com Skkn Ung Dung May Tinh Casio 2011

1 S GIO DC-O TO BN TRE TRNG THPT CHUYN BN TRE ----------

www.VNMATH.com 2 TrongnhiunmquaBGiodcvotocchtrngamytnh Casio v Vinacal vo ging dy trong chng trnh THPT. Hng nm u c t chc cc cuc thi gii ton trn my tnh Casio v Vinacal t cp tnh n cp Quc gia, tuy nhin vic hng dn cho hc sinh vn dng cc loi my tnh b ti mt cch sngtotrongqutrnhhctpbmntonniringvccmntnhinni chungvncnhnch.Nhnchunghcsinhchsdngmytnhmcthc hin cc php tnh n gin m cha ng dng my tnh mc cao hn nh d on kt qu, t duy ton hc da trn cng c my tnh... Quaqutrnhgingdytitchlycmtskinhnghimchonidung ny. Cc vn trnh by trong sng kin kinh nghim l cc chuyn c ng dngtronggingdyvcphbinnngnghiptrongcclnhingh chuynmndoSGDtchctrongccnmhcqua.Bnthntinhnc nhiukinphnhikhchltccngnghiptrongvngoitnh.Sngkin kinhnghimnylstngktcchnlcccchuyncabnthnvitra trong thc tin ging dy cng vi s ng gp nhit tnh ca ng nghip. L do chn ti ca ti xut pht t nhng l do sau: *GipchohcsinhtrungbnhbitcchkimtraktqubngmytnhCasio hoc Vinacal. V d kim tra kt qu ca cc bi tontnh gii hn, o hm, tch phn... iu ny rt c ch khi HS lm cc bi thi TN v H. Nu khng hng dn cho HS nhng th thut ny th cc em s mt nhiu thi gian khi kim tra li ton b qu trnh tnh ton ca mnh. *GipchoHSkh,giicsuynghdngmytnhdonktqu.Vd dngmytnhCasionhmnghimgiiphngtrnhlnggic,pdngtnh chtcahmslintckthpvimytnhgiibtphngtrnh,dngmy tnh tm quy lut dy s... * Gip cho cc bn ng nghip c mt ti liu tham kho trong qu trnh ging dy b mn ton ca mnh. Qua chuyn ny ti hy vng cc bn ng nghip s yuthchhnccngdngmmytnhCasio,Vinacalemlichochngtav truyn s say m ny n cc HS ca mnh. Thc t mt s Thy C khng thch s dng my tnh Casio bi v kt qu ca n a phn l kt qu gn ng, nhng trong chuyn ny cc bn s thy ta c th dng ci gn ng i tm ci ng ( ng dng my tnh Casio hoc Vinacal trong vic gii cc bt phng trnh ) - tinycthpdngrngrichottcgiovindytoncctrng trunghcphthngthamkhovccemhcsinhlp12nthiTtnghipvCao ng - i hc. - Phm vi nghin cu ca ti ny bao gm:www.VNMATH.com 3* Khng trnh by cc vn c bn v my tnh Casio, Vinacal (v cc vn c bn ny c trnh by trong nhiu ti liu ) m ch minh ha cc ng dng c th v c tnh mi trong gii ton. *ngdngmytnhCasio,Vinacaldonnghimgiiphngtrnhlng gic. * ng dng my tnh Casio, Vinacal trong gii cc bt phng trnh phc tp. * ng dng my tnh Casio, Vinacal kim tra kt qu v trong cc dng ton khc. Bn thn nghin cu ti ny nhm mc ch: * Chia s vi ng nghip v cc em hc sinh kinh nghim v ng dng my tnh Casio, Vinacal trong dy v hc mn ton. * Bn thn rn luyn chuyn mn nhm nng cao nghip v s phm. *HngngphongtrovitSKKNcatrngTHPTchuynBnTrevca Cng on ngnh Gio dc pht ng. * SKKN ny khng trnh by li cc chc nng ca my tnh Casio v Vinacal v cc vn ny c ni n trong nhiu ti liu. * SKKN ny cp n mt s vn trong dy v hc b mn ton THPT c tnh chuyn su di dng cc chuyn .* SKKN ny t ra mt vn mi cc bn ng nghip tip tc nghin cu lphthuytiakhnngcamytnhCasiovVinacalmtcchsngto trong vic dy v hc b mn ton THPT. *CcchuynvngdngmytnhCasio,Vinacaltronggiiphngtrnh lnggicvchuynvicgiibtphngtrnhvvicgiiphngtrnhlcc chuyn mi cha c trnh by trn bt k ti liu no v vn ng dng my tnh b ti trong gii ton. www.VNMATH.com 4 Phng php nghin cu SKKN ny da trn c s: * Cc kin thc c bn v my tnh Casio, Vinacal. * Cc kin thc ton hc c bn trongchng trnh THPT. * Mt s k thut bin i i s v ng dng ca my tnh cm tay. Cngvisphttrincacngnghthngtin,ccphnmmtonhcngy cnghtrclcchogiovinvhcsinhtrongvicdyvhcmnton,tuy nhin khng phi hc sinh no cng c iu kin to cho mnh mt my vi tnh v ci t cc phn mm thch hp hc tp b mn ton, hn th na theo quy ch hc sinh khng c em my vi tnh vo phng thi... Trong khi mi hc sinh u c my tnh Casio hoc Vinacal, do vic rn luyn cho hc sinh s dng cc loi my tnh cm tay ny mt cch thnh tho l mt vic lm cn thit. Thc trng hin nay cho thy k nng s dng my tnh cm tay ca hc sinh cn rt yu, a s ch bit dng my tnh thc hin cc php tnh cng, tr, nhn, chia, khai cn v tnh gi tr ca cc hm s lng gic m thi. Do SKKN ny cp n mt vn milgiphcsinhkhaithctiaccchcnngcamytnhCasiov Vinacal trong t duy gii ton. Nu lm tt cng vic ny th cht lng dy v hc mn ton s c nng ln. NGDNGMYTNHCASIOVVINACALDONNGHIMGII PHNG TRNH LNG GIC A. t vn : Khigiiccphngtrnh athctathng dng cch nhmnghimbini phngtrnhyvdngphngtrnhtch.Vyvicgiiphngtrnhbccaoc chuyn v vic gii phng trnh bc thp hn. Trong chuyn ny s minh ha cho vic ng dng t tng ny vo vic gii mt s dng phng trnh lng gic vi s tr gip ca my tnh cm tay. B. Ni dung phng php: giiphngtrnhlnggicbngphngphpny,tastinhnhtheocc bc sau: Bc 1:Tin hnh php th tm mt nghim c bit. Ta th vi cc gi tr c bit sau: 2 3 50; ; ; ; ; ; ; ;6 4 3 2 3 4 6t t t t t t tt . www.VNMATH.com 5Bc 2:Gi s bc 1 tm c nghim 6xt= . Ta tip tc th vi cc gi tr c bit tng ng lin kt vi nghim y. C th: + Th vi gi tr i ca n: 6xt= , nu tha mn phng trnh th ta d on phng trnh c nghim x sao cho 3cos2x = , hay phng trnh c a v dng tch vi mt tha s l(2cos 3) x + Th vi gi tr b vi n : 56xt= , nu tha mn th ta d on phng trnh c nghim x sao cho 1sin2x = , hay phng trnh c a v dng tch vi mt tha s l inx (2s 1) . + Th vi mt gi tr hn ( km ) n t , th vi 76 6xttt + = =( hay th vi 56 6xttt = =) Nu gi tr ny tha mn th ta d on phng trnh c nghim x : 3t anx3= . Hay c th bin i phng trnh v phng trnh tch vi mt tha s( 3 t anx 1) . C. Phng tin dng nhm nghim: C th dng my tnh Casio fx 570 ES tin hnh nhm nghim theo mt trong hai cch sau: Cch 1: Dng chc nngCALC . Chc nng ny c cng dng l tnh gi tr ca mt hm s ti mt im. - Chuyn phng trnh v dng f(x) = 0. Gi s cn th vi gi tr 6xt= , ta thc hin nh sau: - Nhp vo my hm s f(x), nhnphmCALC , my hi x ? ta nhp vo 6t v nhn phm= . th vi cc gi tr khc, ta tip tc nhn phmCALC Cch 2: DngchcnngSOLVE .Chcnngnyccngdngltmnghimca phng trnh trong mt ln cn ca x ch ra. Ta thc hin theo cc bc sau y: - Chuyn my tnh v n v . - Nhp vo phng trnh f(x) = 0. www.VNMATH.com 6-NhnphmSOLVE ,myhinthx?tanhpvogitrmtadonl nghim,chnghn30(300),mysdtmmtnghimtronglncnca300. Tip tc nhn phmSOLVE kim tra nghim khc D. Cc v d minh ha: Gii phng trnh:3cos2 5sin cos sin2 4 x x x x + + = . Gii Phn tch: Thc hin php th, thu c 2 nghim 5,6 6x xt t= = . Do d on phng trnh s c nghim x : 1sinx2= . Vy li gii c trnh by theo hai cch sau: Cch 1 tsinx( 1) t t = s . Ta vit phng trnh cho thnh phng trnh vi n s t: 23(1 2 ) 5 cos 2 cos 4 t t x t x + + =26 (2cos 5) (1 cos ) 0 t x t x + + = (*) Theo d on trn th phng trnh (*) c nghim 112t = . p dng nh l Viet: 1 2 25 2cos 1 cos6 3x xt t t + = =Vy phng trnh cho 1sinx23sin cos 1 x x

=

+ = n y ta d dng ch ra tp nghim ca phng trnh. Cch 2 Phn tch Do d on c 1sinx2=, do nu bin i phng trnh v dng phng trnh tch th s c mt tha s l(2sin 1) x . Vy nn kt hp hai s hng no vi nhau c tha s(2sin 1) x ? Ta c th thy ngay nn kt hp nh sau : cos sin2 cos (1 2sin ) x x x x = .Cn tng(3cos2 5sin 4)? x x + Mt iu chc chn rng c th phn tch tng ny thnh tha s m c mt nhn t l(2sin 1) x . Tht vy : 2 2(3cos 2 5sin 4) 3(1 2sin ) 5sin 4 6sin 5sin 1 x x x x x x + = + = + (2sin 1)(3sin 1) x x = * Vy li gii c trnh by ngn gn nh sau : (cos sin2 ) (3cos2 5sin 4) PT x x x x + + 2cos (1 2sin ) ( 6sin 5sin 1) 0 x x x x + + =cos (1 2sin ) (1 2sin )(3sin 1) 0 x x x x + =(1 2sin )(cos 3sin 1) 0 x x x + =www.VNMATH.com 7Vy phng trnh cho 1sinx23sin cos 1 x x

=

+ = n y ta d dng ch ra tp nghim ca phng trnh. Gii phng trnh:cos3 cos 2 sin2 sin 5cos 3 (1) x x x x x + + + =Phn tch Thc hin php th ta tm c 2 nghim: 23xt= . Vy ta d on phng trnh c nghim x : 1cos2x= . Cch 1 tcos ( 1) t x t = s . Phng trnh (1) tr thnh: 3 2(4 3 ) (2 1) 2 sin sinx 5 3 t t t t x t + + + =3 24 2 (2sin 8) (sin 4) 0 (2) t t x t x + + + =Do th nghim trn nn ta bit PT(2) c nghim 12t = . Thc hin php chia v tri ca (2) cho 1( )2t +. ta c: 21(2) ( ) 4 (2sin 8) 02PT t t x( + + = Vy 2 21 1cos cos(1) 2 24cos 2sin 8 0 2sin sin 2 0x xPTx x x x = = + = + = 1 2cos 2 , ( )2 3x x k ktt = = + eVy phng trnh c nghim : 22 , ( )3x k ktt = + eCch 2 ( Bin i phng trnh v dng phng trnh tch ) Do d on trn nn khi bin i phng trnh v dng phng trnh tch th phi c mt nhn t(2cos 1) x .Vy ta nn kt hp hai s hng no c nhn t(2cos 1) x ? -C th thy ngay , nn kt hpsin2 sinx x + . -(1) (sin2 sinx) (cos3 cos2 5cos 3) 0 PT x x x x + + + = 3 2(sin2 sinx) (4cos 2cos 8cos 4) 0 x x x x + + + =sin2 sin sin (2cos 1) A x x x x = + = +3 24cos 2cos 8cos 4 B x x x = + www.VNMATH.com 8Cng t d on trn nn ta suy ra B c th phn tch thnh nhn t v chc chn c mt tha s l(2cos 1) x . Do ta thc hin php chia B cho(2cos 1) x , ta s c: 2(2cos 1)(2cos 4) B x x = . Vy 2(1) (2cos 1)(sinx 2cos 4) 0 PT x x + + =

2(2cos 1)( 2sin sin 2) 0 x x x + + =

1 2cos 2 , ( )2 3x x k ktt = = + eVy phng trnh c nghim : 22 , ( )3x k ktt = + e Gii phng trnh:2 21 8 12cos cos ( ) sin2 3cos sin (1)3 3 2 3x x x x xtt| | |\ .+ + = + + + +Gii Thaycos( ) cos x x t + = vcos( ) sinx2xt+ = . (1) 6cos cos2 3sin2 9sin 8 0 (2) PT x x x x + + =Nhn thy22x ktt = +l nghim ca phng trnh. Vy ta c sinx = 1. Cch 1 tsinx( 1) t t = s . Ta vit phng trnh (2) thnh phng trnh vi n s t: 22 (9 6cos ) (6cos 7) 0 t x t x + + =Phng trnh bc hai ny c A + B + C = 0. Vy : sin 1(2)6cos 7sin2xPTC xxA=

= = sin 12sin 6cos 7xx x=

+ = M phng trnh :2sin 6cos 7 x x + = v nghim. Vy phng trnh cho c nghim :2 , ( )2x k ktt = + e . Cch 2 ( Bin i phng trnh v phng trnh tch ) (1) (6cos 3sin2 ) (cos 2 9sin 8) 0 PT x x x x + + =26cos (1 sin ) ( 2sin 9sin 7) 0 x x x x + + =76cos (1 sin ) 2(sin 1)(sin ) 02x x x x =(1 sin )(6cos 2sin 7) 0 x x x + =sin 1sin 1 26cos 2sin 7 2xx x kx xtt=

= = +

+ = Vy phng trnh cho c nghim :2 , ( )2x k ktt = + e . www.VNMATH.com 9 Gii phng trnh:sin3 6sin2 9sin cos3 9cos 8 (1) x x x x x + + =Gii -Nhn thy22x ktt = +l nghim phng trnh. Vy ta c sinx = 1. -tsinx ( 1 ) t t = s . Thay : 3 3sin3 3sin 4sin , os3 4cos 3cos x x x c x x x = = . -Phng trnh (1) tr thnh: 3 24 4(cos ) (12 12cos ) (8cos 8) 0 (2) t x t x t x + + + =-Thc hin php chia VT(2) cho ( t 1 ), Ta c : 21(2)4 (4cos 4) 8 8cos 0 (3)tPTt x t x

= + + = Ta c: 2(3) 4sin (4cos 4).sinx 8 8cos 0 (4) x x x + + =Nhn thy2 x k t =l nghim ca (4), vy PT(4) c nghim x sao cho cosx = 1. - t u = cosx, PT(4) tr thnh : 24(1 ) (4 4)sinx 8 8 0 u u u + + =24 (4sin 8) (4 4sin ) 0 u x u x + + = 24 (4sin 8) (4 4sin ) 0 1 1 sinx u x u x u u + + = = v = cos 1 sinx cos 1 x x = v + = . Vy phng trnh c 2 h nghim: 2 , 2 ( )2x k x k ktt t = = + e Gii phng trnh:sin2 sin cos8 cos6 cos7 (1) x x x x x + = + +Nhn xt:Thc hin php th, ta c 23xt= l nghim. Vy nn nhm cc s hng sao cho xut hin tha s chung ( 2cosx + 1). Bi ny khng nn quy v phng trnh vi n s t = cosx v s c phng trnh bc cao. Do ta phi dng phng php phn tch thnh nhn t vi nh hng lm xut hin tha s ( 2cosx + 1). Gii (1) sin (2cos 1) 2cos7 .cos cos7 PT x x x x x + = +sin (2cos 1) cos7 (2cos 1) (2cos 1)(sin cos7 ) x x x x x x x + = + + T y suy ra phng trnh c cc h nghim: 22 , , ( )3 6 4 12 3k kk x x k xt t t t tt + = + = + e = Gii phng trnh:4sin cos 3sin .tan 3tan 3 (1) x x x x x + + =Gii iu kin:cos 0 x =-Thc hin php th c hai nghim 4xt= v 34xt= . Vy cn nhm cc s hng xut hin tha s(t anx 1) + . -(1) (sinx cos ) (3sin 3sin .tan ) 3(tan 1) PT x x x x x + + + = +cos (tan 1) 3sin (tan 1) 3(tan 1) x x x x x + + + = +www.VNMATH.com 10(1 tan )(cos 3sin 3) 0 x x x + + =tan 1cos 3sin 3xx x=

+ =( n y cc em HS c th gii tip c) Gii phng trnh:2(sin2 cos2 ) tan 1 (1) x x x = +Gii iu kin:cos 0 x =Thc hin php th c cp nghim 4xt=v 34xt= . Vy ta d on phng trnh c nghim x sao cho tanx = 1 Cch 1 t t = tanx, phng trnh (1) tr thnh: 22 22 111 12t ttt t| | = + |+ +\ . 3 23 3 0 t t t + = Vy:3 t anx 1 t anx = v = . Kt qu: phng trnh c cc h nghim: 4 3, ( ) k k x x kt tt t + + = = eCch 2 Ta nhm cc s hng xut hin tha s: ( tanx 1) (1) 2sin2 2cos2 2 tan 1 PT x x x = 2tan 1 4sin cos 4cos x x x x = 2tan 11) 01cos 22(tan 1)(4cosxxxx=

=

= Kt qu: phng trnh c cc h nghim: 4 3, ( ) k k x x kt tt t + + = = eE. Bi tp rn luyn: Gii cc phng trnh: 1)sin2 3sin 2cos 3 x x x + =2)sin5 sin2 cos6 cos4 cos x x x x x + + = +3)1 cos cos2 sin2 sin3 sin4 x x x x x + + = + +4)sin cos2 sin cos 1 0 x x x x + + + + =5)cos2 cos cos4 sin3 sin2 x x x x x + = +6)sin2 10cos sin 5 x x x + =7) 4sin cos 1 tan 3sin .tan x x x x x + = + www.VNMATH.com 11 NG DNG MY TNH CASIO V VINACAL GII BT PHNG TRNH. A. t vn : Nhm mc ch trang b cho hc sinh cc phng php gii ton hu hiu gii quyt tt cc dng ton ca thi tuyn sinh i hc, qua qu trnh ging dy nhiu nm cclpcuicpchngticktcmtskinhnghimvvichngdn hc sinh gii quyt cc bi ton v bt phng trnh. Chng ti thy rng vic gii mt bt phng trnh c dng f(x) > 0 ( f(x) > 0, f(x) < 0, f(x)s 0) th phc tp hn nhiu so vi vic gii phng trnh f(x) = 0. Thc cht ca bi ton gii bt phng trnh l quy vvicxtducabiuthcf(x)trnminxcnhDcan.Dovynidungca chuyn ny l quy vic gii cc bt phng trnh v vic gii phng trnh f(x) = 0, sau lp bng xt du ca f(x) v t suy ra tp hp nghim ca bt phng trnh. B. Ni dung phng php: Ni dung ca phng php ny da trn tnh cht sau y: Tnh cht: Gi s hm s f(x) lin tc trn min K ( K c th l (a ; b ) ; [a;b];(a;b];[a;b);(-;a);(-;a];(a;+);[a;+);R).Nu phng trnh f(x) = 0 v nghim trn min K th f(x) khng i du trn K. Phng php gii bt phng trnh: Da trn tnh cht trn ta suy ra phng php gii bt phng trnh dng f(x) > 0 ( f(x) < 0, f(x) > 0, f(x) < 0, f(x) > 0) nh sau : * Tm tp xc nh D ca hm s f(x). * Gii phng trnh f(x) = 0. * Lp bng xt du ca f(x) ( xc nh du ca f(x) trn cc khong con K ca D m f(x) v nghim, ta ch cn xc nh du ca f(x0) vi x0 l mt phn t bt k ca K). Ch : tnh gi tr ca hm s ti mt im mt cch nhanh chng ta c th dng my tnhCasio hay Vinacal. Cch tnh : dng chc nng CALC c minh ha qua v d sau: V d: Tnh gi tr ca hm s y = x2 +3x -12 vi x = 7, x = 8 thc hin nh sau : * Nhp biu thc n : 23 12 ALPHA YALPHA X X ALPHAX = + * Lu biu thc n: CALC* Tnh gi tr ca y vi x = 7 n :7=* Tnh gi tr ca y vi x = 8 n :8 CALC =* Tnh gi tr ca y vi x = 23 n : /2 3b cCALC a =Ni chung khi nhp biu thc y vo xong th ta c th tnh gi tr ca y ti cc im x1, x2, .. y vn m ta quan tm l du ca y ti cc im x1, x2, .. ( cc gi tr ca y ti cc im ny c th l cc gi tr gn ng. iu ny khng nh hng g kt qu nghim ca bt phng trnh). www.VNMATH.com 12C. Cc v d minh ha: a)Gii cc bt phng trnh v t. Gii bt phng trnh : 22401616x xx+ + s+(1) Cch 1: (Phng php c bn) BPT (1) 2 2 2 216 16 40 16 24 (2) x x x x x x + + + s + s Xt cc trng hp sau y : a) Nu x = 0 th BPT(2) lun tha mn. b) Nu x < 0 th BPT(2) 22222222 2 2 2242 6 002 6242416162 60 2 6240( 16) (24 )xxxxxxxxxxxxxx x xx s

| | + > + > < | \ . < <

>+ > 2 22 6 0 2 6 02 6 2 602 6 2 630 2 6 0 2 63 64 24x xx xxx xxx xx x s < s < >> < < < > < < < < > > . Kt hp vi x < 0 ta c x < 0. c) Nu x > 0 thi BPT(2) 2222222 2240 2 6 2 62416 0 30 2 6 0 2 624163 3 64 24xx xxxx xx xxxxx xx > < < + s < s < 0 Quabngxtducaf(x),tasuyraBPT f(x) s 0 c nghim x s 3. Gii phng trnh : 23 2 1 (1)3 2xx xx > Cch 1 iu kin x > 23.BPT(1) 2(3 2) (1 ) 3 2 ( 1)( 2) ( 1) 3 2 0 x x x x x x x x > + > 22112 02 01 0(3 2) (2 )3 2 22( 1)( 2 3 2) 011 032 03 2 23 2 (2 )xxV xxxx xx xx x xxxxx xx x >> >

s > >

>

+ > < s

s

> s s

22221 21 22 1 67 6 21 2231223 113317 6 06xxxxx xx xx xxxxxx xx > >

s s

s s

> s s

+ s s >s s s s s s s

+ > > . Tm li BPT c nghim x > 23. Cch 2: iu kin x > 23. BPT(1) f(x) = 23 2 1 03 2xx xx + > PT f(x) = 0 22 0( 1)( 2 3 2) 0 1 3 2 2 13 2 (2 )xx x x x V x x x Vx x > + = = = = = x = 1 V 2221 37 6 0xxx x< s= + = www.VNMATH.com 14+10xf (x)23++ * Hm s f(x) lin tc v v nghim trn cc khong : (23; 1); (1 ; + ) nn trn tng khong ny f(x) khng i du.Ta c f(56) = 0,108 > 0 ,f(2) = 1 > 0 Qua bng xt du ca f(x), ta suy raBPT f(x) > 0 c nghim x >23 . Gii bt phng trnh :1 1 (1) x x x + >Cch 1 iu kin : -1 sx s 1. BPT(1) 21 0( )1 ( 1 )1 11 0( )1 0x xAx x xx x xxBx x + >

+ > + + > + + >

+ s 1) Xt h (A) : 21 01 ( 1 )x xx x x + >+ > + 2200 10 11 01 01 0 1 0 1 1 101 5 1 51 01 2 2xxxxxx x x x x xxx xxx x s s s s s

+ > s s

s s + > > s s

> +

+ ss s > H (A) 2 21 1 1 11 1 2 1 2 1 2 (2)x xx x x x x x x x x s s s s + > + + s + Nu x = 0 th (2) ng + Nu x < 0 th (2) 2 22 1 2 4(1 ) (2 ) 0 x x x x x > > sv nghim trn [-1 ; 0 ). + Nu x > 0 th (2) 2 22 1 2 4(1 ) (2 ) 0 x x x x x s s >ng trn (0 ; 1]. Vy h (A) c nghim 0 s x s 1. 2) Xt h (B) : 1 01 0xx x+ > + s221 01 11 51 1 1 0021 0 111 52xxx xxxx x x xx xx s s s s

s s s s s

>

+ > s s +>

v nghim Tm li: BPT c nghim 0 s x s 1. Cch 2: iu kin : -1 sx s 1. Xt hm s f(x) =1 1 x x x + vi x e [-1 ; 1 ]. BPT(1) f(x) > 0. www.VNMATH.com 15PT : f(x) = 0 2 2 21 1 1 ( 1 ) 1 1 2 1 2 1 2 x x x x x x x x x x x x x x x + = + + = + + = + + = 20 2 1 2 0 4(1 ) (2 ) 0. x V x x x V x x x = = = = =Th li thy x = 0 l nghim ca phng trnh f(x) = 0. Bng xt du ca f(x) _00xf (x)-11+ *Hmsf(x)lintcvvnghim trncckhong:[-1;0);(0;1]nn trntngnakhongnyf(x)khng i du.Ta c f(12) = - 0,017 < 0 ,f(12) = 0,017 > 0 Qua bng xt du ca f(x), ta suy raBPT f(x) > 0 c nghim 0 s x s 1. b) Gii cc BPT m, logarit Gii BPT : 2 2 1 x 1 x x(x x 1) (x x 1) + + + > + + (*) Cch 1 iu kin : 1 x 01 x 0 > + > -1 s x s 1 (1) * Ta c x2 +x +1 > 0 , x e R. Ta xt 3 trng hp : a) Trng hp : x2 +x +1 = 1 x2 + x = 0 x = 0 V x = -1 tha mn BPT (*). b) Trng hp : x2 +x +1 > 1 x2 + x > 0 x < -1 V x > 0, kt hp vi k (1) ta c0 < x s 1 (2) BPT(*) 21 x 1 x x x 1 x 1 x x (1 x) 2x 1 x 1 x > + + > + + + > +2 22 1 x 2 x 0 4(1 x) x 4x 4 x 0 x 0 loai (do k (2) ). > > > + s = c) Trng hp x2 + x+ 1 < 1 x2 + x < 0 -1 < x < 0 (3) BPT(*) 21 x 1 x x 1 x (1 x) x 2x 1 x x 2 2 1 x 0 (do x < 0) s + s + + + + + s2 22 1 x x 2 4(1 x) x 4x 4 x 0 x 0 loai ( do k (3) ) + > + + > + + s = Tm li: bt phng trnh c nghim x = 0; x = -1. Cch 2: iu kin 1 s x s 1. Xt hm s : 2 2 1 x 1 x xf (x) (x x 1) (x x 1) + = + + + +BPT(*) BPT f(x) > 0 . By gi ta gii PT f(x) = 0. Ta c : 222x x 1 11 x 1 x 0 Vx 1 x 0 Vx 1x 0x ( 1;1] \ {0} x ( 1;1] \ {0}f (x) 0x x 1 1x 11 x 1 x x 2x 1 x 2 1 x x 2 1 x 11 x 1 x x + + = s s = = = = =

e e = + + =

=

= + + + + = + s s

= + www.VNMATH.com 16Th li ta thy PT f(x) = 0 c nghim x = 0 ; x = -1. Hm s f(x) lin tc v v nghim trn cc khong(-1; 0) v (0 ; 1] nn trn cc khong ny f(x) khng i du. Ta c 1f ( )2~- 0,0035 < 0 v1f ( )2 ~ - 0,0147 < 0 T suy ra BPT(*) c nghim x = -1; x = 0. Gii bt phng trnh: 211331 1(1)log (x 1)log 2x 3x 1 >+ + Cch 1 : iu kin : 20 2x 3x 1 1 1 3 3x ( 1 ; 0) (0 ; ) (1 ; ) ( ; ) (2)2 2 2 0 x 1 1 < + =e + < + = Khi BPT(1) 2 23 33 31 1 1 1(3)log (x 1) log (x 1)log 2x 3x 1 log 2x 3x 1> < + + + + Xt 4 trng hp sau : a) Trng hp 1 : -1 < x < 0 . Khi : 2 2332x 3x 1 1 log 2x 3x 1 0x 1 1log (x 1) 0 + > + > + + > Suy ra BPT(3) nghim ng vi mi x : 10 x2< < . c) Trng hp 3: 31 x2< < . Khi : 2 2332x 3x 1 x(2x 3) 1 1 log 2x 3x 1 0x 1 1log (x 1) 0 + = + < + < + >+ > Suy ra BPT(3) nghim ng vi mi x : 31 x2< < . d) Trng hp 4: 3x2> . Khi : 2 2332x 3x 1 1 log 2x 3x 1 0x 1 1log (x 1) 0 + > + > + >+ > BPT(3) 2 2 2 23 3log 2x 3x 1 log (x 1) 2x 3x 1 x 1 2x 3x 1 x 2x 1 + > + + > + + > + + x(x-5)> 0 x > 5. Suy ta BPT c tp nghim : 1 3T (0 ; ) (1 ; ) (5 ; ).2 2= + Cch 2iu kin : 20 2x 3x 1 1 1 3 3x ( 1 ; 0) (0 ; ) (1 ; ) ( ; ) (2)2 2 2 0 x 1 1 < + =e + < + = Xt hm s2 211331 1ln( ) ln( )1 13 3f (x)log (x 1) ln(x 1)log 2x 3x 1 ln 2x 3x 1= = + + + + Vi k (2) th f(x) = 0 2 2 2 2ln 2x 3x 1 ln(x 1) 2x 3x 1 (x 1) 2x 3x 1 x 2x 1 + = + + = + + = + + x2 5x = 0 x = 0 V x = 5 . So vi k (2) th f(x) = 0 x = 5. www.VNMATH.com 17* Hm s f(x) lin tc v v nghim trn cc khong : (-1 ; 0); (0 ; 12);(1 ; 32); (32; 5);(5 ;+ ) nn trntng na khong ny f(x) khng i du.Ta c f(12) = -3,584 < 0 ;f(14) = 7,163 > 0 ; 5f ( )4 =3,594 > 0 ; f(2) = -1 < 0 ;f(6) = 0,016 > 0 +_+5+3212_00xf (x)-11+ Qua bng xt du ca f(x), ta suy raBPT f(x) > 0 c tp nghim 1 3T (0 ; ) (1 ; ) (5 ; ).2 2= + Gii bt phng trnh : 26sincos23 log 2005 03xx| |+ > |\ . Nhn xt: Nu dng my tnh ta s c : 6log 2005 4, 243537... ~D thy :201sincos2 23 3 43 3xx| | | |+ s + = ||\ . \ . T y suy ra bt phng trnh cho v nghim. Nu khng dng my tnh th vic gii phng trnh ny s gp kh khn. BSUNGTHMMTSVNGIITONVISTRGIPCA MY TNH CASIO V VINACAL 1)Dng my tnh chng minh phng trnh bc 3 c 3 nghim phn bit. V d: Cho hm s : 2 46 4 6 y x x x = + + . Chng minh rng hm s c 3 cc tr. Ta c :3' 4 12 4 y x x = + . Ta ch cn chng minh PT y = 0 c 3 nghim phn bit. Dng my tnh ta bit c 3 nghim : 1 2 31,8 ; 0, 3 ; 1, 5 x x x ~ ~ ~ . Sau p dng nh l v hm s lin tc cho hm s g(x) = 34 12 4 x x +trn cc on: [-2 ; -1], [0 ; 1], [1 ; 2] ta c iu phi chng minh. 2) Dng my tnh dy bi nhn dng tam gic. Trong tit hc v nhn dng tam gic cho hc sinh bi ton : Tm gi tr ln nht ca biu thc www.VNMATH.com 18T = cosA + cosB + cosC vi A, B, C l cc gc ca mt tam gic. Yu cu mi em hy tnh gi tr ca T ng vi mt tam gic c th v vit cc kt qu ln bng, t rt ra kt lun :T 32ssau dng l lun chng minh pht hin ny. * Mt th d khc xt bi ton : Nhn dng tam gic ABC bit: Gio vin c th yu cu hc sinh thay vo ng thc trn cc gi tr ca A, B, C c th t rt ra kt lun : 3 3sin sin sin2A B C + + s . Sau dng l lun chng minh bt ng thc ny v ch ra ng thc xy ra khi v ch khi tam gic ABC u. 3)Dng my tnh gii cc dng ton v phng trnh c nghim duy nht. V d: Gii phng trnh: 2 81 2 4 3 2 1 141xx x xx += + + + + Gii iu kin:1 x >Xt cc hm s : 2 8( ) ,1[1; )xfx xx += e++( ) 1 2 4 3 2 1 14, [1; ) g x x x x x = + + + e +Dng o hm, d dng chng minh c trn min [1; ) + hm s f(x) nghch bin v hm s g(x) ng bin. Do phng trnh cho c nhiu nht mt nghim. Dng chc nngSOLVEta tm c nghim x = 5. Vy phng trnh cho c nghim duy nht x = 5. Nhn xt: NukhngrnluynchohcsinhsdngmytnhCasio,Vinacalmtcch thnh tho th th cc em s gp kh khn khi tm ra nghim x = 5. Trong bi ton trn nu ta thay x bi 3x 55 th ta s c phng trnh:

Bng cch l lun nh trn, ta cng chng minh c phng trnh c nhiu nht mt nghim. Sau ta phi nhm mt nghim kt thc bi ton, nu khng s dng my tnh thnh tho th kh m tm c nghim ny!( PT ny c nghim duy nht x = 20 4)Dng my tnh Casio, Vinacal kim tra li kt qu gii hn ca hm s. My tnh Casio v Vinacal khng c chc nng tnh gii hn ca hm s, tuy nhin ta c th d on kt qu ca gii hn qua tng sau: Gi s cn tnh( ) limx a fx, ta dng chc nngCALC tnh gi tr ca hm s f(x) ti cc gi tr ca x rt gn a. Sau y l cc v d minh ha: www.VNMATH.com 19V d 1: Tnh 3227 253 2limxx xx x+ + + Bng phng php gi s hng vng, ta vit: 3 32 22 27 25 ( 7 3) ( 25 3)3 2 3 2lim limx xx x x xx x x x + + + + = + + 32 22 2( 7 3) ( 25 3)3 2 3 2lim limx xx xx x x x + + = + + T y ta d dng tm c gi tr ca gii hn cho l 754. Sau ta dng my tnh kim tra li kt qu, bng cch tnh gi tr ca hm s 327 25( )3 2x xfxx x+ += + ti gi tr ca x rt gn vi s 2, chng hn tnh: f(1,99999999), ta s c kt qu l 0.1300000, y cng chnh l gi tr gn ng ca 754. 5)Dng my tnh Casio, Vinacal kim tra li kt qu ca tch phn. Gistnhtchphn:( )bafx dx},tacktqulm.GVnnhngdnHS dng my tnh kim tra li kt qu sau: -Nhp vo my tnh :( )bafx dx m }, nu my tnh hin th kt qu rt gn vi s 0 th gi tr ca tch phn l m. iu ny gip cho hc sinh t tin hn khi lm bi thi i hc. Nhnthynubitkthpvicdyvhcmntonvistrgipcamy tnh b ti mt cch linh hot th hiu qu thu c s rt tt. Chng ti thc nghim phng php trn cc lp12 n thi i hc, Vi cc bi tp v gii bt phng trnh , ccdngtonvphngtrnhlnggiccnghimcbittngtnhccvd nu trn, nu dng phng php truyn thng th khng n 30% hc sinh cho li gii ng, nhng nu ng phng php trong cc chuyn trn th a s cc em gii c ddng.Quavicngdngphngphpnycngiphcsinhvndngkinthc gii tch soi sng mt dng ton i s, rn k nng s dng thnh tho my tnh b ti, y cng l mt yu cu c B gio dc ra. Cn rt nhiu dng ton m nu gii bng phng php bnh thng s gp nhiu kh khn. Bit khai thc nhng th mnh m my tnh em li s gip cho hc sinh d dng nh hng v lm cho cng vic hc ton bt nng n hn. Cng vi cc phng phptruynthngvgiibtphngtrnh,phngtrnhlnggicmhcsinh bit, hy vng vic b sung thm cc chuyn ny s lm phong ph thm kinh nghim giitonchohcsinh,gpphnhnhthnhlngamm,yuthchbmn,t hng cc em vo vic nghin cu tm ra nhng ng dng mi, khng hi lng vi www.VNMATH.com 20nhngkinthcbitmlunlunctinhthntmtisngtotmnhtmra kin thc mi. Qua thc t ging dy chng ti thy rng vn no m gio vin quan tm v truyn th cho hc sinh bng lng say m v nhit tnh ca mnh th s cun ht cc em vo con ng nghin cu. a my tnh cm tay vo ging dy trong chng trnhphthngkhngphilvnmi,nhngthctchothycnnhiuThy C cha quan tm ng mc v vn ny. Vi SKKN ny hy vng gp phn thc hinttchocaBGDlamytnhvothctgingdyphthngvbi dng hng nm c nhiu hc sinh t gii cao trong cc k thi hc sinh gii v gii ton trn my tnh Casio, Vinacal cp tnh v khu vc. Ti vit SKKN ny nhm mc ch chia s vi ng nghip v cc em hc sinh nhng kinh nghim m bn thn tch ly c trong qu trnh ging dy. Cc chuyn c trnh by trong SKKN ny th hin cc tng mi, mong mun khai thc vsdngmytnhcmtaymtcchththiuqutrongcngvicgingdyv hc tp b mn ton. Nhng vn c trnh by trong SKKN ny l nhng gi , hy vng rng qu ng nghip s tip tc nghin cu a ra ngy cng nhiu cc ththutngdngmytnhcmtaysaochoththiuqu.Nulmttcngvic ny s gip cho vic hc ton ca hc sinh c nh nhng hn v gip cho cc em t kt qu tt trong cc k thi tt nghip v i hc. SKKN ny c th trin khai ng dng nh chuyn bi dng cho cc hc sinh lp 10, 11, 12, cc hc sinh n tp thi tt nghip v i hc. Trong iu kin hin naymihcsinhucmytnhcmtaynnvicrnluynchohcsinhctduy gii ton vi s tr gip ca my tnh l mt vic lm kh thi. t c hiu qu cao trong cng vic th gio vin cn phi c tinh thn nghin cu v sng to, c nh vy gio vin mi pht hin ra cc vn mi trong ng dng v y chnh l yu t quan trng thu ht s quan tm ca hc sinh. QuaSKKNnytimunchiasviccbnngnghipmtskinhnghim mtitchlyctrongqutrnhgingdymntonvbidngchoi tuyn Quc gia v gii ton trn my tnh Casio, Vinacal. Hy vng qu Thy C s lng ghp ni dung v k thut gii ton vi s tr gip ca my tnh cm tay vo bi ging ca mnh. Chng ti mong nhn c s trao i, gp cho chuyn t cc anh ch ng nghipvccemhcsinh.HyvngSKKNnysgpphnnngcaochtlng dy v hc ton trng THPT. www.VNMATH.com 21 Thnh ph Bn Tre, ngy 20 thng 3 nm 2011. Ngi vit Nguyn Vn Qu www.VNMATH.com 22 Trang Phn m u 2 Phn ni dung 4 CHUYN 1: NG DNG MY TNH CASIO V VINACAL D ON NGHIM GII PHNG TRNH LNG GIC 4 CHUYN 2: NG DNG MY TNH CASIO V VINACAL GII BT PHNG TRNH 11 CHUYN 3: B SUNG THM MT S VN GII TON VI S TR GIP CA MY TNH CASIO V VINACAL 17 Phn kt lun 20 1) Cc ti liu hng dn s dng my tnh Casio ca BGD 2) Sch gio khoa 3) Tp ch Ton hc v tui tr

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