Tài liệu Học ôn luyện theo cấu trúc đề thi môn toán (nxb đại học quốc gia) vũ thế hựu, 297 trang

510.76 rS. VO THE HirU - NGUYEN VINH CAN H419V • • HOC a ON LUYIN T H E O C A U T R U C D E THI MON I TS. VU THE HlfU - NGUYEN VINH CAN nioc & ON LUYEN T H E O C A U T R U C D E THI THi; VIEN TINH BINH THU*N ON THI DAI HOC Ha NQI N H A X U A T B A N D A I HQC QUOC G I A H A N O I H O C vA NHA XUAT B A N DAI HQC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba TrUng - Ha Npi Dien thoai: Bien tap - Ciie ban: (04) 39714896; Hanh ctiinii: (04)3 9714899; Tong Bien tap: (04) 39714897 Fax: (04) 39714899 Chiu trdch nhiem xuat ban: Gidm doc - Tong bien tap: T S . P H A M T H I T R A M Nha sach H O N G A N Che ban: THAI VAN Sica bai: H O N G SON Bien tap: Trinh bay bia: THAI HOC Thj^c hi?n lien kit: Nha s a c h H O N G A N SACH LIEN KET O NLUYEN THEO CAU TRUC D ET H I M O N TOAN THPT IVla so: 1L - 65DH2013 In 2.000 cuon, I

1). M 5 i each sap thuf til eac phan tiif ciia t a p hop A duoc goi l a m o t hoan vi ciia n p h a n tuf eua A . Dinh li : So hoan v i khac nhau ciia n p h a n til bang : Pn = n ( n - l ) ( n - 2 ) . . . 2 . 1 = n ! 3. C h i n h hrfp M o t tap hop A hOTu h a n gom n phan tuf (n > 1) va so nguyen k (0 < k < n). M o i tap hop eon eua A gom k phan til sSp theo mot thuf tiT nhat d i n h dLfgrc goi la mot chinh hap chap k cua n phan tuf. Dinh li : So c h i n h hop chap k ciia n p h a n tuf bang : A;; = n ( n - l ) ( n - 2)...(n - k + 1) = (n - k ) ! 4. (Quy irde : 0! = 1). T o hofp Cho t a p h o p A hufu h a n c6 n p h a n tuf ( n > 1) va so nguyen k (0 < k < n). M o i tap hop con gom k p h a n tuf ciia A (khong t i n h thuf tU eac p h a n tuf) g o i l a m o t to hop chap k cua n p h a n tuf. n' A'' Dinh li : So to hop chap k cua n p h a n tuf l a : C'' = = —(n-k)!k! k! He qua: Cl=C:=l; 0^= Cr''; C ^ = C!; + C ^ \ HQC va on luy$n theo CTDT mon Toan THPT S 5 1. a) BAI TAP C h o cac chuf so 2 , 3, 4 , 5, 6, 7. C o b a o n h i e u so' t\i n h i e n c6 h a i chuf so ducfc t a o n e n til t a p h o p c a c chuf so d a cho. b) C o bao n h i e u so t i i n h i e n c6 h a i chuf so k h a c n h a u difOc tao n e n tiT t a p hcrp chuf so d a cho. CHI a) DAN D e t a o m o t so c6 h a i chuf so t a t h i f c h i e n h a i c o n g d o a n : 1. C h o n m o t chuf so l a m chuf so h a n g chuc : c6 6 k e t qua c6 t h e . 2. C h o n m o t chuf so l a m chuf so h a n g d o n v i : c6 6 k e t qua c6 t h e . T h e o q u y t ^ c n h a n so k e t q u a t a o t h a n h cac so c6 h a i chuf so tii t a p hgfp 6 chCif so d a cho l a : n = 6 x 6 = 3 6 so'. b) L a p l u a n g i o n g n h i f c a u a ) n h u i i g liTu y sir k h a c b i e t so vdfi trifofng h o p t r e n of cho so d u g c t a o t h a n h c6 h a i chuf so k h a c n h a u . D o do t a c6 k e t q u a n h u sau : 1. C h o n m o t chuf so l a m chuf so' h a n g chuc : c6 6 k e t q u a c6 t h e . 2. C h o n m o t chuf so l a m chuf so h a n g d o n v i : c6 5 k e t q u a c6 t h e ( v i chuf so n a y p h a i k h a c chuf so h a n g chuc d a c h o n trifdrc do). T h e o q u y t a c n h a n : so cac so c6 h a i chuf so k h a c n h a u difcfc t a o t h a n h tCr t a p h o p 6 chuf so d a cho l a : n ' = 6 x 5 = 3 0 so. Cdch khac : M o i so c6 h a i chuf so t a o t h a n h tiT 6 chuT so d a cho l a m o t t a p hop c o n s^p thuf tiT g o m h a i p h a n tuf tiT 6 p h a n tuf d a cho. D o do so cac so C O h a i chuf so k h a c n h a u t a o t h a n h tiT 6 chuf so d a cho l a so c h i n h hop c h a p 2 cua t a p hop 6 p h a n tuf. n = Ag = 6.5 = 30 so. 2. C h o t a p h o p cac cha so 0, 1, 2, 3 , 4, 5, 6. C o bao n h i e u so t U n h i e n c6 4 chuf so k h a c n h a u tCmg d o i tii t a p b) C o bao n h i e u so t u n h i e n c6 4 chuf so tii t a p h o p cac chuf so d a cho. a) hop cac chuf so d a cho. CHI a) DAN Co 6 each c h o n chuf so h a n g n g h i n (chuf so d a u t i e n p h a i k h a c 0), 7 e a c h c h o n chff so h a n g t r a m , 7 e a c h c h o n chuf so h a n g chuc v a 7 e a c h c h o n chuf so h a n g d o n v i . T h e o q u y tSe n h a n : so each t a o t h a n h so t i f n h i e n 4 chuf so tii t a p h o p 7 chiJ so d a cho l a : N = 6 x 7 x 7 x 7 b) = 2 0 5 8 so. C o 6 e a c h c h o n chuf so h a n g n g h i n , k h i c h o n x o n g chuf so h a n g nghin c o n l a i 6 chuf so k h a c vdi chuf so h a n g n g h i n d a c h o n . V a y c6 6 e a c h chon chuf so h a n g t r a m . K h i d a c h o n chuf so h a n g nghin va hang t r a m , e o n l a i 5 ehOf so k h a c v d i cac chuf so d a c h o n . D o do eo 5 e a c h 6 :S; IS. Vu The Hi/u - NguySn Vinh Cin chpn chOf so h a n g chuc. Tifcfng tir, c6 4 each chon chOr so h a n g don v i . Theo quy tac n h a n . So cac so txi n h i e n c6 4 chOf so khac nhau tCfng doi difcfc tao t h a n h tix t a p hop 7 chuf so da cho l a : N ' = 6 X 6 X 5 X 4 = 720 so. Cdch lap luan khdc : M o i so tiT n h i e n c6 4 chOf so khac nhau tao t h a n h tCr tap hop 7 chOf so da cho l a m o t c h i n h hgrp chap 4 ti^ t a p hgfp 7 chuf so m a cac c h i n h hgfp nay k h o n g c6 chuT so 0 or dau. Do do so cac so CO 4 chijf so khac nhau tiT 7 chijf so l a : N' = - Ag = 7 X 6 X 5 X 4 - 6 X 5 X 4 = 720 so. 3. Mot to hoc sinh c6 10 ngUofi xep thijf tif thanh hang 1 de vao lorp. Hoi a) Co bao n h i e u each de to xep h a n g vao l(Jp. b) Co bao n h i e u each de to xep h a n g vao Idfp sao cho h a i b a n A v a B eua to luon d i canh nhau va A dufng tri/dtc B . CHI D A N a) So each xep h a n g bang so hoan v i ciia 10 p h a n tiif. N i = 10! = 3628800 each. b) Coi h a i b a n A va B n h i i m o t ngudi. Do do so each xep h a n g ciia to de vao 16p t r o n g do h a i b a n A v a B d i l i e n nhau bang so hoan v i cua 9 phan tijf. N2 = 9! = 362880 each. 4. Co bao nhieu each xep 6 ngiicfi ngoi vao m o t ban a n 6 cho t r o n g cac triiofng hcfp sau : a) S^p 6 ngiTofi theo h a n g ngang ciia m o t ban a n d a i . b) S4J) 6 ngiTori ngoi vong quanh m o t b a n a n t r o n . CHI D A N a) M o i each ngoi theo h a n g ngang l a m o t hoan v i cua 6 p h a n tijf. So' each sap xep l a : 6! = 720 each. b) Gia suf 6 ngifofi a n diTOc d a n h so thijf t i f la : 1, 2, 3, 4, 5, 6 v a m o t each sap xep theo b a n t r o n n h i i h i n h . 2 5 1 3 6 4 (1) 5 1 3 6 4 2 (2) 1 3 6 4 2 5 (3) 3 6 4 2 5 1 (4) 6 4 2 5 1 3 (5) 4 2 5 1 3 6 (6) Neu t a eat b a n t r o n a v i t r i giCfa 2 va 4 r o i t r a i d a i theo b a n ngang t h i t a CO hoan v i (1) tUofng ijfng m o t each xep ngiiofi ngoi theo ban a n dai. TiTofng txi c a t of v i t r i giufa 5 va 2. N h u vay m o t each sap xep theo ban t r o n tiiOng ufng vdri 6 each s a p xep theo b a n d a i . Do do so each Hoc va on luyen theo CTDT mon Toan THPT 7 5. x e p 6 ngiTofi n g o i q u a n h b a n a n t r o n l a : N = — = 120 e a c h . 6 M o t t o CO 15 ngifofi g o m 9 n a m v a 6 nOf. C a n l a p n h o m cong t a e eo 4 ngUdri. H o i eo b a o n h i e u e a c h t h a n h l a p n h o m t r o n g m o i trifcfng h o p sau d a y : a) N h o m c6 3 n a m v a 1 nur. b ) So n a m v a nCf t r o n g n h o m b a n g n h a u . c) P h a i CO i t n h a t m o t n a m . CHI DAN 9 8.7 a) So e a c h c h o n 3 n a m t r o n g so 9 n a m l a : Cj! = " = 84 1.2.3 So e a c h c h o n 1 niJ t r o n g so 6 nOr l a : Cg = 6 So each t h a n h l a p n h o m g o m 3 n a m v a 1 nCf (theo quy tSc n h a n ) l a : N i = C^C^ = 5 0 4 e a c h . b ) So e a c h l a p n h o m g o m 2 n a m v a 2 nuf l a : N2= C^C^ = — ' ' 1.2 1.2 = 540 each. c) So e a c h t h a n h l a p n h o m 4 ngu'ofi t r o n g do c6 i t n h a t 1 n a m l a : 1 nam, 3 nO h o a c 2 n a m , 2 nuf hoae 3 n a m , 1 nur h o a c 4 n a m . — Cg.Cg + .Cg 4~ Cg.Cg "I" Cg _ 6.5.4 9.8 6.5. 9.8.7 ^ 9.8.7.6 = 9. + + .6 + = 1350 each. 1.2.3 1.2 1.2 1.2.3 1.2.3.4 Ghi chu : C u n g eo t h e l a p l u a n n h i f sau : C a t o CO 15 ngiTcfi. So e a c h l a p n h o m 4 n g U d i t u y y l a : ^ 4 ^ ^ — 1^4 .x1 3o. 1^2 ^ ^ g g g ^ ^ ^ ^ _ 1 5 . ~ 1.2.3.4 1.2.3.4 So e a c h l a p n h o m 4 ngUofi t o a n nuf l a : C ^ = Cg = 6.5. = 15 e a c h . 1.2 So e a c h l a p n h o m 4 n g i / d i eo i t n h a t 1 n a m l a : N = CJs - C^ = 1365 - 15 = 1 3 5 0 e a c h . T r o n g m a t p h a n g eo n d i e m p h a n b i e t ( n > 3 ) t r o n g do eo d i i n g k 6. d i e m n S m t r e n m o t d i i d n g t h S n g (3 < k < n ) . H o i c6 bao n h i e u t a m g i a c n h a n cac d i e m d a cho l a d i n h . CHIDAN Cuf 3 d i e m k h o n g t h S n g h a n g t a o t h a n h m o t t a m g i a c . So cac t a p h o p c o n 3 d i e m t r o n g n d i e m l a : C^. So cac t a p c o n 3 d i e m t r o n g k d i e m t r e n diTcfng t h i n g l a : C^. So t a m g i a c c6 3 d i n h l a cac d i e m d a cho l a : N = C^ - Cl t a m g i a c . 8 ; TS. Vu The Hi/u - Nguyen VTnh Can 7. a) Co b a o nhieu so t i i n h i e n l a so chan c6 6 chiif so doi m o t khac nhau va chuf so dau t i e n la chOf so le. b) Co bao nhieu so t i i n h i e n c6 6 chuf so doi mot khac nhau, trong do c6 dung 3 chuf so le, 3 chuf so chSn (chuf so dau t i e n phai khac 0). CHI D A N a) So can t i m c6 d a n g : x = a^agaga^agag t r o n g do a i , ae l a y cac chOf so 0, 1, 2, 8, 9 vdfi a i ?i 0, ai aj v d i 1 < i ?i j < 6. - V i X la so chSn nen ae c6 5 each chon tiT cac chuT so 0, 2, 4, 6, 8. - V i a i la chuT so le nen c6 5 each chon tiT cac chuf so 1, 3, 5, 7, 9. Con l a i a2a3a4a5 l a m o t chinh hop chap 4 eiia 8 chuf so con l a i s a u k h i da chon ae va a i . Theo q u y t^c n h a n , so cac so can xac d i n h l a : N i = S.S-Ag = 5.5.8.7.6.5 = 42000 so. b) M o t so theo yeu c a u de b a i gom 3 chuf so tii tap X i = |0; 2; 4; 6; 81 va 3 chuf so tCr t a p hop X2 = I I ; 3; 5; 7; 91 ghep l a i va loai d i cac day 6 chuf so CO chuf so 0 dufng dau. So each lay 3 chuf so thuoc t a p X i la : Ci? = 10 each. So each lay 3 p h a n tuf thuoc X2 l a : Cg = 10 each. So' each ghep 3 p h a n tuf l a y txi X i v o i 3 p h a n tuf l a y tii X2 l a : C^C^ = 10.10 = 100 each. So' day so' eo thuf t i f eiia 6 p h a n tuf diioc ghep l a i l a : 100.6! = 72000 day. Cac day so c6 chuf so 0 a dau g o m 2 chiJ so khac 0 ciia X i va 3 chuf so' ciia X2 : So cac day so nhif t r e n la : C 4 . C 5 . 5 ! = 7200 day. So cac so theo yeu cau de b a i la : N2 - C ^ C ^ 6 ! - C ^ C ^ 5 ! = 72000 - 7200 = 64800 so. 8. M o t hop diing 4 v i e n b i do, 5 v i e n h i t r a n g va 6 v i e n b i vang. NgLfofi ta chon r a 4 v i e n b i t i f hop do. H o i c6 bao nhieu each l a y de t r o n g so b i j a y r a k h o n g dii ca 3 mau. CHI D A N Cdch 1 : So each chon 4 v i e n b i k h o n g d u 3 mau b a n g so' each chon 4 v i e n b a t k i trir d i so each chon 4 v i e n c6 ca 3 mau. N = Cjg - (C^ .C^ .C^ + C^ .C\ + Cl .C\) = 645 each. Cdch 2 : So each chon 4 v i e n b i k h o n g d u 3 m a u bang so each chon 4 v i e n m o t m a u (4 do, 4 t r a n g va 4 vang) cong v d i so each chon 4 v i e n hai mau ( 1 do, 3 t r a n g hoae 2 do, 2 t r a n g hoac 3 do, 1 t r a n g hoae 1 do, 3 v a n g hoac 2 do, 2 v a n g hoae 3 do, 1 v a n g hoae 1 t r S n g , 3 v a n g hoae 2 trSng, 2 v a n g hoac 3 t r a n g , 1 vang). N = c : +Ct +CI+ ClCl + ClCl + C^C^ + ... + C^C^ = 645 caeh. Hoc va on luyen theo C T D T m o n loan T H P T SI 9 9. Co 15 n a m va 15 nuT k h a c h du l i c h dijfng t h a n h vong t r o n quanh ngon lijfa t r a i . H o i c6 bao n h i e u each xep de k h o n g eo triTcfng hop hai ngi/6i eCing gidfi canh nhau. CHI DAN ThiTe h i e n sap xep bang each d a n h so 30 cho t r e n di/orng t r o n tii 1 den 30 va cho n a m dufng so le nuT dufng cho so chSn hoac ngi/gc l a i (2 each). Co 15! each sSp n a m dufng trong cae cho so' le (hoac chSn) va 15! each sdp nuf dufng t r o n g cae cho so ch^n (hoac le). V i diidng t r o n 30 cho nen m o i each sSp xep nao do xoay tua 30 cho theo dung t r a t tiT do ta cung chi eo mot each sap t r e n dirofng t r o n (xem b a i so 4). Do do so each sSp xep theo difcfng t r o n 30 k h a c h du l i c h theo yeu cau 2.(15!)(15!) , , de la : N = = 14!.15! each. 30 10. Chufng m i n h cae dang thufc : a) + + ... + = ——- (1) t r o n g do A^ la c h i n h hop chap 2 eua n. Ag A3 A„ n b) CHI C;; = C;;:; + Cl;}^ + ... + Cl:\) t r o n g do C; la to hop chap r ciia n. DAN a) V(Ji k e N, k > 2 ta c6 : A', = k ( k - 1) ^ = = - ' ^ A^ k(k-l) k-1 k Thay k = 2, 3, n vao (*) ta c6 ve t r a i ciia (1) la : (1 (I V f 1 1^ — — — — — — — +... + [n-1 nj u 2. l 2 3v b) Theo t i n h chat eua to hop ta c6 : = Cn_3 (*) + C^^g Cong ve vdi ve cae dang thufc t r e n ta diTOe : c:;-c::;+c-^3+c::u...+c:-uc: Do C;; = C;::} = l n e n thay C;: d dang thufc cuoi bori C^:} ta dugfc dang thufc (2) can chufng m i n h . 11. Chufng m i n h bat dang thile : t r o n g do k e N, k < 2000, C^ooi + ^ C\Z + CfZ la to hop chap k eua n p h a n tuf. 1 0 t4l TS. Vu The Hi/u - Nguyin Vinh CSn CHI DAN V(Ji 0 < k < 1000 t h i ^2001 k+1 2001 k '^2001 - 2001! (k + l ) ! ( 2 0 0 0 - k ) ! k +1 k!(2001-k)! 2001! 2001-k '-^2001 - '^2001 - , plOOO _ p l O O l •• - '-^2001 ~ ^ 2 0 0 1 pk *-^2001 <1 p k + 1 ^ plOOO ^2001 — ^2001 plOOl ^^2001 -111-k M a t k h a c , v 6 i 1 0 0 0 < k < 2 0 0 0 t h e o t i n h c h a t ciia t o h o p C ' = C;;'^ t a - CIZ'^ • ^ 2 0 0 1 ^ ^-^2001 ~" '^2001 ' "^2001 < Cir, +CIZ\i 0 < 2000 - k < 1000, - '^2001 ' ^2001 0 < 2 0 0 1 - k < 1 0 0 0 t h e o p h a n t r e n d a chufng m i n h . C A C BAI TAP Tl/ GIAI 12. TCr d i e m A d e n d i e m B n g i / d i t a c6 t h e d i q u a C h o a c d i q u a D v a k h o n g CO diTcfng d i t h a n g tii C d e n D . Til A d i t h a n g d e n C c6 2 e a c h , t i r C d i t h a n g d e n B c6 3 e a c h . TCr A d i t h a n g d e n D c6 3 e a c h tix D d i t h a n g d e n B eo 4 e a c h . a) H o i txi A CO b a o n h i e u e a c h d i tdfi B ? b) H o i tCr A d e n B r o i til B trdf v e A A / \ CO b a o n h i e u e a c h ? DS : 3 X^/ a) 18 e a c h b) (18)^ e a c h . 13. 4 D TCr 7 chOf so 0, 1 , 2 , 3, 4, 5, 6 eo t h e g h i dirge b a o n h i e u so tiT n h i e n m o i so' CO 5 chCT so k h a c n h a u tCrng d o i . DS : 2 1 6 0 so. 14. a) b) C h o t a p h o p cac chOf so X = |0; 1 ; 2 ; 3; 4 ; 5; 61. D u n g t a p h o p X eo t h e g h i dufcfe bao n h i e u so tiT n h i e n eo 5 chiT so. D u n g t a p h o p X c6 t h e g h i dirge b a o n h i e u so t i r n h i e n c6 5 chOf so' k h a c n h a u tCrng d o i . c) DCing t a p h g p X c6 t h e g h i dugc b a o n h i e u so tiT n h i e n c6 5 ehCt so k h a c n h a u l a so e h S n . DS 15. : a ) 6.7* so b ) 6 l 5 . 4 . 3 so e) A ^ + 15A^ so. M o t t o h o c s i n h c6 5 n a m , 5 nOf x e p t h a n h m o t h a n g d o c . a) Co b a o n h i e u e a c h x e p k h a c n h a u . b) Co bao n h i e u each x e p h a n g sao cho h a i ngircfi dijfng k e n h a u k h a c g i d i . DS : a ) 1 0 ! e a c h 16. b ) 2(5!)^ e a c h . M o t i g p CO 2 5 n a m h o c s i n h v a 2 0 nuf h o c s i n h . C a n c h o n m o t n h o m c o n g t a c 3 ngiTdi. H o i eo bao n h i e u each c h g n t r o n g m o i t r i r d n g h g p s a u a) B a h o c s i n h b a t k i eua Idp. b) H a i nijf s i n h v a m o t n a m s i n h . Hpc va on luyen theo CTDT mon Toan THPT I J : 1 1 c) B a hoc s i n h c6 i t n h a t m o t nuf. DS 17. : a) C;;^ e a c h b) 25.C^o e a c h c) C'^^ -Cl, each. Co bao n h i e u e a c h p h a n p h o i 7 do v a t cho 3 n g U d i t r o n g cac trUcfng h o p sau : M o i n g u d i i t n h a t m o t do v a t v a k h o n g q u a 3 do v a t . b) M o t ngUofi n h a n 3 do v a t , eon 2 ngUcfi m o i ngUcfi h a i do v a t . a) DS 18. : a) 3.C^C^ e a c h M o t to CO b) S.CtCl + SCl.Cl each. 9 n a m v a 3 nOf. Co bao n h i e u e a c h c h i a t o t h a n h 3 n h o m m o i n h o m 4 ngUofi v a t r o n g b) Co bao n h i e u e a c h c h o n m o t n h o m 4 ngUcfi t r o n g do eo 1 nijf. a) m o i n h o m c6 1 nuf. DS : a) 3.C^ e a c h b) 3.C;;.2C^ = 10080 each. 19. T i m cac so n g u y e n d u o n g x, y t h o a m a n cac d a n g thufe : 6 f)S 20. : X ^ ^ ^ " 5 ^ " 2 ^ • = 8, y = 3. Co bao n h i e u so t U n h i e n chain c6 4 ehuf so d o i m o t k h a c n h a u . DS 21. •.n= Al+ 4.8.8 = 7 6 0 so. C h o d a g i a c d e u 2 n d i n h AiA2...A2n, n > 2 n o i t i e p t r o n g d u d n g t r o n . B i e t r a n g so t a m g i a c c6 d i n h l a 3 t r o n g 2 n d i e m t r e n n h i e u g a p 20 I a n so h i n h ehuf n h a t eo d i n h l a 4 t r o n g 2 n d i n h t r e n . T i m so n . £>S : n = 8. 22. T i m so t U n h i e n n , b i e t r a n g C" + 2C;, + 4 C ' + ... + 2 " C " = 2 4 3 . : n = 5. T r o n g m o t m o n h o c , t h a y g i a o eo 3 0 c a u h o i k h a c n h a u , g o m 5 cau 24. G i a i b a t p h u o n g t r i n h ( v d i h a i a n n , k G N) 23. h o i k h o , 10 c a u h o i t r u n g b i n h v a 15 c a u h o i de. T i r 30 cau h o i do c6 t h e l a p dUcfc bao n h i e u de k i e m t r a g o m 5 cau k h a c n h a u sao cho t r o n g m 5 i de n h a t t h i e t p h a i eo d u b a l o a i cau h o i ( k h o , t r u n g b i n h , de) v a so c a u h o i de k h o n g i t h o n 2. DS:n= 25. Cl,ClCl+C',,C\,Cl+C%C\f = 56785 d l . C h o t a p hcfp A eo n p h a n tuf ( n > 4). B i e t r S n g so t a p h o p eon eo 4 p h a n tuf eua A g a p 2 0 I a n so t a p hofp c o n c6 2 p h a n tuf ciia A . T i m so' t U n h i e n k sao cho so t a p h o p eon eo k p h a n tuf eua A l a I d n n h a t . : n = 18, C^g > C\^' flS o k = 9. 12 ;.'; TS. Vu Th§' Hyu - Nguygn VTnh Can §2. NHI THlfC NIUTCfN K I E N THLTC 1. N h i thufc N i u t c f n (a + b ) " = Cf,a"b° + Cla"-'h + ... + ClJa'^'^b'^ + ... + C > V = Xc;;a"-''b'' k=0 Ydi 2. q u y vide a, b ^ 0, a" = b° = 1 , C° = 1 . Tarn giac P a t c a n Cac h e so' ciia n h i thufc N i u t o r n ufng vdi n = 0, 1 , 2, 3, ... c6 t h e s a p x e p diidfi d a n g t a r n g i a c dtfofi d a y g o i l a t a r n g i a c P a t c a n . 1 n =0 n =1 1 n =2 1 n =3 2 1 n =4 1 3 4 n =6 1 3 1; 4 : 1 6 5 n =5 1 10 10 20 15 6 5 15 1 6 1 T r o n g m 6 i k h u n g t h e h i e n t i n h c h a t t o n g h a i h e so h a n g t r e n so h a n g or h a n g diTdfi h a y C^'^ + Cj^ = bang Cl;^i. BAITAE^ 26. T i m cac so h a n g k h o n g chufa x t r o n g k h a i t r i e n n h i thufc N i u t O n ciia vdfi X > 0. /X J (Trich de TSDH kho'i D nam 2004) CHI D A N Vdfi X > 0, t a 1 CO : \/x = x ^ ; I _ i = x "* —j=r %/x 1 f /X +• /x; = (x'^ + x 7-k _k 7-1 - C ° x 3 + Cix 1 3 X 4 + c?x 7-2 2 ^ X 4 7 + ... + C)x^ x"-* + ... + C l x ' ^ So h a n g k h o n g chufa x l a so h a n g thuT k + 1 t r o n g k h a i t r i e n sao c h o : Hoc va on luyen theo CTOT mon Toan THPT ' 1 3 27. C^'x x"^ =C^x 3 ^ " * =C,'x*' tufc la phai c6 : - - = 0 3k = 4(7 - k) o k = 4 3 4 Vay so hang khong chijfa x trong khai trien la : = 35. Tim so hang chinh giijfa cua nhi thufc NiutOn : (x^ - xy)^*. CHI DAN KJiai trien nhi thufc (x^ - xy)^^ c6 15 so hang, so hang chinh giiJa la so hang thuf 8 c6 dang : C L ( x ^ r " ( - x y ) ^ = -CLx^^xV^ - -3432x^V^ 28. Tim so hang thuf tii cua khai trien nhi thufc a b- a + b2 ' - „a2 A" . Biet a rang he so ciia so' hang thuf ba cua khai trien do bSng 21. CHI DAN Trong cong thufc nhi thufc NiutOn (A + B)" so hang thuf 3 ciia khai trien c6 he so la : C? = 21 o Vay so hang thuf tii cua khai trien 29. ~'^^ = 21 o n = 7 a b^-a'^^' la : b-a • + 7-3 ^b^-aM a(b + a) C? = 35 [b-aj b-a ^ a J Biet rSng tong tat ca cac he so cua khai trien nhi thufc (x^ + 1)" bang 1024. Hay t i m he so ciia so hang chufa x^^ trong khai trien do. CHIDAN (1 + x'r = ci +c\x' +cix' + ... + c y +... + c:y" Cho X = 1 ta dirge : (1 + 1)" = + c;, + Cf, + ... + + ... + C;; = 1024 = 2" = 2'*^ ^ n = 10 Do do he so cua x'^ la : = 6!4! = 210. / 30. 2 Trong khai trien nhi thuTc NiutOn ' nx 14 chufa x\t rSng 5C;;-' = C'l -I 1 ^ , X ^ 0, hay t i m so hang x^ (Trich de TSDH khoi A - 2012) CHI DAN bCr 14 = Ct n(n - l)(n - 2) 5n = 1.2.3 n(n'^ - 3n - 28) = 0 n = 7 TS. Vu The' Hifu - Nguygn VTnh Can Thay n = 7 vao nhi thuTc Niutofn da cho t h i c6 : 1 -c* 12 .2, I-] X X 2 + ... + V f-1 So hang chufa x^ trong khai trien la so' hang thuf k + 1 sao cho : 7-k 12; k .Xy X^ 27-k ^ 2 ( 7 - k ) - k = 5=^k = 3 Vay so hang chufa x la : -C^ 31. CHI 1 7.6.5 1 ^5^_35^3 1.2.3 2' 16 Tim he so cua so' hang chufa x^° trong khai trien nhi thufc NiutOn ciia (2 + x)", biet rang 3"C° - 3""'C;, + 3"'C^ - 3"-''Cl + ... + (-1)"C,'; = 2048. (Trich de TSDH khoi B - 2007) vXy D A N Xet khai trien nhi thiifc Niutcfri : (x - D" = c>" - c^x"-' + c'^x"-' - cf,x"-^ +... + (-1)"c;; Cho X = 3 ta diroc : 2" = 3"c;; -3"-'c;, +3""'cf, -3"-'c^ + ... + (-i)"c;; = 2048 2" = 2048 = 2" => n = 11 Thay n = 11 vao khai trien (2 + x)" ta diioc : (2 + x ) " = 2"c?, + 2^°c;jx +... + 2c;?x^°'+ c;;x" 32. CHI (*) He so cua x^° trong khai trien (*) la : a^o = 2C\\ 22. Khai trien bieu thufc P(x) = x ( l - 2 x f + x^(l + Sx)^** va viet P(x) diTdi dang da thufc vdri luy thifa tang cua x. Hay t i m he so ciia x'' ciia da thufc do. D A N Ta CO : x ( l - 2xy^ = x(C° - 2C^x + 2'C5'x' - 2''C^x' + 2'C5'x' - 2'C^x'^) x ' ( l + Sx)"" = x^(C°o + 3Cj„x + 3'Cfnx' + 3'C;'nx' + + 3*C,'x" +3^C?„x^+... + 3 " ' C ; V ° ) =^ P(x) = C;|x + (C?o - 2C;)x' + ... + (3''C-^„ + 2''C^)x^ +... + 3^"c;°x Vay he so ciia so hang chufa x'' la : as = 1 f) q Q + 2'Ct = 2 7 . ^ ^ ^ ^ + 16.5 = 3320. ' 1.2.3 Hqc va on luyen theo CTDT mon Toan THPT .'' 1 5 33. CAC BAITAP lij GIAI T i m so h a n g k h o n g chijfa x cQa k h a i t r i e n n h i thijfc N i u t o n . if X + — X . j DS : 924. K h a i t r i e n va r u t gon P(x) = (x + 1)^ + (x - 2f t h a n h da thufc v d i luy 35. K h a i t r i e n va r u t gon bieu thufc : P(x) = ( 1 + x f + (1 + x)^ + (1 + xf + (1 + x)^ + (1 + x)^" ta dirgfc : P(x) = aiox^° + agx® + asx** + ... + aix + ao T i n h ag. £>S : a« = 55. 34. thtra giam dan ciia x. T i m he so cua cac so hang chufa x^ va x^. DS : He so ciia x^ la : - 6 2 2 , ciia x^ la : 570. 36. Chufng m i n h vdfi n nguyen diiOng t a c6 : a) cL+cL+... + CL=cL+cL+... + c r . b) c;, + 2Ci + 3Ci +... + nc;; - n2"-'. CHI DAN a) K h a i t r i e n P(x) = (x - 1)^" r o i cho x = 1. b) K h a i t r i e n P(x) = ( 1 + x)". T i m P'(x) r o i t i n h P ' ( l ) . T r o n g k h a i t r i e n n h i thufc 38. Viet k h a i t r i e n Niutcfn, bieu thufc (3x - 1)^'', tU do chufng m i n h rSng : 37. x - Xem thêm -