huong dan su dung may tinh bo tui fx 570es

HNG DAN S DUNG VA GIAI TOAN

TREN MAY TNH

H

N

GD

AN

S

DU

NG

VA

GIA

IT

OA

NT

RE

NM

AY

TN

H

1

NGUYEN TRNG CHANG NGUYEN THE THACH

DUNG CHO CAC LP 10 11 12

(Tai ban lan th nhat)

NHA XUAT BAN GIAO DUC

2

385-2008/CXB/6-800/GD Ma so : PTK03n8-LKH

3

MUC LUC

Trang M au 5 Cai at ban au 6 Gi an toan cho may 7 Mode tnh toan va cai at may 11 Nhap bieu thc va gia tr 15

Hien th ket qua dang , 2 23 Tnh toan c ban 26 S dung tnh lien tiep trong phep tnh 30 S dung bo nh phep tnh va xem lai 31 S dung bo nh may tnh 32 S dung chc nang CALS 35

S dung chc nang SOLVE 37 Cac phep tnh ham 40 Chuyen oi gia tr hien th 49 Toan ve so phc 51 Thong ke 54

Toan trong he em c so N 65 Giai phng trnh 68 Ma tran 70 Bang so t mot ham 74 Toan Vect 77

Hang so khoa hoc 79 Thong tin k thuat 83 Nang lng 92 Phu luc hng dan 93

5

M AU

K hieu MATH ch nh dang toan.

K hieu LINE ch nh dang dong phan nhap, xuat.

Cac phm an c at trong o vuong

K hieu SHIFT , ALPHA ch rang phm nay c an trc phm chc nang.

V du :

la chc nang chnh, an trc tiep

mau vang, an sau

D mau o an sau

Phm mau tm (nh i) an trc tiep trong chng trnh a goi (nh CMPLX )

Phm mau xanh luc ( nh HEX ) an trc tiep trong chng trnh a goi (nh la BASE N).

Cac ch trong ngoac sau phm an dung e giai thch y ngha cua phm

V du : ( ) 1 1(sin ) :

co y ngha la an e goi chc nang (arcsin)

Khi menu hien len, muon chon chc nang nao th ta an so ghi trc chc nang ay.

6

V du : Trong menu SETUP (goi bang phm ) An 3 e chon Deg hoac an 2 e chon dang dong khi nhap, xuat. Neu an tiep phm ta c trang menu ke Hai phm , lam hien cac trang menu cung loai. Hoat ong cua con tro c ch ra bi T, S, W, X. Hng dan nay ch dung muc ch minh hoa, oi khi co khac oi vi tng may tren thc te. Noi dung nay co the thay oi ma khong can bao trc. Trong bat k trng hp nao cong ti CASIO se khong chu trach nhiem vi ai ve nhng ton that phat sinh bi viec mua hoac s dung san pham va tai lieu nay. Va hn the na, cong ti CASIO se khong chu trach nhiem ve bat k khieu nai g oi vi viec s dung tai lieu nay bi mot ben khac.

S dung ban phu luc ri Khi thay co ghi chu Phu luc th phai xem them bang phu luc kem hng dan nay. V du khi thay #021> th tra cu muc #021 cua phu luc.

Deg : Ch n v o goc la o. Rad : Ch n v o goc la radian.

CAI AT BAN AU Phai thc hien thao tac sau e cai ban au (mac nh) cho may : Phai an nh sau e xoa tat ca d lieu nh hien hanh. 9 (CLR) 3 (ALL) (Yes) Xem them phan cai at may ban au ve MODE (phan sau) Xem them ve so nh (phan sau)

Y ZU

V

7

GI AN TOAN CHO MAY Phai oc cac ieu nay trc khi s dung may va gi lai e nghien cu ve sau

! Can than Dau hieu nay thong bao co the gay ton thng hoac hong may neu khong chu y.

Pin Sau khi thao pin ra khoi may, hay cat vao ni an toan, xa tam tay tre em e tranh tre em nuot phai. Neu tre em bat ng nuot phai pin, hay a ngay en bac s.

Khong c sac pin lai, hay lay pin ra khi b yeu. Khong c bo pin vao cho nong hay ot pin. S dung pin khong ung cach de ro r va h hong cac vat e gan, co the gay hoa hoan hay thng tch. Luon at pin ung cc khi lap vao may.

Ch s dung ung loai pin ghi trong hng dan.

Huy may tnh

Khong c huy may tnh bang cach ot bo v khi lam the mot so linh kien co the gay no mot cach bat ng tao rui ro hoa hoan va thng tch.

CAN THAN KHI S DUNG Luon an phm ON khi s dung may.

Tham ch khi may van hoat ong bnh thng, hay nen thay pin t nhat 2 nam mot lan. Pin chet co the ro r gay h hong va tnh toan sai. Khong c e pin het nang lng trong may.

8

Pin kem theo may co the b giam nang lng trong qua trnh van chuyen va bao quan. V the nen thay pin sm hn tuoi tho pin. Pin yeu co the lam cho noi dung bo nh b hong hoac hoan toan b mat i. Hay luon gi so lieu quan trong bang van ban. Tranh s dung va e may trong moi trng nhiet o thai qua. Nhiet o qua thap co the gay nen cham hien th hay hoan toan khong hien th va lam giam tuoi tho cua pin. Tranh e may tiep xuc trc tiep vi anh sang mat tri, gan ca so, lo si hay bat c ni nao nhiet o cao. o nong co the gay bien mau, bien dang vo may va h hong cac mach ien t ben trong. Tranh s dung va cat may nhng ni co o am cao va bui bam. Can than khong c e may b nc ri vao hay at ni co o am cao va bui bam. Nhng ieu kien nh vay co the gay h hong mach ien ben trong. Khong c lam ri may hay tac ong manh vao may. Khong c van hay be cong may. Tranh bo may vao tui quan hay nhng ni chat choi cua quan ao v no co the lam van va cong may. Khong c thao th may ra. Khong c an phm bang au but bi hay vat nhon. Dung vai mem, kho e lau sach ben ngoai may. Neu may b d, lau sach bang vai hi am vi mot t bot giac trung tnh. Vat that rao trc khi lau sach. Khong c s dung chat pha sn, ben zen hay cac tr chat de bay hi e lam sach may. Neu lam nh vay se mat i lp in va co the lam hong vo may.

TRC KHI S DUNG MAY TNH

M nap may Lat may lai (pha lng len tren : thay ro c sau lo inh oc) dung ngon tay cai ay may len e lay nap ra.

9

e mat phm quay len, at nap pha di va ay len cho sat lai.

M va tat may An e m may An (OFF) e tat may

Chnh o tng phan An (SETUP) 6 (W CONT X) Man hnh chnh o tng phan hien len. Dung phm W, X e chnh o sang toi. Khi at mc o va y th an AC

[ ] [ ]W X

CONTRAST

LIGHT DARK

Cung co the chnh o tng phan bang W, X khi menu MODE hien len Ghi chu : Khi chnh o tng phan ma khong thay hieu ng tren man hnh la pin a yeu, hay thay pin ngay.

Man hnh hien th Man hnh LCD co 31 96 iem e hien cac k hieu.

K hieu hien th V du : CMPLX D

10

K hieu Y ngha

S Va an phm SHIFT. Neu an tiep mot phm khac na th k hieu nay lan

A Va an phm ALPHA. Neu an tiep mot phm khac na th k hieu nay lan

M Co so nh M c dung

STO Va an SHIFT STO (chuan b nhap gia tr vao ten bien)

RCL Va an phm RCL (chuan b goi gia tr a gan trc)

STAT ang mode thong ke STAT

CMPLX ang mode so phc

MAT ang mode ma tran

VCT ang mode vect

D Mac nh n v o goc la o

R Mac nh n v o goc la radian

G Mac nh n v o goc la grad

FIX Co chon so ch so le thap phan

SCI Co chon so ch so hien len dang thap phan

Math ang dang math

T, S Co dong d lieu hng ang ch Disp Con ket qua tiep theo

Ghi chu : Vi vai bai toan phc tap, ket qua hien cham, trong khi ch i phep tnh ang thc hien, man hnh khong hien g

11

MODE TNH TOAN VA CAI AT MAY

Mode tnh toan: Yeu cau Mode chon

Tnh toan chung COMP

Toan so phc CMPLX

Thong ke va hoi quy STAT

He em c so N BASE N

Giai phng trnh EQN

Ma tran MATRIX

Lap bang so theo bieu thc TABLE

Toan vect VECTOR

Cach chon mode: (1) An e hien menu

1 : COMP 2 : CMPLX 3 : STAT 4 : BASE N 5 : EQN 6 : MATRIX 7 : TABLE 8 : VECTOR

(2) An so tng ng trc ten mode muon chon

V du : e chon mode CMPLX, an 2

Cai at may : An (SETUP) e hien menu cai at cho tnh toan va hien th. Man hnh gom hai trang, chuyen nhau bang T S

1 : MthIO 2 : LineIO 3 : Deg 4 : Rad 5 : Gra 6 : Fix 7 : Sci 8 : Norm

12

U V 1 : ab/c 2 : d/c 3 : CMPLX 4 : STAT 5 : Disp 6 : W CONT X

(Xem them phan chnh o tng phan khi dung W CONT X) Xac nh dang nhap / xuat

Dang An

Math (MthIO)

Linear (LineIO)

dang Math, phan so, so vo t va cac bieu thc c ghi giong sach giao khoa. dang Line, phan so va cac bieu thc c ghi chung mot dong.

D Math S 4 25 3+

2215

Dang Math

D S 4 5 2 3+ 22 15

Dang Line

Xac nh n v o goc

n v chon An o (Deg)

Radian (Rad)

Grad (Gra)

13

Xac nh dang so hien th

Goc o90 goc2= = goc 100 grad

Dang so hien th An

Co an nh so ch so le thap phan (Fix) 0 9

Co an nh so ch so hien th (Sci) 0 9

Dang thng (Norm) 1 (Norm 1) hay 2 (Norm 2)

V du hien th ket qua Fix : Xac nh tiep t 0 en 9 e an nh so ch so phan le thap phan. Ket qua hien th c ghi tron so cuoi V du : 100 7 = 14.280 (Fix3) 14.29 (Fix2)

Sci : Xac nh tiep t 0 en 9 e an nh so ch so hien len. Ket qua hien th ghi tron so cuoi

V du : 1 7 = 1.4286 110 (Sci 5) 1.429 110 (Sci 4) Norm : Co hai dang (Norm1, Norm 2) ghi so dang thng trong gii han an nh, ngoai gii han th ghi thanh na 10 Norm 1 : 2 1010 x 10 < Norm 2 : 9 1010 x 10 < V du : 1 200 = 35 10 (Norm1) 0.005 (Norm 2)

Xac nh hien th phan so va hon so

Dang so hien th An

Dang hon so T (ab/c) Dang phan so T (d/c)

14

Xac nh dang hien th so phc

Dang so hien th An

Dang ecac (CMPLX) (a+bi)

Dang toa o cc (CMPLX) (r ) Xac nh dang hien th bang thong ke

Thao tac sau lam hien hay mat cot tan so

Dang hien th An

Hien cot tan so (STAT) (ON)

An cot tan so (STAT) (OFF)

Xac nh dang hien th dau cach phan le so thap phan

Dang hien th An

Dau cham (Dot) (Disp) (Dot)

Dau phay (Comma) (Disp) (Comma)

S xac nh nay ch co tac dung dong ket qua. Khi nhap van phai dung dau cham (.) e ngan cach phan nguyen va phan le thap phan.

Cai at ban au Thc hien thao tac sau e lap cai at ban au (CLR) (SETUP) (Yes) Chi tiet cai at Trang thai ban au Mode COMP Dang xuat/nhap MathIO n v goc o Hien th so Norm 1 Hien th phan so d/c Dang so phc a+bi Hien th thong ke OFF Dau cach phan le thap phan Dot (.) Muon bo qua cai at, an (Cancel).

15

NHAP BIEU THC VA GIA TR

Nhap bieu thc tnh theo nh dang chuan May cho phep nhap bieu thc tnh nh ghi tren giay roi an e thc hien. May t ong thc hien theo th t u tien cua cac phep cong, tr, nhan, chia, ham so va dau ngoac.

V du : 2 (5+4)2(3) = LINE 2 5 4

2 3 2(5 4) 2 ( 3)

24+

Nhap ham co dau ngoac Khi nhap cac ham sau th co dau m ngoac t ong i kem. Khi nhap bien xong ta phai ong ngoac

1 1 1

1 1 1 ^ ^

3

sin( , cos( , tan( , sin ( , cos ( , tan ( , sinh( , cosh(,

tanh( , sinh ( , cosh ( , tanh ( , log( , ln( , e ( , 10 ( ,

( , ( , Abs( , Pol( , Rec( , ( , d / dx( , ( , P( , Q( , R(,

arg( , Conjg( , Not( , Neg( , det( , Trn( , Rnd( .

V du : sin 30 = LINE

An sin 30 Ghi chu : Thao tac nay co khac trong dang Math (xem nhap trong dang Math)

Bo qua dau nhan Ta co the bo qua (khoi an) dau nhan () trong cac trng hp sau

Trc dau m ngoac : 2(5+4) ch ghi 2(5+4)

Trc ham co m ngoac : 2sin(30) ch ghi 2sin(30)

D SSin(30)

0.5

16

Trc k hieu m au (gom ca dau so am) : 2h123 ch ghi 2h123 Trc ten bien, hang hay so ngau nhien : 2A, 2 ch ghi 2A, 2

Ghi chu quan trong : May ES khong danh u tien cho phep nhan tat nen neu ghi 32A th may hieu la 32A va thc hien t trai sang phai (khac vi mot so ho may khac)

V du : Ghi 32 th may ES hieu 32 = 3

2 = 4.71238898

Dau ong ngoac cuoi cung cua bieu thc Mot hay nhieu dau ong ngoac cuoi cung (se an tiep theo) co the bo qua (khoi an).

Hien bieu thc dai May co the hien th toi a 14 k t. Khi nhap k t th 15 th bieu thc b ay qua trai va co dau W hien len ben trai (chng to ben trai con co k t). Khi a con tro ve trai neu lam khuat k t ben phai th ben phai co hien dau X.

So lng k t nhap (Byte) Ta co the nhap toi a 99 byte cho bieu thc n. C ban moi

phm la mot byte, moi ham gom hai phm an nh 1sin (an ) la mot byte. Trong dang Math moi mau ghi s dung hn 1 byte (xem them phan nhap dang Math). Thong thng con tro la vach ng hay ngang nhap nhay

dong bieu thc. Khi con 10 k t tr xuong th con tro chuyen thanh hnh vuong e bao cho biet. Gap dau hieu bao nay cua con tro hay ket thuc nhap tai iem thch hp va an .

Chnh sa bieu thc Nhn vao bieu thc nhap, neu thay sai ta co the chnh sa bang cach ghi chen hay ghi e

17

Ghi chen hay e khi nhap che o ghi chen, k t mi c chen ben trai con tro. che o e, k t mi an se thay the k t tai con tro. Mac nh la che o chen, ta co the oi sang che o e neu can.

che o chen con tro dng ng (|), che o e con tro nam ngang ( ). Mac nh ban au Line la che o chen, ta co the thay oi

bang cach an (INS). Trong dang Math, ta ch co the che o chen, an (INS)

cung khong sang c che o e (Xem them phan gia tr cua ham). May t ong thay oi sang che o chen khi ta thay oi dang

xuat/nhap t Line sang Math.

Thay oi k t va ham va nhap V du : chnh sa t 369 13 thanh 369 12

LINE

369 13 D

369 13|

D

369 1|

D

369 12|

Xoa k t va ham khi nhap

V du : chnh sa t 369 12 thanh 369 12

LINE che o e

369 12 D

369 12|

D

369 |12

18

D

369 |12

Che o e :

369 12 D

369 12 _

D

369 12

D

369 12

Chnh phep tnh V du : chnh sa t cos(60) thanh sin(60)

LINE che o chen

cos 60 ) D

cos(60)|

D

60)

D

sin(|160)

Che o e :

60 D

cos(60) _

D

c os(60)

D

sin(60)

19

Nhap chen vao phep tnh Dung che o chen cho thao tac nay. Dung phm hay a con tro en ni can chen roi an k t muon chen.

Hien th cho sai Khi an ma thay may bao loi, an , con tro se hien ngay cho co loi, ta chnh sa ngay

V du : Khi nh an 14102 = ma lai an sai 1402 = Thao tac nay trong che o chen nh sau

LINE

14 0 2

An hay D14 0| 2

cho co loi 1 D

14 0|0 2

D

14 10 22.8

Neu an th phep toan b xoa.

Nhap dang Math Khi nhap dang Math, ta co the cho nhap va hien th phan so va mot vai ham so giong nh trong sach. Ghi chu : Vai bieu thc hay k hieu cong thc ln hn mot dong. Chieu

cao toi a co the en hai man hnh (31 iem 2). Vt qua gii han nay, khong nhap c.

DMath ERROR [AC] : Cancel [W][X] : Goto

20

Co the ghep ham va dau ngoac n vao. Neu ghep ham va ngoac n qua nhieu th may khong thc hien c, gap trng hp nay phai chia phep tnh thanh cac phan nho e tnh rieng.

Nhap ham so va k hieu trong dang Math Cot Byte cho biet so byte s dung trong bo nh

Ham hoac k hieu Phm an Byte

Phan so

9

Hon so

13

alog b log 6 ^10 x log (10 ) 4

^e x ln (e ) 4

Can bac hai 4

Can bac ba ( )3 9 Bnh phng, lap phng , 3(x ) 4

Nghch ao 5

Luy tha x 4

Can so x ( ) 9

Tch phan 8 Gia tr ao ham d

dx

6

Tnh tong log

=

8

Gia tr tuyet oi hyp (Abs) 4

Ngoac n (hoac) 1

21

Cac v du nhap dang Math Nhng thao tac sau c dung trong dang Math.

Phai chu y en vung anh hng (do kch thc) cua con tro hien th tren man hnh khi nhap d lieu trong dang Math.

V du 1 : Nhap 32 1+ MATH

2 x 3 3|D Math

2

1 3D Math

2 1|+

V du 2 : Nhap 1 2 3+ + MATH

+ 2 D Math

1 2|+

3 D Math

1 2 3|+ +

V du 3 : Nhap 221 2

5 + =

MATH

1

2 5

2 Khi an va nhan ket qua trong dang Math, mot phan cua bieu thc khi nhap vao co the b che khuat. Neu can xem lai bieu thc a nhap vao, an va sau o an

2D Math

21 25

9825

+

S

22

Them k hieu vao bieu thc tnh Trong dang Math, co the a mot bieu thc a nhap (k hieu, bieu thc co dau ngoac, v. v..) vao trong mot ham

V du : a (2 + 3) cua bieu thc 1 + (2 + 3) + 4 vao trong MATH

D Math

1 (2 3) 4+ + +

Di chuyen con tro en ay

SHIFT (INS)

Xem hnh dang con tro thay oi ay

An D Math

1 (2 3) 4+ + +

(nhom (2+3) c a vao dau )

Neu con tro ang ben trai cua mot gia tr ac biet hay phan so (co dau m ngoac) th ca khoi nay c a vao ham c dung. Neu con tro ben trai cua mot ham th ham nay c a vao

k hieu chc nang c dung.

Bieu thc goc 1 + |(2 + 3) + 4

Ham Phm an Bieu thc tao mi

Phan so

| (2 3)1 4++ +

alog b log |1 log+ ((2 + 3)) + 4

Can so ( )x |1 (2 3) 4+ + +

D Math 1 + (2+3)+4

23

Bieu thc goc 1 + | (X + 3) + 4

Ham Phm an Bieu thc tao mi

Tch phan | (X 3)dx 41 + ++ ao ham d

dx

d1 (|(X 3)) x

dx+ + =

Toan log

=

x1 (|(X 3)) 4

=+ + +

Cach them nay cung dung c cho cac ham sau

(10 ) , (e ), , x ,

( )3 , (Abs) HIEN TH KET QUA CAC DANG , 2 , ... (DANG SO VO T)

Khi nhap/xuat trong dang MthIO th ta co the nhan c ket qua dang va 2 (so vo t) An e nhan ket qua dang so vo t.

An e nhan ket qua dang so thap phan.

Ghi chu Khi nhap/xuat trong dang LineIO phep toan luon cho ket qua thap phan du an hay ieu kien hien th dang (gom ca so vo t) giong nh viec chuyen oi S D (xem them phan chuyen S D) V du : 2 8 3 2+ = MATH

1. 2 8

D Math S2 8+

3 2

24

2. 2 8

V du 2 : sin(60) = 32

MATH

60

V du 3 : 1 1sin (0.5)6

=

MATH

1(sin ) 0.5 =

Cac phep tnh dang (bao gom so vo t) dung c cho

a. Phep tnh so hoc vi k hieu 2 3 1( ), x , x , x . b. Ham lng giac c. Tnh suat cua so phc d. So phc dang cc (r ) Nhng dang sau luon cho ket qua dang can

n v goc Gia tr nhap Gii han

o Boi so cua o15 6x 9 10< Rad Boi so cua

12 x 20<

Gra Boi so cua 503

x 10000<

Khac vi an nh nay ket qua se hien bang so thap phan.

D Math S2 8+

4.242640687

D Math Ssin(60

32

D Math S1sin (0.5

16

25

Ghi chu Trong tnh toan so phc (CMPLX), cac ieu kien sau c ap dung cho phan thc va phan ao. Ket qua dang can ch c hai so hang (mot so nguyen cung c tnh la mot so hang) va gom cac dang sau :

a b, d a b, a b d ec f

Moi he so phai nam trong pham vi quy nh sau : 1 a < 100, 1 < b < 1000, 1 c < 100 0 d < 100, 0 e < 1000, 1 f < 100 V du :

2 3 4 8 3 = Dang can thc 35 2 3 = 148.492424 ( 105 2)=

Dang thap phan

150 2 8.48528137425

= Dang thap phan

2 (3 2 5) 6 4 5 = Dang can thc 23 (5 2 3) 35.32566285 = ( 115 46 3)=

Dang thap phan

10 2 15 3 3 45 3 10 2+ = + Dang can thc 15 (10 2 3 3) + = 290.02743207 ( 45 3 150 2)= +

Dang thap phan

2 3 8 3 3 2+ + = + Dang can thc 2 3 6 5.595754113+ + = Dang thap phan

Phan gach di ch cho a gay ra ket qua dang thap phan.

26

Nhng l do a lam cho ket qua hien th dang thap phan Gia tr vt qua pham vi cho phep. Nhieu hn hai so hang trong ket qua. Ket qua tnh hien th dang can thc c quy ong mau so

a b d e a b d ec f c

++

c' la boi so chung nho nhat cua c va f

Ket qua c quy ong mau so van hien dang can thc khi cac so a, c va d vt qua pham vi cua cac so a, c va d V du :

3 2 10 3 11 211 10 110

++ = Ket qua hien th dang thap phan khi co ket qua trung gian cha hn hai so hang.

V du : (1 2 3)(1 2 3)+ + = 4 2 6 = 8.898979486 Neu co mot so hang khong bieu th c bang hay phan so th ket qua se hien dang thap phan

V du : log 3 + 2 =1.891334817

TNH TOAN C BAN

Phan nay se trnh bay tnh toan vi so hoc, phan so, phan tram va o, phut, giay.

Tat ca cac muc tnh toan trong phan nay eu c thc hien Mode COMP ( )

Phep tnh so hoc Dung cac phm , , , e thc hien phep tnh so hoc

27

V du : 7 8 4 5 = 36

LINE

7 8 4 5 D

7 8 4 536

S

May t ong tnh theo th t u tien cac phep tnh (Xem them phan Th t u tien cac phep tnh)

So ch so le thap phan va so ch so hien len Ta co the an so ch so le thap phan hay so ch so hien len phan ket qua

V du : 1 6 LINE

Cai ac mac nh (Norm 1) D

1 60.1666666667

S

An nh 3 so le thap phan (Fix 3) D FIX

1 60.167

S

An nh 3 so dang na 10 (Sci3) 1

D SCI1 6

0.167 10

S

(Xem them phan an nh dang so hien len)

Bo qua dau ong ngoac cuoi Ta co the bo qua dau ong ngoac cuoi e an ngay dau thc hien phep tnh

V du : (2 + 3) (4 1) = 15

LINE

2 3 4 1 + D

(2 3) (4 115

S

28

Toan phan so Ta co the chon cach nhap/xuat phan so theo hai dang sau

Dang Phan so Hon so

Math 73

7 3

T

123

(SHIFT

2 X 1 T 3 )

Line 7 3

7 3

2 1 3

2 1 3

Tuy theo cai at ban au, may se hien dang phan so hay hon so Ket qua luon hien dang phan so toi gian Phu luc

2 1 73 2 6+ =

1 2 113 1 44 3 12+ = (dang hon so ab/c)

1 14 32 2

= (dang hon so ab/c) Neu tong cac k t (so nguyen, t, mau, dau cach) ln hn 10 th ket qua hien len la so thap phan. Ket qua mot phep tnh gom phan so va so thap phan luon la so thap phan.

Chuyen oi nhau gia phan so va hon so

An phm S D b d(a )c c e oi cach ghi phan so va

hon so.

29

Chuyen oi nhau gia phan so va so thap phan

D

3 21.5

S S D

D

3 23 2

S

Dang cua phan so tuy thuoc vao dang cai at ban au (phan so hay hon so). Khong the oi mot so thap phan ra dang hon so neu tong cac k t cua hon so ln hn 10.

Ve chi tiet cua phm S D xin xem them phan chuyen oi S D.

Toan phan tram Nhap gia tr va an (%) e lam gia tr nay tr thanh phan tram.

Phu luc

2% = 0.02 2( )100

150 20% = 30 20150100

Tnh : 660 la may phan tram cua 880 (75%)

Tnh : 2500 + 15% cua 2500 (2875)

Tnh : 3500 25% cua 3500 (2625)

Giam 20% cua (168 + 98 + 734) th thanh bao nhieu ? (thanh 800)

(500 + 300) la may phan tram cua 500 ? (160%)

40 tr thanh 46 va 48 la a tang bao nhieu phan tram (oi vi 40) (15%, 20%)

30

Toan ve o (gi), phut, giay May co the thc hien cac phep toan nay va chuyen oi thanh so thap phan hay ngc lai.

Nhap gia tr Dung pha nhap o (gi), phut, giay

Phu luc nhap o2 0 30 Phai nhap ay u cac phan o (gi), phut du gia tr cua no la 0

Cac phep tnh Thc hien cac phep toan Cong, tr cac gia tr ve o, phut, giay. Nhan, chia gia tr ve o, phut, giay vi so thap phan Tnh t so hai gia tr o, phut, giay

Phu luc o o2 20 30 39 30 3 00 00 + = Chuyen oi nhau gia o, phut, giay va so thap phan

Dung phm e chuyen oi nhau gia gi, phut, giay va so thap phan Phu luc oi 2.255 ra dang o, phut, giay.

S DUNG TNH LIEN TIEP TRONG PHEP TNH Co the dung dau hai cham (:) e noi 2 hay nhieu bieu thc va thc hien tng phan t trai qua phai khi an

V du : e tnh lien tiep hai phep tnh 3 + 3 va 3 3

LINE

3 3

(:) 3 3 D3 3 : 3 3|+

+D Disp

3 36

S

31

Disp bieu th ay la mot ket qua trung gian cua mot phep tnh lien tiep

D

3 39

S

S DUNG BO NH PHEP TNH VA XEM LAI

Bo nh phep tnh ghi moi bieu thc a nhap vao, thc hien va ket qua cua no.

Ch co the s dung bo nh phep tnh tnh trong Mode COMP (MODE 1)

Goi noi dung bo nh phep tnh An e lay tng d lieu t bo nh phep tnh. Bo nh phep tnh cho xem ca bieu thc phep tnh va ca ket qua cua no.

V du :

LINE 1 1

2 2 3 3

+D

2 24

ST

+D

1 12

T

Lu y rang noi dung bo nh phep tnh se b xoa khi tat may tnh, an phm ON , chuyen sang mode phep tnh hoac dang nhap/xuat hoac thao tac bat c hoat ong reset nao.

+D

3 34

S

32

Bo nh phep tnh la co han. Khi phep tnh ang thc hien lam cho bo nh ay, phep tnh th nhat t ong b xoa e lay cho cho phep tnh mi.

Chc nang hien lai Khi mot ket qua tnh toan ang hien th, co the an phm sau o an hoac e bien tap bieu thc a dung ma a tnh toan trc o. Neu ang s dung dang LINE, co the hien th bieu thc trc o bang cach an phm hoac ma khong can an trc.

Phu luc

S DUNG BO NH MAY TNH

Ten bo nh Mieu ta Bo nh Ans Lu lai ket qua phep tnh cuoi cung. Bo nh oc lap Ket qua phep tnh co the cong hoac tr

vi bo nh oc lap. Hien th M ch ra co d lieu trong bo nh oc lap.

Cac bien so Sau bien so A, B, C, D, X va Y co the dung e lu cac gia rieng.

Phan nay s dung Mode COMP ( ) e hng dan s dung bo nh nh the nao.

Bo nh (Ans) Mieu ta ve bo nh (Ans) Noi dung bo nh Ans c cap nhat bat c khi nao lam mot phep tnh s dung mot trong cac phm sau : , , ,

RCL, RCL (STO). Bo nh co the gi ti 15 ch so. Noi dung bo nh Ans khong thay oi neu co loi trong viec va thc hien phep tnh. Noi dung bo nh Ans van con ngay ca khi an phm , thay oi mode phep tnh, hoac tat may.

33

Dung bo nh Ans e thao tac mot so phep tnh

V du : Lay ket qua cua 3 4 roi chia cho 30

LINE 3 4

(Tiep tuc) 30 D

Ans 300,4

S

An t ong nhap vao lenh Ans

Vi thao tac tren, can thc hien phep tnh th 2 ngay sau phep tnh th nhat. Neu can goi noi dung bo nh Ans sau khi an , an tiep .

Nhap noi dung bo nh Ans vao mot bieu thc V du : e thao tac phep tnh sau ay 123 + 456 = 579 789 579 = 210

LINE

1 2 3 4 5 6 +D

123 456579

S

7 8 9 D

789 Ans210

S

Bo nh oc lap (M) Co the lam phep tnh cong them hoac tr i ket qua trong bo nh oc lap. Ch M hien th khi bo nh oc lap co lu mot gia tr.

D3 4

12

S

34

Mieu ta chung ve bo nh oc lap Sau ay la tom tat mot so thao tac co the s dung bo nh oc lap.

Y ngha An phm

Them gia tr hoac ket qua hien th cua bieu thc vao bo nh oc lap

Bt i gia tr hoac ket qua hien th cua bieu thc t bo nh oc lap

(M)

Goi noi dung bo nh oc lap gan nhat RCL (M)

Cung co the chuyen bien so M vao mot phep tnh, yeu cau may tnh s dung noi dung bo nh oc lap tai v tr o. Di ay la cach an phm e chuyen bien so M. (M) Ch M hien pha tren ben trai khi co mot gia tr nao o khac 0 c lu trong bo nh oc lap. Noi dung bo nh oc lap van con ngay ca khi an phm AC thay oi mode tnh toan, ke ca khi tat may.

Cac v du s dung bo nh oc lap Neu ch M hien th th thao tac Xoa bo nh oc lap trc khi thc hien cac v du nay. V du : 23 + 9 = 32 23 + 9 M+ 53 6 = 47 53 6 M+ 45 2 = 90 45 2 (M)

99 3 = 33 99 3 M+ (Cong) 22 RCL (M)

Xoa bo nh oc lap An 0 RCL (STO) M+. ieu o se xoa bo nh oc lap va lam cho ch M lan.

35

Cac bien A, B, C, X, Y

Mieu ta chung ve bien Ban co the cho mot gia tr hoac mot ket qua vao bien

V du : Cho ket qua cua 3 + 5 vao bien A

3 + 5 RCL (STO) (A)

S dung thao tac sau khi ban muon kiem tra noi dung cua bien

V du : e goi noi dung cua bien A

RCL (A)

Di ay cho biet a bien vao trong bieu thc nh the nao

V du : Nhan noi dung cua bien A vi noi dung cua bien B (A) (B)

Noi dung cua bien van con ngay ca khi an phm AC thay oi mode phep, ke ca khi tat may.

Phu luc

Xoa noi dung cua toan bo nh S dung cac thao tac sau e xoa noi dung cua bo nh Ans, bo nh oc lap va tat ca cac bien. An phm 9 (CLR) 2 (Memory) (Yes) e huy hoat ong xoa ma khong can lam g khac, an (Cancel) thay cho .

S DUNG CHC NANG CALC

Chc nang CALC cho phep ta nhap bieu thc vi bien, sau o nhap gia tr bien e tnh.

Chc nang CALC s dung c trong mode COMP (MODE 1) va mode CMPLX (MODE 2)

36

Thao tac thc hien : Cac loai bieu thc : Bieu thc cha bien

V du : 2X + 3Y, 5B + 3i, 2AX + 3BY + C Tnh lien tiep

V du : X + Y : X(X+Y) Bieu thc co dau va cha bien

V du : Y = 2X, A = 2X X 3+ +

V du ve s dung phm V du : LINE

3 (A) D

3 A

Yeu cau nhap gia tr A

Gia tr A hien hanh

5 D

3 A15

S

(hoac ) D

A ?5

S

10 D

3 A30

S

Thoat an .

Neu bieu thc co hn mot bien, may yeu cau nhap tng bien.

DA ?

0

37

Phu luc Tnh 2a en 5a biet n 1 na a 2n+ = + (vi 1a 1)= (Ket qua 2 3a 3,a 7,= = 4 5a 13,a 21)= = * 1 gan 1 cho 1a ; *2 gan 1 cho n *3 gia tr cua 2a *4 gan gia tr cho 2a *5 gan 2 cho n *6 gia tr cua 3a

*7 gia tr cua 4a *8 gia tr cua 5a

CHC NANG SOLVE (COMP)

Chc nang SOLVE dung phng phap Newton e tm nghiem gan ung cua phng trnh Chc nang SOLVE ch dung mode COMP ( ) Cu phap : Phai theo cu phap sau cho an so can tm V du : Y = X + 5, Y (Tm Y). XB = C + DB (Tm B) Cu phap cho ham Logarit Y = X log(2) (Khi an so la X th khoi ghi nh Y = X log(2 co ngha la Y = X 10log 2 theo an X) Y = X log(2, Y (an la Y c ghi ro) Y = X log(2,Y) (th an lai la X c hieu ngam) Ngoai cac nhan manh nh tren, cac phng trnh eu theo an X V du : Y = X + 5, X = sin(M), X + 3 = B + C XY + C (Viet gon cua XY + C = 0) Chc nang SLOVE khong dung c vi cac phng trnh cha tch phan, ao ham, chc nang (, Pol(, Rec( hay tnh lien tiep. Co thong bao khi phng trnh khong co bien.

38

Thao tac mau cho chc nang SOLVE V du : Giai 2y ax b= + theo x khi y = 0, a = 1, b = 2. MATH S D (Y) (=) (A) (X) (B)

(,) (X) 2D Math

Y AX B,X= +

(SOLVE) D Math

Y ?0

yeu cau nhap gia tr Y gia tr Y hien hanh

0 D Math

A ?0

1 D Math

B?0

2 D Math

Solve for X0

gia tr X hien hanh

2

D Math

Y AX B,XX 1,414213562L R 0

= += =

Thoat SOLVE, an Ghi chu : Lenh SOLVE co the khong at ket qua do gia tr ban au (gia tr d oan cua an) khong thch hp. Neu muon ta thay oi gia tr ban au roi giai lai.

39

Lenh SOLVE cung co the khong at ket qua du phng trnh co nghiem. Lenh SOLVE (giai bang phng phap Newton) ch hien 1 lan 1 mot nghiem du phng trnh co nhieu nghiem. Lenh SOLVE (giai bang phng phap Newton) co the giai c cac phng trnh : Co ham tuan hoan

Ham ma o th co doc ng (nh x 1y e , y )x

= = Ham khong lien tuc (y x= , )

Man hnh ket qua

2

Phng trnh

D Math

Y AX B,XNghiemsoAn so X 1,414213562

L R 0

canh trai canh phai

= + =

=

Neu LR la 0 th o chnh xac cao.

Man hnh goi tiep tuc Lenh SOLVE thc hien phep hoi tu t gia tr ban au. Khi khong tm c nghiem ma man hnh hien continue [=] e hoi co muon tiep tuc hay khong? An neu muon tiep tuc, an muon ngng.

Phu luc Giai phng trnh 2y x x 1= + khi y = 3,7,13 va 21. (Nghiem x = 2, 3, 4, 5 khi y = 3, 7, 13, 21 theo th t) *1 gan 3 cho y *2 gan 1 cho x

40

CAC PHEP TNH HAM

Phan nay giai thch viec s dung cac ham cai san cua may tnh.

Cac ham cai san phu thuoc vao mode phep tnh ang dung. Phan nay giai thch ve cac ham cai san ay. Tat ca cac v du

trong phan nay eu mode COMP (MODE 1)

Cac phep tnh can co mot thi gian e hien th ket qua. Khi ang tnh d dang co the cho lenh ngng bang phm

Pi () va logarit t nhien hay logarit nepe (c so e) Co the nhap pi () hoac logarit t nhien vao may tnh Sau ay la cach an phm e co c cac gia tr cua pi() va e = 3.14159265358980 ( x10 ()) e = 2.71828182845904 ( x10 (e))

Dung va e trong moi mode, ch tr Base N

Ham lng giac va lng giac ngc Ham lng giac va lng giac ngc dung trong cac mode

COMP, STAT, EQN, TABLE. Cac ham nay cung dung c trong mode CMPLX (so phc) neu cac ham nay bo tr hp l n v goc ma cac ham lng giac va lng giac ngc mac

nh co san tren may. Trc khi tnh can chon n v goc muon dung. Xem them n v goc mac nh

Phu luc sin 30 = 0,5, 1sin 0.5 30 =

Hyperbolic va hyperbolic ngc An phm e hien th menu cua ham hypebolic. An phm tng ng e goi ham muon nhap

Phu luc sinh 1 = 1.175201194, 1cosh 1 0 =

41

oi n v goc (Ket qua theo n v goc mac nh cua may)

Sau khi a nhap gia tr vao may, an (DRG ) e hien th cac n v goc ang nhap nh menu xuat hien di ay. An phm so lien he e ghi n v goc cua gia tr nhap vao. May tnh se t ong chuyen sang n v goc mac nh tren may.

o1 : r2 :

g3 :

V du : Chuyen gia tr sau sang o :

Goc 2 radian = goc o90 , goc 50 grad = goc o45

Thao tac sau vi n v goc mac nh cua may tnh la o

LINE

x10 () 2 (DRG ) 2 r( ) 50 (DRG ) 3 g( )

Phu luc cos ( radian) = 1, cos(100 grad) = 0. 1cos ( 1) 180 = 1cos ( 1) =

Cac ham mu va ham logarit Ham mu va ham logarit co the dung nh ham lng giac

Vi ham logarit log (vi c so m dung cu phap)

log(m, n).

r

D

( 2)90

S

g

D

5045

S

42

Neu ch nhap mot gia tr n, may hieu la logarit thap (v du log(100) = 2) ln(ham logarit t nhien vi c so e.

Cung co the dung phm log khi nhap bieu thc vi dang log(m,n) dang Math. Xin xem them chi tiet

Phu luc lu y phai nhap c so (c so m) khi dung phm log e nhap.

Phu luc en *1 C so 10 (logarit thap phan) se c dung neu khong co c so nao c nhap vao.

Cac ham luy tha va cac ham can Ham luy tha va ham can co the dung mode COMP (Tnh

toan thong thng), STAT (Thong ke va hoi quy), EQN (Giai phng trnh), MATRIX (Ma tran), TABLE (Lap bang) va VECTOR (Toan vect)

Ham bnh phng 2(x ) , lap phng 3(x ) , nghch ao 1(x ) dung c trong so phc (CMPLX).

x , 3 (, (, Dung c trong Mode CMPLX khi cac ham nay ch anh hng en rieng tng phan thc va ao (hay mot va giac so), (khong anh hng toan bo mot so phc)

Phu luc en

Phep tnh tch phan May thc hien phep tnh tch phan theo phng phap

Gauss Kronrod bang cu phap (f (x),a, b, tol) f(x) : ham bien x (neu khong co cha x th la hang so) a : Can di b : Can tren tol : Gii han (khi nhap xuat Line)

43

Ta co the bo qua phan tol. Mac nh la 51 10 (, d / dx(, Pol(, Rec( va (khong dung c trong f(x), a, b hoac tol Phep tnh tch phan ch thc hien Mode COMP

Ket qua tch phan la so am khi f(x) am va a x b V du : 2(0,5X 2, 2,2) = 5.333333333 Co bao loi khi nhap co sai sot.

Phai chon n v o goc la Radian khi thc hien phep tnh tch phan cac ham lng giac. Can mot thi gian e may thc hien phep tnh tch phan.

Khi chon gii han tol nho th ket qua chnh xac hn nhng lai mat nhieu thi gian. Phai chon tol la 141 10 hay ln hn Khong nhap c tol dang Math.

Co sai so ln va co the co thong bao loi cho vai ham so oi dau tren khoang tnh tch phan. An e ngng tnh tch phan

Hoan thien phep tnh tch phan e c gia tr kha chnh xac Khi ham f(x) tuan hoan hay oi dau tren [a, b], ta chia ra tng oan dng, am rieng roi tnh tng oan nho va cong lai.

b

af (x)dx =

c

af (x)dx +

b

c( f (x)dx)

44

Khi f(x) la ham dao ong tat dan, ta chia ra nhieu oan nho, tnh rieng va cong lai.

1xb

a af (x)dx f (x)dx= + 2

1

x

xf (x)dx ......+ +

4

b

xf (x)dx

Phu luc

(ln(x),1,e) 1= (tol bo qua) 72

1( ,1,5,1 10 ) 0,8x

= ao ham:

Ta co the tnh gan ung gia tr ao ham cua mot ham so tai iem a ch nh theo cu phap d/dx(f(x), a, tol) f(x) : ham so theo bien x, khong cha x th la hang so a : iem tnh ao ham tol : gii han (ch nhap, xuat Line)

Co the bo qua gia tr tol, gia tr mac nh la 101 10 (,d / dx (, Pol(, Rec( va (khong dung c trong f(x), a, tol. Ch thc hien c phep tnh gia tr ao ham tai iem trong mode COMP.

45

Phai chon n v o goc la Radian khi thc hien phep tnh ao ham cac ham lng giac. Co bao loi hien len khi phep tnh gia tr ao ham khong thc hien c. Khi chon gii han tol nho th ket qua chnh xac hn nhng lai mat nhieu thi gian. Phai chon tol la 141 10 hay ln hn Khong nhap c tol dang Math.

Ket qua thieu chnh xac va co the co bao loi do :

iem tnh ao ham la iem gian oan. Tai iem cc iem tnh thuoc vung cc ai hay cc tieu iem tnh la iem uon Tai iem tnh ham so khong co ao ham Tai iem ma ket qua phep tnh gan zero An e ngng

Phu luc

Cho f(x) = sin(x), tnh f ( )2 (bo qua tol)

Tnh 2 12d (3x 5x 2,2,1 10 ) 7dx

+ =

Tnh tong Vi (, ta co the tnh tong cua bieu thc f(x) khi xac nh pham vi cua x (f(x), a, b) = f(a) + f(a + 1) ++ f(b) f(x) : ham so bien x (neu khong cha x th la hang so) a : gia tr bat au b : gia tr cuoi

a, b phai la so nguyen va 10 101 10 a b 1 10 < < Bc nhay cua phep tnh c xac nh la 1

46

(,d / dx(, Pol(, Rec( va (khong dung c trong f(x), a hay b) An e ngng

Phu luc

Tnh (X 1,1,5) 20 + =

Chuyen oi gia toa o e cac vuong goc va toa o cc

Toa o e-cac vuong goc (Rec) Toa o cc (Pol)

Chuyen oi gia cac toa o c thc hien trong mode COMP, STAT, MATRIX va VECTOR oi ra toa o cc t toa o ecac vuong goc Pol(X,Y) X : ch gia tr x Y : ch gia tr y

Ket qua tnh toan c hien th trong o o180 180 < Ket qua tnh toan s dung n v goc mac nh Ket qua tnh toan r c gan vao x, c gan vao y oi ra toa o ecac vuong goc t toa o cc Rec(r, ) r: ch gia tr r cua toa o cc : ch gia tr cua toa o cc Gia tr a nhap la gia tr goc, phu hp vi cai at n v goc cua may tnh. Ket qua phep tnh x c gan vao bien X, y c gan vao Y.

47

Neu thao tac chuyen oi toa o ben trong mot bieu thc thay cho mot bieu thc ng mot mnh, phep tnh khi thc hien ch s dung gia tr au tien (ca gia tr r hoac gia tr X) tao ra bi chuyen oi.

V du : Pol( 2, 2) + 5 = 2 + 5 = 7. Phu luc ti

Cac ham khac Phan nay giai thch viec s dung cac ham sau ay nh the nao

!, Abs(, Ran#, nPr, nCr, Rnd(

Cac ham nay co the dung cung vi mot mode nh ham lng giac. Rieng Abs (va Rnd(co the dung trong so phc (mode CMPLX)

Giai tha (!) Chc nang nay ch tnh c giai tha cua 0 hoac so nguyen dng

Phu luc (5+3)! = 40320

Phep tnh gia tr tuyet oi (Abs) Khi ang thc hien phep tnh mot so thc chc nang nay cho ket qua la gia tr tuyet oi.

Phu luc Abs (27) = 5

So ngau nhien (Ran#) Chc nang nay sinh ra mot so ngau nhien vi 3 ch so nho hn 1.

Phu luc Sinh ra nhng so ngau nhien 3 ch so. Cac gia tr thap phan 3 ch so ngau nhien c chuyen sang cac gia tr so nguyen 3 ch so bang cach nhan vi 1000. Lu y rang cac gia tr ay ch la cac v du. Cac gia tr ma may tnh sinh ra se khac.

48

Chnh hp (nPr) va To hp (nCr) Cac chc nang nay giup thc hien cac phep tnh chnh hp va to hp n va r phai la so nguyen trong day so 0 r n < 101 10 Phu luc Tnh 10 4,10 4P C

Chc nang lam tron (Rnd) Chc nang nay lam tron gia tr ket qua hien th cua bieu thc theo cai at trc.

So ch so hien th va cai at : Norm 1 hoac Norm 2 Phan hien th c lam tron thanh 10 so.

So ch so hien th cai at : Fix hoac Sci Gia tr se c lam tron en ch so a ghi. V du : 200 7 14 = 400 LINE

200 7 14 D

200 7 14400

S

(nh 3 ch so le thap phan)

(Fix) D Fix

200 7 14400.000

S

(Phep tnh c thc hien ben trong 15 ch so)

200 7 D Fix

200 728.571

S

14 D Fix

Ans 14400.000

S

49

Phep tnh o c lam tron sau ay

200 7 D Fix

200 7 1428.571

S

(Lam tron gia tr en so ch so a ghi) (Rnd) (kiem tra ket qua a lam tron)

1 4 D Fix

Ans 14399.994

S

Cac v du

Phu luc

2

0(sin X cosX) dX

+ = (Bo qua tol)

Xac nh gia tr 5

x 0

1e ( )X !=

CHUYEN OI GIA TR HIEN TH

Co the dung thao tac trong phan nay e chuyen oi mot gia tr hien th thanh so dang ky thuat hoac chuyen oi gia dang chuan va dang thap phan

Phu luc

, Chuyen gia tr 1.234 ra dang 3a 10 va 6a 10

S dung chuyen oi S D Co the s dung phm S D e chuyen mot gia tr dang thap phan (D)

50

Cac dang thc ho tr cho chuyen oi S D Chuyen oi S D co the c s dung e chuyen oi cac ket qua phep tnh thap phan a hien th thanh mot trong nhng dang sau

Lu y Khi chuyen t dang thap phan sang dang S, may tnh se t quyet nh dang S e dung. Khong the ch nh dang S Phan so : Cai at dang thc phan so hien th hien hanh la mot phan so hay hon so. : Di ay la cac dang . ay thc s ch co trong dang thc toan n (n la so nguyen) dc hoac ba

c (tuy thuoc vao cai at dang thc hien th phan so)

Chuyen oi sang dang thc , phan so b gii han en ket qua va ham lng giac ngc thng bieu hien bang radian.

Sau khi co mot ket qua tnh toan trong dang , co the chuyen sang dang thap phan bang cach an phm S D. Khi ket qua phep tnh ban au dang thap phan, no khong the chuyen sang dang .

V du ve chuyen oi S D Lu y la chuyen oi S D can mot thi gian e thc hien. V du : Phan so So thap phan

5 6

D Math5b

56

S

* An phm S D e thay oi gia 2 dang. S D 0.8333333333

S D 56

51

Phu luc phan so so thap phan so thap phan

TOAN VE SO PHC

May thc hien c cac phep tnh so phc sau : * Cong, tr, nhan, chia * Tnh modun, acgumen * Nghch ao bnh phng, lap phng * So phc lien hp

Tat ca phep tnh trong phan nay the hien trong Mode COMP ( )

Phu luc

(1 + 3i) 3 1(2i) i2 2

=

Nhap so phc : Trong mode CMPLX, phm chuyen thanh phm nhap ky hieu i (ao), dung i e ghi a + bi. V du sau la nhap

2 + 3i

2 3 +CMPLX D Math

2 3i

Co the nhap so phc theo dang cc (r ) V du sau la nhap 5 30

5 ( ) 30 CMPLX D Math

5 30

n v giac so phai theo n v goc cai at ban au

52

Dang ket qua hien th May co the hien th ket qua theo dang toa o ecac hay dang cc. Co the chon dang theo cai at ban au. (Xem them phan cai at dang so phc)

V du ve hien ket qua theo dang a + bi

V du 1: 2 ( 3 i) 2 3 2i + = + = 3.464101615 + 2i MATH

3

* Trong dang LINE, phan thc va phan ao c hien thanh hai dong khac nhau

V du 2 : 2 45 1 i = + (n v goc la o) MATH

2 ( ) 4 5 +

CMPLX D Math

2 451 i

S

V du hien ket qua theo dang cc (r ) V du 1: 2 ( 3 i) 2 3 2i 4 30 + = + = (n v goc la o) MATH

2 3

* Trong dang LINE, suat va giac so c hien hanh hai dong khac nhau

++

CMPLX D Math

2 ( 3 i)

2 3 2i

S

+

CMPLX D Math

2 ( 3 i)4 30

S

53

V du 2: 1 i 2 45+ = (n v goc la o) MATH

1 i +

CMPLX D Math

1 i

2 45

S

Giac so hien trong pham vi o o180 180 <

An nh dang ket qua hien th Ta co the an nh dang ket qua hien th nh sau e ket qua hien th dang toa o e-cac, ta an 2 (CMPLX) 4 ( a bi) + e ket qua hien th dang cc, ta an 2 (CMPLX) 3 (( r )) Phu luc

1 i( 2 45)+ = = 1.414213562 45

So phc lien hp Thao tac sau cho ta so phc lien hp 2 (CMPLX) 2 (Conjg)

Phu luc Xac nh so lien hp cua 2+3i

Suat va giac so (Abs, arg) Thao tac sau cho ta suat ( Z ) , va giac so (arg) trong mat phang

Gauss cho so phc Z = a + bi hyp (Abs) ; 2 (CMPLX) 1 (arg)

Phu luc Tm suat va giac so cua 2 + 2i *1 suat *2 giac so

54

THONG KE (STAT)

Tat ca phep tnh trong phan nay the hien

trong Mode STAT ( )

Chon mot kieu tnh thong ke

Trong Mode STAT, hien th man hnh chon kieu tnh thong ke

Cac kieu tnh thong ke Phm Muc menu Tnh thong ke

1 1VAR Bien n

2 A+BX Hoi quy tuyen tnh

3 + 2CX Hoi quy bac hai

4 lnX Hoi quy logarit

5 e ^ X Hoi quy so mu c so e

6 ^A B X Hoi quy so mu c so B 7 ^A X B Hoi quy luy tha 8 1/X Hoi quy nghch ao

Nhap d lieu mau

Hien th man hnh STAT

Man hnh STAT xuat hien sau khi ban truy nhap Mode STAT t mot mode khac. S dung menu STAT e chon mot kieu tnh thong ke t mot man hnh STAT khac.

An (STAT) (Data)

55

Man hnh STAT Co hai dang thc man hnh STAT, phu thuoc vao kieu tnh thong ke la chon.

STAT D1 X23

STAT D1 X Y23

Thong ke bien n Thong ke 2 bien Dong th nhat cua man hnh STAT cho thay gia tr cua mau th nhat hoac gia tr cho cap th nhat cua mau.

Cot FREQ (tan so) Neu m hang muc hien th thong ke tren man hnh cai at cua may tnh, mot cot mang ten FREQ se hien cung tren man hnh STAT. Co the s dung cot FREQ e ch tan so (so lan ma cung mot mau xuat hien trong nhom d lieu) cua moi gia tr mau.

Quy tac nhap d lieu mau tren man hnh STAT

D lieu nhap c chen vao o ni co con tro. S dung cac phm con tro e di chuyen con tro gia cac o

STAT D1 X Y23

Con tro

Cac gia tr va bieu thc nhap tren man hnh STAT giong nh nhap trong Mode COMP vi dang Line

An khi d lieu ang nhap xoa d lieu a nhap hien hanh Sau khi nhap mot gia tr, an . No se nhan gia tr va hien th ti 6 ch so trong o a nhap hien hanh.

56

V du : Nhap gia tr 123,45 vao o X1

(Chuyen con tro en o X1)

STAT D1 X Y23

123.45

Gia tr nhap vao hien th ay

Nhap mot gia tr lam cho con tro chuyen xuong mot o.

Cac chu y khi nhap

So dong trong man hnh STAT (so gia tr d lieu co the nhap) phu thuoc vao loai thong ke va cai at hien th thong ke a chon

Hien th thong ke

Loai thong ke

OFF (tat) (Khong co cot FREQ)

ON (Bat) (Co cot FREQ)

Bien n 80 dong 40 dong

Bien oi 40 dong 26 dong

Cac kieu nhap sau khong thc hien c tren man hnh STAT

Cac hoat ong M+, SHIFT, M+ (M)

Chuyen vao cac bien (STO)

Cac chu y ve lu tr d lieu mau

D lieu nhap vao se b xoa t ong bat c luc nao khi chuyen t Mode STAT sang mode khac chuyen cai at hien th thong ke (lam cho cot FREQ an hay hien) tren man hnh cai at.

57

Chnh d lieu mau

Thay d lieu trong o (1) Tren man hnh STAT, chuyen con tro en o muon chnh. (2) Nhap d lieu va bieu thc mi, sau o an

Chu y !

Lu y phai thay toan bo d lieu hien co cua o bang nhap so mi. Khong the chnh tng phan cua cac d lieu hien co.

Xoa mot dong (1) Tren man hnh nhap STAT, chuyen con tro en dong muon xoa

(2) An

Chen vao mot dong (1) Tren man hnh nhap STAT, chuyen con tro en dong muon chen vao (2) An 1 (STAT) 3 (Edit)

(3) An 1 (Ins)

Chu y Lu y la viec chen se khong thc hien c neu so lng dong cho phep toi a danh cho man hnh nhap STAT a dung het. Xoa toan bo noi dung nhap STAT (1) An 1 (STAT) 3 (Edit) (2) An 2 (DelA) Khi o, toan bo d lieu mau tren man hnh STAT se b xoa

Lu y

Lu y rang ch co the thc hien thao tac Chen vao mot dong va Xoa toan bo noi dung STAT khi man hnh STAT hien th.

58

Man hnh phep tnh STAT. Man hnh phep tnh STAT la e thc hien phep tnh thong ke vi d lieu nhap bang man hnh STAT. An phm khi man hnh nhap STAT c hien th chuyen sang man hnh phep tnh STAT. Man hnh phep tnh STAT cung dung dang Line bat ke dang nhap/xuat hien hanh ang cai at tren man hnh.

S dung menu STAT Khi man hnh nhap STAT hoac man hnh tnh toan STAT hien th, an (STAT) e hien th menu STAT. Noi dung cua menu STAT phu thuoc vao kieu hoat ong thong ke hien hanh s dung bien n hay bien oi

1 : Type 2 : Data 1 : Type 2 : Data 3 : Edit 4 : Sum 3 : Edit 4: Sum 5 : Var 6 : MinMax 5 : Var 6 : MinMax 7 : Distr 7 : Reg

Kieu Y ngha

1 : Type Hien th man hnh chon kieu thong ke

2 : Data Hien th man hnh nhap STAT

3 : Edit Hien th menu phu Edit noi dung man hnh STAT

4 : Sum Hien th menu phu Sum cua cac lenh tnh tong

5 : Var Hien th menu phu Var cua cac lenh tnh gia tr trung bnh, o lech tieu chuan v.v

6 : MinMax Hien th menu phu MinMax cua cac lenh tnh gia tr cc ai va cc tieu

59

Cac muc trong menu STAT Menu bien n

Kieu Y ngha

7 Distr Hien th menu con cua lenh tnh cac phan phoi bnh thng (xem them menu con ve phan phoi)

Menu bien oi Kieu Y ngha

7 Reg Hien th menu phu Reg cua cac lenh cua phep tnh hoi quy phan phoi thong thng Ve chi tiet xem them Cac lenh tnh hoi quy tuyen tnh (A+BX) va Cac lenh tnh hoi quy bac hai 2( CX )+

Cac lenh tnh thong ke bien n (1VAR) Di ay la lenh xuat hien tren menu phu khi chon (Sum),

(Var), (MinMax) hoac (Distr) tren menu STAT trong kieu tnh thong ke bien n.

Xem Phu luc ve cong thc tnh s dung cho tng lenh Menu phu Sum ( (STAT) (Sum)

Kieu Y ngha

1: 2x Tong bnh phng cua d lieu mau 2: x Tong cua d lieu mau Menu phu Var ( (STAT) (Var))

Kieu Y ngha

1: n Tong tan so

2 : x Trung bnh cua d lieu mau

n3 : x o lech tieu chuan n n 14 : x o lech tieu chuan mau n 1

60

Menu phu MinMax ( (STAT) (MinMax))

Kieu Y ngha

1 minX Cc tieu

2 maxX Cc ai

Menu phu Distr (SHIFT STAT 7 (Distr))

P( Q( R( t

Menu nay giup chung ta tnh cac gia tr phan phoi tieu chuan bnh thng. Bien t c tnh t gia tr trung bnh x va o lech tieu chuan nx nhan c t d lieu man hnh nhap

Phan phoi tieu chuan bnh thng

= nX xX tx

Phu luc Thong ke bien n

Chon bien n (1VAR) va nhap d lieu sau : { }1,2,3,4,5,6,7,8,9,10 (FREQ m) Nhap d lieu cho cac so sau, s dung chen va xoa : {1,2,3,4,5,6,7,8,9,10} (FREQ m) Nhap d lieu FREQ theo cac so sau {1,2,1,2,2,2,3,4,2,1} (FREQ m) Cac v du t en tat ca eu s dung d lieu nh v du . Tnh tong bnh phng cua d lieu mau va tong cua d lieu mau

61

Tnh so mau, trung bnh va o lech chuan

Tnh cc ai va cc tieu

Tm gia tr phan phoi tieu chuan bnh thng cho d lieu a nhap v du .

Gia tr phan phoi xac suat P(t) tai x = 3

Gia tr phan phoi xac suat R(t) tai x = 7

Cac lenh khi tnh hoi quy tuyen tnh (Ax+B) Vi hoi quy tuyen tnh, hoi quy c thc hien phu hp vi phng trnh mau sau :

y = A + BX

Sau ay la cac lenh trong menu phu xuat hien khi chon 4 (Sum), 5 (Var), 6 (MinMax) hoac 7 (Reg) tren menu STAT trong hoi quy tuyen tnh thuoc kieu tnh thong ke.

Xem Phu luc e biet them ve cong thc phep tnh s dung cho tng lenh.

Menu phu Sum ( 1 (STAT) 4 (Sum)) Kieu Y ngha

1: 2x Tong bnh phng cua d lieu X 2: x Tong d lieu X 3: 2y Tong bnh phng cua d lieu Y 4: y Tong d lieu Y 5: xy Tong xy 6: 3x Tong cac lap phng cua d lieu X 7: 2x y Tong 2x y 8: 4x Tong d lieu X luy tha 4

62

Menu phu Var (SHIFT 1 (STAT) 5 (Var) Kieu Y ngha

1 : n So cac mau

2 : x Trung bnh cua d lieu X

3 : nx o lech tieu chuan n cua d lieu X 4 : n 1x o lech tnh chat mau n 1 cua d lieu X 5 : y Trung bnh cua d lieu Y

6 : ny o lech tieu chuan n cua d lieu Y 7 : n 1y o lech tieu chuan mau n 1 cua d lieu Y

Menu phu MinMax ( (STAT) 6 (MinMax))

Kieu Y ngha

1:minX Gia tr cua cc tieu d lieu X

2:maxX Gia tr cua cc ai d lieu X

3:minY Gia tr cua cc tieu d lieu Y

4:maxY Gia tr cua cc ai d lieu Y

Menu phu Reg ( (STAT) 7 (Reg))

Kieu Y ngha

1: A He so A

2: B He so B

3: r He so tng quan r

4: x Gia tr c lng cua x

5 y Gia tr c lng cua y

63

Phu luc Hoi quy tuyen tnh en Cac v du en eu s dung d lieu nhap trong v du

*1 Gia tr c tnh (y = 3 x ?) = *2 Gia tr c tnh (y = 2 y ?) =

Cac lenh c dung khi tnh hoi quy bac hai

2( CX )+ Vi hoi quy bac hai, c thc hien theo phng trnh mau di ay

y = A + BX + 2CX

Xem Phu luc e biet them ve cach dung cac lenh

Menu phu Reg ( 1 (STAT) 7 (Reg))

Kieu Y ngha

1:A He so A

2:B He so B

3:C He so c

14 : x Gia tr gia tr c lng cua 1x

5: 2x Gia tr gia tr c lng cua 2x

6: y Gia tr gia tr c lng cua y

Hoat ong cua menu phu Sum (sums), menu phu Var (so mau trung bnh, o lech chuan) va menu phu MinMax (gia tr ln nhat, gia tr nho nhat) giong phep hoi quy tuyen tnh.

Phu luc Tnh hoi quy bac hai en

Cac v du en eu s dung d lieu nhap trong v du

64

Cac phan khac

Ve chi tiet cac cong thc cua lenh bao gom cac kieu hoi quy, tham khao cac cong thc a ghi

Phu luc t en

Kieu tnh thong ke Phng trnh mau Cong thc tnh

Hoi quy logarit (lnX) y = A + BlnX

Hoi quy mu e (e^X) BXy Ae=

Hoi quy mu ab (A*B^X) Xy AB=

Hoi quy luy tha (A*X^B) ny A X=

Hoi quy nghch ao (1/X) By AX

= +

Phu luc So sanh cac ng cong hoi quy V du di ay dung d lieu trong cac v du ; So sanh he so cho logarit, hoi quy mu e, so mu ab, luy tha va nghch ao (FREQ : Tat)

Phu luc Cac kieu khac cua phep tnh hoi quy en Cac lenh bao gom trong menu phu Reg co the can thi gian lau e thc hien cac phep tnh logarit, hoi quy mu e, mu ab, luy tha khi co nhieu mau d lieu.

65

TOAN TRONG HE EM C SO N (BASE N)

Mode BASE N giup ta thc hien cac phep tnh so hoc, so am, toan logic trong he em nh phan, bat phan, thap phan, thap luc phan.

Cac phep toan sau c thc hien trong mode BASE N (an MODE 4)

Cai at cho he em va so nhap Dung cac phm sau cho BASE N Khi chon he em th k hieu tng ng hien len ben phai

Phm He em Dau hieu hien

DEC Thap phan Dec

HEX Thap luc phan Hex

BIN Nh phan Bin

OCT Bat phan Oct

K hieu he em hien hanh hien th dong th hai cua man hnh Cai at mac nh ban au khi vao BASEN la Dec

Nhap gia tr Trong BASEN ta ch dung c cac ch so cua he em hien hanh Bao loi hien len khi ta nhap cac ch so ngoai he em hien hanh (nh nhap 2 he BIN) Khong nhap c phan so hay ham mu trong BASEN. Neu tnh toan ra so thap phan, may t ong cat bo phan le.

Nhap gia tr HEX Cac phm A, B, C, D, E, F la phm so trong HEX.

66

Bang pham vi gia tr

He em Pham vi

BIN So dng 0000000000000000 x 0111111111111111 So am 00000000000 x 1111111111111111

OCT So dng 00000000000 x 17777777777 So am 20000000000 x 37777777777

DEC 2147483648 x 2147483647

HEX So dng 00000000 x 7FFFFFFF So am 80000000 x FFFFFFFF

Pham vi nay la day 16 bit hep hn day 32 bit

Bao loi hien len khi ket qua tnh toan ngoai pham vi nay

Xac nh he em ngay khi nhap

Trong BASEN, may cho phep nhap so cua he em khac e len he em a cai trc. Muon dung cach nay phai an (BASE) e hien menu (hai trang) va ta an so lien he can thiet ghi trc cac he em

An phm Xac nh so trong he

1 (d) Thap phan (C so 10)

2 (h) Thap luc phan (C so 16)

3 (b) Nh phan (C so 2)

4 (o) Bat phan (C so 8)

Thao tac sau nham nhap so 3 cua he thap phan tren man hnh nh phan

BIN (BASE) (d) d3

so 3 trong DEC

67

Phu luc

Tnh 2 21 1+ trong BIN Tnh 8 87 1+ trong OCT Tnh 16 161F 1+ trong HEX oi 1030 sang BIN, OCT, HEX

Tnh 10 165 5+ (ra ket qua) trong BIN

Tnh so am va toan logic Muon tnh ve so am va toan logic, an (BASE) e hien menu BASE va dung cac lenh tng ng.

Phm an Yeu cau

1 (and) Nhap and (e thc hien phep AND)

2 (or) Nhap or (e thc hien phep OR)

3 (xor) Nhap xor (e thc hien phep XOR)

4 (xnor) Nhap xnor (e thc hien phep XNOR)

5 (Not) Nhap Not (e thc hien phep NOT)

6 (Neg) Nhap Neg e lay Neg (so am, so bu 2) Phep lay Neg (so am) trong BIN, OCT, HEX da vao phep bu 2 cua BIN, roi chuyen lai theo c so a chon. So am trong DEC th mang dau tr ()

Phu luc T v du en la cac phep tnh ve so am trong he em nh phan (BIN) nen khi thc hien phai an trc

BIN

68

GIAI PHNG TRNH (EQN)

Tat ca phep tnh trong phan nay eu c

thc hien Mode EQN (MODE 5)

Cac loai phng trnh Menu phng trnh xuat hien khi an phm MODE 5 (EQN) va vao Mode EQN

Phm

1 n n na X b Y c+ = He phng trnh tuyen tnh 2 an so

2 n n n na X b Y c Z d+ + = He phng trnh tuyen tnh 3 an so

3 2aX bX c 0+ + = Phng trnh bac 2

4 3 2aX bX cX d 0+ + + = Phng trnh bac 3

An phm MODE 5 (EQN) e vao Mode EQN. Viec o se xoa toan bo Mode EQN hien hanh va hien th menu phng trnh tren

Nhap cac he so S dung man hnh nhap he so e nhap cac he so cua mot phng trnh. Man hnh nhap he so cho thay cac o nhap ma moi he so yeu cau bi kieu phng trnh a chon.

He phng trnh bac nhat 2 an Phng trnh bac 3

69

Quy tac nhap va chnh cac he so

D lieu nhap c chen vao cac o co con tro. Khi nhap d lieu vao mot o, con tro se dch sang o tiep theo ve ben phai Khi he phng trnh tuyen tnh 3 an so hoac phng trnh bac 3 c chon, cot d se khong hien th tren man hnh he so hien th lan au. Cot d se xuat hien khi dch con tro en cho no, lam cho man hnh oi cho.

Cac gia tr va bieu thc nhap vao man hnh he so giong nh nhap trong Mode COMP vi dang Line.

An khi ang nhap th xoa d lieu a nhap hien hanh.

Sau khi nhap xong d lieu, an . Nhap c va hien th en 6 ch so trong o la chon hien hanh

e thay oi noi dung cua cac o, s dung cac phm con tro e di chuyen con tro ti o sau o nhap d lieu mi.

a gia tr cac he so ve 0

Co the xoa tat ca cac he so ve 0 bang cach an phm tren man hnh nhap he so.

Cac chu y khi nhap d lieu man hnh nhap he so

Ve c ban, cac chu y khi nhap d lieu man hnh nhap he so giong nh chu y nhap d lieu man hnh nhap STAT. iem khac nhau duy nhat la lu y au tien cua chu y nhap d lieu man hnh STAT khong ap dung vi man hnh nhap he so. Ve chi tiet, xin xem them Cac chu y trong luc nhap man hnh STAT

Hien th nghiem Moi lan an lai hien ra mot nghiem tiep theo neu co.

An khi nghiem cuoi cung c hien th th tr ve man hnh nhap he so

70

Trong trng hp cac phng trnh tuyen tnh, co the dung va e hien th gia nghiem X va Y (va Z).

Khi co nhieu nghiem cho mot phng trnh bac 2 hay 3, co the dung va e hien th gia 1 2X ,X va 3X . So lng nghiem phu thuoc vao phng trnh thc te.

An khi mot nghiem c hien th, se tr ve man hnh nhap he so

Dang hien th cua nghiem phu thuoc vao dang nhap/xuat va dang hien th so phc cai at cua man hnh cai at may.

Lu y rang khong the chuyen cac gia tr en k hieu k thuat khi ma mot nghiem phng trnh c hien th.

Phu luc en

MA TRAN (Matrix)

Ta co the nhap ten Mat A, Mat B, Mat C vao bo nh ma tran. Khi tnh toan, ma tran ket qua mang ten MatAns

Vao chng trnh tnh ma tran, an

Thiet lap va quan l ma tran Thiet lap va lu ma tran

(1) Trong Mode ma tran, an (Matrix) (Dim)

Man hnh hien

Matrix ?1 : MatA 2 : MatB3 : MatC

Man hnh nay ch hien khi vao Mode Matrix

71

(2) An hay e chon ma tran A, B, C va ta c man hnh chon kch thc ma tran.

(3) An 1 en 6 e chon kch thc thch hp. Sau khi chon kch thc (so dong, so cot) ta co mot khung ma tran c chon hien len

(4) Nhap cac phan t cua ma tran

Nhap giong nh khi nhap he so cua phng trnh (xem them cach nhap cac he so) Neu muon thay oi ma tran khac th bat au t bc 1

Chep noi dung mot ma tran vao ma tran khac (1) Dung man hnh nhap e hien ma tran muon chep hay hien man hnh MatAns V du muon chep noi dung ma tran A vao ma tran B, ta an

(Matrix) (Data) (MatA)

(2) An RCL (STO) (co ch STO hien len)

(3) Xac nh ten ma tran muon chep vao

Ten ma tran Phm an Matrix A (MatA) Matrix B (MatB) Matrix C Hyp (MatB)

An (MatB) e chep noi dung vao ma tran B va noi dung ma tran B va chep vao hien len.

72

Thc hien tnh toan An khi co man hnh chon hay nhap ma tran e vao man hnh tnh toan.

Man hnh ma tran ket qua (MatAns)

Man hnh ma tran ket qua (MatAns) ch ket qua cua phep toan ma tran

Khong the chnh c noi dung cua MatAns

Tr ve man hnh tnh toan, an

Khi MatAns hien len, ta co the an tiep phm phep tnh

(nh +, ) va dung MathAns nh mot ma tran so hang e s dung MathAns nh la mot ma tran so hang (Xem them phan dung MathAns e thc hien chuoi tnh toan)

Menu ma tran Trong Mode Matrix, menu sau hien len khi an (Matrix)

Mau chon Yeu cau

1 Dim Chon kch thc (cot, dong) cho MatA, MatB, MatC

2 Data Tm noi dung cua MatA, MatB, MatC

3 MatA Nhap MatA

4 MatB Nhap MatB

5 MatC Nhap MatC

6 MatAns Nhap MatAns

7 Det Nhap det (Tnh nh thc

8 Trn Nhap Trn (Tm ma tran chuyen v

73

Phu luc

Nhap MatA = 2 11 1 .

MatC = 1 0 10 1 1

Chep MatA = 2 11 1 vao MatB va chnh thanh

MatB = 2 11 2

Cac v du sau dung ma tran a nhap va (MatA, MatB, MatC)

MatA + MatB (Cong hai ma tran)

MatA MatB, MatB MatA MatA MatB (Nhan hai ma tran)

3 MatA

Tnh nh thc MatA

Tm ma tran chuyen v cua MatC

Tm ma tran nghch ao cua MatA

(Dung phm e nhap 1. Nh rang khong dung c

x ay)

Tm suat cua MatB (Abs (MatB)

(Dung phm (Abs))

Tnh Mat 2A , Mat 3A cua MatA

(Dung phm , 3(x ) . Nh rang khong dung c

x ay).

74

BANG SO T MOT HAM

Tat ca cac phep tnh trong phan nay c

thc hien Mode TABLE (MODE 4)

nh ham tao ra bang so Thao tac sau ay nh ham tao ra bang so vi cai at nh sau

Ham : 2 1f (x) x2

= +

Gia tr ban au : 1. Gia tr cuoi : 5, Bc nhay : 1.

LINE

(1) An (TABLE) D

f (X)

(2) Nhap ham

(3) Sau khi nhap ham, an .

Khi o may se yeu cau nhap gia tr ban au

D

Start ?1

gia tr ban au mac nh la 1

Neu gia tr ban au khong phai la 1, an 1 e nh ro gia tr mac nh ban au cho v du

(4) Sau khi a nh ro gia tr au, an

Khi o se hien th man hnh nhap gia tr cuoi

D

End ?5

gia tr mac nh cuoi la 5

Xac nh gia tr cuoi

75

(5) Sau khi nh ro gia tr cuoi, an

Khi o hien th man hnh gia tr bc nhay

D

Step ?1

gia tr bc nhay mac nh ban au la 1

nh ro gia tr bc nhay

Ve chi tiet nh ro gia tr au, cuoi va bc nhay, xem quy tac gia tr au, cuoi va bc nhay

(6) Sau khi nh ro gia tr bc nhay, an

DX F(X)

1 1 1..52 2 4.53 3 9.5 1

An tr ve man hnh nhap ham

Cac chc nang ho tr Ngoai bien X, cac bien khac (A, B, C, D, Y) va bo nh oc lap (M) tat ca eu c coi la cac gia tr (bien hien hanh chuyen vao bien hay lu trong bo nh oc lap).

Ch co bien X la co the c s dung nh bien cua ham

Phep bien oi toa o khong the s dung lam ham tao ra bang so

Lu y rang phep tnh tao ra bang so lam cho cac noi dung cua bien X thay oi

1

76

Quy tac ve gia tr ban au, gia tr cuoi va bc nhay Dang Line luon c dung e nhap gia tr

Co the nh ro cac gia tr hoac bieu thc phep tnh (phai sinh ra ket bang so) gia tr au, cuoi va bc nhay

nh ra mot gia tr cuoi ma nho hn gia tr ban au se gay ra loi, do o bang so se khong c tao ra

Cac gia tr ban au, cuoi va bc nhay se sinh ra mot gia tr toi a la 30 gia tr x cho bang so c tao ra. Lap ra mot bang so s dung cac gia tr ban au, cuoi va bc nhay nhieu hn 30 gia tr x se gay ra loi

Lu y

Mot so ham va gia tr ban au, cuoi va bc nhay can nhieu thi gian e tao ra bang so

Man hnh bang so Man hnh bang so cho thay cac gia tr x tnh toan s dung cac gia tr ban au, cuoi va bc nhay, cung nh cac gia tr thu c khi moi gia tr x c thay the trong ham f(x)

Lu y la ch s dung bang so e xem cac gia tr. Noi dung cua bang khong the chnh

An phm e tr ve man hnh nhap ham.

Cac chu y ve man hnh bang so (Mode TABLE) Lu y la viec thay oi cac cai at dang nhap / xuat (dang Math hoac dang Line) tren man hnh cai at Mode TABLE se xoa ham tao lap bang so

77

TOAN VECT (VECTOR)

Dung VctA, VctB va VctC e ghi ten cac vect trong bo nh. Vect ket qua cua phep tnh c ghi la VctAns

Toan vect c thc hien Mode VECT

( )

Thiet lap va quan ly vect

Thiet lap va lu vect vao bo nh

(1) Trong mode Vect an (VECTOR) (Dim)

Hien man hnh chon vect

Man hnh vect ch hien trong mode VECTOR

(2) An , hay e xac nh ten vect muon chon

Hien man hnh xac nh chieu cua vect

(3) An hay e xac nh chieu cua vect

An e chon vect 3 chieu (trong khong gian), hay e chon 2 chieu (mat phang)

Sau khi an nh chieu th man hnh nhap tng ng hien ra

(4) Dung man hnh nay e nhap cac thanh phan so (toa o)

Cach nhap cung giong nh nhap he so phng trnh (xem them phan nhap he so phng trnh)

Neu muon tao vect khac, hay tr lai bc 1

78

Chep toa o cua vect nay sang vect khac

Co the chep toa o cua mot vect (hay cua VctAns) sang mot vect khac. Ve c ban viec sao chep cung giong nh ma tran (xem them phan sao chep ma tran)

Thc hien tnh toan e thc hien toan vect, an man hnh nhap

Vect Ans (VctAns) ch ket qua cua phep tnh vect va thc hien

Khong the chnh noi dung cua VctAns

Chuyen ve man hnh tnh toan vect, an

Menu vect Bang sau la menu vect (hien trong mode vect) sau khi an 5 (VECTOR)

Mau chon Yeu cau

1:Dim Goi VctA, VctB, VctC e an nh chieu (mat phang hay khong gian) cho cac vect nay

2:Data Goi VctA, VctB, VctC e hien toa o va chnh sa toa o

3:VctA Nhap VctA

4:VctB Nhap VctB

6:VctAns Nhap VctAns

7 Dot Nhap dau (e lay tch vo hng 2 vect)

79

Phu luc Nhap VctA = (1, 2) va VctC = (2, 1, 2)

Chep VctA = (1, 2) vao VctB roi sa thanh VctB = (3,4)

Cac v du sau s dung so lieu va (VctA, VctB, VctC)

VctA+VctB (Cong vect)

3 VctA (nhan so thc vi vect)

VctB 3 VctA (phep tnh co dung VctAns)

VctA VctB (Tch vo hng)

VctA VctB (Tch hu hng)

Suat cua VctC

Tm goc (theo o) hp bi A = (1, 0, 1) va

B = (1, 2, 0) va vect n v nJG

vuong goc vi VctA, VctB.

*1 (A B)cosA B = vi = 1 (A B)cos

A B

2* (A B)nA B

JG =

HANG SO KHOA HOC

May lu 40 hang so khoa hoc e dung trong tnh toan thong thng. Co the dung hang so khoa hoc trong moi phep toan tr trong BASE N.

Muon goi mot hang so khoa hoc, an (CONST). May hien menu bang so khoa hoc. Nhap 2 ch so tng ng vi ten hang so can goi khi o k hieu va gia tr cua hang so hien len man hnh.

80

Bang sau la cac hang so cai san (kem k so e goi)

Hang so Ma so

Khoi lng proton (mp) 01

Khoi lng neutron (mn) 02

Khoi lng electron (me) 03

Khoi lng muon (m) 04 Ban knh Bohr 0(a ) 05

Hang so Plank (h) 06

Manheton hat nhan (N) 07 Manheton Bohr (B) 08 hbar (= ) 09 Hang so cau truc tinh te () 10 Ban knh electron (re) 11

Bc song compton (c) 12 T so t c (p) 13 Bc song compton proton (cp) 14 Bc song compton neutron (cn) 15 Hang so Rydberg (R ) 16

n v khoi lng nguyen t (u) 17

Momen t proton (p) 18 Momen t electron (e) 19 Momen t neutron (n) 20 Momen t muon ( ) 21

81

Hang so Faraday (F) 22

ien tch c ban (e) 23

Hang so Avogaro (NA) 24

Hang so Boltzmann (k) 25

The tch mol kh l tng (Vm) 26

Hang so mol kh (R) 27

Van toc anh sang 0(C ) 28

Hang so phong xa 1C 29

Hang so phong xa 2C 30

Hang so Setefan Boltzmann () 31 Hang so ien moi chan khong 0( ) 32 Hang so t tham 0( ) 33 Lng t t thong 0( ) 34 Gia toc chuan cua trong lc (g) 35

Lng t dan ien 0(G ) 36

Tr khang ac trng cua chan khong 0(z ) 37

Nhiet o Celsius (t) 38

Hang so hap dan (G) 39

Atmophe chuan (atm) 40

Cac gia tr da theo ISO 1992 va CODATA 1998. Ve chi tiet xem phu luc

Phu luc va

Thc hien cac v du trong Mode COMP ( )

82

OI N V

May cai san oi n v o lng. Co the oi n v trong Mode tr Mode BASEN va lap bang.

e goi lenh oi n v, an (CONV). Man hnh hien menu oi n v. An 2 ch so tng ng vi lenh oi n v ta can

Xem phu luc cho biet lenh oi n v vi k so tng ng

Cach tnh n v oi da theo An ban chuyen e NIST 811 (1955)

Phu luc en

Thc hien cac v du tren trong Mode COMP ( )

83

THONG TIN K THUAT

Th t u tien cac phep tnh May tnh thc hien cac phep tnh th t u tien sau C ban cac phep tnh c thc hien t trai qua phai. Bieu thc co dau ngoac n c u tien cao nhat. Phep tnh thc hien theo u tien sau:

1. Ham Pol(, Rec( (, d / dx(, (

P(, Q(, R(

sin(, cos(, tan(, 1 1 1sin (, cos (, tan (,

sinh(, cosh(, tanh(, 1 1 1sinh (, cosh (, tanh (

log(, ln(, 3e ^ (,10^ (, (, (

arg(, Abs(, Conjg( Not(, Neg( Det(, Trn( Rnd(

2. Cac ham c tao bi gia tr luy tha, khai can luy tha: 2 3 1x , x , x , x !, , o, r, g, ^(, x (

Phan phoi thng : X t Phan tram : %

3. Phan so : ba / c

4. Dau: () (dau am) d, h, b, o (c so n)

84

5. : cm X in, v.v Phep tnh c lng thong ke : 1 2x, y, x , x

6. Chnh hp, to hp : nPr, nCr, cua so phc dang cc

7. Nhan vo hng : (Dot)

8. Phep nhan, chia : , Phep nhan tat : Dau nhan c bo qua ngay trc , e, bien, hang so (2, 5A, A, 3mp, 2i,), ham co dau ngoac n (2 (3), Asin(30)) Lu y rang may ES khong danh u tien cho phep nhan tat, ngha la phep nhan tat hay phep nhan thng eu co u tien bang nhau (3 5A hay 3 5 A eu giong nhau ve th t u tien)

9. Phep cong, tr : +,

10. and

11. or, xor, xnor Neu mot phep tnh cha mot gia tr am, can ong gia tr am o trong ngoac n. Chang han, khi bnh phng 2 can phai nhap 2( 2) . V 2x la u tien 2 cao hn v tr cua dau am () u tien 4. V du :

22 4 = 2 2( 2) 4 = Phep nhan, chia va phep nhan tat la cung mc o u tien (u tien 8). Do o, khi ca hai yeu to cung hien dien th phep tnh c thc hien t trai qua phai. Neu a dau ngoac n vao se lam thay oi phep u tien va ket qua tnh toan se khac nhau

85

V du:

1 2 i 11 2i i2

=

1 2 i 11 (2i) i2

=

Gii han Stack (nhom) May tnh nay s dung cac khu vc nh goi la stack e lu tam thi cac gia tr day phep tnh, cac lenh va chc nang u tien thap hn. Stack so co 10 mc va stack lenh co 24 mc nh hnh sau ay

Loi Stack ERROR hien len khi phep tnh ang thc hien co mot trong 2 stack b vt qua han

Pham vi cua phep tnh, so ch so va o chnh xac Cac day phep tnh, so ch so dung cho phep tnh va o chnh xac phu thuoc vao kieu phep tnh ang thc hien.

86

Pham vi cua phep tnh va o chnh xac

Day phep tnh 991 10 en 999,999999999 10 hoac 0 So ch so cho tnh toan ben trong

15 ch so

o chnh xac Nhn chung, 1 tai ch so th 10 cho mot phep tnh n. o chnh xac mu 1 cua so a trong na 10 . Sai so tnh don t nhng phep tnh lien tiep

Pham vi cac ham nhap

Ham Pham vi

DEG 90 x 9 10 < RAD 0 x 157079632.7 < sin x GRA 100 x 1 10 DEG 90 x 9 10 < RAD 0 x 157079632.7 < cos x GRA 100 x 1 10 DEG giong nh sin x, tr khi x = (2n 1) 90 RAD giong nh sin x, tr khi x = (2n 1) /2 tan x GRA giong nh sin x, tr khi x = (2n 1) 100

1sin x 1cos x

0 x 1

1tan x 990 x 9.999999999 10 sinh x

cosh x 0 x 230.2585092

87

1sinh x 990 x 4.999999999 10 1cosh x 991 x 4.999999999 10

tanhx 990 x 9.999999999 10 1tanh x 10 x 9.999999999 10

log/lnx 990 x 9.999999999 10< x10 999.999999999 10 x 99.99999999

xe 999.999999999 10 x 230.2585092 x 1000 x 1 10 < 2x 50x 1 10<

1/x 100x 1 10 ; x 0< 3 x 100x 1 10< x! 0x69 (x la mot so nguyen)

nPr 0 n < 1 1010 , 0 r n (n, r la cac so nguyen) 1 { } 100n!/(n r)! 1 10 <

nCr 0 n < 101 10 , 0 r n (n, r la cac so nguyen) 1 n!/r! < 1001 10 hoac 1 n!/(n r)! < 1001 10

Rec(r,) 0 r 999.999999999 10 100a , b,c 1 10< o

oHJJ 100x 1 100< oi so thap phan o phut giay o0 0 0 x o9999999 59 59

88

y^(x )

100x 0 : 1 10> < ylogx < 100 x = 0 : y > 0

x < 0 : y = n, m2n 1+ (m, n la cac so nguyen)

x y

y > 0 : x 0, 1001 10 < 1/xlogy < 100 x = 0 : y > 0

y < 0 : x = 2n + 1, 2n 1m+

(m 0, m, n la cac so nguyen) b / ca

Tong so so nguyen, t so va mau so phai la 10 k t hoac t hn (bao gom ca cac dau cach)

o chnh xac ve c ban giong nh mieu ta trong Pham vi cua phep tnh va o chnh xac tren

Cac chc nang y 3x^(x ), y, , x ! , nPr, nCr vi nhng phep tnh ben trong lien tiep co the tch luy sai so ln ket qua cuoi cung Co the co sai so ln vung can cua nhng iem ac biet hay iem uon

Bao loi Co bao loi khi mot phep tnh co ket qua vt pham vi an nh, luc nhap so lieu sai quy tac hoac co thao tac loi tng t

Khi bao loi hien ra Sau ay la cach x l chung khi gap bao loi Dung hay e hien th man hnh nhap bieu thc trc khi co bao loi. Con tro se nam tai v tr loi (xem them Hien th v tr loi) An e xoa bieu thc phep tnh, nhap lai bieu thc ung neu can. Trong trng hp nay, may khong lu phep tnh goc trong bo nh.

89

Math ERROR (Loi phep tnh) Nguyen nhan Ket qua trung gian hoac cuoi cung cua phep tnh vt qua pham vi an nh tren D lieu nhap vt qua pham vi an nh cho phep hoac cac ham

Phep tnh ang thc hien cha mot thao tac bat hp l (nh chia cho 0)

Cach x l Kiem tra cac gia tr nhap, giam ch so th, lam lai

Khi s dung bo nh oc lap hoac bien, phai nh rang cac bien phai nam trong pham vi cho phep cua ham

Stack ERROR (Loi ve nhom) Nguyen nhan Phep tnh ang thc hien vt qua kha nang a an nh ve nhom

Cach x l n gian hoa bieu thc e va vi kha nang cua nhom

Chia phep tnh thanh hai phan hoac nhieu hn

Syntax ERROR (Loi cu phap) Nguyen nhan Co sai sot ve cu phap ang dung

Cach x l Xem lai va ieu chnh

Argument ERROR (Loi Argument) Nguyen nhan Co sai sot ve Argument

Cach x l Xem lai va ieu chnh

90

Dimension ERROR (Ch xay trong Ma tran va Vect) Nguyen nhan Nhap cac yeu to cho ma tran va vect trong khi cha an nh kch thc hay chieu Thc hien phep tnh hai ma tran hoac vect ma so chieu cha thch hp

Cach x l An nh lai Dim cho vect hay ma tran trc khi nhap cac yeu to Chon Dim thch hp cho ma tran va vect trong phep toan

Variable ERROR (Ch xay ra trong chc nang SLOVE) Nguyen nhan Khong co bien x trong phng trnh

Cach x l Nhap lai phng trnh phai cha x

*Cant Solve Error (Ch xay ra trong chc nang SLOVE) Nguyen nhan May khong tm c nghiem

Cach x l Kiem tra lai phng trnh

Nhap lai gia tr au va giai lai

Insufficient MEM Error (Loi thieu bo nh) Nguyen nhan Khong u bo nh e thc hien

Cach x l Giam bt pham vi bang bang cach oi gia tr au, gia tr cuoi va bc nhay

91

Time Out Error Nguyen nhan Khong thoa man ieu kien e phep tnh ao ham hay tch phan thc hien

Cach x l Tang gia tr tol (co the lam giam o chnh xac) hay kiem lai bieu thc nhap

Trc khi xac nh la may tnh truc trac Thc hien cac bc sau ay khi loi xay ra luc tnh toan hoac khi cac ket qua tnh toan khong phai la ap so ung nh mong muon. Neu bc trc khong sa c th chuyen sang bc tiep theo.

Lu y la nen ghi nhng d lieu quan trong khi thc hien cac bc kiem tra cac bieu thc tnh toan e chac rang khong con cha loi nao ca.

(1) Chac rang la s dung mode ung cho phep tnh ang thc hien.

(2) Neu cac bc tren khong at c van e at ra th an phm . Khi o may tnh se kiem tra xem cac chc nang cua may

tnh co thc hien chnh xac hay khong. Neu may tnh tm thay ieu bat thng, no se t ong cho gia tr ban au cua mode phep tnh va xoa noi dung lu. Ve chi tiet cua viec cai at gia tr ban au xem them Cac mode phep tnh va cai at may tnh

(3) Cho gia tr ban au tat ca cac mode va cai at nh sau :

(CLR) (Setup) (Yes)

92

NANG LNG

May tnh nay c cung cap nang lng bi pin AAA (R03(um4)

Thay pin Cac con so b m tren man hnh hien th cho biet pin yeu. Khi may pin yeu neu tiep tuc s dung co the lam may tnh sai. Hay thay pin cang sm cang tot khi ch cha b m nhieu. Ngay ca trng hp may ang s dung bnh thng, hay thay pin t nhat 2 nam 1 lan.

Chu y ! Lay pin ra khoi may tnh se lam noi dung bo nh oc lap va cac gia tr ghi trong cac bien b xoa.

(1) An e tat may tnh

(2) Thao inh oc mat sau va lay nap gi pin ra

(3) Lay pin cu ra

(4) Lap pin mi vao theo ung cc (+) va cc am ()

(5) ong nap va xiet lai inh oc

(6) An lan lt cac phm sau :

(CLR) (ALL) (Yes)

Can phai lam theo cac thao tac tren, ng bo qua thao tac nao ca

Chc nang t ong tat May t ong tat sau khoang 6 phut khong thao tac. An e m lai.

136

Chu trach nhiem xuat ban : Chu tch HQT kiem Tong Giam oc NGO TRAN AI

Pho Tong Giam oc kiem Tong bien tap NGUYEN QUY THAO

Chu trach nhiem noi dung : Pho Tong Giam oc kiem Giam oc NXBGD tai TP. Ho Ch Minh

VU BA HOA

Bien tap noi dung va tai ban : O LNH

NGUYEN HU KHOI

Bien tap k thuat : O VAN SAC THIEN AN

Trnh bay ba : O VAN SAC

Sa ban in : C VIEN

Che ban : HOANG LONG

n v lien doanh in va phat hanh Cong ti co phan XNK Bnh Tay

huong dan su dung may tinh bo tui fx 570es

Documents