16287126-Matematika-I-dio

8/6/2019 16287126-Matematika-I-dio

1/51

Via tehnika kola

Doboj

Vjebe

Prof.Vesna Mii

Doboj 2003


2/51

Stepen

aaaaan =

n-inilaca

mnmn aaa += Proizvod stepena istih bazamnmn aaa =: Dijeljenje stepena istih baza

( )( ) +

+

T

T

yT

yT

yxx

yxx

,,

,,

Stepenovanje stepena

mnn m aa = Korjenovanje korijena

( )

n mn

m

n

n

aa

a

aa

a

=

=

=

1

01

0

( )

( ) ( ) ( ) 12100110

1000

11100

121

1000

11:

100

121011,0:21,1

16

81

2

3

3

72

7

3

7:

2

7

3

12:

2

13

:

30532532

:

2733

63232

43

2

2

2222

4

4

4444

2

524322432

32525

6 43 333

==

=

=

=

=

=

=

==

==

==

==

==

=

++

++++

bcbacbacbacba

abaabbabaabba

xxxx

xxxxx

aaaa

aaaaa

aaa

pn pmn m

n pmpn m

aa

aa

=

=

abba

xxxx

xxx

xxx

nn ==

=

==

==

333

3

2

2

3

13 2

6 423 223 2


3/51

( (

12512

51

12

17

12

98

4

3

3

2

43

32

3333333333333

22222222.6

752

125812550-205020-85225104.5

xxxxxxxxx =====

=+=+=+++=++

+

x xx xxx xxx x

xx xxx xxx

x xx

x

x

x

xxxxxxxxxx

xxxxxx

xxxxxxxxxxxxx

xxx

xxxxxxxxxx

6 436 10636 106 636 102 2

2 3 362 232232

6 436

43

6

133

6

1

2

111

2

1

6

1

2

11

2

2

2

1

32

::

:::II.Rj.

::I.Rj..7

+++

+

+++

====

===

===

==

=

LogaritamLogaritam pozitivnog realnog broja xpo bazi a je eksponentykojim treba stepenovati bazu a da bi se dobio brojx.

100:

log

= =

axD

xayx ydefinicija

a

4

122

4

1log

01log12log

8238log

22

22

32

==

====

a - baza

( ) ( )

( )

3

7

4

3

4

1

3

4

4

3

7

4

7

3

4

4

3

17

36

36

119

4

7

6

8

4

3

1736

36139594

7

4:

100

8108

10046410

8

54

5

3

1736

413

959

7

4:

25

2

100

108

251

10064

100

125

5

4:

10

8

5

4:5,02,1

17

22

4

13

9

56

7

4:

25

208,1

25

164,0

25,15

4:8,0

.4

8228321664321664321664.3

:::2.

1.

5425

1

4

1

2

12,025,05,0

221

21

13

3

3

1

3

1

3

1

=++=++=+

+=

=+

+

=+

+

=+

+

=+=+=+=+

====

===

m nmn

m

nnm

m

nnm

m

mn

m

mnnm m

nm m aaaaaaaa

xxxxxx


4/51

x - numerus

xnx

yxy

x

yxyx

n loglog

logloglog

logloglog

=

=

+=

718281828,2

n

x n

xx

+=

=

11

lnlog

lim

4

16

16216log.8 2

=

=

==

x

x

xx

81

1

27

1

274

3-27log.9

3 4

4

3

=

=

==

x

x

xx

5

5

1

100

100

1002,0log.10

=

==

x

xx

32

23

2

1

2

1

4

1

8

1

4

1log.11

23

8

1

=

=

=

=

=

x

x

x

x

x

( ) ( )

x xx xxxx xx x

xx

x

x

xx

x

xx

x

x

x

xx

x

x

x

x

x xx x

aaaaaaa

aaa

aaaaaaaa

6 33166 96424182 323 243

6x 33166

3316

6

962422

6

32312112

2

32

3

2129

2

32

3

243

2 323 243

::II.Rj.12.

:::.12

++

+++++

+

==

===

====

Linearne jednaineBxA =

I. 0A ABx = -jedinstveno rjeenje

II. 0=A Bx =0 i.) 0=Bii.) 0B

i.) 0=B , 00 = x jednaina je neodreena ( )Rx ii.) 0B 00 = Bx 00 nemogue, nema rjeenja

( )0: = BBAB

A


5/51

0;:

= BDBD

BCAD

D

C

B

Asabiranje razlomaka

0; = BD

BD

AC

D

C

B

Amnoenje razlomaka

BC

AD

C

D

B

A

D

C

B

A

BCBC

AD

D

CB

A

D

C

B

A

==

==

:

0;:

dijeljenje razlomaka

bax += jedinstveno rjeenje

I.bax

abba

+=+ 022

II. 00

022

=

=+

x

abba

neodreeno rjeenje( )Rx

( ) ( )bababa += 22 razlika kvadrata

( ) 222 2 bababa += kvadrat binoma

( )

( )

( ) ( )( ) ( )

( ) ( )654233245677

43223455

2233

32233

222

33

2

babbabababaababa

babbabaababa

babababa

babbaaba

IIIIIIIII

++++=+

++++=

+=

++=

+=

( ) ( )

( ) ( )

( ) ( ) ( )( ) ( )

22

22

2222

3322

3232

22

/1

1

1

111.13

baba

bababax

bababababax

baxbabxxa

abxbxbaxa

abxabx

bxbaxa

ax

bxb

bx

axa

x

bx

a

b

x

ax

b

a

x

b

a

b

x

a

b

a

+++=

++=+

+=+

=+

=+

=+

=

+

=

+


6/51

1615201561

15101051

14641

1331

121

11

1

( ) 543223455 510105 babbababaaba +++++=+

( ) ( )=

=+n

k

kknnk

nbaba

0

Binomna formula

bx 8= jedinstveno rjeenje

I.bx

ab

8

02

=+

II.00

02=

=+x

ab neodreeno rjeenje ( )Rx

( ) ( )

( )2

2

2

22

22

2

22

2

22

211

211

2-1/21

2

1

1

1

1

1

2

1

1

1

1.14

ax

aaxaxx

aaxxaxx

xaxa

a

x

ax

x

ax

xa

a

ax

ax

=

=++

=++

=+

+

=

++

2ax = jedinstveno rjeenje

( )

( )

( )

( ) ( )

( ) ( )

( ) ( ) ( )( )

( )( ) 3

22

2

2

3

2

2

23

2

113

112113

11

3

1

1213

1

12

1

3

1

13.15

+ ++=+ +++

++

+=

+

+

+

+

++

+=

+

+

aaaab

aaxaxaab

a

a

a

ab

aa

xa

a

xab

aa

xa

a

ab

a

ax

a

ab

( ) ( ) ( )

( ) ( ) ( )ababbabx

abbaxbx

babbxaxabbx

babxbax

b

bx

ba

ax

++=++=+

+=+

+=+

=

++

2:/282

8162

164442

4422

2

4

4

2.13

2

2


7/51

( ) ( )

( ) ( )

( ) ( )

( )

( ) 2

223

3

223

3

223

2

223

3

223

2

2

1

363

1

363

1

363

1

1212363

1

363

1

12113

+

+++

+

+++

=

+

+++

=+

+++++

+

+++=

+

++

aa

aabbaba

a

aabbaba

x

a

aabbaba

aa

aaaabbaba

x

a

aabbaba

aa

aaabx

I. ( )011

++

= aa

ax jedinstveno rjeenje

II. ( ) 036301

363 2232

223

=+++=+

+++aabbabaaa

aabbaba

00 == AB

A

B

Anije definisan za 0=B

00 =x neodreeno ( )Rx

( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

0;2

2

1

/11

111

111.16

2

3

22332222223

22223

2222222

2

2222222

232223433

=

=

+++++=++

+++=++

+++=

++++

=++

+++

+

=++

+

+

aa

x

aax

xaaxxaxaaxaaxxaxaxx

aaxxxaaxaxx

xaaxxxaxaxxa

xaaxxxaxaaxx

xxaxaaxxx

a

aaxxxa

x

xxaxxaaxx

a

xa

x

za 000 == xa neodreeno rjeenje ( )Rx

Sistem linearnih jednainaDeteminante

22018456365

431221

22

11 ==== bababa

ba

123132213312231321

33

221

33

221

33

221

333

222

111

cbacbacbacbacbacba

ba

bac

ca

cab

cb

cba

cba

cba

cba

++=

=+=


8/51

rjeenja

123123123321321321

33

22

11

333

222

111

bacacbcbabacacbcba

ba

ba

ba

cba

cba

cba

++=

+++

231312123132213321

333

222

111

cbacbacbacbacbacba

cba

cba

cba

++=

amnmvrstan kolona

434241

333231

232221

14

434241

343231

242221

13

434241

343331

242321

12

444342

343332

242322

11

44434241

34333231

24232221

14131211

aaa

aaa

aaa

a

aaa

aaa

aaa

a

aaa

aaa

aaa

a

aaa

aaa

aaa

a

aaaa

aaaa

aaaa

aaaa

+=

=

++

==

==

==

=++= =+

zzzz

yyyy

xxxx

zyx

zyx

zyx

342

1223

32.17

( ) ( ) ( ) ( ) ( ) ( )

111246466

143

221

113

1131641244

131224122431221122

142

223112

=+++=

=

=+=

=++= =

x


9/51

33981236212

342

123

312

2291229122

132

213

132

=+++==

=+=

=

z

y

311

33

211

22

111

11

=

=

=

=

=

=

=

=

=

zz

yy

xx

3;2;1 === zyx rjeenja jednaine ima u kolonama koeficijente uz xyzx u prvoj koloni umjesto koeficijenata uz x ima slobodne lanovey u drugoj koloni umjesto koeficijenata uz y ima slobodne lanovez u treoj koloni umjesto koeficijenata uz z ima slobodne lanove

Kvadratna jednaina

02 =+++ cbxaxacbD 42 += - Diskriminanta

a

acbbx

2

42

2,1

= Rjeenja kvadratne jednaine

a

Dbx

22,1

=

RxD 2,10

RxD 2,10 2,1x - konjugovano kompleksan broj

( ) ( )212 xxxxacbxax =++ Rastavljanje kvadratnog trinoma na

linearne inioce

acxx

a

bxx

=

=+

21

21

Vietove formule

3

2

9

4

49

049.18

2,1

2

2

2

=

=

=

=

x

x

x

x


10/51

( )

5

3-0

0355

3

010

33

03552

05493

035035.19

22

1

12,1

2,1

22

==

=+==

=

=+

=

=+=+

xx

xx

xx

xxx

xxxx

( ) ( ) ( )

bx

bx

bxbxbxbxb

bxbxbxb

xb

xb

b

xb

== +=+

+=+

+=

+

2,1

22

222

22

2

2.20

( )

( ) ( )

( ) ( )

aab

bt

bab

a

ab

babat

abbabat

ab

abbabat

tbaabt

1

2

1

2

2

2

2

2

14

01.21

21

2,1

2

2,1

2

====++

=

+=

++=

=++

ix

ix

ix

x

xx

+==

=

=

=+

1

1

2

22

2

12442

022.22

2

1

2,1

2,1

2

2

4

4

044.23

2,1

21

21

2

==+=+

=+

x

xx

xx

xx

( ) 044.24 2 =++ kxkx sastaviti jednainu ija su rjeenja

22

11

21,

21

xz

xz +=+= .

( ) ( )

48

44

18

44

8

44

8

161684

42

4444

2

1

2,1

2

2,1

2

2,1

kkkx

kkx

kkx

kkkk

x

kkkx

=++

=

=++

=

+=

+++

=

++

=

( )

0243

38

08

3

8

4

2

1

31

21

2

2

1

=+

++

=

+

+=+=

=+=

k

kz

k

zzkz

k

kzz

k

k

kz

z

( ) 024384

024338

2

2

=++

=+

kzkkz

kkzzkzzk


11/51

25. Neka sux1 i x2 korijeni (rjeenja jednaine) ( ) 0532 2 =+ mxmx gdje je Rm sastaviti jednainu iji su korijeni :

a) 211 2 xxy += i 212 2xxy +=

b) u dobijenoj jednaini odrediti m tako da je 21 5xx =

c) za odgovarajue vrijednosti m odrediti odgovarajue vrijednostix1ix2.

( )

( ) ( )

4512

5-21

2

1

2

512

2

5

4

73

14

4

4

73

4

73-

4

49143

22

254963.)

2

1

2

1

2,1

2

2,1

2

2,1

=+=+=

=

+=

=

+=

==+

=

=

+=

+

=

mmm

y

mmy

mmmx

mmx

mmx

mmmx

mmmmxa

( )

313

399

14010

2

145

1

33

5582

2

1

54

5.) 12

=

==

=

==

=

=

=

m

m

mm

mm

m

m

mm

m

m

yyb

( )

1

3

3

2

032

0642

051312

1.)

2

1

21

21

2

2

2

==

==+

=+

=+

==

x

x

xx

xx

xx

xx

xx

mzac

0123

0246

03

2

3

42

053

133

3

132

3

13

2

2

2

2

=

=

=

=+

=

xx

xx

xx

xx

mza

( )

16

42

3

1

6

42

6

42

6

13442

2

1

2,1

2,1

=

+

=

=

=

=

=

x

x

x

x


12/51

Sistemi linearne i kvadratne jednaine

0

0

22

11

=+++++

=++

feydxcybxyax

cybxa

Rjeava se metodom zamjene ili supstitucije, tako da se iz linearne

jednaine jedna nepoznata izrazi preko druge i uvrsti u kvadratnujednainu.

082

01523819

01684641236278

042223

4412932

0422

23

2

2332

2

32

232

04232.26

2

2

22

22

22

22

=+

=+

=++

=+

++

++

=+

++

+

+=

==++

xx

xx

xxxxx

xxxxx

xx

xx

xy

xy

yxyx

2

4

8

2

2

1

21

21

==

==+

x

x

xx

xx

( )

4

2

223

52

243

2

1

=+

=

=+

=

y

y

A (-4,-5) B(2,4)

2

4

328

13945

945

13

459.27

2,1

2

2

22

22

22

22

==

=

=+

=

=+

=+

y

y

y

yy

yx

yx

yx

3

9

4945

2,1

2

2

==

=

x

x

x

7203616

180369

18094

204

18094

204.28

2

2

22

22

=+

==+

==

=

=+

=

vu

vu

vu

vu

vy

ux

yx

yx

6

6

36

36

90025

2

1

2

==

=

==

x

x

x

u

u

2

2

4

36204

20436

2

1

2

2

2

==

=

=

=

y

y

y

y

y

A (-3,-2) , B (-3,2)

C (3,2) , D (3,-2)

R={(-6,-2);(-6,2);(6,-2);(6,2);}


13/51

4

3

1

42

344497

0374

03

74

:

16251625

03

74

0374

1625

:/0374.28

2

1

2,1

2

22

22

2

22

=

=

=

=+

=+

=

=+=+

=+=+

=+

=+

t

t

t

tt

tt

ty

xsmjena

yxyxyxyx

x

y

y

x

xy

y

xy

xy

xy

x

yxyx

xyyxyx

9

34

9

34

33

4

27

16

1627

16962445

164

32

4

35

1625

2

1

2

2

222

222

22

iy

iy

iy

y

y

yyy

yyy

yxyx

=

=

=

=

=

=+

=+

=+

iix

iix

27

316

9

34

3

4

33

43

934

2

1

==

==

1

1

1

1625

2,1

2,1

2

222

=

==

=+

x

y

y

yyy

( ) ( )

= iiiiR

9

34,

27

316;

9

34,

3

3;1,1;1,1

II4

3

4

3

I

1

yx

y

x

yx

y

x

=

=

=

=


14/51

Reciprone jednaine treeg stepena

( ) ( )( ) ( )( ) ( ) ( )

( ) ( )[ ]( ) ( )

3

4

24

32

24

725

4

3

24

18

24

72524

725

122

121242525

1

01225121

0371121

01371112

0137112

111037371212

012373712.29

3

2

3,2

2

3,2

2

2

2

3

2323

23

==+

=

==

=

=

=

==+

=++

=++

=

++==+

=+

x

x

x

x

x

xxx

xxxx

xxxxx

xxx

xxxxxxx

xx

= ;

3

4;

4

3;1R

2

5

4

10

4

37

14

4

4

37

4

37

22

524497

0572

09742

21

21

:smjena091

71

2

027

972

:/027972

2

1

2,1

2,1

2

2

2

2

2

2

2

2

2234

==+

=

==

=

=

=

=+

=+

=

+

==

+=+

+

+

=++

=++

t

t

t

t

tt

tt

tx

x

tttx

xx

xx

x

xx

xx

xxxxx

01

11

2 =+

=

+

xx

xx

0252

2

51

2 =+

=

+

xx

xx


15/51

2

31

2

31

2

31

2

11411

2

1

2,1

2,1

ix

ix

ix

x

+=

=

=

=

24

8

4

35

21

42

435

4

35

22

224255

4

3

4,3

4,3

==+

=

===

=

=

x

x

x

x

+

= ;2;2

1;

2

31;

2

31 iiR

Iracionalne jednaine

Iracionalna jednaina je jednaina u kojoj se nepoznata nalazi pod

korijenom.

[ )

( ) ( )

3

279

10252

102521

/521

21292-12

921121

/32-1

,2

21

0201:321.30

22

2

2

2

==

+=+=+

=+

+=+

=+++++

=++

+

+=++

x

x

xxxx

xxxx

xxx

xxx

xxxx

xx

x

xx

xxDxx

[ )

( )

( ) ( )

[ )+==

==

++=

++=+

+=+

=+++=+

+=+++

+

+++=+

,7:119

11

0992

12982

127142

/17142

2-:/2271422

1427571422

5771422142

,7:

577

0507014257142.31

2

1

2

22

2

2

Dxx

x

xx

xxx

xxxx

xxx

xxx

xxxxx

xxxxx

xD

xxx

xxxxxx


16/51

Trinomna jednaina

03 =++ cbxax nn

( ) ( )

2

31

2

3112

11411

1

011

01

1

8

1

089

089.32

3

2

3,2

1

2

3

33

2

1

2

336

ix

ix

x

x

xxx

x

x

t

t

tt

txxx

+=

=

=

==++

=

=

==

=+

==+

( ) ( )

31

312

322

12

41442

042

2

0422

08

8

6

5

6,5

6,5

2

4

2

3

3

ix

ix

ix

x

xx

x

xxx

x

x

+=

=

=

=

=++

==++

=

=

++

= ;31;31;2;2

31;

2

31;1 ii

iiR

Eksponencijalne jednaine( ) ( )

( ) ( )xxgf

gfaa xx ==

5

3412

88

8

18

125,08.33

3412

3412

3412

=+=+

=

=

=

++

+

+

x

xx

xx

xx

xx

( ) ( ) ( )

5

1364

33

8-:/313313

3333

3399.34

1364

213264

13134464

13132232

==

=

=

=

=

+

+

x

xx

xx

xx

xxxx

xxxx

( ) ( )

7

7

8

2

8772

7172212

77222.35

2

2

22

1221

++

=

=+=++

+=++

xx

xx

xx

xxxxx

3

03

172

72

00

33

==

==

=

x

x

xx

( )0,,0

log

==

=

bax

ba

ba

xx

axx b


17/51

7

9

97

012

97

012

1093661212

06

5

4

3

2

11

1.36

06

5

4

3

2

1

1

6 54 311

=

=

=+

=++++=+

+

+

+

=

=+

+

x

x

x

xxxx

xxx

x

aaaaa

aaaa

xxx

x

xxxx

( )0,01log >>== xaxayx ya( ) ( )

( )[ ] ( )

5

3

915

412934

23log34log

23log234loglog.37

22

2

=

=+=+

=+=++

x

x

xxxx

xxx

xxx

( )

10

54414412

04125

log

log124log5

log124loglog5

12log

4log5.38

2,1

2

2

=

=+

==+

=+

=+

t

tt

tx

xx

xxx

xx

210

20

10

812

5

2

10

4

10

812

10

812

2

1

2,1

==+

=

==

=

=

t

t

t

( )

100

10

100

10

5

2log

2

22

51

52

1

1

==

=

=

=

x

x

x

x

x

Nejednaine


18/51

x2

x1

T

x1

x2 x

1=x

2=

a>0 min

D>0

( ) ( )

( )

21

21

21

0

,0

,,0

xxxxy

xxxy

xxxy

===

( )

a

bcx

aza

a

bcx

aabcax

cbax

>+

>+

>+

,8

9

8

9

98

361246612312416

12/124

11

3

11

2

1.39

xx

x

xxxxxx

xxx

2 cbxaxy ++= kvadratna fukcija (parabola)


19/51

x1

x2

D>0

( )( ) ( )

21

21

21

0

,,0

,0

xxxxy

xxxy

xxxy

===+

a


20/51

2 3 +

-2 - - + +-3 - - - +

+ + - +

[ ]3,2x

2 6 +

-2 - - + +

12-2x + + + - - - + 0 -

( ) [ )+ ,62,x

1

0

-1

2

2

O=2r

r =1O=2

k (0,1)

poluprenikcentar

1cossinsin

cos

cos

sin

22 =+

=

=

ctg

tg

osnovni

trigonometrijski

indetitet

02

212

02

210

12

10

.42

+

x

x

x

xx

x

x

x=2 taka prekida

x=6 nula izraza

Kompleksni brojevi


21/51

cos

sin

{ }

osu-xzaMtatabrojjeM,cos0M

osu-yzaMtatabrojjeM,sin0M

Mkruznicatrig.krakII

0XkrakI,

xx

yy

T

=

==

=

M

b

c

katetasuprotnakatetanalegla

katetanalegla

katetasuprotna

hipotenuza

katetanaleglacos

hipotenuza

katetasuprotnasin

=

=

=

=

ctg

tg

brojagkompleksnooblikalgebarskiyixZ

jedinicaimaginarna1

1

1

01

2

2

+==

=

=

=+

i

Rx

x

x

),(

231

yxyixz

iz

=+=+=

a x

b z (a,b)

yi

{ }

{ }zb

iza

biaz

Im

jedinicaimaginarna1Re

===

+=

( )

biaz

Raia

aa

+==+

=0

0,

- Konjugovano kompleksan broj ima isti realan, a suprotan imaginarni

dio

Trigonometrijska krunica k (0,1) je krunica u centru poluprenika jedan

na koju se moe preslikati brojna prava.Definicija :

Sinus (kosinus) je trigonometrijska funkcija koja svakom realnom broju

pridruuje broj jednak ordinati (apcisi) take koja odgovara broju natrigonometrijskoj krunici.


22/51

b

z (a,b)

yi

=z

a

btg

a

b

=

=

=

cos

sin

Modul

kompleksnog broja

iz

b

a

+===

sincos

sin

cos

trigonometrijski oblik

kompleksnog broja

ii =+ sincos

( ) sincossincos ini n +=+

Eulorova formula

Moavrova formula

( )

+=

=

===

2sin

2cos1

211,0

iz

iz

z

378470533

4

3

4

52516943

43.43

22

1

==

=

==+=+=

+=

arctg

tg

iz

( )7053sin7053cos5 += iz


23/51

( ) iiiiiii 22331131311.44 3211233 =+=+=

( )

( ) ( ) 111

1

11

1

224

23

22

======

==

=

iii

iiiii

i

i

( ),...3,2,1,0 Zk

77319315419

9

17

1917

9

17

442

289

442

289

81361

17

9

17

19

17

9Im;

17

19Re

17

9

17

19

17

919

116

14520

4

4

4

5

4

5.45

22

==

=

==+

=

+

=

==

=

=+

=

+

=+

=

arctgarctg

z

ZZ

iiii

i

i

i

i

i

iz

( ) ( )

22

50

433

50

343

50

334

100

638Im

50

433

100

836Re

100

638836

100

863836

86

86

86

3

43

2

1

2

3

43

6sin

6cos

.46

+

+=

+=

+=

=

=

++=

++=

++

+

=

+=

+

iiZ

Z

iiii

i

i

i

i

i

i

i

i

( )

( )

2

1

4

1

2

10

2

1Im

0Re

2

1

4

2

2

2

2

1

121

1

1

1

1.47

22

2

2

==

+=

==

==

=

=

=

=

Z

Z

iii

i

iiiz

iz

k22

1arcsin

2

12

1

sin

+==

=

( ) ( )

3

132.48

2

+

i

ii

iii

ii

i

k

k

k

k

==

=

=

+

+

+

34

24

14

4

1

1


24/51

( ) ( ) ( ) ( )

( )

631125365

3arcsin

5

3sin

52591634

3410

1339927

3

3

3

139

3

41239

3

1343

3

13144

22

=

=

=

==+=+=

=++

=

+

+=

+++

=+

=

+

iii

i

i

i

i

i

ii

i

ii

i

ii

01615

1516

154

4

82

15

2815

/8i-15.49

24

2

2

22

22

22

2

=+

=

=

=

==

+=

+=

yy

yy

yy

yx

ixyi

yx

yxyixi

yix

( )ii

x

x

y

y

t

t

tt

ty

=

==

==

==

==

=+

=

4815

41

4

41

4

1

1

1

16

01615

:smjena

2

1

2,1

2

2

1

2

2

( )

( )

( )

( ) ( )

!55

4

525

2

112

2

11

8

2626

22

22

22

3

44

26

4411

217

211

217

54121

?2

211

7.50

22

2

22

2

Dokazano

z

iii

i

i

i

i

i

i

i

i

i

i

z

fzDokazati

izz

zf

z

z

=

=

+=

+=++

=++

=

=+

=

+

+

=+=+=

=

+=

=

51. f(z)

=z2-5(1+i)z+17i

Dokazati da su 4+i i 1+4i nule polinomaf(z)

( )

( ) +

+

T

T

yT

yT

yxx

yxx

,,

,,


25/51

=+===

==

+

=

+=

+=

+=

=

4524

1

2

2

2

2

112

2

2

2

2

2

2

2

2

1

12

012.52

22

1

3

3

3

karctgarctg

iz

iz

iz

iz

1217sin

1217cos

3

224sin

3

224cos1

4

3sin

4

3cos

3

12

4cossin3

12

4cos1

12sin

12cos

3

024sin

3

024cos1

2sin2cos

4sin

4cos1

2

32

1

31

0

30

1

iz

iz

iz

iz

iz

iz

nki

nkz

iz

nnk

+=

++

+=

+=

+

+

+

=

+=

++

+=

+++=

+=

z0

z1

z2

0

53. Odrediti a i b tako da kompleksan broj 1+i bude rjeenje jednaine

067 =++ bazz

( ) ( )

18

888

0888

01167

==

+==+

=++++

ab

iabi

biai

biai

( )

( )

( )

( ) iii

ii

iii

ii

karctg

z

iz

8

2

3sin

2

3cos81

46sin

46cos21

882

2

2

2281

47sin

47cos21

2

41

1

211

1

6

66

7

77

=

+=+

+=+

=

=+

+=+

+==

=+==

+=


26/51

( ) ( )

( ) ( )[ ] ( )( ) ( )[ ] ( )

( ) ( )

( ) ( )

00002

10

2

1

112

1112

1144

211

44

21

12

1

2

11

2

1

2

1

412111

412111

2

1

2

1

2

1

2

1

2

1

2

1

44

4

2224

2224

44

4

=+=

+

+=

=+ ++ += + + + +=

=

+

+

+

+

+=+

=+=+=+

===

+

++

+

+

+=+

+

+

ii

iiii

iiiiff

iii

iii

iiiiiiff

n

nn

n

nn

nnnn

nn

22122

22221

2

2

2

241

4sin

4cos41

iz

iz

iz

iz

=

=+

=+

+

=+

(

(

(

(

( 32213sin32213cos137

1 4 6 01433sin

7

14601433cos13

50159sin50159cos137

1 0 8 01433sin

7

10801433cos13

04107sin04107cos137

7201433sin

7

7201433cos13

4156sin4156cos137

3601433sin

7

3601433cos13

844sin844cos137

1433sin

7

1433cos13

24421433

3

2arctg13232323.56

14144

14143

14142

14141

14140

2211

7

iiz

iiz

iiz

iiz

iiz

ziziz

( ) ( )( )

( ) ( )( )41316sin41316cos13

7

21801433sin

7

21801433cos13

48264sin48264cos13

718201433sin

718201433cos13

146

146

145

145

+=

+++

=

+=

+++=

iz

iz

iz

iz

8

13sin

8

13cos

4

62sin

4

62cos1

8

9sin

8

9cos

4

42sin4

42cos1

8

5sin

8

5cos

4

22sin

4

22cos1

8sin

8cos

4

2sin4

2cos1

22

1.57

43

42

41

40

4

iiz

iiz

iiz

iiz

kiz

+=

++

+=

+=

+++=

+=

++

+=

+=

+=

+===

( ) ( )( )

( ) ( )( )

( ) ( )( )15263sin15263cos103

7203371sin

3

7203371cos10

15143sin15143cos103

3603371sin

3

3603371cos10

1523sin1523cos103

3371sin3

3371cos10

1845337131031.58

662

661

660

3

+=

++

+=

+=

++

+=

+=

+=

===+=

iiz

iiz

iiz

arctgziz

54. Ako je ( )

nn

n

iif

+

+=

2

1

2

1gdje je n prirodan broj.Dokazati da je

( ) ( ) 04 =++ nn ff

55. Odrediti kompleksan broj z ako se zna da je ( ) 41;4

1arg =+=+ zz

.


27/51

59. Odrediti skup taaka u kompleksnoj ravni za koje je realni dio

012

2Re =

z

iz.

yixz +=


28/51

60. Dati su kompleksni brojevi it

zt

itz +=+=2

cos2

sinsin2

221 .Odrediti

tako jez1=z2 .

61. Odrediti kompleksan broj, ako vrijedi{ } iiizizz 25Im +=+++=

62. z = 2-2i

z 7 = ?

kt

kt

tttt

ttttt

tt

22

242

2

2

2

tsin

2

cos:/2

cos2

sin2

cos22

2

2cos

2sin2

22sinsin

2cossin

2

2

+=

+=

=

=

=

+=

=

( )

( ) iiyx

yx

iiz

iz

21

51

2

5

22

=++=++

=+

=+

1

45

1

21

2

==

==+

x

x

y

y

iz

iz

+=+=

1

1

2

1


29/51

( ) iiiz

k

102410242

2

2

22102447sin47cos22.63

242

2arccos

22

2arccos

22844

77

=

+

=

+=

+===

==+=

Matematika indukcija

I korak : Provjeri da li je formula tana za n=1.

II korak : Pretpostavimo da je formula tana za n = k (n=1).

III korak : Treba dokazati da je formula tana za n=k+1.

( )

( )

( )

( )

( )

( ) ( ) ( )

( ) ( )

( )( )( )

( )( )

( )

( )( )

Tk

k

k

k

k

k

kk

k

k

k

kk

kk

k

k

kk

kk

k

k

kkk

k

k

k

kkkk

knn

n

nn

Tkkk

knf

T

n

nn

nn

n

n

nkkn

k

k

n

k

n

k

2

1

2

1

2

1

21

1

2

1

21

12

2

1

21

12

2

1

21

1

1

11

1

21

1

1

1...

43

1

32

1

21

1

1zaT11

1....

43

1

32

1

21

1jedadokazatiTreba:korakIII

1

1

1

1...

43

1

32

1

21

1

.zaTjedakaPretpostav:korakII

2

1

2

1

11

1

111

1

1:korakI

111....

431

321

211.64

11

1...21

2

2

1

11

++

=++

++

=++

+

++

=++++

++=

++++

++

=++

++

+++

=++

++

++

+

+

+=+

=+

++

+

+

+=

+++

+

+

=

=

+=

+

=+=+++++

+=

+=+++

+

==

( ) ( )=

=n

k

nnk1

223 1212.65


30/51

( ) ( )

( ) ( )

( ) ( )( ) ( )

( ) ( )( ) ( ) ( )( )( )

( )

( ) ( )( )

Tkkkkkkkk

kkkkkkkkkkkkk

kkkkkk

kkkk

knnnn

kkk

knf

n

nnn

kk

161182161182

14228442161282

142121212...531

112111212...531

.1zaT1212...531jedadokazatiTreba:korakIII

1212...7531


11

1121112

1:korakI

1212...7531

234234

2232342324

223

12

3333

2233333

223333

2233333

223

2233333

22

++++=++++

++++++++=++++

++++=++++++

++=++++++

+==++++

=+++++

==

=

==+++++

66. Dokazati da je za svaki prirodni broj n , 210 1 +n djeljiv sa 3.

( )

( ) ( ) Tpp

kn

Zpp

knf

n

Z

k

k

k

k

n

6103633101821010

21010

210

1zaT210jedadokazatiTreba:korakIII

3210


T3:3

3:/21

210

210

1:korakI

3:/210

1

1

11

1-n

1

0

11

1

+

==+

+

+

+=+

=+

=

++

+

=+

67. Dokazati da je za svaki prirodan broj n, 98322

+

nn

djeljivo sa 64.

( )Zppk

knf

T

n

n

k

n

=

=

=

+

+

+

64983


64:64

64:/17-81

64:/173

64:/9183

1:korakI

983

22

4

212

22


31/51

( ) ( )

( )

( )

( ) T64:/1964

64:/164649

64:/6464983964:/17893

64:/9883

64:/9183

.1zaT983jedadokazatiTreba:korakIII

64

22k

22

222

212

22n

Z

p

k

k

k

kp

kp

kkk

k

k

knn

+

+

++

++

+

++++

++

+

+=

Binomna formula( ) ( )

( )( )

( )

( )( )

( ) ( ) 2112

67

1234512

1234567

1212345

1234567

1512

56

!461234

123456

nad!!

!

12012345!5

faktorijel1231!

72

75

64

=

=

=

==

==

=

=

===

knknk

n

nnnn

nk

nk

( ) ( )n knnk =

( ) ( )=

=+n

k

kknnk

nbaba

0

Binomna formula

( ( 10 == nnn Po definiciji

Binomni koeficijenti ( )( )!!

!

knk

nnk

= .

Opti lan binomnog razvoja ( kknnkk baB + =1( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) nnnnnnnnnnnn

nnnnnnnnn

babababa

bababababa

0111

222

333

333

222

111

00

+++++

++++=+

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

aab

a

b

a

b

aab

abb

ab

aaaaa

aaaaaa

+=

=+

+

+

=

+++++=

=+++++=+

33 2

333

2

3

32

2

3

31

3

3

30

3

3

5432

543223455

331

1111.68

1590270405243

353103103533.67

=

5

10

11.69


32/51

( ) ( ) ( ) ( ) ( ) ( )

59049,010000059049

10000015010001000050000100000

100000

1

2000

1

100

1

10

1

2

11

100000

1

10000

115

1000

1110

100

1110

10

1151

10

11

10

11

10

11

10

11

10

11

10

11

505

5

415

4

325

3

235

2

145

1

055

0

==++=

++=++=

=

+

+

+

+

+

=

70. Nai etvrti lan razvoja ( )743 xx + .

( )( ) ( ) 1224 33344373 351234123

1234567xxxxxxx =

=

71. Odrediti koeficijent lana koji sadri ( )733 2 zz + na (k+1) mjestu

biti e ( ) kkk zz 377

( ) 35571234123

1234567

3

182321

6/332

7

73

332

7

==

=

==+

=+

=

k

kk

kk

zzz

kk

72.10

+x

a

a

xodrediti srednji lan?

( ) 25245

78910

1234512345

123456789105

5

5

555105 =

=

=

x

a

a

x

x

a

a

x

73. Ako je u razvoju binoman

xxx

+4

1koeficijent II lana je za 44

manji od koeficijenata III lana.Odrediti lan koji ne sadrix .( )

11

4

2

1

2

2

1

11

8

0883

882

2/2

144

+

==

=

=+

=+

xxx

Zn

Zn

nn

nnn

nnn

Traimo lan koji ne sadrix.

( ) ( )4

4

1111 1

x

xxk

k


33/51

( ) ( ) ( ) bcacabcbacbacbacba 2222222 +++++=++++=++

( )[ ] ( ) ( )[ ] ( )[ ]( ) ( )

18208268208

4244168164

2848241681644241

42414241

424412122121

2345678

8765476543

654325432432

432432

24322242242

+++++=

=+++++

+++++++++++=

=++++++=

=+++=+++=+

xxxxxxxx

xxxxxxxxxx

xxxxxxxxxxxxxx

xxxxxxxx

xxxxxxxxxxx

3

3311

0811222

042

1111

01

042

11

11

4

11

==

=+

=+

=

=

k

k

kkk

kkk

xxxx

xxx

k

k

k

kk

74. Za kojex je peti lan razvoja jednak 700?

( )

( ) ( ) ( )

10

10

10log

10log

700log70

700log1234123412345678

700log

log

10

44

4484

844

==

=

==

=

=

x

x

x

xx

xx

xx

xx

xx

x

x

75. ( )4221 xx +

76. Dokazati da je 6132 +n djeljivo sa 7.

( )Zpp

kn

T

n

k

n

=+

=

+

=+

7613

.zaTtvrdnjajedaoetpostavimPr:korakII

25

7:175

7:/613

1:korakI

7:/613

2

12

2


34/51

( )

( )

Tp

p

kn

Z

k

k

k

1441697

71447169

7:/1008613169

7:/613

7:/613

.1zaTtvrdnjajedadokazatiTreba:korakIII

2

22

12

+

+

+

+=

+

+

77. Dokazati da li je 122 1211 ++ + nn djeljivo sa 133?

( )

( )

( )

( ) Tpp

kn

Zpp

kn

n

Z

k

k

k

p

kk

kk

kk

kk

nn

11144133

133:/11133133144

133:/111331211144

133:/144121111

133:/1211


1331211

zaiTtvrdnjajedaimoPretpostav:korakII

T23

133:/3059

133:/17281331

133:/1211

133:/1211

1:korakI

1211

2

2

2

133

122

122

11221

122

33

11221

122

+

+

+++

++

++++

++

++

++

+

+

+

+==+

=

++

+

=+

78. Dokazati da li je 122 54 ++ + k djeljivo sa 21?

( )21:/54

.1zaiTtvrdnjajelidadokazatiTreba:korakIII

2154

.zaTtvrdnjajedaimoPretpostav:korakII

T21:21

21:/516

21:/54

1:korakI21:/54

11211

121

11211

121

+++

+

+

+

+

+==+

=

++

=+

kk

kk

nn

kn

p

kn

n


35/51

( )

( ) TpZ

k

k

p

kk

kk

42521

21:/4215425

21:/5544

1

1

21

121

2121

+

++

+

+

+

79. ( ) 1!1!1

+==

nkkn

k

( )

( )

( )

( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) Tkk

kkk

kkk

kkkk

kkkkk

kn

kkkkn

T

n

nnn

k

1!21!2

1!212!1

1!2111!1

1!2!111!1

1!11!11!!33!22!11


1!1!!33!22!11.zaTtvrdnjajedaimoPretpostav:korakII

11

121

1!11!11

1:korakI

1!1!!33!22!11

1!1

+=++=++

+=++++=++++

++=++++++++=

+=++++=

==

+==

+=++++

+

80. ( )72 ba +( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

7654

322233

777

676

575

474

3273

2272

371

370

7

1484280

560672448108

248

163264108

2

bbababaa

babaabaaa

bbaabbaa

babaabaaa

ba

++++

++++=

++++

++++

=+

81. ( ) ?02,1 5 =

( )

( ) ( ) ( ) ( ) ( )

1040808032,101000000000

21104080803

01000000000

32800080000040000000100000000001000000000

01000000000

32

100000000

161

1000000

81

10000

41

50

111

100

21

100

10202,1

55

54

253

352

451

5

555

==+++++

=

=+++++=

=

+=

=

82. Odrediti 28 lan razvoja ( ) 3085 yx + .

28 lan je ( ( ) ( ) 2733027 85 yx .


36/51

83. Diskutovati sistem

mmzymx

zymx

mzyx

=++=++

=

1

133

1

( )

1

1

133333

1

33

11

1

11

1

3311

2,1

2

2222

==

=++=

=

=

m

m

mmmmmm

mm

m

m

mz

yx

mm

mm

Da bi sistem imao jedinstvena rjeenja m mora biti razliit 1m .

( ) ( )

( )

( )( )

( ) ( )

( )( )( ) ( )

rjeenja.ajedinstven113

353

33

21

13

13

13

353

33

2

113

12

133333

1

33

11

353333

1

31

11

12639333333

11

331

11

2

23

2

2

2

23

2222

232232

22

=

+=

+

==

=

=

+=

=

+

=+

=

=

=++=

=

+=+++=

=

==+++=

=

zm

mmmy

m

mx

m

mzz

m

mmmyy

m

m

mm

mmxx

mmmmmm

mm

m

m

z

mmmmmmmmmm

mmm

m

m

y

mmmmmmmmmm

mm

m

x

( )

( )( )

3130

35310

230

10II

110I

==

===

z

y

x

im

mm

Nemogue, nema rjeenja.

3130

35310

010

01.)

=

+==

==

z

y

x

mii

Neodreeno, ima beskonano rjeenja

84.Diskutovati sistem.


37/51

=

3

2

9

4

y

x

a

a

Sistem je neodreen ima beskonano

rjeenja

Sistem je neodreen, ima beskonano mnogo

rjeenja

Sistem je nemogu, nema rjeenja

6

3

6

2

+=

+

=

ay

ax

Jedinstvena rjeenja

39

24

=+=+

ayx

yax

36

9

4 2 =

= a

a

a

( )

( )

( )

( ) ( )

( )

( )( )

( ) 62

062

0

6.)

6

036

0II

6

3

6

20I

6

3

66

63

6

2

66

62

6318339

2

621223

42

2

+=

==

==

=+

=+

=+

=

+

=

=

+=

+

=

=

==

=

==

=

aa

xx

ai

a

a

ay

ax

aaa

ayy

aaa

axx

aaa

y

aaa

x

00

030

0

00

020

==

==

=

y

yy

240240

0

6.)

= =

==

x

x

aii


38/51

Matrice

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

=

=

=

=

=

=

=

=

=

=

+++

+++

+++

2221

12113333

2321

13112332

3222

31211331

3231

12113223

3331

13112222

3332

13121221

3231

22213113

3331

23112112

3332

23221111

333231

232221

131211

11-1

11-1

111

aa

aaA

aa

aaA

aa

aaA

aa

aaA

aa

aaA

aa

aaA

aa

aaA

aa

aaA

aa

aaA

aaa

aaa

aaa

A

( )[ ]mnij

AA

A =det

11

110

2110030210

20

3100310220

21

320

000

2121130111

10

3110030120

10

320

000

1030202102

10

2021120111

10

211

1det

100

210321

333231

232221

131211

===+===+==

===+=

==+=

=

========

=

=

AAA

AAA

AAA

A

A

=

=

=

100

010

001

100

010

001

1

1

100

010

001

1A

Aij

85. Izraunati ( ) EAxEA += 2

=

=

100

010

001

101

432

210

E

A


39/51

( ) ( )

( )

( )

=

=

=

=

=

==

===

=

==

==

=

===

==

=

=++=

=

+=

=

+

=

12

1

2

1613

12

1

2

1

633

36618

633

6

1

201

442211

411

1206211

6

1

411

1206

211

6

12

412

121

01

121

01

12

1242

220

11

226

11

42

241

211

10

211

10

41

62242

101

412

212

2det

2

201

442

211

101

412

212

100

010

001

101

432

210

100

010

001

2

101

432

210

1

332313

322212

312111

1

x

x

EA

AAA

AAA

AAA

EA

EAEAx

x

x

86.Izraunati

9452

16324

743

=++=++

=++

zyx

zyx

zyx

4564454202424

452

324

143

det

9

16

7

452

324

143

=++=

=

=

A

z

y

x


40/51

1024

437

52

4316

52

24

534

1310

42

1310

42

34

1032

1411

45

147

45

32

332313

322212

312111

======

======

======

AAA

AAA

AAA

=

10716

51010

10117

45

11A

=

+

+

=

=

2

1

3

45

9011211245

451607045

9017649

9

16

7

10716

51010

10117

45

1

z

y

x

z

y

x

z

y

x

87. Izraunati

2

1

=++

=++=++

zyx

zyx

zyx

( )( ) ( )

( )

( ) ( ) ( )

+=

+=++=++=

=

+=+=++=

=

=

111

1111

1

1

111

1;1;2

1122311

11

11

11

1

11

11

11

2232322

2

33

2

x

x

z

y

x


41/51

( )

( )

( ) ( )( )

( ) ( ) ( )

( )

( ) ( )( ) ( )( )

( )( ) ( )( )

2-1jeakorjeenjaaJedinstven2

1

112

1

12

1

112

1

2

1

112

111

1121

11

1

11

11

1

11

11

222

2

2224224

2

2323

2

=+

=

=

+

=+

=

=

=+

+=

=

=+=++=

=

=++=

=

zz

yy

xx

z

y

( ) ( )

0

2jejernemoguejerjeenje

03

0

21

11

0

2

1121

11

3

2

21

11

1jeAko

=

==

=

=

+

==

+=

=

y

zyx

( ) ( )

0300

00

100

00

1-00.)

nemogujesistem

-3000

20200

0

3

200.)

2.)

1.)0I

==

==

=

=

====

=

=

===

zz

yy

xxii

zz

yy

xxi

xii

xi

Sistem je neodreen ima beskonano rjeenja.88.Izraunati

02

22

1

=++=++

=++

zyx

zayx

zyax

=

0

2

1

112

21

11

z

y

x

a

a


42/51

Homogeni sistem, svi slobodni lanovi jednaki su nuli.

Trivijalno rjeenje

Kod homogenog sistema je

=

0

2

11A

z

y

x

( )

11

12

12

121

12

1

2121

12

12

13

12

21

22

110

11

112

11

2

212214

112

21

11

det

2332313

322212

312111

22

======

======

======

=++=

=

aa

aAa

aAa

aA

aa

Aaa

AA

aa

AAaa

A

aaaaa

a

A

( )

( )

( )

( )

( )

( )

( )

rjeenjaaJedinstven

2

45

2

122

1

2

24212

4232

2

0

2

1

1221

2123

202

2

1

1221

2123

202

2

1

2

2

2

2

2

22

22

1

=

+

+

=

=

=

a

aa

aa

z

y

x

a

aaa

aa

a

aaa

aa

aa

az

y

x

aaa

aa

aa

aA

rjeenjanemaanprotivrjejesistem

00

10.)

2044

0det

2,1

2

=

==+

=

i

aaa

A

0

0

0

=++=++=++

azyx

zayx

zyax


43/51

( ) 13226

54

6.1

=+++

=++=++

zyax

zyax

zyx

1

1

1.2

=

=+=+

x-y-az

zayx

zyax

12323

522

42.3

=++

=++=++

zyx

zyx

zymx

32

125

22.4

=+=+=+

bzyx

yx

zax

0136

032

052.5

=+=+=+

zyx

zyx

zyx

Zadaci za vjebu :

Vektori

( ) ( ) ( )

( ) ( ) ( )211221122112

111

222

122112211221

22

11

22

11

22

11

222

111

yxyxkzxzxjzyzyi

zyx

zyx

kji

ab

yxyxkzxzxjzyzyi

yx

yx

kzx

zx

jzy

zy

izyxzyx

kji

ba

+==

+=

=+==

(( )( )( )

( ) ( ) ( )

( )

0

0,0,0

000193329632361

639

213

639

23

2

32

=

=

++=+++==

+=

+=

++=

+=

ba

baba

kji

kji

ba

kiib

kjia

kjib

kjia

89. Izraunati povrinu paralelograma kjibkjia 62343 +=+=( ) ( ) ( ) ( )

7499364362

3,6,2362132343634261

222 ==++=++=

=++=+++=

ba

kjikjiba

90.Ako supi qbilo kakvi vektori dokazati da vrijedi relacija.( ) ( ) pqqpqp =+ 2

pqpqpqqqpqqppp =+=+==

2

00

91.Dati su vektori ( )2,1,1=a i ( )1,2,1=b .Odrediti ugao izmeu vektora.


44/51

( )6,4,24

062

6

0212

2

=

=

=+

=

=+

=

z

y

y

z

z

x

( ) ( ) ( )( )

( ) kba

ba

23

602

1arccos,

2

1

6

3

66

221

121211

1.2.12,1,1,cos

222222

+===

==

++=

++++

=

92. Dati su vektori ( ) ( ) ( )10,2,3,1,1,1,3,3,2 === cba izraunati kizpostavljenog uslova ( cbka + .

( ) ( ) ( ) ( )( ) ( ) ( )( )

( )

( ) ( )( ) ( ) ( )

2

3015

010306236

03103223

0

10,2,3

3,3,2

332332

0

0

212121

==

=+++=++

=++

=+=+

+++=+++=+

=++=

k

k

kkk

kkk

cbkacbka

c

kkkbka

kkkjkikjikkjibka

zzyyxx

baba

b b bojekcijaaPr

a

boja

Pr abojaba aPr=

a

bab

b

boj

ba

ba

a

=

=

Pr

Pr

93. Dati su vektori ( ) ( ) ( )2,2,1,1,1,5,1,1,1 === cba .Odrediti vektor( )zyxz ,,= iz postavljenih uslova. bzaz , i .2Pr =zojc

63

026

3222

05

0

2

0

0

==+

==+

=++

=

=

=

x

zx

zyx

zyx

zyx

czc

bz

az

Vektorski proizvod


45/51

b h

a

( )ba,

Uvrstimo u (1).

( )

( )babaP

haP

,sin

1

=

=

baP

=

( )( )222

111

,,

,,

zyxb

zyxa

==

222

111

zyx

zyx

kji

ba

=

C 3 4 -3D(x,y,z)

A(6,2,3) B(0,-1,5)

1,2

3,

2

3

113

233236

12

3

2

3

2

2

2

3

2

6

===

=+=+=+

=+=+=+

zyx

zyx

zyx

( ) ( )4,1,931,21,63 ==AD

( )4,1,9 =AD

( )

1

,sin

0

0

=

=

v

babavba

94. Izraunati povrinu paralelograma kojeg ine take A(6,2,3),

B(0,-1,5) i taka C(3,4,-3).?=ABCDP

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) 492401214214

21,42,14612636418

623

236

6,2,333,24,63

2,3,635,21,60

222 ==++==

=++=

=

==

==

ACABP

kji

kji

ACAB

AC

AB

ABCD


46/51

( ) [ ] pedaparalelopi,, Vcbacba ==

a

b

c

[ ] ( )iKomplanarn,,0,, cbacba

=

a

bc

Zapremina trostrane

prizme

[ ]cbaV ,,

2

1=

a

bc

[ ]cbaV ,,

6

1=Zapremina

trostrane piramide

Mjeoviti proizvod

( )

( )

( )

[ ]

=

==

=

333

222

111

333

222

111

,,

,,

,,

,,

zyx

zyx

zyx

cba

zyxc

zyxb

zyxa

95. Pokazati da su vektori cba ,, komplanarni (lee u istoj ravni).

( )

( )

( )1,4,4

1,5,1

2,1,3

==

=

c

b

a


47/51

[ ] 0112408415144

151

213

,, =++++=

=cba

cba ,, lee u istoj ravni,

odnosno komplanarni su.

96.Izraunati zapreminu prizme ako su nam poznata tjemena.( ) ( ) ( ) ( )2,3,2;1,1,6;2,2,2;3,2,2 SCBA

( ) ( )

( ) ( )

( ) ( )

10606

1

6

1

608020

550

214

140

5,5,032,23,22

2,1,431,21,26

1,4,032,22,22

===

=+=

=

=+==

=+==

=+==

VV

V

ASc

ACb

ABa

97. Izraunati ( ) cba , ako je 2,4, === babac i ( ) = 30,ba .

( )

( ) 164

2

18

30sin24

,sin

===

=

=

==

cccbac

ba

ba

bababac

98. Dati su vektori ( ) ( ) ( ) ( )1,2,3,1,1,1,2,1,1,1,1,1 dcba .Izraunati sljedeeformule.

( ( ( ( ( ( ??;?;?;?;?; ====== dcbadcbacbacbababa( ) ( )

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( )4,1,3132321111

213

4121121

111

211

111

2,1,3111212

211

111

22112111112,1,11,1,1

=++++=

=

=++++=

=

=++=

=

=+=+==

kji

kji

cba

cba

kji

kji

ba

ba


48/51

( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )5,1,3162341121

213

72231221131,2,12,1,3

1,2,1323121

123

111

=+++=

=

=++=++==

=+++=

=

kji

kji

dcba

dcba

kji

kji

dc

99. Sastaviti jednainu ravni koja prolazi kroz presjek ravni023 =+ zyx i 032 =++ zyx i normalna je na ravan

0222 =++ zyx .( )

( )

7

17

11

117

32

23

0

032

023

0:

0222

032023

==

==+=

==++=+

=+++=++

=++=+

y

x

x

yx

yx

z

zyx

zyx

DCzByAx

zyx

zyxzyx

7

3

7

1

17

132

12

03

2

=

=

==+

=+=

=

x

y

y

yy

yx

yx

z

=

= 2,

7

1,

7

3;0,

7

1,

7

11BA

( )

022

01428

0222

022

027

2

7

8

0222

0222022

027

1

7

3

07

1

7

11

=+==++

=+

=

=++=++=+

=+++

=+

CBA

CBA

DCA

CBA

CBA

DCA

zyxCBA

DCBA

DBA


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CB

CBC

CD

DCC

CA

CA

=

=+

=

=++

=

=

022

32

2

5

022

2

32

2

3

1510

05223

2/0

2

5

2

3

=+

=+

zyx

CCzCyCxC

05223 =+ zyx Jednaina ravni .


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