Tài liệu Onluyentheocautrucdethimontoan phan01

510.76 6-454L d TS. VO THE HlTU - NGUYEN VINH CAN TIICO C n U TRUC DC Till MON TS. VU THE HlTU - NGUYEN VINH CAN ON LUVEN m THEO cAu TDUC DEI THI M6]\N ON THI TOTNGHIEP THPT QUOC GIA (2 trong 1) lUHA XUAT BAN DAI HOC QUOC GIA HA NOI GIJII TICH Hoc va On luy§n theo CTDT mfln Toan THPT E I 3 IK tX yYENflEl.fllllSOTOHIfPVflXIJCSU'' §1. HOAN VI, CHINH HdP, TO HOP KIEN THLTC 1. Quy tac C Q n g va quy t ^ c n h a n a) Quy tdc cong : Neu tap hc(p A c6 m phan tuT, tap hop B c6 n phan tuf va giufa A va B khong CO phan tuf chung t h i c6 m + n each chon mot phan tuf thuoc A hoac thuoc B . b) Quy tdc nhdn : De hoan thanh mot cong viec A phai thuc hien hai cong doan. Cong doan I CO m each thifc hien, cong doan I I c6 n each thiTc hien t h i c6 m . n each de hoan thanh cong viec A . Tong quat, de hoan thanh cong viec A phai qua k cong doan. Cong doan thuf i (1 < i < k) c6 m i each t h i t h i c6 m i . m 2 . . . m k each de hoan thanh cong viec A . 2. Hoan vi Mot tap hop A hufu han c6 n phan tuf (n > 1). Moi each sap thuf tii cac phan tuf cua tap hop A dage goi la mot hoan vi cua n phan tuf cua A . Dinh li : So hoan v i khac nhau cua n phan tuf bang : P„ = n(n - l)(n - 2)...2.1 = n! 3. C h i n h hofp Mot tap hop A hufu han gom n phan tijf (n > 1) va so nguyen k (0 < k < n). Moi tap hop con cua A gom k phan tuf sap theo mot thuf t i i nhat dinh dLfOe goi la mot chinh hop chap k cua n phan tuf. Dinh li : So chinh hop chap k cua n phan tuf bang : A;; = n ( n - l ) ( n - 2 ) . . . ( n - k + l ) = - 4. ^, (n-k)! (Quy \x6c : 0! = 1). Tohtfp Cho tap hop A hufu han c6 n phan tuf (n > 1) va so' nguyen k (0 < k < n). Moi tap hop eon gom k phan tuf eiia A (khong tinh thuf tiJ cac phan tuf) goi la mot to hap chap k cua n phan tuf. n! - A'' Binh li : So to hop chap k cua n phan tuf la : Cj^ =( n - k ) ! k ! k ! Hequd: Cl^Cl=l; ik-i C); = Cr" ; C);,, = C^ + C^ HQC fln luy?n theo CTDT mOn ToSn THPT 0 5 BAI TAP 1. Cho cac chff so 2, 3, 4, 5, 6, 7. a) Co bao n h i e u so tir n h i e n c6 h a i chiJ so dirge tao n e n tCr t a p hgp eae chCr so da eho. b) Co bao n h i e u so tir n h i e n c6 h a i ehuf so khac nhau dugc tao n e n ttf tap hgj) cac chiJ so da cho. CHI DAN a) 1. 2. b) 1. 2. De tao m o t so c6 h a i chuf so t a thiTe h i e n h a i eong doan : Chgn m o t chuf so l a m ehuf so h a n g ehuc : eo 6 k e t qua c6 the. Chgn m o t chuf so l a m ehuf so h a n g don v i : eo 6 k e t qua c6 the. Theo quy t&c n h a n so' k e t qua tao t h a n h cac so' c6 h a i ehuf so tCr tap hgp 6 chuf so da eho la : n = 6 x 6 = 36 so. L a p luan giong n h u eau a) n h i m g liTu y sir khac biet so vdi trirdng hgp t r e n d cho so dirge tao t h a n h eo h a i chCT so khac nhau. Do do t a c6 k e t qua nhif sau : Chgn m o t chuf so l a m ehOf so h a n g ehuc : eo 6 k e t qua eo t h e . Chgn m o t ehuf so l a m chuf so hang don v i : eo 5 k e t qua eo the (vi chQ so nay p h a i khac chij; so h a n g ehue da ehgn triTdrc do). Theo quy tie n h a n : so eae so eo h a i chuf so khac nhau dirge tao t h a n h tCr t a p hgp 6 eha so da cho la : n' = 6 x 5 = 3 0 so. Cdch khac : M 6 i so eo h a i ehiJ so tao t h a n h tCr 6 chuf so da eho la mot tap hgp con sap thuf t i i gom h a i p h a n tuf tu" 6 p h a n tuf da cho. Do do so cac so C O h a i chuf so khac nhau tao t h a n h tCr 6 ehuf so da eho la so e h i n h hgp chap 2 cua tap hgp 6 p h a n tuf. n = = 6.5 = 3 0 so'. 2. Cho t a p hgp cac chuf so 0, 1, 2, 3, 4, 5, 6. a) Co bao nhieu so tir n h i e n c6 4 chuf so tir tap hgp cac ehuf so da eho. b) Co bao n h i e u so tiT n h i e n eo 4 ehuf so khac nhau tiTng doi tiT tap hgp eae chuf so da eho. C H I DAN a) Co 6 each ehgn chuf so hang n g h i n (chuf so dau t i e n phai khac 0), 7 each ehgn chiT so' h a n g t r a m , 7 each chgn chuf so' h a n g ehuc va 7 each ehgn ehuf so h a n g don v i . Theo quy t^c n h a n : so each tao t h a n h so tiT n h i e n 4 chuf so tCr t a p hgp 7 chij" so da eho l a : N = 6 x 7 x 7 x 7 = 2058 so. b) Co 6 each chgn chuf so' h a n g n g h i n , k h i ehgn xong ehC? so' h a n g n g h i n con l a i 6 chOr so khae v d i ehuf so h a n g n g h i n da chgn. Vay c6 6 each chgn ehuf so h a n g t r a m . K h i da chgn chiT so h a n g n g h i n va hang t r a m , eon l a i 5 ehu: so khac v d i cac chuT so da chgn. Do do eo 5 each 6 a TS. Vu Thg' Huu - Nguyen Wnh C?n chon chOf so h a n g chuc. Ttfang t i i , c6 4 each chon chOf so h a n g dcfn v i . Theo quy tac n h a n . So cac so t u n h i e n c6 4 chijr so khac nhau tCrng doi ducfc tao t h a n h til t a p hop 7 chOf so da cho l a : N ' = 6 X 6 X 5 X 4 = 7 2 0 so. Cdch lap luan khdc : M o i so t i i n h i e n c6 4 chOr so khac nhau tao t h a n h tCr t a p hop 7 chCf so da cho l a m o t c h i n h h o p chap 4 tCr t a p h o p 7 chuf so ma cac c h i n h hop n a y k h o n g c6 chOf so' 0 o dau. Do do so' cac r so C O 4 chOr so khac nhau til 7 chuf so l a : N' = = 7 X 6 X 5 X 4 - 6 X 5 X 4 = 7 2 0 so. 3. Mot to hoc sinh c6 10 ngufofi xep thCf td thanh hang 1 de vao Icfp. H o i a) Co bao n h i e u each de to xep h a n g vao Idfp. b) Co bao nhieu each de to xep h a n g vao Idp sao eho h a i b a n A va B cua to luon d i canh nhau va A dufng triTdfc B. CHI DAN a) So' each xep h a n g bang so' hoan v i cua 10 p h a n tuf. N i = 10! = 3628800 each. b) Coi h a i b a n A va B n h i i m o t ngifdi. Do do so each xep h a n g cua t o de vao Idp t r o n g do h a i b a n A va B d i l i e n n h a u bang so h o a n v i cua 9 p h a n tuf. N2 = 9! = 362880 each. 4. Co bao nhieu each xep 6 ngUc/i n g o i vao m o t b a n a n 6 cho t r o n g cac trirdng hop sau : a) Sip 6 ngiidi theo h a n g ngang cua m o t b a n a n d a i . b) Sip 6 ngLfcfi ngoi vong quanh m o t b a n a n t r o n . CHI DAN a) M o i each ngoi theo h a n g ngang l a m o t hoan v i cua 6 p h a n tuf. So each sap xep la : 6! = 720 each. b) Gia sijf 6 ngiicfi a n dirge d a n h so thuf t u l a : 1, 2, 3, 4, 5, 6 v a m o t each sap xep theo b a n t r o n nhiT h i n h . 2 /^^^T^K^ ^ 5 1 3 6 4 5 1 3 6 4 2 (2) ^\ 3 6 4 2 5 J. j (1) (3) Neu thi dai. ban (4) 6 4 2 5 1 3 (5) 4 \ ^ _ 3 _ ^ 3 6 4 2 5 1 (6) 2 5 1 3 6 t a cat b a n t r o n d v i t r i giufa 2 v a 4 r o i t r a i d a i theo b a n ngang t a C O hoan v i (1) tiTcfng lirng m o t each xep ngudi n g o i theo b a n a n Tirong t i i cat d v i t r i giufa 5 va 2. N h i i vay m o t each sSp xep theo t r o n tifOng ufng v d i 6 each sap xep theo b a n d a i . Do do so each Hpc v4 On luy$n theo CTDT mfln Toan THPT £3 7 xep 6 ngiicri ngoi quanh b a n a n t r o n l a : N = — = 120 each. 6 5. M o t to CO 15 ngiJcfi gom 9 n a m va 6 niJ. C a n lap n h o m cong tac c6 4 ngiTcfi. H o i c6 bao n h i e u each t h a n h l a p n h o m t r o n g m o i triTofng hop sau day : a) N h o m c6 3 n a m va 1 nvC. b) So n a m va nOf t r o n g n h o m bSng nhau. c) P h a i CO i t n h a t m o t n a m . CHI D A N 9 87 a) So each chon 3 n a m t r o n g so 9 n a m l a : Cg = " = 84 1.2.3 So each chon 1 nuf t r o n g so 6 nuf la : Cg = 6 So each thanh lap nhom gom 3 nam va 1 nuf (theo quy tac nhan) la : Ni = C^C^ = 504 each. b) So each l a p n h o m gom 2 n a m va 2 nuf la : N 2 = C ^ C ^ = — — = 540 each. ' ' 1.2 1.2 c) So each t h a n h l a p n h o m 4 ngudi t r o n g do c6 i t n h a t 1 n a m la : 1 n a m , 3 nuf hoSe 2 n a m , 2 nOr hoSc 3 n a m , 1 nuf hoSc 4 n a m . N = C^.C^+C^C^+C^C^+C^ ^ 6.5.4 9.8 6.5. 9.8.7 ^ 9.8.7.6 = 9. + + .6 + = 1350 each. 1.2.3 1.2 1.2 1.2.3 1.2.3.4 Ghi chu : Cung c6 the l a p l u a n nhiJ sau : Ca to CO 15 ngUdi. So each l a p n h o m 4 ngUcJi t u y y l a : ^4 15.14.13.12 , , Ci5 = — — = 1365 each. 1.2.3.4 6 5 So each l a p n h o m 4 ngudi t o a n nuf la : Cg = Cg = — ^ = 15 each. So each l a p n h o m 4 ngiTdi c6 i t n h a t 1 n a m la : N = C^5 - C^ = 1365 - 15 = 1350 each. 6. T r o n g mSt p h a n g c6 n d i e m p h a n b i e t ( n > 3) t r o n g do c6 dung k d i e m n k m t r e n m o t d i T d n g t h a n g (3 < k < n ) . H o i c6 bao n h i e u t a m giae n h a n eac d i e m da cho la d i n h . CHI D A N C i i 3 d i e m k h o n g t h a n g h a n g tao t h a n h m o t t a m giae. So eac t a p hdp con 3 d i e m t r o n g n d i e m la : C^. So eac t a p con 3 d i e m t r o n g k diem t r e n dirdng t h a n g la : C^. So t a m giae c6 3 d i n h la eac d i e m da cho l a : N = C ^ - C ^ t a m giae. 8 £ 3 TS. Vu ThS' Hi/u - Nguyen V i n h C?n 7. a) Co bao n h i e u so tu n h i e n l a so chSn c6 6 chOr so doi m o t khac nhau v a chiJ so dau t i e n l a chui so le. b) Co bao nhieu so tuT n h i e n c6 6 chuf so doi m o t khac nhau, trong do c6 dung 3 chCif so le, 3 chuf so chSn (chuf so dau t i e n phai khac 0). CHI D A N a) So can t i m c6 dang : so 0, 1, 2, 8, 9 vdi x = ajagaga^Hgag aj ;t 0, ai aj vdi t r o n g do 1< i ai, ae l a y cac chijr ;t j < 6. - V i X la so chSn nen ag c6 5 each chon tii cac chuf so 0, 2, 4, 6, 8. - V i a i la chuf so le nen c6 5 each chon tif cac chuf so 1, 3, 5, 7, 9. Con l a i a 2 a 3 a 4 a 5 l a m o t c h i n h hop chap 4 cua 8 chuf so con l a i sau k h i da chon ae va a i . Theo quy tSc n h a n , so cac so can xac d i n h l a : N i = 5.5.Ag = 5.5.8.7.6.5 = 42000 so. b) M o t so theo yeu cau de b a i gom 3 chuf so tCr t a p X i = {0; 2; 4; 6; 81 va 3 chuf so tii t a p hop X2 = { 1 ; 3; 5; 7; 9) ghep l a i va l o a i d i cac day 6 chuf so CO chuf so 0 dufng dau. So each l a y 3 chuf so thuoc t a p X i l a : C 5 = 10 each. So each l a y 3 p h a n tijf thuoc X2 l a : C 5 = 10 each. So each ghep 3 p h a n tuf l a y tii X i v d i 3 p h a n tuf l a y tii X2 l a : C^.C^ = 10.10 = 100 each. So day so eo thuf tir cua 6 p h a n tuf diicfc ghep l a i l a : 100.6! = 72000 day. Cac day so c6 chuf so 0 of dau gom 2 chuf so khac 0 cua X i v a 3 chijf so cua X2 : So cac day so n h u t r e n l a : C ' . C g . S ! = 7200 day. So cac so theo yeu cau de b a i l a : N2 = C ^ . C ^ . 6 ! - C ^ C ^ . 5 ! = 72000 - 7200 = 64800 so. 8. M o t hop diing 4 v i e n b i do, 5 v i e n b i t r S n g va 6 v i e n b i v a n g . NgUcfi ta chon r a 4 v i e n b i tii hop do. H o i c6 bao n h i e u each l a y de t r o n g so b i l a y r a k h o n g du ca 3 m a u . CHI D A N Cdch 1 : So each chon 4 v i e n b i k h o n g du 3 m a u b a n g so each chon 4 v i e n b a t k i trCr d i so each chon 4 v i e n eo ca 3 m a u . N = C\, -{ClC\.C\+Cl.C\.C\+Cl.C\.C\) = 645 each. Cdch 2 : So each chon 4 v i e n b i k h o n g du 3 m a u b a n g so each chon 4 v i e n m o t m a u (4 do, 4 t r a n g va 4 vang) eong v d i so each chon 4 v i e n h a i m a u (1 do, 3 t r a n g hoac 2 do, 2 t r a n g hoac 3 do, 1 t r S n g hoac 1 do, 3 v a n g hoSe 2 do, 2 v a n g hoac 3 do, 1 v a n g hoSc 1 t r a n g , 3 v a n g hoac 2 t r a n g , 2 v a n g hoac 3 t r a n g , 1 vang). N=C^ + + + C^C^ + ClCl + C^C^ +... + C^C^ = 645 each. Hoc va On luyen theo CTBT mfln ToSn THPT S 9 DAN CHI Co 15 n a m va 15 nuf khach du l i c h duTng t h a n h vong t r o n quanh ngon lufa t r a i . H o i c6 bao nhieu each xep de k h o n g c6 triTdng hop h a i ngiTdi cung g i d i canh nhau. 9. ThiTc h i e n sAp xep b k n g each danh so" 30 cho t r e n diTdng t r o n tii 1 den 30 va cho n a m duTng so le nuf dufng cho so chkn hoSc ngiioc l a i (2 each). Co 15! each sSp n a m dufng t r o n g eae ch6 so le (hoac ehSn) va 15! each sap nuf dufng t r o n g cac eho so ehSn (hoftc le). V i dudng t r o n 30 cho n e n m o i each sap xep nao do xoay tua 30 eho theo diing t r a t tiT do t a cung chi c6 m o t each sap t r e n ducfng t r o n (xem b a i so 4). Do do so each sSp xep theo diiofng t r o n 30 k h a c h du l i c h theo yeu cau „ 2.(15!)(15!) ^ ., . , de la : N = = 14!.15! each. 30 10. Chufng m i n h cac dSng thuTc : 1 1 = ——- (1) t r o n g do A^ la c h i n h hop chap 2 cua n . a) -ir + —7r + -. + A„ n b) c;; = c;;:'i + c;;:'^ +... + c;::; (2) trong do c;; la to hop chap r cua n . CHI DAN a) V d i k e N , k > 2 t a eo : A^ = k ( k - 1) J_ k(k -1) k-1 - ~ (*) T h a y k = 2, 3, n vao (*) ta c6 ve t r a i cua (1) l a : (1 1^ (1 1^ 1 _ n-1 1^ _^ + — +... + ( 1 — n n 2j ^n-l nj U 3j b) Theo t i n h chat cua to hop ta c6 : u lr-l Cn-2 = Cn.3 + C;;_3 Cong ve v d i ve eae dang thufc t r e n t a dirgfc : c : ; = c - + C - U C - 3 + ... + C - + C : Do C[ = C[:\ 1 n e n thay C; d dSng thufc euoi boi C^:} t a dUdc dang thiJc (2) can chufng m i n h . 11. Chufng m i n h b a t dSng thufc : C^QOI + C'OM ^ C^^i + t r o n g do k e N , k < 2000, ^Zl la to hop chap k cua n phan tuf. £ 2 TS. Vu ThS' Huu - Nguyen Vinh C$n 10 CHI DAN V d i 0 < k < 1000 t h i CLi C ^ 2001! k!(2001-k)! pk ^ pk+1 ^ pk+2 '-^2001 - '-^2001 - ^ 2 0 0 1 (k + l ) ! ( 2 0 0 0 - k ) ! 2001! ^ ^ plOOO _ p l O O l - ••• - '-'2001 " '-^2001 k+ 1 2001-k ^ p k *-^2001 <1 p k +1 ^2001 pi ^ p 11000 4- -ilOOl i ^2001 '-^2001 - ^ 2 M a t khac, v d i 1000 < k < 2000 theo t i n h chat cua to hop c6 : C^oo, 4- C^-. = + C--- = C"""^ t a < C - ° + C^ol v i 0 < 2000 - k < 1000, 0 < 2001 - k < 1000 theo p h a n t r e n da ehufng m i n h . OAQJiMJiMUmk 12. TCr diem A den d i e m B ngtrdi t a c6 the d i qua C hoac d i qua D v a k h o n g C O difdng di t h a n g tCr C den D. TCr A d i t h S n g den C c6 2 each, tCr C d i t h a n g den B c6 3 each. TCr A d i t h ^ n g den D c6 3 each tCr D d i t h a n g den B c6 4 each. a) H o i tCr A c6 bao n h i e u each di t d i B ? 2 b) H o i tCr A den B r o i tCr B t r d ve A A / \ C O bao n h i e u each ? DS : a) 18 each 3 \ ^ / 4 b) (18)2 13. D TCr 7 chOf so 0, 1, 2, 3, 4, 5, 6 c6 the ghi dirge bao n h i e u so tu n h i e n m o i so' C O 5 chuT so' khac nhau tCrng doi. £)S : 2160 so. Cho tap hop cac chuf so X = {0; 1; 2; 3; 4; 5; 6}. a) Dung tap hop X c6 the ghi dufOc bao nhieu so tU n h i e n c6 5 chuf so. b) Dung tap hop X c6 the ghi dirge bao n h i e u so tU n h i e n c6 5 chuf so khac nhau tCrng doi. c) D u n g tap hgp X c6 the g h i difgc bao n h i e u so t i f n h i e n c6 5 chur so khac nhau la so chSn. c) A^ + 15AI so. b) 6 l 5 . 4 . 3 so BS : a) 6.7^ so 14. 15. M o t to hoc s i n h c6 5 n a m , 5 nuf xep t h a n h m o t h a n g doc. a) Co bao n h i e u each xep khac nhau. b) Co bao nhieu each xep hang sao cho hai ngircri dufng ke nhau khac gidi. DS : a) 10! each b) 2(5!)' each. 16. M o t Idp C O 25 n a m hoc s i n h va 20 nff hoc s i n h . C a n chon m o t n h o m cong tac 3 ngiTdi. H o i c6 bao nhieu each chon t r o n g m o i triTdng hop sau a) Ba hoc sinh bat k i cua Idp. b) H a i nu" sinh va m o t n a m s i n h . HQC 6n luy$n theo CTDT mOn Toan THPT S 11 c) Ba hoc s i n h c6 i t n h a t m o t nuf. DS : a) C% each b) 25.C^o each c) C^^ - C^^ each. 17. Co bao n h i e u each p h a n phoi 7 do v a t cho 3 ngirdi t r o n g cac t r u 6 n g hop sau : a) M o t ngLfcfi n h a n 3 do v a t , con 2 ngiicfi m o i nguf6i h a i do v a t . b) M o i ngirdi i t n h a t m o t do v a t va k h o n g qua 3 do v a t . DS : a) S^.Cl each b) 3.C^C,' + 3C^.C^each. 18. M o t to CO 9 n a m va 3 nOr. a) Co bao nhieu each chon m o t n h o m 4 ngUofi t r o n g do c6 1 nuf. b) Co bao nhieu each chia to t h a n h 3 n h o m m o i n h o m 4 ngiroti va trong m o i n h o m c6 1 nuf. DS : a) 3.C^ each b) 3.C^.2C^ = 10080 each. 19. T i m cac so nguyen dirong x, y thoa m a n eac dang thufc : DS : 20. X = 8, y = 3. Co bao n h i e u so t i i n h i e n ehSn c6 4 chuf so doi m o t khac nhau. DS 21. -.11= Al+ 4.8.8 = 760 so. Cho da giac deu 2 n d i n h A i A 2 . . . A 2 n , n > 2 n o i t i e p t r o n g difc/ng t r o n . B i e t r a n g so t a m giac c6 d i n h l a 3 t r o n g 2 n d i e m t r e n nhieu gap 20 I a n so h i n h chuf n h a t c6 d i n h la 4 t r o n g 2 n d i n h t r e n . T i m so n . DS :n = 8. 22. T i m so t\i n h i e n n , b i e t r a n g C° + 2C;, + 4C^ + ... + 2"C;; = 243. 5 S : n = 5. 23. Giai bat phiTdng t r i n h (vdi h a i a n n , k ^"^^ 24. G N) < eoA"::' (n-k)! T r o n g m o t m o n hoc, t h a y giao c6 30 eau h o i khac nhau, gom 5 cau h o i k h o , 10 cau h o i t r u n g b i n h va 15 eau h o i de. Tii 30 cau h o i do eo the l a p difcfe bao nhieu de k i e m t r a gom 5 cau khac nhau sao cho t r o n g m o i de n h a t t h i e t p h a i eo du ba loai cau h o i (kh o, t r u n g b i n h , d i ) va so cau h o i de k h o n g i t h o n 2. DS:n^ 25. + C?,C;„C^ + C%C\f = 56785 d l ClClCl Cho t a p h d p A eo n p h a n tijf ( n > 4). B i e t r k n g so t a p hdp con c6 4 p h a n tuf cua A gap 20 I a n so t a p h a p con c6 2 p h a n tuf cua A . T i m so tiT n h i e n k sao cho so t a p hap con eo k p h a n tuf cua A l a lorn n h a t . £)S : n = 18, Cf^ > C\^^ o k = 9. 12 El TS. VO Thg' Hi;u - NguySn V n C?n Th §2. N H I T H l f C NIUTOfN KIEN T H l f C i 1. N h i thuTc N i u t t f n (a + b ) " = C ° a " b ° + C\a"-'h + ... + C;:a"-''b'' + ... + C > " b " = ^ C ' a " - ' ' b ' ' V d i q u y Lfdc a, b ^ 0, a° = b° = 1 , 2. = 1. Tarn giac P a t c a n Cac h e so c u a n h i thufc N i u t o n ufng v d i n = 0, 1 , 2 , 3, ... c6 t h e sSp x e p dudi d a n g t a r n g i a c d u d i d a y g o i l a tarn gidc 1 n = 0 1 n =1 1 2 1 n = 2 3 1 n = 3 n = Patcan. 4 1 n = 6 10; 15 6 l i 1 4 ; 6 |5 1 n = 5 ;3 4 1 1 10 120 1 5 6 15 T r o n g m o i k h u n g t h e h i e n t i n h c h a t t o n g h a i h e so h a n g t r e n b a n g so h a n g d h a n g d u d i h a y C^-' +0^= iBAIIAP 26. C^,. A T i m cac so h a n g k h o n g chufa x t r o n g k h a i t r i e n n h i thufc N i u t o n c i i a \ vdi (Trich X > de tuyen 0. sink DH kiwi D nam 2004) CHI D A N Vdi 1 -0, t a C O : ^/x = x^ ; - p . = x X > 7-1 ^ 7-k + ... + 7-2 2 = ( x 3 + x " M ' = C ? X 3 + C ^ X 3 x"^ + C , ' X 3 X ' + - ± = C^X 3 k x " * + ... + C?X 4 So h a n g k h o n g c h i i a x l a so h a n g thuf k + 1 t r o n g k h a i t r i e n sao c h o : ' Hoc va 6n luygn theo CTDT m6n Toan THPT £3 13 C^X x"" = 3 k 7-k C^X 7-k k 3 "4 tufc la p h a i c6 : = C^x" - ^ = 0 3k = 4(7 - k ) o Vay so h a n g k h o n g chura x t r o n g k h a i t r i e n la : k = 4. =35. D A N CHI T i m so h a n g c h i n h giijfa cua n h i thufc Niutcfn : (x^ - xy)^". 27. K h a i t r i e n n h i thufc (x^ - xy)^'* c6 15 so hang, so h a n g c h i n h giufa la so h a n g thuf 8 c6 dang : Cl,(x^)"-^(-xy)^ = -Clx'\xy 28. = -3432x^V^ T i m so h a n g thuf tir ciaa k h a i t r i e n n h i thiJc a b-a + b^-a^ . Biet r a n g he so cua so h a n g t h i i ba cua k h a i t r i e n do b k n g 2 1 . CHI DAN T r o n g cong thufc n h i thufc N i u t o n (A + B)° so h a n g thuf 3 cua k h a i t r i e n c6 he so la : C!; = 21 <^ n ( n _ - l ) ^ V a y so h a n g thuf tiX cua k h a i t r i e n - 35 b-a n = 7 b^-a^^' h-a +• la : a(b + a) b-a 29. B i e t rSng tong t a t ca cac he so cua k h a i t r i e n n h i thufc (x^ + 1)" bkng 1024. Hay t i m he so cua so hang chufa x^^ t r o n g k h a i t r i e n do. CHI DAN (1 + x^)" = C° + C^x^ + C^x* + ... + C^x^^ + ... + C > ^ " Cho X = 1 t a dircfc : (1 + 1)" = C° + C", + C' + ... + C^ + ... + C" n n = 1024 = 2" = 2^° ^ n = 10 10! Do do he so cua x'^ la : Cf„ = = 210. '° 6!4! 30. ^nx^ T r o n g k h a i t r i e n n h i thufc N i u t o n 1 , X ^ 0, hay t i m so' hang 14 chufa x ^ b i e t rSng SC;;-^ = (Trich de tuyen sink DH khoi A CHI DAN n ( n - l ) ( n - 2) 5Cr' = Cl o 5n = 1.2.3 <:> n(n^ - 3n - 28) = 0 <=> n = 7 2012) 14 rS TS. Vu The' Huu - Nguy?n Vinh CJn Thay n = 7 vao nhi thufc Niuton da cho t h i c6 : (,,1 2 \ 1 X v2. + ... + v2. v2. + . . . - C7 1 So hang chufa x^ trong khai trien la so hang thuf k + 1 sao cho : f 2 S}-^ r 1 A'' ^ 2(7 - k) - k = 5 =^ k = 3. X 27-k v2. 4 : -C^ rx^^ r r 5 l i . r<3 Vay so hang chufa x la 31. 3 7.6.5 1.2.3 1 2' 16 Tim he so' cua so' hang chufa x^° trong khai trien nhi thufc Niuton cua (2 + xf, biet rang 3"c° - 3"-'c;, + 3"-'c^ - 3"-'c^ +... + (-irc;; = 2048. (Trich de tuyen sink DH khdi B - 2007) CHI D A N Xet khai trien nhi thufc Niuton : (x - ir = c°x" - c;,x"-* + c^x"-' - c^x"-3 +... + (-D^C;; Cho X = 3 ta di/oc : 2" = 3"C"„ - 3"-^C;, + 3"-'C^ - 3"-'C^ + ... + (-1)"C;; = 2048 2" = 2048 = 2 " n = 11 Thay n = 11 vao khai trien (2 + x)" ta diTgfc : (2 + x)^^ = 2''c'i, + 2'°cjix+... + 2c;;x'°'+ c;;x" (*) He so cua x^° trong khai trien (*) la : a^^ = 2CJi = 22. 32. Khai trien bieu thufc P(x) = x(l - 2x)^ + x^(l + 3x)^° va viet P(x) dudi dang da thufc v6i luy thtra tang cua x. Hay tim he so cua x^ cua da thufc do. CHI D A N Ta CO : x(l - 2x)' = x(C° - 2C^x + 2'C5'x' - 2'C^x^ + 2*C^x'' - 2'C^x^) x^Cl + 3x)^° = x^(C?o + 3C;oX + 3^C?oX^ + 3^C?oX^ + + 3'C^oX^ + 3'C%x' + ... + 3^''C;°x^°) ^ P ( x ) = C°x + ( C ° o - 2 C ^ ) x 2 + . . . + (3^C?o+2^C5*)x^+... + 3^'^C;°x^^ Vay he so cua so hang chufa x^ la : as = 3'C^„ + 2^Ct = 2 7 . ^ ^ ^ + 16.5 = 3320. 1.2.3 HQC V4 On luygn theo CTDT mOn Toan THPT E3 1 5 I 33. T i m so h a n g k h o n g chiJa x cua k h a i t r i e n n h i thufc N i u t o n . 1 X + — 34. x_ DS : 924. K h a i t r i e n va r i i t gon bieu thufc : P(x) = (1 + x)^ + (1 + x)^ + (1 + x)^ + (1 + x)'' + (1 + x)^" ta diioc : P(x) = aiox^° + agx^ + agx^ + ... + aix + ao T i n h as. DS : as = 55. 35. K h a i t r i e n va r u t gon P(x) = t x + if + (x - 2)^ t h a n h da thufc v d i luy thiia giam dan cua x. T i m he so cua cac so hang chufa x^ va x^. DS : He so cua x^ la : - 6 2 2 , cua x^ la : 570. 36. Chufng m i n h v d i n nguyen ducfng t a c6 : a) Cl+Cl+... b) + Cl^Cl+Cl+... + Cl:-\ C > 2C^ + 3C^ + ... + nC;; = n2"-\ CHI DAN a) K h a i t r i e n P(x) = (x - 1)^" r o i cho x = 1. b) K h a i t r i e n P(x) = (1 + x)". T i m P'(x) r o i t i n h P ' ( l ) . 37. Viet k h a i t r i e n Niutdn, bieu thufc (3x - 1)^^, tiT do chufng m i n h r&ng : 31 6 p 0 olSpl ^16 ql4p2 '16 Ql3p3 '16 28 38. T r o n g k h a i t r i e n n h i thufc X\/x + X . Pl6 '16 _ Ol6 A" . H a y t i m so h a n g k h o n g 1^ phu thuoc X , b i e t rSng : C;; + C""' + C""' - 79. £>S : a = 792. 39. T i m he so cua so h a n g chufa x^^ t r o n g k h a i t r i e n f 1 — + x^ biet u c L i + c L i + ... + C L . , = 2 - - i . DS :a = 210. 40. T i m he so cua so' h a n g chufa x^ t r o n g k h a i t r i e n f 1 I— — + Vx^ biet c:;:i-c::,3 = 7(n+3). D S : a = 495. 1 6 £2 TS. Vu ThS' H^u - Nguygn Vinh CSn §3. X A C S U A T 1. P h e p thijf n g a u n h i e n , k h o n g g i a n m a u M o t phep thuf ( t h i nghiem) c6 the lap l a i so I a n t u y y vdti cac dieu k i e n co b a n gio'ng nhau nhiTng k h o n g the xac d i n h chSc chfin, k e t qua nao t r o n g m o i I a n thifc h i e n m a chi c6 the n o i k e t qua do thuoc m o t tap hop xac d i n h t h i t a goi la phep thii ngdu nhien. Tap hop t a t ca cac k e t qua c6 the c6 cua phep thuf ngSu n h i e n goi la khong gian mdu cua phep thijf do. B i e n co n g a u n h i e n M o t phep thijr ngau n h i e n T co k h o n g gian mSu la E, m o i t a p hop A e E bieu t h i mot bien co ngdu nhien ( l i e n quan t d i T ) . B i e n co ngfiu n h i e n , chi gom m o t p h a n tijf cua E duoc goi la bien co so cap. B i e n co dac b i e t gom m o i p h a n tuf cua E la bien co chdc chdn. B i e n co k h o n g chiJa p h a n tuf nao cua E la bien co khong the co, k i h i e u 0 . H a i b i e n C O A , B ma A n B = 0 t h i A va B dirge goi la hai bien co xung khdc. X a c s u a t c i i a b i e n co n g a u n h i e n Phep thuf ngau n h i e n c6 k h o n g gian m a u E gom n bie'n co so cap c6 k h a n a n g xuat h i e n dong deu (dong k h a nang). B i e n co ngau n h i e n A gom k b i e n co sc( cap (cua E) t h i xac suat cua bien co ngau nhien A, 2. 3. ki hieu P(A) la t i so: 4. a) b) c) d) 5. P(A) = n C a c tinh chat ciia xac suat B i e n co' ngau n h i e n A bat k i t a deu co 0 < P(A) < 1. P(0) = 0, P(E) = 1. A va B la h a i b i e n co' xung khac (tufc A n B = 0 ) t h i P(A u B) = P(A) + P(B) Neu A va B la h a i b i e n co bat k i t h i P(A ^ B ) = P(A) + P(B) - P(A n B). Neu A va A la h a i b i e n co' ngau n h i e n ddi lap (tufc la A u A = E, A n A = 0) t h i P(A) = 1 - P(A). B i e n co d p c l a p v a q u y t a c n h a n x a c s u a t H a i b i e n co ngau n h i e n A va B cCing l i e n quan v d i m o t phep thuf ngau n h i e n la doc lap vai nhau neu viec xay r a hay k h o n g xay r a cua b i e n co' nay k h o n g a n h htfdng t d i k h a n a n g xay r a cua b i e n co k i a . Quy tdc nhdn xac suat Neu h a i b i e n co ngau n h i e n A va B doc lap v d i nhau t h i P(A n B) = P(A).P(B). 41. a) b) CHI a) b) 42. CHI 43. CHI 18 ^ BAI TAP Tung mot dong tien dong chat va can doi ba Ian. Khong gian mau co bao nhieu phan tuf ? Goi A la bien co, trong ba Ian tung co dung mot Ian xuat hien mat sap. DAN K i hieu S neu dong tien xuat hien mSt sap va N neu dong tien xuat hien mat ngufa. Ket qua tung dong tien ba Ian bieu t h i bang day 3 chuf cai S hoSc N . Nhii vay khong gian mau gom 8 phan tuf. E = INNN; NSN; SNN; NNS; NSS; SNS; SSN; SSSl. Bien co A ba Ian tung dong tien co dung mot Ian xuat hien mSt sap bieu t h i bdi tap hop A = {SNN; NSN; NNS|. Gia thiet dong tien la can doi va dong chat neu cac ket qua cua phep thuf la dong kha nang. Khong gian mSu co 8 phan tuf. Bien co A co 3 3 phan tuf, do do xac suat cua A la : P(A) = —. 8 Trong mot hop co 4 vien bi mau do, 3 vien bi mau xanh (cac vien bi chi khac nhau ve mau sSc). Lay ngau nhien curig mot liic 3 vien bi. Tinh xac suat de trong 3 vien bi lay ra co dung hai vien bi mau do. DAN Khong gian mSu co bien co scr cap (co C 7 tap hop con 3 phan tuf trong 7 phan tuf), moi each lay 3 vien bi la lay 1 tap hop do. So each lay 2 vien bi do trong 4 vien bi do la C 4 each. So each lay 1 vien bi xanh trong 3 vien bi xanh la C 3 . So each lay 3 vien bi co 2 vien bi do, 1 vien bi xanh la C 4 . C 3 each. Xac suat trong 3 vien bi lay ra co 2 vien bi do la : P(A) = = —. 35 Chon ngau nhien mot so tir nhien co 3 chuf so. Tinh xac suat de so dtfcfc chon la mot so c h i n co 3 chuf so khae nhau. DAN Goi A la bien co so dtroc chon co 3 chuf so khae nhau la so chSn. Khong gian mau E la so cac so co 3 chOr so (9 each chon chuf so hang trSm, 10 each chon cha so hang chuc, 10 each chon chuf so hang don vi) la : 9 X 10 X 10 = 900 so. So cac so CO 3 chuf so khac nhau la so tan cung la 0 la : 9.8 = 72 so (9 each chon chuf so hang tram, 8 each ebon chuf so hang ehuc) So cac so chSn co 3 chur so khdc nhau co cha so hang dcfn vi khac 0 la 8.8.4 = 256 so. TS. Vu The' Huu - Nguy§n VTnh C?n ; ' ~ (4 each chon chuf so h a n g ddn v i , 8 each chon chuT so h a n g t r a m , 8 each chon chuf so h a n g chuc). So cac so C O 3 chOr so khac nhau l a so chkn l a : n = 72 + 256 = 328. Xac suat cua A l a : P(A) = — « 0,3644. 900 44. M o t to hoc sinh co 10 ngtrdi gom 6 n a m v a 4 nuT, chon ngSu n h i e n m o t nhom 3 ngiiofi cua to. T i n h xac suat xay r a m o t trLfcfng hop dudfi day : a) Ca ba ngiTdi diioc chon deu l a n a m . b) Co i t n h a t m o t t r o n g ba ngiidi diTOc chon l a n a m . CHI D A N a) K h o n g gian m a u co : C^Q = ^^'^'^ = 120 p h a n tuf. 1.2.3 Co Cg = ^"^'^ = 20 each chon 3 ngiTdi deu l a n a m . Xac suat b i e n co 3 1.2.3 .-,3 ngi/cfi dtfOc chon deu l a n a m l a : P(A) = C?o 20 1 120 6 b) Goi B l a b i e n co 3 ngiTofi diTOc chon co i t n h a t 1 n a m . B i e n co d o i l a p cua B l a 3 ngiTdi dtfoe chon deu l a nuf : — 1 — 2Q P(B) = ^ = — =^ P ( B ) = 1 - P ( B ) = — . C?o 30 30 45. Cho 8 qua can co k h o i liTOng I a n iMt l a 1kg, 2 k g , 3 k g , 4 k g , 5 k g , 6 k g , 7kg, 8kg. Chon ngau n h i e n 3 qua can. T i n h xac suat de t o n g k h o i luong ba qua can di/crc chon k h o n g virot qua 9kg. CHI D A N So each chon 3 qua can t r o n g 8 qua can (so p h a n tuf cua k h o n g gian mau) l a : C o = ^"'^'^ = 56 each. ^ 1.2.3 A la bien co tong khoi luong 3 qua can diTcfc chon khong qua 9kg. Cac bien C sq cap thuan loi cho A (thuoc tap hop A) co 7 bien co la : O ( 1 ; 2; 6), ( 1 ; 3; 5), (2; 3; 4), ( 1 ; 2; 3), ( 1 ; 2; 4), ( 1 ; 2; 5), ( 1 ; 3; 4) Xac suat cua A : P(A) = — = 0,125. 56 46. Tung m o t I a n h a i con siic sac dong chat can doi. a) T i n h xac suat b i e n co' t o n g so' cham t r e n h a i con sue s^c b a n g 8. b) T i n h xac suat b i e n co t o n g so cham t r e n h a i con sue sac l a m o t so le hoSe m o t so chia h e t cho 3. CHI DAN K h o n g gian m a u co 36 p h a n tuT (6 x 6 = 36 eSp so ( i ; j ) v d i i , j nguyen dirong 1 < i < 6; 1 < j < 6). HQC fa. On luy?n theo CTDT mCn Toan THPT S 19 a) Cac bien co so cap thuan Igi bien co A (tong so cham bang 8) la : (2; 6), (6; 2), (3; 5), (5; 3), (4; 4). Xac suat cua A l a :P(A) = — . 36 b) B i e n co B : t o n g so cham l a so le hoac chia h e t cho 3. Goi B i l a b i e n co t o n g so cham b k n g so le, B2 l a b i e n co t o n g so cham la m o t so chia h e t cho 3, t h i B = B i u B2 P(B) = P ( B i ) + P(B2) - P ( B i n B2) B i xay r a k h i m o t con sue sac n a y m S t chSn, m o t con n a y m a t le, co 18 b i e n CO so cap : ( 1 ; 2), ( 1 ; 4), ( 1 ; 6), (3; 2), (3; 4), (3; 6), (5; 2), (5; 4), (5; 6), (2; 1), (2; 3), (2; 5), (4; 1), (4; 3), (4; 5), (6; 1), (6; 3), (6; 5). B2 x a y r a vdi 12 b i e n so so cap : ( 1 ; 2), (2; 1), ( 1 ; 5), (5; 1), (2; 4), (4; 2), (3; 3), (6; 6), (3; 6), (6; 3), (4; 5), (5; 4). B i e n co B i n B2 tong so' cham le va chia h e t cho 3 gom 6 bien co' so cap : ( 1 ; 2), (2; 1), (3; 6), (6; 3), (4; 5), (5; 4). 47. P(B) = P ( B i ) + P(B2) - P ( B i n B2) = — + — - — = — = 36 36 36 36 3 H a i x a t h u cung b ^ n ( m o t each doc lap) vao m o t muc t i e u m 6 i ngiTdi m o t v i e n dan. Xac suat b a n t r i i n g dich t r o n g m o t I a n bSn cua nguTdi thur n h a t l a 0,9; cua ngudi thur h a i l a 0,7. T i n h xac suat t r o n g m o i t r i r d n g h o p sau : a) Ca h a i v i e n deu t r i i n g dich. b) i t n h a t co m o t v i e n t r i i n g dich. c) C h i CO m o t v i e n t r i i n g . C H I D A N a) Goi A i la bien co ngLTcri thijf nhat ban t n i n g dich, A2 la bien co ngUcfi thtf hai ban t n i n g dich. A la bien co ca hai vien deu t n i n g dich t h i A = A i n A2. Do hai ngu6i ban doc lap v d i nhau nen A i va A2 la doc lap n e n P(A) = P ( A i n A2) = P(Ai).P(A2) = 0,9.0,7 = 0,63. b) Goi B l a b i e n co co i t n h a t m o t v i e n d a n t r i i n g dich : B = A i u A2 P(B) = P ( A i u A2) = P(Ai) + P(A2) - P ( A i n A2) = 0,9 + 0,7 - 0,63 = 0,97 c) Goi C l a b i e n co, h a i ngLfdi bSn m 6 i ngudi m o t v i e n c h i co m o t v i e n t r i i n g dich. B i e n co C xay r a k h i ngiidi thiir n h a t t r i i n g dong t h d i ngtrdi t h i i h a i t r u g t hoSc ngudi t h i i n h a t triiOt, ngtfcfi t h i i h a i t r i i n g . P ( C ) = P ( A i A 2 u A i A 2 ) = P(AiA2) + P ( A i A 2 ) - P ( A i A 2 n A i A g ) 2 0 53 TS. Vu ThS' Hi^u - Nguygn Vinh CSn Do A i , A2 doc lap t h a n h thuf A^, A2 doc l a p , A i va A2 doc l a p va b i e n CO A1A2 n A1A2 = 0 cho n e n P(C) = P(Ai)P(A2) + P(Ai)P(A2) - 0 = 0,9.0,3 + 0,1.0,7 = 0,34. 48. M o t 16 h a n g c6 30 san p h a m , t r o n g do c6 3 phe p h a m . Ngirdi t a chia n g i u n h i e n 16 h a n g t h a n h 3 p h a n , m o i p h a n 10 san p h a m . a) T i n h xac suat de m o i p h a n c6 diing m o t phe p h a m . b) T i n h xac suat de c6 i t n h a t m o t p h a n c6 dung m o t phe p h a m . CHi DAN a) Ta thirc h i e n chia n h u sau : L a y ngSu n h i e n 10 san p h a m t r o n g 30 san p h a m t a c6 p h a n thuf n h a t . T r o n g 20 san p h a m con l a i lay ngau n h i e n 10 san p h a m de c6 p h a n thuf h a i . Con l a i la 10 san p h a m p h a n thuf 3. S6' each chia n h i i vay b k n g CH.cH each. So' each chia de p h a n thuf n h a t eo m o t san p h a m xau la : C3C27 (lay m o t san p h a m xau ghep v d i 9 san p h a m to't t r o n g 27 san p h a m t6't). So' each chia de p h a n thuf h a i c6 m o t san p h a m xau la : C2C18. S6' each chia de m o i p h a n c6 m o t san p h a m xau la : CgCgy.CgCig. Xac suat de m 6 i p h a n eo dung m o t san p h a m xau la : P(A) = ^^^f-^,f^« = 0,246. plOplO '-'30^20 b) Goi B l a b i e n eo' t r o n g 3 p h a n cd i t n h a t m o t p h a n c6 m o t phe p h a m . Neu goi ( i , j , k ) la so' san p h a m xau theo thuf t u cua cac p h a n thuf n h a t , thiJ h a i , thuf ba v d i i , j , k nguyen dtfctng v d i 0 < i , j , k < 3. K h i do B xay r a ufng v d i cac bo ( 1 ; 1; 1), ( 1 ; 2; 0), ( 1 ; 0; 2), (2; 0; 1), (2; 1; 0), (0; 1; 2), (0; 2; 1). B i e n c6' doi lap cua B xay r a tucfng lifng v d i cac bo (3; 0; 0), (0; 3; 0), (0; 0; 3). Ro r a n g t i n h P(B) = 1 - P(B) t h o n g qua t i n h P(B) dan g i a n han. So each chia de p h a n thuf n h a t eo 3 san p h a m x a u , cac p h a n con l a i deu t o t la : C^.C^^.C^g. So' each chia de p h a n thuf h a i eo 3 san p h a m x a u , cac p h a n khae deu la san p h a m t o t la : C^^.C^.Cjy. S6' each chia de h a i p h a n dau cac san p h a m deu to't, p h a n thuf ba c6 3 san p h a m xau la : C27.Cj°. *-^3'^27*^20 + '-^27'-'3'-^17 + ^^^> = ^27^17 T^w^o '-'30*-^20 P(B) = 1 - P ( B ) « 0,911. HQC 6n luy§n theo CTDT men To^n THPT E 21