m1_2_zavrsni_1dio_1314_130_140 (2)

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     z  ∈ C  z 5 + (1 + i) 10 z  = 0.    BX  +  C  = 2B + AX,  A = 1 2 2 0 3 4 0 0 1 , B = 1 2  −3 0 4 2 0 0 2   C  = 3 0 1 0 2 4 0 0 1 .    A(1, 1, 1), B(2, 4, 3)   C (1, 0, 4).    c        f  ( x) = log  x 2 3x + 2 x 1  +  e x x 2 9 .      lim x→∞ x( √ x 2 + 1 x).        n  c R    A =  1  1 c  1              (f  ◦ f 1 ) (x)    f  ( x) = e ln x  g (x) = l n(e x )

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Page 1: M1_2_zavrsni_1dio_1314_130_140 (2)

7/24/2019 M1_2_zavrsni_1dio_1314_130_140 (2)

http://slidepdf.com/reader/full/m12zavrsni1dio1314130140-2 1/2

z ∈C

z 5 + (1 + i)10 z = 0 .

BX + C = 2B + AX,

A =1 2 20 3 40 0 1

, B = −1 2 −30 4 20 0 2

C = −3 0 10 2 40 0 1

.

A(1, 1, 1), B (2, 4, 3)

C (1, 0, 4).

c

f (x) = log x 2 −3x + 2

x −1 +

ex

x 2 −9.

limx →∞

x(√ x 2 + 1 −x).

n

c ∈R

A = 1 −1c 1

(f ◦f − 1 ) (x)

f (

x) =

eln x g

(x

) = ln (ex

)

Page 2: M1_2_zavrsni_1dio_1314_130_140 (2)

7/24/2019 M1_2_zavrsni_1dio_1314_130_140 (2)

http://slidepdf.com/reader/full/m12zavrsni1dio1314130140-2 2/2

z = 4√

32(cos

3 π

2 +2 kπ

4 + i sin

3 π

2 +2 kπ

4 ), k = 0 , 1, 2, 3

X = −12 −2 −4

0 6 60 0 3

vc = √ 13114

D f = ⟨2, 3⟩∪⟨3, + ∞⟩

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