Môn Toán - Học và ôn luyện theo cấu trúc đề thi: Phần 1

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Phần 1 tài liệu Học và ôn luyện theo cấu trúc đề thi môn Toán trình bày các kiến thức phần Đại số và giải tích bao gồm: Đại số tổ hợp và xác suất, phương trình và bất phương trình đại số, phương trình lượng giác, phương trình, bất phương trình mũ và logarit,... Mời các bạn cùng tham khảo nội dung chi tiết.

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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 ) ! (Quy irde : 0! = 1). 4. 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
BAI TAP 1. C h o cac chuf so 2 , 3, 4 , 5, 6, 7. a) 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 DAN a) 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. a) 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. b) 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 hop cac chuf so d a cho. CHI DAN a) 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 = 2 0 5 8 so. b) 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
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 5. 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^ = — = 540 each. ' ' 1.2 1.2 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 . 6. 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 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) C;; = C;;:; + Cl;}^ + ... + Cl:\) t r o n g 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 t r a i ciia (1) la : (1 (I V f 1 1^ — — — — — — — +... + u 2. l 2 3v [n-1 nj 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 : C^ooi + ^ C\Z + CfZ t r o n g do k e N, k < 2000, 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 2001! (k + l ) ! ( 2 0 0 0 - k ) ! k +1 k+1
c) B a hoc s i n h c6 i t n h a t m o t nuf. DS : a) C;;^ e a c h b) 25.C^o e a c h c) C'^^ -Cl, each. 17. 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 : a) 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 . b) 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 . DS : a) 3.C^C^ e a c h b) S.CtCl + SCl.Cl each. 18. M o t to CO 9 n a m v a 3 nOf. a) 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. b) 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 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 ^ ^ ^ " 5 ^ " 2 ^ • f)S : X = 8, y = 3. 20. 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 •.n= Al+ 4.8.8 = 7 6 0 so. 21. 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. 23. 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) 24. 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 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= Cl,ClCl+C',,C\,Cl+C%C\f = 56785 d l . 25. 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 . flS : n = 18, C^g > C\^' 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 q u y vide a, b ^ 0, a" = b° = 1 , C° = 1 . 2. 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 . n =0 1 n =1 1 1 n =2 1 2 1 n =3 1 3 3 1; n =4 1 4 6 4 : 1 n =5 5 10 10 5 1 n =6 6 15 20 15 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 bang so h a n g or h a n g diTdfi h a y C^'^ + Cj^ = 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 1 I _ i Vdfi X > 0, t a CO : \/x = x ^ ; —j=r = x "* %/x f 1 7-1 1 7-2 2 /X +• = (x'^ + x - C ° x 3 + Cix 3 X 4 + c?x ^ X 4 /x; 7-k _k 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
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 : C L ( x ^ r " ( - x y ) ^ = -CLx^^xV^ - -3432x^V^ A" a b2 ' - „a2 28. Tim so hang thuf tii cua khai trien nhi thufc + . 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 a b^-a'^^' Vay so hang thuf tii cua khai trien la : b-a • + 7-3 ^b^-aM a(b + a) C? = 35 [b-aj ^ a J b-a 29. 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 : = = 210. 6!4! / 2 -I 30. Trong khai trien nhi thuTc NiutOn ' nx 1 ^ , X ^ 0, hay t i m so hang 14 x^ chufa x\t rSng 5C;;-' = C'l (Trich de TSDH khoi A - 2012) CHI DAN n(n - l)(n - 2) bCr = Ct 5n = 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 t h i c6 : 2 1 -c* I-] + ... + 12 .2, X V X f-1 So hang chufa x^ trong khai trien la so' hang thuf k + 1 sao cho : 7-k k X^ 27-k ^ 2 ( 7 - k ) - k = 5=^k = 3 12; .Xy 1 7.6.5 1 ^5^_35^3 Vay so hang chufa x la : -C^ 1.2.3 2' vXy 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 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" (*) He so cua x^° trong khai trien (*) la : a^o = 2C\\ 22. 32. 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. CHI 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 : 1 f) q Q as = + 2'Ct = 2 7 . ^ ^ ^ ^ + 16.5 = 3320. ' 1.2.3 Hqc va on luyen theo CTDT mon Toan THPT .'' 1 5
CAC BAITAP lij GIAI 33. 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 . X + — if . X j DS : 924. 34. 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. 35. 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 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 ) . 37. Viet k h a i t r i e n Niutcfn, bieu thufc (3x - 1)^'', tU do chufng m i n h rSng : 28 \ 38. T r o n g k h a i t r i e n n h i thufc 15 . H a y t i m so' hang khong x
§3. XAC SUAT K I E N THCTC 1. 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. 2. 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. 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 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, ki hieu P(A) la t i so: P(A) = - . n 4. C a c tinh chat cua xac suat a) B i e n 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 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). d) 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). 5. 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 ^ Hoc va on luyen theo CTDT mon Toan THPT . 1 7 7/15 X23' Ji
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 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 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 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 C'i 20 1 ngiicfi dU'ge chon deu l a n a m l a : P(A) = 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. 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. CHI 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) Xac suat ciia A : P(A) = — = 0,125. 56 46. 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 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 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. 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) = 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) 20 , . ' , TS. Vu The Huu - l\lguy§n Vinh Can