
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 :
1
f)
q Q
as

= +
2'Ct
=
27.^^^^
+ 16.5 =
3320.
'
1.2.3
Hqc va on luyen
theo
CTDT mon Toan THPT .'' 15
CAC
BAITAP
lij
GIAI
33. Tim
so
hang khong chijfa
x
cQa
khai
trien
nhi
thijfc Niuton.
if
X
+ — .
X
j
DS

:
924.
34.
Khai
trien
va rut gon
bieu thufc
:
P(x)
=
(1
+ xf +
(1
+
x)^
+
(1
+ xf +
(1
+
x)^
+
(1
+
x)^"
ta
dirgfc
:
P(x)
=

aiox^°
+
agx®
+
asx**
+ +
aix
+ ao
Tinh
ag.
£>S
: a« =
55.
35.
Khai
trien
va
rut gon P(x)
= (x +
1)^
+ (x - 2f
thanh
da
thufc
vdi luy
thtra
giam
dan
ciia
x. Tim he

so
cua cac
so
hang chufa
x^ va x^.
DS
:
He
so
ciia
x^ la
:
-622,
ciia
x^ la
:
570.
36.
Chufng minh vdfi
n
nguyen diiOng
ta
c6 :
a)
cL+cL+
+
CL=cL+cL+
+
cr.
b) c;, + 2Ci + 3Ci

+
+ nc;;
-
n2"-'.
CHI
DAN
a)
Khai
trien
P(x)
= (x -
1)^" roi cho
x = 1.
b)
Khai
trien
P(x)
=
(1
+
x)". Tim P'(x) roi
tinh
P'(l).
37.
Viet khai
trien
Niutcfn, bieu thufc
(3x
-
1)^'', tU

do
chufng minh rSng
:
38.
Trong khai
trien
nhi
thufc
x<Jx
+
X
28
\
15
. Hay tim
so'
hang khong
phu
thuQC X, biet rSng
:
C;; + C;;' + C"
' =
79.
DS
:a =
792.
39. Tim
he so
ciia
so

hang chufa
x
trong khai
tri§n
DS
:a =
210.
40. Tim
he so
ciia
so
hang chufa
x^
trong khai
trien
C-j-C-
=7(n + 3).
DS
:a =
495.
+
X
Vx
biet
biet
16
;;'.
TS. Vu The HUu - Nguygn Vinh Can
§3. XAC
SUAT

KIEN
THCTC
1.
2.
3.
Phep
thijf
ngSu
nhien,
khong gian mau
Mot
phep
thuf
(thi nghiem) c6 the lap lai so Ian tuy y
vdfi
cac dieu
kien
co ban giong nhau
nhiftig
khong the xac
dinh
chSc
chSn, ket qua
nao trong moi Ian thifc hien ma chi co the noi ket qua do
thuoc
mot
tap
hop xac
dinh
thi ta goi la

phep
thii
ngdu nhien. Tap hop tat ca
cac ket qua co the co cua
phep
thijf
ngau nhien goi la khong gian mdu
ciia
phep
thijf
do.
Bien
co ngau
nhien
Mot
phep
thuf
ngau nhien T co khong gian mau la E, m6i tap hop A c
E
bieu thi 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. Bien co khong
chufa phan tuf nao ciia E la bien co khong the co, ki hieu 0. Hai bien
CO
A, B ma A n B = 0 thi A va B difofc goi la hai bien co xung khdc.

Xac
suat
ciia
bien co' ngau
nhien
Phep
thuf
ngau nhien co khong gian mau E gom n bien co scf cap co
kha
nang xuat hien
dong
deu
(dong
kha nang). Bien co ngau nhien A
gom k bien co' so cap (ciia E) thi xac sudt cua bien co ngdu nhien A,
ki
hieu P(A) la ti so:
P(A) =
n
4. Cac
tinh
chat cua xac suat
a) Bien co ngau nhien A bat ki ta deu co 0 < P(A) < 1.
b) P(0) = 0, P(E) = 1.
c) A va B la hai bien co xung
khSc
(tufc A n B = 0) thi
P(A u B) = P(A) + P(B)
Neu A va B la hai bien co bat ki thi
P(A L^B) = P(A) + P(B) - P(A n B).

d) Neu A va A la hai bien co' ngau nhien ddi lap
(tufc la A u A = E, A n A = 0) thi P(A) = 1 - P(A).
5.
Bien
co dpc lap va quy t^c
nhan
xac suat
Hai
bien co ngau nhien 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 tdi kha nang xay ra cua bien co kia.
Quy tdc nhan xac sudt
Neu hai bien co ngau nhien A va B doc lap
vdfi
nhau thi
P(A n B) = P(A)^P^
7/15 X23' Ji
Hoc va on
luyen theo CTDT

mon
Toan THPT
.17
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)
Ki 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 thi bang day 3
chuf cai S
hoac
N. Nhu vay khong gian mau gom 8 phan tii.
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 thi 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 mlu co 8 phan
tuf.
Bien co A co 3
3
phan tuf, do do xac
suat
cua A la : P(A) = —.
8
42.
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
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 C4
each.
So
each
lay 1 vien bi
xanh trong 3 vien bi xanh la C3. So
each
lay 3 vien bi co 2 vien bi
do,
1 vien bi xanh la
C4.C3
each.
Xac
suat
trong 3 vien bi lay ra co 2 vien bi do la : P(A) =
—i-^
= —.
C^^ 35
43.
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 each 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
hang don vi,
8
each
chon chiJ
so
hang tram,
8
each
chon chuf
so
hang chuc).
So cae
so co 3
chOf so khac nhau
la so
chSn
la
: 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
nhom 3 ngiiofi eiia to. Tinh xac suat xay ra mot tri/cfng hop diidi day
:
a)
Ca ba
nguofi diioc chon deu
la
nam.
b)
Co it
nhat mot trong ba ngUoti

duoe
chon
la
nam.
CHI
DAN
10 9
8
a) Khong gian mau
co
:
C^^ = —'—^
=
120 phan
tijf.
1.2.3
Co
C'l =
^'^'^
=
20
each
chon
3
ngi/ofi deu
la
nam. Xac suat bien
co
3
1.2.3

ngiicfi
dU'ge
chon deu
la
nam
la
:
P(A)
=
C'i
20 1
Cl
120 6-
b) Goi
B
la
bien
co
3
ngUdi
dirge chon
co it
nhat
1
nam. Bien
co
doi lap
eiia
B
la

3
nguofi
difgc
chon deu
la
niJ
:
—
C'* 1 — 29
P(B) = —L =
_ ^
P(B) =
1
-
P(B) =
—.
C^o
30 30
45.
Cho
8
qua can
co
khoi
lu'Ong
Ian liigt
la
1kg, 2kg, 3kg, 4kg, 5kg, 6kg,
7kg, 8kg. Chon
ngau

nhien
3
qua can. Tinh
xac
suat
de
tong
khoi
lu'Ong ba qua can dirge ehgn khong virgt qua 9kg.
CHI
DAN
So
each
ehgn
3
qua can trong
8
qua can
(so
phan tif eiia khong gian
mau)
la
:
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) Tinh xac suat bien
co
tong
so
cham tren hai con sue sSc bang
8.
b) Tinh xac suat bien
co
tong
so

cham tren hai con sue
sac la
mot
so
le
hoac
mot
so
chia het eho
3.
CHI
DAN
Khong gian mau
eo
36 phan
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 bien co sof cap thuan
Igfi
bien co A (tong so cham
bang
8) la : (2; 6),
(6; 2), (3; 5), (5; 3), (4; 4). Xac
suat
cua A la : P(A) = —.
36
b)
Bien co' B : tong so cham la so le
hoac
chia het cho 3.
Goi Bi la bien co' tong so' cham
bang
so' le, B2 la bien co tong so cham
la
mot so chia het cho 3, thi
B = BiuB2
P(B) = P(B,) + P(B2) - P(Bi n B2)

Bi xay ra khr mot con sue sic nay mat chin, mot con nay mat le, co
18 bien co scf 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 xay ra
vdfi
12 bien 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).
Bien co Bi n B2 tong so cham le va chia het cho 3 gom 6 bien co sof
cap :
(1;
2),
(2;
1),
(3;
6),
(6;
3),
(4;
5),
(5;
4).
P(B) = P(Bi) + P(B2) - P(Bi n B2) = — + — - — = — =
36 36 36 36 3
47. Hai xa
thii
cCing
ban (mot
each
doc lap) vao mot muc tieu moi

ngir&i
mot
vien dan. Xac
suat
bSn
triing
dich trong mot Ian bSn ciia ngUdi
thuf
nhat la 0,9; cua ngiicfi
thuf
hai la 0,7. Tinh xac
suat
trong moi
trilofng
hgfp sau :
a) Ca hai vien deu trung dich.
b)
it nhat co mot vien trung dich.
c) Chi CO mot vien
triing.
CHI
DAN
a) Ggi Ai la bien co ngiTcfi thOf nhat ban tning dich, A2 la bien co ngiTcfi
thijf
hai
ban tning dich. A la bien co ca hai vien deu tning dich thi A = Ai n A2. Do
hai
ngiicfi ban doc lap vcfi nhau nen Ai va A2 la doc lap nen
P(A) = P(Ai n
A2)

= P(Ai).P(A2) =
0,9.0,7
= 0,63.
b)
Goi B la bien co co it nhat mot vien dan tning dich :
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) Goi C la bien co, hai
ngiidi
ban moi
ngiidi
mot vien chi co mot vien
tning
dich. Bien co C xay ra khi
ngiidi
thii
nha't tning dong
thcfi
ngiiofi
thuf
hai trUcft
hoac
ngiTofi
thijf
nha't
triigft,
ngifdi
thijf
hai

tning.
P(C) =
P(AiA2^AiA2)
=
P(AiA2)
+
P(AiA2)-P(AiA2
n A1A2)
20 ,.', TS. Vu The Huu -
l\lguy§n
Vinh
Can
Do Ai, A2 doc lap
thanh
thuf
Ai, A2 doc lap, Ai va A2 doc lap va bien
CO Aj A2 n Ai A2 =0 cho nen
P{C)
=
P(Ai )P(A2) + P(Ai )P(A2)
-
0
=
0,9.0,3
+
0,1.0,7
=
0,34.
48. Mot 16 hang c6 30 san pham, trong do c6
3

phe pham. Ngiic/i ta chia
ngau nhien 16 hang
thanh
3
phan, moi phan 10 san pham.
a) Tinh xac suat de moi phan c6 diing mot phe' pham.
b) Tinh xac suat de c6 it nhat mot phan c6 dung mot phe pham.
CHI
DAN
a) Ta thyc bien chia nhxi sau
:
Lay ngau nhien 10 san pham trong 30
san pham ta c6 phan
thuf
nhat. Trong 20 san pham con lai lay ngau
nhien 10 san pham de c6 phan
thuf
hai. Con lai la 10 san pham phan
thuf
3. So'
each
chia
nhii
vay
bSng
C^^.Cgo
each.
So
each
chia de phan

thuf
nhat eo mot san pham xau la
:
C3C27
(lay
mot san pham xau
ghep
v6i
9
san pham t6't trong 27 san pham to't).
So
each
chia
de
phan
thijf
hai c6 mot san phaim xau
la :
CgCig.
S6
each
chia de moi phan c6 mot san pham xau la
:
C3C27.C2C18.
Xac suat de moi phan eo dung mot san pham xau la
:
plp9 plp9
P(A)
=
^^'^"f;

•'^ =
0,246.
*^:i()^20
b) Goi
B
la bien c6 trong
3
phan c6 it nhat m6t phan eo mot phe pham.
Neu goi (i, j,
k) la
s6' san pham xau
theo
thuf
tii ciia cac phan
thuf
nhat,
thuf
hai,
thuf
ba vdfi i,
j,
k
nguyen
dUOng
vdri
0 <
i,
j,
k <
3. Khi do

B
xay ra 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). Bien e6' doi lap ciia
B
xay ra tUOng ufng vdfi cac bo
(3; 0; 0), (0; 3; 0), (0; 0; 3). Ro
tinh
P(B) = 1
-
P(B) thong qua
tinh
P(B)
don gian hcfn.
So'
each
chia de phan
thuf
nhat c6
3
san pham xau, cac phan con lai
deu tot la
:
Ci^.C^^.C^i;.
So
each
chia de phan
thuf
hai eo
3
san pham xau, cac phan

khae
deu
la
san pham tot la
:
Cl^^.Cl.C]
So
each
chia de hai phan dau cac san pham deu t6't, phan
thuf
ba c6
3
san pham xau la
:
C27.C}".
_
pSpiT
plO
,
plOplipV
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.
a)
b)
0
50.
a)
b)
0
d)
51.
52.
53.
a)
b)
c)
54.
a)
b)
Tung
dong
tien
can doi

dong
chat
bon
Ian.
Khong gian m^u
c6 bao
nhieu phan
tijf.
Tinh
xac
suat bien
co
dong
tien
nay mat sap
diing
3
Ian.
Tinh
xac
suat bien
co
dong
tien
nay mat sap it
nhat
2
Ian.
a)
16

phan tuf
.
^
.
-
1
DS
b) P(A)
=
c)
P(B) =
—.
16
Mot
hop
dung
7
qua cau do,
5
qua cau
xanh
(cCing
kich thudc
chi
khac
mau s&c).
Chon
ngau nhien
2
qua cau.

Tinh
xac
suat
:
Hai
qua cau
dixgc
chon
mau
do.
Hai
qua cau
diicfc
chon
mau xanh.
Hai
qua cau
duofc
chon
mot do, mot
xanh.
Hai
qua cau
dUOc
chon
co
it
nhat
mot qua
do.

DS
: a)
P(A)
=
pipi
c)
P(C)
=
b)
P(B)
C2
d)
P(D)
= 1 -
12
Xep ngau nhien
5
chijr
cai
B, G, N, O, O
thanh
mot day
ngang. Tinh
xac suat
de
duoc
chCf
BOONG.
1
1 1

~
5.4.3 " 60'
DS
:
P(A)
=
Mot
dot x6 so
phat hanh
20.000
ve,
trong
do co
mot
giai
dac
biet,
100 giai nhat,
200
giai nhi,
1000
giai
ba va
5000
giai
tu.
Tinh
xac
suat
de

mot
ngUcfi
mua
3 ve
triing
mot
giai nhat
va 2
giai tU.
DS
:
P(A)
= ^
0,0009.
"^20000
Co
hai
hop
dung
cac qua cau
ciing kich thudc, trong lUOng
hop
thuf
nhat dung
7
cau
trSng,
3
cau den. Hop
thuf hai dUng

6
cau
trSng
4
cau
den.
Rut
ngau nhien tU
moi hop mot qua cau.
Tinh
xac
suat
:
Hai
qua cau
rut
ra deu
mau trdng.
Hai
qua cau
riit
ra
cCing mau.
Hai
qua cau
riit
ra
khac
mau.
DS

: a)
P(A)
=
0,42
b)
P(B)
=
0,54
c)
P(C)
=
0,46.
Co
hai
to hoc
sinh.
To I co 9
nam,
6
nijf.
To H co 7
nam,
8
nU.
Chon
ngau nhien
moi
to
mot
ngUoti. Tinh

xac
suat
:
Hai
ngUcfi
dugc
chon
co
diing
mot
nU.
Hai
ngUcfi
dugfc
chon
co
it
nhat
mot
nam.
DS
: a)
P(A)
=
114
225
b)
P(B)
=
177

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
KIEN
THCTC
A(x).B(x)
=
0 o

1. Phifcfng
trinh
b^c nhat mpt an
c6
the
diia
ve
dang
:
ax
+
b
=
0;a,
b e
E,
a ;t 0
CO
mot nghiem duy nhat
x
=
-
—.
a
2. Phvfcfng
trinh
tich
mpt an
c6
dang

:
A(x).B(x)
=
0 (1)
Tap nghiem ciia
(1)
la
tap nghiem ciia phiTcfng
trinh
A(x)
= 0
hcfp
vdi
tap nghiem ciia phiTOng
trinh
B(x)
=
0.
"A(x)-0
B(x)-0'
3. Bat phiicfng
trinh
bac nhat mpt an
c6
the difa
ve
dang
:
ax
+

b
>
0
hay
ax
+
b
> 0; a,
b e
M,
a
?t
O
• Neu
a > 0,
ax
+
b
>
0
CO
tap nghiem
x
>
-
—
a
• Neu
a <
0, ax

+
b
>
0
CO
tap nghiem
x
< —.
a
4. Dau cua nhi thtfc bac nhat
ax
+
b
(a
0)
Nhi
thufc
bac
nhat
: f
= ax
+
b
(a
?t
0)
Gia tri
x
=
-

—
tai
do
f
c6
gia tri bang
0
goi
la
nghiem cua nhi thufc
a
•
Ta
CO
dinh li
ve
dau ciia nhi thufc nhir sau
:
Dinh
li
+
Vdi cac gia
tri
x
<
-
—
thi
f
= ax

+
b
c6 dau
trai
vdi dau ciia he
so
a.
a
+
Vori
cac
gia
tri
x
>
—
thi
f
=
ax
+
b
c6
dau ciing dau vdfi
he
so a.
a
1. Giai
cac
phiTofng

trinh
:
,
10-x 20-X 30-X _
a)
+ +
=
3
(1)
100 110 120
Hoc
va on
luyen theo
CTOT
mon Toan
THPT
'.
23
b)
- - X + 1 = 0
c)
x'*_+ 2x^
+ 5x2 + 4x - 12 = 0
CHI
DAN
(2)
(3)
a) (1) o
10-X
100

-1
20-X
^
110
-1
30-X
120
-1
=
0
-90-X
-90-X -90-X ^
C5>
+ + = 0
100
(90+
x)
110
+
•
120
1
100
110 120
=
Oc^ X =
-90.
b) x'^
- x' - X + 1 = 0 « x'(x - 1) - (x - 1) = 0
o

(X
-
l)(x'^
- 1) = 0 »
x''
-
1
= 0
x-l
= 0
<=> X = ±1.
c)
xS2x^
+
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\
X
+ -
23
>0
Vx)
(Bof

vi x^ + x + 6 =
x
+ 2 = 0
Tap nghiem
1-2;
11.
2. Giai, bien luan
theo
tham
so m
phiiong
trinh
:
m^x
- 3m = 9x +
m^
(1)
CHI
DAN
(1)
o
(m^
- 9)x =
m^
+ 3m
+
Neu
m^
- 9 0,
nghia

la
vdi
m ?i ±3
thi phiiong
trinh
(1) c6
nghiem
:
X
=
m
+
3m
m
m^-9
m-3
+
Neu m = 3,
phiTOng
trinh
(1) c6
dang
: 9x - 9 = 9x + 9
Phiiong
trinh
nay v6
nghiem.
+
Neu m = -3
thi phifong

trinh
(1) c6
dang
: 9x + 9 = 9x + 9
Phiforng
trinh
v6
dinh (mpi
so
thiic
deu la
nghiem ciia phuong
trinh).
3.
Cho
phiiong
trinh
: a^x + b = ax + ab (1)
trong
do a, b la cac
tham
so
thiTc. Giai, bien luan
theo
a, b
phu'ong
trinh
tren.
CHI
DAN

(l)oa(a
- l)x = b(a - 1)
+
Neu a(a - 1) ;^ 0,
tufc
la
neu
a 0 va a ;^ 1
thi
(1) c6
nghiem duy nhat
^
^
(a-l)b
^ b
a(a-l)
a
+
Neu a = 0,
phuong
trinh
(1) c6
dang
: Ox = -b
Trong trudng
hop nay c6 hai
kha nang
:
-
Neu b ;^ 0

(tijfc
la a = 0, b ;^ 0)
phiiong
trinh
v6
nghiem.
-
Neu b = 0
(tufc
la a = 0, b = 0) moi so
thiic deu
la
nghiem.
24 TS Vu The Hi/u - Nguygn Vinh Can
+
Neu a = 1,
phUofng
trinh
(1) c6
dang
: Ox = 0,
phifdng
trinh
c6 tap
nghiem
la
moi
so
thiTc.
4.

Giai
cac
phiiorng
trinh
:
a)
(X -
1)-^
+ (x + 2f = (2x +
1)'^
(1)
b) 9a_x^
-
ISx'^
- 4ax + 8 = 0
(2),
a la
tham
so.
CHI
DAN
a)
(1)« (2x +
l)[(x
-
1)^
- (x -
l)(x
+ 2) + (x + 2f] = (2x + if
o (2x

+
l)[(x
-
1)-
- (x -
IXx
+ 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
nghiem
: x =
—,
x =
±
—
a
3
2
+
Neu
a = 0, (2) c6 cac
nghiem
: x = ±
—.
3
5.
Giai
cac bat
phLfomg
trinh
diidi
day va
bieu dien
tap
nghiem ciia

no
tren
true
so.
a)
3(x + 2) - 1 >
2(x
- 3) + 4 b) (x -
2)^
+ 1 +
x'"^
>
2x^
- 3x - 4.
CHIDAN
-10
a)
3x + 5 > 2x - 2
M%«Mt
>
o3x
- 2x >-2 - 5 o X >-7
Tap nghiem
dugc
bieu dien
d
hinh
ben. -^'"^
°
b)

o
x^
- 4x + 4 + 1 +
x^
>
2x^
- 3x - 4 -4x + 3x > -4 - 5
ci>
—X
^ —9 X ^ — — 9 ^
\////////////////////
_X
•
^>>>>>>»m>»>>>>,
Bieu
dien tap nghiem
d
hinh
ben.
Hinh
h
6.
Giai
va
bien luan theo tham
so m
bat phifofng
trinh
:
x-l

x +
1
^ ,,,
x+
>
(m
+
l)x
(1)
m
m
CHI
DAN
2
Dieu
kien
xac
dinh
m
=i'i
0.
Chuyen
ve,
rut gon
ta
di/dc
:
(m
+
2)x

> —
m
+
Neu
m 0 va m > -2
(thi
m + 2 > 0)
bat phifdng
trinh
c6
nghiem
2
x
>
m(m +
2)
2
+
Neu
m < -2
(tat nhien
m 0)
thi
(1) c6
tap nghiem
x <
m(m
+ 2)
+
Neu

m = -2
bat phiiang
trinh
trd thanh
Ox > -1
Moi
so
thuc deu
la
nghiem ciia (1).
3X
b
7.
Giai,
bien luan bat phi/cfng
trinh
vdi tham
so a va b: x + 1 > 1—(1)
b
a
Hoc
va on
luy?n
theo
CTDT
mon
Toan
THPT
•.'.
25