The document defines: 1) Powers and exponentiation, including rules for multiplying, dividing, and raising powers of numbers. 2) Examples are provided to illustrate the rules. 3) It notes that care must be taken with negative bases and even/odd exponents. The summary provides the high level definition of powers/exponentiation and notes some key rules and examples are given to illustrate, highlighting the need to consider signs with negative bases. It does not include details of the specific examples or problems shown in the document.

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
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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.
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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
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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
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( 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
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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
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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
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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
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−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
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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
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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
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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
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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
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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)
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