Trang 1 Su tm : Tng Duy Khoa Nickhocmai :balep Vic su tm khng th khng thiu st Mong cc bn c gi thc mc,gp hoc chuyn quaemail duykhoatang@gmail.com bi vit thm phong ph hn. Trang 2 I.CC BI TON V : PHP NHN TRN MN HNH Bi 1:Tnh chnh xc tng S = 1.1! + 2.2! + 3.3! + 4.4! + ... + 16.16!. Gii:V n . n! = (n + 1 1).n! = (n + 1)! n! nn: S = 1.1! + 2.2! + 3.3! + 4.4! + ... + 16.16! = (2! 1!) + (3! 2!) + ... + (17! 16!) S = 17! 1!. Khng th tnh 17 bng my tnh v 17! L mt s c nhiu hn 10 ch s (trn mn hnh). Nn ta tnh theo cch sau: Ta biu din S di dng : a.10n + b vi a, b ph hp khi thc hin php tnh, my khng b trn, cho kt qu chnh xc. Ta c : 17! = 13! . 14 . 15 . 16 . 17 = 6227020800 . 57120 Li c: 13! = 6227020800 = 6227 . 106 + 208 . 102 nnS = (6227 . 106 + 208 . 102) . 5712 . 10 1 = 35568624 . 107 + 1188096 . 103 1 = 355687428096000 1 = 355687428095999. Bi 2:Tnh kt qu ng ca cc tch sau: a)M = 2222255555 . 2222266666. b)N =20032003 . 20042004. Gii: a)t A = 22222, B = 55555, C = 666666. Ta c M = (A.105 + B)(A.105 + C) = A2.1010 + AB.105 + AC.105 + BC Tnh trn my: A2 = 493817284 ; AB = 1234543210 ; AC = 1481451852 ; BC = 3703629630 Tnh trn giy: A2.10104938172840000000000 AB.105123454321000000 AC.105148145185200000 BC3703629630 M4938444443209829630 b)t X = 2003, Y = 2004. Ta c: N = (X.104 + X) (Y.104 + Y) = XY.108 + 2XY.104 + XY Tnh XY, 2XY trn my, ri tnh N trn giy nh cu a) Kt qu:M = 4938444443209829630. N = 401481484254012. Bi tp tng t:Tnh chnh xc cc php tnh sau: a)A = 20!. b)B =5555566666 . 6666677777 c)C = 20072007 . 20082008 d)10384713 e)201220032 II. TM S D CA PHP CHIA S NGUYN a) Khi cho s b hn 10 ch s: S b chia = s chia . thng + s d (a = bq + r) (0 < r < b) Trang 3 Suy ra r = a b . q V d : Tm s d trong cc php chia sau:1)9124565217 cho 123456 2)987896854 cho 698521 b) Khi cho s ln hn 10 ch s: Phng php:Tm s d ca A khi chia cho B ( A l s c nhiu hn 10 ch s) -Ct ra thnh 2 nhm , nhm u c chn ch s (k t bn tri). Tm s d phn u khi chia cho B. -Vit lin tip sau s d phn cn li (ti a 9 ch s) ri tm s d ln hai. Nu cn na tnh lin tip nh vy. V d: Tm s d ca php chia 2345678901234 cho 4567. Ta tm s d ca php chia 234567890 cho 4567: c kt qu s d l : 2203 Tm tip s d ca php chia 22031234 cho 4567. Kt qu s d cui cng l 26. Bi tp: Tm s d ca cc php chia: a)983637955 cho 9604325 b)903566896235 cho 37869. c)1234567890987654321 : 123456 c) Dng kin thc v ng d tm s d. * Php ng d:+ nh ngha: Nu hai s nguyn a v b chia cho c (c khc 0) c cng s d ta ni a ng d vi b theo modun c k hiu(mod ) a b c + Mt s tnh cht: Vi mi a, b, c thuc Z+ (mod ) a a m (mod ) (mod ) a b m b a m (mod ); (mod ) (mod ) a b mb c m a c m (mod ); (mod ) (mod ) a b m c d m a c b d m (mod ); (mod ) (mod ) a b m c d m ac bd m (mod ) (mod )n na b m a b m V d 1: Tm s d ca php chia 126 cho 19 Gii: ( )236 2 312 144 11(mod19)12 12 11 1(mod19)= = Vy s d ca php chia 126 cho 19 l 1 V d 2: Tm s d ca php chia 2004376 cho 1975 Gii: Bit 376 = 62 . 6 + 4 Ta c: 24 212 348 42004 841(mod1975)2004 841 231(mod1975)2004 231 416(mod1975)2004 416 536(mod1975) Vy Trang 4 606262.3 362.6 262.6 42004 416.536 1776(mod1975)2004 1776.841 516(mod1975)2004 513 1171(mod1975)2004 1171 591(mod1975)2004 591.231 246(mod1975)+ Kt qu: S d ca php chia 2004376 cho 1975 l 246 Bi tp thc hnh: Tm s d ca php chia : a)138 cho 27 b)2514 cho 65 c)197838 cho 3878. d)20059 cho 2007 e)715 cho 2001 III. TM CH S HNG N V, HNG CHC, HNG TRM... CA MT LU THA: Bi 1: Tm ch s hng n v ca s 172002 Gii: ( )210002 2000 100021000200017 9(mod10)17 17 9 (mod10)9 1(mod10)9 1(mod10)17 1(mod10)= Vy 2000 217 .17 1.9(mod10) . Ch s tn cng ca 172002 l 9 Bi 2: Tm ch s hng chc, hng trm ca s 232005. Gii + Tm ch s hng chc ca s 232005 123423 23(mod100)23 29(mod100)23 67(mod100)23 41(mod100) Do :( )520 4 52000 1002005 1 4 200023 23 41 01(mod100)23 01 01(mod100)23 23 .23 .23 23.41.01 43(mod100)= = Vy ch s hng chc ca s 232005 l 4 (hai ch s tn cng ca s 232005 l 43) + Tm ch s hng trm ca s 232005 14520 42000 10023 023(mod1000)23 841(mod1000)23 343(mod1000)23 343 201(mod1000)23 201 (mod1000) Trang 5 510020002005 1 4 2000201 001(mod1000)201 001(mod1000)23 001(mod1000)23 23 .23 .23 023.841.001 343(mod1000)= Vy ch s hng trm ca s 232005 l s 3 (ba ch s tn cng ca s 232005 l s 343) III. TM BCNN, UCLN My tnh ci sn chng trnh rt gn phn s thnh phn s ti gin A aB b=T p dng chng trnh ny tm UCLN, BCNN nh sau: + UCLN (A; B) = A : a + BCNN (A; B) = A . b V d 1: Tm UCLN v BCNN ca 2419580247 v 3802197531 HD: Ghi vo mn hnh : 24195802473802197531 v n =, mn hnh hin 711 UCLN: 2419580247 : 7 = 345654321 BCNN: 2419580247 . 11 = 2.661538272 . 1010 (trn mn hnh) Cch tnh ng: a con tr ln dng biu thc xo s 2 ch cn 419580247 . 11 Kt qu : BCNN: 4615382717 + 2.109 . 11 = 26615382717 V d 2: Tm UCLN ca 40096920 ; 9474372 v 51135438 Gii: n 9474372 . 40096920 = ta c : 6987. 29570. UCLN ca 9474372 v 40096920 l 9474372 : 6987 = 1356. Ta bit UCLN(a; b; c) = UCLN(UCLN(a ; b); c) Do ch cn tm UCLN(1356 ; 51135438). Thc hin nh trn ta tm c:UCLN ca 40096920 ; 9474372 v 51135438 l : 678 Bi tp: Cho 3 s 1939938; 68102034; 510510. a)Hy tm UCLN ca 1939938; 68102034. b)Hy tm BCNN ca 68102034; 510510. c)Gi B l BCNN ca 1939938 v 68102034. Tnh gi tr ng ca B2. IV.PHN S TUN HON. V d 1: Phn s no sinh ra s thp phn tun hon sau: a)0,(123) b)7,(37) c)5,34(12) Gii:Ghi nh: 1 1 10, (1); 0, (01); 0, (001)9 99 999= = =... a) Cch 1:Ta c 0,(123) = 0,(001).123 = 1 123 41.123999 999 333= = Cch 2: t a = 0,(123) Ta c 1000a = 123,(123) . Suy ra 999a = 123. Vy a = 123 41999 333= Trang 6 Cc cu b,c (t gii) V d 2: Phn s no sinh ra s thp phn tun hon 3,15(321) Gii: t 3,15(321) = a. Hay 100.000 a = 315321,(321) (1) 100 a = 315,(321) (2) Ly (1) tr (2) v theo v, ta c 999000a = 315006 Vy 1665052501999000315006= = aBi 3: Tnh 2 2 20,19981998... 0, 019981998... 0, 0019981998...A = + +Gii t 0,0019981998... = a. Ta c:1 1 12.100 102.111100Aa a aAa| |= + + |\ .= Trong khi : 100a = 0,19981998... = 0,(0001) . 1998 = 19989999 Vy A = 2.111.999911111998=V. TNH S L THP PHN TH N SAU DU PHY. V d 1:Tm ch s l thp phn th 105 ca php chia 17 : 13 Gii: Bc 1: + Thc hin php chia 17 : 13 = 1.307692308 (thc cht my thc hin php tnh ri lm trn v hin th kt qu trn mn hnh) Ta ly 7 ch s u tin hng thp phn l: 3076923 + Ly 1,3076923 . 13 = 16,9999999 17 - 16,9999999 = 0,0000001 Vy 17 = 1,3076923 . 13 + 0.0000001 (ti sao khng ghi c s 08)??? Khng ly ch s thp cui cng v my c th lm trn. Khng ly s khng v17 = 1,30769230 . 13 + 0,0000001= 1,30769230 . 13 + 0,0000001 Bc 2:+ ly 1 : 13 = 0,07692307692 11 ch s hng thp phn tip theo l: 07692307692 Vy ta tm c 18 ch s u tin hng thp phn sau du phy l: 307692307692307692 Vy 17 : 13 = 1,(307692) Chu k gm 6 ch s. Ta c 105= 6.17 + 3 (105 3(mod6) ) Vy ch s thp phn th 105 sau du phy l ch s th ba ca chu k. chnh l s 7 V d 2: Tm ch s thp phn th 132007 sau du phy trong php chia 250000 cho 19 Gii: Trang 7 Ta c 250000 171315719 19= + . Vy ch cn tm ch s thp phn th 132007 sau du phy trong php chia 17 : 19 Bc 1:n 17 : 19 = 0,8947368421.Ta c 9 ch s u tin sau du phy l894736842 + Ly 17 0, 894736842 * 19 = 2 . 10-9 Bc 2: Ly 2 : 19 = 0,1052631579.Chn s hng thp phn tip theo l: 105263157 + Ly 2 0,105263157 * 19 = 1,7 . 10-8 = 17 . 10-9 Bc 3: Ly 17 : 19 = 0,8947368421. Chn s hng thp phn tip theo l+ Ly 17 0,0894736842 * 19 = 2 . 10-9 Bc 4:Ly 2 : 19 = 0,1052631579.Chn s hng thp phn tip theo l: 105263157 ... Vy 17 : 19 = 0, 894736842105263157894736842105263157 ... = 0,(894736842105263157) . Chu k gm 18 ch s. Ta c ( )6693 2007 3 66913 1(mod18) 13 13 1 (mod18) = Kt qu s d l 1, suy ra s cn tm l s ng v tr u tin trong chu k gm 18 ch s thp phn. Kt qu : s 8 Bi tp: Tm ch s thp phn th 2007 sau du phy khi chia: a)1 chia cho 49 b)10 chia cho 23 VI. CC BI TON V A THC Mt s kin thc cn nh: 1.nh l Bezout S d trong php chia f(x) cho nh thc x a chnh l f(a) H qu: Nu a l nghim ca f(x) th f(x) chia ht cho x a 2.S Hor n Ta c th dng s Hor n thm kt qu ca php chia a thc f(x) cho nh thc x a. V d: Thc hin php chia (x3 5x2 + 8x 4) cho x 2 bng cch dng s Hor n. Bc 1: t cc h s ca a thc b chia theo th t vo cc ct ca dng trn. Bc 2: Trong 4 ct trng dng di, ba ct u cho ta cc h s ca a thc thng, ct cui cng cho ta s d. -S thnht ca dng di = s tng ng dng trn a = 2 -58-41 Trang 8 -K t ct th hai, mi s dng di c xc nh bng cch ly a nhn vi s cng dng lin trc ri cng vi s cng ct dng trn Vy (x3 5x2 + 8x 4) = (x 2)(x2 3x + 2) + 0 * Nu a thc b chia l a0x3 + a1x2 + a2x + a3 , a thc chia l x a, ta c thng l b0x2 + b1x + b2 d l r. Theo s Hor n ta c: Bi 1: Tm s d trong cc php chia sau: a)x3 9x2 35x + 7 cho x 12. b)x3 3,256 x + 7,321 cho x 1,1617. c)Tnh a x4 + 7x3 + 2x2 + 13x + a chia ht cho x + 6 d) 5 3 26, 723 1,857 6, 458 4, 3192, 318x x x xx + ++ e)Cho P(x) = 3x3 + 17x 625 + Tnh P(2 2 ) + Tnh a P(x) + a2 chia ht cho x + 3 Bi 2 :Cho P(x) = x5 + ax4 + bx3 + cx2 + dx + f .Bit P(1) = 1 , P(2) = 4 , P(3) = 9 , P(4) = 16 , P(5) = 15 .Tnh P(6) , P(7) , P(8) , P(9) Gii:Ta c P(1) = 1 = 12; P(2) = 4 = 22 ; P(3) = 9 = 32 ; P(4) = 16 = 42 ; P(5) = 25 = 52 Xt a thc Q(x) = P(x) x2. D thy Q(1) = Q(2) = Q(3) = Q(4) = Q(5) = 0. Suy ra 1; 2; 3; 4; 5 l nghim ca a thc Q(x). V h s ca x5 bng 1 nn Q(x) c dng: Q(x) = (x 1)(x 2)(x 3)(x 4)(x 5). Vy ta c Q(6) = (6 1)(6 2)(6 3)(6 4)(6 5) = P(6) - 62 Hay P(6) = 5! + 62 = 156. Q(7) = (7 1)(7 2)(7 3)(7 4)(7 5) = P(7) 72 Hay P(7) = 6! + 72 = 769 Bi 3: Cho Q(x) = x4 + mx3 + nx2 + px + q . Bit Q(1) = 5 , Q(2) = 7 , Q(3) = 9 ,Q(4) = 11 . Tnh cc gi tr ca Q(10) , Q(11) , Q(12) , Q(13)Hng dnQ(1) = 5 = 2.1 + 3; Q(2)= 7 = 2.2 + 3; Q(3) = 9 = 2.3 + 3 ; Q(4) = 11 = 2.4 + 3 Xt a thc Q1(x) = Q(x) (2x + 3) Bi 4 : Cho P(x) = x5 + ax4 + bx3 + cx2 + dx + e .Bit P(1) = 3 , P(2) = 9 , P(3) = 19 , P(4) = 33 , P(5) = 51 . Tnh P(6) , P(7) , P(8) , P(9) , P(10) , P(11) .Bi 5:a = 2 -58-41 1 -320 aa1 a2 a3 a0 b0 rb1 b2 a0 ab0 + a1 ab1 + a2 ab2 + a3 Trang 9 Cho P(x) = x4 + ax3 + bx2 + cx + d. C P(1) = 0,5 ; P(2) = 2 ; P(3) = 4,5 ;P(4) = 8. Tnh P(2002), P(2003) Bi 6: Cho P(x) = x4 + ax3 + bx2 + cx + d. Bit P(1) = 5; P(2) = 14; P(3) = 29; P(4) = 50. Hy tnh P(5) , P(6) , P(7) , P(8) Bi 7: ChoP(x) = x4 + ax3 + bx2 + cx + d. Bit P(1) = 0; P(2) = 4 ; P(3) = 18 ; P(4) = 48. Tnh P(2007) Bi 8 : Cho P(x) = x5 + 2x4 3x3 + 4x2 5x + m .a)Tm s d trong php chia P(x) cho x 2,5 khi m = 2003 . b)Tm gi tr ca m P(x) chia ht cho x 2,5c)P(x) c nghim x = 2 . Tm m .Bi 9: Cho P(x) = 4 322 5 73 x x x + + . a)Tm biu thc thng Q(x) khi chia P(x) cho x 5. b)Tm s d ca php chia P(x) cho x 5 chnh xc n 3 ch s thp phn. Bi 10: Tm s d trong php chia a thc x5 7,834x3 + 7,581x2 4,568x + 3,194 chox 2,652. Tm h s ca x2 trong thc thng ca php chia trn. Bi 11: Khi chia a thc 2x4 + 8x3 7x2 + 8x 12 cho x 2 ta c thng l a thc Q(x) c bc l 3. Hy tm h s ca x2trong Q(x) Bi 12: Cho a thc P(x) = 6x3 7x2 16x + m . a)Tm m P(x) chia ht cho 2x + 3 b)Vi m tm c cu a ) , hy tm s d r khi chia P(x) cho 3x 2 v phn tch P(x) thnh tch ca cc tha s bc nht c)Tm m v n Q(x) = 2x3 5x2 13x + n v P(x) cng chia ht cho x 2 . d)Vi n tm c trn , hy phn tch Q(x) ra tch ca cc tha s bc nht. Bi 13:Cho P(x) = x4 + 5x3 4x2 + 3x + m v Q(x) = x4 + 4x3 - 3x2 + 2x + n . a)Tm cc gi tr ca m v n P(x) vQ(x) cng chia ht cho x 2 . b)Vi gi tr ca m v n tm c , chng t rngR(x) = P(x) Q(x) ch c mt nghim duy nht Bi 14 :Cho f(x) = x3 + ax2 + bx + c . Bit :f |.|

\|31 = 1087 ; f |.|

\|21= 53; f |.|

\|51 = 50089 . Tnh gi tr ng v gn ng caf |.|

\|32 .Bi 15:Xc nh cc h s a, b, c ca a thc: P(x) = ax3 + bx2 + cx 2007 sao cho P(x) chia cho (x 13) c s d l 1, chia cho (x 3) c s d l l 2, v chia cho (x 14) c s d l 3 (Kt qu ly vi hai ch s hng thp phn) Bi 16: Xc nh cc h s a, b, c, d v tnh gi tr ca a thcQ(x) = x5 + ax4 + bx3 + cx2 + dx 2007 ti cc gi tr ca x = 1,15; 1,25; 1,35; 1,45 Trang 10 VII. MT S BI TON V DY S Bi 1:Cho dy s a1 = 3; an + 1 = 331n nna aa++. a)Lp quy trnh bm phm tnh an + 1

b)Tnh an vi n = 2, 3, 4, ..., 10 Bi 2:Cho dy s x1 = 12; 3113nnxx ++= . a)Hy lp quy trnh bm phm tnh xn + 1 b)Tnh x30 ; x31 ; x32 Bi 3: Cho dy s 141nnnxxx++=+ (n > 1) a)Lp quy trnh bm phm tnh xn + 1 vi x1 = 1 v tnh x100. b)Lp quy trnh bm phm tnh xn + 1 vi x1 = -2 v tnh x100. Bi 4: Cho dy s 21 24 51nnnxxx++=+(n > 1) a)Cho x1 = 0,25. Vit quy trnh n phm lin tc tnh cc gi tr ca xn + 1 b)Tnh x100 Bi 5: Cho dy s ( ) ( )5 7 5 72 7n nnU+ =vi n = 0; 1; 2; 3; ... a)Tnh 5 s hng u tin U0, U1, U2, U3, U4 b)Chng minh rng Un + 2 = 10Un + 1 18Un . c)Lp quy trnh bm phm lin tc tnh Un + 2theo Un + 1 v Un. HD gii: a)Thay n = 0; 1; 2; 3; 4 vo cng thc ta cU0 = 0, U1 = 1, U2 = 10, U3 = 82, U4 = 640 b)Chng minh: Gi s Un + 2 = aUn + 1 + bUn + c. Thay n = 0; 1; 2 v cng thc ta c h phng trnh:

2 1 03 2 14 3 21010 8282 10 640U aU bU c a cU aU bU c a b ca b c U aU bU c= + + + = = + + + + = + + = = + + Gii h ny ta c a = 10, b = -18, c = 0 c) Quy trnh bm phm lin tc tnh Un + 2 trn my Casio 570MS , Casio 570ES a U1 vo A, tnh U2 ri a U2 vo B 1 SHIFT STO A x 10 18 x 0 SHIFT STO B, lp li dy phm sau tnh lin tip Un + 2 vi n = 2, 3, ... x 10 18 ALPHA A SHFT STO A (c U3) x 10 18 ALPHA B SHFT STO B (c U4) Bi 6: Cho dy s 3 5 3 522 2n nnU| | | |+ = + || ||\ . \ . vi n = 1; 2; 3; ... a)Tnh 5 s hng u tin U1, U2, U3, U4 , U5 b)Lp cng thc truy hi tnh Un + 1 theo Un v Un 1. Trang 11 c)Lp quy trnh bm phm lin tc tnh Un + 1 trn my Casio Bi 7:Cho dy s vi s hng tng qut c cho bi cng thc

3 2) 3 13 ( ) 3 13 (n nnU += vin = 1 , 2 , 3 , ... k , . . . a) Tnh 8 7 6 5 4 3 2 1, , , , , , , U U U U U U U U b) Lp cng thc truy hi tnh 1 + nUtheo nUv 1 nU c) Lp quy trnh n phm lin tc tnh 1 + nUtheo nUv 1 nUBi 8: Cho dy s{ }nU c to thnh theo quy tc sau: Mi s sau bng tch ca hai s trc cng vi 1, bt u t U0 = U1 = 1. a)Lp mt quy trnh tnh un. b)Tnh cc gi tr ca Un vi n = 1; 2; 3; ...; 9 c)Chaykhngshngca dychiahtcho4?Nucchovd.Nukhng hy chng minh. Hng dn gii: a) Dy s c dng: U0 = U1 = 1, Un + 2 = Un + 1 . Un + 1, (n =1; 2; ...) Quy trnh tnh Un trn my tnh Casio 500MS tr ln: 1 SHIFT STO A x 1 + 1 SIHFT STO B. Lp li dy phm x ALPHA A + 1 SHIFT STO A x ALPHA B + 1 SHIFT STO B b) Ta c cc gi tr ca Un vi n = 1; 2; 3; ...; 9 trong bng sau: U0 = 1U1 = 1U2 = 2U3 = 3U4 = 7 U5 = 22U6 = 155U7 = 3411U8 = 528706U9 = 1803416167 Bi 9: Cho dy s U1 = 1, U2 = 2, Un + 1 = 3Un + Un 1. (n > 2) a)Hy lp mt quy trnh tnh Un + 1 bng my tnh Casio b) Tnh cc gi tr ca Un vi n = 18, 19, 20 Bi 11: Cho dy s U1 = 1, U2 = 1, Un + 1 = Un + Un 1. (n > 2) c)Hy lp mt quy trnh tnh Un + 1 bng my tnh Casio d) Tnh cc gi tr ca Un vi n = 12, 48, 49, 50 S cu b)U12 = 144, U48 = 4807526976, U49 = 7778742049 , U49 = 12586269025 Bi 12: Cho dy s sp th t vi U1 = 2, U2 = 20 v t U3 tr i c tnh theo cng thc Un + 1 = 2Un + Un + 1 (n > 2). a)Tnh gi tr ca U3 , U4 , U5 , U6 , U7 , U8 b)Vit quy trnh bm phm lin tc tnh Un c)S dng quy trnh trn tnh gi tr ca Un vi n = 22; 23, 24, 25 Trang 12 III. MT S BI TON V LIN PHN S. Bi 1: Cho 12305102003A = ++ . Vit li 11111...onnA aaaa= +++ + Vit kt qu theo th t | | | |0 1 1, ,..., , ...,...,...,...n na a a a=Gii: Ta c 12 12.2003 24036 4001 130 3 30 30 1 315 2003520035 20035 20035102003 4001A = + = + = + = + + = ++

1313054001= ++.Tip tc tnh nh trn, cui cng ta c: 131151133121112112A = +++++++ Vit kt qu theo k hiu lin phn s| | | |0 1 1, ,..., , 31, 5,133, 2,1, 2,1, 2n na a a a=Bi 2: Tnh gi tr ca cc biu thc sau v biu din kt qu di dng phn s: 311213145A =+++ ; 101716154B =+++ ; 20032345879C =+++ p s: A) 2108/157 ; B) 1300/931 ; C) 783173/1315 Ring cu C ta lm nh sau: Khi tnh n 2003: 1315391. Nu tip tc nhn x 2003 = th c s thp phn v vt qu 10 ch s.V vy ta lm nh sau: 391 x 2003 = (kt qu 783173) vy C = 783173/1315. Bi 3: a) Tnh 1111111111111 1A = +++++++b) 1313131313133B = +++ Trang 13 c) 11121314151617189C = ++++++++d) 19283746556473829D = ++++++++ Bi 4:a) Vit quy trnh tnh: 3 11712 51 231 11 312 117 72002 2003A = + ++ ++ ++ + b) Gi tr tm c ca A l bao nhiu ? Bi 5: Bit 2003 1712732111abcd= +++++. Tm cc s a, b, c, d. Bi 6: Tm gi tr ca x, y. Vit di dng phn s t cc phng trnh sau: a)41 11 41 12 31 13 24 2x x+ =+ ++ ++ + ; b) 1 11 21 13 45 6y y=+ ++ + Hng dn: t A = 11112134+++ ,B = 11413122+++

Ta c 4 + Ax = Bx. Suy ra 4xB A=. Kt qu 844 1255681459 1459x = = .(Tng t y = 2429) Trang 14 Bi 7:Tm x bit: 3 3819783382007838383838383838181 x=++++++++++ Lp quy trnh n lin tc trn fx 570MS, 570ES. 381978 : 382007 = 0.999924085 n tip phm x-1 x 3 8 v n 9 ln du =. Ta c: 11Ansx=+. Tip tc n Ans x-1 1 =Kt qu : x = -1,11963298hoc 1745760908336715592260478921| | |\ . Bi 8: Thi gian tri t quay mt vng quanh tri t c vit di dng lin phn s l: 1365141713151206++++++ . Da vo lin phn s ny, ngi ta c th tm ra s nm nhun. V d dng phn s 13654+th c 4 nm li c mt nm nhun. Cnnudnglinphns 1 7365 36512947+ =+thc29nm(khngphil28 nm) s c 7 nm nhun. 1) Hy tnh gi tr (di dng phn s) ca cc lin phn s sau: a) 136514173+++ ; b) 13651417135++++ ; c) 13651417131520+++++ 2)Kt lun v s nm nhun da theo cc phn s va nhn c. I V.L ai kep Nien khoan Trang 15 Bai toan mau: Givao ngan hang soti en l aa ong, vil aisuat hang thang l ar% trong n thang. Tnh cavon l an l aiA sau n thang? -- Gi ai-- GoiA l ati en von l an l aisau n thang ta co: Thang 1 (n = 1): A = a + ar= a(1 +r) Thang 2 (n = 2): A = a(1 + r) + a(1 + r)r= a(1 + r)2 Thang n (n = n): A = a(1 + r)n 1 + a(1 + r)n 1.r= a(1 + r)n Vay A = a(1 + r )n(* ) Trong o:a tien von ban au, rlai suat (%) hang thang, n sothang, A tien von lan lai sau n thang. Tcong thc (* ) A = a(1 + a)n ta tnh c cac ail ng khac nh sau: 1) Al nanl n(1 r)=+;2)nAr 1a= ; 3) na(1 r) (1 r) 1Ar+ + =;4) nAra(1 r) (1 r) 1=+ + (l ntrongcongthc1l aLogari tNepe,trenmayf x-500MSvaf x-570MS phm l n an trc ti ep) Vdu:Motsoti en58.000.000gi ti etki emtheol ai suat0,7%thang. Tnh cavon l an l aisau 8 thang? -- Gi ai-- Ta co: A = 58000000(1 + 0,7%)8 Ket qua: 61 328 699, 87 V du: Mot ngico58 000 000 muon givao ngan hang ec 70 021 000. Hoiphaigiti et ki em bao l au vilaisuat l a0,7% thang? -- Gi ai-- Sothang toithi eu phaigil a: ( )70021000l n58000000nl n 1 0, 7%=+ Ket qua: 27,0015 thang Vay toithi eu phaigil a27 thang. V du: Soti en 58 000 000 giti et ki em trong 8 thang th l anh vec 61 329 000. Tm l aisuat hang thang? -- Gi ai-- Laisuat hang thang: 861329000r 158000000= Ket qua: 0,7% V du: Moithang giti et ki em 580 000 vil aisuat 0,7% thang. Hoisau 10 thang th lanh vecavon l an l ail abao nhieu? Trang 16 --Gi ai -- Soti en l anh cagoc lan lai : ( )10 10580000.1, 007. 1, 007 1 580000(1 0, 007) (1 0, 007) 1A0, 007 0, 007 + + = =Ket qua: 6028055,598 Vdu:Muonco100000000sau 10thangthphai gi quyti etki eml abao nhi eu moithang. Vil aisuat gil a0,6%? -- Gi ai-- Soti en gihang thang: ( ) ( )( )10 10100000000.0, 006 100000000.0, 006a1, 006 1, 006 11 0, 006 1 0, 006 1= =+ + Ket qua: 9674911,478 Nhan xet: Can phan bi et rocach giti en ti et ki em:+ Gi soti en a mot lan -----> l ay cavonl anl aiA. + Gihang thang soti ena -----> l ay cavon l an l aiA. Can phantch cac baitoan mot cach hp l yec cac khoang tnh ung an. Cothesuyl uanetmracaccongthct1)->4)tngt nhbaitoan mau Cac baitoan vedan socung cotheap dung cac cong thc tren ay. V.Tm a thc thng khi chia a thc cho n thc Bai toan mau: Chi a a thc a0x3 + a1x2 + a2x + a3 cho x c ta sec thng l amot a thc bac haiQ(x) = b0x2 + b1x + b2 vasod r. Vay a0x3 + a1x2 + a2x + a3 = (b0x2 + b1x + b2)(x-c) + r = b0x3 + (b1-b0c)x2 + (b2-b1c)x + (r + b2c). Ta l aicocong thc truy hoiHorner: b0 = a0; b1= b0c + a1; b2= b1c + a2; r = b2c + a3. Tng tnh cach suy l uan tren, ta cung cos oHorner etm thng vasod khichi a a thc P(x) (tbac 4 trl en) cho (x-c) trong trng hp tong quat. V du: Tm thng vasod trong phep chi a x7 2x5 3x4 + x 1 cho x 5. -- Gi ai-- Ta co: c = - 5; a0 = 1; a1 = 0; a2 = -2; a3 = -3; a4 = a5 = 0; a6 = 1; a7 = -1; b0 = a0 = 1. Qui trnh an may (fx-500M S vafx-570 M S) ( ) 5 SHI FT STO M 1 ALPHA M 0 ALPHA M 2ALPHA M ( ) 3 ALPHA M 0 ALPHA M 0ALPHA M 1 ALPHA M ( ) 1 + = = + = + = + = + = + =( -5) ( 23)( -118) ( 590 ) ( -2950 )( 14751) ( -73756) Trang 17 Vay x7 2x5 3x4 + x 1 = (x + 5)(x6 5x5 + 23x4 118x3 + 590x2 2590x + 14751) 73756. VI .Phan tch a thc theo bac cua n thc Apdungn-1l an dangtoan2.4tacothephantchathc P(x)bacntheo x-c: P(x)=r0+r1(x-c)+r2(x-c)2++rn(x-c)n. V du: Phan tch x4 3x3 + x 2 theo bac cua x 3. -- Gi ai-- Trcti enthchi enphepchi aP(x)=q1(x)(x-c)+r0theosoHornerec q1(x) var0. Sau ol aiti ep tuc tm cac qk(x) vark-1 ta c bang sau: 1-301-2x4-3x2+x-2 310011q1(x)=x3+1, r0 = 1 313928q2(x)=x3+3x+1,r1= 28 31627q3(x)=x+6, r0 = 27 319q4(x)=1=a0, r0 = 9 Vay x4 3x3 + x 2 = 1 + 28(x-3) + 27(x-3)2 + 9(x-3)3 + (x-3)4. Vdu:Tmtatca cacso t nhi enn(1010sns2010)saocho na 20203 21n = +cung l asotnhi en. -- Gi ai-- V 1010s ns 2010 nen 203,5 ~41413 s an s62413~ 249,82.V an nguyen nen 204s ns 249. Ta coan2 = 20203 + 21n = 21.962 + 1 + 21n. Suy ra: an2 1 = 21(962+n), hay (an - 1)(an + 1) = 3.7.(962+n). Do o,( )( )2n n na 1 a 1 a 1 = +chi a het cho 7. Chng to(an - 1) hoac (an + 1) chi a het cho 7. Vay an = 7k + 1 hoac an = 7k 1. *Neu an = 7k 1 thido 204s n =7k-1 s 249 => 29,42s ks 35,7. Do k nguyen nen{ } k 30; 31; 32; 33; 34; 35 = . V 2na 1 7k(7k 2) = chi a het cho 21 nen k chl a: 30; 32; 33; 35. Ta co: k30323335 n1118140615571873 an 209223230244 * Neuan=7k+1thi do204sn=7k-1s249=>29,14sks35,57.Dok nguyennen{ } k 30; 31; 32; 33; 34; 35 = .V 2na 1 7k(7k 2) = + chi ahetcho21nenk chl a: 30; 31; 33; 34. Ta co: Trang 18 Nh vay ta cotat ca8ap so. V du: Tnh A = 999 999 9993 -- Gi ai-- Ta co: 93=729; 993= 970299; 9993=997002999; 99993= 99992.9999=99992(1000-1)= 999700029999. Tota coquy l uat: 3n 1 chso n 1 chso nchso9nchso999...9 99...9 7 00...0 299...9 = Vay 999 999 9993 = 999 999 997 000 000 002 999 999 999. VII.Kim tra mt s l nguyn t hay hp s? Cslni dungnhlsau: al mt s nguyntnun khngchia ht cho mi s nguyn t khng vt qua Xut pht t c s , ta lp 1 quy trnh bm phm lin tip kim tra xem s a c chia ht cho cc s nguyn t nh hnahay khng! Nhn xt: Mi s nguyn t u l l (tr s 2), th nn ta dng php chia a cho cc s l khng vt qua . Cch lm: 1.Tnha . 2.Ly phn nguyn b ca kt qu. 3.Ly s l ln nht c khng vt qu b. 4.Lp quy trnh c A a A B A 2 A Gn s l c vo nh A lm bin chy. Dng lnh 1. B l mt bin cha. Dng lnh 2. A l mt bin chy. IFT SH =... Lp 2 DL trn, n du=v quan st n khi A = 1 th dng. 5.Trong qu trnh n= :-Nu tn ti kq nguyn th khng nh a l hp s. -Nu khng tn ti kq nguyn no th khng nh a l s nguyn t. VD1: Xt xem 8191 l s nguyn t hay hp s? 1.Tnh8191 c 90,50414355 2.Ly phn nguyn c 90. k30323335 n1118140615571873 an 209223230244 Trang 19 3.Ly s l ln nht khng vt qu n l 89. 4.Lp quy trnh: 89 A 8191 A B A 2 A IFT SH =... 5.Quanstcckt qutathyukhng nguyn,cho nnkhng nh 8191 l s nguyn t. VD2: Xt xem 99 873 l s nguyn t hay hp s? 1. Tnh99873c 316,0268976. 2. Ly phn nguyn c 316. 3. Ly s l ln nht khng vt qu n l 315. 4. Lp quy trnh: 315 A 99 873 A B A 2 A IFT SH =... 5. Quan st mn hnh thy c kt qu nguyn l 441, cho nn khng nh 99 873 l hp s. 5.6-Phn tch mt s ra tha s nguyn t? Nhn xt: Cc s nguyn t u l s l (tr s 2) Cch lm:TH1: Nu s a c c nguyn t l 2, 3 (Da vo du hiu chia ht nhn bit). Ta thc hin theo quy trnh: a C 2 A (hoc 3 A) C : A B B : A C IFT SH == My bo kq nguyn ta nghi 2 (hoc 3)l mt SNT. Cc kq vn l s nguyn th mi ln nh th ta nhn c 1 TSNT l 2 (hoc 3). Tm ht cc TSNT l 2 hoc 3 th ta phn tch thng cn li da vo trng hp di y VD1: Phn tch64 ra tha s nguyn t? M t quy trnh bm phm ngha hoc kt qu Trang 20 64 C 2 A C : A B B : A C IFT SH ====Gn Gn Kq l s nguyn 32. Ghi TSNT 2 Kq l s nguyn 16. Ghi TSNT 2 Kq l s nguyn 8. Ghi TSNT 2 Kq l s nguyn 4. Ghi TSNT 2 Kq l s nguyn 2. Ghi TSNT 2 Kq l s nguyn 1. Ghi TSNT 2 Vy 64 = 26 VD2: Phn tch 540 ra tha s nguyn t? M t quy trnh bm phm ngha hoc kt qu 540 C 2 A C : A B B : A C 3 A C : A B B : A C C : A B Gn Gn Kq l s nguyn 270. Ghi TSNT 2 Kq l s nguyn 135. Ghi TSNT 2 Nhn thy 135 2 nhng 1353 ta gn: Kq l s nguyn 45. Ghi TSNT 3 Kq l s nguyn 15. Ghi TSNT 3 Kq l s nguyn 5. Ghi TSNT 3 Thng l B = 5 l 1 TSNT. Vy 540 = 22335 TH2: Nu a l s khng cha TSNT 2 hoc 3. Quy trnh c minh ho qua cc VD sau y. VD3: Phn tch 385 ra tha s nguyn t? M t quy trnh bm phm ngha hoc kt qu 385 C 3 A C : A B A + 2 A IFT SH =Gn Gn Lp dng lnh 1 Lp dng lnh 2 Lp 2 DL trn. Kq l s nguyn 77.Chng t C A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 5 Trang 21 V/ B:A C A + 2 A IFT SH = = Kq l s nguyn 11.Chng t B A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 7 V/ C:A B A + 2 A IFT SH = = = Kq l s nguyn 1. (qu trnh kt thc) Chng t C A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 11 Vy 385 = 5.7.11. VD3: Phn tch 85 085 ra tha s nguyn t? M t quy trnh bm phm ngha hoc kt qu 85085 C 3 A C : A B A + 2 A IFT SH = =(2 ln du= ) Gn Gn Lp dng lnh 1 Lp dng lnh 2 Lp 2 DL trn. Kq l s nguyn 17 017.Chng t C A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 5 V/ B:A C A + 2 A IFT SH = Kq l s nguyn 2431.Chng t B A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 7 V/ C:A B A + 2 A IFT SH = = = Kq l s nguyn 221.Chng t C A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 11 V/ B:A C A + 2 A Trang 22 IFT SH =Kq l s nguyn 17.Chng t B A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 13 V/ C:A B A + 2 A IFT SH == Kq l s nguyn 1. (Dng li y) Chng t C A, A l 1 s nguyn t. Khi ta n AC ri ghi SNT l 17 Vy 85 085 = 5.7.11.13.17 Bi tp: Phn tch cc s sau ra tha s nguyn t: a) 94 325(527311) b) 323 040 401.(7921913271 VIII.Phn tch a thc f(x) thnh nhn t. C s:1.Nu tam thc bc hai ax2 + bx + c c 2 nghim l x1, x2 th n vit c di dng ax2 + bx + c = a(x-x1)(x-x2). 2.Nu a thc f(x) = anxn + an-1xn-1+... + a1x + a0 c nghim hu t pqth p l c ca a0, q l c ca a0. 3.cbit:Nuathcf(x)=anxn+an-1xn-1+...+a1x+a0ca1=1th nghim hu t l c ca a0. 4.Nu a thc f(x) c nghim l a th athc f(x) chia ht cho (x-a). VD1: Phn tch a thc f(x) =x2 + x - 6 thnh nhn t? Dng chc nng gii phng trnh bc hai ci sn trong my tm nghim ca f(x) ta thy c 2 nghim l x1 = 2; x2 = -3. Khi ta vit c: x2 + x - 6 = 1.(x-2)(x+3) VD2: Phn tch a thc f(x) = x3+3x2 -13 x -15 thnh nhn t? Dngchcnnggiiphngtrnhbc 3cisntrongmytmnghim ca f(x) ta thy c 3 nghim l x1 = 3; x2 = -5; x3 = -1. Khi ta vit c: x3+3x2 -13 x -15 = 1.(x-3)(x+5)(x+1). VD3 :Phn tch a thc f(x) = x5 + 5x4 3x3 x2 +58x - 60 thnh nhn t? Trang 23 Nhn xt: Nghim nguyn ca a thc cho l (60). Ta c (60) = { 1; 2; 3; 4; 5; 6; 10; 12; 15; 20; 30; 60} Lp quy trnh kim tra xem s no l nghim ca a thc:Gn: -1 X Nhp vo my a thc:X5 + 5X4 3X3X2 +58X -60 ri n du =my bo kq -112 Gn tip:-2 X /= / my bo kq -108 Gn tip: -3 X/ = / my bo kq 0 Do vy ta bit x = -3 l mt nghim ca a thc cho, nn f(x) chia ht cho (x+3). Khi bi ton tr v tm thng ca php chia a thc f(x) cho (x-3). Khi ta c f(x) = (x+3)(x4+2x3-9x2+26x-20) * Ta li xt a thc g(x) = x4+2x3-9x2+26x-20 Nghim nguyn l c ca 20.Dng my ta tm c (20) = { 1; 2; 4; 5; 10; 20} Lp quy trnh kim tra xem s no l nghim ca a thc g(x):Gn: -1 X Nhp vo my a thc:x4+2x3-9x2+26x-20 ri n du= my bo kq -96 Gn tip:-2 X / = / my bo kq -148 Gn tip:-4 X /= / my bo kq -180 Gn tip: -5 X /= /my bo kq 0 Do vy ta bit x = -5 l mt nghim ca a thc cho, nn f(x) chia ht cho (x+5). Khi bi ton tr v tm thng ca php chia a thc f(x) cho (x+5). Khi ta c g(x) = (x+5)(x3-3x2+6x-4) *Tiptcdngchcnnggiiphngtrnhbc3tmnghimnguyncaa thch(x) = x3-3x2+6x-4 Kt qu, l a thc h(x) c nghim l x = 1 nn chia h(x)cho (x-1) ta c: h(x) = (x-1)(x2-2x+4) Ta thy a thc (x2-2x+4) v nghim nn khng th phn tch thnh nhn t. Vy f(x) = (x+3)(x+5)(x-1)(x2-2x+4)