1. STEPENOVANJE
Proizvod a ⋅ a ⋅ ... ⋅ a = a n naziva se n -tim stepenom broja. Ako je a ∈ R , a ≠ 0 i neka je n ∈ N
n − puta
Po definiciji je:
0
⎛4⎞
1) a 0 = 1 → primer: 50 = 1, (−3) 0 = 1, ⎜ ⎟ = 1
⎝7⎠
1 1 1 1 1
2) a − n = n → primer: 3− 2 = 2 = , 5−3 = 3 =
a 3 9 5 125
______________________________________________________________________
Još važe sledeća pravila:
3) a m ⋅ a n = a m+n → primer: 32 ⋅ 35 = 32+5 = 37
4) a m : a n = a m−n → primer: 710 : 7 6 = 710−6 = 7 4
5) ( a m ) n = a m⋅ n → primer: (23 ) 5 = 23⋅5 = 215
6) ( a ⋅ b) = a n ⋅ b n → primer: (12 ⋅11) 5 = 125 ⋅112
n 2
⎛a⎞ an ⎛7⎞ 72
7) ⎜ ⎟ = n → primer ⎜ ⎟ = 2
⎝b⎠ b ⎝4⎠ 4
−n n −2 2
⎛a⎞ ⎛b⎞ ⎛2⎞ ⎛3⎞ 32 9
8) ⎜ ⎟ =⎜ ⎟ → primer ⎜ ⎟ = ⎜ ⎟ = 2 =
⎝b⎠ ⎝a⎠ ⎝3⎠ ⎝2⎠ 2 4
O čemu treba voditi računa?
Treba paziti na zapis: (−5) 2 = (−5)(−5) = 25 , dok − 52 = −5 ⋅ 5 = −25 . Uopšteno važi:
(− a) paran = a paran
(− a) neparan = − a neparan
Dakle, paran izložilac ‘’uništi’’ minus.
www.matematiranje.com
1

2. ZADACI
( 27 : 25 ) ⋅ 23
1) Izračunati:
24 : 22
(27 : 25 ) ⋅ 23 27 −5 ⋅ 23 22 ⋅ 23 22+3 25
= = = 2 = 2 = 25 − 2 = 23 = 8
24 : 22 24 − 2 22 2 2
_____________________________________________________________
35 ⋅ 93
2) Izračunati:
27 2 ⋅ 3
35 ⋅ 93 35 ⋅ (32 )3 35 ⋅ 36 35
= 3 2 1 = 6 1 = 1 = 35−1 = 34 = 3 ⋅ 3 ⋅ 3 ⋅ 3 = 81
27 ⋅ 3 (3 ) ⋅ 3
2
3 ⋅3 3
______________________________________________________________
( x 4 )3 ⋅ x 3 : x 5
3) Izračunati: =
( x 5 : x 2 )3
( x 4 ) 3 ⋅ x 3 : x 5 x12 ⋅ x 3 : x 5 x12+ 3−5 x10
= = 3 3 = 9 = x10−9 = x1 = x
( x 5 : x 2 )3 ( x 5− 2 ) 3 (x ) x
_______________________________________________________________
3n +1 ⋅ 3n + 2
4) Izračunati:
32 n + 4
3 n +1 ⋅ 3 n + 2 3 n +1+ n + 2 3 2 n + 3
= 2 n + 4 = 2 n + 4 = Pazi pa zagrade zbog minusa
32n+ 4 3 3
1 1
= 3( 2 n + 3) −( 2 n + 4 ) = 32 n +3− 2 n − 4 = 3−1 = 1 =
3 3
_______________________________________________________________
www.matematiranje.com
2

3. 5) Izračunati 0,5−1 + 0,25−2 + 0,125−3 + 0,0625−4
0,5−1 + 0,25−2 + 0,125−3 + 0,0625−4 =
−1 −2 −3 −4
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞
⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ =
⎝2⎠ ⎝4⎠ ⎝8⎠ ⎝ 16 ⎠
1 2 3 4
⎛ 2 ⎞ ⎛ 4 ⎞ ⎛ 8 ⎞ ⎛ 16 ⎞
⎜ ⎟ +⎜ ⎟ +⎜ ⎟ +⎜ ⎟ =
⎝1⎠ ⎝ 2⎠ ⎝1⎠ ⎝ 1 ⎠
21 + 4 2 + 83 + 16 4 = 2 + 16 + 512 + 65536 = 66066
_________________________________________________________________
6) Izračunati 1−1 + 2 −2 + 3−3 + (−1) −1 + (−2) −2 + (−3) −3
1−1 + 2 −2 + 3−3 + (−1) −1 + (−2) −2 + (−3) −3 =
1 1 1 1 1 1
+ 2+ 3+ + + =
1 2 3 (−1) (−2) (−3) 3
1 2
1 1 1 1 1 1 2 1
1+ + −1+ − = + = =
4 27 4 27 4 4 4 2
_________________________________________________________________
−4 2 −2
⎛1⎞ ⎛3⎞ ⎛5⎞
7) Ako je a = 5 ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ i b = 10 3 ⎜ ⎟
3
nadji a ⋅ b −1
⎝4⎠ ⎝2⎠ ⎝3⎠
−4 2 4
⎛1⎞ ⎛3⎞ ⎛4⎞ 3
2
53 ⋅ 44 ⋅ 32
a = 53 ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ = 53 ⋅ ⎜ ⎟ ⋅ 2 =
⎝4⎠ ⎝2⎠ ⎝1⎠ 2 22
53 ⋅ (22 ) 4 ⋅ 32
= 2
= 53 ⋅ (22 )3 ⋅ 32 = 53 ⋅ 26 ⋅ 32
2
−2 2
⎛ 3 ⎞ 10 ⋅ 3 (5 ⋅ 2)3 ⋅ 32 53 ⋅ 23 ⋅ 32
3 2
⎛5⎞
b = 10 ⋅ ⎜ ⎟ = 103 ⋅ ⎜ ⎟ =
3
= = = 5 ⋅ 23 ⋅ 32
⎝ 3⎠ ⎝ 5⎠ 52 52 52
Konačno: izračunati i a ⋅ b −1
1
a ⋅ b −1 = 53 ⋅ 26 ⋅ 32 ⋅
= 52 ⋅ 23 = 25 ⋅ 8 = 200
5⋅ 2 ⋅3
3 2
______________________________________________________________
www.matematiranje.com
3

4. ⎛ ⎛ 5 x −5 ⎞ −2 ⎛ y −1 ⎞ −3 ⎞
8) Izračunati ⎜ ⎜ − 2 ⎟ ⋅ ⎜ −1 ⎟ ⎟ : 10 x 2 y −3
⎜ ⎜ 2 y ⎟ ⎜ 5x ⎟ ⎟
⎝⎝ ⎠ ⎝ ⎠ ⎠
⎛ ⎛ 5 x −5 ⎞ −2 ⎛ y −1 ⎞ −3 ⎞
⎜⎜ ⎟
⎜ 2 y − 2 ⎟ ⋅ ⎜ 5 x −1 ⎟ ⎟ : 10 x y =
2 −3
⎜ ⎟ ⎜ ⎟
⎝⎝ ⎠ ⎝ ⎠ ⎠
⎛ 5−2 ⋅ x10 y3 ⎞
⎜ − 2 4 ⋅ −3 3 ⎟ : 10 x 2 y −3 =
⎜ ⎟
⎝ 2 ⋅y 5 ⋅x ⎠
(5− 2+3 ⋅ x10−3 ⋅ y 3− 4 ⋅ 2 2 ) : 10 x 2 y −3 =
(51 ⋅ x 7 ⋅ y −1 ⋅ 4) : 10 x 2 y −3 =
20 7 − 2 −1−( −3)
x y = 2 x 5 y −1+ 3 = 2 x 5 y 2
10
__________________________________________________________
1 −3 1 − 4
10 + 10
9) Ako je 10 =
x 2 2 Odrediti x.
55 ⋅10 −7
1 −3 1 − 4 1 1
10 + 10 +
2 2 = 2000 20000 = Izvučemo gore zajednički
55 ⋅10 −7
55
10000000
1 ⎛ 1⎞ 1 11 11
⎜1 + ⎟ ⋅
2000 ⎝ 10 ⎠ 2000 10
= = 20000 =
55 55 55
10000000 10000000 10000000
11⋅10000000 11⋅10000000 10000000
= = = 100 = 102
20000 ⋅ 55 11⋅100000 100000
sada je 10 x = 10 2 , dakle x = 2
________________________________________________________
10)
a) A ⋅10 −5 = 0,2 ⋅ 0,008 b) B ⋅10 −6 = 0,04 ⋅ 0,006
A ⋅10 −5 = 2 ⋅10 −1 ⋅ 8 ⋅10 −3 B ⋅10 −6 = 4 ⋅10 − 2 ⋅ 6 ⋅10 −3
A ⋅10 −5 = 16 ⋅10 − 4 B ⋅10 −6 = 24 ⋅10 −5
16 ⋅10 − 4 24 ⋅10 −5
A= B=
10 −5 10 −6
A = 16 ⋅10 −4−( −5) B = 24 ⋅10 −5+ 6
A = 16 ⋅10 −4+5 B = 24 ⋅10
A = 16 ⋅10 B = 240
A = 160 4

5. Ovde smo koristili zapisivanje realnog broja u sistemu sa osnovnim 10. Ovo je dobra
opcija kada je broj ‘’glomazan’’.
Primeri:
1) Brzina svetlosti je približno c ≈ 300000000m / s a mi je ‘’lakše’’ zapisujemo
c ≈ 3 ⋅108 m / s , 108 -znači da ima 8 nula iza jedinice!!!
1 1 1 2
2) = = ⋅10 −5 = ⋅10 −5 = 2 ⋅10 −1 ⋅10 −5 = 2 ⋅10 −6
500000 5 ⋅10 5
5 10
−5 −5
3) 0,000069 = 6,9 ⋅10 ≈ 7 ⋅10
4) Površina zemlje je 510083000km 2 ali mi zapisujemo ≈ 5 ⋅108 km 2
⎛ 3a − x 2a − x ax ⎞ a−x
11) Izračunati ⎜ − − 2x ⎟ : x
⎜ 1− a−x 1+ a−x a −1 ⎟ a − a−x
⎝ ⎠
⎛ 3a − x 2a − x ax ⎞ a−x
⎜ 1− a−x − 1+ a−x − a2x −1 ⎟ : a x − a−x =
⎜ ⎟
⎝ ⎠
⎛ 3 2 ⎞ 1
⎜ ax x ax ⎟ x
⎜ − a − 2x ⎟ : a =
1
⎜ 1− x 1+ x 1 a −1 ⎟ x 1
a − x
⎝ a a ⎠ a
⎛ 3 2 ⎞ 1
⎜ x x ax ⎟ ax
⎜ xa − a − ⎟: =
⎜ a −1 a x + 1 a2 x −1 ⎟ a2 x −1
⎜ x ⎟
⎝ a ax ⎠ ax
⎛ 3 2 ax ⎞ 1
⎜ x − x − x ⎟ : 2x =
⎝ a − 1 a + 1 (a − 1)(a + 1) ⎠ a − 1
x
3(a x + 1) − 2(a x − 1) − a x a 2 x − 1
⋅ =
(a x − 1)(a x + 1) 1
3a x + 3 − 2a x + 2 − a x (a − 1)(a + 1)
x x
⋅ =
(a x − 1)(a x + 1) 1
= 3+ 2 = 5
www.matematiranje.com
5

6. ⎛ x − x −2 x − x −1 ⎞ 1 − x −1
12) Izračunati ⎜ x + x −1 + 1 1 + x − 2 + 2 ⋅ x −1 ⎟ : 1 + x −1
⎜ −2 − ⎟
⎝ ⎠
⎛ x − x −2 x − x −1 ⎞ 1 − x −1
⎜ −2 −1
− −2 −1 ⎟
: −1
=
⎝ x + x +1 1+ x + 2 ⋅ x ⎠ 1+ x
⎛ 1 1 ⎞ 1
⎜ x − x2 x−
x ⎟ 1− x
⎜ 1 1 − : =
1 2⎟
⎜ 2 + +1 1+ 2 + ⎟ 1+ 1
⎝x x x x⎠ x
⎛ x3 − 1 x2 − 1 ⎞ x −1
⎜ ⎟
⎜ x2 − 2 x ⎟: x =
⎜ 1+ x + x
2
x +1+ 2x ⎟ x +1
⎜ ⎟
⎝ x2 x2 ⎠ x
⎛ ( x − 1) ( x 2 + x + 1) x( x − 1) ( x + 1) ⎞ x −1
⎜ − ⎟: =
⎜ x2 + x + 1 ( x + 1) 2 ⎟ x +1
⎝ ⎠
⎛ x − 1 x( x − 1) ⎞ x − 1
⎜ − ⎟: =
⎝ 1 x +1 ⎠ x +1
( x − 1)( x + 1) − x( x − 1) x + 1
⋅ =
x +1 x −1
( x − 1) [ ( x + 1) − x ] x + 1
⋅ = x +1− x = 1
x +1 x −1
www.matematiranje.com
6

7. ⎛ an a −n ⎞ ⎛ a n a −n ⎞
13) Izračunati A = ⎜ + ⎟−⎜
⎜ 1 − a −n 1 + a −n ⎟ ⎜ 1 + a −n 1 − a −n ⎟
+ ⎟
⎝ ⎠ ⎝ ⎠
⎛ a n
a −n
⎞ ⎛ a n
a −n
⎞
A=⎜ −n
+ −
−n ⎟ ⎜ −n
+ −n ⎟
⎝ 1− a 1+ a ⎠ ⎝ 1+ a 1− a ⎠
⎛ 1 ⎞ ⎛ 1 ⎞
⎜ an n ⎟ ⎜ an n ⎟
A=⎜ + a ⎟−⎜ + a ⎟
⎜ 1 − 1n 1 + 1n ⎟ ⎜ 1 + 1n 1 − 1n ⎟
⎝ a a ⎠ ⎝ a a ⎠
⎛ a n
1 ⎞ ⎛ 1 ⎞
⎜ ⎟ ⎜ an ⎟
A = ⎜ n1 + n
n
a ⎟−⎜ + nan ⎟
⎜ a −1 a +1 ⎟ ⎜ a +1 a −1 ⎟
n
⎜ n ⎟ ⎜ n ⎟
⎝ a an ⎠ ⎝ a an ⎠
⎛ a 2n 1 ⎞ ⎛ a 2n 1 ⎞
A=⎜ n + n ⎟−⎜ n + n ⎟
⎝ a −1 a +1 ⎠ ⎝ a +1 a −1 ⎠
a 2 n (a n + 1) + 1(a n − 1) a 2 n (a n − 1) + 1(a n + 1)
A= −
(a n − 1)(a n + 1) (a n + 1)(a n − 1)
a 3n + a 2 n + a n − 1 − (a 3n − a 2 n + a n + 1)
A=
(a n − 1)(a n + 1)
a 3n + a 2 n + a n − 1 − a 3n + a 2 n − a n − 1
A=
(a n − 1)(a n + 1)
2a 2 n − 2 2(a 2 n − 1) 2(a n − 1)(a n + 1)
A= = n = n =2
(a n − 1)(a n + 1) (a − 1)(a n + 1) (a − 1)(a n + 1)
www.matematiranje.com
7