Mot So de Thi Toan Tren May Tinh Casio Cua Cac Tinh Thanh Tren CA Nuoc

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MT S THI TON TRN MY TNH CASIO CA CC TNH THNH TRN C NC A.BC NINH 1. thi Gii ton trn my tnh Casio nm 2003 2004 (THPT)Tr 5 2. thi Gii ton trn my tnh Casio nm 2004 2005 (THPT)Tr 7 3. thi Gii ton trn my tnh Casio nm 2004 2005 (THCS)Tr 9 B.CN TH 1. thi Gii ton trn my tnh Casio nm 2001 2002 (THCS - LP 6)Tr 11 2. thi Gii ton trn my tnh Casio nm 2001 2002 (THCS - LP 7)Tr 13 3. thi Gii ton trn my tnh Casio nm 2001 2002 (THCS - LP 8)Tr 15 4. thi Gii ton trn my tnh Casio nm 2001 2002 (THCS - LP 9)Tr 17 5. thi Gii ton trn my tnh Casio nm 2001 2002 (THPT - LP 10)Tr 19 6. thi Gii ton trn my tnh Casio nm 2002 2003 (THPT - LP 12)Tr 21 7. thi Gii ton trn my tnh Casio nm 2002 2003 (THCS - LP 9)Tr 23 8. thi Gii ton trn my tnh Casio nm 2004 2005 (THPT - LP 12)Tr 24 C.NG NAI 1. thi Gii ton trn my tnh Casio nm 1998(THCS)Tr 26 2. thi Gii ton trn my tnh Casio nm 1998(THPT)Tr 28 D.H NI 1. thi Gii ton trn my tnh Casio nm 1996(THPT- LP 10 - CP TRNG)Tr 30 2. thi Gii ton trn my tnh Casio nm 1996(THPT- LP 10 Vng 1)Tr 32 3. thi Gii ton trn my tnh Casio nm 1996(THPT- LP 11-12 CP TRNG)Tr 34 4. thi Gii ton trn my tnh Casio nm 1996(THCS- Vng Chung kt)Tr 35 5. thi Gii ton trn my tnh Casio nm 1996(THPT- Vng 1)Tr 37 6. thi Gii ton trn my tnh Casio nm 1996(THPT- Vng Chung kt)Tr 39 7. thi Gii ton trn my tnh Casio nm 2004(THPT- B tc)Tr 41 E.HI PHNG 1. thi Gii ton trn my tnh Casio nm 2002-2003(THCS- Lp 8)Tr 42 2. thi Gii ton trn my tnh Casio nm 2002-2003(THPT- Lp 11)Tr 44 3. thi Gii ton trn my tnh Casio nm 2002-2003(THPT- Lp 10)Tr 46 4. thi Gii ton trn my tnh Casio nm 2003-2004(THCS- Lp 9-Vng 2)Tr 48 5. thi Gii ton trn my tnh Casio nm 2003-2004(THCS- Lp 9-Vng 1)Tr 51 6. thi Gii ton trn my tnh Casio nm 2003-2004(THCS- Chn i tuyn)Tr 53 7. thi Gii ton trn my tnh Casio nm 2006-2007(THPT)Tr 55 F.H CH MINH 1. thi Gii ton trn my tnh Casio nm 1996(THCS- VNG 1)Tr 56 2. thi Gii ton trn my tnh Casio nm 1996(THCS- VNG CHUNG KT)Tr 58 3. thi Gii ton trn my tnh Casio nm 1997(THPT- VNG 1)Tr 60 4. thi Gii ton trn my tnh Casio nm 1997(THPT- VNG CHUNG KT)Tr 62 5. thi Gii ton trn my tnh Casio nm 1998(THCS)Tr 63 6. thi Gii ton trn my tnh Casio nm 1998(THPT)Tr 64 7. thi Gii ton trn my tnh Casio nm 2003(THCS)Tr 66 8. thi Gii ton trn my tnh Casio nm 2004(THPT- CHN I TUYN)Tr 67 9. thi Gii ton trn my tnh Casio nm 2004(THCS)Tr 68 10. thi Gii ton trn my tnh Casio nm 2004(THPT)Tr 69 11. thi Gii ton trn my tnh Casio nm 2005(THCS-CHN I TUYN)Tr 70 12. thi Gii ton trn my tnh Casio nm 2005(THPT-CHN I TUYN)Tr 71 13. thi Gii ton trn my tnh Casio nm 2005(THCS)Tr 72 14. thi Gii ton trn my tnh Casio nm 2005(THPT)Tr 73 15. thi Gii ton trn my tnh Casio nm 2006(THCS-CHN I TUYN)Tr 74 16. thi Gii ton trn my tnh Casio nm 2006(THPT-CHN I TUYN)Tr 75 17. thi Gii ton trn my tnh Casio nm 2006(THCS)Tr 76 18. thi Gii ton trn my tnh Casio nm 2006(THPT)Tr 77 19. thi Gii ton trn my tnh Casio nm 2006(THPT-BT)Tr 78 20. thi Gii ton trn my tnh Casio nm 2007(THPT- LP 11)Tr 79 21. thi Gii ton trn my tnh Casio nm 2007(THCS)Tr 80 22. thi Gii ton trn my tnh Casio nm 2007(THPT- LP 12)Tr 81 23. thi Gii ton trn my tnh Casio nm 2008(THPT)Tr 82 24. thi Gii ton trn my tnh Casio nm 2008(THCS)Tr 83 G.HO BNH 1. thi Gii ton trn my tnh Casio nm 2003 2004Tr 84 2. thi Gii ton trn my tnh Casio nm 2004 2005Tr 86 3. thi Gii ton trn my tnh Casio nm 2005 2006Tr 87 4. thi Gii ton trn my tnh Casio nm 2006 2007Tr 88 5. thi Gii ton trn my tnh Casio nm 2007 2008Tr 89 H.HU 1. thi Gii ton trn my tnh Casio nm 2004 (Lp 8)Tr 90 2. thi Gii ton trn my tnh Casio nm 2004 (Lp 9)Tr 95 3. thi Gii ton trn my tnh Casio nm 2004 (Lp 11)Tr 100 4. thi Gii ton trn my tnh Casio nm 2004 (Lp 12)Tr 105 5. thi Gii ton trn my tnh Casio nm 2005 (Lp 8)Tr 110 6. thi Gii ton trn my tnh Casio nm 2005 (Lp 9)Tr 118 7. thi Gii ton trn my tnh Casio nm 2005 (Lp 11)Tr 126 8. thi Gii ton trn my tnh Casio nm 2005 (Lp 12-BT)Tr 137 9. thi Gii ton trn my tnh Casio nm 2005 (Lp 12)Tr 145 10. thi Gii ton trn my tnh Casio nm 2006 (Lp 8)Tr 154 11. thi Gii ton trn my tnh Casio nm 2006 (Lp 9)Tr 161 12. thi Gii ton trn my tnh Casio nm 2006 (Lp 11)Tr 169 13. thi Gii ton trn my tnh Casio nm 2006 (Lp 12-BT)Tr 177 14. thi Gii ton trn my tnh Casio nm 2006 (Lp 12)Tr 184 15. thi Gii ton trn my tnh Casio nm 2007 (Lp 8)Tr 194 16. thi Gii ton trn my tnh Casio nm 2007 (Lp 9)Tr 203 17. thi Gii ton trn my tnh Casio nm 2007 (Lp 11)Tr 110 18. thi Gii ton trn my tnh Casio nm 2007 (Lp 12-BT)Tr 218 19. thi Gii ton trn my tnh Casio nm 2007 (Lp 12)Tr 226 I.KHNH HO 1. thi Gii ton trn my tnh Casio nm 2000-2001 (Lp 9)Tr 233 J.NINH BNH 1. thi Gii ton trn my tnh Casio nm 2007-2008 (THCS)Tr 235 K.PH TH 1. thi Gii ton trn my tnh Casio nm 2003-2004 (THCS- LP 9)Tr 241 2. thi Gii ton trn my tnh Casio nm 2003-2004 (THPT- LP 12)Tr 243 3. thi Gii ton trn my tnh Casio nm 2004-2005 (THPT- LP 9)Tr 245 4. thi Gii ton trn my tnh Casio nm 2004-2005 (THBT- LP 12-D B)Tr 248 5. thi Gii ton trn my tnh Casio nm 2004-2005 (THBT)Tr 251 6. thi Gii ton trn my tnh Casio nm 2004-2005 (THPT)Tr 254 7. thi Gii ton trn my tnh Casio nm 2004-2005 (THCS)Tr 257 L.QUNG NINH 1. thi Gii ton trn my tnh Casio nm 2004-2005 (THCS)Tr 260 2. thi Gii ton trn my tnh Casio nm 2004-2005 (THPT)Tr 265 3. thi Gii ton trn my tnh Casio nm 2005-2006 (THCS)Tr 272 4. thi Gii ton trn my tnh Casio nm 2005-2006 (THBT)Tr 281 5. thi Gii ton trn my tnh Casio nm 2005-2006 (THPT)Tr 288 M. THI NGUYN 1. thi Gii ton trn my tnh Casio nm 2002 (THBT)Tr 292 2. thi Gii ton trn my tnh Casio nm 2002 (THPT)Tr 293 3. thi Gii ton trn my tnh Casio nm 2003 (THBT)Tr 294 4. thi Gii ton trn my tnh Casio nm 2003 (THCS 1)Tr 296 5. thi Gii ton trn my tnh Casio nm 2003 (THCS 2)Tr 297 6. thi Gii ton trn my tnh Casio nm 2003 (THPT)Tr 299 7. thi Gii ton trn my tnh Casio nm 2004 (THBT)Tr 300 8. thi Gii ton trn my tnh Casio nm 2004 (THPT)Tr 301 N.THANH HO 1. thi Gii ton trn my tnh Casio nm 1996 (LP 10)Tr 303 2. thi Gii ton trn my tnh Casio nm 1996 (LP 11-12)Tr 305 3. thi Gii ton trn my tnh Casio nm 2004-2005 (LP 9)Tr 307 4. thi Gii ton trn my tnh Casio nm 2007-2008 (THCS)Tr 311 5. thi Gii ton trn my tnh Casio nm 2007-2008 (HSG)Tr 313 CNG TY CP XNK BNH TY (BITEX) BAN QUN TR TRANG WEB WWW.BITEX.EDU.VN S GIO DC O TO BC NINH THI GII TON TRN MY TNH CASIO 2004 Thi gian 150 pht ------------------------------------------------------------- ( kt qu tnh ton gn nu khng c quy nh c th c ngm hiu l chnh xc ti 9 ch s thp phn ) Bi 1 : Cho hm sf(x) = a, Tnh gn ng n 5 ch s thp phn gi tr hm s ti x = 1 +b, Tnh gn ng n 5 ch s thp phn gi tr cc s a , b sao cho ng thng y =ax +b l tip tuyn ca th hm s ti im c honh x = 1 + Bi 2 : Tnh gn ng n 5 ch s thp phn gi tr ln nht ca hm s f(x)= trn tpcc s thc S={x: } Bi 3 : Cho ;vi0 n 998 , Tnh gn ng gi tr nh nht [ ] Bi 4 : Tnh gn ng n 5 ch s thp phn gi tr ca im ti hn ca hm sf(x) = trn on[0;2 ] Bi 5 : Trong mt phng to Oxy , cho hnh ch nht c cc nh (0;0) ; (0;3) ; (2;3) ; (2;0) c di n v tr mi bng vic thc hin lin tip 4 php quay gc theo chiu kimng h vi tm quay ln lt l cc im (2;0) ; (5;0) ; (7;0) ; (10;0) . Hy tnh gnng n 5 ch s thp phn gi tr din tch hnh phng gii hn bi ng cong do im(1;1) vch ln khi thc hin cc php quay k trn v bi cc ng thng : trc Ox ; x=1; x=11 Bi 6 : Mt bn c vung gm 1999x1999 mi c xp 1 hoc khng xp qun c no .Tm s b nht cc qun c sao chokhi chn mt trng bt k , tng s qun c tronghng v trong ct cha t nht l 199 Bi 7 : Tam gic ABC c BC=1 , gc . Tnh gn ng n 5 ch s thp phn gi trkhong cch gia tm ng trn ni tip v trng tm ca tam gic ABC. Bi 8 : Tnh gn ng n 5 ch s thp phn gi tr cc h s a, b ca ng thng y=ax+b ltip tuyn ti M(1;2) ca Elp =1 bit Elp i qua im N(-2; ) Bi 9 : Xt cc hnh ch nht c lt kht bi cc cp gch lt hnh vung c tng din tch l1 ,vic c thc hin nh sau : hai hnh vung c xp nm hon ton trong hnh ch nhtm phn trong ca chng khng ln nhau cc cnh ca 2 hnh vung th nm trn hocsong song vi cc cnh ca hnh ch nht . Tnh gn ng khng qu 5 ch s thp phngi tr nh nht din tch hnh ch nht k trn Bi 10 : Cho ng congy =, m l tham s thc. a, Tnh gn ng n 5 ch s thp phn gi tr ca m tim cn xin ca th hm sTo vi cc trc to tam gic c din tch l 2b, Tnh gn ng n 5 ch s thp phn gi tr m ng thng y=m ct th ti haiim A, B sao cho OA vung gc vi OB HT UBND TNH BC NINH THI HC SINH GII THPT S GIO DC O TO Gii ton trn MTT CASIO nm 2004 2005 Thi gian : 150pht ----------------------------------------------------------------- Bi 1 ( 5 im ) Trong cc s sau 2; ; ;6 3 4 3 s no l nghim dng nh nht ca phng trnh : 2sin sin 2 cos 2cos x x x + = + xBi 2 ( 5 im ) Gii h :22log 4.3 67. log 5.3 1xxxx + =+ = Bi 3 ( 5 im ) Cho a thc : ( )3 22 5 1 f x x x x = +a, Tnh ( gn ng n 5 ch s thp phn ) s d ca php chia f(x) cho 12x + b, Tnh ( gn ng n 5 ch s thp phn ) nghim ln nht ca phng trnh : f(x) = 0 Bi 4 ( 5 im )

Bi5 ( 5 im ) 1. Tm tt c cc cp s t nhin (x,y) sao cho x l c ca v y l c ca2. Chng minh rng phng trnh c nghim t nhin khi v ch khi a=3 Tm tt c cc cp s t nhin (x,y) l nghim ca phng trnh3. Tm tt c cc b s t nhin (x,y,z) l nghim ca phng trnh :Bi 6 ( 5 im ) : T mt phi hnh nn chiu cao12 3 h =v bn knh yR=5 2 c th tin c mt hnh tr cao nhng y hp hoc hnh tr thp nhng y rng . Hy tnh ( gn ng 5 ch s thpphn ) th tch ca hnh tr trong trng hp tin b t vt liu nht .Bi 7 ( 5 im ) : Cho hm sy=c th (C) , ngi ta v hai tip tuyn ca th ti im chonh v ti im cc i ca th hm s . Hy tnh ( gn ng 5 ch s thp phn )din tch tam gic tao bi trc tung v hai tip tuyn cho. Bi 8 ( 5 im ) Hy tnh ( gn ng 4 ch s thp phn ) l nghim ca phng trnh:

Bi 9 ( 5 im ) Hy tnh ( gn ng 4 ch s thp phn )Bi 10 ( 5 im ) Tm ch s hng n v ca s HT CHN I TUYN TRUNG HC C S (S GIO DC BC NINH NM 2005) Bi 1 : 1.1: Tm tt c cc s c 10 ch s c ch s tn cng bng 4 v l lu tha bc 5 ca mt s t nhin. S : 1073741824, 2219006624, 4182119424 , 7330402241.2 : Tm tt c cc s c 10 ch s c ch s u tin bng 9 v llu tha bc nm ca mt s t nhin. S : 9039207968 , 9509900499 Bi 2 :2.1. Tm s c 3 ch s l lu tha bc 3 ca tng ba ch s ca n. S : 512 2.2. Tm s c 4 ch s l lu tha bc 4 ca tng bn ch s c n. S : 2401 2.3. Tn ti hay khng mt s c nm ch s l lu tha bc 5 ca tng nm ch s ca n ? S : khng c s no c 5 ch s tho mn iu kiu bi Bi 3 : 3.1. Cho a thc bc 4f(x) = x4+bx3+cx2+dx+43 c f(0) = f(-1);f(1) = f(-2) ; f(2) = f(-3) . Tm b, c, d S : b = 2 ; c = 2 ; d = 1 3.2. Vi b, c, d va tm c, hy tm tt c cc s nguyn n sao cho f(n) = n4+bn3+cn2+n+43 l s chnh phng. S : n = -7 ; - 2 ; 1 ; 6 Bi 4 : T th trn A n Bc Ninh c hai con ng tovi nhau gc 600 . Nu i theo ng lin tnh bn tri n th trn B th mt 32 km ( k t th trn A), sau r phi theo ng vung gc v i mt on na th s n Bc Ninh.Cn nu t A i theo ng bn phi cho n khi ct ng cao tc th c ng na qung ng, sau r sang ng cao tc v i nt na qung ng cn li th cng s n Bc Ninh .Bit hai con ng di nh nhau. 4.1. Hi i theo hng c on ng cao tc n Bc Ninh t th trn A thi nhanh hn i theo nglin tnh bao nhiu thi gian( chnh xc n pht), bit vn tc xe my l 50 km/h trn ng lin tnh v 80 km/ h trn ng cao tc. S :10 pht 4.2. Khong cch t th trn A n Bc Ninh l bao nhiu mt theo ng chim bay. S :34,235 km Bi 5 : Vi n l s t nhin, k hiu an l s t nhin gn nht can.Tnh 2005 2 1 2005... a a a S + + + =. S :598652005 = SBi 6 : 6.1. Gii phng trnh : 22333 1 5 33 5 355 5 9x xx xxx x ++ + = + + +S : ( )22 5 32 , 1 = x ; ( )5 22 5 36 , 5 , 4 , 3 = x 6.2. Tnh chnh xc nghim n 10 ch s thp phn. S : ; ; 618033989 , 11 x 381966011 , 12 x ;850650808 , 04 , 3 x 7861511377 , 06 , 5 xBi 7 : 7.1. Trc cn thc mu s : 3 39 3 2 2 12 += MS :1 2 9 723 6+ + + = M7.2 Tnh gi tr ca biu thc M ( chnh xc n 10 ch s) S : 533946288 , 6 = MBi 8 : 8.1 Cho dy s,11 0= = a a1211++=nnnaaa Chng minh rngvi mi 0 1 31221= + ++ + n nnna a a a0 n8.2. Chng minh rng vi mi1 13 + =n n na a a 1 n8.3.Lp mt quy trnh tnh aiv tnh ai vi i = 2 , 3 ,,25 Bi 9 : 9.1. Tm tt c cc cp s t nhin (x,y)sao cho x l c ca y2+1 v y l c ca x2+1 9.2. Chng minh rng phng trnh x2 + y2 axy + 1 = 0 c nghim t nhin khi v ch khi a = 3. Tm tt c cc cp s t nhin ( x, y, z ) l nghim ca phng trnh x2 + y2 3xy + 1 = 0 9.3 .Tm tt c cc cp s t nhin ( x, y, z ) l nghim ca phng trnh x2(y2 - 4) = z2 + 4 S : , y = 3 , na x =12 3 =n na a zBi 10 : Cho mt s t nhin c bin i nh mt trong cc php bin i sau Php bin i 1) : Thm vo cui s ch s 4 Php bin i 2) : Thm vo cui s ch s 0 Php bin i 3) : Chia cho 2 nu ch s chn Th d: T s 4, sau khi lm ccphp bin i 3) -3)-1) -2) ta c 140 14 1 2 4) 2 ) 1 ) 13 ) 3 10.1. Vit quy trnh nhn c s 2005 t s 4 10.2. Vit quy trnh nhn c s 1249 t s 4 10.3. Chng minh rng, t s 4 ta nhn c bt k s t nhin no nh 3 phpbin s trn. HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN TH THCS, lp 6, 2001-2002 Bi 1: Tnh 1 3 5 7 9 11 13 152 4 8 16 32 64 128 256A = + + + + + + +Bi 2: So snh cc phn s sau: 19 1919 191919 19191919; ; ;27 2727 272727 27272727 Bi 3: Tnh 1994 1993 2 1993 19941994 2121211992 1992 1994 19931993 1994 434343B = ++ Bi 4: Tm v lm trn n su ch s thp phn: 3 0, 4 0, 09 (0,15 2, 5) (2,1 1, 965) (1, 2 0, 045)0, 32 6 0, 03 (5, 3 3, 88) 0, 67 0, 00325 0, 013C = + + + Bi 5: Tm x v lm trn n ch s thp phn th nm:

13 7 7 1 11, 4 2, 5 2 4 0,1 70, 5 528 784 180 18 2 2A = + Bi 6: Tm x v lm trn n bn ch s thp phn: 1 1 1 1 1... 140 1, 08 [0, 3 ( -1)] 1121 22 22 23 23 24 28 29 29 30x + + + + + + = Bi 7: Mt ao c c 4800 con c gm ba loi: trm , m, chp. S m bng3 s trm, s chp bng5 s m. Tnh s lng mi loi c trong ao. 7 7 Bi 8: Tm cc c chung ca cc s sau: 222222;506506;714714;999999 Bi 9: S 19549 l s nguyn t hay hp s? Bi 10: Chia s 6032002 cho 1905 c s d lr . Chia cho 209 c s d l. Tmr . 1 1r2r2Bi 11: Hi c bao nhiu s gm 5 ch s c vit bi cc ch s 1,2,3 v chia ht cho 9? Bi 12: Tnh din tch hnh thang c tng v hiu hai y ln lt l 10,096 v 5,162; chiu cao hnh thang bng 23 tch hai y.Bi 13: Tnh: 1111111111111 1+++++++ Bi 14: Tnh tng din tch ca cc hnh nm gia hnh thang vhnh trn ( phn mu trng ). Bit chiu di hai y hnh thang l 3m v 5m, din tch hnh thang bng 220m Bi 15: Tnh din tch phn hnh ( mu trng ) gii hn bi 4 hnh trn bng nhau c bn knh l 12cm . HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHCS, lp 7

Bi 1: So snh cc phn s sau: 19 1919 191919 19191919; ; ;27 2727 272727 27272727 Bi 2: Tm x v lm trn n nm ch s thp phn:

13 7 7 1 11, 4 2, 5 2 4 0,1 70, 5 528 784 180 18 2 2A = + Bi 3: Tm x v lm trn n bn ch s thp phn: 3 0, 4 0, 09 (0,15 2, 5) (2,1 1, 965) (1, 2 0, 045)0, 32 6 0, 03 (5, 3 3, 88) 0, 67 0, 00325 0, 013C = + + + Bi 4: Tnh: 12121212121212122++++++++ Bi 5: Dn s nc ta nm 1976 l 55 triu vi mc tng 2,2 %. Tnh dn s nc ta nm 1986. Bi 6: Tnh : 2 3 47 22 5 2 16 773 2 16 17 4 3 15 20h ph g h ph gh ph g h ph gD + = + Bi 7: Tm s nguyn dng nh nht tha: chia 2 d 1, chia 3 d 2, chia 4 d 3, chia 5 d 4, chia 6 d 5, chia 7 d 6, chia 8 d 7. Bi 8: Vit quy trnh tm phn d ca php chia 19052002 cho 20969. Bi 9: |Cho x = 1,8363. Tnh 5 4 23 2 351 x x x xCx+ +=+ Bi 10: Tm thi gian xe p ht qung ng ABC di 186,7km. Bit xe i trn qung ng AB = 97,2km vi vn tc 16,3lm/h v trn qung ng BC vi vn tc 18,7km/h. Bi 11: Hi c bao nhiu s gm 6 ch s c vit bi cc ch s 2, 3, 7 v chia ht cho 9? Bi 12: Tm mt s gm ba ch s dngxyzbit tng ca ba ch s bng php chia 1000 cho xyzBi 13: Mt ngi ngi s dng xe c gi trj ban u l 10triu. Sau mi nm, gi tr ca xe gim 10% so vi nm trc . 1) Tnh gi tr ca xe sau 5 nm. 2) Tnh s nm gi tr ca xe nh hn 3 triu. Bi 14: Tam gic ABC c y BC = 10, ng cao AH = 8. Gi I v O ln lt l trung im ca Ah v BC. Tnh din tch cc tam gic IOA v IOC. Bi 15: Tnh din tch phn hnh ( mu trng ) gii hn bi 4 hnh trn bng nhau c bn knh l 9cm . HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHCS, lp 8, 2001-2002 Bi 1: So snh cc phn s sau: 19 1919 191919 19191919; ; ;27 2727 272727 27272727 Bi 2: Tnh 2 24100, 6 1, 256 1 3 25 3551 5 1 1 5 2 50.61 6 3 225 9 4 17 + + Bi 3: Tm x v lm trn n bn ch s thp phn: 1 1 1 1 1... 140 1, 08 [0, 3 ( -1)] 1121 22 22 23 23 24 28 29 29 30x + + + + + + = Bi 4: Tnh: 1313131313133+++ Bi 5: Tm cc c chung ca cc s sau: 222222;506506;714714;999999 Bi 6: Chia s 19082002 cho 2707 c s d lr . Chia cho 209 c s d l. Tmr . 1 1r2r2Bi 7: Hi c bao nhiu s gm 6 ch s c vit bi cc ch s 2, 3, 5 v chia ht cho 9? Bi 8: Vit quy trnh tm phn d ca php chia 19052002 cho 20969. Bi 9: Tm s nguyn dng nh nht tha: chia 2 d 1, chia 3 d 2, chia 4 d 3, chia 5 d 4, chia 6 d 5, chia 7 d 6, chia 8 d 7, chia 9 d 8, chia 10 d 9. Bi 10: Tam gic ABC c y BC = 10. ng cao AH = 8. Gi I v O ln lt l trung im AH v BC . Tnh din tch ca tam gic IOA v IOC. Bi 11: Phn tch a thc thnh nhn t4 3 2( ) 2 13 14 2 Px x x x x = + + 4Bi 12: Tm mt s gm ba ch s dngxyzbit tng ca ba ch s bng php chia 1000 cho xyzBi 13: Mt ngi b bi vo hp theo quy tc: ngy u 1 vin, mi ngy sau b vo s bi gp i ngy trc . Cng lc cng ly bi ra khi hp theo quy nguyn tc: ngy u v ngy th hai ly mt vin, ngy th ba tr i mt ngy ly ra s bi bng tng hai ngy trc 1) Tnh s bi c trong hp sau 10 ngy. 2) s bi c trong hp ln hn 1000 cn bao nhiu ngy? Bi 14: Cho hnh thang vung ABCD( ) AB CD , F l im nm gia CD, AF ct BC ti E. Bit . Tnh din tch tam gic BEF.1, 482; 2, 7182; 2 AD BC AB = = =Bi 15: Tnh din tch phn hnh ( mu trng ) gii hn bi 4 hnh trn bng nhau c bn knh l 13cm . HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHCS, lp 9, 2001-2002 Bi 1: Tnh ( lm trn n 6 ch s thp phn): 4 3 5 6 7 8 9 11 2 3 4 5 6 7 8 9 1 A = + + + + 00Bi 2: Tnh 2 24100, 6 1, 256 1 3 25 3551 5 1 1 5 2 50.61 6 3 225 9 4 17 + + Bi 3: Tnh ( lm trn n 4 ch s thp phn): 9 8 7 6 5 4 39 8 7 6 5 4 3 2 C =Bi 4: Tm phn d ca php chia a thc: 5 4 3 2(2 1, 7 2, 5 4, 8 9 1) ( 2, 2) x x x x x x + Bi 5: Tm cc im c ta nguyn dng trn mt phng tha mn: 2x + 5y = 200 Bi 6: Phn tch a thc thnh nhn t 4 3 2( ) 2 15 26 120 Px x x x x = + +Bi 7: Mt ngi b bi vo hp theo quy tc: ngy u 1 vin, mi ngy sau b vo s bi gp i ngy trc . Cng lc cng ly bi ra khi hp theo quy nguyn tc: ngy u v ngy th hai ly mt vin, ngy th ba tr i mt ngy ly ra s bi bng tng hai ngy trc 1) Tnh s bi c trong hp sau 15 ngy. 2) s bi c trong hp ln hn 2000 cn bao nhiu ngy? Bi 8: Vit quy trnh tm phn d ca php chia 26031913 cho 280202. Bi 9: Tnh ( cho kt qu ng v kt qu gn ng vi 5 ch s thp phn): 11121314151617189++++++++ Bi 10: Tm s nguyn dng nh nht tha: chia 2 d 1, chia 3 d 2, chia 4 d 3, chia 5 d 4, chia 6 d 5, chia 7 d 6, chia 8 d 7, chia 9 d 8, chia 10 d 9. Bi 11: Tm nghim gn ng vi su ch s thp phn ca 22 3 3 1, 5 x x 0 + = Bi 12: S no trong cc s 33; ; 3;1, 87 l nghim ca phng trnh 4 3 22 5 3 1, 5552 0 x x x + =Bi 13: Cho 20cotA=21. Tnh 2Asin os2Acos sin 23A cBA=+ Bi 14: Cho tam gic ABC c AH l ng cao. Tnh di BH v CH bit .3; 5; 7 AB AC BC = = =Bi 15: Tnh din tch phn hnh nm gia tam gic v cc hnh trn bng nhau c bn knh l 3cm ( phn mu trng ) HTS GIO DC O TO THI TUYN HC SINH GII MY TNH B TICN THTHPT, lp 10, 2001-2002 Bi 1: Tm x ( , pht, giy), bit 18v tanx = 0,7065193280 270ox < Tnh gi tr gn ng vi 5 ch s thp phn ng cao ca hnh chp Bi 8 Cho tam gic ABC c bn knh ng trn ngoi tip v ni tip ln lt l 3,9017 cm v 1,8225 cm.Tm khong cch gia hai tm ca hai ng trn.Bi 9 Cho tam gic ABC c gc A= ,ni tip trong ng trn c bn knh R=5,312 cm ' 24 80 ; ' 15 350 0= BA,Tnh din tch tam gic ABC B,Tnh bn knh r ca ng trn ni tip tam gic ABC Bi 10 Cho hnh t din ABCD,gi M,N ln lt l trung im ca AB v AD,P l im trn CD sao cho PD=2PC.Mt phng (MNP) chia t din ABCD thnh hai phn .Tnh t s th tch phn cha nh A v phn cha nh B ca t din. thi casio nm 2005-2006 Bi 1: Cho cosx=0,765 v 00

9STOX,ALPHAX,ALPHA=,ALPHAX+1, ALPHA:,72ALPHAX-(3ALPHAX^5-240677),bm=lintip.KhiX=32thckt qu ca biu thc nguyn y = 5. Thay x = 32 vo phng trnh (*), gii pt bc 2 theo y, ta c thm nghim nguyn dng y2 =4603. ( )( )32; 5;32; 4603x yx y= == = Li gii Kt qu x = 32 0,5 0,5 2 4.1 Sau 4 nm, bn Chu n ngn hng: A= 4 3 22000000(1.03 1.03 1.03 1.03) 8618271.62 + + + NmthnhtbnChuphigp12m(ng).Gi 1 0.03 1.03 q = + =Sau nm th nht, Chu cn n: 112 x Aq m = Saunmthhai,Chucnn: ( )2212 12 12 ( 1) x Aq mq m Aq mq = = +...Saunmthnm,Chucnn . 5 4 3 2512 ( 1) x Bq m q q q q = + + + +Gii phng trnh: ,tac 5 4 3 2512 ( 1) 0 x Bq m q q q q = + + + + =156819 m = Cch gii Kt qu cui cng ng 0,5 0,5 4 4.2 Thng th nht, sau khi gp cn n: A = 5000000 -100000 = 4900000 (ng). 4900000 STO A, 100000 STO B, th: Thng sau gp: B = B + 200000 (gi tr trong nh B cng thm 20000), cn n: A= A1,007 -B. Thc hin qui trnh bm phm sau: 4900000 STO A, 100000 STO B, 1 STO D, ALPHA D, ALPHA =, ALPHA D+1, ALPHA : , ALPHA B, ALPHA =, ALPHA B + 20000, ALPHA : , ALPHA A, ALPHA =, ALPHA A1,007 - ALPHA B, sau bm = lin tip cho n khi D = 19 (ng vi thng 19 phi tr gp xong cn n: 84798, bm tip =, D = 20, A m. Nh vy ch cn gp trong 20 thng th ht n, thng cui ch cn gp : 847981,007 = 85392 ng. Cch gii Kt qu cui cng ng 0,5 0,5 2 5 32013' 18"cbaaaABCD a = 3,84 ; c = 10 (cm) 2 22 cos 7.055029796 b a c ac D = + 2 222cos 0, 68773889942a bBa= n0133 27' 5" ABC 15.58971171ABCDS 0,5 0,5 2 .27.29018628; 4.992806526SH MHSH IHMH MS= = =+=R(bnknhmtcu ni tip). Thtchhnhcu(S1): 3343521.342129( )V Rcm =. 28, 00119939 SM 6, 27; MH IK I = = H0,5 0,5 6 Khong cch t tm I n mt phng i qua cc tip im ca (S1) vi cc mt bn ca hnh chp: 24.866027997IHd EISH IH= = = Bn knh ng trn giao tuyn:2 21,117984141 r EK R d = = Din tch hnh trn giao tuyn: 274, 38733486( ) S c m0,5 0,5 2 F l s l, nn c s ca n khng th l s chn. F l s nguyn t nu n khng c c s no nh hn 106.0047169 F = . gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, 11237ALPHA D, bm = lin tip (my 570ES th bmCALCsaumibm=).Nut3chon 105phpchiakhngchn,thktlunFls nguyn t. Qui trnh bm phm Kt qu: F: khng nguyn t 0,5 0,5 (1897, 2981) 271 UCLN = . Kim tra thy 271 l s nguyn t. 271 cn l c ca3523. Suy ra: ( )5 5 5 5271 7 11 13 M = + +Bm my tnh. 5 5 57 11 13 549151 A = + + =gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, 549151ALPHAD,bm=lintip,phpchia chn vi D = 17. Suy ra: 17 32303 A = Bngthutgiikimtrasnguyntnhtrn,ta bit 32303 l s nguyn t. 0,5 7 Vy cc c nguyn t ca M l: 17; 271;323030,5 SKI720MHDABCSE KIMHTa c: 1 2345103 3(mod10); 103 9(mod10);103 3 9 27 7(mod10);103 21 1(mod10);103 3(mod10); = Nh vy cc lu tha ca 103 c ch s tn cng lin tip l: 3,9,7,1 (chu k 4). 2006 2(mod10) , nn c ch s hng n v l 9. 20061030,5 0,5 8 1 23 45 629 29( 1000); 29 841(mod1000);29 389(mod1000); 29 281(mod1000);29 149(mod1000); 29 321(mod1000);Mod ( )210 5 220 240 8029 29 149 201(mod1000);29 201 401(mod1000);29 801(mod1000); 29 601(mod1000);= 100 20 8029 29 29 401 601 1(mod1000); = ( )202000 100 202006 2000 629 29 1 1(mod1000);29 29 29 1 321(mod1000);= = Ch s hng trm ca P l 3. 1,0 2 Gii thut: 1 STO A, 0 STO D, ALPHA D, ALPHA =, ALPHA D + 1, ALPHA : , ALPHA A, ALPHA =, ALPHA A + (-1)D-1 x ((D-1)D2. Sau bm = lin tip, theo di s m D ng vi ch s ca uD, ta c: 4 5 6113 3401 967; ;144 3600 1200u u u = = = ; 1,0 9 200, 8474920248; u u25 0,8895124152; u30 0.8548281618 1,0 2 u10 = 28595 ;u15 = 8725987 ;u21 = 98848794231,0 S10 = 40149 ;S15 = 13088980 ;S20 = 49424397110,5 10 1STOA,2STOB,3STOM,2STOD,ALPHAD, ALPHA=, ALPHA D+1, ALPHA : , ALPHA C, ALPHA =,ALPHA3ALPHAA,+,2ALPHAB,ALPHA:, ALPHA M, ALPHA =, ALPHA M + ALPHA C, ALPHA : ALPHA A, ALPHA =, ALPHA B, ALPHA : , ALPHA B, ALPHA =, ALPHA C, ALPHA : ,ALPHA D, ALPHA=, ALPHA D+1, ALPHA : , ALPHA C,ALPHA=,ALPHA2ALPHAA,+,3ALPHAB, ALPHA : , ALPHA M, ALPHA =, ALPHA M + ALPHA C, ALPHA : ALPHA A, ALPHA =, ALPHA B, ALPHA : ,ALPHAB,ALPHA=,ALPHAC,saubm=lin tip, D l ch s, C l uD , M l SD 0,5 2 Bi 2: TX: R. Y' = 13*x^2-14*x-2/(3*x^2-x+1)^2 ( )22213 14 2'3 1x xyx x = +, 1 2' 0 1.204634926; 0.1277118491 y x x = = = 1 20.02913709779; 3.120046189 y y = =1 23.41943026 d M M = =Y"=-6*(13*x^3-21*x^2-6*x+3)/(3*x^2-x+1)^3 Bi 3:0.4196433776 x ( )3 2326(13 21 6 3)"3 1x x xyx x += +, 1 2 3" 0 1.800535877; 0.2772043294; 0.4623555914 y x x x = = = = 1 2 30.05391214491; 1.854213065; 2.728237897 y y y = = =Bi 4: 83 17;13 13C 16.07692308; 9.5ADC ABCS S Din tch hnh trn ngoi tip ABCD: ( )58.6590174ABCDS Bi 5: Sau 4 nm, bn Chu n ngn hng: A=4 3 22000000(1.03 1.03 1.03 1.03) 8618271.62 + + + Nm th nht bn Chu phi gp 12m (ng). Gi1 0.03 1.03 q = + =Sau nm th nht, Chu cn n: 112 x Aq m = Sau nm th hai, Chu cn n:( )2212 12 12 ( 1) x Aq mq m Aq mq = = +... Sau nm th nm, Chu cn n 5 4 3 2512 ( 1) x Bq m q q q q = + + + + . Gii phng trnh, ta c 5 4 3 2512 ( 1) 0 x Bq m q q q q = + + + + = 156819 m =Bi 6: .27.29018628; 4.992806526SH MHSH IHMH MS= = =+: bn knh mt cu ngoi tip. Th tch hnh cu (S1):.521.342129 V =Bn knh ng trn giao tuyn: 24.866027997 74.38734859IHr SSH IH= = = HT S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 12 BTTH - Nm hc 2005-2006 Thi gian: 120 pht (Khng k thi gian giao ) Ngy thi: 03/12/2005. Ch :- thi gm4 trang - Th sinh lm bi trc tip vo bn thi ny. - Nu khng ni g thm, hy tnh chnh xc n 10 ch s. im ton bi thi Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng thi ghi) GK1 Bng sBng ch GK2 Bi 1:Cho hm s 222 3 5( )1x xf xx+ =+c th (C). Ga s ng thng y = ax + b tip tuyn ca th hm s ti im trn (C) c honh 305 x = . Tnh gn ng cc gi tr ca a v b. S lc cch gii: Kt qu: Bi 2: Cho hm s 222 3 5( )1x xf xx+ =+c th (C). Xc nh to ca cc im un ca th (C) ca hm s cho. S lc cch gii: Kt qu: im un U1:11xy im un U2:22xy im un U3:33xy Bi 3: Tmnghimgnng(,pht,giy)caphngtrnh5sin3 6cos3 7 x x + = trong khong( ) 1900; 2005 . S lc cch gii: Kt qu: 1x 2x Bi 4: Tnh gn ng gi tr ln nht v gi tr nh nht ca hm s: sin 2cos 1( )2 cosx xf xx+ +=+trn on[ ] 0; 4S lc cch gii: Kt qu: Bi 5: TrongmtphngvihtoOxy,chotamgicABCbitccnh .( ) ( ) ( 1;1, 4; 2, 2; 3 A B C ) 5.1 Tnh gn ng s o (, pht, giy) ca gc nBACv din tch tam gic ABC. S lc cch gii: Kt qu: 5.2 Tnh to tm v din tchhnh trn ngoi tip tam gic ABC. S lc cch gii: Kt qu: Bi 6: ChohnhchptgicuS.ABCDccnhy12, 54( ) a cm = ,cccnhbn nghing vi y mt gc. 072 =Tnh th tch v din tch xung quanh ca hnh chp S.ABCD. S lc cch gii: Kt qu: Bi 7: Tnh gn ng gi tr ca a v b nu ng thng y ax b = +i qua im(5; 4) M v l tip tuyn ca hypebol 2 2116 9x y = . S lc cch gii: Kt qu: 11ab 22ab Bi 8: Tnh gn ng cc nghim ca phng trnh3 4cos 2 5xx x = + S lc cch gii: Kt qu: Bi 9: Bitathcchiahtchoccnhthc 4 3 2( ) 11 Px x ax bx cx = + + + 1; 2; 3 x x x + . Tnh cc h s v cc nghim ca a thc P(x)., , abcS lc cch gii: Kt qu: a = ;b = c =; x1 = x2 = ; x3 = x4 = Bi 10:Trong mt phng vi h to Oxy, cho hai ng trn c phng trnh: ( )( )2 212 22: 2 4 1: 6 8 16C x y x yC x y x y0,0+ + + =+ + = 10.1 Tnh gn ng to cc giao im A v B ca hai ng trn. 10.2 Tnh di cung nh pABca ng trn( )1CS lc cch gii: Kt qu: UBND TNH Tha Thin Huk thi chn hoc sinh gii tnh S Gio dc v o tolp 12 BTTH nm hc 2005 - 2006 Mn :MY TNH B TIp n v thang im: BiCch giip s im TP im ton bi 1,179874664 a 1,0 1 0, 4941280673 b 1,0 2 Tnh c( )( )( )3 23223 21 9 7"1x x xf xx +=+ 3 2"( ) 0 3 21 9 7 0 f x x x x = + =0.5 0.5 2 Gii phng trnh c: 1 237, 364344451; 0, 4094599913;0, 7738044428x xx Dng chc nng CALC tnh c: 1 232, 273258339; 2, 942905007;3, 830353332y yy 0.5 0.5 2 t 32xt tg = , phng trnh tng ng: 213 10 1 0 t t + =0,5 Gii phng trnh ta c: 1 20, 6510847396; 0,1181460296 t t 0,5 Suy ra nghim tng qut ca phng trnh: ( )0 00 022 2' 42" 1204 29' 31" 120x kkx k + +Z0,5 3 22.04502486 ShiftSTO A ; 4.492022533 Shift STO B ; -1 STO D (bin m); ALPHA, D, ALPHA, CALC (=), ALPHA, D + 1; ALPHA, : ;... D=D+1 : A+120D : B+120Dsau n lin tip= ng vi k = 16, ta c 2 nghim ca phng trnh trong khong (1900 ; 2005) l:0 01 21942 2' 42"; 1924 29' 31" ; x x 0,5 2 ( )22cos 3sin 1'( )2 cosx xf xx +=+ Giipt:trn on [0 ; 4], ta c: '( ) 0 2 cos 3sin 1 0 f x x x = +=1 20, 8690375051; 3, 448560356 x x 0,50 4 1 21,154700538; 1,154700538 y y 0,50 2 So snh vi, ta c: (0) 1; (4) 0, 7903477515 f f = [ ][ ] 0;40;41,154700538;1,154700538( )( )Max f xMin f x 0,50 5 l0cos 0, 4280863447 115 20' 46" A A 1 1. sin2 2ABCS AB AC A = =9 Phng trnh ng trn c dng: Tm ng trn (ABC) l: 83 73;38 38I Din tch hnh trn ngoi tip tam gic ABC: 258, 6590174( ) S c m1,0 0,5 0,5 2 Chiu cao ca hnh chp: 0272 27, 290186282aSH tg = Th tch khi chp ( )211430, 4751523V a h cm = 3 0,5 0,5 6Trung on ca hnh chp: = + 2228, 001199394ad SHDin tch xung quanh ca hnh chp: ( )21.4 . 702, 27008072xqS a d cm = 0,5 0,5 2 ng thng y = ax + b i qua im M(5; 4) nn: 5 4 B a = p dng iu kin tip xc: ( )22 216 9 5 4 9 40 25 0 a a a a = + + =0,5 0,5 1 20, 7523603827; 3, 692084062 a a 0,5 7 1 20, 2381980865; 14, 46042031 b b 0,5 2 Dng chc nng SOLVE gii phng trnh: 3 4 cos 2 5 0xx x =Vi gi tr u X = 0, ta c mt nghim: 10, 414082619 x 0,5 0,58 Vi gi tr u X = 1, ta c mt nghim: 21.061414401 x 1,0 2 9 Gii h pt: 43 2108 4 2 11 23 3 3 11 3a b ca b ca b b + = + + = + + = 4 35;6253256abc= == 1,0 2 11( ) ( 1)( 2)( 3)6P x x x x x = + 0,5 Cc nghim ca a thc l: 1 2 3 4111; 2; 3;6x x x x = = = =0,5 ( )( )2 212 222 2: 2 4 1 0,0 : 6 8 162 4 1 01524 + + + =+ + = + + + = = C x y x yC x y x yx y x yy x{ 215 5161524x xy x + = = 0 1,0 10 Gii phng trnh ta c: 1 20, 9873397172; 0, 01266028276 x x 1 21, 775320566; 3, 724679434 y y + Gc n1,15244994( ) AIB Rad + di cung nh p: 2, 304599881 ABl 0,5 0,25 0,25 2 Bi 2: TX: R. Y' = 13*x^2-14*x-2/(3*x^2-x+1)^2 ( )22213 14 2'3 1x xyx x = +, 1 2' 0 1.204634926; 0.1277118491 y x x = = = 1 20.02913709779; 3.120046189 y y = =1 23.41943026 d M M = =Y"=-6*(13*x^3-21*x^2-6*x+3)/(3*x^2-x+1)^3 Bi 3:0.4196433776 x ( )3 2326(13 21 6 3)"3 1x x xyx x += +, 1 2 3" 0 1.800535877; 0.2772043294; 0.4623555914 y x x x = = = = 1 2 30.05391214491; 1.854213065; 2.728237897 y y y = = =Bi 4: 83 17;13 13C 16.07692308; 9.5ADC ABCS S Din tch hnh trn ngoi tip ABCD: ( )58.6590174ABCDS Bi 5: Sau 4 nm, bn Chu n ngn hng: A=4 3 22000000(1.03 1.03 1.03 1.03) 8618271.62 + + + Nm th nht bn Chu phi gp 12m (ng). Gi1 0.03 1.03 q = + =Sau nm th nht, Chu cn n: 112 x Aq m = Sau nm th hai, Chu cn n:( )2212 12 12 ( 1) x Aq mq m Aq m q = = +... Sau nm th nm, Chu cn n 5 4 3 2512 ( 1) x Bq mq q q q = + + + + . Gii phng trnh, ta c 5 4 3 2512 ( 1) 0 x Bq mq q q q = + + + + = 156819 m =Bi 6: .27.29018628; 4.992806526SH MHSH IHMH MS= = =+: bn knh mt cu ngoi tip. Th tch hnh cu (S1):.521.342129 V =Bn knh ng trn giao tuyn: 24.866027997 74.38734859IHr SSH IH= = =. HT S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 12 THPT - Nm hc 2005-2006 Thi gian: 120 pht (Khng k thi gian giao ) Ngy thi: 03/12/2005. Ch :- thi gm5 trang - Th sinh lm bi trc tip vo bn thi ny. - Nu khng ni g thm, hy tnh chnh xc n 10 ch s. im ton bi thi Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng thi ghi) GK1 Bng sBng ch GK2 Bi 1:Cho cc hm s 22 42 3 5 2sin( ) ; ( )1 1 cosx x xf x gxx x+ = =+ +. 1.1 Hy tnh gi tr ca cc hm hp v( ( )) g f x ( ( )) f gxti 35 x = . S lc cch gii: Kt qu: ( ) ( )35 g f ( ) ( )35 f g 1.2 Tm cc nghim gn ng ca phng trnh( ) ( ) f x gx =trn khong( ) 6; 6 S lc cch gii: Kt qu: Bi 2: Cho hm s 222 5( )3 1x xy f xx x += =3 + . 2.1Xcnhimccivcctiucathhmsvtnhkhongcchgiacc im cc i v im cc tiu . S lc cch gii: Kt qu: im C: 11xy im CT: 22xy 2.2 Xc nh to ca cc im un ca th hm s cho. S lc cch gii: Kt qu: im un U1:11xy im un U2:22xy im un U3:33xy Bi 3: Tm nghim dng nh nht ca phng trnh ( ) ( )3 3sin cos 22x x x = + . S lc cch gii: Kt qu: Bi 4: TrongmtphngvihtoOxy,chohnhthangcnABCDbitccnh .( ) ( ) ( 1;1, 4; 2, 2; 3 A B D ) 4.1 Xc nh to ca nh C v tm ng trn ngoi tip hnh thang ABCD.S lc cch gii: Kt qu: 4.2 Tnh din tch hnh thang ABCD v din tch hnh trn ngoi tip n. S lc cch gii: Kt qu: Bi 5:5.1 Sinh vin Chu va trng tuyn i hc c ngn hng cho vay trong 4 nm hcmi nm 2.000.000 ng np hc ph, vi li sut u i3%/nm. Sau khi tt nghip i hc, bn Chu phi tr gp hng thng cho ngn hng s tinm (khng i) cng vi li sut 3%/nm trong vng 5 nm. Tnh s tin hng thng bn Chu phi tr n cho ngn hng (lm trn kt qu n hng n v). mS lc cch gii: Kt qu: 5.2 B bn Bnh tng cho bn y mt my tnh hiu Thnh Ging tr gi 5.000.000 ng bngcchchobntinhngthngviphngthcsau:ThngutinbnBnh cnhn100.000ng,ccthngtthngthhaitri,mithngnhncs tin hn thng trc 20.000 ng. Nu bn Bnh mun c ngay my tnh hc bng cchchnphngthcmuatrgphngthngbngstinbchovilisut 0,7%/thng,th bn Bnh phi tr gp bao nhiu thng miht n ? S lc cch gii: Kt qu: Bi 6: ChohnhchptgicuS.ABCDccnhy12, 54( ) a cm = ,cccnhbn nghing vi y mt gc. 072 =6.1 Tnh th tch hnh cu (S1) ni tip hnh chp S.ABCD. S lc cch gii: Kt qu: 6.2 Tnh din tch ca hnh trn thit din ca hnh cu (S1) ct bi mt phng i qua cc tip im ca mt cu (S1) vi cc mt bn ca hnh chp S.ABCD. S lc cch gii: Kt qu: Bi 7: 7.1 Hy kim tra s F =11237 c phi l s nguyn t khng. Nu qui trnh bm phm bit s F l s nguyn t hay khng. + Tr li: + Qui trnh bm phm: 7.2 Tm cc c s nguyn t ca s: . 5 51897 2981 3523 M = + +5S lc cch gii: Kt qu: Bi 8: 8.1 Tm ch s hng n v ca s: 2006103 N =8.2 Tm ch s hng trm ca s:200729 P =S lc cch gii: Kt qu: Bi 9: Cho 2 2 2 21 2 31 ... .2 3 4nnu1ni= + + + 1 i (= nunl,i 1 = ununchn,nls nguyn).1 n 9.1 Tnh chnh xc di dng phn s cc gi tr:u u . 4 5 6, ,9.2 Tnh gi tr gn ng cc gi tr:. 20 25 30, , u u u9.3 Nu qui trnh bm phm tnh gi tr ca nuu4 = ----------------------u5 = -----------------------u6 = ------------------------ u20 u25 u30 Qui trnh bm phm: Bi 10: Cho dy s xc nh bi: nu++++ = = = +11 2 212 31; 2;3 2nnn nu uu u uu un, nu , nu n l nchn10.1 Tnh gi tr ca 10 15 21, , u u u10.2 Gi l tng ca s hng u tin ca dy s nS n ( )nu . Tnh. 10 15 20, , S S Su10 = u15 = u21=S10 = S15 = S20 = Qui trnh bm phm tnh un v Sn: UBND TNH Tha Thin Huk thi chn hoc sinh gii tnh S Gio dc v o tolp 12 THPT nm hc 2005 - 2006 Mn :MY TNH B TIp n v thang im: BiCch giip s im TP im ton bi 1.1 i n v o gc v Radian Gn 35cho bin X, Tnh 222 31X X 5XY+ =+ 1, 523429229 Y v STO Y, Tnh 42sin( ) ( ( )) 1.9977467361 cosYg Y g f xY= = +. ( ( )) 1, 784513102 f gx 1,0 1 1.2 Dng chc nng SOLVE ly cc gi tr u ln lt l -6; -5; -4; ...,0;1; ...; 6 ta c cc nghim: 1 23 45, 445157771; 3, 751306384;1, 340078802; 1, 982768713x xx x 1,0 2 2.1 TX: R. ( )22213 14 2'3 1x xyx x = +, 1 2' 0 1.204634926; 0.1277118491 y x x = = = 1 20.02913709779; 3.120046189 y y = =1 23.41943026 d M M = = 0.5 0.5 2 ( )3 2326(13 21 6 3)"3 1x x xyx x += +, 1 23" 0 1.800535877; 0.2772043294;0.4623555914y x xx= = == 1 230.05391214491; 1.854213065;2.728237897y yy= == 0.5 0.5 2 0.4196433776 x 1,0 3 Nu cch gii ng: + a v ( ) ( )3 3cos cos 222x x x = + + Rt 3 214k x x = + 0,5 0,5 2 83 73;13 13C 16.07692308; 9.5ADC ABCS S 0,50 4 Din tch hnh trn ngoi tip ABCD: ( )58.6590174ABCDS TmngtrnngoitiptamgicABDcngl ng trn ngoi tip hnh thang ABCD: Tm ng trn (ABCD) l: 83 73 194; ;38 38 19I Din tch hnh trn ngoi tip hnh thang ABCD: 258, 6590174( ) S c m0,50 0,50 2 5.1 Sau 4 nm, bn Chu n ngn hng: A= 4 3 22000000(1.03 1.03 1.03 1.03) 8618271.62 + + + NmthnhtbnChuphigp12m(ng).Gi 1 0.03 1.03 q = + =Sau nm th nht, Chu cn n: 112 x Aq m = Saunmthhai,Chucnn: ( )2212 12 12 ( 1) x Aq mq m Aq mq = = +...Saunmthnm,Chucnn . 5 4 3 2512 ( 1) x Bq m q q q q = + + + +Gii phng trnh: ,tac 5 4 3 2512 ( 1) 0 x Bq m q q q q = + + + + =156819 m = Cch gii Kt qu cui cng ng 0,5 0,5 2 5 5.2 Thng th nht, sau khi gp cn n: A = 5000000 -100000 = 4900000 (ng). 4900000 STO A, 100000 STO B, th: Thng sau gp: B = B + 200000 (gi tr trong nh B cng thm 20000), cn n: A= A1,007 -B. Thc hin qui trnh bm phm sau: 4900000 STO A, 100000 STO B, 1 STO D, ALPHA D, ALPHA =, ALPHA D+1, ALPHA : , ALPHA B, ALPHA =, ALPHA B + 20000, ALPHA : , ALPHA A,ALPHA=,ALPHAA1,007-ALPHAB,sau bm=lintipchonkhiD=19(ngvi thng 19 phi tr gp xong cn n: 84798, bm tip =,D=20,Am.Nhvychcngptrong20 thngthhtn,thngcuichcngp: 847981,007 = 85392 ng. Cch gii Kt qu cui cng ng 0,5 0,5 .27.29018628; 4.992806526SH MHSH IHMH MS= = =+= R (bn knh mt cu ni tip). Th tch hnh cu (S1): 3343521.342129( )V Rcm = 28, 00119939 SM 6, 27; MH IK I = = H0,5 0,5 6 Khong cch t tm I n mt phng i qua cc tip im ca (S1) vi cc mt bn ca hnh chp: 24.866027997IHd EISH IH= = = Bn knh ng trn giao tuyn:2 21,117984141 r EK R d = = Din tch hnh trn giao tuyn: m 2 274, 38733486( ) S c0,5 0,5 F l s l, nn c s ca n khng th l s chn. F lsnguyntnunkhngccsnonh hn106.0047169 F = . gn 1 cho bin m D, thc hin cc thao tc: ALPHAD,ALPHA=,ALPHAD+2,ALPHA:, Qui trnh bm phm : khng 0,5 ,5 11237 ALPHA D, bm = lin tip (my 570ES thbmCALCsaumibm=).Nut3chon 105phpchiakhngchn,thktlunFls nguyn t. Kt qu: Fnguyn t 07 71 cn l c ca3523. Suy ra: )Bm my tnh. gn 1 cho bin m D, thc hin cc thao tc: A=,ALPHAD+2,ALPHA:, chia snguyntnhtrn,ta bit 32203 l s nguyn t. ,5 ,5 2 (1897, 2981) 271 UCLN = . Kim tra thy 271 l s nguyn t. 2(5 5 5 5271 7 11 13 M = + +5 5 57 11 13 549151 A = + + =ALPHAD,ALPH549151ALPHAD,bm=lintip,phpchn vi D = 17. Suy ra: 17 32303 A = Bngthutgiikimtra Vy cc c nguyn t ca M l: 17; 271;32303 0 0SKI720MHDABCSE KIMHTa c: 1 2345103 3(mod10); 103 9(mod10);103 3 9 27 7(mod10);103 21 1(mod10);103 3(mod10); = Nh vy cc lu tha ca 103 c ch s tn cng lin tip l: 3,9,7,1 (chu k 4). 2006 2(mod10) , nn c ch s hng n v l 9. 20061030,5 0,5 8 1 23 45 629 29( 1000); 29 841(mod1000);29 389(mod1000); 29 281(mod1000);29 149(mod1000); 29 321(mod1000);Mod ( )210 5 220 240 8029 29 149 201(mod1000);29 201 401(mod1000);29 801(mod1000); 29 601(mod1000);= 100 20 8029 29 29 401 601 1(mod1000); = ( )202000 100 202006 2000 629 29 1 1(mod1000);29 29 29 1 321(mod1000);= = Ch s hng trm ca P l 3. 1,0 2 Gii thut: 1 STO A, 0 STO D, ALPHA D, ALPHA =, ALPHA D + 1, ALPHA : , ALPHA A, ALPHA =, ALPHA A + (-1)D-1 x ((D-1)D2. Sau bm = lin tip, theo di s m D ng vi ch s ca uD, ta c: 4 5 6113 3401 967; ;144 3600 1200u u u = = = ; 1,0 9 200, 8474920248; u u25 0,8895124152; u30 0.8548281618 1,0 2 u10 = 28595 ;u15 = 8725987 ;u21 = 98848794231,0 10 S10 = 40149 ;S15 = 13088980 ;S20 = 49424397110,5 1STOA,2STOB,3STOM,2STOD,ALPHAD, ALPHA=,ALPHAD+1,ALPHA:,ALPHAC, ALPHA=,ALPHA3ALPHAA,+,2ALPHAB, ALPHA:,ALPHAM,ALPHA=,ALPHAM+ ALPHA C, ALPHA : ALPHA A, ALPHA =, ALPHA B, ALPHA : , ALPHA B, ALPHA =, ALPHA C, ALPHA : ,ALPHAD,ALPHA=,ALPHAD+1,ALPHA:, ALPHAC,ALPHA=,ALPHA2ALPHAA,+,3 ALPHA B, ALPHA : , ALPHA M, ALPHA=, ALPHA M+ALPHAC,ALPHA:ALPHAA,ALPHA=, ALPHAB,ALPHA:,ALPHAB,ALPHA=,ALPHA C, sau bm = lin tip, D l ch s, C l uD , M l SD 0,5 2 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 8 THCS - Nm hc 2006-2007 Thi gian: 120 pht-Ngy thi: 02/12/2006. Ch :- thi gm3 trang - Th sinh lm bi trc tip vo bn thi ny. - Nu khng ni g thm, hy tnh chnh xc n 10 ch s. im ton bi thi Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng thi ghi) GK1 Bng sBng ch GK2 Bi 1: Tnh gi tr ca cc biu thc: 2 2 43 3 21642 2x xy y x x yAx y x x + = ++ + y khi5;4 5x y = =22, ly kt qu chnh xc. 4 42 2 2 2 2 23 2 16 164 9 6 4 4x y x y x yBx y x xy y x y + = + + + + khi: A = B =a/( . 5; 16) x y = = B b/( 1, 245; 3, 456). x y = = Bi 2: Bit 20062007 1120081111abcdefg= ++++++, , , , , , d e fg. Tm cc s t nhina b c . a = ; b = c = ; d = e = ; f = g = Bi 3: a/ Phn tch thnh tha s nguyn t cc s sau: 252633033 v 8863701824. b/ Tm ch s b sao cho469283866chia ht cho 2007.3658 b Bi 4:Khai trin biu thc ta c a thc gi tr chnh xc ca biu thc: (1521 2 3 x x + +)0Tnh vi2 30 1 2 30... . a a x a x a x + + + +. 0 1 2 3 29 302 4 8 ... 536870912 1073741824 E a a a a a a = + + + Bi 5: Tm ch s l thp phn th11k t du phy ca s thp phn v hn tun hon ca s hu t 20071000029. Bi 6: Tm cc ch s sao cho s567abcdal s chnh phng. Nu qui trnh bm phm c kt qu. Bi 7: Cho dy s: 1 2 3 41 1 1 12 ; 2 ; 2 ; 21 122 2 21 122 21222u u u u = + = + = + = ++ + ++ ++1; ... 1212...122nu = ++ (biu thc c cha tng phn s).nTnh gi tr chnh xc cauu v gi tr gn ng ca. 5 9 1, , u0 15 20, u u u5 = ----------------------u9 = -----------------------u10 = ------------------------ Ch s l thp phn th 11ca 2007 1000029 l: E =Kt qu: Qui trnh bm phm: b/b =a/ 252633033 = 8863701824 = Bi8:Choathcbit 3 2( ) P x ax bx cx d = + + + (1) 27; (2) 125; (3) 343 P P P = = = v .(4) 735 P =u15 = ----------------------u20 = ----------------------- a/ TnhP P (Ly kt qu chnh xc.( 1); (6); (15); (2006). P P b/ Tm s d ca php chia( ) 3 5 P x cho x . S d ca php chia l:r = ( ) 3 5 P x cho x ( 1) ; (6))(15) ; (2006)P PP P = == = Bi 9: Li sut ca tin gi tit kim ca mt s ngn hng hin nay l 8,4% nm i vi tin gi c k hn mt nm. khuyn mi, mt ngn hng thng mi A a ra dch v mi: Nu khch hng gi tit kim nm u th vi li sut 8,4% nm, sau li sut nm sau tng thm so vi li sut nm trc l 1%. Hi nu gi 1.000.000 ng theo dchvthstinsnhnclbaonhiusau:10nm?;15nm?Nuslc cch gii. S tin nhn c sau 10 nm l: S tin nhn c sau 15 nm l: S lc cch gii: Bi 10: Trong mt phng ta cho hnh tht gic ABCDEFG vi cc nh c ta :14 26 63 11 45 15(1;1), 2; , ; 7 , ;5 , 11; , ; 3 , ; 23 5 6 4 7 8A B C D E F G .Tnhdin tch ca hnh tht gic (cho n v trn cc trc ta l cm), kt qu l mt phn s. Ht Ht 2ABCDEFGHS cm = S Gio dc v o to k thi chn hoc sinh gii tnh Tha Thin Hulp 8 thCS nm hc 2005 - 2006 Mn :MY TNH B TIp n v thang im: BiCch gii im TP im ton bi Rtgnbiuthctac: 1( 4 A x xx y= + +) y .Thay 5;4 5x y = =22, ta c: 20 327 36631113 16 1808A = = 0,25 0,5 Rt gn biu thc ta c: ( )3 3 2 22 24 7 18 49 6 4x y xy x yBx xy y +=+ +. 0,5 1 286892( 5; 16)769x y B = = = ( 1, 245; 3, 456) -33.03283776 x B = = 0,5 0,25 2 2 9991; 25; 2; 1; 6. a b c d e f g = = = = = = =2 3 26 2252633033=3 53 3331; 8863701824=2 101 1171 0,5 0,5 3 469283866 chia cho 2007 c s d l 1105. 1105 SHIFT STO A; SHIFT STO B; ALPHA B ALPHA = ALPHA B +1 : ( 100000 ALPHA A +10000 ALPHA B + 3658) 2007. Bm phm = (570MS) hoc CALC v = (570ES). 1 Kt qu tm c l7 b = 1,0 2 4 t ( )302 300 1 2 30( ) ... 1 2 3 P x a a x a x a x x x = + + + + = + +2=. Khi : 2 30 1 2 329 30 1529 30( 2) ( 2) ( 2) ...( 2) ( 2) ( 2) 9E a a a aa a P= + + + ++ + = =Ta c: 9 ;34867 = ; 10 53486784401; 9 59049 =584401 9 4983794649 =59 2058861483E=205886148300000+4983794649E=205891132094649. 1,0 1,0 2 52 1000029 =344.82758620689655172413793103448275862068965517241379310344827586... 1000029 l s hu t c phn tch thp phn v hn tun hon c chu k 28. 611 1(mod 28) ( )3342007 611 11 = ; Vych s l thp phn th 11l: 1. 3 334 311 1 11 (mod 28) 15(mod 28) 2007 1,0 0,5 0,5 6 Qui trnh bm phm: Ta c: 56700000 567 56799999 7529 567 7537 abcda abcda < < < < Gn cho bin m D gi tr 7529; 21: X X X = + . Bm phm = lin tip (570MS) hoc CALC v bm = lin tip, ta tm c: S: 56700900; 56715961; 56761156 1,0 1,0 2 7 Giuta c qui lut v mi lin h gia cc s hng ca dy s: 02 =1 20 11 1 12 ; 2 ;...; 2 ;...kku u uu u u = + = + = +1 Gii thut: 0 SHIFT STO D; 2 SHIFT STO A; ALPHA D ALPHA = ALPHAD+1: ALPHA A ALPHA = 2+ 1ALPHA A. Bm phm = lin tip (570MS) hoc CALC v bm = lin tip (570ES). Kt qu: 5 9 10169 5741 13860; ;70 2378 5741u = = = u u ; . 15 20, 2.414213562 u u 0,5 1,5 2 ( )33 3(1) 27 (2 1 1) ; (2) (2 2 1) ; (3) 2 3 1 . P P P = = + = + = +3( ) (2 1) 0 P x x + = 1; 2;3. x Suy ra: c cc nghim=Do : 3( ) (2 1) ( 1)( 2)( 3) P x x k x x x + = 3( ) ( 1)( 2)( 3) (2 1) P x k x x x x = + + (*) (4) 735 ( ) 1 P gt = k =( 1) 25; (6) 2257; (15) 31975; P P P = = =(2006) 72674124257 P = . 0,25 0,25 1,08 Khai trin P(x) ta c: P(x) =9 63 217 5 x x x + + . S d ca php chia( ) 3 5 P x cho x l: 2453r =0,25 0,25 2 9 1000000SHIFTSTOA;8.4 100SHIFTSTOB;0SHIFT STO D (bin m). ALPHAD=ALPHAD+1:ALPHAAALPHA=ALPHAA (1+AlphaB):ALPHABALPHA=ALPHAB(1+1 100). Bm phm = (570MS) hoc CALC v = (570ES), kt qu:Sau 10 nm: 2321713.76 ng; Sau 15 nm: 3649292.01 ng 1,0 1,0 2 10 DintchhnhagicABCDEFGlhiudintchcahnh vung HIJK ngoi tip a gic. Chia phn hnh vung ngoi a gic thnh cc tam gic vung v hnh thang vung. Ta c din tch phn hnh vung (cnh l 10 cm) ngoi a gic l: 1 14 1 14 26 26 636 7 7 2 11 112 3 2 3 5 5 61 3 63 1 1 455 2 11 112 4 6 2 4 71 45 15 1 15 118571 1 1 32 7 8 2 8 560 + + + + + + + + + + + + = SuyradintchagicABCDEFGl: ( )2 211875 4414310560 560S c = = m 1,0 1,0 2 S Gio dc v o toK thi chn hc sinh gii tnh Tha Thin HuGii ton trn my tnh Casio thi chnh thcKhi 9 THCS - Nm hc 2006-2007 Thi gian: 120 pht- Ngy thi: 02/12/2006. Ch :- thi gm4 trang - Th sinh lm bi trc tip vo bn thi ny. - Nu khng ni g thm, hy tnh chnh xc n 10 ch s. im ton bi thi Cc gim kho (H, tn v ch k) S phch (Do Ch tch Hi ng thi ghi) GK1 Bng sBng ch GK2 Bi 1: Tnh gi tr ca cc biu thc: 5 0 4 033 7 0 3 0235, 68 cot 23 35' os 69 43'62, 06 69 55' sin 77 27'g cAtg= . Lm trn n 5 ch s l thp phn. 4 42 2 2 2 2 23 2 16 164 9 6 4 4x y x y x yBx y x xy y x y + = + + + + khi: A B =a/( . 5; 16) x y = = B b/( 1, 245; 3, 456). x y = = Bi 2: a = ; b = c = ; d = e = ; f = g = a/Bit 20062007 1120081111abcdefg= ++++++, , , , , , a b c d e fg. Tm cc s t nhin. b/Cho dy s 1 1 1 11 1 1 12 4 8 2n nu = . Tnh(chnh xc) v (gn ng) 5u10 15 20, , u u u Bi 3: a/ Phn tch thnh tha s nguyn t cc s sau: 252633033 v 8863701824. b/ Tm cc ch s sao cho s567abcdal s chnh phng. a/ 252633033 = 8863701824 = b/ Cc s cn tm l: Bi 4:Khai trin biu thc ta c a thc gi tr chnh xc ca biu thc: (1521 2 3 x x + +)0Tnh vi2 30 1 2 30... . a a x a x a x + + + +. 0 1 2 3 29 302 4 8 ... 536870912 1073741824 E a a a a a a = + + + E = Bi 5: Tm ch s l thp phn th11k t du phy ca s thp phn v hn tun hon ca s hu t 20071000029. Ch s l thp phn th 11ca 2007 1000029 l: Bi6:Tmccstnhin(2000 60000) n n < < saochovimisth 354756 15na = + ncng l s t nhin. Nu qui trnh bm phm c kt qu. n = Qui trnh bm phm: Bi 7: Cho dy s: 1 2 3 41 1 1 12 ; 2 ; 2 ; 21 122 2 21 122 21222u u u u = + = + = + = ++ + ++ ++1; ... 1212...122nu = ++ (biu thc c cha tng phn s).nTnh gi tr chnh xc cauu v gi tr gn ng ca. 5 9 1, , u0 15 20, u uu5 = ----------------------u9 = -----------------------u10 = ------------------------ Bi8:Choathcbit 3 2( ) P x ax bx cx d = + + + (1) 27; (2) 125; (3) 343 P P P = = = v .(4) 735 P =a/ TnhP P (Ly kt qu chnh xc).( 1); (6); (15); (2006). P P b/ Tm s d ca php chia( ) 3 5 P x cho x . u15 = ----------------------u20 = ----------------------- S d ca php chia( ) 3 5 P x cho x l:r =( 1) ; (6))(15) ; (2006)P PP P = == = Bi 9: Li sut ca tin gi tit kim ca mt s ngn hng hin nay l 8,4% nm i vi tin gi c k hn mt nm. khuyn mi, mt ngn hng thng mi A a ra dch v mi: Nu khch hng gi tit kim nm u th vi li sut 8,4% nm, sau li sut nm sau tng thm so vi li sut nm trc l 1%. Hi nu gi 1.000.000 ng theo dchvthstinsnhnclbaonhiusau:10nm?;15nm?Nuslc cch gii. S tin nhn c sau 10 nm l: S tin nhn c sau 15 nm l: S lc cch gii: Bi10:Cho3ngthng 1 2 3( ) : 3 2 6 ; ( ) :2 3 15; ( ) : 3 6 d x y d x y d x y = + = + =1( ) d3) d).Hai ng thng v(ct nhau ti A; hai ng thng v(ct nhau ti B; hai ng thng(v ct nhau ti C. 1( ) d2d) (d)2d3a) Tm ta ca cc im A, B, C (vit di dng phn s). Tam gic ABC l tam gic g? Gii thch. b) Tnh din tch tam gic ABC (vit di dng phn s) theo on thng n v trn mi trc ta l 1 cm. d) Tnh s o ca mi gc ca tam gic ABC theo n v o (chnh xc n pht). V th v in kt qu tnh c vo bng sau: Ht S Gio dc v o to k thi chn hoc sinh gii tnh Tha Thin Hulp 9 thCS nm hc 2005 - 2006 Mn :MY TNH B TIp n v thang im: BiCch giiim TP im ton bi 3, 01541 A 0,75 Rt gn biu thc ta c: ( )3 3 2 22 24 7 18 49 6 4x y xy x yBx xy y +=+ +. 0,5 1 286892( 5; 16)769x y B = = = ( 1, 245; 3, 456) -33.03283776 x B = = 0,50 0,25 2 a/ 9991; 25; 2; 1; 6. a b c d e f g = = = = = = = 1,0 2 b/0 SHIFT STO X; 1 SHIFT STO A; ALPHA X ALPHA = ALPHA X+1: ALPHA A ALPHA = ALPHA A ( 1 12X ). Bm phm = lin tip (570MS) hoc CALC v bm = lin tip (570ES). Kt qu: 5 1015 209765; 0.2890702984;327680.2887969084; u 0.2887883705 u= u u 1,0 2 a) 3 26 2252633033=3 53 3331; 8863701824=2 101 1171 0,5 0,5 3 b) Ta c: 56700000 567 56799999 7529 567 7537 abcda abcda < < < 0) sao cho 25+213+217+2n l mt s chnh phng.sao cho (25+213+217+2n ) l mt s chnh phng. Cch tnh Quy trnh bm myKt qu Bi 8. Tm cc cp s (x, y) nguyn dng l nghim ng ca phng trnh:5 23 19(72 ) 240677 x x y = . Cch tnh Quy trnh bm myKt qu Bi 9. Cho tam gic nhn ABC c A = 6701535; B=7803515.GiAH,BI,CKlnltlcc ngcaocatamgic.Tnhtsdintchtam gic HIK v din tch tam gic ABC (chnh xc n 9 ch s thp phn). Cch tnhKt qu 5 Bi10.ChongtrntmObnknhr=5cmtipxcngoivibangtrn bngnhau.Bangtrnnyimttipxcvinhau(nhhnhv).Tnhdin tch phn c t mu en (chnh xc n 8 ch s thp phn). Cch tnhKt qu 6 ..Ht.. S GIO DC O TO THI TUYN HC SINH GII MY TNH B TIPH TH BC THCS, lp 9, 2003-2004 Bi 1:1) Vit quy trnh bm phm tnh gi tr ca biu thc 22 53 13 x xAx+ = 2) p dng quy trnh tnh A khi 1 1; ;2 3x x x13= = =Bi 2: Khi dng my tnh Casio thc hin php tnh chia mt s t nhin cho 48, c thng l 37, s d l s ln nht c th c c ca php chia . Hi s b chia l bao nhiu?Bi 3: Tnh chnh xc tng1 1! 2 2! 3 3! ... 16 16! S = + + + + Bi 4: Tnh bng my tnh. C th dng kt qu tnh c tng m khng s dng my tnh. Em hy trnh by li gii tnh tng S. 2 2 21 2 3 ... 10 A = + + + +2214 62 2 22 4 6 ... 20 S = + + + +Bi 5: Tnh s o cc gc ca tam gic ABC bit rng:21 A B C = ==. Bi 6: Cho sa( tch ca 17 s t nhin lin tip, bt u t s 1). Hy tm c s ln nht ca a, bit c s : a) L lp phng ca mt s t nhin. b) L bnh phng ca m s t nhin.1.2.3...17Bi 7: Tam gic ABC c 4 5; cos5 13A B = = cos . Tnh ln nht ca gc C ( , pht, giy)Bi 8: Thc hin php chia s 1 cho s 23 ta c mt s thp phn v hn tun hon. Hy xc nh s ng th 2004 sau du phy.Bi 9: Trc cn thc mu s ri dng my tnh tnh gi tr ca biu thc 3 322 2 2 4B =+ + vi chnh xc cng cao cng tt. Bi 10: Hai hnh vung ng tm c cc cnh song song vi di theo th t l 3 cm v 4 cm. Hnh vung bn trong c quay qanh tm mt gc cho n khi cc cnh ca n nm trn cc cnh ca hnh vung ln. Tnh gc ( 45 )o o ox x 1) ta u nhn c mt s d l r. Tnh d v r ? 4.2 Chia 8xcho 12x +ta c thng lq v d l. Li chiaqcho 1( ) x x1r1( )12x +ta c thng l v d l. Hy xc nh. 2( ) q x2r2rBi 5: Trong hnh v 1, ABH v CDE l cc tam gic u c din tch ln lt l 232 3 cmv 28 3 cm ; BCFG l mt hnh vung c din tch 32cm . Cho di on AD gim 12,5 % kchthc ca n trong khi cc di AB v CD vn khng thay i. 2Tnh din tch ca hnh vung gim i bao nhiu phn trm? Bi 6: Hnh 2 m t mt bi c ng knh AB = 3 cm. Trong c bn vin bi, trong mt vin c ng knh AC = 2cm, mt vin c ng knh BC = 1 cm, hai vin cn li c ng knh d. Tnh di ng knh d chnh xc n 0,0001 cm? Bi 7: Bit rng athc chia ht cho 3 2( ) 8 4 42 45 f x x x x = +2( ) ( ) gx x r = . Hy xc nh r ? Bi 8: Cho 8.1 Vit quy trnh n phm tnh 3 3( ) ( 3 1) f x x x = + +0( ) f vi 3 31 5 1 52 2+ = +8.2 Bng php ton, hy chng minh kt qu trn l ng . Bi 9: Biu din gi tr ca 6663331 121 1x xx xPx xx x + + = + + + di dng hn s bit (11 6 2) 11 6 2 (11 6 2) 11 6 25 2 5 25 1x+ +=+ + + Bi 10: Mt tp hp cc s t nhin lin tip bt u t s 1 c vit trn bng. Nu ngi ta xa i mt s th trung bnh ca nhng s cn li l 73517. Tm s b xa ? Tnh gi tr gn ng. HT (Thi Khu vc, B GD v T, Lp 12 THBT, 2004-2005, d b) Quy c : Khi tnh gn ng ch ly kt qu vi 5 ch s thp phn. Bi 1: Cho tam gic ABC c BC = a = 9,357 cm, AC = b = 6,712 cm, AB = c = 4,671 cm 1.1 Tnh gc C theo n v (chnh xc n pht). 1.2 Tnh bn knh ng trn ngoi tip ca tam gic. 1.3 Tnh din tch ca tam gic. Bi 2: Cho tam gic ABC c M l trung im ca BC, 900< A0) v p dng quy trnh bm phm gii pt cosx +3 sin x =2Bi 4:4.1. Cho cos= 0,7316 v -900 < 0 em nhn vi 6, nhng do nhm ln li em chia 6. Vit quy trnh n phm tnh sai s phn trm phm phi v cho bit sai s phn trm l bao nhiu?(Sai s phn trm =100sai sogiatr ung Bi 5: Cho dy s u1, u2,. un theo quy lut: u1 =2; u2 = 3; un+1 = 4un + 5un-1 , 3, n n NHy vit quy trnh bm phm lin tc ( cc kt qu khng phi ghi ra giy) tnh cc s hng theo u10 , u13 v u15 . Bi 6: Cho x = 0,123456789101112998999, trong ta vit du phy cc s t 1 ti 999 lin tip nhau. Vy ch s th 2005 bn phi du phy l bao nhiu? Bi 7: Mt b hc sinh chn tru gip gia nh mt a im C cch mt con sui SE l 4 km (hnh 2). Bn mun tm cho con tru con sui ri tr v trang tri v tr H. Hi qung ng ngn nht m bn c th hon thnh cng vic ny l bao nhiu km?8 kmE S 7 km 4 km C H EHnh 2 (cc kch thuc cho trong hnh 2) Bi 8: Cho X, Y, Z l tp hp m cc phn t ca chng l ngi, i mt khng giao nhau. Tui trung bnh ca nhng ngi trong cc tp hp X, Y, Z, , , X YX Z Y Zc cho trong bng di y: Tp hpXYZ X Y X Z Y ZTui Tr.Bnh3723412939,533 Hy tm tui trung bnh ca ngi trong tp hp X Y Z . Bi 9: T mt tm tn hnh vung c cnh bng 2a = 4010cm ngi ta mun ct ra c hnh khai trin ca mt hnh chp u SMNPQ sao cho cc nh ca hnh vung c gn li nh S ca hnh chp. Hi phi ct b i bn ming tn c dng bn tam gic cn bng nh nhau MAB, NBC, PCD v QDA vi cc cnh y AB, BC, CD v DA t tm tn hnh vung nh th no c hnh chp SMNPQ c th tch ln nht? Hy tnh th tch theo a vi c hnh xc cng cao cng tt . Bi 10: Cho hm s f(x) = 4x3 3xc th l ng cong (C) v ng thng (d): y = 12 10.1. Bit rng ng thng (d) ct ng cong (C) ti 3 im phn bit x1, x2, x3 . Tm honh giao im ca (C) v (d) vi chnh xc cng cao cng tt. 10.2. Cc tip tuyn ti xi (i = 1,2,3) ct ng cong (c) ln lt ti Ni (i = 1,2,3). Chng minh rng cc im Ni(i = 1,2,3) thng hng. HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TIPH THLp 12 THPT, 2004-2005

Bi 1: Cho tam gic ABC c9, 357 ; 6, 712 ; 4, 671 BC a cmAC b cmAb c cm = = = = = =1.1 Tnh gc C theo n v ( chnh xc n pht). 1.2 Tnh bn knh ng trn ngoi tip ca tam gic. 1.3 Tnh din tch ca tam gic. Bi 2: Cho tam gic ABC c M l trung im ca BC,90 180 , sin 0, 6153o oA A < < =; 17, 2 ; 14, 6 AB cmAC = = cm. Tnh: 2.1 tgA 2.2 di cnh BC. 2.3 Din tch ca tam gic ABC. 2.4 di ca trung tuyn AM. Bi 3: Xy dng quy trnh bm phm gii phng trnh lng gic v p dng quy trnh gii phng trnh osx+bsinx=c (a>0) accos 3sin 2 x x + =Bi 4:4.1 a) Gi sl cc s nguyn khi chia cho D c s d ln lt l. Chng minh rng tch s khi chia cho D cng c cng s d. b)p dng: Tm s d ca php chiacho 13. 4.21000 hnh lp phng c cnh bng 1 dm c lp rp li vi nhau to thnh hnh lp phng c cnh bng bng 10 dm. Ta sn hnh lp phng ln ny ri li tch ra thnh 1000 hnh lp phng nh c. Trong cc hnh lp phng nh ny, c bao nhiu hnh lp phng c t nht mt mt c sn. 1 2N ,N1 2r ,r1 2 1N N ,r r21002Bi 5: ng h ch qung ng ca xe t i c o bng s vng quay ca bnh xe. Trong chuyn i, ng h trn xe cho bit khong cch i l 724,2048 km. Khi tr v, vn chi xe , nhng c thay bnh xe c ng knh ln hn nn ng h trn xe ch 708,11136 km. Hy tnh vi di chnh xc cng cao cng tt tng bn knh ca bnh xe nu bn knh lc u ca bnh xe l 38,1 cm. Bi 6: Vi n l s t nhin, xt pt5(14n+3)x2 (147n + 29)x + 3(21n + 4) =0 6.1 Gii pt tm x di dng phn s khi n = 2005 6.2 Chng minh rng pt cho khng th c nghim nguyn vi mi gi tr ca s t nhin n. Bi 7: Cho pt x4 ( 2 + 5 )x3 + ( 10 3 )x2 3 10= 0 cmr pt c 4 nghim phn bit xi (i =1,2 , 3, 4) v tnh gi tr ca biu thcM = 1 2 31 1 1 120052 2 2 24x x x x + + + Bi 8: Mt khung mi bng thp c dng nh hnh 1 dng lp tn chng dt cho mt sn thng ABCD ca ngi nh c din tch l 80,2341 m2. Hy tnh din tch phn tn m bn mua v lp. Bit rng cc mi tn c lp u nghing u trn sn ABCD mt gc 4403517 v 0,034% din tch tn hao ph cho cc phn lp v phn ct b. DC AB Bi 9:Cho f(x) = x12 x11 + 3x10 + 11 x3 x2 + 23x + 30 v g(x) = x3 + 2x + m.Hy xc nh gi tr nguyn ca m sao cho f(x) = g(x) . q(x)x R Bi 10: Cho pt Tx2 +3 x + 16 = 0, trong T = sin62A + sin62B+ sin62C vi A, B, C l 3 gc ca tam gic ABC. 10.1 Gii pt cho khi tam gic ABC l tam gic u (tm x vi chnh xc cng cao cng tt) 10.2 Cmr T364vi mi tam gic ABC. HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TIPH THchn i tuyn THCS, 2005

Bi 1: Tnh 1)4444 88 2)444444 888 3) 44444444 8888 4)4444444444 88888 5)N244...44 8...8nn

Bi 2: Tm tt c cc s c dng34 5 x ychia ht cho 36.Bi 3:3.1Tnh gn ng gi trj ca biu thc 4 4 424 3 5 2 25 125A = + 3.2 Rt gn biu thc A, sau tnh gn ng gi tr ca A. Bi 4: K hiu, trong 1nnS x x = +2n1 2, x xl nghim ca phng trnh bc hai 4.1 Lp mt cng thc truy hi tnh 28 1 x x + = 01 nS + theo v nS1 nS , 4.2 Lp mt quy trnh tnh tnh trn my Casio. 4.3 Tnh theo quy trnh trn v tnh theo cng thc nSnS'nS'1 2(4 15) (4 15)n n nnS x x = + = + +n vi1, 2,...,11. n =Bi 5:5.1 Nu mt quy trnh tm thng v phn d ( nu c ) ca php chia cc s 10000100001 v 1000001000001 cho 37 trn my tnh Casio c 10 ch s. 5.2 Trong cc s sau, s no chia ht cho 37: 10101; 1001001; 100010001; 10000100001; 1000001000001; 100000010000001. 5.3 Vi gi tr no ca n th a thc 21n nx x + +chia ht cho tam thc bc hai 21 x x + + . 5.4 Vi gi tr no ca n th s c dng( vi n ch s 0 gia hai ch s 1) chia ht cho 37. N N10...010...01n nBi 6: Mt sinh vin c gia nh gi tit kim vo ngn hng s tin l 20.000.000 ( hai mi triu) vi li sut tit kim l 0,4%/thng. 6.1 Hi sau 5 nm ( 60 thng ) s tin trong s s l bao nhiu. 6.2 Nu mi thng anh sinh vin rt ra mt s tin nh nhau vo ngy ngn hng tnh li th hng thng anh ta rt ra bao nhiu tin ( lm trn n trm ng) sau ng 5 nm s tin va ht. 6.3 Nu khng gi tit kim m hng thng anh sinh vin vn s dng mt s tin nh nhau sau 5 nm s tin va ht th hng thng anh ta b thit bao nhiu. 6.4 Nu gi tit kim m hng thng nah sinh vin khng rt tin ra ( t tc c khng cn tr gip ca gia nh) th sau bao nm trung bnh mt thng anh ta c thm bao nhiu tin so vi gi tit kim m hng thng rt tin ra. Bi 7: Mt hng xe ti c hai gara, mt c t thnh ph v mt c t sn bay. Hng c 105 chic xe v mun t bao nhiu xe mi gara tha mn nhu cu th trng, Sau khi nghin cu tnh hnh cc nm trc, cng ty quyt nh mi thng iu 70% s xe gara thnh ph i sn bay s tr v 30% s li sn bay; 25% s xe gara sn bay s tr v v 75% li thnh ph. Gi s nxv nyl s cc xe taxi cc gara thnh ph v sn bay vo u thng th n. Sau mt thng, s xe 1 nx + gara thnh ph s l s xe xut pht t v quay v, tc l,70%nxcng vi s xe sn bay n v li , tc l75%ny . Nh vy, 10, 7 0, 75n n nx x y+ = + . Tong t, s xe gara sn bay u thng th n+1 s l tng s xe t thnh ph v li sn bay, tc l30%nxv s xe t sn bay v quay tr v, tc l25%ny . Nh vy, 10, 3 0, 25n n ny x y+ = + . 7.1 Gi s hng khai trng ngy 1.1.2004 vi 1 160, 45 x y = = . Lp quy trnh bm phm trn my Casio fx-500MS tnh s xe nxv nymi thng ti mi gara theo cng thc truy hi: 7.2 Tnh 110, 7 0, 75, 1, 2, 3....0, 3 0, 25n n nn n nx x yny x y= + == +nxv nyvi n = 1, 2, 3,,11,12. 7.3 Sau mt nm, hng thy tnh hnh kinh doanh cha ph hp vi thc t: 40% s ngi i t thnh ph ra sn bay khn gtr v thnh ph v s lng xe khng p ng nhu cu hnh khch, v vy hng mua thm 10 xe t na v iu chnh s xe cho ph hp. Hy vit phng trnh iu ng xe v cho bit s xe mi gara. Bi 8:8.1 Tm s dvkhi chia3v31 2( ); ( ) d n d n3( ) d n ;5n n5n n+cho 13 vi n = 0,1,2,,15. 8.2 Vi gi tr no ca n th3chia ht cho 13.5n+nBi 9:9.1 Tm tt c s c hai ch s 1 2a asao cho 21 2 1 2( ) a a a a = +9.2Tm tt c s c bn ch s 1 2 3 4a a a asao cho 21 2 3 4 1 2 3 4( ) a a a a a a a a = +9.3Tm tt c s c su ch s 1 2 3 4 5 6a a a a a asao cho 21 2 3 4 5 6 1 2 3 4 5 6( ) a a a a a a a a a a a a = +Bi 10: Cho mt na ng trn bn knh 1S112r =ni tip na ng trn S ng knh AB =2 ( tip xc vi na ng trn S v tip xc vi AB). Dy cc ng trnbn knh c xc nh nh sau: tip xc vi na ng trn S, vi on thng AB v vi ng trn. K hiu 1SkSkr1 nS + nS1nnar= . 10.1 Lp mt h thc lin h gia v 1,na a+ n 1 na . 10.2 Lp mt quy trnh tnh. Tnh cc vi n = 2,,10. 10.3 Chng minh rng l s chnh phng, cn nana2na2 1 na+ l hai ln s chnh phng vi mi n l s t nhin . HT Phch nh km thi chnh thc lp 9 THCS . Bng A S Gio dc- o toqung ninh ---------------- K thi cp tnh gii ton trn My Tnh casio bc trung hc nm hc 2004 - 2005 ------------- @ ------------- Lp : 9 THCS . Bng A Thi gian thi: 150 pht (khng k thi gian giao ) Ngy thi: 20/01/2005 H v tn th sinh: ................................................................................................Nam (N) ..................... S bo danh: .....................................................................................................................................................Ngy, thng, nm sinh: ................................................ Ni sinh: ................................ ............................. Hc sinh lp: ..................... Trng THCS: ............................................................................................. Huyn (TX, TP): ........................................................................................................................ H v tn, ch k ca gim th S phch Gim th s 1: ................................................................. Gim th s 2: ................................................................. (Do Ch tch hi ng chm thi ghi) Quy nh : 1)Th sinh phi ghi y cc mc phn trn theo hng dn ca gim th.2)Th sinh lm bi trc tip vo bn thi c phch nh km ny. 3)Th sinh khng c k tn hay dng bt c k hiu g nh du bi thi, ngoi vic lm bi thi theo yu cu ca thi. 4)Bi thi khng c vit bng mc , bt ch; khng vit bng hai th mc. Phn vit hng,ngoicchdngthcgchcho,khngctyxobngbtccchgkc bt xo. Ch c lm bi trn bn thi c pht, khng lm bi ra cc loi giy khc. 5)Tri vi cc iu trn, th sinh s b loi. S Gio dc- o toqung ninh ---------------- K thi cp tnh gii ton trn My Tnh casio bc trung hc nm hc 2004 - 2005 ------------- @ ------------- thi chnh thc Lp : 9 THCS . Bng A Thi gian lm bi: 150 pht (khng k thi gian giao ) Ngy thi: 20/01/2005 Ch : - thi ny c :03 trang - Th sinh lm bi trc tip vo bn thi ny. im ca ton bi thi H v tn, ch k cc gim kho S phch(DoCh tchH chm ghi )Bng sBng ch ......................................................

...................................................... Quy nh : 1)Th sinh ch c dng my tnh:Casio fx-220, Casio fx-500A, Casio fx-500MS v Casio fx-570MS. 2)Nu trong bi khng ni g thm, hy tnh chnh xc n 9 ch s thp phn. Bi 1:Cho ba s :a = ' 51 12 sin 2' 51 12 sin 3 100 ;b = 015 cos 3- 1 ; c = 15 3 2315.....333231+ + + +Hy so snh cc s a; b; c v in kt qu vo trng. Bi 2: Tm s d trong php chia717 : 2005 v in kt qu vo trng.

r= ....................................... Bi 3:TnhM = 1234567890 6789012345v in kt qu vo trng. Trang 1Bi 4:Gii h phng trnh:v in kt qu vo trng. == +263 3xyy x Bi 5: Tm gi tr ca xvit di dng phn s t phng trnh sau v in kt qu vo trng.5 + 9181716+++x = 6171819+++x x= ....................... Bi 6:Trong cc s t nhin c dng5 1 3 2 1 t z y x , tm s ln nht chia ht cho 2005 v in kt qu vo trng. Bi 7:Cho tam gic ABC vung ti A, c AB = 4,6892cm ; BC = 5,8516cm . a) Tnh s o gc B (theo , pht, giy). b) Tnh di ng cao AH v di ng phn gic trong CI ca tam gic ABC. (chnh xc n 5 ch s thp phn) B .................................. AH ................................. CI .................................. Bi 8 : Cho a thc P(x) = x3 + ax2 + bx + c Bit rng: P(1945) = 1945 ; P(1954) = 1954 ; P(1975) = 1975. a) Tnh P(2005).b) t Q(x) = P(x) + m. Tm gi tr ca m a thc Q(x) chia ht cho (x - 2005,05)(chnh xc n 5 ch s thp phn). P(2005) = ..................................... m ..................................... Trang 2 Bi 9:Dy s {Un} c cho nh sau: U0 = U1 = 2 ;Un+2 = Un+1.Un + 1 vi n = 0, 1, 2, 3, ..... a) Hy lp mt quy trnh bm phm lin tc tnh Un vi n 2.(nu r dng cho loi my no) b) Tnh cc gi tr U2, U3, U4, U5, U6, U7, U8. Qui trnh bm phm: U2 = .................................U3 = .................................U4 = ................................. U5 = .................................U6 = .................................U7 = ................................. U8 = ................................................................................................................................. Bi 10:Cho hnh thang ABCD vung ti B v C, c AB < CD, AB = 12,35cm, BC = 10,55cm v ADC = 570. a) Tnh din tch hnh thang ABCD. (chnh xc n 5 ch s thp phn) b) Tnh t s gia din tch tam gic ADC v din tch tam gic ABC. (chnh xc n 5 ch s thp phn) Hnh v SABCD ..................................... SADC : SABC ..................................... -------------------- Ht ----------------------- Trang 3 s gd-t qung ninh hng dn chm thi HSG gii ton trn my tnh casio lp 9 - bng a . nm hc 2004-2005 BiKt quim chi titTng im 1 b < a < c 55 2r = 1167 55 3M = 8.381.496.645.950.602.050 55 4 (x1 ; y1) (1,259921050 ; 1,587401052 ) (x2 ; y2) (1,587401052 ; 1,259921050 ) 2,5 2,5 5 5 x - 933431 = - 343193 55 6s cn tm l :192939145 55 7 B 36044'26'' AH 2,80504 cm CI 3,91575 cm 1,5 1,5 2,0 5 8P(2005) = 93805 m - 94124, 90263 2,5 2,5 5 9a) Qui trnh bm phm: - Vi fx-500A:2min2+1 =(cho U2 )v lp i lp li dy phm: SHIFTX MMR+1= (ln th n cho Un+2) - Vi fx-500MS: 2SHIFTSTOA2+1SHIFTSTOB(c U2)v lp i lp li dy phm: ALPHAA+1SHIFTSTOA (c U3, U5,...) ALPHAB+1SHIFTSTOB (c U4, U6, ...) b) U2 = 5 ; U3 = 11 ; U4 = 56 ; U5 = 617; U6 = 34553;U7 = 21319202vU8 = 736.642.386.707 (Ring U8, nu ch tnh bng my th trn mn hnh nn phi kt hp vi tnh bng tay) 2,5 1,5 1,0 5 10 a) SABCD 166,43284 cm2

b) SADC : SABC 1,55476 2,5 2,5 5 s gio dc-o toPhch nh km thi chnh thc lp 12 THPT . Bng B S Gio dc- o toqung ninh K thi cp tnh gii ton trn My Tnh casio bc trung hc nm hc 2004 - 2005 ------------- @ ------------- Lp :12THPT . Bng B Thi gian thi:150 pht (khng k thi gian giao ) Ngy thi: 20/01/2005 H v tn th sinh: ................................................................................................Nam (N) ..................... S bo danh: .....................................................................................................................................................Ngy, thng, nm sinh: ................................................ Ni sinh: ................................ ............................. Hc sinh lp: ..................... Trng THPT: ............................................................................................. H v tn, ch k ca gim th S phch Gim th s 1: ................................................................. Gim th s 2: ................................................................. (Do Ch tch hi ng chm thi ghi) Quy nh : 1)Th sinh phi ghi y cc mc phn trn theo hng dn ca gim th.2)Th sinh lm bi trc tip vo bn thi c phch nh km ny. 3)Th sinh khng c k tn hay dng bt c k hiu g nh du bi thi, ngoi vic lm bi thi theo yu cu ca thi. 4)Bi thi khng c vit bng mc , bt ch; khng vit bng hai th mc. Phn vit hng,ngoicchdngthcgchcho,khngctyxobngbtccchgkc bt xo. Ch c lm bi trn bn thi c pht, khng lm bi ra cc loi giy khc. 5)Tri vi cc iu trn, th sinh s b loi. S Gio dc- o toqung ninh K thi cp tnh gii ton trn My Tnh casio bc trung hc nm hc 2004 - 2005 ------------- @ ------------- thi chnh thc Lp : 12THPT . Bng B Thi gian lm bi: 150 pht (khng k thi gian giao ) Ngy thi: 20/01/2005 Ch : - thi ny c :04 trang - Th sinh lm bi trc tip vo bn thi ny. im ca ton bi thi H v tn, ch k cc gim kho S phch(DoCh tchH chm ghi )Bng sBng ch ......................................................

...................................................... Quy nh : 1)Th sinh ch c dng my tnh:Casio fx-220, fx-500A, fx-500MS v fx-570MS. 2)Ccktqutnhtongnng,nukhngcyucucth,cngmnhl chnh xc n 5 ch s thp phn. Bi 1: Tnh gn ng cc gi tr ca a th hm s y = x3 + 3ax2 + (a + 2)x - a + 2 c im un nm trn trc honh. Tm tt cch giiKt qu Bi 2: Hm s y = 132+ +xm x mx(vi m 0, m - 23) c th (C). Tim cn xin ca(C)ctcctrctotiAvB.Tmgnngccgitrcamtamgic OAB c din tch bng 6 (n v din tch). Tm tt cch giiKt qu Trang 1Bi 3:Tnh gn ng (, pht, giy) cc nghim ca phng trnh : sinx + cox = 2 25 5 Tm tt cch giiKt qu x ......................... + k3600

x ......................... + k3600 Bi 4:Cho parabol (P) : y2 = 2x v hai im A(2 ; 2 2 ) , B(1 ; 0). Tnh gn ng to im C thuc parabol (P) sao cho tam gic ABC vung ti B. (Tnh chnh xc n 9 ch s thp phn) Tm tt cch giiKt qu Bi 5: p dng cng thc tnh gn ng nh vi phn :f(x0+x) = f(x0) + f'(x0). x tm gn ng cc gi tr sau:326;cos610 . (chnh xc n 9 ch s thp phn). Tm tt cch giiKt qu 326 ................................... cos610 ............................... Trang 2 Bi 6:Cho tam gic ABC c BC = 15,637cm; CA = 13,154cm; AB = 12,981cm.1) Tnh gn ng (, pht, giy) s o gc A ca tam gic. 2) Tnh gn ng bn knh R ca ng trn ngoi tip tam gic. Tm tt cch giiKt qu A

....................................... R

........................................ Bi 7: Tnh gn ng gi tr ln nht v gi tr nh nht ca hm sy = xx x2 cos 2 41 2 cos 3 2 sin 2+ + Tm tt cch giiKt qu max y ................................ min y ................................ Bi 8 :Tnh gn ng nghim ca phng trnh :ex + x = 3 Tm tt cch giiKt qu x ....................................... Trang 3 Bi 9:Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v D, cnh bn SA vung gc vi y, mt bn SBC to vi mt y gc 45027'36'', c AB = 2AD= 2DC = 6,912cm.Tnh gn ng din tch xung quanh ca hnh chp S.ABCD. Tm tt cch giiKt qu

Sxq ..................................... Hnh v ------------------------ Ht ------------------------- Trang 4 s gd-t qung ninh hng dn chm thi HSG gii ton trn my tnh casio lp 12 - bng b . nm hc 2004-2005 Bi Tm tt cch gii p s im tng phnim ton bi 1Tm c tung im unyun = 2a3 - a2 - 3a + 2. imun Ox 2a3- a2 - 3a +2 = 0 a = 1 hoc a = (-1 17 ) a = 1 a - 1,28078 a 0,78078 1,5 2,5 2,0 6 2Tim cn xin l: y = mx + m + 3 TCX Ox = A(-(m+3)/m ; 0) TCX Oy = B(0 ; m+3) SOAB = (m+3)2/2m SOAB = 6 (m+3)2/2m = 6 m = 3 hoc m= - 9 6 2 m = 3 m - 17,48528 m - 0,51472 1,5 3,0 2,5 7 3 PT cho: sinx + cox = (5 -5 )/2 2 sin(x+450) = (5 -5 )/2 2x 32044'27''x 57015'33'' 2,5 2,5 5 4 ph.tr ng thng d qua B v AB l x + 2 2 y - 1 = 0. d (P) ti 2 im C1 v C2 viyC1 = - 4 2+ 6 v yC2 = - 4 2- 6 T tm c xC1 = 1 - 2 2 yC1

v xC2 = 1 - 2 2 yC2

C1( 0,029437251 ; 0,343145750 ) C1( 33,970562750 ;- 11,656854250 ) 3,0 3,0 6 5 * Tnh 326 : chn x0 = 27 ; x = - 1 * Tnh cos610: Ta c : cos610= cos(600+10) = cos(/3+/180) => chn x0 = /3 ; x = /180 326 2,962962963 cos610 0,111457394 2,5 2,5 5 6a) cosA = (b2 + c2 - a2)/2bcDng MOD 4vi fx500A hoc MOD MOD MOD 1 vi fx500MS b) R = abc/4S=> R = abc/4 ) )( )( ( c p b p a p p A 73029'44'' (8,154487085) R 8,15449 2,5 2,5 5 7 T y = xx x2 cos 2 41 2 cos 3 2 sin 2+ + suy ra 2sin2x + (3-2y)cos2x = 4y + 1 (*) (*) c nghim 22 + (3-2y)2 (4y+1)2 T tm c max y v min y max y 0,46837 min y -2,13504 2,5 2,5 5 8Gii phng trnh x = ln(3-x) bng phng php lp => tm c x x 0,792059968 1,5 3,5 5 9 t AD = DC = a; g(SBC, ABCD) = AB = 2a; SCA = = 45027'36''; a = 3,456cm. Cc tam gic SAB, SBC, SCD, SDA vung => Sxq= .22a) 2 2 1cos22 2 (2 tg tg tg + + + +Thay s, tnh c din tch x.q hnh chp. (53,2312642) Sxq 53,23126 cm2 2,5 3,5 6 Bi Tm tt cch gii p s im tng phnim ton bi 7 T y = xx x2 cos 2 41 2 cos 3 2 sin 2+ + suy ra 2sin2x + (3-2y)cos2x = 4y + 1 (*) (*) c nghim 22 + (3-2y)2 (4y+1)2 T tm c max y v min y max y 0,46837 min y -2,13504 2,5 2,5 5 8Gii phng trnh x = ln(3-x) bng phng php lp => tm c x x 0,792059968 1,5 3,5 5 9 t AD = DC = a; g(SBC, ABCD) = AB = 2a ; SCA = = 45027'36''; a = 3,456cm.Cc tam gic SAB, SBC, SCD, SDA vung => Sxq= .22a) 2 2 1cos22 2 (2 tg tg tg + + + +Thay s, tnh c din tch x.q hnh chp. (53,2312642) Sxq 53,23126 cm2 2,5 3,5 6 s gd-t qung ninh. Phch nh km thi chnh thc lp 9 THCS . Bng A S Gio dc- o toqung ninh ---------------- K thi cp tnh gii ton trn My Tnh casio bc trung hc nm hc 2005 - 2006 ------------- @ ------------- Lp : 9 THCS . Bng A Thi gian thi: 150 pht (khng k thi gian giao ) Ngy thi: 11/01/2006 H v tn th sinh: ................................................................................................Nam (N) ..................... S bo danh: .....................................................................................................................................................Ngy, thng, nm sinh: ................................................ Ni sinh: ................................ ............................. Hc sinh lp: ..................... Trng THCS: ............................................................................................. Huyn (TX, TP): ........................................................................................................................ H v tn, ch k ca gim th S phch Gim th s 1: ................................................................. Gim th s 2: ................................................................. (Do Ch tch hi ng chm thi ghi) Quy nh : 1)Th sinh phi ghi y cc mc phn trn theo hng dn ca gim th.2)Th sinh lm bi trc tip vo bn thi c phch nh km ny. 3)Th sinh khng c k tn hay dng bt c k hiu g nh du bi thi, ngoi vic lm bi thi theo yu cu ca thi. 4)Bi thi khng c vit bng mc , bt ch; khng vit bng hai th mc. Phn vit hng,ngoicchdngthcgchcho,khngctyxobngbtccchgkc bt xo. Ch c lm bi trn bn thi c pht, khng lm bi ra cc loi giy khc. Khng lm ra mt sau ca ca t thi. 5)Tri vi cc iu trn, th sinh s b loi. S Gio dc- o toqung ninh ---------------- K thi cp tnh gii ton trn My Tnh casio bc trung hc nm hc 2005 - 2006 ------------- @ ------------- thi chnh thc Lp : 9 THCS . Bng A Thi gian lm bi: 150 pht (khng k thi gian giao ) Ngy thi: 11/01/2006 Ch : - thi ny c :05 trang - Th sinh lm bi trc tip vo bn thi ny. im ca ton bi thi H v tn, ch k cc gim kho S phch(DoCh tchH chm ghi )Bng sBng ch ......................................................

...................................................... Quy nh : 1)Th sinh ch c dng my tnh:Casio fx-220, Casio fx-500A, Casio fx-500MS v Casio fx-570MS. 2)Cc kt qu tnh ton gn ng, nu khng c yu cu c th, c qui nh l chnh xc n 9 ch s thp phn. Bi 1:Tnh gn ng gi tr ca cc biu thc sau: 1.1)A =3) 1 2 4 (32+ +x xx vi x = 2 1+ 1.2) B = 2 0 3 0 2 0 3 03 0 3 02cos 55 .sin 70 10cotg50 .cotg653cos 48.cotg70 p s: A ............................................... ; B ............................................... Bi 2:Cho s a = 1.2.3...16.17 (tch ca 17 s t nhin lin tip, bt u t s 1). Hy tnh c s ln nht ca a bit rng s l lp phng ca mt s t nhin. Tm tt cch gii:p s: Trang 1 Bi 3:K hiu M =21315171+++ + 43568791+++;N = ba11715131++++ 3.1)Tnh M, cho kt qu di dng phn s. p s: 3.2) Tm cc s t nhin a v b bit rng: 116763655 = N. Tm tt cch gii:p s: Bi 4: Cho :x1003 + y1003 = 1,003 v x2006 + y2006 = 2,006. Hy tnh gn ng gi tr biu thc:x3009 + y3009.

Tm tt cch gii:p s: Trang 2 Bi 5:Xt cc s thp phn v hn tun hon : E1 = 0,29972997... vi chu k l (2997) ; E2 = 0,029972997... vi chu k l (2997) E3 = 0,0029972997... vi chu k l (2997). 5.1) Chng minh rng sT = 13E + 23E + 33E l s t nhin. Tm tt cch gii: 5.2) S cc c nguyn t ca s T l: A. 1B. 2C. 3D. 5E. 11 (Tr li bng cch khoanh trn ch ci ng trc p s ng). Bi 6:Cho ng trn (I ; R1) v ng trn (K ; R2) tip xc ngoi vi nhau ti A. Gi BC l mt tip tuyn chung ngoi ca hai ng trn, B thuc ng trn (I ; R1), C thuc ng trn (K ; R2). Cho bit R1 = 3,456cm v R2 = 4,567cm. 6.1)Tnh gn ng di BC (chnh xc n 5 ch s thp phn).6.2) Tnh gn ng s o gc AIB v gc AKC (theo , pht, giy). 6.3) Tnh gn ng din tch tam gic ABC (chnh xc n 5 ch s thp phn). V hnh. Tm tt cch gii cu 6.3) p s: Trang 3 Bi 7:7.1)Bit a thc Q(x) = x4 + mx3 - 44x2 + nx - 186 chia ht chox + 2 v nhn x = 3 l nghim. Hy tnh gi tr ca m v n ri tm tt c cc nghim cn li ca Q(x). Tm tt cch gii:p s: 7.2)Cho a thc P(x) = x4 + ax3 + bx2 + cx - 12035. Bit rng: P(1) = 2; P(2) = 5 ; P(3) = 10, hy tnh gn ng gi tr biu thc:P(9,99) - P(9,9). Tm tt cch gii:p s: Bi 8:Cho dy s {Un} nh sau:Un =( )n6 2 5+ +( )n6 2 5vi n = 1, 2, 3, ..... 8.1) Chng minh rng Un+2 + Un = 10Un+1 vi n = 1, 2, 3, ..... Tm tt cch gii: Trang 4 8.2) Lp mt quy trnh bm phm lin tc tnh Un+2 vi n 1.(nu r dng cho loi my no) Qui trnh bm phm: 8.3) Tnh U11 ; U12 . p s:

Bi 9:Cho tam gic ABC vi ng cao AH. Bit gc ABC = 450, BH = 2,34cm, CH = 3,21cm.9.1) Tnh gn ng chu vi tam gic ABC. (chnh xc n 5 ch s thp phn) V hnh :p s: 9.2) Tnh gn ng bn knh ng trn ni tip tam gic ABC. (chnh xc n 5 ch s thp phn) Tm tt cch gii:p s: -------------------- Ht ----------------------- Trang 5 s gd-t qung ninh hng dn chm thi HSG gii ton trn my tnh casio lp 9 - bng a . nm hc 2005-2006 BiTm tt cch giiKt quCho im1 A - 0,046037833 B -36,8228381162,5 2,5 2Vit c a = 215.36.53.72.11.13.17. => s phi tm l: 215.36.53 2985984000 2,5 2,5 3.1 M = 284626871 2,5 3.2 Tnh c N =1/(365511676) =...= 11191715131++++ T suy ra a v b a = 9 ; b = 11 2,0 0,5 4t a = x1003 ; b = y1003=> cn tnha3+b3 . Bin i c: a3+b3 = (a+b)(3(a2+b2)-(a+b)2)/2 T tnh c a3+b3 2,513513487 2,5 3.5 5.1 C 10000E1 = 2997,29972997... = 2997 + E1 => E1 = 2997/9999 => 333/1111 Tng t, tnh c E2= 333/11110 ;E3 = 333/111100 Bm my theo quy trnh: 3 : 333 ab/c 1111 + 3 : 333 ab/c 11110 + 3 : 333 ab/c 111100 =suy ra gi tr ca T T = 1111 1,0 1,0 1,5 0,5 5.2p s B l ng.1,0 6.1 BC 7,94570 cm 2,0 6.2 AKC 8202'25'' AIB 97057'35'' 1,0 0,5 6.3C SABC = SIBCK - (SAIB + SAKC) Tnh SAKC theo y AK, ng cao h t C Tnh SAIB theo y AI, ng cao h t B Tnh SIBCK theo 2 y KC, IB v ng cao IK Bin i, c SABC = 2R1R22 1.R R/(R1 + R2). Thay s, tnh ra SABC . SABC 15,63149 (cm2) 1,0 1,0 0,5 7.1T gi thit => Q(-2) = Q(3) = 0 => tm m, n T gi thit => Q(x) c 2 nghim nguyn => Q(x) = (x+2)(x-3)(x2+7x-31)Dng my gii ph/tr bc 2 => 2 nghim cn li. m = 6; n = -11 x2 = -2 x3 3,076473219x4 -10,076473219 1,0 0,5 0,75 0,75 BiTm tt cch giiKt quCho im 7.2Xt F(x) = P(x) - (x2+1). T g/th => F(1) = F(2) = F(3) = 0 => F(x) = (x-1)(x-2)(x-3)(x+m).Tnh F(0) ri suy ra m = 2006. T tnh c P(9,99) - P(9,9). P(9,99) - P(9,9) 34223,33546. 1,0 2,0 8.1 t a =( 6 2 5+ ) ; b =( 6 2 5 ) => a2 - 10a + 1 = 0 ; b2 - 10b + 1 = 0 => an(a2 - 10a + 1) = 0 ; bn(b2 - 10b + 1) = 0 => ... => Un+2 + Un = 10Un+1(pcm !) 2,0 8.2a) Qui trnh bm phm: - Vi fx-500A: - Vi fx-500MS:Tnh tay c U1 = 10; U2= 98. 98SHIFTSTOA10-10 SHIFTSTOB (c U3) Dng con tr lp i lp li dy phm v tnh Un : 10 - ALPHAASHIFTSTOA (c U4, U6,...) 10 - ALPHABSHIFTSTOB (c U5, U7, ...) 2,5 8.3U11 = 89.432.354.890U12 = 885.289.046.402 0,5 1,0 9.1Nu c AH = BH; BC = BH + HC;AB = BH. 2 ; AC = 2 2CH AH+Chu vi tam gic ABC = 2p = AB + BC + AC Thay s, tnh ra kt qu. 2p 12,83163 (cm) 1,0 2,0 9.2 Nu c r = SABC : p p = (AB+BC+CA)/2 ; SABC = AH.BC/2 T tnh c r r 1,01211 (cm) 1,5 1,5 Cc ch : 1. Nu trong yu cu tm tt cch gii nhng hc sinh ch cho kt qu ng vi p nthvnchoimphnktqu.Nuphntmttcchgiisainhngktqungth khng cho im c cu hoc bi .2. Trng hp hc sinh gii theo cch khc: - Nu ra kt qu khng ng vi p n th khng cho im.- Nu ra kt qu ng vi p n th gim kho kim tra c th tng bc v cho im theo s thng nht ca c t chm. 3. Vi bi 8.2) nu hc sinh vit quy trnh bm phm khc, gim kho dng my kim tra, nu ra kt qu ng th cho im ti a. s gio dc v o to Bi 5:Tm x, y nguyn dng, x 1 tha mn:y = 31 9 + x + 31 9 x . Tm tt cch gii:p s: 5 t a = 31 9 + x ; b = 31 9 x=> a3+b3 = 18; ab = 382 x v y = a+b=> y3 = 18 + 3aby => y(y2-3ab) = 18 => y {1;2;3;6;9;18}. Th trn my => p s. x = 81; y = 3 2,0 3,0 HT Phch nh km thi chnh thc lp 12 B tc THPT . S Gio dc- o toqung ninh K thi cp tnh gii ton trn My Tnh casio b tc trung hc nm hc 2005 - 2006 ------------- @ ------------- Lp :12 B tc THPT . Thi gian thi:150 pht (khng k thi gian giao ) Ngy thi: 11/01/2006 H v tn th sinh: ................................................................................................Nam (N) ..................... S bo danh: .....................................................................................................................................................Ngy, thng, nm sinh: ................................................ Ni sinh: ................................ ............................. Hc sinh lp: ..................... Trng THPT: ............................................................................................. H v tn, ch k ca gim th S phch Gim th s 1: ................................................................. Gim th s 2: ................................................................. (Do Ch tch hi ng chm thi ghi) Quy nh : 1)Th sinh phi ghi y cc mc phn trn theo hng dn ca gim th.2)Th sinh lm bi trc tip vo bn thi c phch nh km ny. 3)Th sinh khng c k tn hay dng bt c k hiu g nh du bi thi, ngoi vic lm bi thi theo yu cu ca thi. 4)Bi thi khng c vit bng mc , bt ch; khng vit bng hai th mc. Phn vit hng,ngoicchdngthcgchcho,khngctyxobngbtccchgkc bt xo. Ch c lm bi trn bn thi c pht, khng lm bi ra cc loi giy khc. 5)Tri vi cc iu trn, th sinh s b loi. S Gio dc- o toqung ninh K thi cp tnh gii ton trn My Tnh casio b tc trung hc nm hc 2005 - 2006 ------------- @ ------------- thi chnh thc Lp : 12 B tc THPT .Thi gian lm bi: 150 pht (khng k thi gian giao ) Ngy thi: 11/01/2006 Ch : - thi ny c :04 trang - Th sinh lm bi trc tip vo bn thi ny. im ca ton bi thi H v tn, ch k cc gim kho S phch(DoCh tchH chm ghi )Bng sBng ch ......................................................

...................................................... Quy nh : 1)Th sinh ch c dng my tnh:Casio fx-220, fx-500A, fx-500MS v fx-570MS. 2) Cc kt qu tnh ton gn ng, nu khng c yu cu c th, c qui nh l chnh xc n 5 ch s thp phn. Bi 1:Tm gn ng ta cc im un ca th hm s y = x4 - 10x2 + 9. p s: Bi 2:Tnh gn ng ta im cc tiu ca th hm s y = 0,71x3 + 0,88x2 - 4,72x + 5. p s: Bi 3:Tnh gn ng (, pht, giy) cc nghim ca phng trnh: 3sinx + 5cos3x = 4 2 Tm tt cch giiKt qu Trang 1Bi 4:Trong mt phng vi h ta Oxy cho elp (E) : 19 252 2= +y x v ng thng (d):y = -3 (x - 4). Tnh gn ng to cc giao im M v N ca (d) vi elp (E). Tm tt cch giiKt qu Bi 5:Cho hm sf(x) =x tg22 + vi th (C). Tnh gn ng cc gi tr ca cc h s a v b ng thng y = ax + b l tip tuyn ca th (C) ti tip im c honh x = 7. Tm tt cch giiKt qu Bi 6:Trong mt phng vi h ta Oxy cho ba imA(1;2) , B(3;-2) , C(8;5).Tnh gn ng (, pht, giy) s o gc A v din tch ca tam gic ABC. Tm tt cch giiKt qu Trang 2 Bi 7:Tnh gn ng gi tr ln nht v gi tr nh nht ca biu thc F(x) =3 cos2x -5 cosx. Tm tt cch giiKt qu Bi 8 :Tm gn ng ta im M trn th hm s y = 222 +xx x sao cho tip tuyn vi th ti im M ct cc trc ta ti A v B to thnh tam gic vung cn OAB (vi O l gc ta ). Tm tt cch giiKt qu Bi 9:Cho hnh chp t gic u S.ABCD c cnh y bng 5,55cm, gc gia cnh bn v mt phng y bng 300.9.1) Tnh gn ng th tch hnh chp S.ABCD. V hnh, tm tt cch gii.Kt qu Trang 3 9.2) Tnh gn ng (, pht, giy) gc gia hai mt phng (SAB) v (ABCD). V hnh, tm tt cch gii.Kt qu Bi 10:Tnh gn ng nghim ca phng trnh :2x5 - 2cosx + 1 = 0 (chnh xc n 9 ch s thp phn) Tm tt cch giiKt qu

------------------------ Ht ------------------------- Trang 4 s gd-t qung ninh hng dn chm thi HSG gii ton trn my tnh casio lp 12 b tc thpt . nm hc 2005-2006 Bi Tm tt cch gii p s im tng phnim ton bi 1Tm c 2 im un U1(944;35 ) ; U2(944;35 )x1 1,29099 y1 - 4,88889 x2 - 1,29099 y2 - 4,88889 2y' = 0 ti x1; x2 vi x1 < x2

Do y c h s a = 0,71 > 0 nn y t cc tiu ti x2 => yCT = y(x2) xCT 1,13173 yCT 1,81452 3 PTr cho cos(x - ) = cos nhn, cos =5/ 34 ; sin = 3/ 34 ; v cos = 4 2 / 34 . Tnh ; , nh vo my ri suy ra x. x1 450 + k3600

x2 5038'33'' + k1200

4 Gii h:19 252 2= +y x; y = -3 (x - 4) c 2 nghim: x1 = 5/2 ; y1 = 23 3 v:x2 = 65/14 ; y2 = 143 9 C hai giao im l M(x1;y1) v M'(x2;y2) x1 2,5 y1 2,59808 x2 4,64286 y2 - 1,11346 5 C a = f(/7) b = f'(/7) - (/7).f(/7) a 0,39710 b 1,31574 6a) cosA = (b2 + c2 - a2)/2bcDng MOD 4vi fx500A hoc MOD MOD MOD 1 vi fx500MS b) S =) )( )( ( c p b p a p p A 88033'37'' SABC 224,8228261 224,82283 cm2. 7 C F(x) =3 cos2x -5 cosx = 2 3 cos2x - 5 cosx -3 . Xt hm s f(t) = 2 3 t2 -5 t -3 . vi -1 t 1. Dng o hm hoc tnh cht tam thc bc hai, tm c fmin = f(-1) =3+5v fmax = -29/4. 3 fmin -2,09289 fmax 3,96812 Bi Tm tt cch gii p s im tng phnim ton bi 8Tip tuyn vi th ti M phi song song vi ng thng y = x hoc y = - x. tip tuyn c h s gc k = 1 Tm c 2 tip tuynSuy ra c 2 im M tha mn : M1(2+ 2 ; 5+3 2 ) v M2(2- 2 ; 5-3 2 ) M1(3,41421 ; 9,24264) M1(0,58579 ; 0,75736) 9 9.1) K SH (ABCD) => SAH = 300

tnh c SH ri => VS.ABCD =6 .5,553/18 9.2) Gi E l trung im AB => SEH l gc gia 2 mt phng (SAB) v (ABCD). Tnh c tgSEH =2 / 3ri suy ra SEH VS.ABCD 2,47346 SEH 39013'53'' 10Xt hm s f(x) = 2x5 - 2cosx + 1 Thy f(0) = -1 v f(/4) 0,183481134 => phng trnh c nghim (0 ; /4) (1) x = 521 cos 2 x Dng phng php lp, tm c xTnh, chng hn trn my fx - 500MS: Vo MODE MODE MODE 2Khai bo x0 = 0:0 = Khai bo biu thc:5SHIFTx ((COS ANS - 1) : 2 ) v thc hin dy lp:= ... =ra kt qu

x 0,747506599 s gd-t qung ninh. S Gio dc v o toqung ninh ---------------- K thi thi tuyn My Tnh casio tnh qung ninhnm hc 2005 - 2006 ------------- @ ------------- Lp 12 THPT.Ngy thi th : 07/3/2006 Thi gian lm bi: 150 pht (khng k thi gian giao ) im ca ton bi thi H v tn, ch k cc gim kho S phch(DoCh tchH chm ghi )Bng sBng ch ......................................................

...................................................... Quy nh : 1)Th sinh ch c dng my tnh:Casio fx-220, Casio fx-500A, Casio fx-500MS, ES; Casio fx-570MS, ES. 2)Cc kt qu tnh ton gn ng, nu khng c yu cu c th, c qui nh l chnh xc n 5 ch s thp phn. Bi 1:Cho hm s y = 11+xx(C). M0(x0; y0) (C); l tip tuyn ca (C) ti M0. Tnh gn ng ta im M0 bit hnh thang to bi ng thng v cc ng thng x = 1; x = 2; y = 0 c din tch ln nht. Tm tt cch gii. p s: Bi 2:Cho a thc f(x) vi h s hu t, l s thc tha mn : [ ] = = 5 ) ( ) (533 f fK hiu fn(x) = f(f(...(f(x)...)) (n ln). Hy tnh T = [f2004()]4 + [f2005()]3 - f2006() Tm tt cch gii. p s: Bi 3:Tm ch s tn cng ca s T = 3 + 45 + 59 + 613 + ... + 5042005. Tm tt cch gii. p s: Bi 4: Trong h trc ta Oxy cho parabol (P) : y = x2 - 2x v elip (E) :11 922= +y x. a) Tnh gn ng ta A, B, C, D l giao im ca (P) v (E). b) Tnh gn ng bn knh v ta tm ng trn i qua 4 im A, B, C, D. Tm tt cch gii. p s: Bi 5: ng b bin ca mt hn o c dng elp (E) vi cc bn trc OA v OB OA = 5km, OB = 3km. B bin trong t lin l mt ng thng ct tia OA ti C, ct tia OB ti D. Bit OC = 8km, OD = 4,5 km, tnh gn ng khong cch ngn nht t o vo b. Tm tt cch gii. p s: Bi 6:Trong mt phng cho hnh vung ABCD c nh c cnh bng 2cm. Gi s (O) l hnh trn ngoi tip ABCD. Ngi ta thc hin php di on AB nh sau: - A, B di chuyn ngc chiu kim ng h. - A, B di chuyn nhng vn thuc (O). on thng AB c di lin tc n trng vi on CD sao cho A trng vi C, B trng vi D. Tnh din tch ca hnh do on thng AB qut ra khi di ch . Tm tt cch gii. p s: Bi 7: Tm s nguyn dng n nh nht tha mn: n c ng 144 c s dng phn bit v tn ti 10 s nguyn lin tip trong s cc c s dng ca n. Tm tt cch gii. p s: Bi 8:Tnh gn ng di nh nht ca cnh mt hnh vung sao cho c th t vo trong n 5 hnh trn bn knh r = 2 m khng c hai ng trn no chm ln nhau. Tm tt cch gii. p s: Bi 9: Cho tam gic ABC c na chu vi p = 13,5cm, BAC = 370, ABC = 640. Tnh gn ng din tch tam gic ABC. Tm tt cch gii. p s: Bi 10:Cho dy s {Un}n=1 nh sau:Un = ln(ln(...(lnn)...))( n ln). Dy trn c hi t khng ? Tm tt cch gii. p s: -------------------- Ht ----------------------- S GIO DC O TO THI TUYN HC SINH GII MY TNH B TITHINGUYNchn i tuyn THBT, lp 12, 2002-2003 Bi 1: Tm mt nghim gn ng ca phng trnhcos tan3 0 x x + =Bi 2: Tm mt nghim gn ng ca phng trnh 52 sin(3 1) 2 0 x x x + =Bi 3: Tnh 2(sin15 sin 75 )o o+Bi 4: Tnh khong cch gia hai nh khng lin tip ca mt ng gic u ni tip trong ng trn bn knh R = 5,12345 cm. Bi 5: Cho hai ng trn c cc phng trnh tng ng 2 25 6 1 0 x y x y + + + =v . Hy tm ta giao im ca chng. 2 22 3 2 x y x y + + = 0Bi 6:Cho x = 0,36. Tnh 23 524 55log 2(log ) 3log 212(log 2 ) 4log 22x x xMx x+ +=+ Bi 7: Tm ta giao im ca hyperbol 2 214 9x y =v parabol 25 y x = . Bi 8: Cho ng trn c phng trnh 2 24 6 171 0 x y x y + + =Hy xem xt v tr ca im so vi ng trn ( nm trong, nm ngoi hay nm trn ng trn). (2, 000001; - 3, 000123) ABi 9: Trong khng gian cho bn im (1, 2; 0, 23; 1, 756), ( 7, 247; 3,14386; 4,12), A B . Hy tnh n 7 ch s sau du phy th tch ca t din ABCD. (5, 245; 4, 567; 3, 421), (6, 512; 4, 35; 7,18) C D Bi 10: Cho tam gic ABC c15, 432 ; 7, 675 ; 18 AB cmBC cm CA = = cm = . Hy tnh din tch tam gic v ba gc tng ng ( , pht, giy). HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TITHINGUYN chn i tuyn THPT, lp 12, 2002-2003 Bi 1: Tm mt nghim gn ng ca phng trnh 4 27 2 x x x 0 + + =Bi 2: Tm mt nghim gn ng ca phng trnh- 3 0xe x + =Bi 3: Tm n 28( 1)! 5, 5 10 n n + !Bi 4: Tnh khong cch gia hai nh khng lin tip ca mt ng gic u ni tip trong ng trn bn knh R = 5,12345 cm. Bi 5: Tm h s ca 50xtrong a thc nhn c sau khi khai trin: 2 3 1000(1 ) 2(1 ) 3(1 ) ... 1000(1 ) x x x x + + + + + + + +Bi 6:Cho x = 0,36. Tnh 23 524 55log 2(log ) 3log 212(log 2 ) 4log 22x x xMx x+ +=+ Bi 7: Phng trnh Fermat l phng trnh: 1 2 1 2... ...n nn nnx x x x x x = + + + ( 1) Pht biu bng li: Tm cc s c n ch s sao cho tng ly tha bc n ca cc ch s bng fchnh s y. Trong cc s sau y, s no l nghimh ca phng trnh (1): 157; 301; 407; 1364; 92727; 93064; 948874; 174725; 4210818; 94500817;472378975. Bi 8: Cho ng trn c phng trnh 2 24 6 171 0 x y x y + + =Hy xem xt v tr ca im so vi ng trn ( nm trong, nm ngoi hay nm trn ng trn). (2, 000001; - 3, 000123) ABi 9: Trong khng gian cho bn im (1, 2; 0, 23; 1, 756), ( 7, 247; 3,14386; 4,12), A B . Hy tnh n 7 ch s sau du phy th tch ca t din ABCD. (5, 245; 4, 567; 3, 421), (6, 512; 4, 35; 7,18) C D Bi 10: S2l nguyn t hay hp s? 111 HT S GIO DC O TO THI TUYN HC SINH GII MY TNH B TITHINGUYN THBT, lp 12, 2003 Bi 1: Tm mt nghim gn ng ca phng trnh 22sin -1 0 x x + =Bi 2: Cho( ) 3 12 2002x xPx = x. Tnh P(1,0012). Bi 3: Tnh gc A ( radian) ca tam gic ABC bit9, 357 ; 6, 712; 4, 671 BC cm CA AB cm = = =Bi 4: Tm thng v s d ca php chia 3456789 cho 23456.Bi 5: Tm thng v s d ca php chia 9 5 22 3 4 1 x x x x + + +cho4,12345 x + Bi 6:Tnh 2 0,2log 3 log 0, 5