CC TH THUT CASIO
(Tc gi : Bi Th Vit, 11 Ton 2, THPT Chuyn Thi Bnh, Thi Bnh)
Th thut 1: Khai trin a thc h s nguyn hoc h s l phn s nh(Ci ny p
dng rt nhiu trong vic gii ton)
a) H s nguynNi dung: Ta nn nh mt iu nh sau:Gi s khi khai trin a
thc th a thc c dng:anxn+an1xn1+...+a1x+a0Tix=10th a thc c gi tr
lanan1...a1a0Tix=100th a thc c gi tr lan0an10...0a10a0
Tix=1000th a thc c gi tr lan00an100...00a100a0...Chc bn s kh hiu
v ci ny ! Nhng hy n phm trn CASIO v lm theo cc bc sau l bn s hiu
ngay:Bc 1: Nhp a thc :9X3+2X2+7X+1Bc 2: n CALC, my hiX?Bc 3: Nhp10v
n nt=. Bn s thy kt qu l9271. n tip "=", my hiX?Bc 4: Nhp100v n nt=.
Bn s thy kt qu l9020701. n tip "=", my hiX?Bc 5: Nhp1000v n nt=. Bn
s thy kt qu l9002007001Vy chc bn hiu, nu khng hiu Comment bn di
Nhng nu nhng h s l s nguyn m th sao ? Li tm hiu tip nh !
Bc 1: Nhp a thc :9X32X27X+1Bc 2: n CALC, my hiX?Bc 3: Nhp10v n
nt=. Bn s thy kt qu l8731. n tip "=", my hiX?Bc 4: Nhp100v n nt=.
Bn s thy kt qu l8979301. n tip "=", my hiX?Bc 5: Nhp1000v n nt=. Bn
s thy kt qu l8997993001. n tip "=", my hiX?Bc 6: Nhp10000v n nt=.
Bn s thy kt qu l8999799930001
Nhn xt: Nu s bn nhp l10x(tc l s100...0vixs0), hy chia kt qu thnh
cc khongxch s t phi sang tri. VD:8997993001th
l8|997|993|001hoc8999799930001th l8|9997|9993|0001Gi gi tr khong
tht(tn) lktth ta c:+ Nukc nhiu s9th h sat=10xkt+ Nukc nhiu s0th h
sat=kt
P/s: Mnh ni hi kh hiu v lng vng, tt nht l nn c lun cch lm bn
di:
Cch lm:Cch 1: (Ch p dng cho cc bi c h s3).VD cn khai
trin2(x+1)2(x1)7(x2+1)8
Bc 1: Nhp a thc nXvi cc h s nguyn v khng qu cng
knh.(VD2(X+1)2(X1)7(X2+1)8)Bc 2: n CALC, my hiX?, n1000v n=Bc 3: My
hin ra kt qu l mt s c nhiu ch s, tch ra tng 3 ch s mt t phi sang
tri(VD: My hin1994997983th ta tch1|994|997|983)Bc 4: Ta ln lt tm h
sa0,a1,...bng cch sau:Nhm 3 ch s thk(tnh t phi sang tri) c gi tr
lMk, ch s hng trm caMkl s9th chng t h s caxks lMk1000(s m), v gi tr
ca nhm thk+1s c gi tr lMk+1+1(tng thm 1)Nhm 3 ch s thk(tnh t phi
sang tri) c gi tr lMk, ch s hng trm caMkl s0th chng t h s caxks
lMk(s dng)(VD: Nhm 1:|983|th h sa0l17v thm 1 vo nhm 2Nhm 2:|997|th
h sa1l3+1=2v thm 1 vo nhm 3Nhm 3:|994|th h sa2l6+1=5v thm 1 vo nhm
4Nhm 4:|001|th h sa3l1+1=2)Bc 5: in kt
qu:2(x+1)2(x1)7(x2+1)8=2x35x22x17Bc 6: Th li cho chc n
!(n2(x+1)2(x1)7(x2+1)8(2x35x22x17), gn gi trx=1,2,3,4,...m thy kt
qu lun =0 th chc l chnh xc)
Nhn xt: Cch ny khng hay lm, nu lm quen th chc nhn h s cc nhm l s
bit c ngy kt qu trin.
Cch 2: p dng cho bc cao, h s nguyn (Bc cng ng cao qu, h h)VD cn
khai trin2(x+1)3(x1)27(x2+1)28Bc 1: Gn gi trx=1000hoc10000nu
thch.(Tix=1000th kt qu l1,994995982x1015)Bc 2: Nhn vo gi tr sau du
phy, xem xt s bn cnh n ! Nu s bn cnh l9th h s bc cao nht l h s sau
du phy cng 1, nu l s0th d nguyn.(Sau du phy l s1, cnh n l s9, suy
ra h s bc cao nht (bc 5) l2)Bc 3: Vit li a thc, sau tr i bc cao nht
va tm.(2(x+1)3(x1)27(x2+1)282x5)Bc 4: Chox=1000th kt qu l bc a thc
s gim, tip tc lm nh bc 2(Tix=1000th gi tr nhn c l5,004017998x1012.
Do bc h t15xung12nn a thc c h s bc 4 khc 0.Sau du phy l s5, cnh n l
s0nn h s bc 4 l5n tip2(x+1)3(x1)27(x2+1)282x5+5x4Gnx=100th kt qu
l4179813, tch thnh4|17|98|13|ta c h s bc3l4, h s bc2l18, h s bc nht
l2, h s t do l13)Bc 5: Ghi kt
qu:2(x+1)3(x1)27(x2+1)28=2x55x44x318x2+2x13Bc 6: Th li
Nhn xt: Lm nhiu mi quen, ch ci ny kh ni lm. Cng hay ch nh ?
b) H s l phn s:Bc 1: Tm c chung ln nht cc mu m ta d on chng s gp
mt trong h s sau khi phn tchBc 2: Vit a thc, c c phn s, tt c a thc
nhn vi c chung ln nht va tm cBc 3: Lm nh phn a)
P/s: Ai khng hiu c comment
Th thut 2: Phn tch phng trnh bc 4 thnh nhn t (Ci ny mnh post
li)
i vi phng trnh bc 4 dngf(x)=ax4+bx3+cx2+dx+eta chia lm 2 mng
ln:*** u tin l phng trnhf(x)c nghim, ta xt:- Nu trong trng hp bn
phi i thi, kim tra th bn nn s dng my tnh CASIOfxm gii nh, sau y l
hng dn gii phng trnh bc 4 bng Casio :+Trng hp 1: Bn ly my tnh, vit
phng trnh bc 4 ca bn vo, n Shift + Solve v sau n "=" gii phng trnh
bc 4 :@@1: Nu my tnh hin raX=mt s nguyn c th no hoc l s v hn c tun
hon (VD:1,3333333...)
th bn n AC, sau n RCL + X th my s hin ln chnh xc nghim ca bn (s
nguyn hoc phn s ti gin).
Khi f(x)c mt nhn t l(xX)(vi X l nghim bn va tnh c).
Sau bn s phn tch thnh(xX)(mx3+nx2+px+q).
Khi dng my tnh gii nghim phng trnh bc 3 nh bng cch vo Mode Mode
Mode 1 ri ln lt ghi h s ca n vo nh.
T bn nhn c tt c cc nghim caf(x)gm X v 3 ngim ca phng trnh bc 3 .
. .
@@2: Nu my tnh hin raX=mt s v hn khng tun hon, bn chuyn sang
Trng hp 2(Ci ny mi kh)
+Trng hp 2:( Ci ny l cng thc b mt y):
Khi tm c 1 nghim ca phng trnh bc 4 , bn chuyn d liu sang A bng
cch n Alpha X Shift Sto A
Sau bn vit li phng trnh bc 4 , n Shift + Solve, my hin tipX?bn
nhp 100 vo, n "=", n "=" gii.
Khi my s tnh mt nghim na khc vi nghim ban u.
Bn chuyn d liu nghim va tm c sang B bng cch n Alpha X Shift Sto
B.
Sau bn vit li phng trnh bc 4 , n Shift + Solve, my hin tipX?bn
nhp -100 vo, n "=", n "=" gii.
Khi my s tnh mt nghim na khc vi nghim ban u.
Bn chuyn d liu nghim va tm c sang C bng cch n Alpha X Shift Sto
C (Th l ).
Ci ny l xong n: n Alpha A + Alpha B ri "=", nu kt qu l s nguyn
hoc phn s th bn n tip Alpha A Alpha B ri "=" tnh c tch ca 2 s .
Khi y p dng nh l Vit o ta cf(x)c mt nhn t lx2(A+B)x+AB(Hay
cha).
Cn nu A+B khng l s nguyn hoc s v hn c tun hon (Tc l phn s y) th
Bn lm tng t vi tng B+C, C+A t tm c nhn t caf(x)
Ni khng bng lm, bn hy lm theo v d sau, chc bn s hiu:
x4+3x34x211x+5=0Ta n phm trn my tnh CASIO nh sau:Vit
PTx4+3x34x211x+5=0trn my tnh CASIO fx-570MS hoc fx-570ES.n shift +
SOLVEMy hi X?n 10 = (Nu l my fx-570ES th khng cn lm tip, i vi my
fx-570MS th n tip Shift SOLVE)
Sau mt hi, my hin X=1,791287847
n AC,
n Alpha X Shift STO
A_______________________________________________________________Vit
li phng trnh :x4+3x34x211x+5=0
n shift + SOLVE
My hi X?
n -10 = (Nu l my fx-570ES th khng cn lm tip, i vi my fx-570MS th
n tip Shift SOLVE)
Sau mt hi, my hin X= - 2,791287847
n AC,
n Alpha X Shift STO
B______________________________________________________
Vit li phng trnh :x4+3x34x211x+5=0
n shift + SOLVE
My hi X?
n -1 = (Nu l my fx-570ES th khng cn lm tip, i vi my fx-570MS th
n tip Shift SOLVE)
Sau mt hi, my hin X= 0,4142135624
n AC,
n Alpha X Shift STO
C________________________________________________________________Nhn
xt:n Alpha B + Alpha C =
My hin : -2,377074285
n Alpha C + Alpha A =
My hin : 2,20550141
n Alpha A + Alpha B =
My hin : -1_____________________________
Chng t trong cc tng A+B, B+C, C+A th ch thy A+B nguyn (hoc l mt
s v hn tun hon)
p tip Alpha A x Alpha B =
My hin : -5
Chng t A, B l nghim ca phng trnh bc 2 n x :x2(A+B)x+AB=0
M A+B= -1, A.B= -5
Suy ra A, B l nghim ca phng trnhx2+x5=0
M A, B cng l nghim ca phng trnh:x4+3x34x211x+5=0
Suy rax4+3x34x211x+5khi phn tch nhn t c mt nhn t lx2+x5
Suy rax4+3x34x211x+5=(x2+x5)(ax2+bx+c)
T ta phn tch thnh nhn t c
Bi tp p
dng:x4+3x34x211x+5=0x4+12x3+21x224x+5=0x46x3132x2+885x+500=010x4+27x316x245x+28=010x4+27x3+245x2+306x+1288=0x4+9x3+20x2+9x+1=0
Th thut 3: Phn tch a thc bc cao thnh nhn t (Tng qut ca th thut
2)
Nhn xt: i khi ta thy nhng bi phng trnh v t m ch cn nhn l thy bnh
phng ln ra phng trnh bc cao cho n lnh ( = Bc ng cng - Nguyn Cng
Hoan) nhng chnh vic khai trin n, phn tch thnh nhn t khin chng ta
nn. Nhng phng php sau y s gip ch phn no iu .
Ni dung: Trc tin, cn xc nh bc ca a thc, khi phn tch thnh nhn t
ta s kim tra xem c thiu nhn t no khng !
VD:(x2+1)2(x2+5x+4)21x336x27x+2c bc l6Sau , xc nh khong cha nghim
ca phng trnh, ging nh phng trnh bc 4
Cch lm:Cch 1: p dng cho nhng bi m nhn t ca n l a thc bc < 3Bc
1: Nhp a thc:(x2+1)2(x2+5x+4)21x336x27x+2Bc 2: Gii nghim phng trnh,
choXl im gia khong
nghimVD:0.414213562,2.414213562,1.618033988,0.6180339880Bc 3: C tm
xem cc nghim y l nghim ca phng trnh bc 2 hay bc 3 no
?VD:x2x1=0vx2+2x1=0Bc 4: Vit lun ra v rng PT tng ng
vi(x2x1)(x2+2x1)(...)vi ... l mt tam thc bc 2 c dngax2+bx+c. Quan
trng by gi l tma,b,cBc 5: V h s bc cao nht phng trnh bc 6 l 1
nna=1, h s t do bng6nnc=6Bc 6: Vit ra my tnh nh
sau:(x2+1)2(x2+5x+4)21x336x27x+2(x2x1)(x2+2x1)(x2+Ax+6),ABc 7: n
Shift + Solve gii phng trnh trn theoA. u tin choX=1,2,3,...m khi
gii, ta lun cA=4, do b=4Bc 8: Vit tip(x2+4x+6)Bc 9: Th li
Nhn xt: Cch ny hi hn ch
Cch 2: (Mt s bi ton khi bnh phng gii phng trnh bc cao, li ra mt
tam thc bc 2 nhn vi mt a thc bc 4 hoc bc 3, cch ny vn gn ging cch 1
nhng n gip chng ta tm c nhn t phng trnh cn li. Cch ny p dng th thut
1.)
VD: Gii phng trnh(x2+1)2(x2+5x+4)21x326x217x8=0
Bc 1: Tm cc nghim phng trnh, thy phng trnh c ng 2 nghim v t ta c
nhn t(x2x1)(Nh cch 1)Bc 2: Ta s tm nt nhn t bc 4 cn li, cch lm nh
sau:Vit ln my tnh:(x2+1)2(x2+5x+4)21x326x217x8x2x1Bc 3: Chox=1000th
ta c kt qu l1,006013008x1012Chng t h s bc 4 l1Bc 4: Vit
tip(x2+1)2(x2+5x+4)21x326x217x8x2x1x4Chox=1000ta c6013008004nn ta c
phng trnh bc 4 l:x4+6x3+13x2+8x+4Bc 5: Vit
:(x2x1)(x4+6x3+13x2+8x+4)=0Bc 6: Chng minh phng trnh bc 4 kia v
nghim (Xem th thut 4)Bc 7: Kt lun (Ci ny nhiu ngi thiu)
Nhn xt: Th thut ny lm mt i tr c, t duy con ngi nn khng khuyn co
dng cch ny...
Th thut 4: Chng minh phng trnh bc 4 v nghim: (Post li bi mnh
post)
Thm mt phng php "t" ca mnh, l cch chng minh phng trnh bc 4 v
nghim ! (Ai khng hiu g c pmmmm nha, nhng cng hi au u
y)_________________________Xt PTf(x)=x4+ax3+bx2+cx+dvid>0va,b,cl
cc h s.Khi bn gii mi ci ny m khng ra nghim (Can't solve), bn hy
chng minh phng trnh v nghim
V d 1:Gii phng trnh:x46x3+16x222x+16=0
Cch 1:Cch n may: chnh lf(x)phn tch thnh 2 ci bc 2 cng vi mt h s
t do khng m,ging nhf(x)=x46x3+16x222x+16
Khi f(x)=(x22x+3)(x24x+5)+1>0
[?] Vy ti sao li c th phn tch thnh ci ny, l cu hi kh ?
Cch lm y l tf(x)=(x2+ax+b)(x2+cx+d)+eSuy
raf(x)=x4+(a+c)x3+(d+ac+b)x2+(bc+ad)x+bd+eng nht vi a thc ban u
lf(x)=x46x3+16x222x+16Ta c:a+c=4d+ac+b=16bc+ad=22bd+e=16
T d dng suy raa=2,b=3,c=4,d=5,e=1nh phng php m(V y l cch n may
m)
Cch 2:(Cch ny o nht, by gi tui mi pht hin ra)
Cng t:A=f(x)=x46x3+16x222x+16
Ta s chng minhf(x)>0bng cch tx=ya4, mt i h s cay3
tx=y+32
Biu thc cho tr
thnh:A=y4+5y22y+6116=y42my2+m2+(2m+52)y2y+6116m2
(Ch ny kh o, nhng hay)
Cn tmm>52 PT(2m+52)y2y+6116m2v nghim (khi n mi >0)
Th=52
C nhiumtha mn lm, VD:m=0hocm=1hocm=1l p mt nht
Chn mt ci v lm !
Gi s:
a)m=1thA=(y2+1)2+32(y13)2+17548Suy
raA=(x23x+134)2+32(x116)2+17548>0
b)m=0thA=y4+52(y15)2+29780Suy
raA=(x32)4+52(x1710)2+29780>0
c)m=1thA=(y21)2+72(y17)2+419112Suy
raA=(x23x+54)2+72(x2314)2+419112
_______________________
Nhn xt:Nhng cc bn cng khng nn li dng n qu, ging nh minhtuyb nhn
xt:
"Mnh cng chia s cht ch ny :Khi raA=y4+5y22y+6116th trc khi chn h
smthch hp nh trn nn kim tra xem tam thc bc hai5y22y+6116c v nghim
hay khng:+) Nu v nghim(
