1

2

HNG DN S DNG

V GII TON TRN MY TNH CASIO

FX-570VN PLUS

3

4

Mc Lc Trang

Phn I : Hng dn s dng Danh mc my tnh c php mang vo phng thi 2

Thng tin quan trng 6

Cn thn khi s dng my tnh 7

Trc khi tnh ton 8

Tnh ton c bn 10

Tnh ton s thp phn tun hon 15

Php ton c nh 17

Php chia c d 19

Ly tha s nguyn t 19

Php tnh vi cc hm 20

Tnh c Chung Ln Nht V Bi Chung Nh Nht 25

Tnh phn nguyn Int v Intg 25

Gii phng trnh H phng trnh 25

Tnh ton bt phng trnh 30

T s RATIO 32

Tnh ton phn phi Dist 32

Thng k Hi qui 34

Thng tin k thut 40

Th t u tin ca php ton 41

Phm vi nhp 42

Mt s ch nng khc 44

Ton s phc 45

5

H m c s N 47

o hm 50

Tch phn 50

Ma trn 51

Ton Vect 54

Bng s t hai hm 57

Hng s khoa hc 59

Chuyn i o 60

Phn II: Gii ton

A. Gii ton t lp 6 n lp 12

Mt s kin thc cc lp 6-7-8-9 61

Ton lp 10 64

Ton lp 11 80

Ton lp 12 102

B. Gii mt s bi thi tt nghip THPT 130 C. Gii mt s bi ton l ha sinh 150

6

Thng Tin Quan Trng

Cc hin th minh ha nh (cc nhn phm) c nu trong bng hng dn s

dng ny ch c dng trong mc ch minh ha, c th khc vi cc mc thc

t khi biu din

Khi u my tnh tay

Thc hin th tc sau khi bn mun khi u my tnh tay v tr v trng thi

ban u. Lu rng thao tc ny cng xa i tt c cc d liu hin thi trong b

nh my tnh tay

(All) (yes)

Thn Trng V An Ton

Pin ngoi tm tay tr nh

Ch dng kiu pin chuyn dng cho my tnh ny c nu trong ti liu

ny

B V Cng

Trc khi dng my tnh hy trt v cng xung b n ra ri gn v cng

vo ng sau my tnh.

Bt v tt ngun

n bt my tnh

Nhn (off) tt my tnh

T ng tt ngun

My tnh ca bn s t ng tt ngun nu bn khng thc hin thao tc trong 10

pht. Nu diu ny xy ra bn n bt my tnh bn tr li

iu chnh tng phn hin th

Hin th mn hnh CONTRAST bng vic thc hin thao tc sau:

(SETUP) ( CONTRAST ). Tip dng v iu

chnh tng phn. Sau khi thit lp ng nh bn chn th bn n

iu quan trng : nu iu chnh tng phn m vn khng c c, iu

c th l ngun pin b yu ri. Hy thay pin.

Nhn phm

7

Nhn phm hay tip theo sau l phm th hai s thc hin chc nng

thay phin ca phm th hai. Chc nng thay phin c ch ra bi ch c in

trn phm ny .

Bng sau ch ra ngha cc mu khc nhau ca ch trn phm chc nng thay

phin

Nu ch nhn ca phm c mu: Ngha l:

Vng Nhn ri n phm ny nhp vo hm p dng c

Nhn ri n phm ny a vo bin, hng hay k hiu p dng c( tr sau phm v )

Mu tm Vo chng trnh CMPLX nhp chc nng ny

Mu xanh lc Vo chng trnh BASE-N nhp chc nng ny

Cn Thn Khi S Dng My Tnh

Lun n phm khi s dng my

Thm ch khi my hot ng bnh thng, hy nn thay pin t nht 3 nm 1 ln

Pin cht c th r r gy h my v tnh ton sai. Khng c pin ht nng

lng trong my

Pin km theo my c th b gim nng lng trong qu trnh vn chuyn v lu

kho. V th nn thay pin sm hn tui th ca pin.

Pin yu c th lm cho ni dung b nh b h hng hoc hon ton b mt i. Hy

lun gi s liu quan trng bng vn bn.

Trnh s dng v my trong mi trng nhit cao.

Nhit qu thp c th gy nn chm hin th hay hon ton khng hin th v

lm gim tui th ca pin. Trnh my tip xc trc tip vi nh sng mt tri,

gn ca s, l si hay bt c ni no c nhit cao. nng c th gy bin

mu, bin dng v my h hng cc mch in t bn trong

Trnh s dng v ct my ni c m cao v bi bm. Cn thn khng c

my b nc ri vo hay t ni c m cao v bi bm. Nhng iu kin

nh vy c th gy h hng mch bn trong

Khng c lm my ri hay tc ng mnh vo my

Khng c b vn cong my. Trnh b my vo ti qun hay nhng ni cht

chi ca qun o v n c th vn lm cong my.

Khng tho my ra

8

Khng c n phm bng u bt bi hay vt nhn.

Dng vi mm, kh lau sch bn ngoi my. Nu my b d, lau sch bng vi

hi m vi mt t bt git trung tnh. Vt tht ro khi lau sch. Khng s dng

cht pha sn, benzene hay cc tr cht d bay hi lm sch my. Nu lm nh

vy s b mt i lp in v c th lm hng v my

Trc Khi Tnh Ton

Mode Trc khi tnh ton, bn phi chn ng Mode theo bng ch dn di y:

Php tnh Hy thc hin thao tc phm

Tnh ton chung (COMP)

Ton s phc (CMPLX)

Tnh ton thng k v hi quy (STAT)

H m c s N (BASE-N)

Gii phng trnh (EQN)

Ton ma trn (MATRIX)

Bng s (TABLE)

Ton Vct (VECTOR)

Gii bt phng trnh (INEQ)

Tnh t s (RATIO) (RATIO)

Tnh phn phi (DIST)

n mode ta c mn hnh ci t cho my, theo hng dn trn mn hnh ta la

chn ci dt hay vo chc nng thch hp

Trong hng dn ny tn ca mode cn vo thc hin php tnh c ghi

bng tiu chnh ca mi phn.

V d: gii phng trnh (EQN)

Sa li li nhp

Khi ta mun sa li ta dng phm duy chuyn con tr n ch cn

chnh.

Mun xa s m ta cn xa th dng phm duy chuyn con tr n pha

sau s m ta cn xa ri ta n phm

Khi mun chn thm mt s hay mt php tnh th ta dng phm duy

chuyn con tr n ch cn chn ri ta thm s hay php tnh vo

9

Hin th li biu thc v kt qu va tnh

Sau khi mi ln tnh ton, my lu biu thc v kt qu tnh ton vo b nh. Ta

n phm hin th li mn hnh trc (biu thc v kt qu va tnh), ta n

th mn hnh trc na hin li.

Khi mn hnh c hin li ta dng phm hoc chnh sa php tnh hoc

tnh li( k c mn hnh ang tnh).

n con tr hin th dng u ca biu thc. Nu bn mun chnh sa th

dng phm duy chuyn con tr chnh sa

n phm mn hnh my tnh s khng b xa b nh

Lu : khi con tr dng ng th ta ang ch d ghi chn. Khi con tr nm

ngang (n Shift INS LineIO) ta c ch ghi .

Lu : b nh mn hnh b xa khi

Ta n

Lp li mode v ci t ban u ( (yes))

i mode

Tt my

nh v tr sai

n hay sau khi c thng bo li, con tr nhp nhy lin sau k t li.

Kt ni nhiu biu thc

Dng du: ( ) kt ni 2 hay nhiu biu thc li vi nhau.

V d: Tnh 3 + 2 v ly kt qu nhn 4

Dng a

Mn hnh ch hin ra 10 ch s v hai ch s m . Gi tr ln hn c hin ra

dng a . Vi s thp phn ta chn mt trong 2 dng ca a .

thay i dng hin ta n v n tip mt s tng ng vi s la chn

ca ta.

10

Fix Sci Norm

6 7 8

n v n tip (Norm 1) hoc ( Norm 2)

Norm 1: a vo dng a nhng s x c:

Norm 2: a vo dng a nhng s x c:

Ch : Tt c cc v d trong ti liu ny iu l Norm 1

Tr v trng thi ban u (ALL) (Yes)

u Quan Trng

Nu bn thc hin mt tnh ton bao gm c php chia, php nhn trong c

php nhn b b qua, th du ngoc s t ng chn thm nh cc v d bn di:

Du nhn s b b qua ngay trc mt du ngoc m hoc sau mt du

ngoc ng

6 2(2 1) my s hin th 6 (2(2 1) 6 (1 2)A my s hin th 6 ( (1 2)A

Du nhn s b b qua ngay trc mt bin s hoc hng s

6 2 my s hin th 6 (2 ) 2 2 2 my s hin th 2 (2 2)

yTnh ton c bn Thc hin Mode COMP

Php tnh thng thng

Vo COMP Mode n (COMP)

S m trong php tnh phi t trong du ngoc sin -1,23 1,23

Nu s m l s m th khi t trong ngoc

V d1: Tnh

n my ( )

11

V d 2: Tnh

n my

C th b qua du trc

Ton v phn s

Phn s

Php tnh phn s

V d 1:

Cch n my tnh

V d 2 : Cng hai hn s

Cch n my

Mun i v dng hn s ta n ta c kt qu

V d 3: Cch n gin phn s

Cch n my

V d 4: Cng phn s v s thp phn

Cch n my

Mun i v dng s thp phn ta n

12

Cch i phn s phn s thp phn

V d1: Cch i phn s thnh s thp phn

Cch n my

ta c kt qu 1,5

V d 2: i s thp phn thnh phn s

0,6 =

Cch n my

ta c kt qu

i hn s phn s

V d 1: i hn s thnh phn s

1 =

Cch n my

ta c kt qu

V d 2 i phn s c t ln hn mu thnh hn s

Cch n my

Tnh phn trm

V d 1: Tnh 12% ca 2000 ( c kt qu l 240)

n my 2000 12 (%)

V d 2: Tnh 600 l bao nhiu phn trm ca 800 ( ta c kt qu l 75%)

n my 600 800 (%)

13

V d 3: Tnh phn trm 2500 + 15% ca 2500 (ta c kt qu l 2875)

n my 2500 2500 15 (%)

V d 4: Tnh phn trm 3500 - 25% ca 3500 (ta c kt qu l2625)

n my 3500 3500 25 (%)

Php tnh v , pht, giy(hay gi, pht giy)

Ta thc hin php tnh trn (hay gi), pht, giy.

Trc khi n my ta cho mn hnh my tnh v hin D bng cch n:

(Deg)

V d 1 Tnh tng , pht, pht, giy

n my 2 20 30 0 39 30

V d 2 i s thp phn ra , pht, giy

ta c kt qu

V d 3: Tnh

n my

Fix, Sci, Norm

Ta c th ci t mn hnh n nh ch s l, thp phn,nh s dng chun

(a bng cch sau:

n c mn hnh

:MthIO :LineIO

:Deg :Rad

:Gra :Fix

:Sci :Norm

Fix

14

n chn (Fix) n nh s ch s l t 0 n 9

Ty theo ta mun hin ln bao nhiu ch s l c tnh trn m ta chn

V d: 1 ta mun lm trn 4 ch s trong php chia

Cch n my ta s c kt qu lm trn 4 ch s

Nu ta ly kt qu

Cch n my ta c kt qu l 8

V d: 2 ta mun lm trn 6 ch s trong php chia

Cch n my ta s c kt qu lm trn 6 ch s

Nu ta ly kt qu

Cch n my ta c kt qu l 8

y tuy my lm trn s nhng b nh ca my vn lu 15 ch s

Nu dng phm

Ta vn dng cch lm trn 4 ch s trong php chia nhng khi

th nhn li th kt qu li khng chnh xc

Cch n my (Rnd)

SCi

n (Sci) n nh ch s ca a trong a (s nguyn ca a t 1 n 9). n

t 0 n 9 n nh ch s ca a

V d : tnh 5 chia 500 vi 4 ch s 5 =

Cch n my ta c kt qu

Norm

n v tip n (Norm1), hoc (Norm2)

Norm1: a vo dng a nhng s x c :

15

V d:

n my

Norm2: a vo dng a nhng s x c

Tt c cc v d trong ti liu ny iu l Norm1

Tnh ton s thp phn tun hon

My tnh ca bn ang dng s thp phn tun hon khi bn a vo gi tr. Kt

qu tnh ton cng c th c hin th bng dng thp phn tun hon bt k khi

no p dng c

a vo s thp phn tun hon

Khi a vo s thp phn tun hon , ta n ( )

a gi tr ca s thp phn tun hon 0,909090 0,(90) . Thc hin thao tc

nh sau : ( )

V d : Tnh tng ca hai s thp phn tun hon : 1,(021) 2,(312) cho kt qu:

10

3

Ch n my: ( ) ( )

Ta n chuyn kt qu v dng s thp phn tun hon: 3,(3)

Lu : Bn c th xc nh ti 14 v tr thp phn cho chu k thp phn tun

hon. Nu bn a vo hn 14 v tr thp phn gi tr ny s b x l nh s thp

phn kt thc v khng phi l s thp phn tun hon.

iu kin hin th s thp phn tun hon

Nu kt qu tnh ton tha mn cc iu kin sau, n

s hin th n nh gi tr

thp phn tun hon

Tng s ch s c dng trong phn s c hn s (k c s nguyn, t s,

mu s v k hiu phn tch) phi khng qu 10.

16

Kch c d liu ca gi tr c hin th nh s thp phn tun hon phi

khng ln hn 99bytes. Tng gi tr v du chm thp phn yu cu mt byte,

v tng ch s thp phn tun hon yu cu mt byte. Chng hn s sau y

yu cu 8byte(4 byte cho gi tr, 1 byte cho du chm thp phn 3byte cho thp

phn tun hon): 0,(123)

Lu : bit thng tin v vic chuyn dng thc hin th ca kt qu tnh

ton khi OFF c la cho thit t Rdec trn menu thit t

V d v s thp phn tun hon

V d 1: 26

0,(3) 0,(45) 0,(78)33

Cch n my :

V d 2:

411,(6) 2,(8) 4,(5)

9 Cch n my

Chuyn kt qu tnh ton

Khi hin th t nhin c la, mi ln n s chuyn kt qu tnh ton

c hin th sang dng phn s, thp phn dng cn thc v dng thp phn

ca n, hay dng thc v dng thp phn ca n

V d 1: 1

6 0,52359877566

( )

V d 2: ( 2 2) 3 6 2 3 5,913591358

Cch n my

Php ton c nh

Thc hin Mode COMP ( n )

Sau khi nhn th gi tr va nhp hay kt qu ca biu thc c t ng gn vo phm

Phm cng c gn vo kt qu ngay sau khi n , , hay v tip theo l mt ch ci

Gi kt qu l phm

Phm lu kt qu n 15 ch s chnh v 2 ch s m.

Phm khng c gn kt qu khi php tnh c li

17

Tnh lin tip

Kt qu sau khi n c th s dng trong php tnh k tip Kt qu ny c th s dng nh mt trong cc hm mu A ( 2 3 1, , , !x x x x ),

, ,^( ), , , , Pr, ,y xx n nCr v 0'"

S Nh c Lp M

Mt s c th nhp vo s nh M, thm vo s nh , bt ra t s nh . S c lp M tr thnh tng cui cng

S nh c lp c gn vo M

xa s nh c lp M ta n

V d 23+9=32 n my 23 9

53- 6 =47 53 6

-45 2= - 90 45 2

Tng -11 ta c kt qu -11

B Nh Tr Li (Ans)/ B Nh Tr Li Trc PreAns

Kt qu tnh ton cui cng thu c l thu trong b nh Ans (tr li). Kt qu tnh

ton thu c trc kt qu tnh ton cui cng c lu trong b nh PreAns(b nh

trc). S hin th kt qu ca tnh ton mi s chuyn ni dung b nh Ans hin ti

n b nh PreAns v lu kt qu tnh ton mi trong b nh Ans. B nh PreAns

ch s dng trong chng trnh COMP. Ni dung b nh PreAns s c xa bt c

khi no my nhp vo chng trnh khc t COMP

V d1 : chia kt qu 3 4cho 30.

Ta thc hin nh sau : 3 4 12 . Ri ta thc hin tip tc php chia cho 30. Mn hnh

hin ra 2

305

Ans

Cch n my:

18

V d 2: i vi 2 1k k kT T T ( dy s Fibonacci). Xc nh dy s t 1T ti 5T . Tuy

nhin cn lu rng 1 21; 1T T

Gii: Vi 1 1T th ta n 1= my hin ra l 1( 1 1T Ans ) . 2 1T th ta n 1 my

hin ra l 1. Vy ta c 2 11;Pr 1Ans T eAns T .

3 2 1 1 1 2T T T Cch n my (PreAns) .

Ta n c gi tr ca 4 3 2 2 1 3T T T v n ta c gi tr ca

5 4 3 3 2 5T T T

Bin nh

C 9 bin nh ( A, B, C, D, E, F, M, X v Y) c th dng gn s liu , hng s, kt qu v cc gi tr khc

V d: mun gn 15 vo A, ta n 15 Mun xa gi tr nh ca A, ta n :

Mun xa tt c cc bin nh thi ta n (All) (yes)

V d

n my 192

Php Chia C D Bn c th dng chc nng R tm thng v s d trong php chia.

Tnh thng v s d ca php chia 8 3

Ta ghi vo my 8 3 my hin th ra kt qu 2; 2R ( Vi 2: l thng 2R : l s

d ca php chia)

Cch n my: ( )R

Lu :

Ch c gi tr thng ca php tnh ton R l c lu tr trong b nh Ans

Gn kt qu ca mt php chia c d cho mt bin s th ch gn c gi tr

s thng. V d : 10 3cho ra kt qu 3; 1R mun gn vo bin x th ta ch

gn c gi tr thng l 3

Nu tnh ton ca R l mt phn ca tnh ton nhiu bc th ch c s

thng c dng lm kt qu cho cc bc tip theo. V d

( )R my cho ra kt qu l 12. My thc hin php tnh l

10 2 12

19

Thao tc ca phm v phm b mt tc dng khi kt qu php chia c

d vn cn hin th trn mn hinh

Trng hp php chia c d tr thnh php chia khng ghi s d

Nu mt trong cc iu kin sau y tn ti khi thc hin thao tc ca php chia

c s d, th tnh ton c x l theo php chia bnh thng ( khng ghi s d)

Khi s b chia hay s chia l mt gi tr qu ln

V d : 20000000000 ( )R 17. Th s c tnh nh l 20000000000 17

Khi thng khng phi l mt s nguyn dng, s d khng phi l s

nguyn dng hay gi tr m

V d : ( )R c tnh nh l

Ly Tha S Nguyn T Trong chng trnh COMP, bn c th ly tha s cho mt s nguyn c ti

10 ch s thnh tha s nguyn t ti ba ch s.

V d 1: Ly tha s nguyn t ca 1014

Ta c tha s nguyn t ca 1014 l 42 5 13

Cch n my (FACT)

Khi thc hin ly tha s nguyn t vi mt gi tr cha tha s nguyn t c

nhiu hn ba ch s, phn khng th c ly tha s s c bao du ngoc

bn trn hin th

V d 2: Phn tch tha s nguyn t 4104676

My tnh phn tch 4104676 thnh 22 (1026169)

Cch n my : 4104676 (FACT)

Cch n sau khng ly c tha s nguyn t

Nhn (FACT) hay

Nhn bt k mt trong cc phm sau y: hay

Dng menu thit t thay i thit t n v gc (Deg, Rad, Gra) hay

thit t ch s hin th (Fix, Sci, Norm)

Lu : Bn s khng th thc hin ly tha s nguyn t trong gi tr

thp phn, phn s hay kt qu tnh ton gi tr m c hin th. C lm

nh vy s gy ra li ton hc ( Math ERROR)

Bn s khng thc hin c vic ly tha s nguyn t trong mt kt

qu tnh ton c dng Pol, Rec, R c hin th

Php tnh vi cc hm

Vo Mode COMP (n ) khi mun thc hin cc php ton c bn .

Mt vi php ton cho kt qu hi chm

20

Phi ch kt qu hin ln mi bt u php tnh k tip

Hm lng gic Hm lng gic ngc Phi n nh n v o gc(, radian, hay grad)

n mn hnh hin ra

:MthIO :LineIO

:Deg :Rad

:Gra :Fix

:Sci :Norm

n tip s n v c chn

(gc

V d 1: tnh

n my (Deg)

V d 2 tnh rad) =0,5

n my (rad)

V d 3: = (radian)

n my (rad)

V d 4: = 18,

n my my (Deg)

0.741

n tip 18,

Ghi ch l hm arcsin, arccos, arctan.

Hm Hype Hm Hype ngc Ta n mn hnh hin ra

:sinh

:tanh

: cos

:cosh

:

:

V d 1: tnh sinh 3,6 = 18,28545536

21

n my

V d 2: tnh 30 = 4,094622224

n my

Logarit thp phn-logarit t nhin-logarit ngc (hm m)

logarit c s bt k

My k hiu log : logarit thp phn ( logarit c s 10)

ln : logarit t nhin ( logarit nepe)

log : logarit c s bt k

V d 1: Tnh log 1,23 = 0,08990511144

n my

V d 2: tnh ln90 = ( =4,49980967

n my

Tnh lne =1

n my

V d 3: Tnh =22026,46579

n my ( )

V d 4: Tnh =31,6227766

n my ( )

Tnh

n my

V d 5: tnh =8,965784285

n my

Cn bc hai, cn bc ba, cn bc n, bnh phng, lp phng,

nghch o , giai tha , s ngu nhin, s , v t hp , chnh

hp

V d 1: Tnh =

Cch n my

V d 2: Tnh

Cch n my ( ) ( )

V du 3: Tnh

Cch n my

V d 4: Tnh 123 + 3 =1023

Cch n my

22

V d 5: tnh

n my ( 3x )

V d 6 : Tnh =12

Cch n my

V d 7 : tnh 10! = 3628800

Cch n my

Hin th mt s ngu nhin gia 0,000 v 0,999

Cch n my ( #Ran ) 0,611

Lu : v n hin th mt s ngu nhin nn mi ln n my, my s cho ra mt

kt qu khc nhau khng bit trc c.

V d 9 : Tnh

n my

V d 10: C bao nhiu s gm 5 ch s khc nhau c chn trong cc ch s t

1 n 7 ? ( Ta c kt qu 2520)

Cch n my (nPr)

V d 11: C bao nhiu cch thnh lp nhm 4 ngi trong 10 ngi ? ( kt qu

210)

Cch n my (nCr)

i n v o gc n mn hnh hin ra

:MthIO :LineIO

:Deg :Rad

:Gra :Fix

:Sci :Norm

Tip theo ta n s th t ty theo n v m ta mun i

V d: Ta i 4,25 radian ra

Do ta i t radian sang nn ta chn my

n (DRG) (r)

4,25r

243,5070629

23

n tip

i ta ( ta -cc v ta cc) Kt qu c hin ra theo th t x, y

V d 1: i ta cc ( ) ra ta -cc (x,y) (Deg)

Ri nhp (Rec) Kt qu x= 1

Y= 1,732050808

V d 2: i ta - cc(x=1,y= ) sang ta cc ( ) ( radian)

Ri ta nhp (Pol) Kt qu

K hiu k thut V d 1: Chuyn 56088 ra dng

56088

Cch n my 56088

V d 2: Chuyn 0,08125 ra dng

0,08125 =

Cch n my 0.08125

Tnh tng

Vi ta c th tnh tng gi tr mt biu thc f(x) khi xc nh phm vi ca x.

f(x), a, b)= f(a) + f(a+1) ++ f(b)

f(x) : Hm s bin x ( nu khng cha x th l hng s )

a : Gi tr bt u

b : Gi tr cui

a, b phi l s nguyn v

Bc nhy ca php tnh c xc nh l 1

v khng dng c trong f(x), a hay b

n ngng

V d : Tnh tng dy s Tnh tng M =

Cng thc tng qut

Ghi vo mn hnh my tnh 20

1

3x

x

v n

24

Kt qu M= 5230176601

Cch n my: ( ) (X)

Php Tnh Tch

Xc nh tch s ca ( )f x vt qu min cho cng thc tnh l:

( ( )) ( ) ( 1) ( 2) ... ( )b

x a

f x f a f a f a f b . C php hin th t nhin l:

( ( ))b

x a

f x ,trong c php a vo hin th tuyn tnh l : ( ( ), , )f x a b . Vi a,b l

hai s nm trong min 10 101 10 1 10a b .

V d: Tnh 5

1

( 1) 720x

x

Cch n my : ( ) (X)

Lu : Cc hm sau khng c dng trong ( )f x : Pol, Rec, R . Cc hm sau

khng c dng trong ( )f x , a hay b , , / , ,d dx

Tnh c Chung Ln Nht V Bi Chung Nh Nht

GCD xc nh c chung ln nht ca hai gi tr

V d : Xc nh c chung ln nht ca hai s 28 v 35

Ta c c chung ln nht ca 28 v 35 l 7

Cch n my : (GCD) (,)

LCM xc nh bi chung nh nht

Xc nh bi chung nh nht ca hai s 9 v 15

Ta c bi chung nh nht ca hai s 9 v 15 l 45

Cch n my: (LCM) (,)

Tnh phn nguyn Int v Intg

(3,5) 3Int : Cch n phm (Int)

( 3,5) 3Int : Cch n phm (Int)

25

Tnh bng Ingt

(3,5) 3Intg : Cch n phm (Intg)

( 3,5) 4Ingt : Cch n phm (Intg)

Gii phng trnh H phng trnh

Mode EQN gip ta gii phng trnh bc hai mt n, bc ba mt n, h phng trnh

2 n v h phng trnh 3 n

Vo Mode EQN ta n

Gn mt kt qu cho mt bin nh

Trong mt gi tr ca kt qu php tnh cn hin th, n (STO) (A) gn

kt qu ny cho bin A

Lu

Bn c th gn gi tr mt kt qu khi gii phng trnh cho bt k bin s sn

c no(A,B,C,D,E,F,X,Y,M).

Bn c th gn mt kt qu cho mt bin s ngay c khi n l mt s phc.

Lu rng s phc c gn cho mt bin ch c chp nhn t chng

trnh EQN n chng trnh CMPLX. Nhp vo bt c chng trnh no khc

s lm cho phn o c gn cho bin b xa.

Phng trnh bc 2 v phng trnh bc 3 mt n Phng trnh bc 2 c dng

Ta n vo Mode mn hnh hin ra

: COMP :CMPLX

: STAT : BASE-N

:EQN : MATRIX

: TABLE : VECTOR

Ta chn (Mode EQN) mn hnh hin ra

: X Dng cho gii h phng trnh bc nht hai n

: X Dng cho gii h phng trnh bc nht ba n

26

: Dng cho gii phng trnh bc hai mt n

: a Dng cho gii phng trnh bc ba mt n

Gii phng trnh bc hai ta n

V d 1: gii phng trnh bc hai 73 - 47x - 25460

Phng trnh ta c 2 nghim thc phn bit

Kt qu

My cng cho kt qu v ta nh ca Parabol (P) 273 47 25460y x x nh

sau:

X Value47

146Minimum

Y Value 25467,56507Minimum

Cch n my

Nhp cc h s

(nghim 1x ) (nghim 2x ) (

honh nh Parabol (P) ( tung nh Parabol (P) )

Khi nhp s b sai ta n tr li mn hnh nhp h s v dng phm

duy chuyn con tr ti s m ta cn chnh sa.

Ta n phm xem nghim k tip. Dng phm xem i xem li cc

nghim

Lu : khng nhp c s phc vo h s

V d 2 cho phng trnh

Ta c kt qu dng nghim phc l:

Cch n my

27

Nhp cc h s

y l s phc dng a + bi, nu gi s phc dng cc ta s c

2 2 120x

i vi lp 11 tr xung khi xut hin nghim phc ta kt lun l phng trnh

v nghim

Cho phng trnh

Ta c nghim kp x = - 2

Cch n my

Nhp cc h s

Nghim kp my tnh ch hin mt ln

Phng trnh bc ba c dng

gii phng trnh bc 3 ta n

V d 1: gii phng trnh

Ta c phng trnh c 3 nghim thc :

Cch n my

Nhp cc h s

Nu phng trnh ch c 1 nghim thc, th my s cho ra 1 nghim thc v 2

nghim phc (dng a+bi hay dng , nu nghim thc s m my s ghi

r ( nu my ch Deg))

V d: gii phng trnh:

Ta gii phng trnh trn ta c kt qu ghi dng a+bi

Cch n my

Nhp cc h s

28

Nu ci r ( ch Deg) th cc nghim c ghi nh sau

Cch chuyn qua dng cc

Cch n my (CMPLX) ( )

Ta phi hiu nghim l mt s thc

H phng trnh bc nht 2 n, 3 n

H phng trnh bc nht hai n my ghi dng:

gii phng trnh bc nht hai n s

Ta n vo m hnh my s hin ra cc cc dng :

: COMP :CMPLX

: STAT : BASE-N

:EQN : MATRIX

: TABLE : VECTOR

Ta chn phm (Mode EQN) mn hnh hin ra

: X Dng cho gii h phng trnh bc nht hai n

: X Dng cho gii h phng trnh bc nht ba n

: Dng cho gii phng trnh bc hai mt n

: a Dng cho gii phng trnh bc ba mt n

Chn gii h phng trnh 2 n

V d : Gii h phng trnh sau

Do phng trnh ny khng l dng ca my, khi gii bi ny bng my tnh casio

fx570VN PLUS. u tin ta phi phi chuyn n v dng ca my c dng nh

sau :

29

Sau khi a v dng ca my, ta nhp vo my v c hai nghim ca phng

trnh:

Cch n my

Nhp cc h s

V d cho h phng trnh :

Tng t nh cch nhp vo my nh h phng trnh trn my hin ra mn

hnh No-Slution (phng trnh v nghim), Infinite Sol v s nghim.

H phng trnh bc nht ba n

H phng trnh ba n my c dng

gii phng trnh bc nht ba n s

Ta n vo m hnh my s hin ra cc cc dng :

: COMP :CMPLX

: STAT : BASE-N

:EQN : MATRIX

: TABLE : VECTOR

Ta chn phm (Mode EQN) mn hnh hin ra

: X Dng cho gii h phng trnh bc nht hai n

: X Dng cho gii h phng trnh bc nht ba n

: Dng cho gii phng trnh bc hai mt n

30

: a Dng cho gii phng trnh bc ba mt n

Chn gii h phng trnh 3 n

V d: gii phng trnh:

Ta c nghim ca h

Cch n my

Lu : h phng trnh khng nhp c s phc, nu nhp s phc my s bo

Infinite Sol

Tnh Ton Bt Phng Trnh

Bn c th dng th tc sau gii bt phng trnh bc hai, bc ba:

n (INEQ) vo chng trnh gii bt phng trnh INEQ

Trn menu s hin ra , la kiu gii bt phng trnh :

la chn kiu bt phng trnh Hy n phm

Bt phng trnh bc hai : 2 2aX bX c

Bt phng trnh bc ba : 3 2aX bX cX d

Trn menu s xut hin cc kiu bt phng trnh. Ta chn t n la

chn cc kiu bt phng trnh m ta mun gii.

V d 1: Gii bt phng trnh sau: 2 2 3 0x x

n (INEQ) ( 2aX bX c ) mn hnh hin ra :

: 2 0aX bX c

: 2 0aX bX c

: 2 0aX bX c

: 2 0aX bX c

Chn phm ( 2 0aX bX c )

Gii bt phng trnh 2 2 3 0x x ta c nghim 3 1x

Nghim trong my c hin th nh trn y l hin th tuyn tnh

31

Nhp cc h s :

V d 2: gii bt phng trnh 3 22 3 0x x

n (INEQ) ( 3 2aX bX cX d )( bt phng trnh bc ba)

Mn hnh hin ra :

: 3 2 0aX bX cX d

: 3 2 0aX bX cX d

: 3 2 0aX bX cX d

: 3 2 0aX bX cX d

Chn phm ( 3 2 0aX bX cX d )

Gii bt phng trnh 3 22 3 0x x ta c nghim 3

0,2

x x

Nhp cc h s

Hin Th Nghim c Bit

All Real Numbers xut hin trn mn hnh nghim khi nghim bt phng

trnh u l thc. Tc l nghim ng vi mi s thc R

V d 3: 2 0x

(INEQ) (2aX bX c ) (

2 0aX bX c )

Nhp cc h s my bo: All Real Numbers

No- Solution xut hin trn mn hnh khi khng c nghim cho bt phng

trnh nh : 2 0x

T S RATIO

Chuong trnh RATIO cho php bn xc nh gi tr X trong biu thc t s

a b X d hay a b c X . Khi 3trong 4 gi tr a,b,c,d c bit

Vo chng trnh RATIO n (RATIO)

V d tnh ton chng trnh RATIO

V d 1:

tnh X trong t s 1 2 10X

32

Vo chng trnh tnh RATIO: n (RATIO) ( a b X d )

Nhp cc h s : ta c kt qu X=5

V d 2

Tnh X trong t s 1 2 10 X

Vo chng trnh tnh RATIO: n (RATIO) ( a b c X )

Nhp cc h s : ta c kt qu X=20

Trong chng trnh RATIO khng thc hin c cc thao tc : , (M-),

(STO), Pol, Rec, R , , /d dx v a cu lnh cng khng thc hin c

trong chng trnh RATIO

Tnh Ton Phn Phi

Bn c th dng cc bc sau y thc hin by loi phn phi khc nhau:

n phm (DIST) vo chng trnh DIST

Cc loi ton s xut hin trn menu

Chn php tnh n phm trn my

Mt xc sut bnh thng : Normal PD

Phn phi ly tha bnh thng : Normal CD

Phn phi tch ly bnh thng nghch o : Inverse Normal

Xc sut nh thc : Binomial PD

Phn phi tch ly nh thc : Binomaial CD

Xc sut Poisson : Poisson PD

Phn phi tch ly Poisson : Poisson CD

V d v tnh ton trong chng trnh DIST

tnh mt xc sut bnh thng khi 36, 2, 35x .

(DIST) ( Normal PD)

My hi Normal PD X? nhp 36X v n

My hi Normal PD: ? Nhp 2 v n

My hi Normal PD: ? Nhp 35 v n

Ta c mt xc sut bnh thng 0,1760326634p

33

n hay tr v mn hnh a vo x

tnh xc sut nh thc cho d liu : {10,11,12,13,14}khi N=15 v p=0,6

(DIST) (Binomial PD)

Chn (LIST) Nu xc nh d liu dng dng thc tham s n (Var)

Nhp cc h s

My hi Binomial PD: N? Nhp N=15 v n

My hi Binomial PD: P? Nhp p=0,6 v n

Mn hnh hin ra

1

2

3

4

5

X

10

11

12

13

14

Ans 0,1859

0,1267

0,0633

0,0219 34,7 10

Bpd

10

n ta c xc sut nh thc ca 10 = 0,18594

n ta c xc sut nh thc ca 11 = 0,12678

n ta c xc sut nh thc ca 12 = 0,063388

n ta c xc sut nh thc ca 13 = 0,021942

n ta c xc sut nh thc ca 14 = 34,7018 10

Thng k Hi qui

Dng phm vo STAT n

Chn Mt Kiu Thng K

Trong Mode STAT, hin th mn hnh chn kiu tnh thng k

Cc kiu tnh thng k

Phm

Mc menu

Tnh thng k

34

1 1-VAR Bin n

2 A+BX Hi quy tuyn tnh

3 + C Hi quy bc hai

4 lnX Hi quy logarit

5 Hi quy s m c s e

6 Hi quy c s m c s B

7 Hi quy ly tha

8 1/X Hi quy nghch o

Nhp d liu mu

Hin th mn hnh STAT

Mn hnh STAT xut hin sau khi bn truy nhp Mode STAT t mt mode khc .

S dng menu STAT chn mt kiu tnh thng k t mt mn hnh t nt mn

hnh STAT khc.

n (STAT) (Data)

Mn hnh STAT

C hai dng thc mn hnh STAT, ph thuc vo kiu tnh thng k la chn.

Thng k bin n Thng k hai bin

Dng th nht ca mn hnh STAT cho thy gi tr ca mu th nht hoc gi

tr cp th nht ca mu

Ct FREQ ( tn s )

Nu m hng mc hin th thng k trn mn hnh ci t ca my tnh , mt ct

mang tn FREQ s hin cng trn mn hnh STAT.

C th s dng ct FREQ ch ci tn s (s ln m cng mt mu xut hin

trong nhm d liu ) ca mi gi tr mu

Quy tc nhp d liu mu trn vo mn hnh STAT

- D liu nhp c chn vo c con tr . S dng cc phm con tr duy

chuyn con tr gia cc

STAT D

1

2

3

3

STAT

D

1 Y

2

3

STAT D

1 Y

2

35

Con tr

- Cc gi tr v biu thc nhp trn mn hnh STAT ging nh nhp trong

Mode COMP vi dng Line

n khi d liu ang nhp xa d liu nhp hin hnh

Sau khi nhp mt gi tr, n . N s nhn ra gi tr v hin th ti su ch s

trong nhp hin hnh

V d: Nhp gi tr 123,45 vo

Chuyn con tr n

Nhp mt gi tr lm cho con tr duy chuyn xung 1

Cc ch khi nhp

S dng trong mn hnh STAT ( s gi tr d liu c th nhp) ph thuc vo loi

thng k v ci t hin th ca thng k chn

Hin th thng k

Loi thng k

OFF ( tt) ( khng c ct FREQ)

ON ( bt ) ( c ct FREQ)

Bin n 80 dng 40 dng

Bin i 40 dng 26 dng

Cc kiu nhp sau khng thc hin c trn mn hnh STAT

Cc hot ng M+,SHIFT, M+, (M-)

STAT D

1 Y

2 3

123,45

36

Chuyn vo cc bin (STO)

Ch v lu d liu mu

D liu nhp vo s b xa t ng bt c lc no khi chuyn t Mode STAT

sang Mode khc chuyn ci t hin th thng k ( lm cho ct FREQ n hay

hin) trn mn hnh ci t

Chnh d liu mu

Thay d liu trong

1) Trn mn hnh STAT, chuyn con tr n mun chnh

2) Nhp d liu v biu thc mi , sau n

Ch

Lu phi thay ton b d liu hin c ca bng nhp s mi . Khng th

chnh tng phn ca cc d liu hin c

Xa mt dng

1) Trn mn hnh nhp STAT , chuyn con tr n dng mun chn xa

2) n

Chn vo mt dng

1) Trn mn hnh nhp STAT , chuyn con tr n dng mun chn vo

2) n (STAT) 3 (Edit)

3) n 1( Ins)

Ch l vic chn khng thc hin c nu lng dng cho php ti a

dnh cho mn hnh nhp STAT dng ht.

Xa ton b ni dung nhp STAT

(1) n (STAT) 3( Edit)

(2) n 2 ( Del A)

Khi , ton b d liu mu trn mn hnh STAT s b xa

Lu

Lu rng ch c th thc hin thao tc chn vo mt dng v xa ton b

ni dung STAT khi mn hnh STAT hin th.

Mn hnh php tnh STAT

Mn hnh php tnh STAT l thc hin php tnh thng k vi d liu nhp

bng mn hnh STAT. n phm khi mn hnh nhp STAT c hin th

chuyn sang mn hnh php tnh STAT

Mn hnh php tnh STAT cng dng dng Line bt k dng nhp/ xut hin hnh

ang ci t trn mn hnh.

37

Thu ly gi tr thng k vo d liu

thu ly gi tr thng k, n khi trong b son tho thng k v th ri nh

bin thng k ( bn mun. Cc bin thng k c h tr v cc

phm bn phi n nh chng c nu di y. Vi cc tnh ton thng k

bin n , cc bin c nh du bng du (*) l c sn

Sum:

( STAT) ( Sum) ti

S cc khong mc , trung bnh lch chun khng gian mu

: . lch chun mu:

( STAT) ( Var) ti

H s hi qui : A , B h s tng quan : r , gi tr c lng

( STAT) ( Reg) ti

H s hi qui cho hi quy bc hai : A, B, C . Gi tr c lng ,

( STAT) ( Reg) ti

khng phi l bin. Chng l cc ch lnh c kiu nhn mt i

ngay trc chng .

Gi tr ti thiu : min , minY, gi tr ti a: max , maxY

( STAT) ( MinMax) ti

Lu : khi tnh ton thng k mt bin c la , bn c th a vo hm v

lnh thc hin tnh ton phn s chun t menu xut hin khi bn thc hin

thao tc phm sau: ( STAT) ( Distr) .

V d 1

a d liu vo bin n x={1,2,2,3,3,3,4,4,5}, dng trong ct FRQ xt

nh s lp cho tng khon mc } v tnh gi

tr trung bnh v lch chun khng gian mu

(STAT) (ON)

(STAT) (1-VAR)

Nhp xong ta n (STAT) (Var) )

Ta c kt qu gi tr trung bnh 3

Ta tnh lch chun khng gian mu

n (STAT) (Var)

38

Ta c kt qu lch chun khng gian mu 1,154700538

V d 2

Tnh ton cc h s tng quan hi qui tuyn tnh v hi qui logarit choi d liu

bin i sau v xc nh cng thc hi qui cho tng quan mnh nht:

(x,y)=(20,3150), (110, 7310), (200, 8800), (290, 9310) ly kt qu n s thp

phn th 3

(STAT) (OFF)

(Fix)

(STAT) (A+BX)

20 110 200 290

3150 7310 8800 9310

(STAT) (Reg) (r)

Ta c h s tng quan hi qui tuyn tnh : 0,923

(STAT) (Type) (lnx)

(STAT) (Reg) (r)

Ta c h s tng quan hi quy logarit 0,998

(STAT) (Reg) (A)

(STAT) (Reg) (B)

Kt qu : h s tng quan hi qui tuyn tnh : 0,923

H s tng quan hi qui logarit: 0,998

Cng thc hi qui logarit y= -3857,984+2357,532lnx

Tnh gi tr c lng

Da trn cng thc tnh hi qui thu c bng tnh ton thng k bin i, gi tr

c lng ca y c th c bng tnh ton theo gi tr x cho. Gi tr x tng

3857,984

2357,532

39

ng ( hai gi tr trong trng hp hi qui bc hai ) cng c th tnh ton

cho gi tr ca y trong cng thc hi qui

V d 3

xc nh gi tr c lng cho y khi x=160 trong cng thc hi qui c to ra

bi hi qui logarit ca v d 2 ly kt qu n s thp phn th 3.

(Fix)

(STAT) (lnx)

20 110 200 290

3150 7310 8800 9310

160 (STAT) (Reg) ( )

Ch : iu quan trng : tnh ton h s hi qui , h s tng quan v cc gi tr

c lng c th tn thi gian ng k khi c s ln cc khon mc d liu

Thng tin k thut

Khi c vn

Nu kt qu php tnh bt thng hoc c thng bo v li, hy thc hin cc

bc sau :

1) n tr li ci t li Mode ban u.

2) Kim tra li cng thc ang dng

3) Nhp Mode ng v thc hin li php tnh. Nu cc bc trn khng chnh

ng bi ton th n my t kim tra li thao tc v xa tt c cc d liu

bt thng trong b nh my. Nn ghi cc d liu quan trng gi li

Thng bo li

My b ng khi c li thng bo hin ln. n v chnh sa li. Sau

y l chi tit ca cc li

Math ERROR (li v tnh ton)

L do:

Kt qu php tnh ngoi kh nng ca my

Thc hin php tnh ngoi phm vi nhp ca hm.

8106,898

40

Thc hin thao tc bt hp l (nh chia cho 0)

Cch sa

Kim tra phm vi cho php ca gi tr nhp

Ch cc gi tr thuc vng nh ang s dng

Stack ERROR (li v nhm php tnh)

L do

Nhm hoc php tnh vt tri qu kh nng

Cch sa

n gin biu thc tnh ton khng cho n vt qu kh nng

Chia php tnh ra 2 hoc nhiu hn .

Syntax ERROR (li c php)

L do

Thao tc ton sai

Cch sa

Dng phm tm ch sai v chnh li

Agr ERROR( li argument)

L do

Sai argument

Cch sa:

Dng phm tm ch sai v chnh li

Dimension ERROR :(li Ma trn hay Vector)

Ma trn hay vector dng tnh ton m a vo my khng xc nh dng, ct

ca n.

Bn ang c gng tnh ton ma trn vec-t c chiu khng c php tnh ton

cho kiu tnh ton

Th t u Tin Ca Php Ton Th t u tin ca php tnh thc hin nh sau:

1) Pol(x,y) , Rec(

P(,Q(, R(

2) Vi cc hm mu A

Vi cc hm ny , gi tr nhp trc , phm hm n sau

41

K hiu k thut phn phi thng thng

,

i n v gc

i n v

3) Ly tha ,cn s :^(

4)

5) Dng nhn tt vi , vi tn s nh , vi tn bin :

6) Hm mu B

Vi cc hm ny , phm hm n trc , gi tr nhp sau : ,

sin, cos, tan,

,( - ),d, h, b, o, Neg, Not, Det, Trn, Abs, Conjg.

7) Dng nhn tt vi cc hm s mu B: 8) Chnh hp, t hp: nPr, nCr, 9) Nhn tch v hng (.)(Dot)

10)

11)

12) And 13) Xnor, xor, or

Cc php ton tng t sau c thc hin t phi sang tri :

)

Cc php tnh khc c tnh t tri sang phi

Cc php ton trong ngoc c thc hin trc

Phm vi nhp S tnh ton bn trong 15

chnh xc ch s th 10.

Hm Phm vi nhp

sinx

DEG 0

RAD

GRA 0

cosx

DEG

RAD

GRA 5.000000009

42

tanx

DEG Nh sinx, ngoi tr

RAD Nh sinx, ngoi tr

GRA Nh sinx, ngoi tr

sinhx

coshx

tanhx

Logx/lnx

nPc

nCr

Chuyn i thp phn h su mi

Tuy nhin :

43

Tuy nhin :

Ton b s nguyn , t s mu s phi l 10 ch s hay t hn k c du chia

Randln#(a,b)

Mt s chc nng khc

Chc nng CALC

Chc nng CALC gip ta lu tm thi mt biu thc v tnh ngay gi tr ca biu

thc theo mi gi tr ca bin ( ch)

Gi tr ca bin c nhp theo yu cu ca tng bin

V d: tnh vi ( kt qu y=58)

V khi kt qu y=76)

Nhp biu thc ,n y x x

Sau khi xong biu thc:

Lu biu thc n

nhp 7 vo x? n

Nhp 8 vo x? n

Biu thc b xa i khi bt u cc thao tc khc, i Mode hay tt my.

Chc nng SOLVE

SOLVE dng phng php Newton tm nghim xp x ca phng trnh

Lu

Chc nng SOLVE ch dng c trong chng trnh COMP( )

Vi lnh SOLVE ta c th tm nghim phng trnh bc nht hay cao hn (

nhng mi ln tm ch c mt nghim)

V d: Gii phng trnh bc nht mt n sau:

44

Nu gp phng trnh dng ny m ta c a v phng trnh bc nht dng

Ax+B=0 gii th mt rt nhiu thi gian, vi my tnh Casio fx570ES PLUS, ta

gii nh sau cho nhanh:

Ta ghi vo mn hnh

Cch n my

Sau khi nhp xong vo my

n my hi X? Ta c th cho x bt k. y ta cho x=1 chng hn

n ta c kt qu x=-1.4492

Ngoi bin x ta cng c th cc phng trnh bng bin A,B,C,D nhng khi

nhp v ta phi bo bin:

V d: gii phng trnh bc nht sau:

Ta ghi vo mn hnh 3A+3=30,A ( bo bin A)

Cch n my

Sau khi nhp xong n ta c kt qu A=9

Lu : nu biu thc khng ghi = 0 th my cng coi nh c du =0.

Ton s phc

n ( CMPLX) vo ton s phc,

Trong mode CMPLX, phm chuyn thnh phm nhp k hiu ( o )

dng ghi a+bi. V d sau l nhp 2+3i

Cch nhp

45

C th nhp s phc theo dng cc ( )

V d sau l nhp

Cch n ( )

Dng kt qu hin th

My c th hin th kt qu theo dng ta - Cc hay dng cc. C th chn

dng ci t ban u

V d v hin th kt qu theo dng a + bi

Ci t hin th dng a + bi

(CMPLX)

(a+bi)

V d 1 ( n v

o gc l )

Cch n my

Trong dng LINE, phn thc v phn o thc hin thnh 2 dng khc nhau

V d 2 ( n v gc l )

( )

V d hin th kt qu theo dng cc ( )

(CMPLX)

( )

V d 1 ( ) ( n v gc l )

Cch n my

Trong dng LINE, phn sut v gic s thc hin thnh 2 dng khc nhau

V d 2 ( n v gc o l )

46

Gic s hin trong phm vi

S phc lin hp ( conjg)

V d:

Kt qu hin th dng ta -Cc :

Cch n my:

(CMPLX)

(a+bi)

(Conjg)

Kt qu hin th dng ta cc:

Cch n my:

(CMPLX)

( )

(Conjg)

Sut v gic s (Abs, arg)

Thao tc sau cho ta sut ( ), v gic s (arg) trong mt phng Gauss cho s

phc Z=a + bi

(Abs); (CMPLX) (arg)

Tm sut v gic s ca

Sut :

Gic s

(CMPLX) (arg)

47

H m c s N

Vo Base n (Base-N)

Ngoi cc thao tc vi s h thp phn ta c th thao tc vi s nh phn , bt

phn v thp lc phn

Ta c th n nh mt h m trc ri nhp cc s h m khc c xc nh

tn h

Khng thc hin c cc hm khoa hc ( cng nh cc phn l v ) h

m c s N ( nu nhp s c phn t l, my t ng ct b cc phn l )

My thc hin cc php ton logic nh and ,or ,xor ,xnor ,not v Neg

Khi chn h m th th k hiu tng ng hin ln bn phi

Phm H m Du hiu hin

DEC Thp phn Dec

HEX Thp lc phn Hex

BIN Nh phn Bin

OCT Bt phn Oct

K hiu m hin hnh hin th dng th hai ca mn hnh

Ci t mc nh ban u khi vo BASE-N l Dec

Nhp gi tr

Trong BASE-N ta ch dng c cc ch s ca h m hin hnh

Bo li hin ln khi ta nhp cc s ngoi h m hin hnh ( nh nhp 2 h

BIN)

Nhp gi tr HEX

Cc phm A, B, C, D, E, F l phm s trong HEX

Bng phm vi gi tr

H m Phm vi

BIN S dng S m

OCT S dng S m

DEC

HEX S dng S m 80000000 FFFFFFFF

Phm vi ny l dy 16bit hp hn dy 32bit

Bo li hin ln khi kt qu tnh ton ngoi phm vi ny

48

Xc nh h m khi nhp

Trong BASE-N, my my cho php nhp s ca h m ln h m ci

trc. Mun dng cch ny th phi n (BASE) hin menu (hai

trang) v ta n s lin h cn thit ghi trc cc h m

n phm Xc nh s trong h

(d) Thp phn (c s 10)

(h) Thp lc phn ( c s 16 )

(b) Nh phn ( c s 2 )

(o) Bt phn ( c s 8 )

V d 1: Tnh h nh phn

Cch n my

(BASE-N) (BIN)

V d 2: H bt phn chia cho h thp phn

Cch n my

(BASE-N) (OCT ) (d)

Tnh s m v ton logic

Mun tnh v s m v ton logic, n (BASE) hin menu BASE

v dng cc lnh tng ng

n phm Yu cu

(and) Nhp and ( thc hin php tnh AND)

(or) Nhp or( thc hin php tnh OR)

(xor) Nhp xor( thc hin php tnh XOR) (xnor) Nhp xnor( thc hin php tnh XNOR) (Not) Nhp Not( thc hin php tnh NOT) (Neg) Nhp Neg ly Neg (s m, s b 2)

Php ly Neg ( s m ) trong BIN, OCT, HEX da vo phn php b 2 ca BIN, ri chuyn li theo c s chn . S m trong DEC th mang du ( - )

Cc php tnh v s m trong h m nh phn ( BIN) trc khi thc hin phi n (BIN)

49

o Hm Thc hin MODE COMP ( n )

Ta c th tnh gn ng gi tr o hm ca mt hm s ti mt im ch nh

theo c php

d/dx(f(x),a,tol)

f(x) : hm s theo bin x, khng cha x th l hng s

a: l im tnh o hm

tol : ( ch nhp, xut Line )

C th b qua gi tr tol, gi tr mc nh l

,d/dx (,Pol(,Rec( v khng dng c trong f(x),a,tol

Phi chn n v l gc Radian khi thc hin php tnh o hm cc hm lng

gic

C bo li hin khi php tnh gi tr o hm khng thc hin c

Khi chn gi tr tol nh th kt qu chnh xc hn nhng li mt nhiu thi gian .

Phi chn tol l hay ln hn

Khng nhp tol dng Math.

Kt qu thiu chnh xc v c th c bo li do:

- im tnh o hm l im gin on

- Ti im cc

- im tnh thuc vng cc i hay cc tiu

- im tnh l im un

- Ti im tnh hm s khng c o hm

- Tai im m kt qu php tnh gn zer

n ngng

V d 1: cho hm s . Tnh kt qu ti ?

n my

( kt qu l 7 )

Lu : trn my tnh ny ta ch tnh c bin x, cn cc bin khc nh A,

B, C th my khng tnh c.

Tch Phn

Thc hin MODE COMP n ( )

Bn yu t cn nhp tnh tch phn l hm s theo bin x, cn a, b v s n

( my chia trong quy tc Simson ).

C php

50

hm s, a, b,n)

V d: tnh

n my

Ghi ch

Chn n l s nguyn t 1 n 9 hay b qua cng c

My cn mt thi gian tnh ton

Khi ang tnh tch phn th mn hnh khng hin ra g c

Ma trn

My thit lp c ma trn 3 dng x 3 ct vi cc php tnh +, -, x, chuyn

v, nghch o, tch v hng, tnh nh thc v sut ca ma trn

Ta c th nhp tn Mat A , Mat B, , Mat C vo b nh ma trn . Khi

tnh ton , ma trn kt qu mang tn MatAns

Vo chng trnh ma trn , n

Phi thit lp mt hay nhiu ( ti a 3 ) ma trn trc khi tnh ton

Lp ma trn

n ( MATRIX ) (Dim) v xc nh tn Ma trn cn lp (A, B hay

C) tip theo l xc nh s dng, s ct . Cui cng l nhp tng phn t ca

ma trn vo a ch dng ct

Thot khi mn hnh ma trn th n

Menu ma trn

Trong Mode Matrix, menu sau hin ln khi n (Matrix)

51

Mu chn Yu cu

(Dim) Chn kch thc ( ct, dng) cho MatA,MatB,MatC

(Data) Tm ni dung ca MatA, MatB, MatC

(MatA) Nhp MatA

(MatB) Nhp MatB

(MatC) Nhp MatC

(MatAns) Nhp Mat Ans

(Det) Nhp det( Tnh nh thc

(Trn) Nhp Trn( Tm ma trn chuyn v

Cng, tr v nhn ma trn

Lp ma trn A, B ri tnh A B

A B A B

Nhp ma trn A

(Matrix) (Dim) (MatA) (3 2)

Nhp ma trn B

(Matrix) (Dim) (MatB) (2 3)

Tnh A B

(Matrix) (MatA) (Matrix) (MatB)

(Matrix) (MatC)

Tnh nh thc ca ma trn vung

V d: Tnh nh thc ca ma trn

kt qu l 73

(Matrix) (Dim) (MatA) (3 3 : 3dng, 3ct)

52

Tnh nh thc ma trn A

(Matrix) (det ) (Matrix ) (MatA)

Bo li khi ta nhp ma trn khng vung

Chuyn v mt ma trn

V d: Chuyn ma trn

thnh

Nhp ma trn B

(Matrix) (Dim) (MatB) (2 3)

(Matrix) ( Trn) (Matrix) (MatB)

Nghch o mt ma trn

V d: nghch o mt ma trn vung:

thnh

(Matrix) (Dim) (MatC) (3 3)

(Matrix) (MatC)

Khng thc hin c vi ma trn khng vung hay ma trn c nh thc bng 0.

Lp sut ca mt ma trn

Lp sut ca ma trn ( kt qu trn)

Kt qu:

53

Cch nhp ma trn nh nhng phn trn.

Tnh Abs Mat C (Abs) (MATRIX) (MatC)

Ton Vect

Dng VctA, VctB, VctC ghi cc vect trong b nh. Vect kt qu ca

php tnhc ghi l VctAns

Ton vect c thc hin Mode VECT ( )

Menu vect

Bng sau l menu vect ( hin trong mode vect ) sau khi n

Mu chn Yu cu

( Dim ) Gi vect VctA, VctB, VctC n nh Dim ( mt phng hay khng gian ) cho vect ny

(Data ) Gi VctA, VctB, VctC hin ta v chnh sa ta

( VctA ) Nhp VctA

( VctB ) Nhp VctB

( VctC) Nhp VctC

(VctAns ) Nhp VctAns

(( Dot ) Nhp du ghi tch v hng 2 vect

Nhp Vect

nhp vect , n (VECTOR) (Dim) ri xc nh Vect nhp

(A,B,hay C) ri nhp Dim (s chiu) v tip theo l cc thnh phn s ta .

Thot khi mn hnh n

Cng tr Vect

V d: Cng hai vect A=(1, - 2, 3) vi vect B=(4, 5, - 6)

Ta c kt qu l (5, 3, -3)

Cch n my

Nhp vect A Dim 3: (VECTOR) (Dim) (VctA) (3)

54

Nhp ta :

Nhp vect B Dm3 : (VECTOR) (Dim) (VctB) (3)

Nhp ta :

Cng hai vect: (VECTOR) (VctA) (VECTOR) (VctB)

Bo li xut hin khi VctA v VctB c s chiu ( Dim) khc nhau

Tch mt s thc vi vect

V d: Vct (-7, 8, 9). Tnh Vct 5C = (-35, 40, 45)

Cch n my

Nhp Vect C Dim 3: (VECTOR) (Dim) (VctC) (3)

Nhp ta :

5 x Vct C (VECTOR) (VctC)

Tch v hng hai vect ( du ( Dot))

V d : Tnh tch v hng hai vect A=(1, - 2, 3) vi vect B=(4, 5, - 6). Ta c

kt qu =

Nhp vect Dm 3 : (VECTOR) (Dim) (VctA) (3)

Nhp ta :

Nhp vect B Dim 3 : (VECTOR) (Dim) (VctB) (3)

Nhp ta :

Tch v hng vect A v vect B

(VECTOR) (VctA) (VECTOR) (Dot) (VECTOR)

(VctB)

C bo li khi s chiu (Dim) ca khc nhau.

Tch hu hng hai vect

V d : Tnh tch hu hng ca hai vect A=(1, - 2, 3), vect B=(4, 5, - 6).

Ta c kt qu

Nhp vect A Dim 3: (VECTOR) (Dim) (VctA) (3)

55

Nhp ta :

Nhp vect B Dim 3: (VECTOR) (Dim) (VctB) (3)

Nhp ta :

Tnh tch hu hng

(VECTOR) (VctA) (VECTOR) (VctB)

Tnh Mun ca mt vect

Cho vect A=(1, - 2, 3). Tnh Mun ca vect A ( ). Kt qu

Nhp vect A Dim 3 : (VECTOR) (Dim) (VctA) (3)

Nhp ta :

Tnh Mun : (Abs) (VECTOR) (VctA)

V d : tnh gc ( bng ) ca hai vect v ta php

vect ca mt phng nhn lm cp vect ch phng

Vect A Dim 3 (VECTOR) (Dim) (VctA) (3)

Nhp ta

Vect B Dim 3 (VECTOR) (Dim) (VctB) (3)

Nhp ta

Tnh gc

(VECTOR) (VctA) (VECTOR) (Dot) (VECTOR)

(VctB) (Abs) (VECTOR) (VctA) (Abs)

(VECTOR) (VctB)

Ta c kt qu

56

Tnh

(VECTOR) (VctA) (VECTOR) (VctB)

(Abs) (VECTOR) (VctA) (VECTOR) (VctB)

Ta c kt qu (-0,66666, 0,33333, -0,66666)

Bng s t hai hm

Tt c cc php tnh trong phn ny c thc hin

Dng bin x a vo hai hm ( )f x v ( )g x

Nu a bin s x vo hai hm mt dng ( )f x v mt dng ( )g x

Hy a bin x vo bng cch n (X) khi sinh ra bng s . Bt k bin no khc X iu x l nh mt hng s

Nu s dng mt s n th ch a mt hm vo dng thc ( )f x

Cc hm khng th dng trong hm ny l : Pol, Rec, , / , ,d dx

p li li nhc xut hin, hy a vo cc gi tr bn mun dng, n sau mi gi tr

Vi li nhc a vo

Start ? a vo gii hn ca X (gii hn thp =1)

End ?

a vo gii hn ca X (gii hn cao =5) Lu : Hy chc chn rng gi tr End lun ln hn gi tr start

Step ?

a vo bc tng ( mc nh =1) Lu : Step xc nh cch gi tr Start phi tun t tng ln khi bng s c sinh ra. Nu bn xc nh Start =1 v Step=1 s tun t c gn cc gi tr 1,2,3,4 sinh ra bng s cho ti khi gi tr End c t ti

a vo gi tr Step ri n sinh ra v hin th bng s tng ng vi cc tham bin bn xc nh

n khi mn hnh bng s c hin th s tr li mn hnh a vo hm bc hai

V d : sinh ra mt bng s cho hm 21

( )2

f x x v hm 21

( )2

g x x .

Trong min 1 1x c tng theo bc ca 0,5

n (TABLE)

57

Chn hm :

(TABLE) ( ( ), ( )f x g x ) : chn hai hm ( ), ( )f x g x

Nhp hm ( )f x : (X)

Nhp hm ( )g x : (X)

My hi Start? Nhp

My hi End? Nhp

My hi Step? Nhp

My hin ra bng kt qu :

1

2

3

4

5

X

-1

-0,5

0

0,5

1

F(x)

1,5

0,75

0,5

0,75

1,5

G(x)

0,5

-0,25

-0,5

-0,25

0,5

1

Lu : S ln nht ca cc dng trong bng s sinh ra s ph thuc vo vic ci t ca bng menu thit lp trn 30 dng th s c h tr ci t

( )f x , trong khi 20 dng c h tr ci t ( ), ( )f x g x

Bn c th dng mn hnh bng s ch xem cc gi tr . Ni dung bng khng th c sa i

Thao tc sinh bng s lm cho ni dung ca X b thay i

iu Quang Trng : Hm bn a vo cho vic sinh bng s b xa i bt k khi no bn hin th menu thit lp trong chng trnh TABLE v chuyn gia hin th t nhin v hin th tuyn tnh.

Hng S Khoa Hc

Vo mode COMP n

My tnh ca bn c 40 hng s khoa hc c th dng trong bt c chng trnh no tr chng trnh BASE-N. Tng hng khoa hc u c hin th nh mt k hiu duy nht ( nh ) c th dng bn trong cc tnh ton.

a mt hng s khoa hc vo bn trong tnh ton n (CONST) v ri a vo mt s c hai ch s tng ng vi hng bn mun

58

V d1 : a hng s khoa hc 0C ( tc nh sng trong chn khng ), v

hin th gi tr ca n

n my : (CONST) ( 0C )

My tnh hin th kt qu 0 299792458C

V d 2: tnh 0

0 0

1C

Ta c kt qu 0 299792458C

Cch n my:

(CONST) ( 0 ) (CONST) ( 0 )

Bng m hng s da trn cc gi tr CODATA (2010) khuyn co

01: (mp) khi lng protin 02: (mn) khi lng neutron

03: (me) khi lng in t 04: ( m )khi lng muon

05: ( 0a ) bn knh Bohr 06: (h) hng s Planck

07: ( N ) t tnh ht nhn 08: ( B ) t tnh Bohr

09: ( ) hng s Planck, hp l ha 10: ( ) cu trc mn

11: (re) bn kn in t c in 12: ( c ) chiu di sng Compton

13: ( p ) t l t hi chuyn proton 14: ( cp ) chiu di sng proton Compton

15: ( cn ) chiu di sng neutron compton 16: ( R ) hng Rydberg

17: (u) n v khi lng nguyn t 18: ( p ) momen t proton

19: ( e ) Momen t in t 20: ( n ) momen t neutron

21: ( ) momen t muon 22: (F) hng Faraday

23: (e) in tch s cp 24: (NA) hng s Avogadro

25: (k) hng s Boltzmann 26: (Vm) khi lng phn t gam ca kh l tng (273,15k, 100Kpa

27: (R) hng kh phn t gam 28: ( 0C ) vn tc nh sng trong chn khng

29: ( 1C ) hng pht x th nht 30: ( 2C ) hng pht x th hai

31: ( ) hng Stefan- Boltzmann 32: ( 0 ) hng in t

33: ( 0 ) hng t 34: ( ) lng t t thng

35: (g) gia tc hp dn chun 36: ( 0G ) lng t dn in

37: ( 0Z ) tr khng c trng ca chn khng 38: (t) nhit Celsius

39: (G) hng s hp dn Newton 40: (atm) p sut chun

Chuyn i o

59

Cc lnh chuyn i o c sn ca my tnh lm cho vic chuyn i cc gi tr t n v ny sang n v khc thnh n gin. Bn c th dng cc lnh chuyn i o trong bt c chng trnh tnh ton no ngoi tr chng trnh BASE-N v TABLE

a mt cu lnh chuyn i vo trong mt tnh ton, nhn (CONV) v ri a vo mt s hai ch s tng ng vi cu lnh bn mun

V d : chuyn i 5cm sang inch

Ta c 5 1,968503937cm inch

Cch n my : (CONV)

Bng chuyn i n v . D liu da trn NIST Special publication 811 (2008)

01: in cm 02: cm in 03:ft m 04:m ft

05: yd m 06: m yd 07:mile km 08:km mile

09: n mile m 10: m n mile 11: acre 2m 12: 2m acre

13: gal(US) 14: gal(US) 15: gal(UK) 16: gal(UK)

17: pc km 18:km pc 19:km/h km/s 20: m/s km/h

21: oz g 22: g oz 23: lb kg 24: kg lb

25: atm Pa 26:Pa atm 27:mmHg Pa 28Pa mmHg

29: hp kW 30: kW hp 31: 2/kgf cm Pa 32: Pa 2/kgf cm

33: kgf m J 34:J kgf m 35: 2/lbf in kPa 36: kPa 2/lbf in

37: 0 F 0C 38: 0C 0 F 39: J cal 40:cal J

Mt s kin thc cc lp 6-7-8-9

1. Cc php ton v Bi S Chung Nh Nht V c S Chung Ln Nht Do my ci sn chng trnh tnh trc tp nn ta ch cn nhp cc s cn tnh my s cho ra kt qu. Cch tnh nh sau: V d: Tm USCLN v BSCNN ca 209865 v 283935 Tm USCLN bng cch ghi vo my nh sau: (209865,283935) 12345GCD

Cch n my : Tm BSCNN ta ghi vo my : (209865,283935) 4826895LCM

Cch n my :

2. Php chia c s d A/ Tm thng nguyn v s d ca php chia: V d : Tm s d ca php chia 9124565217 chia cho 123456 Ta tm thng nguyn v s d trc tip trn my nh sau:

60

9124565217 123456 73909, 55713R R ( 73909 l thng nguyn, R=55713 l

s d) Cch n my : B/ Nu s b chia c cho bng dng ly tha qu ln: Th ta dng php ng d (Mod) theo cng thc:

(mod ) (mod )

(mod ) (mod )c ca m p a b m n p

b n p a m p

V d: tm s d ca php chia 3762004 chia cho 1975 Gii: Bit 376 6 62 4 . Ta tnh

22004 841(mod1975)

4 22004 841 231

12 32004 231 416

48 42004 416 536

602004 536 416 1776(mod1975)

622004 1776 841 516

62 3 32004 516 1171

62 6 22004 1171 591

62 6 42004 591 231 246

Kt qu: 3762004 chia cho 175 d 246

3. Chuyn i s thp phn tun hon v hn thnh phn s V d 1 : Hy vit s thp phn tun hon 0,0(8) di dng phn s:

Chnh my v ch COMP

V ghi vo mn hnh :4

0,0(8)45

Cch n my:

V d 2:Chng t rng 0,(37) 0,(62) 1

Ghi vo mn hnh my : 37

0.(37)99

v 62

0.(62)99

Vy ta c : 37 62

199 99

4. Phn tch thnh tha s nguyn t

V d : Phn tch 25725 thnh tha s nguyn t:

Chnh my v ch COMP

Ghi vo mn hnh my : 25725 SHIFT FACT kt qu 3 33 5 7

Cch n my :

5. Cch tnh dy s FIBONACI bng b nh Ans v PreAns

61

Kt qu tnh ton cui cng c lu trong b nh Ans. Kt qu tnh ton trc

c lu trong b nh PreAn

Cho dy s sau: 1;1;2;3;5;8 . Tnh gi tr s hng th 7,8

Ta n 1= ( Lc ny kt qu lu vo b nh Ans)

Ta n 1= (Lc 1 ny lu vo Ans cn 1 kt qu trn lu vo b nh PreAns)

Ghi vo my : Ans+PreAns

n = c s hng th 3 l 2

n = c s hng th 4 l 3

..

V tip tc n = nh vy ta c c s hng th 7 l 13 v s hng th 8 l 21

6. Tnh t s RATIO

V d 1: Tnh gi tr ca x trong phng trnh sau: 6

10 5

x

Chn ch tnh t s trn my (RATIO)

Phng trnh ny c dng :

a b x d nn ta chn

Nhp cc h s 6; 10; 5a b d ta c kt qu 3x

Cch n my :

V d 2: Gii phng trnh 6 3

10 x

Chn ch tnh t s trn my (RATIO)

Phng trnh c dng:

a b c x

Ta nhp cc h s: a=6; b=10 ; c=3 ta c kt qu: 5x

Cch n my

7. Phn tch thnh tha s nguyn t

Phn tch s 25725 thnh thc s nguyn t :

Ghi vo my : 25725 SHIFT FACT ta c kt qu l : 3 33 5 7

Cch n my

8. Bi ton v gi, pht, giy ( , pht, giy )

V d 1: Tnh 2 47 53 4 36 45h ph g h ph g

Chnh mn hnh hnh ch D bng cch n phm

( Nu mn hnh hin D th khi n phn ny )

Ta ghi vo mn hnh 0 0 0 0 0 02 47 53 4 36 45 v n du = ta c kt qu : 0 0 07 24 38 ( c l 247 24 38h ph gi )

Cch n my :

V d 2:

62

Tnh thi gian mt ngi i ht qung ng 100km bng vn tc 17,5km/h

Gii : ghi vo mn hnh

100 17,5v n ta c kt qu 5 42 51h ph gi

9. Bi ton v n thc v a thc

Tnh gi tr ca biu thc :

2 3

2

3 2 5

6

x y xz xyzI

xy xz vi

42,41; 3,17;

3x y z

Gii: Ghi vo mn hnh my 2 3

2

3 2 5

6

X Y XA XYA

XY XA( ta thay bin z=A)

n

My hi X ? nhp X= 2,41

My hi Y ? nhp Y = -3,17

My hi A ? nhp A=4

3

Kt qu 0,7918I

Lp 10

Mnh - Tp Hp

Tm tp hp th i cho sn theo cng thc

V d 1: Cho tp s v hn sau:

A= 2 31

nn N n

nva

Tm s hng ng vi n=35 ca A

63

Nhp cng thc ca tp hp thay n bng A ta ghi vo my: 2( 1)

A

A

Cch n my:

n nhp A=35 v n

Kt qu A=35

1156

V d 2: Cho tap hp so vo han sau:

A=2

*3 2

nn N

n

Tnh s hng ng vi n=15 ca A

Tng t bi trn nhp cng thc s hng tn qut ca A vi n thay bng A

Ri tnh nh sau:

n nhp A=15 v n

Kt qu A=30

47

V d 3 : Cho tp s v hn sau:

A= 2 31

nn N n

nva

Tnh tng 35 s hng u tin

Ta ghi vo my nh sau:

A = A + 1 : B = A (A -1)2 : C = C + B

n nhp A=2 v nhp C=0

n thy A=3 m 1

n c B ( s hng th 1)

n c tng C

64

n thy A=4 m 2..

--------------------------------------------

n thy A=37 m 35

n c B 35 = 37

1296

n c tng C 35

Kt qu tng 35 s hng u tin l : C 35 =3,7921

Hm s

Gi tr ca hm s ti cc gi tr xi

V d 1 : in cc gi tr ca hm s 3 2y x v bng sau:

X -5,3 -4 4

3

2,17 34

7 5 7

Y

Gii

Ta ghi vo mn hnh : 3 2Y X n

My hi X ? n . Ta c kt qu 179

10Y

n tip

My hi X ? n . Ta c kt qu 14Y

n tip

My hi X ? n . Ta c kt qu Y=6

Tip tc n v nhp cc gi tr x ta c bng kt qu

65

X -5,3 -4 4

3

2,17 34

7 5 7

Y 179

10

14 6 451

100

79

7 2 15 7

V d 2: 23,1 2 5

1,32 7,8 3 26,4 7,2

y x x . Tnh gi tr ln nht ca y.

Gii:

Gii phng trnh bc 2

Nhp cc h s : a= 1,32

b= (3,1 2 5) ( 6,4 7,2)

c= 7,8 3 2

Ta c : Y-Value Maximum=-3,541010225

Vy ta c max 3,5410y

Phng trnh v h phng trnh

Gii h phng trnh bc nht mt n

V d 1: Gii phng trnh sau:

2 3 1 6 3 7 15 11

3 5 3 2 4 3 2 3 5x x

Gii:

Ghi vo mn hnh my 2 3 1 6 3 7 15 11

3 5 3 2 4 3 2 3 5X X

n

My hi X ? nhp X =0 (chng hn) v n

Ta c kt qu : x=-1,4492

Gii phng trnh bc hai 2 0ax bx c ( 0a )

66

V d 2: Gii phng trnh bc hai

2 3 2 5 0x x

Gii :

n

Nhp cc h s a=1; b= 3 ; c= 2 5

Ta c kt qu : 1

2

1,4192

3,1512

x

x

Lu : Khi ch biu din nghim di dng ta cc th phng trnh ch

biu din duy nht dng r nu l nghim dng, nu l nghim m

V d 3: Gii phng trnh 4,5649 2,87659 2,4738 5,3143

3,9675 11,9564 7,5379 8,3152

x x

x x(1)

Nu ta quy ng mu s s c phng trnh bc hai ( vi iu kin mu khc

0) . nhanh gn ta gii bi ny bng lnh SOLVE

Ta ghi vo mn hnh my tnh 4,5649 2,87659 2,4738 5,3143

3,9675 11,9564 7,5379 8,3152

X X

X X

V n ta cho X=1 (chng hn ) v n ta c kt qu : x= -1,1539

Sau ta n ta cho X=2 ( chng hn) v n ta c kt qu : x=1,7139

Gii h phng trnh bc nht hai n

V d 4: Gii h phng trnh

5 2 3 7

5, 43 15

x y

x y

Gii:

n

Nhp cc h s ca hai phng trnh trn my sau n ta c kt qu

0,4557

2,6785

x

y

67

Gii h phng trnh bc nht ba n

V d 5: Gii h phng trnh

2x 3y 2z 4

4x 2y 5z 6

2x 5y 3z 8

Gii

n

Nhp cc h s ca h phng trnh vo my v n ta c kt qu

9

4

17

38

8

19

x

y

z

Gii h phng trnh bc hai 2 n

My khng c ci chng trnh gii h phng trnh ny nhng nu ta a v mt

n th ta cng c th tm nghim

V d 6 : Gii h

2 2 8

3

x y

x

y

vi x>0, y>0

Gii

Ta c 33

x xy

y

Ta ghi vo mn hnh 2 2( ) 83

XX . n . Nhp x=1(chng hn) v n ta

c kt qu X=3

68

1Y

V d 7: Gii h 2 2 4

2

x xy y

xy x y

T phng trnh 2 ta suy ra 2

1

xy

x

2

2 2 2 41 1

x xx x

x x(*)

Ta ghi biu thc (*) vo mn hnh my nh sau:

2

2 2 2 41 1

X XX X

X X

Sau khi nhp vo mn hnh ta n ta cho X=1 (chng hn) v n ta c

kt qu x=2. Ta tip tc n ta cho X=0 (chng hn) v n ta c kt

qu x=0.

Vy ta c nghim l (2;0); (0;2)

Tuy nhin ta dng lnh SOVE khng gip ta bit ht nghim hay cha

Thng K

V d: Sn lng la (n v t) ca 40 tha rung th nghim c cng din tch

c trnh by trong bng tn s sau:

Sn lng (x) 20 21 22 23 24

Tn s (n) 5 8 11 10 6

Tm sn lng trung bnh ca 40 tha rung

Gii

Vo ch thng k:

Hin th ct tn s: (STAT) (ON)

Nhp d liu : Ta nhp d liu x trc ri sau nhp tn s:

Nhp x:

69

Sau dng phm duy chuyn v u ct tn s. Ri tip tc nhp cc

gi tr tn s

Tnh gi tr trung bnh

n (STAT) (Var) ( x )

Ta c kt qu 22,1x

Tnh gc v t s lng gic ca mt gc

i n v gia v radian

Hoc theo cng thc :

0d

180

: radian

d:

V d : i 0 '33 45 ra radian

0 '33 45

180 180

d

Cch n my:

Chn mn hnh ch D: (Deg)

70

Nhp tip : ( )

Ta c kt qu3

16(radian)

Hoc dng chng trnh ci sn

Chn mn hnh l Radian ( R ) bng cch n (Rad)

n tip (DRG) ( 0 )

Ta c kt qu3

16(radian)

Gi tr lng gic ca mt gc

V d 1: Tnh 0 0 0 0sin1000 ,cos1000 , tan1000 ,cot1000

Cho mn hnh ch D sau :

n 1000 kt qu 0sin1000 0,9848

a con tr ln sa li

Cos 1000 Kt qu 0cos1000 0,1736

a con tr ln sa li

Tan 1000 Kt qu 0tan1000 5,6713

Tnh 0cot1000

Khi mn hnh hin ra 0tan1000 5,6713

n tip

Kt qu 0cot1000 0,1763

V d 2: Tnh 0 ' ''cot 73 1425

Ghi vo mn hnh ( ch D)

0 0 01 tan73 14 25 v n ta c kt qu 0 ' ''cot 73 14 25 0,3012

V d 3: Cho l mt gc t vi sin 0,4123

71

Tm cos , tan

Ghi vo mn hnh ( ch D )

1cos(sin (0,4123)) 0,911

Du ghi bng phm , phi thm du tr pha trc biu thc trn mn hnh v

ta bit cos 0 )

a con tr ln mn hnh sa dng biu thc thnh

1tan(sin (0,4123)) v n ta c kt qu tan 0,4526

V d 4: Cho l gc t vi sin 0,4123. Tnh ra , pht, giy.

Ghi vo mn hnh (ch D)

0 1180 sin 0, 4123v n

Kt qu 0 ' ''155 392

Ghi ch : Trong php tnh ngc ( tm gc) my ch cho kt qu gc nh tr

chnh m thi

V d 5: cho sin 0,4,cos 0,7 ( , u nhn)

Tm sin(2 3 )

Gii bng cch tnh nhanh

Ghi vo mn hnh 1 1sin(2sin (0,4) 3cos (0,7)) v n

Ta c kt qu : sin(2 3 ) 0,0676

V d 6: Bit 1

tan2 4

tnh 1 cos

1 sin

Gii ( n nhanh)

Ghi vo mn hnh (1+cos(A)) (1-sin(A))

n nhp nhp gi tr ca A l 11

2 tan4

ri n

72

Ta c kt qu 1 cos 32

1 sin 9

V d 7: Tnh 3 3sin cos ,bit 3

cos 25

Gii

Ghi vo mn hnh 3 3(sin( )) (cos( ))A A

n nhp nhp gi tr ca A l 11 3

cos ( )2 5

n

Ta c kt qu 3 3sin cos 0,6261

H Thc Lng Trong Tam Gic

V d 1:Cho tam gic ABC c b=7cm, c=5cm, gc A= 0 ' ''81 4712

a. Tnh .AB AC

b. Tnh din tch S

c. Tnh cnh BC

d. Tnh bn knh ng trn ngoi tip

Gii ( mn hnh ch D)

a. 0 ' ''. 5 7 cos 5 7 cos(81 4712 )AB AC A 5

b. 0 ' '' 21 1

5 7 sin 5 7 sin(81 4712 ) 17,32052 2

S A cm

c. 2 2 0 ' ''5 7 2 5 7 cos(81 4712 ) 8BC cm

d. Tnh bn knh R ca ng trn ngoi tip

Nu mn hnh hin ti l 8 th ta ghi tip 0 0 02sin81 47 12Ans v n

Kt qu R=4,0414cm

V d 2: Ly kt qu v d 1 . Tnh gc B ri kim tra li cng thc.

A

B C

5 7

73

22sin sin 2 sin sin sin

2sin( )

c A BS R A B C

A B

1 2 2 2 0cos ((5 8 7 ) (2 5 8)) 60B

Tnh li 2 0 ' '' 0 0 ' ''5 sin(81 4712 ) sin 60 sin(81 4712 60)S

=17,3205cm2 ( ng kt qu trc)

V d 3: Cho tam gic ABC c ba cnh a, b, c ln lt c di l 8,7,5 (cm).

V ba ng cao ' ' ', ,AA BB CC . Tnh din tch 'S ca tam gic ' ' 'ABC

Gii

Din tch ca tam gic ABC l

( )( )( ) 10 3S p p a p b p c '

2 2 21 (cos cos cos ) 2cos cos cosS

A B C A B CS

2 2 2 1cos

2 7

b c aA

bc

2 2 2 1cos

2 2

a c bB

ac

' 0 1 11 12 10 3 0,5 cos(180 cos ( ) cos (0,5))7 7

S

Kt qu ' 21,9441S cm

H trc ta

V d 1: Cho M(-2;2), N(4;1). Tnh gc MON

Gii

Ta c ( 2;2)A OM

(4;1)B ON

.

cos( , )A B

A BA B

Gi chng trnh VCT(Vect) bng cch n

My hin ra :

: VctA : VctB

: VctC

n chn VectA v n chn mt phng

Nhp ta ca VectA c ta l ( 2;2)a

n sau khi nhp xong VectA ta n

Nhp VectB bng cch sau: (Dim) (VctB) (2)

74

Nhp ta ca VectB c ta l (4;1)b

n sau khi nhp xong VectB ta n

Sau ghi vo mn hnh my tnh ( ch VCT v D)

Cos-1

((VctA VctB) (AbsVctA Abs VctB))

Cch n my:

(VctA) (Dot) (VctB)

(Abs) (VctA) (Abs) (VctB)

Ta c kt qu ca gc MON 0 ' ''( , ) 120 5750a b

Ch : Du ( tch v hng ly Dot)

Abs ghi bng

V d 2: Cho tam gic ABC c ( 4; 3 2), (2 3; 5), (1;3)A B C

a. Tnh gc A

b. Tnh din tch ca tam gic

Gii

a. Gc A nh bi AB AC

cos AAB AC

Nhp VctA = AB nh v d 1 v nhp thng t ta nh cc im A,B(

thc hin php tr ta 2 im B,A ngay khi nhp VctA)

V VctB= AC tng t

Xong ghi vo mn hnh

Cos-1

((VctA VctB) (AbsVctA Abs VctB)) v n

Kt qu A= 0 ' ''61 1028

b. Tnh din tch ca tam gic

S=22 21

. .2

AB AC AB AC

Ghi tip vo mn hnh

20,5 ( )( )VctA VctA VctB VctB VctA VctB v n ta c kt qu

S=28,9233(vdt)

Ch : cng c th tnh AB, AC th 1

sin2

S AB AC A hay tnh ba cnh

ri dng cng thc H-rng hay cng thc tch hu hng hai vect lp

12

ng Thng

V d 1: Tm ta giao im ca hai ng thng sau v gc ca chng

D1: 2x-3y -1=0

75

D2: 5x-2y+4=0

Gii

Ta giao im l ngim ca h phng trnh

2 3 1 0

5 2 4 0

x y

x y

Gii h phng trnh bc nht hai n ta c nghim ca h phng trnh :

14

2 3 1 0 11

5 2 4 0 13

11

xx y

x yy

Tnh gc (D1,D2) nh bi

1 2 1 2

1 22 2 2 2

1 1 2 2

cos( , )a a b b

D Da b a b

Vi :

1 1 2 22; 3; 5; 2a b a b

Nhp vo my n

Ta c kt qu 0 ' ''34 3030

ng Trn

V d: Vit phng trnh ng trn i qua ba im M(1;2), N(5;2)

v P(1;-3)

Gii

Phng trnh ng trn c dng : 2 2 0x y Ax By C

ng trn i qua 3 im M,N,P.

Thay ba im ta ny vo ta c h phng trnh 3 n

2 4 5

10 4 29

2 6 10

A B C

A B C

A B C

Gii h phng trnh 3 n ta c

3

1

2

1

A

B

C

Ba ng Conic

Lp phng trnh

Tnh Tm sai

Tm honh giao im vi ng thng

Vit phng trnh tip tuyn

76

Elip

V d: Vit phng trnh 2 2

2 21

x y

a b. Qua hai im M(

3 133;

4) v im

N(3 11

5;4

)

Gii

a,b l nghim ca h

2

2 2 2

2

2 2 2

3 3 131

4

5 3 111

4

a b

a b

t 2 2

1 1;x y

a b

Ta c h phng trnh

1173 1

16

995 1

16

x y

x y

1

16

1

9

x

y

Vy ta c 22

2

2

1 1

1616

1 1 9

9

aa

b

b

Phng trnh Elip cn tm l 2 2

116 9

x y

Heperbol

V d :Tnh gn ng ta giao im ca ng thng 8 4 0x y vi

Hyperbol 2 2

19 4

x y

Gii

Ta tnh nhanh

Ta c phng trnh ng thng 8 4 0 8 4x y x y

Th vo phng trnh Hyperbol c:

2 28 41

9 4

y y(1)

Ghi phng trnh (1) vo mn hnh my tnh v ta bo bin y: 2 28 4

1,9 4

Y YY v ta n

(Solve) my hin ra Y=0,9121605 ( ghi ra giy v t l y1)

Tip tc n :

77

(Solve) my hin ra Y=0,12427672 ( my t ng lu vo

bin Y ca my). Vy y2=0,12427672

Ghi tip vo mn hnh 8Y-4 v n ta c kt qu x2=-3,00578618

V n nhp Y=0,9421605 v n ta c kt qu x1=3,297284

Vy giao im l A(3,2973; 0,9122) , B(0,1243; -3,0058)

Hyperbol v Parabol

Gi M l giao im c ta dng (x>0;y>0) ca hyperbol 2 2

14 9

x yv

parabol 2 5y x

a. Tnh gn ng ta ca M

b. Tip tuyn ca hyperbol ti M ct parabol tai N khc M. Tnh gn ng

vi 5 ch s thp phn ta N.

Gii

a. Th 2 5y x vo phng trnh hyperbol ta c:

225 1 9 20 36 0

4 9

x xx x

Gii phng trnh bc hai :

12

2

10 2 106

99 20 36 0

10 2 106

9

x

x x

x

2 0x loi. Ta gn trc tip

1

10 2 106

9x vo bin X. (STO) (X)

Ghi vo mn hnh 5X v n ta c kt qu y=4,12251678

Vy M c ta l ( 3,39902892; 4,12251678)

b. Tip tuyn ca hyperbol ti M ct parabol tai N khc M. Tnh gn ng

vi 5 ch s thp phn ta N.

Giao im tip tuyn ti M ca hyperbol vi parabol l nghim ca h

1 1

2

x x y y1

4 9

y 5x

Gii phng trnh 2

1 1x y yy. 14 5 9

Gii phng trnh bc hai

Nhp cc h s a=3,39902892

20

78

H s b=4,12251678

9

H s c= 1c

My cho ra kt qu 1

2

4,12251678

1,42729158

y

y

Ta chn kt qu 2 1,42729158y v gn trc tip trn my .

(STO) (Y) (Do y1trng vi M)

Sau ta ghi vo mn hnh 2y

x5

v n

Kt qu x=0,4074322511

Ta c im N(0,40743;-1,42729)

Lp 11

Hm S Lng Gic

V d 1: Cho hm s sin(3 )6

y x

a. Tnh y khi x c cc gi tr l: ; ; ; ; ;5 7 11 9 7 5

b. Tnh x khi y c gi tr l : 3

0,3;0,7;4

. Bit x thuc khong ( ;2 2

)

Gii

a. Tnh y khi x c cc gi tr l: ; ; ; ; ;5 7 11 9 7 5

Ta ghi vo mn hnh sin(3 )6

Y X ( ch Radian)

n my hi X? ta n 5

Kt qu 0,6691y

Li n my hi X? ta n 7

Kt qu 0,9556y

Tip tc n my hi X? ta n 11

Kt qu 0,9819y

n my hi X? ta n 9

Kt qu 1

2y

79

V ln lt c cc kt qu : x , y 0.73317

x , y 0.97815

b. Tnh x khi y c gi tr l : 3

0,3;0,7;4

. Bit x thuc khong ( ;2 2

)

Ghi vo mn hnh sin(3 ) 0,36

X

n (Solve) my hi Y? n 0,3 ri nhp 0,2 cho X ri n Kt qu: x=0,276 n (Solve) my hi Y? n ri nhp 1 cho X ri n Kt qu : x=1,1202 n (Solve) my hi Y? n ri nhp -1 cho X ri n Kt qu: x=-0,9742 Sau d x1bng bao nhiu na , ta vn c 3 gi tr ca x nh trn trong

khong ( ;2 2

) ng vi y=0,3

Gii tng t vi y=0,7 v y=3

4

Vi y=0,7 ta c cc gi tr ln lt ca x l

x = -1.1311 ; x = 0.4330 ; x = 0.9633

Vi y=3

4 ta c cc gi tr ln lt ca x l

x = -1.1554 ; x = 0.4572 ; x = 0.9390

V d 2: Cho hm s y= sin 3 cos 26 5

x x

a. Tnh y khi x c cc gi tr l ; ; ; ; ;5 7 11 9 7 5

b. Tnh x khi y c gi tr 0,3 . Bit x thuc khong ( ;

2 2)

Gii

a. Ghi vo mn hnh sin 3 cos 26 5

X X ( mn hnh ch Radian)

80

n my hi X? ta n 5

Kt qu y= 0,1399

Li n my hi X? ta n 7

Kt qu 0,0084y

Tip tc n my hi X? ta n 11

Kt qu 0,0164y Tip tc ta c :

0,74199

x y

0,77797

x y

0,66915

x y

b. Tnh x khi y c gi tr 0,3 . Bit x thuc khong ( ;

2 2)

Ghi vo mn hnh my tnh sin 3 cos 26 5

X X Y

n (Solve) my hi Y? n 0,3 ri nhp 0,2 cho X ri n

Kt qu: x= 36,3573 10

n (Solve) my hi Y? n ri nhp 1 cho X ri n Kt qu : x=0,8166 n (Solve) my hi Y? n ri nhp -1 cho X ri n Kt qu: x=-0,7627 n (Solve) my hi Y? n ri nhp -2 cho X ri n Kt qu: x=-1,4703 Sau d x1bng bao nhiu na , ta vn c 4 gi tr ca x nh trn trong

khong ( ;2 2

) ng vi y=0,3

V d 3: Cho ng trn c hai ng knh AB,CD vung gc vi O. I, J l trung im ca OC,OD. ng AJ ko di ct ng trn ti M . Tnh gc AJM bng , pht, giy

Gii

A

81

Gi bn knh ng trn l R

AJ=5

2

R

V 1 11 2

tan cos2 5

A A

4

5

RAM

1

1

1tan A

2

3cos A cos 2A

5

MJ2 = AJ

2 + AM

2 2AJ.AM.cosA

R 41

MJ2 5

2 2 2

5 41 16

JA JM AM 4 20 5cos J2AJ.MJ 5 41

2 .2 2 5

0J 88 12'36"

J I

M

B

O D C

1

C D

M

J I

O

1

B

b

b

b

82

Ghi ch: c th tnh ra gc A1 bng , pht, giy, khi bit 11

tan2

A ri suy

ra gc A, cch tnh s gn hn nhng c th lm ta e ngi nh hng ti kt

qu cui cng

V d 4: Tnh A = tan 90 tan 270 tan 630 + tan 810

Gii

Ghi vo mn hnh ( ch D)

A = tan 90 tan 270 tan 630 + tan 810 v n kt qu A=4

Phng Trnh Lng Gic

Gii phng trnh lng gic c bn dng sin-1,cos-1,tan-1

V d 1 : Gii phng trnh lng gic sau:

a) sinx = 1

4 b) sin(x-20

0) =

2

2

c) sin3x = 1

2 d) sin (2x -10

0) =

3

2

Gii

a) n (sin-1)

Kt qu -1402839 . Vy 1 nghim l -1402839 + 360K

Nghim cn li: 180 (19402839)

Vy nghim cn li l: 19402839 + 360K.

b) n 20 + (sin-1) 2 2

Kt qu 65000. Vy 1 nghim l 65000 + 360K

Nghim cn li: 20+180 (sin-1) 2 2

Kt qu 155000. Vy nghim cn li l: 155000 + 360K.

c) n (sin-1) (30) 3 (10000)

Kt qu 10000. Vy 1 nghim l 10000 + 120K

83

Nghim cn li: 60 (50000)

Kt qu 50000.Vy nghim cn li l: 50000 + 120K.

d) n 10 + (sin-1) 3 2 (70) (35)

Kt qu 35000. Vy 1 nghim l 35000 + 180K

Nghim cn li:

10 180 2 (sin-1

) 3 2 2

Kt qu 65000. Vy nghim cn li l: 65000 + 180K

Gii Phng Trnh Bc Nht Vi Sinx v Cosx

V d 2: Hy biu din sinx + 5 2 5 cosx ra dng csin (x + )

n (pol) 1 (,) 5 2 5

r 3.236067977, 1.256637061

An (Y). Kt qu 2

Y5

.

Vy sinx + 5 2 5 cosx = 3.236067977sin2

x5

.

V d 3: Gii cc pt:

a) 3 cos x sin x 2

b) cos3x sin 3x =1

Gii ( Radian)

a) An (pol) 3 (,) 1

r 2, 0.5235987756

An tip (Y). Kt qu Y6

.

Vy 3 cos x sin x 2 2sin x 26

sin x 16

Gii tng t nh v d 1 ta c nghim ca pt:

84

2x k2

3

4x k2

6

b) An (pol) 1 (,) 1

r 1.414213562, 0.785398163

n tip

(x) . Kt qu x 2

(Y). Kt qu Y4

.

Vy 2 sin 3x 14

cos3x - sin 3x =1

1

sin 3x4 2

Gii tng t nh v d 1 ta c nghim ca pt:

k2x

6 3

k2x

3 3

Tm nghim pt lng gic bt k.

V d 4: Tm nghim ca phng trnh sau trong khong ( /4, 5 /4 )

1 1 10

cos x sin xsin x cos x 3

(1)

Gii

( Radian)

Ghi vo mn hnh biu thc (1) v n (Solve).

My hi x nhp vo 2 ri bm . Kt qu x= 2.9458(radian)

V d 5: Tm mt nghim gn ng thuc ( 0 , 180o ) ca phng trnh: 9sinx

+ 6cosx +3sin2x + cos2x = 8 (2)

85

Gii

( D )

Ghi vo mn hnh biu thc (2) v n (solve)

Nhp 80 cho x ri bm . Kt qu x = 8905659.

( trong khong ny phng trnh ch c 1 nghim)

V d 6: Tm nghim pt:

3cos3x 4x + 2 = 0 (3)

Gii ( radian)

Ghi vo mn hnh biu thc (3) v n (solve)

Nhp 1 cho x ri bm . Kt qu x = 0.5163 (radian).

(Phng trnh ch c 1 nghim)

T Hp

Hon v :Pn=n! Dng phm

V d1: C bao nhiu cch xp 5 bn vo 1 chic bn di.

Gii

S cch xp l P5

An 5 (x!) (120)

Vy c 120 cch.

Chnh hp !

( )!

r

n

n

n rA .Dng phm

86

V d 2: C bao nhiu s nguyn dng gm 6 ch s khc khng v khc nhau

i mt .

Gii

S cc s cn tm l s cc chnh hp chp 6 ca 9 ch s (1,2,..,9): 59A

An 9 (nPr) 5 (15120)

T hp !

!( )!

r

n

n

r n rC

!

!( )!

r

n

n

r n rC .Dng phm

V d 3: C bao nhiu cch chn 3 trong s 10 vin bi ging nhau.

S cch chn l: 310C

An 10 (nCr) 3 (120)

V d 4: C bao nhiu dy nh phn 10 bit ( dy gm10 k t x1 m x1 ch l0

hay1) trong t nht 3 k t 0 v 3 k t 1 ).

Gii

Gi k l s k t 0 th 10 - K l s k t 1.

iu kin k 3 v 10 k 3 3 k 7. Vy c k10C dy nh phn 10 bit c k k

t 0 v 10 - k k t 1.

Vy s dy cn tm l 7

k

10

k 3

C = 912 dy.

Cch tnh 7

k

10

k 3

C nh sau:

An ( )10 (nCr) (x) 3 7

Kt qu: 7

k

10

k 3

C = 912.

87

V d 5: Mt t hc sinh gm 9 nam v 3 n. Gio vin chn 4 hc sinh i trc th

vin. Hi c bao nhiu cch chn nu:

a) Chn hc sinh no cng c.

b) Trong c ng 1 hc sinh n c chn.

c) Trong c t nht 1 hc sinh na c chn.

Gii

a) C 412C 495

b) C 1 33 9C C 252

c) C 4 312 9C C 369

(Cch tnh 412C : Ghi vo mn hnh 12C4 v n )

(Cch tnh 1 33 9C C : Ghi vo mn hnh 3C1 9C3 v n )

(Cch tnh 4 312 9C C : Ghi vo mn hnh 12C4 9C3 v n )

V d 6: C 4 pho tng xp vo 4 trong 6 v tr khc nhau. Hi c my cch sp

xp ?

Gii

C 46A 360 cch.

Ghi 6P4 v n

Nh Thc Newton

V d: Vit h s 5 s hng u tin theo ly tha tng dn ca cc a thc sau:

a) (2x 1)13 b) (2 3x)10

88

c) (1 2

3x)

9

Gii

a) (2x 1)13 = 13

k 13 k k

13

k 0

C (2x) ( 1)

Nhn (7:TABLE)

Nhp vo f(X)= 13 X X13CX(2) ( 1) ri n

Sau nhp 1 vo Star?, 5 vo End?, 1 vo Step?

Sau n my s hin cc gi tr ca F(X) ln lt l 8192, -53248, 159744,

-292864 (lu : gi tr ny my vit tt -2.105 nn phi a con tr vo v tr

), 366080.

y chnh l 5 h s u tin theo ly tha tng dn

b) (2 3x)10 = (3x 2)10=10

k 10 k k

10

k 0

C (3x) ( 2)

Kt qu: 59049, -393660, 1180980, -2099520, 2449440.s

c) (1 2

3x)

9 = (

2

3x + 1)

9 =

9k 9 k k

9

k 0

2C ( x) (1)

3

Kt qu: 512 256 512 1792 448

, , , ,19683 729 243 243 27

Lu : Mn hnh s hin th kt qu i dng rt gn. Do , hin th ng

dng kt qu bn cn a con tr vo ngay v tr cn xem.

Xc Sut

V d 1: Mt bnh ng 16 vin bi gm 7 trng, 6 en v 3 . Ly ngu nhin 10

vin bi. Tnh xc sut c 5 vin trng, 3vin en v 2 vin .

89

Gii

5 3 2

7 6 3

10

16

C C C 45

C 286

( Ghi vo mn hnh 7C5 6C3 3C2 16C10 v n )

V d 2: Chn ngu nhin mt v x s c 5 ch s. Tnh xc sut s ca v

khng c ch s 1 hoc khng c ch s 5.

Gii

Gi A,B ln lt l cc tp hp cc s ko cha s 1 v khng cha s 5.

Ta c:

A B A B A B =95 + 9

5 85

5 5

5

2.9 8P 0.8533

10

Dy s - Cp s cng Cp s nhn

Vi my tnh Casio fx570VN PLUS, cc bi ton b tnh s hng th n, tng hay

tch n s hng u tin ca mt dy s c tnh mt cch d dng

V d 1: Vit 10 s hng u tin ri tnh tng S10 v tch P10 ca 10 s hng y ca

mt dy s c s hng tng qut 3

3n

nun

Gii

Ghi vo mn hnh A=A+1:B=3

3A

A

Vi A l bin m

B l gi tr ca s hng

90

n

My hi A? nhp A=0 v n

My hin A=A+1=1 v ta n (m n=1)

My hin B=3

3A

A=3 n ( 1 3u )

Li tip n

My hi A ? n

My hin A=A+1=2 v n (m n=2)

My hin B=3

3A

A=3 n (

1

9

8u )

Li tip tc n

n my hi A? n

My hin A=A+1=10 v n ( m n=10)

My hin B=3

3 59049

1000

A

A

Vy ta c 10 s hng u tin l 10

59049

1000u

Tnh tng S10

Ta ghi vo mn hnh 10

31

3( )

x

x x v n ta c kt qu 10 116,9492S

Tnh tch P10

Ghi vo mn hnh my tnh 10

31

3( )

x

x xn ta c kt qu P10=3650731,65

91

V d 2: Cho dy cp s cng 10 11

3; ; ;4...3 3

Khng dng cng thc, hy s dng my tnh fx570VN PLUS tnh:

a. S hng th

b. Tng 12 s hng v tch 12 s hng u tin

Gii

Ghi vo my tnh D=D+1:A=A+1

3:B=A+B:C=CA

D: Bin m

A: (S hng trc u1)

B: tng

C: tch

n

Nhp D=0

A=8

3

B=0

C=1

V n nhiu ln cho n khi hin D=12 th A,B,C l kt qu phi tm

Kt qu : u12=20

3

S12=58

P12=113540038,4

V d 3: Cho cp s nhn 80

60;40;3

Khng s dng cng thc, hy s dng my tnh Casio fx570VN PLUS tnh

gn ng

a. S hng th 20

b. Tng 20 s hng v tch 20 s hng u tin

Gii

92

Ghi vo mn hnh my : D=D+1:A=A2

3:B=B+A:C=CA

D: Bim m

A: S hng trc u1

B: Tng

C: Tch

n

Nhp D=0

A=90

B=0

C=1

V n nhiu ln cho n khi hin D=12 th A,B,C l kt qu phi tm l

Kt qu : u20=0,0271

S20=179,4959

P20=127,5516

Ghi ch : Nu u bi ch cho dy s 80

60;40;3

m khng ni r l cp

s nhn th ngi gii c th ngh n dy s vi s hng tng qut l

280 40

3n

nu

nv s i n bi ton khc

V d 4: Tm s hng th 29 v tnh tng 29 s hng u tin ca dy s Fibonaci

Gii

Cch 1 : Dng cng thc tng qut ca dy :