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HNG DAN S DUNG VA GIAI TOAN
TREN MAY TNH
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NGUYEN TRNG CHANG NGUYEN THE THACH
DUNG CHO CAC LP 10 11 12
(Tai ban lan th nhat)
NHA XUAT BAN GIAO DUC
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385-2008/CXB/6-800/GD Ma so : PTK03n8-LKH
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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 23Tnh 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
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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 phmchc nang.
V du :
la chc nang chnh, an trc tiep
mau vang, an sauD 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 trnha goi (nh la BASE N).
Cac ch trong ngoac sau phm an dung e giai thch y ngha cuaphm
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 ghitrc chc nang ay.
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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 keHai phm , lam hien cac trang menu cung loai.
Hoat ong cua con tro c ch ra bi , , , .
Hng dan nay ch dung muc ch minh hoa, oi khi co khac oivi 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 trachnhiem vi ai ve nhng ton that phat sinh bi viec mua hoac sdung san pham va tai lieu nay. Va hn the na, cong ti CASIOse khong chu trach nhiem ve bat k khieu nai g oi vi viec sdung tai lieu nay bi mot ben khac.
S dung ban phu luc riKhi thay co ghi chu Phu luc th phai xem them bang phu luckem 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 AUPhai 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)
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GI AN TOAN CHO MAYPhai oc cac ieu nay trc khi s dung may va gi lai enghien cu ve sau
! Can thanDau hieu nay thong bao co the gay ton thng hoac hong mayneu khong chu y.
Pin Sau khi thao pin ra khoi may, hay cat vao ni an toan, xa tamtay 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 cbo pin vao cho nong hay ot pin.
S dung pin khong ung cach de ro r va h hong cac vat egan, 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 motso linh kien co the gay no mot cach bat ng tao rui ro hoa hoanva 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 nenthay 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.
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Pin kem theo may co the b giam nang lng trong quatrnh van chuyen va bao quan. V the nen thay pin smhn tuoi tho pin.
Pin yeu co the lam cho noi dung bo nh b hong hoachoan toan b mat i. Hay luon gi so lieu quan trongbang van ban.
Tranh s dung va e may trong moi trng nhiet o thaiqua.
Nhiet o qua thap co the gay nen cham hien th hay hoan toankhong hien th va lam giam tuoi tho cua pin. Tranh e may tiepxuc 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 dangvo may va h hong cac mach ien t ben trong. Tranh s dung va cat may nhng ni co o am cao va buibam. Can than khong c e may b nc ri vao hay at nico o am cao va bui bam. Nhng ieu kien nh vay co the gayh 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 quanhay nhng ni chat choi cua quan ao v no co the lam van vacong 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, lausach 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 haycac tr chat de bay hi e lam sach may. Neu lam nh vay se mati lp in va co the lam hong vo may.
TRC KHI S DUNG MAY TNH
M nap mayLat may lai (pha lng len tren : thay ro c sau lo inh oc)
dung ngon tay cai ay may len e lay nap ra.
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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 ( CONT )
Man hnh chnh o tng phan hien len. Dung phm , echnh o sang toi. Khi at mc o va y th an AC
[ ] [ ]
CONTRAST
LIGHT DARK
Cung co the chnh o tng phan bang , khi menu MODEhien 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 thMan hnh LCD co 31 96 iem e hien cac k hieu.
K hieu hien th
V du : CMPLX D
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K hieu Y ngha
S Va an phm SHIFT. Neu an tiep mot phm khac nath k hieu nay lan
A Va an phm ALPHA. Neu an tiep mot phm khacna th k hieu nay lan
M Co so nh M c dung
STO Va an SHIFT STO (chuan b nhap gia tr vao tenbien)
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 oR 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
, 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
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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 MATRIXLap 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 N5 : 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 hienth. Man hnh gom hai trang, chuyen nhau bang
1 : MthIO 2 : LineIO
3 : Deg 4 : Rad
5 : Gra 6 : Fix
7 : Sci 8 : Norm
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1 : ab/c 2 : d/c
3 : CMPLX 4 : STAT
5 : Disp 6 : CONT (Xem them phan chnh o tng phan khi dung CONT )
Xac nh dang nhap / xuat
Dang An
Math (MthIO)
Linear (LineIO)
dang Math, phan so, so vo t va cac bieu thc c ghi giongsach giao khoa.
dang Line, phan so va cac bieu thc c ghi chung motdong.
D Math 4 25 3
+
2215
Dang Math
D 4 5 2 3+
22 15
Dang Line
Xac nh n v o goc
n v chon An
o (Deg)
Radian (Rad)
Grad (Gra)
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Xac nh dang so hien th
Goc o90 goc2
= = goc 100 grad
Dang so hien th AnCo 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 lethap 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
1
10
(Sci 5)1.429 110 (Sci 4)
Norm : Co hai dang (Norm1, Norm 2) ghi so dang thngtrong 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 (ab/c)
Dang phan so (d/c)
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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 keThao 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 lethap phan.
Cai at ban auThc hien thao tac sau e lap cai at ban au
(CLR) (SETUP) (Yes)Chi tiet cai at Trang thai ban auMode COMPDang xuat/nhap MathIOn v goc oHien th so Norm 1Hien th phan so d/cDang so phc a+biHien th thong ke OFFDau cach phan le thap phan Dot (.)
Muon bo qua cai at, an (Cancel).
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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 ane thc hien. May t ong thc hien theo th t u tien cua cacphep cong, tr, nhan, chia, ham so va dau ngoac.
V du : 2 (5+4)2(3) =LINE
2 5 4
2 32(5 4) 2 ( 3)
24
+
Nhap ham co dau ngoacKhi 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 nhaptrong dang Math)
Bo qua dau nhanTa 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 Sin(30)
0.5
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Trc k hieu m au (gom ca dau so am) : 2h123 ch ghi2h123
Trc ten bien, hang hay so ngau nhien : 2A, 2 ch ghi2A, 2
Ghi chu quan trong :May ES khong danh u tien cho phep nhan tat nen neu ghi32A th may hieu la 32 A 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 thcMot hay nhieu dau ong ngoac cuoi cung (se an tiep theo) cothe bo qua (khoi an).
Hien bieu thc daiMay co the hien th toi a 14 k t. Khi nhap k t th 15 thbieu thc b ay qua trai va co dau hien len ben trai (chng to
ben trai con co k t). Khi a con tro ve trai neu lam khuat kt ben phai th ben phai co hien dau .
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 chuyenthanh hnh vuong e bao cho biet. Gap dau hieu bao nay cuacon tro hay ket thuc nhap tai iem thch hp va an .
Chnh sa bieu thcNhn vao bieu thc nhap, neu thay sai ta co the chnh sa bangcach ghi chen hay ghi e
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Ghi chen hay e khi nhap che o ghi chen, k t mi c chen ben trai con tro. cheo e, k t mi an se thay the k t tai con tro. Mac nh la cheo chen, ta co the oi sang che o e neu can.
che o chen con tro dng ng (|), che o e con tro namngang ( ).
Mac nh ban au Line la che o chen, ta co the thay oibang 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 dangxuat/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 13D
369 13|
D369 1|
D369 12|
Xoa k t va ham khi nhapV du : chnh sa t 369 12 thanh 369 12
LINE che o e
369 12D
369 12|
D
369 |12
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D369 |12
Che o e :
369 12 D369 12_
D369 12
D369 12
Chnh phep tnhV du : chnh sa t cos(60) thanh sin(60)
LINE che o chen
cos 60 )D
cos(60)|
D60)
Dsin(|160)
Che o e :
60 Dcos(60)_
Dcos(60)
Dsin(60)
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Nhap chen vao phep tnh
Dung che o chen cho thao tac nay. Dung phm hay acon tro en ni can chen roi an k t muon chen.
Hien th cho saiKhi an ma thay may bao loi, an , con tro se hien ngaycho 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 hayD
14 0| 2
cho co loi
1 D14 0|0 2
D14 10 2
2.8
Neu an th phep toan b xoa.
Nhap dang MathKhi nhap dang Math, ta co the cho nhap va hien th phan sova mot vai ham so giong nh trong sach.
Ghi chu :
Vai bieu thc hay k hieu cong thc ln hn mot dong. Chieucao toi a co the en hai man hnh (31 iem 2). Vt qua
gii han nay, khong nhap c.
DMath ERROR[AC] : Cancel[][] : Goto
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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 tonglog
=
8
Gia tr tuyet oi hyp (Abs) 4
Ngoac n (hoac) 1
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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 3
D Math
2 1|+
V du 2 : Nhap 1 2 3+ + MATH
+ 2D Math
1 2|+
3
D Math
1 2 3|+ +
V du 3 : Nhap221 2
5 + =
MATH
1 2 5
2
Khi an va nhan ket qua trong dang Math, mot phan cuabieu thc khi nhap vao co the b che khuat.
Neu can xem lai bieu thc a nhap vao, an va sau o an
2
D Math
21 25
9825
+
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Them k hieu vao bieu thc tnhTrong 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 trongMATH
D Math1 (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 cdung.
Neu con tro ben trai cua mot ham th ham nay c a vaok 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 Math1 + (2+3)+4
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Bieu thc goc 1 + | (X + 3) + 4
Ham Phm an Bieu thc tao mi
Tch phan | (X 3)dx 41 + ++
ao ham ddx
d1 (|(X 3)) xdx
+ + =
Toan log
=
x
1 (|(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 ketqua 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 quathap phan du an hay ieu kien hien th dang (gom ca so vo t) giong nh viecchuyen oi S D (xem them phan chuyen S D)
V du : 2 8 3 2+ =
MATH
1.
2 8
D Math 2 8+
3 2
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2. 2 8
V du 2 : sin(60) = 32
MATH
60
V du 3 :
1 1
sin (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 cua12 x 20<
Gra Boi so cua 503
x 10000<
Khac vi an nh nay ket qua se hien bang so thap phan.
D Math 2 8+
4.242640687
D Math sin(60
32
D Math 1sin (0.5
16
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Ghi chuTrong tnh toan so phc (CMPLX), cac ieu kien sau c apdung cho phan thc va phan ao.
Ket qua dang can ch c hai so hang (mot so nguyen cungc 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.
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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 soa b d e a b d e
c 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 khicac so a, c va d vt qua pham vi cua cac so a, c va d
V du :3 2 10 3 11 2
11 10 110+
+ =
Ket qua hien th dang thap phan khi co ket qua trung giancha 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 soth 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 tramva o, phut, giay.
Tat ca cac muc tnh toan trong phan nay eu cthc hien Mode COMP ( )
Phep tnh so hoc
Dung cac phm , , , e thc hien phep tnh so hoc
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V du : 7 8 4 5 = 36
LINE
7 8 4 5
D
7 8 4 5 36
May t ong tnh theo th t u tien cac phep tnh (Xem themphan Th t u tien cac phep tnh)
So ch so le thap phan va so ch so hien lenTa co the an so ch so le thap phan hay so ch so hien len
phan ket quaV du : 1 6
LINE
Cai ac mac nh (Norm 1) D
1 60.1666666667
An nh 3 so le thap phan (Fix 3) D FIX
1 60.167
An nh 3 so dang na 10 (Sci3) 1
D SCI1 6
0.167 10
(Xem them phan an nh dang so hien len)
Bo qua dau ong ngoac cuoiTa co the bo qua dau ong ngoac cuoi e an ngay dau thchien phep tnh
V du : (2 + 3) (4 1) = 15
LINE
2 3 4 1+
D(2 3) (4 1
15
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Toan phan soTa co the chon cach nhap/xuat phan so theo hai dang sau
Dang Phan so Hon so
Math 73
7 3
123
(SHIFT
2 1 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 honso
Ket qua luon hien dang phan so toi gian
Phu luc
2 1 7
3 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 10th ket qua hien len la so thap phan.
Ket qua mot phep tnh gom phan so va so thap phan luon laso 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.
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Chuyen oi nhau gia phan so va so thap phan
D3 2
1.5
S D
D
3 2
3 2
Dang cua phan so tuy thuoc vao dang cai at ban au (phan sohay hon so).
Khong the oi mot so thap phan ra dang hon so neu tong cack t cua hon so ln hn 10.
Ve chi tiet cua phm S D xin xem them phan chuyen oiS D.
Toan phan tramNhap gia tr va an (%) e lam gia tr nay tr thanh phantram.
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%)
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Toan ve o (gi), phut, giayMay co the thc hien cac phep toan nay va chuyen oi thanh sothap phan hay ngc lai.
Nhap gia trDung 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 tnhThc 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 phanDung phm e chuyen oi nhau gia gi, phut, giay va sothap phanPhu luc oi 2.255 ra dang o, phut, giay.
S DUNG TNH LIEN TIEP TRONG PHEP TNHCo the dung dau hai cham (:) e noi 2 hay nhieu bieu thc vathc 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 Disp3 3
6
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Disp bieu th ay la mot ket qua trung gian cua mot phep tnhlien tiep
D3 3
9
S DUNG BO NH PHEP TNH VA XEM LAI
Bo nh phep tnh ghi moi bieu thc a nhap vao, thc hien vaket qua cua no.
Ch co the s dung bo nh phep tnhtnh trong Mode COMP (MODE 1)
Goi noi dung bo nh phep tnhAn e lay tng d lieu t bo nh phep tnh. Bo nh phep tnhcho xem ca bieu thc phep tnh va ca ket qua cua no.
V du :
LINE 1 1
2 2
3 3
+
D2 2
4
+
D1 1
2
Lu y rang noi dung bo nh phep tnh se b xoa khi tat maytnh, an phm ON , chuyen sang mode phep tnh hoac dangnhap/xuat hoac thao tac bat c hoat ong reset nao.
+
D3 3
4
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Bo nh phep tnh la co han. Khi phep tnh ang thc hien lamcho bo nh ay, phep tnh th nhat t ong b xoa e lay chocho phep tnh mi.
Chc nang hien laiKhi mot ket qua tnh toan ang hien th, co the an phm sauo an hoac e bien tap bieu thc a dung ma a tnhtoan trc o. Neu ang s dung dang LINE, co the hien thbieu thc trc o bang cach an phm hoac ma khongcan 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 dungbo 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 motphep 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 vathc hien phep tnh.
Noi dung bo nh Ans van con ngay ca khi an phm , thayoi mode phep tnh, hoac tat may.
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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) 30D
Ans 300,4
An t ong nhap vao lenh Ans
Vi thao tac tren, can thc hien phep tnh th 2 ngay sau pheptnh th nhat. Neu can goi noi dung bo nh Ans sau khi an ,an tiep .
Nhap noi dung bo nh Ans vao mot bieu thcV du : e thao tac phep tnh sau ay
123 + 456 = 579 789 579 = 210LINE
1 2 3 4 5 6 +D
123 456579
7 8 9
D
789 Ans210
Bo nh oc lap (M)Co the lam phep tnh cong them hoac tr i ket qua trong bonh oc lap. Ch M hien th khi bo nh oc lap co lu motgia tr.
D3 4
12
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Mieu ta chung ve bo nh oc lap Sau ay la tom tat mot so thao tac co the s dung bo nh oclap.
Y ngha An phmThem gia tr hoac ket qua hien th cuabieu thc vao bo nh oc lap
Bt i gia tr hoac ket qua hien th cuabieu 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 maytnh s dung noi dung bo nh oc lap tai v tr o. Di ay lacach an phm e chuyen bien so M.
(M)
Ch M hien pha tren ben trai khi co mot gia tr nao o khac0 c lu trong bo nh oc lap.
Noi dung bo nh oc lap van con ngay ca khi an phm AC thayoi 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 trckhi 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.
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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 A3 + 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 oimode phep, ke ca khi tat may.
Phu luc
Xoa noi dung cua toan bo nhS dung cac thao tac sau e xoa noi dung cua bo nh Ans, bonh 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 onhap gia tr bien e tnh.
Chc nang CALC s dung c trong
mode COMP (MODE 1) va mode CMPLX (MODE 2)
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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 phmV du :LINE
3 (A)D
3 A
Yeu cau nhap gia tr A
Gia tr A hien hanh
5D
3 A15
(hoac )
D
A ?5
10D
3 A30
Thoat an .
Neu bieu thc co hn mot bien, may yeu cau nhap tng bien.
DA ?
0
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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.
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Thao tac mau cho chc nang SOLVEV du : Giai 2y ax b= + theo x khi y = 0, a = 1, b = 2.
MATH
S D
(Y) (=) (A)(X) (B)
(,) (X) 2
D 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 X
0
gia tr X hien hanh
2
D Math
Y AX B,XX 1,414213562
L 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.
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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 dutrong phan nay eu mode COMP (MODE 1)
Cac phep tnh can co mot thi gian e hien th ket qua. Khiang 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 tnhSau 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 ctrong mode CMPLX (so phc) neu cac ham nay bo tr hp l
n v goc ma cac ham lng giac va lng giac ngc macnh co san tren may. Trc khi tnh can chon n v gocmuon 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 phmtng ng e goi ham muon nhap
Phu luc sinh 1 = 1.175201194, 1cosh 1 0 =
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oi n v goc (Ket qua theo n v goc mac nhcua 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 phmso lien he e ghi n v goc cua gia tr nhap vao. May tnh se tong 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
LINEx10 () 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
g
D
5045
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Neu ch nhap mot gia tr n, may hieu la logarit thap (v dulog(100) = 2)
ln(ham logarit t nhien vi c so e.
Cung co the dung phm log khi nhap bieu thc vi danglog(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 sonao 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 (Giaiphng trnh), MATRIX (Ma tran), TABLE (Lap bang) vaVECTOR (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 hamnay ch anh hng en rieng tng phan thc va ao (hay mot vagiac so), (khong anh hng toan bo mot so phc)
Phu luc en
Phep tnh tch phanMay 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)
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Ta co the bo qua phan tol. Mac nh la 51 10
(, d / dx(, Pol(, Rec( va (khong dung c trong f(x), a, bhoac 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 tnhtch 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 oidau tren khoang tnh tch phan.
An e ngng tnh tch phanHoan 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 tngoan dng, am rieng roi tnh tng oan nho va cong lai.
b
a
f(x)dx =c
a
f(x)dx +b
c
( f (x)dx)
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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)
721( ,1,5,1 10 ) 0,8x
=
ao ham:Ta co the tnh gan ung gia tr ao ham cua mot ham so taiiem 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 hamtol : 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.
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Neu thao tac chuyen oi toa o ben trong mot bieu thc thaycho mot bieu thc ng mot mnh, phep tnh khi thc hien chs dung gia tr au tien (ca gia tr r hoac gia tr X) tao ra bichuyen oi.
V du : Pol( 2, 2) + 5 = 2 + 5 = 7.
Phu luc ti
Cac ham khacPhan 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 lnggiac. Rieng Abs (va Rnd(co the dung trong so phc (modeCMPLX)
Giai tha (!)Chc nang nay ch tnh c giai tha cua 0 hoac so nguyendng
Phu luc (5+3)! = 40320
Phep tnh gia tr tuyet oi (Abs)Khi ang thc hien phep tnh mot so thc chc nang nay choket qua la gia tr tuyet oi.
Phu luc Abs (27) = 5
So ngau nhien (Ran#)Chc nang nay sinh ra mot songau nhien vi 3 ch so nho hn 1.
Phu luc
Sinh ra nhng songau nhien 3 ch so.Cac gia tr thap phan 3 ch songau nhien c chuyen sang cacgia 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 maytnh sinh ra se khac.
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Cac dang thc ho tr cho chuyen oi S DChuyen oi S D co the c s dung e chuyen oi cac ketqua phep tnh thap phan a hien th thanh mot trong nhngdang sau
Lu y
Khi chuyen t dang thap phan sang dang S, may tnh se tquyet nh dang S e dung. Khong the ch nh dang SPhan so : Cai at dang thc phan so hien th hien hanh la motphan 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 bac
(tuy thuoc vao cai at dang thc hien th phan so)
Chuyen oi sang dang thc , phan so b gii han en ket quava ham lng giac ngc thng bieu hien bang radian.
Sau khi co mot ket qua tnh toan trong dang , co thechuyen sang dang thap phan bang cach an phm S D. Khi ket
qua phep tnh ban au dang thap phan, no khong the chuyensang 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 Math
5b56
* An phm S D e thay oi gia 2 dang.S D 0.8333333333
S D 5
6
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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 hientrong Mode COMP ( )
Phu luc
(1 + 3i) 3 1(2i) i2 2
=
Nhap so phc : Trong mode CMPLX, phm chuyen thanh phm nhap kyhieu i (ao), dung i e ghi a + bi. V du sau la nhap
2 + 3i
2 3 + CMPLX D Math2 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
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V du 2: 1 i 2 45+ = (n v goc la o)
MATH
1 i +
CMPLX D Math
1 i 2 45
Giac so hien trong pham vi o o180 180 <
An nh dang ket qua hien thTa co the an nh dang ket qua hien th nh sau
e ket qua hien th dang toa o e-cac, ta an2 (CMPLX) 4 ( a bi) +
e ket qua hien th dang cc, ta an2 (CMPLX) 3 (( r ))
Phu luc
1 i( 2 45)+ = = 1.414213562 45
So phc lien hpThao 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
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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 ke1 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 mauHien th man hnh STAT
Man hnh STAT xuat hien sau khi ban truy nhap Mode STAT tmot mode khac. S dung menu STAT e chon mot kieu tnhthong ke t mot man hnh STAT khac.
An (STAT) (Data)
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V du : Nhap gia tr 123,45 vao o X1
(Chuyen con tro en o X1)
STAT D
1 X Y2
3
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 achon
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 tMode STAT sang mode khac chuyen cai at hien th thong ke(lam cho cot FREQ an hay hien) tren man hnh cai at.
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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 anChu y !
Lu y phai thay toan bo d lieu hien co cua o bang nhap somi. 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 muonxoa
(2) An
Chen vao mot dong
(1) Tren man hnh nhap STAT, chuyen con tro en dong muonchen 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 dongcho 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 dongva Xoa toan bo noi dung STAT khi man hnh STAT hien th.
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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 chuanbnh thng. Bien t c tnh t gia tr trung bnh x va o lechtieu chuan
nx nhan c t d lieu man hnh nhap
Phan phoi tieu chuan bnh thng
=n
X 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 nhv du . Tnh tong bnh phng cua d lieu mau va tong cua d
lieu mau
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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 Tong2x y
8:4
x Tong d lieu X luy tha 4
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Menu phu Var (SHIFT 1 (STAT) 5 (Var)
Kieu Y ngha
1 : n So cac mau
2 : x Trung bnh cua d lieu X3 : 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 X3: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
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Phu luc Hoi quy tuyen tnh en
Cac v du en eu s dung d lieu nhap trongv 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 hai2( CX )+
Vi hoi quy bac hai, c thc hien theo phng trnh mau diay
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 mautrung bnh, o lech chuan) va menu phu MinMax (gia tr lnnhat, 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
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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 modeBASE 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 trTrong 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 hienhanh (nh nhap 2 he BIN) Khong nhap c phan so hay ham mu trong BASEN. Neutnh 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.
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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 hienmenu 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 2cua BIN, roi chuyen lai theo c so a chon. So am trong DEC thmang dau tr ()
Phu luc
T v du en la cac phep tnh ve so am tronghe em nh phan (BIN) nen khi thc hien phai an trc
BIN
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Quy tac nhap va chnh cac he so
D lieu nhap c chen vao cac o co con tro. Khi nhap d lieuvao 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 bac3 c chon, cot d se khong hien th tren man hnh he so hienth lan au. Cot d se xuat hien khi dch con tro en cho no, lamcho man hnh oi cho.
Cac gia tr va bieu thc nhap vao man hnh he so giong nhnhap 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 en6 ch so trong o la chon hien hanh
e thay oi noi dung cua cac o, s dung cac phm con tro e dichuyen 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 trenman 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 sogiong nh chu y nhap d lieu man hnh nhap STAT. iem khacnhau duy nhat la lu y au tien cua chu y nhap d lieu manhnh 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 hnhnhap he so
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Trong trng hp cac phng trnh tuyen tnh, co the dungva 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 nghiemphu thuoc vao phng trnh thc te.
An khi mot nghiem c hien th, se tr ve man hnh nhaphe so
Dang hien th cua nghiem phu thuoc vao dang nhap/xuat vadang 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 thuatkhi 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 matran. 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
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(2) An hay e chon ma tran A, B, C va ta c manhnh chon kch thc ma tran.
(3) An 1 en 6 e chon kch thc thch hp. Sau khi chon kchthc (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 themcach 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 hienman 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.
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Thc hien tnh toan An khi co man hnh chon hay nhap ma tran e vao manhnh tnh toan.
Man hnh ma tran ket qua (MatAns)
Man hnh ma tran ket qua (MatAns) ch ket qua cua phep toanma 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 themphan dung MathAns e thc hien chuoi tnh toan)
Menu ma tranTrong 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
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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 haima 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).
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(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 quytac 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 vaobien 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 bangso
Lu y rang phep tnh tao ra bang so lam cho cac noi dung cuabien X thay oi
1
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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 raloi, do o bang so se khong c tao ra
Cac gia tr ban au, cuoi va bc nhay se sinh ra mot gia trtoi a la 30 gia tr x cho bang so c tao ra. Lap ra mot bangso s dung cac gia tr ban au, cuoi va bc nhay nhieu hn 30
gia tr x se gay ra loiLu y
Mot so ham va gia tr ban au, cuoi va bc nhay can nhieuthi gian e tao ra bang so
Man hnh bang soMan hnh bang so cho thay cac gia tr x tnh toan s dung cacgia tr ban au, cuoi va bc nhay, cung nh cac gia tr thu ckhi 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 cuabang 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 Mathhoac dang Line) tren man hnh cai at Mode TABLE se xoaham tao lap bang so
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Chep toa o cua vect nay sang vect khac
Co the chep toa o cua mot vect (hay cua VctAns) sang motvect 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 thchien
Khong the chnh noi dung cua VctAns
Chuyen ve man hnh tnh toan vect, an
Menu vectBang 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 (matphang hay khong gian) cho cac vect nay
2:Data Goi VctA, VctB, VctC e hien toa o va chnh satoa o
3:VctA Nhap VctA
4:VctB Nhap VctB
6:VctAns Nhap VctAns
7 Dot Nhap dau (e lay tch vo hng 2 vect)
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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 n
vuong goc vi VctA, VctB.
*1(A B)
cos A B
= vi =1 (A B)
cos A B
2* (A B)nA B
=
HANG SO KHOA HOC
May lu 40 hang so khoa hoc e dung trong tnh toan thongthng. Co the dung hang so khoa hoc trong moi phep toan trtrong BASE N.
Muon goi mot hang so khoa hoc, an (CONST). May hienmenu bang so khoa hoc. Nhap 2 ch so tng ng vi ten hangso can goi khi o k hieu va gia tr cua hang so hien len man
hnh.
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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
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OI N V
May cai san oi n v o lng. Co the oi n v trong Modetr Mode BASEN va lap bang.
e goi lenh oi n v, an (CONV). Man hnh hien menuoi 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 sotng 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 ( )
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THONG TIN K THUAT
Th t u tien cac phep tnhMay 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 : t
Phan tram : %
3. Phan so : ba / c
4. Dau: () (dau am)
d, h, b, o (c so n)
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5. : cm in, v.vPhep 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 tienbang nhau (3 5A hay 3 5 A eu giong nhau ve th t utien)
9. Phep cong, tr : +,
10. and11. or, xor, xnor
Neu mot phep tnh cha mot gia tr am, can ong gia tr am otrong ngoac n. Chang han, khi bnh phng 2 can phainhap 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 (utien 8). Do o, khi ca hai yeu to cung hien dien th phep tnhc thc hien t trai qua phai. Neu a dau ngoac n vao selam thay oi phep u tien va ket qua tnh toan se khac nhau
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86/137
85
V du:
1 2 i 11 2i i2
=
1 2 i 1
1 (2i) i2 =
Gii han Stack (nhom)May tnh nay s dung cac khu vc nh goi la stack e lu tamthi cac gia tr day phep tnh, cac lenh va chc nang u tienthap hn. Stack so co 10 mc va stack lenh co 24 mc nh hnhsau ay
Loi Stack ERROR hien len khi phep tnh ang thc hien co mottrong 2 stack b vt qua han
Pham vi cua phep tnh, so ch so va o chnh xacCac day phep tnh, so ch so dung cho phep tnh va o chnhxac phu thuoc vao kieu phep tnh ang thc hien.
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8/6/2019 Sach Fx 570ES
87/137
-
8/6/2019 Sach Fx 570ES
88/137
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)
nPr0 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
o
100x 1 100< oi so thap phan o phut giayo
0 0 0 x o
9999999 59 59
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8/6/2019 Sach Fx 570ES
89/137
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8/6/2019 Sach Fx 570ES
90/137
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 (nhchia 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 bienphai 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 venhom
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 lXem lai va ieu chnh
Argument ERROR (Loi Argument)
Nguyen nhan
Co sai sot ve Argument
Cach x l
Xem lai va ieu chnh
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8/6/2019 Sach Fx 570ES
91/137
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 chathch hp
Cach x l
An nh lai Dim cho vect hay ma tran trc khi nhap cacyeu 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 vabc nhay
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8/6/2019 Sach Fx 570ES
92/137
91
Time Out Error
Nguyen nhan
Khong thoa man ieu kien e phep tnh ao ham hay tch phan
thc hienCach x l
Tang gia tr tol (co the lam giam o chnh xac) hay kiem lai bieuthc nhap
Trc khi xac nh la may tnh truc trac
Thc hien cac bc sau ay khi loi xay ra luc tnh toan hoac khicac ket qua tnh toan khong phai la ap so ung nh mongmuon. Neu bc trc khong sa c th chuyen sang bc tieptheo.
Lu y la nen ghi nhng d lieu quan trong khi thc hien cacbc kiem tra cac bieu thc tnh toan e chac rang khong concha loi nao ca.
(1) Chac rang la s dung mode ung cho phep tnh ang thchien.
(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 tmthay 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 caiat gia tr ban au xem them Cac mode phep tnh va cai atmay tnh
(3) Cho gia tr ban au tat ca cac mode va cai at nh sau :
(CLR) (Setup) (Yes)
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8/6/2019 Sach Fx 570ES
93/137
92
NANG LNG
May tnh nay c cung cap nang lng bi pin AAA (R03(um4)
Thay pinCac con so b m tren man hnh hien th cho biet pin yeu. Khimay pin yeu neu tiep tuc s dung co the lam may tnh sai. Haythay pin cang sm cang tot khi ch cha b m nhieu. Ngay catrng hp may ang s dung bnh thng, hay thay pin t nhat2 nam 1 lan.
Chu y !
Lay pin ra khoi may tnh se lam noi dung bo nh oc lap vacac gia tr ghi trong cac bien b xoa.
(1) An e tat may tnh
(2) Thao inh oc mat sau va laynap 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 tactren, ng bo qua thao tac nao ca
Chc nang t ong tatMay t ong tat sau khoang 6 phut khong thao tac.
An e m lai.
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8/6/2019 Sach Fx 570ES
94/137
93
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8/6/2019 Sach Fx 570ES
95/137
94
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8/6/2019 Sach Fx 570ES
96/137
95
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8/6/2019 Sach Fx 570ES
97/137
96
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8/6/2019 Sach Fx 570ES
98/137
97
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8/6/2019 Sach Fx 570ES
99/137
98
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8/6/2019 Sach Fx 570ES
100/137
99
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8/6/2019 Sach Fx 570ES
101/137
100
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8/6/2019 Sach Fx 570ES
102/137
101
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8/6/2019 Sach Fx 570ES
103/137
102
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8/6/2019 Sach Fx 570ES
104/137
103
-
8/6/2019 Sach Fx 570ES
105/137
104
-
8/6/2019 Sach Fx 570ES
106/137
105
-
8/6/2019 Sach Fx 570ES
107/137
-
8/6/2019 Sach Fx 570ES
108/137
-
8/6/2019 Sach Fx 570ES
109/137
108
-
8/6/2019 Sach Fx 570ES
110/137
109
-
8/6/2019 Sach Fx 570ES
111/137
110
-
8/6/2019 Sach Fx 570ES
112/137
111
-
8/6/2019 Sach Fx 570ES
113/137
112
-
8/6/2019 Sach Fx 570ES
114/137
-
8/6/2019 Sach Fx 570ES
115/137
114
-
8/6/2019 Sach Fx 570ES
116/137
115
-
8/6/2019 Sach Fx 570ES
117/137
-
8/6/2019 Sach Fx 570ES
118/137
117
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8/6/2019 Sach Fx 570ES
119/137
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8/6/2019 Sach Fx 570ES
120/137
119
-
8/6/2019 Sach Fx 570ES
121/137
120
-
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122/137
121
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8/6/2019 Sach Fx 570ES
123/137
-
8/6/2019 Sach Fx 570ES
124/137
123
-
8/6/2019 Sach Fx 570ES
125/137
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8/6/2019 Sach Fx 570ES
126/137
-
8/6/2019 Sach Fx 570ES
127/137
-
8/6/2019 Sach Fx 570ES
128/137
127
-
8/6/2019 Sach Fx 570ES
129/137
-
8/6/2019 Sach Fx 570ES
130/137
129
-
8/6/2019 Sach Fx 570ES
131/137
130
-
8/6/2019 Sach Fx 570ES
132/137
131
-
8/6/2019 Sach Fx 570ES
133/137
-
8/6/2019 Sach Fx 570ES
134/137
133
-
8/6/2019 Sach Fx 570ES
135/137
134
-
8/6/2019 Sach Fx 570ES
136/137
135
-
8/6/2019 Sach Fx 570ES
137/137