Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn A. t vn My tnh in t l mt trong nhng cng c tch cc trongvic dy v hc ton. Nh c my tnh in t m nhiu vn c coi l kh trong dy hc ton ( v d gii phng trnh bc hai, phng trnh ba, phng trnh v t, chui s, cc nh l s hc...) ta c th ging dy cho hc sinh THCS mt cch d dng. Cc quy trnh thao tc trn my tnh in t b ti c th coi l bc tp dt ban u hc sinh dn dn lm quen vi thut ton v lp trnh trn my tnh c nhn. B Gio Dc v o to t chc cho THCS v THPH cc k thi hc sinh gii gii ton trn my tnh casio. Phng Gio Dc v o to Cm xuyn t chc cc k thi hc sinh gii gii ton trn my tnh casio cp huyn v tham gia k thi hc sinh gii gii ton trn my tnhcasiocpTnhsongktqucnkhimtnsovi cchuyn mnh nhCan Lc,Hng lnh, TP H Tnh... Mt s bi d thi ca hc sinh kt qu cn thp, hoc bi lm thiu tnh chnh xc, cch trnh by si sc, ngu hng, cc thut ton trn my tnh cha c vn dng vo bi lm...Vi l do v nim am m ton hc trn my tnh v thc trng qua nhiu nm ging dy v bi dng hc sinh gii, ti mnh dn bin son tp ti liu bi dng HSG gii ton trn my tnh Casio ny lu hnh ni b. Mc ch ca ti liu ngoi hng dn chi tit cc thao tc tnh ton, Cc dng bi tp ton gii bng my tnh cm tay m cn trnh by ngha ton hc ca cc bi ton.V vy nhiu kin thc ton hc ngoi chng trnh vn c a vo.Vic trnh by cc kin thc ton hc, tnh chnh xc kt qu trong tng php tnh c c bit ch trng. Bi l iu c bn v ct li ca vic s dng my tnh. Ngi vit xin c trao i cng bn c qua ti: gii ton trn my tnh casio ti gm ba phn:Phn I: Hng dn s dng my tnh casio Fx:500 MS v Fx:570 MSPhn II: Cc dng bi tp: Gii ton trn my tnh CasioPhn III: Mt s thi Gii ton trn my tnh Casio ( h THCS ) 1Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Trong khi bin son mc d rt c gng song khng th trnh khi nhng thiu st. Rt mong nhn c s gp chn thnh ca qu thy c v bn c. Xin chn thnh cm nCm xuyn, ngy 07/10/2010 Tr-ngNgcBnB. nidungPhn I:H ng dn s dng my tnh casio Fx:500 MS v Fx:570 MS A /. my tnh casio Fx:500 MS I/ Cc phm v cch bm my s dng chung cho c my Fx:500 MS v Fx:570 MS : 1)Cc loi phm:+ Phm trng:Bm trc tip ( v d:5ta n5 =5) + Phm vng: Bm SHIFT + Phm vng (V D:481 , ta bm 4 SHIFT x 81 =481)+Phm :Bm ALPHA + Phm (v d:A, ta bmALPHA A2) M tt my:+ M my:BmON+ Tt my: Bm SHIFT + OFF+ Xo mn hnh khi lm tnh :- Bm AC - BmSHIFTCLR2= -BmSHIFTCLR3=+ kim tra li ta dng cc phm + sa li: -Dng phm >< di chuyn.2Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn -Bm phm DEL xo k t ang nhp nhy- Bm phm SHIFT + IN Schn k t nh stII / . my tnh casio Fx:500 MS: *) Ch Mode:Nhm n nh ngay t u loi hnh tnh ton, loi n v o,dng s biu din kt qu, ch s c ngha,sai s lm trn...ph hp vi gi thit ca bi tona) Bm Mode ( 1 ln) hinh man3 2 1REG SD COMP+ Bm Mode 1 Lm cc php tnh thng+ Bm Mode 2 Lm thng k mt bin+ Bm Mode Lm thng k hai binb) Bm Mode Mode( 2 ln) hinh man

1EQR( gii phng trnh )+ Bm Mode Mode 1 hinh manUNKNO S ( n ) - Bm tip 2 Gii h phng trnh bc nht hai n- Bm tip 3 Gii h phng trnh bc nht ba n+Bm Mode Mode 1 hinh manDegree(bc)- Bm tip 2 Gii phng trnh bc hai mt n- Bm tip 3 Gii phng trnh bc ba mt nc) Bm Mode Mode Mode ( 3 ln) hinh man

3 2 1Gra Ded Deg+ Bm Mode Mode Mode 1 Chn n v o gc l + Bm Mode Mode Mode 2 Chn n v o gc l raian+ Bm Mode Mode Mode 1 Chn n v o gc lgradd) BmMode Mode Mode Mode( 4 ln) hinh man 3 2 1Norm Sci Fix3Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn + Bm Mode Mode Mode Mode 1 C chn s s l thp phn+ Bm Mode Mode Mode Mode 2 C chn hin s dng : a.10n+ Bm Mode Mode Mode Mode 3 C chn s dng thnge) Bm Mode Mode Mode Mode Mode( 5 ln) hinh man 1DispBm tip 1 hinh man 2/1/ c d c ab+ Bm Mode Mode Mode Mode Mode 11 kt qu di dng hn s+Bm ModeModeModeModeMode12ktqudi dng phn s+ Bm Mode Mode Mode Mode Mode 1 hinh man

2 1Comma Dot+ Bm Mode Mode Mode Mode Mode 1 1 Ch du cch phn nguyn v phn thp phn l du (.)+ Bm Mode Mode Mode Mode Mode 1 1 Ch du cch phn nguyn v phn thp phn l du (,)III/. Cch lm mt bi thi Gii ton trn my tnh casio"*Quy nh: 1. Yu cu cc em d thi ch dng my Casio fx 500 MS, Casio fx 570 MS, Casio fx 500 ES, Casio fx 570 ES gii.2. Nu khng qui nh g thm th cc kt qu trong cc thi phivit 10 ch s hin trn mn hnh my tnh. 3. Trnh by bi gii theo cc bc sau :- S lc li gii ( li gii vn tt)- Thay s vo cng thc (nu c)4Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn - Vit quy trnh n phm- Kt qu*Nhn xt : Qua cc thi tnh, khu vc t chc cc nm gn y.Chng ta c th nhn thi Gii ton trn my tnh Casiotheo cc nh hng sau y :1. Bi thi hc sinh gii" Gii ton trn my tnh Casio " phi l mt bi thi Hc sinh gii ton c s tr gip ca my tnh th nghim tm ra cc quy lut ton hc hoc tng tc tnh ton.2. ng sau cc bi ton Gii trn my tnh Casio n cha nhng nh l, thut ton, thm ch c mt l thuyt ton hc ( s hc, dy tru hi...)` 3. Pht huy c vai tr tch cc ca ton hc v my tnh trong gii cc bi ton thc tPhn II: Cc dng bi tp ton gii bng my tnh cm tay I/. Mt s dng ton xc nh s (s hc):1/ . Loi 1. Tnh chnh xc kt qu php tnh:.Ph ng php : Da vo cc tnh cht sau:1) S 8 7 4 3 2 1... a a a a a a =4 3 2 1a a a a. 104+8 7 6 5a a a a2)Tnh cht ca php nhn: ( A + B)( C + D) = AC + AD +BC + BD3)Kt hp tnh trn my v lm trn giy..Mctiu:Chiaslnthnhnhngsnhmkhngtrnmn hnh khi thc hin trn myv d1: tnh chnh xc kt qu ca php tnh sau: A = 12578963 x 14375b) Tnh chnh xc Ac) Tnh chnh xc ca s: B = 12345678925Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn d) Tnh chnh xc ca s: C = 10234563Gii:a) Nu tnh trn my s trn mn hnh nn ta lm nh sau:A = 12578963.14375 = (12578.103 + 963).14375 = 12578.103.14375 + 963.14375* Tnh trn my:12578.14375 = 180808750 12578.103.14375 = 180808750000 * Tnh trn my: 963.14375 = 13843125T ta c: A = 180808750000 +13843125 = 180822593125 Vy A = 12578963 x 14375 = 180822593125 b) B =1234567892=(123450000 + 6789)2= (1234.104)2+ 2.12345.104.6789 + 67892Tnh trn my:123452=152399025; 2x12345x6789=167620410; 67892= 46090521 Vy: B = 152399025.108 + 167620410.104 + 46090521 = 15239902500000000 +1676204100000 46090521 = 15241578750190521d) C = 10234563 = (1023000 + 456)3= (1023.103 + 456)3= 10233.109 + 3.10232.106.456 + 3.1023.103.4562 + 4563Tnh trn my:10233 = 1070599167; 3.10232.456 = 14316516723.1023.4562 = 638155584;4563= 94818816Vy (tnh trn giy): C = 1070599167000000000 6Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 1431651672000000 +638155584000 94818816 = 1072031456922402816Bi tp p dng:Bi 1: Tnh kt qu ng ca cc tch sau: a) M = 2222255555 x 2222266666b) N = 20032003 x 20042004ps:a) M=4938444443209829630b) N= 401481484254012Bi 2:Tnh kt qu ng ca cc php tnh sau: a) A= 1,123456789 - 5,02122003b) B = 4,546879231 + 107,3564177895 ; c) C= 52906279178,48 : 565,432Bi 3: Tnh chnh xc tng: S =1.1! +2.2! +3.3! +4.4! +... + 16.16! *Hng dn: Ta cn.n! = ( n + 1 1).n! =(n + 1).n! n!= (n+1)! n! *p s:S = 355687428095999Bi 4 : a)Tnh bng my tnh: Q = 1 + 22+ 32+ . . . + 102.b)C th dng kt qu tnh tng : K = 22 2 2 220 . ... 6 4 . + + + +m khng dng my tnh .hy trnh by li gii y. p s: a) Q = 385;b) K = 1540Bi 5: Tnh chnh xc ca s A = 21210 23 _ + ,Nhn xt:10 23k+l s nguyn c (k - 1) ch s 3, tn cng l s 47Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn

210 23k _ + ,l s nguyn gm k ch s 1, (k - 1) ch s 5, ch s cui cng l 6* Ta d dng CM c v tnh c kt qu l: A = 111111111111555555555556 2/. loi 2:Tm s d ca php chia ca sacho sb* Phng php:1/.i vi s b chia ti a c 10 ch s: Th s d ca A: B = A - B.1]1

BA (trong 1]1

BA l phn nguyn ca A cho2/.Khi s b chia Aln hn 10 ch s:Khi s b chia Aln hn 10 ch s tangt rathnhhai nhm. Nhm u 9 ch s u( k t b tri). tm c s d nh phn 1). Rivit tip sau s dcn li ti a 9 ch s ri tm s dln hai. Nu cn na th lm lin tip nh vy.*nh l: Vi hai s nguyn bt k a v b, b 0, lun tn ti duynht mt cp s nguyn q v r sao cho:a = bq + r v 0 r < |b|* T nh l trn cho ta thut ton lp quy trnh n phm tm d trong php chia a cho b: aSHIFTSTOA bSHIFTSTOBALPHA A ALPHA B = (1]1

ba) ALPHA B - ALPHA B1]1

ba=(Kqu: r =...)V d1:a) Vit mt quy trnh n phm tm s d khi chia 18901969 cho 3041975 Tnh s db) Tm s d trong php chia: 815 cho 2004Gii:a) Quytrnhnphm: 18901969 SHIFT STO A 3041975 SHIFTSTO B8Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn ANPHA AANPHA B = (6,213716089)SHIFT A- 6 B =(650119)Vy s d l: r = 650119b) Ta phn tch: 815 = 88.87Ta c:881732(mod2004)87 968(mod2004)815 1732 x 968 (mod2004)1232(mod2004) Vy s d l: r = 1232Bi tp p dng:Bi 1:a) Vit quy trnh n phm tm s dkhi chia 3523127 cho 2047.b) Tm s d.Tm thng v s dtrong php chia: 123456789 cho 23456Bi 2: Tm s d trong php chia: a) 987654321 cho 123456789p s: r = 93/. loi 3: Tm UCLN BCNN ca a v b:*Phng php:1.Vi cc s a v b nh hn 10 ch s th ta dng tnh cht rt gn phn s,,,,..bam bm aba Trong (a,; b ) = 1. Khi UCLN (a;b) = m 2. Vi cc s a v b ln hn 10 ch s th ta dng thut ton LE:Tm UCLN(a;b) vi a b ta c thut ton sau :9Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn

0....1 11 23 3 2 12 2 111+ + + + + + n n nn n n nq r rr q r rr q r rr q r br q b aS dcui cng khc 0 l r n chnh l UCLN (a;b) hay : r n = UCLN (a;b) * Ch :BCNN(a;b) = ) ; (.b a UCLNb aV d 1:Tm UCLN ca hai s:a = 24614205, b = 10719433Gii: *C 1: +) Ta c:,,,,..bam bm aba Trong (a,; b ) = 1. Khi UCLN (a;b) = m +) Quy trnh m my:24614205 SHIFTSTOAALPHA A : 10719433 = (1155/503) ALPHA A : 1155 = ( 21311)Vy UCLN(a;b) = 21311* C 2: +)Theo thut ton le tm s d trong php chia s a cho b ta c:+) quy trnh m mylin tc: (Bn c c th d dng lm c v kt qu UCLN(a, b) = 21311)3.Xc nh s c s ca mt s t nhin n *:nh l: Cho s t nhin n, n > 1, gi s khi phn tch n ra tha s nguyn t ta c:10Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 1 21 2... ,ke e ekn p p p vi k, ei l s t nhin v pi l cc s nguyn t tho mn:1 < p1 < p2 n.Bi 13:Tm tt c cc s t nhin: x1 ; x2 ; ... ; x8Sao cho8 2 1... x x x =( )2 1x x4p s H ng dn li gii: Bi 1: p s: - S ln nht l 129304;- S nh nht l 1020344S 2: p s: - S ln nht l 2939475;- S nh nht l: 1030425S 3: p s: a =S 4: p s: x = Bi 5:*S lc li gii::Gi s cn tm l: 1 2 3 4 5 6n a a a a a a .- t 1 2 3x a a a . Khi y 4 5 61 a a a x +v n = 1000x + x + 1 = 1001x + 1 = y2hay (y - 1)(y + 1) = 7.11.13x.16Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Vy hai trong ba s nguyn t 7, 11, 13 phi l c ca mt trong hai tha s ca v tri v s cn li phi l c ca tha s cn li ca v tri.Dng my tnh, xt cc kh nng i n p s:n = 183184 ;328329 ; 528529 ;715716.Bi 6:*S lc li gii:T gi thit, ta c:x = 393.q1 + 210 x -210chia ht cho 393 x = 655.q2 + 210 x -210chia ht cho 655x -210chia ht cho BCNN (393 ; 655) = 1965 x -210 = 1965.k ; (k = 1, 2,...) hay x = 1965k + 210- T gi thit 10000 < x < 15000 10000 < 1965k + 210 < 15000hay 9790 < 1965k < 14790 5 k < 8.Tnh trn my:Vik = 5, ta c: x = 1965.5 + 210 = 10035Vik = 6, ta c: x = 1965.6 + 210 = 12000Vik = 7, ta c: x = 1965.7 + 210 = 13965Vy cc s phi tm l:10035,12000,13965Bi 7: *S lc li gii:V cc s 5, 7, 9 i mt nguyn t cng nhau nn ta phi tm cc ch s x, y, z sao cho579xyzchia ht cho 5.7.9 = 315.Ta c 579xyz= 579000 +xyz= 1838.315 + 30 +xyz30 +xyz chia ht cho 315. V 30 30 +xyz < 1029 nn (Dng my tnh tm cc bi ca 315 trong khong (30 ; 1029):- Nu 30 +xyz= 315 thxyz= 315 - 30 = 285- Nu 30 +xyz= 630 thxyz= 630 - 30 = 600- Nu 30 +xyz= 945 thxyz= 945 - 30 = 91517Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Vy ta c p s sau:x y z2 8 56 0 09 1 5Bi 8: *S lc li gii:- Gi s s cn tm c n + 1 ch s.- T iu kin 1) s dng: 1 2... 6na a a- T iu kin 2), ta c: 1 26 ...na a a= 4.1 2... 6na a a (*)- t 1 2...na a a a , th: 1 2... 6na a a= 10a + 6 1 26 ...na a a= 6.10n + a- Khi (*) tr thnh:6.10n + a = 4.(10a + 6) 2.(10n - 4) = 13a (**)ng thc (**) chng t v tri chia ht cho 13. V (2 ; 13) = 1 nn: 10n - 4 chia ht cho 13.Bi ton quy v: Tm s t nhin n nh nht (10n - 4) chia ht cho 13, khi tm ra s a v s cn tm c dng: 10a + 6.Th ln lt trn my cc gi tr n = 1; 2;... th (10n - 4) ln lt l:6, 96, 996, 9996, 99996,... vsutinchiaht cho13l: 99996.Khi a = 15384 S cn tm l: 153846.Bi 9: *S lc li gii::a) Lp cng thc (2n + 7) : (n + 1) trn my v th ln lt n = 0, 1, 2,...ta c n = 0 v n = 4 th 2n + 7 chia ht cho n + 1.Chng minh vi mi n 5, ta u c 2n + 7 khng chia ht cho n + 1, tht vy:18Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn (2n + 7) M(n + 1) [(2n + 7) - 2(n + 1)] M(n + 1) 5 M(n + 1) n 5.Vy s n cn tm l 0 hoc 4.b) Tng t ta c: n = 4 hoc n = 6.Bi 10:*S lc li gii::Nhn xt:1) n3 c tn cng l 11 th n c tn cng l s 1. Th trn my cc s:11, 21, 31,...81, 91 c duy nht s 71 khi lu tha bc ba c tn cng l 11.2) n3 c tn cng l 111 th n c phi tn cng l s 471. (Th trn my vi cc s: 171, 271, 371,...871, 971 )3) n3 c tn cng l 1111 th n phi c tn cng l s 8471. (Th trn my vi cc s: 1471, 2471, 3471,...8471, 9471 )- Gi s m l s ch s ng gia cc s 111 v 1111:+ Nu m = 3k, k Z+, th: 111 x 103k+4 < n3 = 111...1111< 112 x 103k+4({ { {4 3 43 3111000...000000 111 ... 1111 112000...000000m kk k< 652 n = 652* Quy trnh trn MT Casio fx: 500 MS21Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn (Thut ton:Xt hiu 1,01A- A , gn cho A cc gi tr t nhin: 0, 1, 2,... dng li khi hiu trn chuyn t (-) sang (+))- Gn cho nhAgi tr t nhin u tin:0SHIFT STO A- Lp cng thc tnh hiu 1,01A- A v gn gi tr nh bi s t nhin k tip: 1,01 ANPHA A-ANPHA A :ANPHA A ANPHA = ANPHA A +1- Lp li cng thc trn:=... =Bi ton kt thc khi chuyn t n = 651 sang n = 652.B i 13 :*S lc li gii: Ta c: 10.000.0008 2 1... x x x =( )2 1x x4 99999999 578 6x x 99Ta ghi ln m hnh ( ) 10556001 574 khng tho mn v trx 6 ; x8Dng phm sa v th cc s t57; 58; ...;98; 99. ta c 3 s: 65; 86; 91Vy ta c 3 b s x1 ; x2 ; ... ; x8 l : 654=17850625 ; 864= 54700816 ;914= 68574961 II. a thc:* Kin thc b rung:22Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 1) Cho a thc P (x) bc n: P (x) = an . xn + an-1 . xn-1 + ... + a1. x +a0 (*)Trong : an ; an-1 ; ...a1; a0 /R ; an 0 Khi :an; an-1; an-2; an-3;... ; a1; a0gi cc h sN ux0m P(x0) = 0 th x0 l nghim ca P(x)2) Khi chia a thcP (x) cho (x - ) lun tn ti mt a thc thng Q(x) v s d r. Hay ta lun c:P(x) = Q(x). (x - ) + r

* Ch : (nh l Bezout)1)N u x = l nghim ca P(x) P(x)M(x - ) 2) Nu x0 l nghim nguyn ca P(x) th x0 c ca a0 3) N u tng cc h s bng 0 th P(x) = 0 c nghim l x = 1 ( Hay P(x) M( x - 1) )4) Nu tng cc h s bc chn bng tng cc h s bc l th P(x) = 0 c nghim l x = -1(Hay P(x) M( x + 1) )* S Horner: (i vi a thc mt bin)Khi chia a thc P(x) cho ( x - ) thng l:bn. xn-1 + bn-1. xn-2 + ...+ b2 . x + b1v c s d l: r . Khi ta c s nh sau:anan-1an-2an-3...... a1a0bnbn-1bn-2bn-3....... b1r = b0 Trong :bn = anbn-1 = . bn + an-1bn-2 = . bn-1 + an-2..........................b1 = . bn-1 + a1b0 = . b1 + a0. Khi : 1).P () = b02).Nu P () = 0 thP(x) M(x - )3).Nu P (x) 0 th P (x) : (x - ) c s d l: r = P () V c thng l:bn. xn-1 + bn-1. xn-2 + ... + b2 . x + b11/. Loi 1:Tnhgi trcaaP(x,y,) khi x = x 0, y = y0; *Phng php:1).Tnh trc tip (Thay trc tip cc gi tr ca x, y vo biu thc ritnh kt qu.2). S dng s Horner ( ch s dng khi bi ton yu cu tm th-ng v gi tr ca a thc ti x = ( r = P() = b0 )23Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn *Trn my tnh: 1).- Gn gi tr x0 vo bin nh M. - Ri thc hin quy trnh 2).-Tnh nh vo bin nhAnsV d 1 :Tnh A = 5 3 41 3 2 32 32 4 5+ + + + x x xx x x x khi x = 1,8165Gii:*Cch 1: Tnh nh vo bin nhAnsBm phm: 1. 8165 2 2( 3Ans ^5 2Ans ^4 3Ans x Ans 1) ( 4Ans ^3 Ans x 3Ans 5) + + + + Ktqa: 1.498465582*Cch 2: Tnh nh vo bin nhXBm phm:1 . 8165 SHIFT STO X2 2( 3ALPHA X ^5 2ALPHA X ^4 3ALPHA X x ALPHA X 1) ( 4ALPHA X ^3 ALPHA X x 3ALPHA X 5) + + + + Ktqa: 1.498465582* Ch : Trong cc k thi HSG thng vn hay c dng ton ny. c bit cccucthi cphuyn. Khnnngtnhtondnnsai sthng khng nhiu. Nhng biu thc qu phc tp nn tm cch chia nh bi ton. Trnh tnh trng php tnh vt qu gii hn nh ca my tnh. S dn n kt qu sai ( Kt qu quy trn trn my tnh trong qu trnh thc hin, c trng hp kt qu sai hn). Do vy khng c im trong trng hp ny.Bi tp p dng:Bi 1 :Tnh gi tr biu thc: a. A(x) = 4 3 2x 5x 3x x 1 + + khi x = 1,23456b. 5 4 3 2P(x) 17x 5x 8x 13x 11x 357 + + khi x = 2,185672/.Loi 2: Tm d trong php chia a thc P(x) cho nhi thcax + b*Phng php:Khi chia a thcP (x) cho (ax + b) lun tn ti mt a thc thng Q(x) v s d r. Hay ta lun c:P(x) = Q(x). (ax + b) + r

P(-ba) = r Vy s dtrong php chia P (x) cho (ax + b) l r = P(-ba)24Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn V d 1 : Tm s d trong php chia:P= 14 9 5 4 2x x x x x x 723x 1,624 + + + Gii:tQ(x) = + + + 14 9 5 4 2x x x x x x 723 Khi s d trong php chia:P= 14 9 5 4 2x x x x x x 723x 1,624 + + + l Q(1,624)*Qui trnh bm my (fx-500MS v fx-570 MS)1. 624SHIFT STO XALPHA X ^14 ALPHA X ^9 ALPHA X ^5 ALPHA X ^4 ALPHA X ^2 ALPHA X 723 + + + Kt qu: r = 85,92136979Bi tp p dng:Bi 1: Tm s d trong php chia 5 3 2x 6,723x 1,857x 6,458x 4,319x 2,318 + ++B i 2:Cho ( )4 4 2xP x 5x 4x 3x 50 + + . a) Tm phn d r1, r2 khi chia P(x) cho x 2 v x-3.b) Tm BCNN(r1,r2)?3/. Loi 3: xc nh tham s m a thc P(x)+mchiaht cho nhi thc a.x+ b

*Phng php:Khi chia a thcP (x) + m cho (ax + b) lun tn ti mt a thc thng Q(x) v s d r. Hay ta lun c:P(x) = Q(x). (ax + b) +m + r P (x) + m chia ht cho (ax + b) th:m +r = 0 m =- r

m =- P(-ba)V d 1: Tm a a thc A(x) = 4 3 2x 7x 2x 13x a + + + +chia ht cho x+6.Gii: *S lc li gii:t P(x) = + + +4 3 2x 7x 2x 13xKhi ta c:A(x) = P(x) + a M d khi chia P(x) chox+6 l: r = P(-6)Vy A(x) Mx+6 th r + a = 0 a = - r = - P(-6) *Qui trnh bm my fx-500MS ( ) 6 SHIFT STO X( ) (ALPHA X ^4 +7ALPHA X3x +2ALPHA X2x +13ALPHA X)Kt qu: a = -222V d 2:Cho P(x) = 3x3 + 17x 625. Tm m P(x) + m2 chia ht cho x + 3 ?25Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Gii:*S lc li gii:Ta c: d khi chia P(x) cho x + 3 l:r = P(-3) P(x) + m2 chia ht cho x + 3Th:m2 =- P(-3) = - ( ) ( )33 3 17 3 625 1 + ] => m =t ( ) ( )33 3 17 3 625 1 + ]*Qui trnh bm my fx-500MS: 3( ) ( 3( ( ) 3) 17( ( ) 3) 625) + xKt qu: m =t 27,513632984/. Loi 4: Tm th ng v s dkhi chia a thc cho nthc: *Phng php: S dng s Horneranan-1an-2an-3......a1a0bnbn-1bn-2bn-3.......b1r = b0 Trong :bn = anbn-1 = . bn + an-1bn-2 = . bn-1 + an-2..........................b1 = . bn-1 + a1b0 = . b1 + a0.Khi : 1).P () = b02).Nu P () = 0 thP(x) M(x - )3).Nu P (x) 0 th P (x) : (x - ) c s d l: r = P () V c thng l:bn. xn-1 + bn-1. xn-2 + ... + b2 . x + b1Chng minh :Ta xt a th bc ba: P(x) =a3x3 + a2x2 + a1x + a0chia chox - Ta c:a3x3 + a2x2 + a1x + a0 = (b3x2 + b2x + b1)(x-) + r = b3x3 + (b2-b3)x2 + (b1-b2)x + (r - b1) T ta c cng th truy hi Horner:b3 = a3 b2= b3 + a2 b1= b2 + a1 b0 = r = b1 + a3.26Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn V d 1: Tm thng v s d trong php chia x7 2x5 3x4 + x 1 cho x + 5.GiiTa c: = - 5; a0 = 1; a1 = 0; a2 = -2; a3 = -3; a4 = a5 = 0; a6 = 1; a7 = -1; b0 = a0 = 1.*Qui trnh bm my fx-500MS:Vy:x7-2x5-3x4+x -1 = (x + 5)(x6 -5x5 + 23x4 -118x3 + 590x2-2590x + 14751) - 73756.5/.Loi5: Phn tch athc theo bc camt nthc*Phng php:S dng s Horner cho n lnpdng n-1 ln s dng s Hornerta phn tch c a thc P(x) bc n theo x-:P(x)=r0+r1(x-)+r2(x-)2++rn(x-)n.V d 1 : Phn tch P(x) =x4 3x3 + x 2 theo bc ca x 3.Gii:Thc hin php chia P(x)=q1(x)(x-)+r0 theo theo s Horner ta c q1(x) v r0. Sau tip tc tm cc qk(x) v rk-1 ta c bng sau:Vy x4 3x3+ x 2 =1 + 28(x-3) + 27(x-3)2+ 9(x-3)3 + (x-3)4.6/. Loi 6: Xc nh 1 -3 0 1 -2 x4-3x2+x-23 1 0 0 1 1 q1(x)=x3+1, r0= 13 1 3 9 28q2(x)=x3+3x+1, r1 = 283 1 6 27q3(x)=x+6, r0= 273 1 9 q4(x)=1=a0,r0= 927 + + + + + + ( ) 5SHIFT STO M1 ALPHA M 0ALPHA M 2ALPHA M ( ) 3ALPHA M 0ALPHA M 0ALPHA M 1ALPHA M ( )1(-5)(23)(-118)(590)(-2950)(14751)(-73756)Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn a thc & tnh gi tr mt s gi tr ca a thc khi bit mt s gi tr ca khc ca n :*Phng php:1). Gii h phng trnh t tm c cc h s2). Tm a th ph trc, ri quay li tm a thc.V d 1: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:t A(x) = P(x) - x2 ta c: A(1) = 0 ; A(2) = 0 ; A(3) = 0; A(4) = 0 ; A(5) = 0;Nn theo nh l Bezout ta c: x = 1;2;3;4;5l nghim ca A(x) do ta c:k.( x - 1)(x-2)( x - 3)(x-4)(x - 5) = P(x) - x2=>P(x) = k.( x - 1)(x-2)( x - 3)(x- 4)(x - 5) + x2VP(x) c bc ln nht l: 5v c h s bng 1 nn k = 1Vy P(x) = ( x - 1)(x-2)( x - 3)(x- 4)(x - 5) + x2 => .P(6) = ( 6 - 1)(6-2)(6 - 3)(6-4)(6 - 5) + 62 = 156.P(7) = ( 7 - 1)(7-2)(7 - 3)(7-4)(7 - 5) + 72 = 769.P(6) = ( 8 - 1)(8-2)(8 - 3)(8-4)(8- 5) + 82 = 2584.P(6) = ( 9 - 1)(9-2)(9 - 3)(9-4)(9 - 5) + 92 = 6801V d 2:Cho P(x) = 6x5 + ax4 + bx3 + x2 + cx + 450. Bit a thc P(x) chia ht cho cc nh thc (x - 2) ; (x - 3); (x - 5) . Hy tm cc gi tr a, b, c v cc nghim ca a thc.*HD: Dng chc nng gii hpt ta c kt qu: a = -59; b = 161; c = - 495;x1 = 2; x2= 3; x3=5; x4=3/2; x5=-5/3Bi tp p dng :Bi 1:Cho a thc P(x) = 6x3 7x2 16x + m.a. Tm m P(x) chia ht cho 2x + 3.b. Vi m va tm c cu a hy tm s d r khi chia P(x) cho 3x-2.c. Tm m v n Q(x) = 2x3 5x2 13x + nv P(x) cng chia ht cho x-2.d. Vi n va tm c phn tch Q(x) ra tch cc tha s bc nht.Bi 2: Cho P(x) = x4+ mx3+ nx2+ px + q. Bit Q(1) = 5; Q(2) = 7; Q(3) = 9; Q(4) = 11. Tnh Q(10), Q(11), Q(12), Q(13). 28Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Bi 3 : Cho P(x) = x4 + 5x3 4x2 + 3x + m v Q(x) = x4 + 4x3 3x2 + 2x + n.a. Tm gi tr ca m, n ca cc a thc P(x) v Q(x) chia ht cho x 2.b. Vi gi tr ca m, n va tm c chng t rng a thc R(x) = P(x) Q(x) ch c mt nghim duy nhtBi 4:Cho P(x) = x5 + 2x4 3x3 + 4x2 5x + m.1. Tm s d trong php chia P(x) cho x 2,5 khi m = 20102. Tm gatr m P(x) chia ht cho x 2,53. P(x) c nghim x = 2. Tm m?Bi5. 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 6: Cho a thc P(x) = x10 + x8 7,589x4 + 3,58x3 + 65x + m.a. Tm iu kin ca m P(x) c nghim l: x = 0,3648b. Vi m va tm c, tm s d khi chia P(x) cho (x -23,55) Bi 7 : 1.Cho x=2,1835 v y= -7,0216. Tnh 5 4 3 3 43 2 2 37x y-x y +3x y+10xy -9F=5x -8x y +y2.Tm s drca php chia :5 4 2x -6,723x +1,658x -9,134x-3,2813. Cho 7 6 5 4 3 2P(x)=5x +2x -4x +9x -2x +x +10x-m. T m m P(x) chia ht cho a thc x+2Bi 8:a. Tm m P(x) chia ht cho (x -13) bit P(x) = 4x5 + 12x4 + 3x3 + 2x2 5x m + 7 b. Cho P(x) = ax5 + bx4 + cx3 + dx2 + ex + f bit P(1) = P(-1) = 11; P(2) = P(-2) = 47; P(3)=107. Tnh P(12)?Bi 9: Cho a thc P(x) = x3+ bx2+ cx + d. Bit P(1) = -15; P(2) = -15; P(3) = -9. Tnh:a. cc h s b, c, d ca a thc P(x).b. Tm s d r1 khi chia P(x) cho x 4.c. Tm s d r2 khi chia P(x) cho 2x +3.Bi 10: Cho a thc P(x) = x3 + ax2 + bx + c. Bit P(1) = -25; P(2) = -21; P(3) = -41. Tnh:29Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn a. Cc h s a, b, c ca a thc P(x).b. Tm s d r1 khi chia P(x) cho x + 4.c. Tm s d r2 khi chia P(x) cho 5x +7.d. Tm s d r3 khi chia P(x) cho (x+4)(5x +7).Bi 11:Cho a thc P(x) = x4+ax3 + bx2 + cx + d. Bit P(1) = 0; P(2) = 4; P(3) = 18; P(4) = 48. Tnh P(2010)?Bi 12:Chia P(x) = x81 + ax57 + bx41 + cx19 + 2x + 1 cho x 1 c s d l 5. Chia P(x) cho x 2 c s dl - 4. Hy tm cp (M,N) bit rng Q(x) = x81 + ax57 + bx41 + cx19 + Mx + N chia ht cho (x-1)(x-2)3: Lin phn s: 30Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Cho a, b (a>b) l hai s t nhin. Dng thut ton clt chia a cho b, phn s ab c th vit di dng: 00 00b a 1a abb bb + +V b0 l phn dca a khi chia cho b nn b > b0. Do vy ta c11 100 01b b 1a abb bb + + Tip tc nh vy ta c sau n bc ta c: 00 01n 2nb a 1a a1b ba1...aa + +++.Cch biu din ny gi l cch biu din s hu t di dng lin phn s. Mi s hu t c mt biu din duy nht di dng lin phn s, n -c vit gn l [ ]0 1 na ,a ,...,a.Vn t ra l: hy biu din lin phn s 01n 1n1a1a1...aa+++ v dng ab v ngc li Vi s tr gip ca my tnh ta c th tnh mt cch nhanh chng.* Qui trnh bm my fx-500MS:1). Tnh t di ln trn:Bm ln lt cc phm: b/ c b/ c b/ cn 1 n n 2 0a 1a a a 1a Ans ...a 1a Ans + + + 2). Tnh t trn xung di:Bm ln lt cc phm:+ + + + b/ c b/ c b/ c0 1 2 n 1 na (1a (a 1a ( a ...a 1( a a )))))))))V d 1 :Tnh gi tr ca:1A 112132 +++ Gii: Qui trnh bm trn my fx-500MS *Cch 1:Bm cc phm: b/ c b/ c b/ c b/ c3 1a 2 2 1a Ans 1 1a Ans SHIFT a + + + 23( )16 *Cch 2: Bm cc phm: + + + b/ c b/ c b/ c1 (1a (2 (1a ( 3 (1a 2))))2316 _ ,31Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn V d 2:Bit 15 111711ab++trong a v b l cc s dng. Tm a,b?Gii:Ta c: 15 1 1 1 117 2 1 1171 1 115 115 1572 2 + + ++. Vy a = 7, b = 2.Bi tp vn d ng: B i 1 : Tnh v vit kt qu di dng phn s:5 1A 3 B 74 12 35 12 34 12 35423 + ++ ++ ++ ++B i 2 : Tm cc s t nhin a v b bit: 329 1110513151ab+++B i 3 : Tm gi tr cax, y ca cc phng trnh sau:a. x x41 11 41 12 31 13 24 2+ + ++ ++ +b. y y1 11 21 13 45 6++ ++ +Bi 6: Cho 12A 305102003 ++ Hy vitli A di dng [ ]0 1 nA a ,a ,...,a ?32Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 4.Dy s: 1. Lp quy trnh tnh s hng Dy s cho bi cng thc s hng tng qut:dy s (un) cho bitrong f(n) l biu thc ca n cho trc.Cch lp quy trnh:- Ghi gi tr n = 1 vo nhA :1SHIFT STO A- Lp cng thc tnh f(A) v gn gi tr nh:A = A +1- Lp du bng: = ... =...Gii thch: 1SHIFT STO A : ghi gi tr n = 1 vo nhA f(A) :A = A +1: tnh un = f(n) ti gi trA(khi bm du bng th ln nht) v thc hingn gi tr nhAthm 1 n v: A = A + 1 (khi bmdu bng ln th hai).* Cng thc c lp li mi khi n du=V d 1: Tnh 10 s hng u ca dy s (un) cho bi: 1 1 5 1 5; 1, 2, 3...2 2 5n nnu n 1 _ _+ 1 1 , , ]Gii:- Ta lp quy trnh tnh un nh sau:33un = f(n),n N* Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 1SHIFT STO A( 1 5 ) ( ( ( 1+5 ) 2 ) ANPHA A - ( ( 1 -5 ) 2 ) ANPHA A)ANPHA:ANPHA A ANPHA =ANPHA A +1 =- Lp li phm: =...=... kt qu: u1 = 1, u2 = 1, u3 = 2, u4 = 3, u5 = 5, u6 = 8, u7 = 13, u8 = 21,u9 = 34, u10 = 55.2 . Lp quy trnh tnh s hng Dy s cho bi h thctruy hi dng:dy s (un) cho bi trong f(un) l biu thc ca un cho trc.Cch lp quy trnh:- Nhp gi tr ca s hng u1: a=- Nhp biu thc ca un+1 = f(un) : ( trong biu thc ca un+1 ch no c un ta nhp bngANS)- Lp du bng: =Gii thch:- Khi bm: a=mn hnh hin u1 = a v lu kt qu ny - Khi nhp biu thc f(un) bi phmANS , bm du= ln th nht my s thc hin tnh u2 = f(u1) v li lu kt qu ny.- Tip tc bm du=ta ln lt c cc s hng ca dy s u3, u4...V d 1: Tm 20 s hng u ca dy s (un) cho bi:341n + 1 nu = au = f (u ); n N *'Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 1112, *1nnnuuu n Nu+ +' +Gii:- Lp quy trnh bm phm tnh cc s hng ca dy s nh sau:1=(u1)(ANS +2 ) (ANS +1 )=(u2)=...=- Ta c cc gi tr gn ng vi 9 ch s thp phn sau du phy:u1 = 1 u8 = 1,414215686u2 = 1,5u9 = 1,414213198u3 = 1,4u10 = 1,414213625u4 = 1,416666667u11 = 1,414213552u5 = 1,413793103u12 = 1,414213564u6 = 1,414285714u13 = 1,414213562u7 = 1,414201183u14 =...= u20 = 1,414213562V d 2: Cho dy s c xc nh bi: ( )331313, *n nuu u n N+' Tm s t nhin n nh nht un l s nguyn.Gii:- Lp quy trnh bm phm tnh cc s hng ca dy s nh sau:SHIFT3 3=(u1)ANSSHIFT3 3=(u2)= =(u4 = 3)Vy n = 4 l s t nhin nh nht u4 = 3 l s nguyn.35Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 3. Lp quy trnh tnh s hng Dy s cho bi h thctruy hi dng: Dy s (un) cho bi Cch lp quy trnh:* Cch 1:Bm phm:bSHIFT STO A A +B a+CSHIFT STO BV lp li dy phm:

A+ ANPHA A B+CSHIFT STO A

A+ ANPHA B B+CSHIFT STO B Gii thch: Sau khi thc hin bSHIFT STO A A +B a+CSHIFT STO Btrong nhAl u2 = b, my tnh tng u3 := Ab + Ba + C = Au2 + Bu1 + C v y vo trong nhB, trn mn hnh l: u3 : = Au2 + Bu1 + CSau khi thc hin: A+ ANPHA A B+CSHIFT STO Amy tnh tng u4 := Au3 + Bu2 + C v a vo nhA . Nh vy khi ta c u4 trn mn hnh v trong nhA (trong nhBvn l u3).Sau khi thc hin: A+ ANPHA B B+CSHIFT STO Bmy tnh tng u5 := Au4 + Bu3 + C v a vo nhB . Nh vy khi ta c u5 trn mn hnh v trong nhB(trong nhA vn l u4).Tip tc vng lp ta c dy s un+2 = Aun+1 + Bun + C*Nhn xt:Trong cch lp quy trnh trn, ta c th s dng chc nngCOPY lp li dy lp bi quy trnh sau (gim c 10 ln bm phm mi khi tm mt s hng ca dy s), thc hin quy trnh sau:Bm phm: bSHIFT STO A A +B a+CSHIFT STO B361 2n+2 n+1 nu=a,u b u=A u + B u+ C;n N*'Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn A+ ANPHA A B+CSHIFT STO A A+ ANPHA B B+CSHIFT STO B SHIFT COPY Lp du bng:=... =...* Cch 2: S dng cch lp cng thcBm phm:aSHIFT STO A bSHIFT STO BANPHA C ANPHA =AANPHA B +BANPHA A +CANPHA:ANPHA A ANPHA = ANPHA B ANPHA:ANPHA B ANPHA = ANPHA CLp du bng: =... =...V d : Cho dy s c xc nh bi: 1 2n+2 n+1 nu=1,u 2 u=3u + 4u+ 5;n N* 'Hy lp quy trnh tnh un.Gii:- Thc hin quy trnh:2SHIFT STO A 3+4 1+5SHIFT STO B 3+ ANPHA A 4+5SHIFT STO A 3+ ANPHA B 4+5SHIFT STO BSHIFT COPY =... =...37Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn ta c dy: 15, 58, 239, 954, 3823, 15290, 61167, 244666, 978671...Hoc c th thc hin quy trnh:1SHIFT STO A 2SHIFT STO BANPHA C ANPHA =3ANPHA B +4ANPHA A +5ANPHA:ANPHA A ANPHA = ANPHA BANPHA:ANPHA BANPHA = ANPHA C=... =...ta cng c kt qu nh trn.38Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 4.Dyschobi hthctruyhi vi hsbinthin dng: Dy s (un) cho bi * Thut ton lp quy trnh tnh s hng ca dy:- S dng 3 nh: A : cha gi tr ca nB: cha gi tr ca unC: cha gi tr ca un+1- Lp cng thc tnh un+1 thc hin gnA=A+ 1 vB:= C tnh s hng tip theo ca dy- Lp phm : =V d : Cho dy s c xc nh bi: ( )1n+1 nu=0n u= u +1 ;n N*n+1'Hy lp quy trnh tnh un.Gii:- Thc hin quy trnh:1SHIFT STO A 0SHIFT STO BANPHA C ANPHA =(ANPHA A (ANPHA A +1 ) )

(ANPHA B +1 )ANPHA:ANPHA A ANPHA =ANPHA A +1ANPHA:ANPHA B ANPHA = ANPHA C39{ } ( )1n+1u=a u= ,;n N*nf n u'Trong { } ( ),nf n u l k hiu ca biu thc un+1 tnh theo un v n.Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn =... =... ta c dy: 1 3 5 7, 1, , 2, , 3, ,...2 2 2 2 5.Lp cng thc s hng tng qut: Phng php gii:- Lp quy trnh trn MTCT tnh mt s s hng ca dy s- Tm quy lut cho dy s, d on cng thc s hng tng qut- Chng minh cng thc tm c bng quy npV d 1: Tma2004bit: Gii:- Trc ht ta tnh mt s s hng u ca dy (an), quy trnh sau:1SHIFT STO A 0SHIFT STO BANPHA C ANPHA = ANPHA A(ANPHA A +1 ) ( (ANPHA A + 2) (ANPHA A + 3) ) (ANPHAB + 1)ANPHA:ANPHAAANPHA =ANPHA A +1ANPHA:ANPHA B ANPHA = ANPHA C - Ta c dy: 1 7 27 11 13 9, , , , , ,...6 20 50 15 14 8- T phn tch cc s hng tm quy lut cho dy trn: a1 = 0a2 = 1 5 1.56 30 3.10 d on cng thc s hng tng qut:a3 = 7 2.7 2.720 40 4.10

a4 = 27 3.950 5.10 * D dng chng minh cng thc (1) ng40110( 1)( 1) ; *( 2)( 3)n nan na a n Nn n+ +' + + +)( 1)(2 1)10( 1)nn nan ++(1)vi mi n N* bng quy np.Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn ... 20042003.400920050a 41Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn V d2 : Xt dy s: Chng minh rng s A = 4an.an+2 + 1 l s chnh phng.Gii:- Ta co s s hng u ca dy (an) bng quy trnh:- Ta c dy: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...- Tm quy lut cho dy s:11(1 1)12a+

22(2 1)32a+ d on cng thc s hng tng qut:33(3 1)62a+ 44(4 1)102a+

55(5 1)152a+ * Ta hon ton chng minh cng thc (1)... T : A = 4an.an+2+ 1 = n(n + 1)(n + 2)(n + 3) +1 = (n2+ 3n + 1)2. A l mt s chnh phng.Cch gii khc: T kt qu tm c mt s s hng u ca dy,ta thy:- Vi n = 1 th A = 4a1.a3 + 1 = 4.1.6 + 1 = 25 =(2a2 - 1)2- Vi n = 2 th A = 4a2.a4 + 1 = 4.3.10 + 1 = 121 =(2a3 - 1)2- Vi n = 3 th A = 4a3.a5 + 1 = 4.6.15 + 1 = 361 =(2a4 - 1)2T ta chng minh A = 4an.an+2 + 1 = (2an+1 - 1)2 (*)Bng phng php quy np ta cng d dng chng minh c (*).Bi tp p dng Bi 1: Cho dy s (un), (n = 0, 1, 2,...):

( ) ( )2 3 2 32 3n nnu+ 421 2*21, 32 1;n n na aa a a n N+ ' + )( 1)2nn na+ng vi mi n N*(1)Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn a) Chng minh un nguyn vi mi n t nhin.b) Tm tt c n nguyn un chia ht cho 3.Bi 2: Cho dy s (an) c xc nh bi:

2124 15 60 , *on n naa a a n N+ ' + a) Xc nh cng thc s hng tng qut an.b) Chng minh rng s:( )2185nA a + biu din c di dng tng bnh phng ca 3 s nguyn lin tip vi mi n 1.Bi 3: Cho dy s (un) xc nh bi: 12 10, 11999 ,on n nu uu u u n N+ + ' Tm tt c s t nhin n sao cho un l s nguyn t.Bi 4: Cho dy s (an) xc nh bi: 1 21 15, 112 3 , 2,n n na aa a a n n N+ ' Chng minh rng:a) Dy s trn c v s s dng, s m.b) a2002 chia ht cho 11.Bi 5: Cho dy s (an) xc nh bi:

1 221212, 3,nnna aaa n n Na + ' Chng minh an nguyn vi mi n t nhin.Bi 6: Dy s (an) c xc nh theo cng thc:( )2 3 , *nna n N 1 + 1 ]; (k hiu( )2 3n 1+ 1 ]l phn nguyn ca s( )2 3n+ ).Chng minh rng dy (an) l dy cc s nguyn l.43Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 5. Hnh Hc: 1/. Gii tam gic:* Mt s cng thc: 1/. Cc h thc trong tam gic vung:2/. T s lng gic ca gc nhn:3/ Cc cng thc tnh din tch tam gic: SinC SinB SinA R Sr c P r b P r a P Sr P SRc b aSc b aP c P b P a P P SSinB A c SinA c b SinC b a Sh c h b h a SABCc b a ABCABCABCABCABCc b ABC. . . 2 . / 7) ( ) ( ) ( . / 6. . / 54. .. / 42... ) )( )( .( .. / 3. .21. .21. .21. / 2.21.21.21. / 12 + + * Cc dng ton:V d 1: Tnh cc gc ca tam gic ABC, bit:AB = 4,123; BC = 5,042;CA = 7,415HD: Ta c :S ABC = ) )( )( ( sin . .21sin . .21sin . .21c P b P a P P B a c A c b C b a T ta c:A ; B ;C

Bi 2: Tnh cnh BC, gc B , gc C ca tam gic ABC, bit: AB = 11,52;AC = 19,67 v gc A 54o3512p s: BC =; B ;C Bi 3: Tnh cnh AB, AC, gc C ca tam gic ABC, bit: BC = 4,38; A 54o3512 ; B 101o157p s: AB=;AC = ;C Bi 4:Tam gic ABC c ba cnh: AB = 4,123 ; BC = 5,042 ; CA = 7,415im M nm trn cnh BC sao cho:BM = 2,1421) Tnh di AM?2) Tnh bn knh ng trn ngoi tip tam gic ABM44Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 3) Tnh bn knh ng trn ni tip tam gic ACM.p s: 1)AM =2)R = 3)r=Bi 5: Tam gic ABC c: B 49o27;C 73o52v cnh BC = 18,53.Tnh din tch S ca tam gic ?p s: S = Bi 6: Tam gic ABC c chu vi 58 (cm) ; B 57o18 v C 82o35Tnh di cc cnh AB, BC, CA ?p s: AB = ; BC = ; CA =Bi 7: Tam gic ABC c 90o < A < 180o v sinA = 0,6153 ; AB = 17,2 ; AC = 14,6.Tnh: 1) di cnh BC ? Trung tuyn AM ? 2) Gc B ? 3) Din tch tam gic S = ?p s: BC =; AM = ; B ;S =Bi 8: Tam gic ABC c A 90o ; AB = 7 (cm) ; AC = 5 (cm).Tnh di ng phn gic trong AD v phn gic ngoi AE ?p s: AD =; AE =2 . a gic, hnh trn: 1/. a gic u n cnh, di cnh l a :+ Gc tm:2n (rad),hoc: 360oan()+ Gc nh: 2Ann(rad),hoc 2A .180nn()+ Din tch:cot4 2naS g2/. Hnh trn v cc phn hnh trn:+ Hnh trn bn knh R:- Chu vi:C = 2 R- Din tch: S = R245a AO .ORTi liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn + Hnh vnh khn:- Din tch:S = (R2 - r2) = (2r + d)d+ Hnh qut:- di cung:l = R ;( : rad)- Din tch: 212S R ( : rad)

2360R a (a: )Bi9:Bangtrnccngbnknh3cmi mttipxcngoi (Hnh v)Tnh din tch phn xen gia ba ng trn ?H.Dn: SBi 10:Cho hnh vung ABCD, cnh a = 5,35. Dng cc ng trn tm A, B, C, D c bn knh R = 2a. Tnh din tch xen gia 4 ng trn .H.Dn:Sgch = SABCD - 4Squt Squt = 14SH.trn = 14 R2 Sgch = a2 - 4. 14 R2 = a2 - 14 a2 = a2(1 - 14 ) 6,142441068Bi 11: Cho ng trn tm O, bn knh R = 3,15 cm. T mt im A ngoi ng trn v hai tip tuyn AB v AC (B, C l hai tip im thuc (O) ). Tnh din tch phn gii hn bi hai tip tuyn v cung trn nh BC. Bit OA = a = 7,85 cm.H.Dn:- Tnh : 3,15cos7, 85OB ROA a 46 .ORrd .OR O1O2 A B DC BTi liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 13,15cos7, 85SOBAC = 2SOBA = aRsinSqut = 2 2.2 .360 180R R Sgch = SOBAC - Squt = aRsin- 2.180R 11,16 (cm2)Bi 12: Tnh din tch phn c t m trong hnh trn n v (R = 1)(Xem hnh 1)p s: Bi 13: Tnh t l din tch ca phn c t m v din tch phn cn li trong hnh trn n v (Xem hnh 2)p s:47 A C OHnh 1 Hnh 2Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Phn III:mt s thi ( h THCS ) S GDT QUNG NGI THI GII TON TRN MY TNH CASIO TRNG THCS DTNT BA T BC TRUNG HC NM HC 2008-2009 chnh thcH v tnhc sinh: Lp: cp THCS.Thii gian: 150 pht (khng k thi gian giao )Ngy thi : 13/12/2008.Ch : - ny gm: 4 trang.- Th sinh lm bi trc tip trn bi thi ny.IM TON BI THICH K CA GIM KHO Bng s Bng ch GK1 GK2Quy nh: Nu khng gia thch g thm, hy tnh kt qu chnh xc n 10 ch s.Bi 1: (10 im)Tnh gi tr ca cc biu thc sau v in kt qu vo vung:a) A = ( )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: b) B = 3 3847 8476 627 27+ + c)1C 64 12 122 91 14 4 +++++d)( ) ( )0 0 0 2 0D tg25 15' tg15 27' cotg35 25' cotg 78 15' e) Bitt: cosA = 0,8516 ; tgB = 3,1725 ; sinC = 0,4351. Tnh : E = cotg(A + B C) ?Bi 2: (6 im)48E = A =B =C =D =Ch k ca GT1:Ch k ca GT2:Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Tm gi tr ca x, y, z di dng phn s (hoc hn s) t cc phng trnh sau ri in kt qu vo vung :a) 1 1 1x. 42 1 33 1 23 1 55 1 44 727 69 8 _ + + + + + + + + ,+ +b)y y51 11 41 15 23 3 + ++ +c)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 8z 1 _ _ 1 , , 1 + _ _ 1 + 1 , , ]

Bi 3: (10 im)a) Tm cc s t nhin a v b bit rng:7463 12413073141ab ++++ b) Tm CLN v BCNN ca 170586104 v 157464096.c) Tm s d ca php chia: 987654312987654321 cho 123456789.d) Tmch s hng chc ca 172008 e) Tm s ln nht v s nh nht trong cc s t nhin c dng 5 4 3 2 a b c chia ht cho 13Bi 4: (1im)Cho u1 = 2008; u2 = 2009 v un+1 = un + un-1 vi mi n 2. Xc nh u13 ?49y =a =b = CLN = z = r = BCNN = x = S ln nht l:S nh nht l: Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn Bi 5: (3,5 im )Cho a thc : P (x) = x3 + bx2 + cx + d v cho bit: P(1) = -15; P(2) = -15; P( 3) = -9.a)Lp h phng trnh tm cc h sb, c, d ca P(x). Gii: b, c, d l nghim ca h phng trnh sau: .................................................................................................................................................................................' .................................................................................................................................................................................' b =c =d = b) Tm s drv a thc thng Q(x) trong php chia P (x) cho (x - 13).Bi 6: (1im)Cho a thc : F(x) = x5 + 2x4 3x3 + 4x2 5x + m 2008. Tm gi tr ca m phng trnh F(x) = 0 c mt nghim l x = -1,31208.B i 7 : ( 1,5 im)Cho tam gic ABC vung ti A, bit AC = 3AB . Trn cnh AC ly im D sao cho DC = AB. Tnh tng s o ACB ADB +?Bi 8: (2 im) Cho tam gic ABC c 0A 120 ; AB = 4cm ; AC = 6cm v trung tuyn AM. T B, k BH vung gc vi AC tai H v t M, k MK vung gc vi AC ti K (H, K AC). Tnh di ng trung tuyn AM.50Cch gii:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .AHBMKC461200. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . U13 = r = m = Q(x) = ACB ADB + =B A D CTi liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn in kt qu vo vung:Bi9: (3im)Cho tam gic ABC c AB = 8,91cm ; AC = 10,32cm v 0BAC 72 . (Tnh chnh xc n 3 ch s thp phn). a) di ng cao BH.b) Din tch tam gic ABC.c) di cnhBC. in kt qu vo vung:BH =SABC=BC =Bi 10:(2im)ChohnhthangvungABCD(BC// AD; 0B C 90 ) cAB=12,35cm; BC=10,55cm; 0ADC 57 .a) Tnh chu vi ca hnh thang ABCD.b) Tnh din tch ca hnh thang ABCD.c) Tnh cc gc ca tam gic ADC. ( Lm trn n ) in kt qu vo vung:C ABCD =SABCD=DAC =; DCA = 51Cch gii:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 10,32 C B 8,91 720 H AM = AB 570DHCTi liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn S GDT QUNG NGI THI GII TON TRN MY TNH CASIO TRNG THCS DTNT BA T BC TRUNG HC NM HC 2008-2009 chnh thc P N V BIU IMBi 1: (10 im)Tnh gi tr ca cc biu thc sau v in kt qu vo vung: Mi cu ng 2 ima) A = ( )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: b) B = 3 3847 8476 627 27+ + c)1C 64 12 122 91 14 4 +++++d)( ) ( )0 0 0 2 0D tg25 15' tg15 27' cotg35 25' cotg 78 15' e) Bitt: cosA = 0,8516 ; tgB = 3,1725 ; sinC = 0,4351. Tnh : E = cotg(A + B C) ?Bi 2: (6 im)Tm gi tr ca x, y, z di dng phn s (hoc hn s) t cc phng trnh sau ri in kt qu vo vung : Mi cu ng 2 ima) 1 1 1x. 42 1 33 1 23 1 55 1 44 727 69 8 _ + + + + + + + + ,+ +52E = 0,206600311x = 30116714A = 213 B = 3 C =6734338267331064 D =0,266120976 Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn b)y y51 11 41 15 23 3 + ++ +c)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 8z 1 _ _ 1 , , 1 + _ _ 1 + 1 , , ]

Bi 3: (10 im) Mi cu ng 2 ima) Tm ccc s t nhin a v b bit rng:7463 12413073141ab ++++ b) Tm CLN v BCNN ca 170586104 va 157464096.c) Tm s d ca php chia: 987654312987654321 cho 123456789.d)Tmch s hng chc ca172008 e)Tm s ln nht v s nh nht ntrong cc s tnhin c dng 5 4 3 2 a b c chia ht cho 13Bi 4: (1im)Cho u1 = 2008; u2 = 2009 v un+1 = un + un-1 vi mi n 2. Xc nh u13 ?Bi 5: (3,5 im )Cho a thc : P (x) = x3 + bx2 + cx + d v cho bit: P(1) = -15; P(2) = -15; P( 3) = -9.a)Lp h phng trnh tm cc h sb, c, d ca P(x).( 2 im)53y =418363a = 3 b = 7CLN = 13122008 z = 6r = 9BCNN = 2047033248 4S ln nht l: 5949372S nh nht l: U13= 468008Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn G ii : b, c, d l nghim ca h phng trnh sau: 3 23 23 21 .1 .1 152 .2 .2 153 .3 .3 9b c db c db c d + + + + + + '+ + +

b+c+d=-164b+2c+d=-239b+3c+d=-36' b = -3 c = 2 d = -15b) Tm s drv a thc thng Q(x) trong php chia P (x) cho (x - 13). (1,5 im)Bi 6: (1im)Cho a thc : F(x) = x5 + 2x4 3x3 + 4x2 5x + m 2008. Tm gi tr ca m phng trnh F(x) = 0 c mt nghim l x = -1,31208.Bi 7: ( 1,5 im)Cho tam gic ABC vung ti A, bit AC = 3AB . Trn cnh AC ly im D sao cho DC = AB. Tnh tng s o ACB ADB +?Bi 8: (2 im) Cho tam gic ABC c 0A 120 ; AB = 4cm ; AC = 6cm v trung tuyn AM. T B, k BH vung gc vi AC tai H v t M, k MK vung gc vi AC ti K (H, K AC). Tnh di ng trung tuyn AM. in kt qu vo vung:Bi9: (3im)Cho tam gic ABC c AB = 8,91cm ; AC = 10,32cm v 0BAC 72 . (Tnh chnh xc n 3 ch s thp phn). 54AM = 2,645751311cmCch gii:.Ta c 0 0 0180 120 60 BAH NnAH = AB. cos04.cos 60 2 BAH cmMt khc: BH//MK (gt) m MB = MCSuy ra KH = KC 6 242 2 2HC AC AH + + cm vMK = 12BH( v MK l ng trung bnh caBCH ) = 0 01 1sin .4.sin 60 2.sin 602 2AB BAH Do 2 2 2 0 22 (2.sin 60 ) AM AK MK + += 2,645751311 cm AHBMKC461200r = 1701m = 1,985738113Q(x) = x2 + 10x +132 ACB ADB + =450Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn a) di ng cao BH.b) Din tch tam gic ABC.c) di cnh BC Mi cu ng 1 im ien ket qua vao o vuong:BH = 8,474 cm SABC= 43,726cm2BC =11,361cmBi 10:(2 im)ChohnhthangvungABCD(BC// AD; 0B C 90 ) cAB=12,35cm; BC=10,55cm; 0ADC 57 .a) Tnh chu vi ca hnh thang ABCD.b) Tnh din tch ca hnh thang ABCD.c) Tnh cc gc ca tam gic ADC. ( Lm trn n )Gii: a) Ta c AD = 010, 55sin57sinAHD ;DH = AH. cotgD = 10,55.cotg570

(1 )Nn CABCD = 2AB + BC +DH +AD = 2.12,35 + 10,55 +10,55.cotg570

+ 010, 55sin57=54,68068285 cmb) SABCD = 0( ). (12, 35 12, 35 10, 55.cot 57 ).10, 55166, 43284432 2AB CD BC g + + + cm2 (0,5 )c) Ta c : tg10, 5512, 35AHDCAHC Suy ra 041 DCA . Do 0 0180 ( ) 82 DAC D DCA + (0,5 ) in kt qu vo vung:CCH XP GII K THI MTCT CASIO CP TRNGC ABCD = 54,68068285 cm SABCD= 166,4328443 cm2DAC =820; DCA = 410 55Cch gii:a) Ta cBH =AB SinBAC = 8,91.sin720 = 8,474 cmb) SABC = 12AC.BH = 1210,32.8.474 = 43,726 cm2c) Ta cAH = AB. cos = 8,91.cos720 = 2,753 cm Suy ra HC = AC AH = 10,32 2,753 = 7,567cm Do BC = 2 2 2 28, 474 7, 567 11, 361 BH HC + + cm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 10,32 C B 8,91 720 H AB 570 DHC Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn NM HC : 2008 2009- Gii nht: T 36 n 40 im- Gii nh: T 32 n 36 im- Gii ba: T28 n 32 im- GiiKK: T 2028 imC. Kt lun v kin ngh :ng dng ca my tnhtrong vic gii ton l mt vn ln, i hi ngi hc phi c tnh sng to, c tduy tt v k nng vn dng l thuyt mt cch linh hot. Chnh v l , trong qu trnh ging dy, ngi gio vin cn chun b chu o, t m, r rng tng th loi bi tp c th hc sinh hiu su bn cht v cch vn dng. Xy dng cho cc em nim am m, hng th trong hc tp, tn trng nhng suy ngh, kin v sng to ca cc em. Cn thng xuyn kim tra, nh gi kt qu hc tp, b sung thiu st kp thi, dy su, dy chc v kt hp nhun nhuyn, lgic gia cc bi khc nhau.Trong qu trnh bin son: Ccdngtonthi HSGgiitontrn my tnh casio khng ch gip cho hc sinh yu thch hc b mn ton, m cn l c s gip cho bn thn c thm kinh nghim trong ging dy v ngi bin son nhn ra cc nhn xt c trng sau :56Ti liu bi d ng"Gii ton trn my tnh Casio " bin son: Tr ng Ngc Bn 1. My tnh in t gip cng c cc kin thc c bn tng tc lm ton.2.My tnh in t gip m rnh cc kin thc ton hc3.My tnh in t gip lin kt kin thc ton hc vi thc t Mc d rt c gng khi bin son, song khng th trnh khi thiu st v cu trc lgc, ngn ng v kin thc khoa hc. V vy mt ln na ti rt mong nhn c nhng gp kin gp chn thnh ca thy c v bn c. Xin chn thnh cm n! Cm xuyn, Ngy 07 thng 10 nm 2010 Ngi bin sonTrng Ngc Bn57