Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T1

M U hngtabitrngmytnhCasiolloimyrttinli cho hc sinh t trung hc n i hc. V my gii quyt hu ht cc bi ton trung hc v mt phn i hc. gip hc sinh c bit l hc sinh THCS c th s dng c loi my tnh cm tay kiu khoa hc ni chung, loi my Casio fx 570 MS ni ring.Ngoi nhng ti liu hng dn s dng v gii ton c, khi hc sinh mua my . Hc sinh c nhng ti liu th ch c th bit chcnngcbnccphmvtnhtonccphptoncbn,m cha c bi tp thc hnh nhiu v k nng gii Ton bng my tnh cm tay. HS t mnhkhm ph nhng kh nng tnh ton phong ph, khai thc cc chc nng ca my gn lin vi vic hc trn lp cngnhtrongcchotngngoikhatonhcthngquathc hnh trn my. Vthtrongqutrnhdyhctrnlp(dyhctchn,dy BDHSG,) . Chng ta cn phi trang b cho hc sinh nm c mt s phng php gii v quy trnh n phm. t , mi hc sinh t mnh gii c cc bi tp tonmt cch ch ng v sng to. ngtrcthctrngtrn,vitinhthnyuthchbmn, munckhmph,munchoccemhcsinhTHCScnhng dng bi tp ton gii bng my tnh cm tay. Ti xin a ra mt s dngbitphcsinhtthchnh,rnluynknnggiiTon bng my tnh cm tay. C Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T2

NI DUNG DNG 1: TM S D CA PHP CHIA CA S A CHO S B a)S d ca s A chia cho s B: ( i vi s b chia ti a 10 ch s )

Cch n: AB mn hnh hin kt qu l s thp phn. a con tr ln biu thc sa li AB phn nguyn ca A chia cho B v n V d: Tm s d ca php chia 9124565217 cho 123456. n: 9124565217123456 My hin thng s l: 73909,45128 a cn tr ln dng biu thc sa li l: 9124565217123456 73909 v n Kt qu: S d: r = 55713BI TP:Tm s d trong cc php chia sau: a) 143946 chia cho 32147KQ: r = 15358 b) 37592004 chia cho 4502005KQ: r = 1575964 c) 11031972 chia cho 101972KQ: r = 18996 d) 412327 chia cho 95215KQ: r = 31467 e) 18901969 chia cho 1512005KQ: r = 757909 b)Khi s b chia A ln hn 10 ch s: NunhsbchiaAlsbnhthngnhiuhn10chs.Tangtra thnh nhm u 9 ch s ( k t bn tri ). Ta tm s d nh phn a). Ri vit tip sau s d cn li l ti a 9 ch s ri tm s d ln hai. Nu cn na th tnh lin tip nh vy. V d: Tm s d ca php chia 2345678901234 cho 4567. Tatmsdcaphpchia234567890cho4567cktqul2203. Tm tip s d ca 22031234cho 4567. Kt qu cui cng l 26. Vy r = 26. S d ca AA BB = x phn nguyn ca (A chia cho B )= -x - = x= Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T3

BI TP: 1) Tm s d r khi chia s 24728303034986074 cho 2003. KQ: r = 401 2) Tm s d r khi chia s 2212194522121975 cho 2005. KQ: r = 1095 c)Tm s d ca s b chia c cho bng dng ly tha qu ln th ta dng php ng d thc theo cng thc sau: . . (mod ) (mod )(mod ) (mod )c ca b mn p a m pb n p a m p V d 1: Tm s d ca php chia 2004376 cho 1975 Gii: Ta c 20042 841 (mod 1975) 20044 8412 (mod 1975) 200412 2313 416 (mod 1975) 200448 4164 536 (mod 1975) 200448 .200412 536. 416 (mod 1975) 200460 1776 (mod 1975) 200462 1776. 841 (mod 1975) 200462 516 (mod 1975) 200462x3 5163 1171(mod 1975) 200462x3x2 11712 (mod 1975) 200462x6 591 (mod 1975) 200462x6+4 591.231 (mod 1975) 2004376 246 (mod 1975) Vy 2004376 chia cho 1975 c s d l 246. V d 2: Tm s d ca php chia 17659427 cho 293 Gii: Ta c 176594 208 (mod 293) 1765943 2083 3 (mod 293) 1765942739 (mod 293) 1765942752 (mod 293) Vy 17659427 chia cho 293 c s d l 52 Bi tp: 1)Tm s d ca php chia 232005 cho 100. Gii: Ta c: 23123 (mod 100) 23229 (mod 100) Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T4

234292 41 (mod 100) (234 )5415 (mod 100) 2320 1 (mod 100) (2320 )1001100 1 (mod 100) 232000 1 (mod 100) 232005 =23200.234.231 1.41.23 (mod 100) 232005 43 (mod 100)Vy 232005 chia cho 100 c s d l 43 2) Tm hai ch s cui cng ca 232005 Gii: Ta gii nh bi 1. Tr li: Hai ch s cui cng ca 232005 l 43 3) Tmch s hng chc ca 232005 Gii: Ta cng gii nh bi 1. Tr li: Ch s hng chc ca 232005 l 4. 4) Tm s d ca php chia 72005 chia cho 10( Tm ch s hng n v ca 72005 ) Gii: Ta c 71 7 (mod 10)72 49 (mod 10)74 1 (mod 10) 72004 = (74)501 1501 1(mod 10) 72005 = 72004 .71 1.7 7(mod 10) Vy:+72005 chia cho 10 l 7. + Ch s hng n v ca 72005 l 7. 5) Tm ch s hng n v ca 172002. Gii: Ta c 71 7 (mod 10)72 49 (mod 10)74 1 (mod 10)(74)500 1500 1(mod 10) 72000 1(mod 10) 72002 172000. 172 1.9 9(mod 10)Vy: Ch s hng n v ca 172002 l 9. 6) Tm hai ch s cui cng ca tngA = 22000 + 22001 + 22002 Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T5

Gii: Ta c A = 22000 ( 1+ 21 + 22 ) = 7. 22000 M ta li c 210 24 (mod 100)(210)5 245 24 (mod 100) 2250 245 24 (mod 100) 21250 245 24 (mod 100) 22000 = 21250.2250.2250.2250 24.24.24.24 76 (mod 100) A = 7. 22000 7.76 32 (mod 100) Vy : Hai ch s cui cng ca tng A l 32 7) Tm hai ch s cui cng ca tngB = 22000 + 22001 + 22002 +22003 + 22004 + 22005 + 22006

Gii: Ta c B = 22000 ( 1+ 21 + 22 + 23 + 24 + 25 + 26) = 127. 22000 B = 127. 22000 127.76 52 (mod 100) Vy : Hai ch s cui cng ca tng B l 52 8) Tm s d ca php chia 19971997 cho 13 Gii: Ta c 19971 8 (mod 13) 19972 12 (mod 13) 19973 12.8 5(mod 13) 19974 1 (mod 13) (19974 )499 1499 1(mod 13) 19971997 = 19971996 . 19971 1.8 (mod 13) Hay19971997 8 (mod 13) Vy s d ca php chia 19971997 cho 13 l 8 9) Tm d trong php chia 21000 cho 25 Gii: Ta c 210 24 (mod 25) 220 1 (mod 25) 21000 1500 1 (mod 25) Vy s d trong php chia21000 cho 25 l 1 10) Tm d trong php chia 21997 cho 49 Gii: Ta c 22 4 (mod 49) 210 44 (mod 49) 220 442 25 (mod 49) Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T6

221 25.2 1 (mod 49) (221 )95 195 1 (mod 49) 21995 1 (mod 49) 21997 = 21995 .22 1.4 4 (mod 49) V y d trong php chia 21997 cho 49 l 4 11) Tm d trong php chia 21999 cho 35 Gii: Ta c 21 2 (mod 35) 210 9 (mod 35) 220 442 25 (mod 35) 230 9.25 29 (mod 35) 216 16 (mod 35) 248 1 (mod 35) 21999= (248)41.231 1.29.223 (mod 35) V y d trong php chia 21999 cho 35 l 23. 12) Tm d khi chiaa) 43624362 cho 11 b) 301293 cho 13 c) 19991999 cho 99 d) 109345 cho 14( r = 1 ) e) 31000 cho 49 f) 61991 cho 28( r = 20) g) 35150 cho 425 h) 222002 cho 1001 i) 20012010 cho 2003 13) a) CMR: 18901930 + 19451975 + 1 7 b) CMR: 22225555 + 55552222 7 Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T7

DNG 2: TM CH S x CA S n = 1 1 0...n na a xa a mvi m N Phng php: Thay x ln lt t 0 n 9 sao cho n m V d:Tm ch s x 79506 47 xchia ht cho 23. Gii: Thay x = 0; 1; 2; ; 9. Ta c 79506147 23 Bi tp: 1)Tmslnnhtvsnh nhttrongccstnhincdng1 2 3 4 x y z chia ht cho 7. Gii: -S ln nht dng 1 2 3 4 x y zchia ht cho 7 s phi l 19293 4 z . Ln lt th z = 9; 8; ;1; 0. Vy S ln nht dng 1 2 3 4 x y zchia ht cho 7 s phi l 1929354. -S nh nht dng 1 2 3 4 x y zchia ht cho 7 s phi l 10203 4 z . Ln lt th z =0; 1; ;8; 9. Vy S nh nht dng 1 2 3 4 x y zchia ht cho 7 s phi l 1020334. 2)Tm s ln nht v s nh nht ca s2 3 4 5 x y zchia ht cho 25. KQ: - S ln nht l: 2939475 -S nh nht l: 1030425. 4)Tm ch s b, bit rng:469283861 6505 bchia ht cho 2005. KQ: b = 9. 5) Tm ch s a bit rng469 8386196505 achia ht cho 2005. KQ: a = 0;1; 2; 3; 4; 5; 6; 7; 8; 96)Hy nu cc bc thc hin trn my tnh v t suy ra phi thm s no vo bn phi s 200 mt ch s c s c bn ch s chia ht cho 7. Hng dn: n =200 7 a . KQ: 2002; 2009. Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T8

DNG 3: TM C V BI CA MT S 1. Tm cc c ca mt s a : Phng php: Gn: A = 0 ri nhp biu thc A = A + 1 : a A n nhiu ln phm Gn: Nhp: a n nhiu ln du V d: Tm ( cc c ) tp hp cc c ca 120 Ta gn: A = 0 Nhp: A = A + 1 : 120 A n nhiu ln phmTa c A = {1;2;3;4;5;6;8;10;12;15;20;30;40;60;120}2. Tm cc bi ca b: Gn: A = -1 ri nhp biu thc A = A + 1 : b x A n nhiu ln phm V d : Tm tp hp cc bi ca 7 nh hn 100. Ta gn: A = -1 Nhp: A = A + 1 : 7 x A n nhiu ln phmTa c: B = {0;7;14;21;28;35;42;49;56;63;70;77;84;91;98} BI TP: 1)Tm cc c ca cc s sau: 24; 48;176. 2)Tm tt c cc bi ca 14 nh hn 150 3.Kim tra s nguyn t: kim tramt s l s nguyn t ta lm nh sau: kt lun s a l s nguyn t ( a > 1) , ch cn chng t rng n khng chia ht cho mi s nguyn tm bnh phng khng vt qu a. V nu mt s a l hp s th n phi c c nh hna0 = ShiftSTOT Alpha A A 1= AlphaAlpha A Alpha : +Alpha = = = = A Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T9

V d: S 647 c phi l s nguyn t khng ? Gii Ta c647= 25,43 Gn: A = 0 Nhp: A = A + 1 : 647 A n 25 ln phm mtrn mn hnh kt qu thng l s thp phn th kt lun 647 l s nguyn t

BI TP: 1)Cc s sau y s no l s nguyn t:197; 247; 567; 899; 917; 929 2) Tm mt c ca 3809783 c ch s tn cng l 9KQ: 19339 3) Tm mt s t nhin x bit lp phng ca n c tn cng l ba ch s 1. HD: Gn : A = 10 Nhp: A = A + 1 : A3 KQ: x = 471 4)Tm cc s a, b, c, d ta c5 a x 7850 bcd = . Gii: S5 al c ca 7850. Bng cch th trn my khi cho a = 0; 1; 2; ....; 9 Ta thy rng a ch c th bng 2. Khi a = 2 th7850 bcd = : 25 = 314 Vy a = 2; b = 3; c = 1; d = 4. = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T10

DNG 4: TM CP NGHIM (x; y) NGUYN DNGTHA MN PHNG TRNH V d:Tm cp s (x; y) nguyn dng sao cho x2 = 37y2 +1. Gii: Ta c x2 = 37y2 +1 nn y < x Suy ra x = 237 1 y + . Do gn: Y = 0, X = 0; nhp Y = Y + 1 : X = 237 1 Y + . Nhn du lin tc cho ti khi X nguyn. KQ: x = 73; y = 12. BI TP: 1) Tm cp s (x; y) nguyn dng sao cho x2 = 47y2 +1. KQ: x = 48; y = 7. 2)Tmcps(x;y)nguyndngthamnphngtrnh ( )234 17 2 161312 x x y + = . Gii: Ta c( )234 17 2 161312 x x y + = ( )32161312 4217xx y = 3161312 4217xx y = 4161312 4217xy x= . Do gn: Y = 0, X = 0; nhp X = X + 1 : Y = 2X - 3161312 417X . Nhn du lin tc cho ti khi Y nguyn. KQ: x = 30; y = 4. = = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T11

DNG 5: TM CLN, BCNN CA HAI S V my ci sn chng trnh n gin phn s thnh phn s ti gin. A aB b=( ti gin ) th CLN (A, B) = A a BCNN (A, B) = A xb V d 1: Tm a) CLN( 209865; 283935 ) b) BCNN(209865; 283935 ) Ghi vo mn hnh 209865 289335 v nMn hnh hin: 1723 a) a con tr ln dng biu thc sa thnh 20986517KQ: CLN( 209865; 283935 ) = 12345 b) a con tr ln dng biu thc sa thnh 20986523 KQ: BCNN(209865; 283935 ) = 4826895 V d 2: Tm CLN( 2419580247; 3802197531) BCNN( 2419580247; 3802197531) Ghi vo mn hnh 2419580247 3802197531v n Mn hnh hin: 711 a) a con tr ln dng biu thc sa thnh 2419580247 7 KQ: CLN( 2419580247; 3802197531) = 345654321 b) a con tr ln dng biu thc sa thnh 241958024711 Mn hnh hin 2661538272 x 1010 y li gp tnh trng trn mn hnh. Mun ghi y s ng, ta a con tr ln dng biu thc xo ch s 2 (u tin ca s A) ch cn 419580247 11 v nMn hnh hin46115382717 Ta c kt qu BCNN( 2419580247; 3802197531) = 26615382717 V d 3: Tm cc c nguyn t ca A = 17513 + 19573 + 23693

Gii:Ghi vo mn hnh 17511957 v n My hin: 17 19 Chnh li mn hnh 175117 v n = x= = =x = = x= Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T12

Kt quCLN(1751, 1957) = 103( s nguyn t ) Th li: 2369 cng c c nguyn t 103 A = 1033(173 + 193 + 233) Tnh tip 173 + 193 + 233 = 23939 Chia 23939 cho cc s nguyn t: Ta c 23939 = 37.647( 647 l s nguyn t ) Vy A c cc c nguyn t 37, 103, 647 Bi tp: 1)Tm BCNN v CLN ca a = 24614205, b = 10719433 KQ: BCNN(a,b) = 12380945115 ; CLN(a,b) = 21311 2)Tm BCNN v CLN ca hai s 168599421 v 2654176. KQ: BCNN(a,b) = 37766270304 ; CLN(a,b) = 11849. 3)Tm cc c nguyn t nh nht v ln nht ca s 2152 + 3142 Gii: Tnh 2152 + 3142 = 144821 ;144821 = 380,553 Gn: A = 0 Nhp: A = A + 1: 144821 A nlin tc thy 144821 = 97.1493 Tip tc kim tra 1493 c phi l s nguyn t khng Ta c1493= 38,639 Gn: A = 0 Nhp: A = A + 1: 1493 A nlintcchotiA=40mkhngthyktquthngls nguyn th 1493 l s nguyn t. Vy 2152 + 3142 = 144821 = 97.1493 c c s nguyn t nh nht l 97, c c s nguyn t ln nht l 1493 = = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T13

DNG 6: TNH GI TR CA BIU THC a) A = 15,25 + 3 1, 061 25%4 2 + KQ: A = 16,72 b) B = 2 2 1 10, 4 0, 259 11 3 57 7 11, 4 1 0, 875 0, 79 11 6 + ++ + +KQ : B = 0,5714 c) C = 11 3 1 21 .4 1, 5 6 .31 7 3 195 1 14 12 56 6 3| | |\ | |+ |\ KQ: C = 930, 86916107 =d) D =( )4 2 40, 8: .1, 25 1, 08 :4 5 25 71, 2.0, 5 :1 5 1 2 50, 64 6 3 .225 9 4 17| | | | ||\ \ + +| | |\ KQ: D = 213 e) E = ( )( )217 0, 65 10, 7 5, 26, 7 7 10, 2 1, 7 + KQ: E = 5,40578 f)F = ( ) ( )2 21986 1992 . 1986 3972 3 .19871983.1985.1988.1989 + KQ: F = 1987. g) G = ( ) ( )222 2649 13.180 13. 2.649.180 + KQ: G = 1. h) H = ( )( )( )( )3: 0, 2 0,1 34, 06 33, 81 .42 426: :2, 5. 0, 8 1, 2 6, 84: 28, 57 25,15 3 21 + + + KQ: H = 172 i) I = 14, 5: 47, 375 26 18.0, 75 .2, 4: 0, 8832 517,81:1, 37 23 :13 6| | |\ KQ: I = 4 k) K = ( )( )2217, 005 4, 505 93, 750,1936: 0, 88 3, 53 7, 5625 : 0, 52 + + KQ: K = 20 l) L = 1 5 5 1 313 2 10 .230 464 27 6 5 43 10 1 21 : 12 147 3 3 7| | + |\ | | | |+ ||\ \ KQ: L = -41 m) M = 3 3 3 3 33 5 4 2 20 25 + KQ: M = 0 (1-11) Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T14

n) N = 3 33 33 354 18200 126 2 6 21 2 1 2+ + + + +KQ: N = 8 p) P = 3 3 59 4 5 9 4 5 13 2 7 + + + KQ: P = 4,5045q) Q = 3 4 8 92 3 4 ... 8 9 + + + + + KQ: 1,91164 HD: Nhp: 9 n:9Ans 8

8Ans7

7Ans6 6Ans5 5Ans4 4Ans3 3Ans2 Ans r) R =( ) ( )1 33 2 1 40, 5 .0, 2 : 3 : .1 :3 25 5 3 3| | | | || \ \ KQ: R = 790, 35111225= (( )50, 59=;( )20, 29= ) u) U = ( )17 6, 35 : 6, 5 9, 8999... .12, 8: 0,1251 11, 2: 3, 6 1 : 0, 25 1, 8333... .15 4 + | |+ |\ KQ: U = 213 HD: Ta c 9,8999 = 9,8(9) = 9,8+ 0,0(9) = 9,8 + 1.0, (9)10 = 9,8 + 1 9 9, 8 1.10 9 10 10= + = 9,9 1,8333 = 1,8(3) = (183 -18)(183 18) 165 1190 90 6= = ==+ ==+ ==+ ==+ ==+ ==+ ==+ == Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T15

DNG 7:TNH GI TR CA LIN PHN S Phng php:C1: Tnh t di ln C2: Tnh t trn xung V d 1:Biu din A ra phn s thng v s thp phn A = 53425242523+++++ Gii: C1: Tnh t di ln n : 35 2 4 2 5 2 4 2 5 3n tip:KQ: A = 4,6099644 = 233 17614382 382=C2: Tnh t trn xung Nhp: 3(5 (2 (4 (2 (5 (2 (4 (2 5 3)))))))) + + + + + V d 2: Biu din A ra phn s thng v s thp phn B = 171313134++++ 1x= x + 1xx+ = 1xx + = 1xx + = 1xx+ = = / b caShiftd/c = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T16

C1: Tnh t di ln n : 43 3 3 7KQ: B = 43 10377 7, 302716901142 142= =C2: Tnh t trn xung Nhp:7 (1 (3 (1 (3 (1 (3 1 4)))))) + + + + BI TP:1)Tnh a) A = 111112++b)B = 121112112 ++++ c) C = 135716++d)D = 131111511292++++ e) E = 201213145+++ f) F = 21516178+++ g) G = 20033254768+++ KQ: A = 35; B = 1411; C = 367117; D = 196274980; E = 1360157; F = 7001807; G = 104156137 1x=+ 1x1x+ = 1x+ = 1x+ = = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T17

2) Biu din biu thc M ra phn s. M = 1 11 15 21 14 31 13 42 5++ ++ ++ + Gii: C1:Tnh tng t nh bi 1v gn kt qu s hng u vo s nh A, tnh s hng sau ri cng li. KQ: M = 98157 C2: Tnh trc tip Nhp:(1 (5 (1 (4 (1 (3 1 2)))))) (1 (2 (1 (3 (1 (4 1 5)))))) + + + + + + + 3)Tnh gi tr cc biu thc sau: a) A = 1 11 15 21 14 32 33 41 12 51262++ ++ ++ ++ ++KQ: A= 6524351222392 b) B = 2004 20051 1215 227 19 455 36 91 14 13 2++ ++ ++ ++ + KQ: B = 222,760422 c) C = 20 2 20051 1 32 5 21 1 53 6 41 77 84 65 8+ ++ + ++ + +++ +KQ: C =312753094 = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T18

DNG 8: BIU DIN PHN S RA LIN PHN S V d:Tnh a, b bit: a) A =329 1110513151ab=+++b) B = 15 111711ab=++ Gii: Ta c329 1 1 1 1 11051 64 1 1 110513 3 3 39 1 1329 3295 5 564 16479 9= = = = =+ + + ++ + ++ Vy a = 7, b = 9 Cch n my gii : Ghi vo mn hnh: 329 1051 v n n tip: ( my hin 364329 ) n tip:3 ( my hin 64329 ) n tip: (my hin 5964 ) n tip:5(( my hin 964 ) n tip:(my hin 719 )KQ: a = 7, b = 9 b) KQ: a = 7, b = 2 BI TP: 1)Vit cc s sau di dng lin phn s a) 1037142b) 1761382 c) 23152d) 69178 Kt qu:

1037 171142313134= ++++

= 1x = -= 1x = - = 1x = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T19

1761 1413821111111131212112= +++++++++ 23 111526111111114=+++++ 69 1117821111121111113=+++++++

2)Vit cc s sau di dng lin phn s a) 19758b) 25735c) 58972d) 119223e) 5231032f) 6781999 Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T20

DNG 9: TM X BIT HOC GII PHNG TRNH BC NHT MT N Phng php: C1: p dng th t thc hin cc php ton gii phng trnh. C2: S dng chc nng SOLVE V d: Tm x, bita) 1 1 1 134 13 5 x = + Gii: C1: Nhp : 1 1 134 13 5+ KQ: 260747x = C2: Nhp c biu thc vo my11 1 134 13 5+ KQ: 260747x = b) 24 5 4 137 9 7 x = +Gii: C1: Nhp: 25 4 137 9 7 +4KQ: 529 176411235 1235x = =Hoc nhp: 4 25 4 137 9 7| | + |\ C2: Nhp biu thc 4 25 4 137 9 7 + KQ: 529 176411235 1235x = =BI TP: 1)Tm x > 0 , bit a) 2 2 21 1 15 12 x= + KQ: 8 60413 13x = =C1: n: 2 21 15 12+= 1x = / b caAlphaX=Alpha ShiftSolve1=ShiftSolve = 1x =X= = / b caAlphaX=Alpha ShiftSolve1=ShiftSolve = 1x = Ans= Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T21

C2: Dng chc nng SOLVE b) 2 2 23 1 48 7 x= + KQ:5, 5539 x = C1: n 32 2 23 1 48 7 x= +C2: Dng chc nng SOLVE c)Tng qut: n m ka b dx c e= +C1: n am kb dc e+C2: Dng chc nng SOLVE d) 2 6 7 31 1 1 315 8 5 x= + + KQ:6, 4549 x = e) 25 3 5 77 3 5 34 6 2 x= + KQ:3, 046996466 x =2) Tm x, bit a) 2 21 3 17 4 x= +KQ:65, 32638963 x =C1: Nhp 2 23 17 4+C2: Dng chc nng SOLVE b) 2 33 1 47 9 x= +KQ:13421, 66085 x =C1: Nhp 2 23 17 4+3 Hoc: 32 23 17 4+C2: Dng chc nng SOLVE c)Tng qut: 2 2na b dc e x= +C1: Nhp2 2b dc e+a n Hoc: a2 2b dc e+ n C2: Dng chc nng SOLVE d) q pna b dc e x m = + = Ans= ( )= Ans= ( ) nShift x = 2x= 1x = = 2x= 1x = x ()Ans== 2x=^= 1x = x () Ans = = ^Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T22

C1: Nhpq pb dc e+a n mC2: Dng chc nng SOLVE e) 3 51 1 45 7 1 x= +KQ:14.736, 22728 x =f) 3 75 1 34 5 1 x= ++KQ:101.897, 5329 x =3) Gii phng trnh a)( ) 0, 0001 2 : 0, 3 .0, 01 11, 2 22, 2 x + = KQ:10.000.000 x =b)3 0, 75 26 .2,8 1, 75 : 0, 05 2357 0, 35x | | + = |\ KQ:4 x =c) 13 2 5 1 1: 2 .115, 2.0, 25 48, 51:14, 7 44 11 66 2 513, 2 0, 8 5 3, 252x| | |\ =| |+ |\ KQ:25 x =d) 1 3 14 : 0, 003 0, 3 .11 2 20 2: 62 17, 81: 0, 0137 13011 1 3 1 203 2, 65 .4: 1, 88 2 .20 5 25 8x| | | | ||\ \ + =| | | | + || \ \ KQ:6 x =e) 3 4 4 10, 5 1 . 1, 25.1,8 : 37 5 7 2 35, 2: 2, 53 1 3 415, 2.3,15 : 2 .4 1, 5.0, 84 2 4x| | | | + ||| | \ \ = || |\ + |\ KQ:903, 4765135 x = f) ( ) ( )( )( )2 23 2 40,15 0, 35 : 3 4, 2 . .1 4 3 53 : 1, 2 3,152 3 12 212, 5 . : 0, 5 0, 3.0, 75 :7 5 17x| | + + + | \ = + KQ:1, 39360764 x = g) 2 3 1 6 3 7 15 113 5 3 2 4 3 2 3 5x x| |+ = | | + \ KQ:1, 4492 x = h)41 11 41 12 31 13 24 2x x+ =+ ++ ++ +KQ: 884 1255681459 1459x = = =^= 1x = = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T23

i) 11 11 21 13 45 6y y+ =+ ++ +KQ: 2429y = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T24

DNG 10: TNH GI TR CA A THC PHN THC Phng php:C1: S dng phm nh A, B, C, D, E, F, X, Y, M, AnS C2: S dng chc nng CALC V d 1: Tnh23 12 y x x = + vi7 x =v khi8 x =Gii: C1: - n:7 ( gn 7 vo bin nh X ) hoc n: (7) - Nhp biu thc: X2 + 3X 12hoc nhp biu thc: Ans2 + 3Ans - 12 - n:KQ: y = 58 C2: - Nhp biu thc: 23 12 y x x = + , n: 312 - Lu biu thc: + n my hi X? n 7 KQ: y = 58 + n my hi X? n 8 KQ: y = 76 V d 2: Tnh I = 2 323 2 56x y xz xyzxy xz ++ vi x = 2,41;y = -3,17;z = 43 Gii: n: 2,41 ( gn x = 2,41 vo nh X ) -3,17( gn y = -3,17 vo nh Y ) 43 ( gn z = 43 vo nh A ) Ghi vo mn hnh: ( 3X2Y 2XA3 + 5XYA) ( 6XY2 +XA) V nKQ: I = -0,7918 BI TP: 1)Tnh gi tr cc biu thc a)A = 25 28 49 x x +vi4; 5; 10 x x x = = = b)B = 3 25 3 6 4 x x x + +vi6; 12; 21 x x x = = =c)C = 3 28 60 150 125 x x x + vi7, 4 x = ;43x=d)D = 3 22 5 3 1 x x x + + vi 2, 23 x = 2)Tnh gi tr ca biu thc STOShiftX = = AlphaXYAlpha=Alphax2 +XAlpha - CALC = CALC = STOShiftX STOShift Y STOShift A = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T25

a) A = 5 4 23 23 2 3 14 3 5x x x xx x x + + +khi 1,8165 x =b) B = 2 3 42 3 411x x x xy y y y+ + + ++ + + +khi1, 8597; 1, 5123 x y = =c) C = 7 5 4 35 4 25 13 4 8 52 14 12 7 2x x x xx x x x + + khi2,1413 x =d) D = 3 3 3 3 3 32 2 2 2 2 2a b ab b c bc c a caa b ab b c bc c a ca + + + + khi 1 3; ; 52 2a b c = = =KQ: E = 7 3)Tnh gi tr ca biu thc a)Cho sin= 0,23456( 00 < < 900 ) . Tnh M = 3 3 23 3 3.(1 )( sin ).cotCos Sin tgcos g + ++KQ: M = 0,05735271223b) Bit Cos2= 0,5678( 00 < < 900 ). Tnh N = 2 3 2 33 3 4.(1 cos ) cos .(1 sin )(1 )(1 cot ) 1 cosSintg g + + ++ + +KQ: N = 0,280749911 c) Cho bit tg = tg350 .tg360 .tg370 tg520 .tg530 ( 00 0x ( hay tam thc bc hai x2 + x + 3 c1 4 3 = = nn v nghim ) Suy ra R(x) ch c duy nht mt nghim x = 2. 10)Cho a thc P(x) = 6x3 7x2 16x + m. a)Vi iu kin no ca m th a thc P(x) chia ht cho 2x + 3. b)Vi m tm c cu a. Hy tm s d r khi chia a thc P(x)cho 3x 2. CALC = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T31

c)Vi m tm c cu a. Hy phn tch thc P(x) ra tch ca cc tha s bc nht. d)Tmm v n hai a thc P(x) = 6x3 7x2 16x + mv Q(x) = 2x3 5x2 13x + n cng chia ht cho x - 2 e)Vi n tm c cu trn, hy phn tch ca cc tha s bc nht. Gii: a) P(x) chia ht cho 2x + 3 th P( 32) = 0m = 12. b)Chia a thc P(x) = 6x3 7x2 16x + 12 cho 3x 2 6-7-1612 23 6-3-180 2-1-6 Ta c P(x) = (3x 2)(2x2 x 6) v s d r = 0 c)P(x)= (3x 2)(2x + 3)(x 2). d) hai a thc P(x) = 6x3 7x2 16x + m v Q(x) = 2x3 5x2 13x + n cng chia ht cho x 2 th P(2) = 0 v Q(2) = 0 Suy ra m = 12, n = 30 e)athcQ(x)=2x35x213x+30chiachox2nnchia Q(x)cho x 2 ta c. Q(x) =(x 2)(2x2 x 15). V 2x2 x 15 = 2x2 6x + 5x 15 = (x 3)2x + 5(x 3) = (x 3)(2x + 5). Vy Q(x) = (x 2)(x 3)(2x + 5) 11)Cho a thcP(x) = x5 + ax4 + bx3 + cx2 +dx + e. Bit P(1) = 1,P(2) = 4, P(3) = 9, P(4) = 16, P(5) = 25. a)Tnh cc gi tr P(6), P(7), P(8), P(9) b)Vit li a thc P(x) vi cc h s l s nguyn. Gii: a) Ta c P(1) = 1, P(2) = 4, P(3) = 9, P(4) = 16, P(5) = 25. Xt a thc Q(x) = P(x) x2. D thy Q(1) = 1, Q(2) = 4, Q(3) = 9, Q(4) = 16, Q(5) = 25. Suy ra 1; 2; 3; 4; 5 l nghim ca a thc Q(x). V h s ca x5 = 1 nn suy ra Q(x) c dng: Q(x) = (x 1)(x 2)(x 3)(x 4)(x- 5) Nn Q(6) = (6 1)(6 2)(6 3)(6 4)(6- 5) = P(6) 62. Suy ra P(6) = 62 + 5! = 156. Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T32

Tng t P(7) = 72 + 6! = 769. P(8) = 82 + 7!2! = 2584. P(9) = 92 + 8!3! = 6801. b)P(x) = (x 1)(x 2)(x 3)(x 4)(x- 5) + x2. P(x) = x5 15x4 + 85x3 284x2 + 274x 120. 12)Cho a thc Q(x) = x4 + mx3 + nx2 + px + q v cho bit Q(1) = 5; Q(2) = 7; Q(3) = 9; Q94) = 11. Tnh cc gi tr Q(10); Q(11); Q(12); Q(13). Gii: Nhn xt: Q(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 P(x) = Q(x) (2x + 3). Ta c P(1) = P(2) = P(3) = P(4) = 0. iu ny chng t 1; 2; 3; 4 l nghim ca a thc P(x). Suy ra: P(x) = (x 1)(x 2)(x 3)(x 4)= Q(x) (2x + 3). Nn P(10) = 9.8.7.6 = Q(10) ( 2.10 + 3). Hay Q(10) = 2.10 + 3 + 9.8.7.6 = 2.10 + 3 + 9!5! = 3047. Tng t: Q(11) = 2.11 + 3 +10!6! = 5065. Q(12) = 2.12 + 3 +11!7! = 7947. Q(13) = 2.13 + 3 +12!8! = 11909. 13)Cho a thc 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 cc gi tr P(6), P(7), P(8), P(9), P(10), P(11). Gii: t Q(x) = 2x2 + 1. Khi Q(1) = 3, Q(2) = 9, Q(3) = 19, Q(4) = 33, Q(5) = 51. iu ny chng t a thc (bc 5) R(x) = P(x) Q(x) c 5 nghim 1; 2; 3; 4; 5. Vy : P(x) = Q(x) + (x 1)(x 2)(x 3)(x 4)(x 5). Do : P(6) = 2.62 + 1 + 5!= 193 P(7) = 2.72 + 1 + 6!= 819 Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T33

P(8) = 2.82 + 1 + 7!2!= 2649 P(9) = 2.92 + 1 + 8!3!= 6883 P(10) = 2.102 + 1 + 9!4!= 15321 P(11) = 2.112 + 1 + 10!5!= 30483 14)ChoathcP(x)bc4chsbccaonhtl1vthamn P(1) = 3; P(3) = 11; P(5) = 27; P(5) = 27; P(7) = 51. Tnh gi tr ca P(-2) + 7 P(6). Gii: Nhn xt: P(1) = 3 = 12 + 2;P(3) = 11 = 32 + 2; P(5) = 27 = 52 + 2; P(7) = 51 = 72 + 2. Xt a thc Q(x) = P(x) ( x2 + 2) Ta c Q(1) = Q(3) = Q(5) = Q(7) = 0. iu ny chng t 1; 3; 5; 7 l nghim ca Q(x). Suy ra Q(x) = (x 1)(x 3)(x 5)(x 7) Nn P(x) = Q(x) + x2 + 2 =(x 1)(x 3)(x 5)(x 7) + x2 + 2 Do P(-2) = 951 v P(6) = 23. Vy: P(-2) + 7P(6) = 951 + 7.23 = 1112. Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T34

DNG 12: DY S I/DysLucas:DysLucas ldystng qut ca dyFibonaci:Ccs hngcantuntheoquylutu1=a;u2=b;un+1=un+un-1vimin2. trong a, b l hai s ty . Vi a = b = 1 th dy Lucas tr thnh dy Fibonaci. Dng 1:u1 = a; u2 = b ( a, b ty ).Tnh: un+1 = un +un-1 vi mi n 2 Phng php: - C1: + n: ba u3 + Lp: a u4, u6 , . . . u5, u7 , . . . - C2: + Gn: D = 2 ( bin m ) A = a( S hng u1) B = b( S hng u2) + Ghi vo mn hnh: D = D + 1: A = B + A : D = D + 1 : B = A + B + n: ta c u3, u4, u5, , un V d 1: Vi u1 = 1; u2 = 3. Tnh: un+1 = un +un-1 vi mi n 2 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, V d 2: Vi u1 = -3; u2 = 4. Tnh: un+1 = un +un-1 vi mi n 2 -3, 4, 1, 5, 6, 11, 17, 28, 45, 73, 118, .. V d 3: Vi u1 = -1; u2 = -5. Tnh: un+1 = un +un-1 vi mi n 2 -1, -5, -6, -11, -17, -28, -45, .. V d 4: Vi u1 = 1; u2 = -5. Tnh: un+1 = un +un-1 vi mi n 2 1, -5, -4, -9, -13, -22, -35, -57, -92, -149, . BI TP: 1)Cho dy s u1 = 144; u2 = 233; .; un+1 = un +un-1 vi mi n 2. Tnh u12, u37, u38, u39. KQ: u12 = 28657; u37 = 4807526976; u38 = 7778742049; u39 = 12586269025 ( tnh bng tay ) 2)Cho u1 = 2002, u2 = 2003 v un+1 = un +un-1 vi mi n 2.Xc nh u5, u10 ? KQ: u5 = 10013, u10 = 110144. STOShiftA + STOShiftM +ALPHAA +STOShiftA+ALPHAM STOShiftM = Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T35

II/ Dy s Fibonaci ( Dy Lucas ) suy rng tuyn tnh c dng: Dng 2: u1 = a; u2 = b (a,bty)vun+1=m.un+n.un-1vi mi n 2. Phng php: - C1: + n: b m n a u3 + Lp:mn u4, u6 , . . . mn u5, u7 , . . . - C2: + Gn: D = 2 ( bin m ) A = a( S hng u1) B = b( S hng u2) + Ghi vo mn hnh: D = D + 1: A = m.B + n.A : D = D + 1 : B = m.A + n.B+ n: ta c u3, u4, u5, , un BI TP: 1)Chou1=2;u2 =3vun+1=4.un+5.un-1vimin2.Xc nh u7, u8?KQ: u7 =13022, u8 = 65103. 2) Cho u1 = 2; u2 = 9 v un+1 = 19.un + 45.un-1 vi mi n 2. Xc nh u5, u10? KQ: u5 = 113.661, u7 = 50.732.586,u8 = 1071961389, u9 = 22650232761 ( tnh bng tay)u10 = 19u9 + 45.u8 = 478592684964. ( tnh bng tay)3)Chou1=30;u2 =4vun+1=19.un+75.un-1vimin2. Xc nh u5, u7? KQ: u5 = 1.019.836, u7 = 508.052.446,4)Chou1=3;u2 =2vun=2.un-1+3.un-2vimin3.Xc nh u21?KQ: u21 = 4358480503. 5) Cho dy s sp xp theo th tvi u1 = 2; u2 = 20 v u3 c tnh theo cng thc un+1 = 2.un + un-1 vi mi n 2. a) Vit quy tnh bm phm lin tc tnh gi tr ca un vi u1 = 2; u2 = 20. b) Xc nh u22, u23, u24, u25? Gii: STOShiftA x STOShiftB +ALPHAAxSTOShiftA+ALPHAB STOShift B = + x x xx Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T36

a) + Gn: D = 2 ( bin m ) A = 2( S hng u1) B = 20( S hng u2) + Ghi vo mn hnh: D = D + 1: A = 2.B + A : D = D + 1 : B = 2.A + B+ n: ta c u3, u4, u5, , un b) u22 = 804.268.156,u23 = 1.941.675.090 u24 = 4.687.618.336,u25 = 11.316.911.762 Ch : u25 = 2.u24 + u23 ( Tnh tay ). 6)Choa1=2000;a2 =2001van+2=2.an+1-an+3vimin1.Xc nh a100?Gii: + Gn: D = 2 ( bin m ) A = 2000( S hng u1) B = 2001( S hng u2) + Ghi vo mn hnh: D = D + 1: A = 2B A + 3 : D = D + 1 : B = 2A B +3+ n: ta c u3, u4, u5, , un KQ: a100 = 16.652 III/ Dy Fibonacoci ( dy Lucus ) suy rng bc hai dng: Dng 3: u1 = a; u2 = b ( a, b ty ) v un+1 = u2n + u21 n vi mi n 2 Phng php: - C1: + n: b a u3 + Lp: u4, u6 , . . . u5, u7 , . . . - C2: + Gn: D = 2 ( bin m ) A = a( S hng u1) B = b( S hng u2) + Ghi vo mn hnh: D = D + 1: A = B2 + A2 : D = D + 1 : B = A2 + B2

+ n: ta c u3, u4, u5, , un = = STOShiftA +STOShift B +ALPHAA + STOShiftA+ALPHA BSTOShift B = 2x2x2x2x2x+ 2xSng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T37

BI TP:1) Cho u1 = u2 = 1 v un+1 = u2n + u21 n vi mi n 2. Thc hin trn my theo qui trnh trn ta c dy: 1, 1, 2, 5, 29, 866, 750797, 563696885111. 2)Cho u1 = u2 = 1 v un+1 = u2n - u21 n vi mi n 2. Xc inh u100? KQ: u100 = -1IV/ Dy Lucas bc ba c dng: Dng 4: u1 = a , u2 = b , u3 = c( a, b, c ty ) un+1 = un + un-1 + un-2 vi mi n 3 Phng php: - C1: + n: b ( a u2 vo nh ) c( a u2 vo nh ) a u4 + Lp: u5, u8 , . . . u6, u9 , . . . u7, u10 , . . . - C2: + Gn: D = 3 ( bin m ) A = a( S hng u1) B = b( S hng u2) C = c ( S hng u3 ) + Ghi vo mn hnh:D = D + 1: A = C + B + A : D = D + 1 : B = A + C + B : D = D + 1 : C = B + A + C + n: ta cu4, u5, u6 ,, un V d: Dy Fibonaci bc ba: u1 = u2 = u3 = 1, un+1 = un + un-1 + un-2 vi mi n 3. Thc hin qui trnh trn ta c dy: 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, STOShiftA+ALPHAB +STOShiftA= STOShift B ABALPHA B +ALPHAA + STOShiftCALPHAA+ALPHAC + STOShiftB ALPHAB+ALPHAC +STOShiftB ALPHABSng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T38

BI TP: 1)Cho u1 = 4, u2 = 7, u3 = 5 v un = un -1 + un -2 + un -3 vi mi n 4. Xc nh u30 ?2)Chou1 = 3,u2=2, u3 =1930vun = un-1 + un-2 - un-3 vi mi n 4. Xc nh u78 ? 3)Chou1 = 7,u2=5, u3 =1954vun = un-1 - un-2 + un-3 vi mi n 4. Xc nh u54 ? 4)Cho u1 = 30, u2 = 4, u3 = 1975 v un = un -1 + un -2 - un-3 vi mi n 4. Xc nh u33 ? 5)Cho u1 = 20, u2 = 11, u3 = 1982 v un = un -1 + un -2 + un -3 vi mi n 4. Xc nh u26 ? V/ Dy Lucas bc ba suy rng c dng: Dng 5: u1 = a , u2 = b , u3 = c( a, b, c ty ) un+1 = m.un + n.un-1 + p.un-2 vi mi n 3 Phng php: - C1: + n: b ( a u2 vo nh ) c( a u2 vo nh ) mnap u4 + Lp: mn p u5, u8 , . . . mn p u6, u9 , . . . mn p u7, u10 , . . . - C2: + Gn: D = 3 ( bin m ) A = a( S hng u1) B = b( S hng u2) C = c ( S hng u3 ) + Ghi vo mn hnh:D = D + 1: A = mC + nB + pA : D = D + 1 : B = mA + nC + pB : D = D + 1 : C = mB + nA + pC + n: ta cu4, u5, u6 ,, un STOShiftA+ALPHAB +STOShiftA= STOShift B ABALPHA B + ALPHAA + STOShiftCALPHAAxx x x x x +ALPHAC + STOShiftB ALPHABx x x +ALPHAA + STOShiftC ALPHACx x x Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T39

V d: u1 = 1 , u2 = 2 , u3 = 3 v un+1=2un+3un-1 +4un-2 vi mi n 3. Thc hin quy trnh trn ta c dy: 1, 2, 3, 16, 49, 158, 527, BI TP: 1)Cho u1 = 4, u2 = 7, u3 = 5 v un = 2un -1 - un -2 + un -3 vi mi n 4. Xc nh u30 ? KQ: u30 = 20929015 2)Cho u1 = 3, u2 = 2, u3 =1945 v un = 3un -1 - 2un -2 + 2008un -3 vi mi n 4. Xc nh u10 ? VI/TnhshngthncadysFibonaccitheocngthc nghim tng qut: 1 1 5 1 52 2 5n nnu | | | |+ = || || \ \ Nhp:1 52 15 25 Bm: my hin X ?Thay X = n th tnh c un . V d: Cho dy s : 3 5 3 52 2n nnu| | | |+ = + || ||\ \ . Tnh u6, u18? KQ: u6 = 322,u18 = 33385282 ALPHA ((( + ) ) X-(( - ) ) ALPHA X ) CALCSng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T40

KT LUN Trnylnhngdngbitpmquaqutrnhnghincu ging dy, tham gia dy bi dng, dy hc t chn, bn thn ti tng hp li c. Tht ra y l nhng bi ton m ta c th bt gp cc sch ton, thi, Vic phn chia cc dng bi tp ny l cho hc sinh d nh, d thc hnh. hc sinh t rn luyn k nng thchnh gii ton bng my tnh cm tay. Vi suy ngh nh vy. Ti tin tng mi hc sinh u t hc, t thc hnh trn my tnh cm tay c kt qu. V kh nng v thi gian c hn nn sng kin ny xin tm dng y. Rt mong s gp ca cc ng ch, ng nghip sng kin ny c pht huy v c m rng hn na. Ba T, ngy 25 thng 4 nm 2008 NGI VIT Trn Ngc Duy

Sng kin ci tin k thut : Cc dng bi tp ton gii bng my tnh cm tay Ngi vit: Trn Ngc DuyGV trng THCS DTNT Ba T41

TI LIU THAM KHO 1.Hngdnsdngvgiiton6,7,8,9,10,11,12cav THPT. 2.Hng dn thc hnh Ton trn MTBT Casio Fx 500MS, Fx 570 MS ca v THPT. 3.Gii ton trn my tnh in t Casio Fx 500MS, Fx 570 MS ca TS T Duy Phng NXBGD. 4.Mt s thi cc cp v thi khu vc .