VO THE HlTU - NGUYEN VINH CAN MON... VU THE HlTU - NGUYEN VINH CAN... b Quy tdc nhdn : De hoan thanh mot cong viec A phai thuc hien hai cong doan.. Tong quat, de hoan thanh cong viec A
Trang 1510.76
6-454L
TS VO THE HlTU - NGUYEN VINH CAN
MON
Trang 2TS VU THE HlTU - NGUYEN VINH CAN
Trang 3GIJII TICH
Hoc va On luy§n theo CTDT mfln Toan THPT E I 3
Trang 4b) 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 thi t h i c6 m i m 2 m k each de hoan
thanh cong viec A
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 :
Trang 5et qu
a c
6 the
et qu
a c
6 the
Theo quy t&
c nhan so' ke
t qu
a ta
o than
h ca
c so' c
6 ha
g liT
u y sir kha
et qu
a e
o the
2 Chg
n mo
t ehu
f s
o la
Theo
quy tie
nha
o d
a ch
o l
a : n' =
6 x
5 = 3 0
so
Cdch khac : M
h tC
r 6 chuf s
i go
m ha
i pha
n tu
f tu
" 6 phan tuf d
a cho D
h tC
r 6 ehuf s
2 cu
a ta
p hg
p 6 phan
tuf
n = = 6.
5 = 30 so'
n e
o 4 ehuf s
CHI DA
N
a) C
o 6 each ehgn chuf s
g tram, 7 each chgn chuf so' han
g ehu
c v
a 7 each
ehgn
ehuf s
o hang don v
i The
o qu
y t^
c nhan : s
o eac
h ta
o th
n 4 chuf s
o tC
r ta
p hg
p 7 chij" s
x 7 = 20 58
so
b) C
o 6 each
chgn chuf so' han
g nghin, k
i 6 chOr s
da chgn Va
y c
6 6
each
chgn ehuf s
o han
g tram Kh
va
hang
tram, eo
n la
i 5 ehu: s
o kha
c vd
u Ng
-u ye
n Wn
h C?
n
Trang 6chon 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 tap hop 7 chOf so da cho la :
N ' = 6 X 6 X 5 X 4 = 720 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 tap hop 7 chCf so da cho l a mot chinh hop chap 4 tCr tap hop
7 chuf so ma cac chinh hop nay khong 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 = 720 so
3 Mot to hoc sinh c6 10 ngufofi xep thCf td thanh hang 1 de vao Icfp Hoi
a) Co bao nhieu each de to xep h a n g vao Idfp
b) Co bao nhieu each de to xep hang 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
C H I D A N
a) So' each xep h a n g bang so' hoan v i cua 10 phan tuf
N i = 10! = 3628800 each
b) Coi hai b a n A va B n h i i mot ngifdi Do do so each xep h a n g cua to de
vao Idp trong do h a i b a n A va B d i l i e n nhau bang so hoan v i cua 9
phan tuf
N2 = 9! = 362880 each
4 Co bao nhieu each xep 6 ngUc/i ngoi 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 hang ngang cua m o t b a n a n dai
b) Sip 6 ngLfcfi ngoi vong quanh mot ban a n t r o n
C H I 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 tuf So each
sap xep la : 6! = 720 each
b) Gia sijf 6 ngiicfi a n dirge danh so thuf t u l a : 1, 2, 3, 4, 5, 6 va m o t each
sap xep theo ban t r o n nhiT h i n h
Neu t a cat b a n t r o n d v i t r i giufa 2 va 4 r o i t r a i d a i theo b a n ngang
t h i t a C O hoan v i (1) tiTcfng lirng mot each xep ngudi ngoi theo b a n a n
dai 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
ban t r o n tifOng ufng v d i 6 each sap xep theo b a n dai Do do so each
Hpc v4 On luy$n theo CTDT mfln Toan THPT £3 7
Trang 76 niJ Ca
n la
p nho
a) Nhom c6 3 nam va
1 nvC
b) S
o nam va nO
f tron
g nho
m bSn
g nhau
c) Ph
ai C
O i
t nha
t m
ot nam
g s
o 6 nuf l
b) S
o eac
h la
p nho
m go
m 2 nam va
h la
p nho
m 4 ngudi tron
g d
o c
6
it nh
at 1 nam
la
: 1
nam, 3 nuf hoS
e 2 nam, 2
nO
r hoS
c 3 nam, 1
nuf hoS
c 4 nam
m 4 ngudi toa
n nu
f
la : C
m 4
ngiTdi
c6
it nh
at 1 nam la :
phan biet (
n mot
diTdng
thang (3 <
m d
a ch
o l
a dinh
CH
I D
AN
Cii 3 diem
khong thang hang tao than
h mo
t tam giae S
k die
m
tren dirdng
than
g l
a : C^ S
o ta
m gia
e c
6 3 dinh la eac die
m d
a ch
o
la: N = C
C^
tam giae
Trang 8-7 a) Co bao nhieu so tu n h i e n la so chSn c6 6 chOr so doi m o t khac nhau
v a chiJ so dau t i e n la chui so le
b) Co bao nhieu so tuT nhien c6 6 chuf so doi mot khac nhau, trong do c6 dung 3 chCif so le, 3 chuf so chSn (chuf so dau tien phai khac 0)
C H I D A N
a ) So can t i m c6 dang : x = a j a g a g a ^ H g a g trong do a i , a e lay cac chijr
so 0, 1, 2, 8, 9 v d i aj ;t 0, a i aj v d i 1 < i ;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 la m o t chinh 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 la :
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 tap hop X2 = {1; 3; 5; 7; 9) 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 : C 5 = 10 each
So each lay 3 phan tijf thuoc X2 la : C 5 = 10 each
So each ghep 3 phan tuf lay tii X i v d i 3 phan tuf lay tii X2 la :
8 M o t hop diing 4 vien b i do, 5 vien b i trSng va 6 v i e n b i vang NgUcfi
ta chon r a 4 vien b i tii hop do H o i c6 bao nhieu each lay de trong so
bi lay r a khong du ca 3 mau
C H I D A N
Cdch 1 : So each chon 4 vien b i khon g du 3 mau bang so each chon 4
vien bat k i trCr d i so each chon 4 vien eo ca 3 mau
N = C\, -{ClC\.C\+Cl.C\.C\+Cl.C\.C\) = 645 e a c h
Cdch 2 : So each chon 4 vien b i khon g du 3 mau bang so each chon 4
vien m o t mau (4 do, 4 t r a n g va 4 vang) eong v d i so each chon 4 v i e n hai mau (1 do, 3 t r a n g hoac 2 do, 2 t r a n g hoac 3 do, 1 trSng hoac 1
do, 3 vang hoSe 2 do, 2 vang hoac 3 do, 1 vang hoSc 1 t r a n g , 3 vang hoac 2 t r a n g , 2 vang 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
Trang 9ngon
lufa trai
CH
I DA
Vi dudn
g tron
n
(xem bai s
o 4)
10 Chufn
g min
b) c;
; = c;;:'i + c;;:'^ +
CH
I DA
N
a) Vd
i k
e N, k > 2 ta
eo : A^ = k(k
- 1)
J_
k(k -1)
Thay k = 2, 3, n vao (*
) t
a c
6 v
e tr
ai cu
a (1) l
a :
_^
1 _ n-1
^ -l nj ^n + + 3j — U + 2j — u
b) The
o tin
C;;_3
11
Cong
ve vd
+
+ C-+
g
thiJc (2) ca
n chufn
g min
h
Chufng min
M ^ C^^i +
^Zl
trong d
o k
e N, k < 2000, l
a t
o ho
p cha
p k cua n phan
tuf
10
£2 TS V
u ThS ' Hu
u Nguyen Vin
-h C$
n
Trang 1012 TCr diem A den diem B ngtrdi t a c6 the d i qua C hoac d i qua D va
khong C O difdng di thang tCr C den D TCr A di thSng den C c6 2 each,
tCr C di t h a n g den B c6 3 each TCr A di t h ^ n g den D c6 3 each tCr D di
thang den B c6 4 each
a) Hoi tCr A c6 bao nhieu each di t d i B ? 2
TCr 7 chOf so 0, 1, 2, 3, 4, 5, 6 c6 the ghi dirge bao nhieu so tu n h i e n
moi 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 nhien c6 5 chuf so
b) Dung tap hop X c6 the ghi dirge bao nhieu so tU n h i e n c6 5 chuf so
khac nhau tCrng doi
c) Dung tap hgp X c6 the ghi difgc bao nhieu so tif n h i e n c6 5 chur so
khac nhau la so chSn
BS : a) 6.7^ so b) 6l5.4.3 so c) A^ + 15AI so
15 M o t to hoc sinh c6 5 nam, 5 nuf xep t h a n h mot hang doc
a) Co bao nhieu 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 nam hoc sinh va 20 nff hoc sinh Can chon m o t n h o m
cong tac 3 ngiTdi Hoi c6 bao nhieu each chon trong moi triTdng hop sau
a) Ba hoc sinh bat k i cua Idp
b) H a i nu" sinh va mot nam sinh
HQC 6n luy$n theo CTDT mOn Toan THPT S 1 1
Trang 11o eac
h c)
t ch
o 3 ngirdi tron
g ca
c tru6n
g
hop sau :
a) Mo
t ngLfcf
i nha
n 3
do va
t, co
n 2 ngiicfi
b) Mo
i ngird
i i
t nha
DS
: a
) S^.Cl
each b) 3.C^C,
' + 3C^.C^each
18
M
ot to
C O
9 na
m v
a 3 nOr
3 nho
m mo
i nho
m 4 ngiroti v
a tron
g
mo
i nho
^ = 10080
each
19 Tim cac s
o nguye
n diron
g x, y thoa ma
X
= 8, y = 3
o do
i mo
t kha
c nhau
DS
-.11=
Al+
4.8.8 = 760
A iA 2
.A 2n ,
so ta
n die
m tren nhieu gap
20
Ian
so hin
h chu
f nha
t c
6 din
h l
a 4 trong 2
n din
h tre
22 Tim
so t\i
nhie
n n, bie
t ran
g C
° + 2C;, + 4C^
+
+ 2"C;; =
243
5S : n = 5
23
Giai bat
p hiT dn g
trin
h (vd
i h
ai an n, k
t mon hoc, tha
m 5
cau
hoi kho, 1
0 ca
u ho
i trung bin
m tr
a go
m 5 cau khac nha
u sa
o ch
o
trong mo
i d
e nh
at thie
g binh
,
di) v
l
25
Cho ta
p hd
p A
eo n phan
tijf
(n > 4) Bie
t rkng
so ta
20 Ia
tuf
cua A T
im
so
tiT
nhie
n k sao ch
A
l
a lor n
nhat
£)S : n = 18, Cf
^ >
C\^^
o k = 9
12 E
l TS V
O T
h g' Hi;
u Ng
-u yS
n V
T nh C?
n
Trang 12Cac h e so cua 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 giac d u d i d a y g o i l a tarn gidc Patcan
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 cho :
' Hoc va 6n luygn theo CTDT m6n Toan THPT £3 13
Trang 13k 7-k
= 4(
7 k) o
kha
i trie
n l
a : =35
27 Ti
CHI
D
AN
Kh
ai trie
n nh
i thuf
c (x
^ xy)^'* c
-6 1
5 s
o hang, s
g :
Cl,(x^)"-^(-xy)^
= -Clx'\xy = -3432x
^V^
28 Ti
i trie
n nh
i thiJ
c
a b^
-a
b-a Bie
t
rang
ba cu
a kha
i trie
N
Trong cong thufc nh
i thuf
c Niuto
n (
A + B)° s
21 <
^ n(n_-l) ^
ai trie
b-a
+ • b^
CH
I DA
N
(1 + x^)" =
C° + C^x^ +
C^x* + + C^x^^ +
+ C
= 102
4 = 2" = 2^° ^
'° 6!4
!
30
Trong kha
i trie
n n
hi thuf
c Niuto
^nx^ 1
14
, X ^ 0,
hay ti
- 2)
1.2.3
<:> n(n
^ 3n
- 28) =
0 <
= > n = 7
Trang 14Thay n = 7 vao nhi thufc Niuton da cho thi c6 :
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
Trang 15Tim
so han
g khon
g chiJ
a x cua kha
i trie
n n
hi thuf
c Niuton
i trie
n v
a riit gon bieu thufc :
P(x) = (1 + x)^ + (1 + x)^ + (1 + x)^ + (1 + x)'' + (1 + x)^
"
ta diio
c : P(x) = aiox^° +
agx^ + agx^ +
+ aix +
ao
Tin
h as
i trie
n v
a ru
t go
n P(x) = t
x +
if +
(x 2)^
-than
h v
di n nguyen
ducfng t
a c
6 :
a) Cl+Cl+ + Cl^Cl+Cl+
+ Cl:-\
b) C> 2C^ + 3C^ + + nC;; = n2"-\
CH
I DA
N
a) Kh
ai trie
n P(x) = (x - 1)^"
ro
i ch
o x = 1
b) Kh
ai trie
n P(x) = (1 + x)"
Tim P'(x)
ro
i tin
h P'(l)
37.
Viet kha
i trie
n Niutdn, bie
h r&n
g :
3
16 p0 o lS pl q l4 p2 Q l3 p3
Pl
6 _ O l6 '16 '16 '16 ^16
38.
Trong kha
i trie
n nh
1^
Ha
y ti
i trie
n — + x
^
u
cL
i + c
Li +
+ CL
,=
-i
2-DS :a = 210
he s
o cu
a so' han
g chuf
a x
^ tron
g kha
i trie
n — + Vx
^
c:;:i-c::,3
= 7(n+3)
DS : a = 495
biet
biet
16 £2
TS V
u ThS ' H^
u Nguygn Vin
-h CS
n
Trang 16§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
Mot phep thuf ( t h i nghiem) c6 the lap l a i so I a n tuy y vdti cac dieu kien co ban gio'ng nhau nhiTng khong the xac d i n h chSc chfin, k e t qua nao trong moi Ian thifc h i e n ma chi c6 the noi k e t qua do thuoc mot
tap hop xac dinh t h i ta goi la phep thii ngdu nhien Tap hop t a t ca cac k e t qua c6 the c6 cua phep thuf ngSu n h i e n goi la khong gian mdu
cua phep thijf do
2 B i e n co n g a u n h i e n
Mot phep thijr ngau nhien T co khong gian mSu la E, moi tap hop A e
E bieu t h i mot bien co ngdu nhien (lien quan t d i T) B i e n co ngfiu nhien, chi gom mot phan tijf cua E duoc goi la bien co so cap B i e n co dac biet gom moi phan tuf cua E la bien co chdc chdn B i e n co khong
chiJa phan tuf nao cua E la bien co khong the co, k i hieu 0 H a i bien
CO A, B ma A n B = 0 t h i A va B dirge goi la hai bien co xung khdc
3 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 khong gian mau E gom n bie'n co so cap c6 kha nang xuat h i e n dong deu (dong kha nang) B i e n co ngau n h i e n A
gom k bien co sc( cap (cua E) t h i xac suat cua bien co ngau nhien A,
a) Bien co' ngau n h i e n A bat k i ta deu co 0 < P(A) < 1
b) P(0) = 0, P(E) = 1
c) A va B la hai bien co' xung khac (tufc A n B = 0 ) t h i
P(A u B) = P(A) + P(B)
Neu A va B la hai bien co bat k i t h i
P(A ^ B ) = P(A) + P(B) - P(A n B)
d) Neu A va A la hai bien 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)
5 B i e n co dpc l a p v a quy t a c n h a n x a c s u a t
Hai bien co ngau n h i e n A va B cCing lien quan v d i m o t phep thuf ngau
nhien la doc lap vai nhau neu viec xay ra hay khong xay r a cua bien
co' nay khong anh htfdng t d i kha nang xay ra cua bien co k i a
Quy tdc nhdn xac suat
Neu hai bien co ngau nhien A va B doc lap vdi nhau t h i
Trang 17b) Go
i A
la bie
n co, tron
g b
a Ia
n tun
g co dung mo
E = INNN; NSN
p co 3 o A n c Bie phan tuf o 8 u c n mS g gia Khon a nang g kh a don thuf l
3 — P(A) = la : a A t cu c sua o xa o d , d phan tuf
8 vien b , 3 u do i ma vien b o 4 p c t ho g mo 42 Tron
, mo
i eac
h la
y 3 vien b
i l
a la
y 1 tap ho
p do S
o eac
h
lay 2 vien b
i d
o tron
g 4 vien b
i xan
h l
a
C3
So eac
h la
y 3 vien b
i xan
h la
C 4.
i d
o l
a : P(A) = = —
g g do o han f s n chu h cho 0 eac , 1 g chuc o han a s n ch h cho 0 eac trSm, 1
Trang 18(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 la : P(A) = — « 0,3644
900
44 M o t to hoc sinh co 10 ngtrdi gom 6 n a m va 4 nuT, chon ngSu nhien mot nhom 3 ngiiofi cua to T i n h xac suat xay r a mot 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 trong ba ngiidi diTOc chon l a n a m
45 Cho 8 qua can co k h o i liTOng I a n iMt la 1kg, 2kg, 3kg, 4 k g , 5kg, 6kg,
7kg, 8kg Chon ngau n h i e n 3 qua can T i n h xac suat de tong k h o i luong ba qua can di/crc chon k h o n g virot qua 9kg
C H I 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 bien co' tong so' cham t r e n h a i con sue s^c bang 8
b) T i n h xac suat b i e n co tong 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 het cho 3