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5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
1/32
M-1- rjeenja i kompletni postupakPOTENCIJE
Rjeenja svih zadataka s kompletnim postupkom i uputama
1. 2 3
put a
2 se ponavlja petput a p a u eksponentpi emo
Koristimo pravila:
Uputa: Prebrojite koliko se puta ponavlja isti faktor i taj broj stavite u eksponent:
, , ...
1) 2 2 2 2 2
n
n
a a a a a a a a a a a a
= = =
( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
5
5
5
3
6 3 2 6 3 2
3
4
grupiramo iste faktore...
itamo: dva na petu2
1) 2 2 2 2 2 2
2)
3)
4)
5)
6)
x x x x
x y x y x z x y x x z x x x x x x y y y z z x y z x y z
xy xy xy xy
x y x y x y x y x y
x x x
y y y
=
=
=
= = =
=
+ + + + = +
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
4
3 2
3
7)
8)
x x
y y
x y x y x y x y x y
x y x y x y x y x y x y x y
a b a b a b a b
c c c c
=
+ + =
= + + = +
=
2.
( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( )
( )
na eksponent
na
2 3
put a
2 3
2
parni
3
neparni
4
Zakljuak:
Koristimo pravila:
1 1
1
, , ...
1) 5 5 5 25 2) 2 2 2 2 8 3) 3 3 3 3 3 81
4) 1 1 1 15) 1 1 1 1 1
6) 1 1 1 1 1 1
n
n
a a a a a a a a a a a a
=
= = =
= = = = = =
= =
= =
= =
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
eksponent
3
4
Imamo paran broj "minusa" paje produk t p ozi tivan bro j
Imamo neparan broj "minusa" pa e i umnoak biti negativno tj. imati e predznak minus
1
7) 2 2 2 2 8
8) x x x x x
=
= =
=
4x=
4
7
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
2/32
8
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za poduku
tel-01-4578-431
2 3
put a
2
3
5
2
3
Koristimo pravila: , , ...
1 1 1 19)
3 3 3 9
3 3 3 3 2710)
5 5 5 5 125
2 2 2 2 2 2 3211)
3 3 3 3 3 3 243
2 2 2 2 2 412)
3 3 3 3 3 9
2 213)
3
n
n
a a a a a a a a a a a a
= = =
= =
= =
= =
= = =
=
4
2
3
4
2 2 2 2 2 8
3 3 3 3 3 3 27
3 3 3 3 3 8114)
4 4 4 4 4 256
4 4 4 4 4 1615)5 5 5 5 5 25
4 4 4 4 4 4 4 6416)
5 5 5 5 5 5 5 125
4 417)
5 5
= =
= =
= = + =
= = =
=
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )
2
2
3
3
2
4 4 4 4 4 4 4 256
5 5 5 5 5 5 5 625
18) 0,2 0,2 0, 2 0,04
19) 0,2 0,2 0,2 0,2 0,2 0,04
20) 0,2 0,2 0,2 0,2 0,04 0, 2 0,008
21) 0,2 0,2 0,2 0,2 0, 2 0,2 0,2 0,008
22) 2,5 2,5
= + =
= =
= = + =
= = =
= = =
= ( ) ( )2,5 2,5 2,5 6,25 = + =
2.
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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M-1- rjeenja i kompletni postupakPOTENCIJE
2.
( ) ( ) ( ) ( ) ( )
) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 3
put a
3
4
2 3 4 5
Prebrojimo min
Koristimo pravila: , , ...
23) 2,5 2,5 2,5 2,5 2,5 2,5 2,5 15,625
24 2,5 2,5 2,5 2,5 2,5 2,5 2,5 2,5 2,5 39,0625
25) 1 1 1 1 1 1 1 1
n
n
a a a a a a a a a a a a
= = =
= = =
= = + =
+ + + = + +
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
use -imaih paran broj pa e umnoakbit i poz iti van bro j
20 30 45
1 2 3 4
5 2
4 5
1 1
26) 1 1 1 1 1 1 1
27) 2 2 2 2 2 4 8 16 4 16 2 8 10
28) 2 3 2 2 2 2 2 3 3 32 9 23
29) 3 2 3 3 3 3 2 2 2 2 2 81 32
= + =
+ + = + + =
+ + + = + + = + =
= = =
= = =
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )
3 5
2 3
2 22 3
2 2
49
30) 2 2 2 2 2 2 2 2 2 2 8 32 40
1 1 1 1 1 1 1 1 1 1 2 131)
4 2 4 4 2 2 2 16 8 16 16
32) 0,2 0,2 0,2 0,2 0,2 0,2 0,2
0,04 0,008 0,032 0,001024
+ = + =
+ = + = = =
+ = + =
= = =
9
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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10
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za poduku
tel-01-4578-431
3. ( ) ( )
( ) ( )
( )
( )
je potpuno isti izraz...
ili taj isti zadatak na malo dui ali sigurn
Koristimo pravila:
5
1) 2 3 2 3 5 5 il i
5
2) 2 4 1 2 4 7
n n n n n n
c a d a c d a c a d a c d a
c a d a c d a c a d a c d a
x
x x x x x
x
x x x x x
+ = + =
+ = + =
+ = + = =
+ + = + + =
( )
iji nain :
Dakle:
Praksa je pokazala da velika veina aka radi istu greku: uzimate da je: to nije tono!!
dakle vi kada raunate u glavi greite na ovaj nain :
2 4 1 2 4 1 2 4 7 1
0
2
x x x x x x x x x x
x x
x x
+ + = + + = + + = =
=
+ ( )
( )
( )
( )
ili to nije tono!!!
Jednom zauvj ek treba zapamt it i pa to u zadat ku t reba izg ledat i ovako:
ili taj
4 6 2 4 0 2 4 6
1
2 4 1 2 4 1 2 4 7
3) 7 2 7 2 5 5
4) 9 2 3 9 2 3 1 9 9
x x x x x x x
x x
x x x x x x x x
a a a a a
y y y y y y y
+ = + + = + + ==
+ + = + + = + + =
= = =
+ = + = =
( )
( ) ( )
( )
( )
isti zadatak na malo dui ali sigurniji nain :
9 2 3 9 2 3 1 9 2 3 1 9
5) 2 3 5 7 2
2 7 3 5 2 2 1 7 3 5 2 8 6 8 6
6) 2 3 2 3 1 6
7) 2 4 2 4 1 1
y y y y y y y y y y
x a x a x a
x x x a a a x a x a x a
xy xy xy xy xy
ab ab ab ab ab
+ = + = + =
+ + + =
= + + + = + + + = + = +
+ + = + + =
+ = + =
( )
( ) ( ) ( )
2 2 2 2 2 2
2 2 2
2 2 2
2
2
2 Ovo je potpuno isti izraz
8) 4 2 9 4 2 9 3 3
9) 3 4 5 2 8 3 8
2 3 8 4 3 5 8
1 2 3 8 4 3 5 8
4 4 6
4 4 6
ab
xy xy xy xy xy xy
y x y z x y y z y x y
y y y z z x y x y x y
y z x y
y z x y
y z x y
=
+ = + = =
+ + + =
= + + + =
= + + + =
= + + =
= + +
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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11
M-1- rjeenja i kompletni postupakPOTENCIJE
( ) ( )( ) ( )
( )
( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
3 3 3 3 3 3
2 2 2 2 2
ili taj isti zad
Koristimo pravila:
10) 7 2 4 7 2 4 9 9
11) 2 3 2 3 1
12) 3 4 3 4 1 6
n n n n n n
c a d a c d a c a d a c d a
c a d a c d a c a d a c d a
xy xy xy xy xy xy
x y x y x y x y x y
x y x y x y x y x y
+ = + =
+ = + =
+ = + = =
+ = + = + =
+ + + + = + + = +
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
2 3 2 3 2 3
2 3 2 3 2 3
2 3
2 3
2 3
2 2
atak na malo dui ali sigurniji nain :
3 4 3 4 1
3 4 1
6
13) 5 2 2 7 3 3
5 2 3 2 7
5 2 3 2 7
6 9
6 9
14) 3 3 5 4 5 1
3
x y x y x y x y x y x y
x y
x y
x y z x y z x y z
x y x y x y z z
x y z
x y z
x y z
x x x x
+ + + + = + + + + =
= + + =
= +
+ + + =
= + + + =
= + + + =
= + =
= +
+ + ==( )( )
( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( )
2
2
2
ili taj isti zadatak na malo dui ali sigurniji nain :
4 3 5
1 3 5
3 5
15) 3 1 6 1 1 3 6 1 1 2 1
3 1 6 1 1 3 1 6 1 1 1
3 6 1 1
2 1
x x
x x
x x
x y x y x y x y x y
x y x y x y x y x y x y
x y
x y
+ =
= + =
= +
+ + + + = + + = +
+ + + + = + + + + =
= + + =
= +
3.
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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12
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za poduku
tel-01-4578-431
5 3 5 3 8
2 7 2 7 9
2 2 3
2 5 2 2 5 2 2 5 2 3 7
2 6 4 8 2 6 4 8 20
4 7 4 7 4 7 11
Koristimo pravila: : :
1) 2 2 2 2
2) 5 5 5 5
3) 3 3 3 3
4) 7 7 7 7 7
5) 2 2 2 2 2 2
6) 3 2 3 2 6 6
nn m n m n m n m n m n m
m
x x x x x
m m m m m m m
aa a a a a a a a a
a
a a a a a a
+
+
+
+
+ +
+ + +
+
= = = =
= =
= =
= =
= = =
= =
= = =
2 3 2 3 5
3 5 1 3 5 1 3 5 9 1
2 4 6 2 4 6 12
2 2 5 2 2 2 6 5 3 4 12 15 315223 3 2 6 6 62
2 2 2 3 2 2 2 3 2 2 2 3 2 5
3 6 3 6 9
4 2
Pazi
7)
8)
9)
10)
11)
12)
13)
a a a a a a a
x x x x
x x x x x x x x x x
x x x x x
x x x x x x x
x x x x x x x
a a a a
a a
+
+ +
+ +
+ + + ++ +
+ + + + + + + + + +
+
= =
= = = =
= =
= = = =
= = =
= =
4 2 1 4 2 1 7 1
4 2 4 2 61
4 2 4 2 0 6
2 5 7 3 2 5 7 3 17
3 2 2 3 2 2 6 2
2
Pazi
Dosta esto radite ovakve greke:
to nije tono jer je:
14)
15)
16)
x x x x x x x
x y x x
a a a a a a a a
a a a a aa a
a a a a a
a a a a a a
a a a a a
a a a
+ +
+
+ +
+ + +
+ + + + +
+
= = = =
= = = = =
= =
= =
3 2 3 2 3 4 4
3 4 1 3 4 1 3 4 817) 2 6 3 2 6 3 36 36
y x y x x y x x x y y x ya a a
x x x x x x x x
+ + + + + + + + + +
+ +
= = =
= = =
4.
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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M-1- rjeenja i kompletni postupakPOTENCIJE
4.
( )
2 5 2 3 6 3 2 2
2 5 2 1 3 1 1 6 3 2 2
7 2 1 3 1 1 6 3 4
7 7 3 4 3 4
Koristimo pravila: : :
18) 2 3 4 2
2 3 4 22 3 4 2
2 4 3 2
2 4
nn m n m n m n m n m n m
m
aa a a a a a a a a
a
x x x y x y x x x y y
x x x y y x x x yx x y x x y
x x x y x y
+
+ +
+ + +
= = = =
+ + =
= + + =
= + + =
= + + =
= ( )7 3 4
7 3 4
7 3 4
2 3 4 2 3 4 9
2 1 3 2 2 1 3 2 2 3 1 2
To je potpuno isti izraz... uobiajeni zapis je ovaj zadnji...
3 2
2 5
2 5
1 1 1 1 1
19) 2 2 2 2 2
20) m m m m m m
x x y
x x y
x x y
x x x x
+ +
+ + + + + + + +
+ + =
= + =
= +
= =
= = = 5 3
2 4 3 2 2 4 3 2 2 3 2 4 5 2
33 3 3 3 2 2 72 91 1 2 7 2 1 8 9 9
2 7 22 2 2 2 2
2 3 2 2 3 2 2 3 2 6 2
2 3 2 2
21)
22)
23)
24)
m
m m m m m m m
mm m m m m m mm
m n m n m n m n m n m n m m m n n n m n
m n m n m
x
x x x x x
x x x x x x x
x x x x x x
x x x
+
+ + + + +
+ + ++ + + + + + + +
+ + + + + + + + + + +
= = =
= = = = =
= = =
2 3 2 2 2 3 2 2 7 4
2 3 3 5 2 3 3 5 2 3 3 5 5 2
3 1 4 7 3 1 4 7 4 3 1 7 5 2 6 5 2 6
2 2
25)
26) 2 3 2 3 6 6 6
27) 2 5 2 5 10
n m n m n m n m m m n n n m n
m n m n m n m n m m n n m n
m n m n m n m n m m n n m n m n
x y x y x y x y
x x x
a a a a a
a a a a a a
a a a
+ + + +
+ + + + +
+ + + + + + + +
+ + + + +
= = =
= = =
= = = =
= =
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 3 2
3 2 2 4 3 2 2 4 3 2 2 4 3 2 2 4 5 2
2 2 1 2 1 3
3 4 3 4 7
10
3 32 9 2 9 2 3 328)
3 4 3 4 3 2 2 2 2
29)
30)
x x y y x y
m n m n m n m n m n m n m m n n m n
a a
a a a a a a a
x y x y x y x y x y x y
x y x y x y x y
+ + + +
+ + + + + +
+
+
=
= = = =
= = =
+ + = + = +
13
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
4.
( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
1 2 2 3 4 1 2 2 3 4
2 3 1 2 4 6 1
1 1 1 1 22 2 2 2 2 2
2 3
Koristimo pravila: : :
31) 1 1 1 1
1 1
32)
33)
nn m n m n m n m n m n m
m
m m m m m m
m m m m
m n m n m n m n m n m n
aa a a a a a a a a
a
x y x y x y x y
x y x y
x y x y x y x y x y x y
x y x y x
+
+ + + + + +
+ + + +
+
= = = =
+ + + = + =
= + = +
+ = + + = + = +
+ +( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
2 1 3 2 2 1 3 3
2 2 1 3 3
2 2 1 3 3
2 1
3 2 4 3 2 3 2 4 3 22 2 2 2 2
32
34)
m m m m
m m
m m
m m
x y x y x y y x x y x y x y y x
x
y x y x y x y x y x y
x y x y
x y x y
x y x y
ab ab ab ab ab
c c c c c
ab
c
+ +
+ +
+
+ + +
+
= + + =
= + =
= + =
= +
= =
=
( ) ( )
4 2 2 3
62
4 2 2 4 2 2 1 4 2 2 1 6 3 6 3
56 4 6 4 5 6 5 4 1 11 5
2 3 4 3 4 2 2 3 3 4 4 2 2 3 3
1 1 135) 9 9 3 3 3 3
3 3 3
136) 25 5 5 5 5
5 5
37) 2 3 2 3 6
x x x y y y y
x y
ab
c
x y x y x x y y x y x y x y
yxx y x y y x x y x y
x y z x y z x x y y z z x y
+ +
+ +
+ +
+ +
=
=
= = = =
= = =
= =
( ) ( ) ( )
( ) ( )
4 4 2 5 7 6
2 4 3 2 2 4 2 3 2 2 2 2
2 4 2 3 2 2 2 1 2
6 5 4 3 2
2 2 3 3 2 2 2 3 2 3
1 2 2 1
6
38) 3 2 7 3 2 7
3 2 7
3 2 7
1 139) 2 2 2 2
2 2
2
z x y
x x x x x x x x x x x x x x
x x x x x
x x x x x
xy x y xy x y xy x y xy xy xy x y
x x y y
+
+ + + +
=
+ = + + + =
= + =
= +
+ = + =
= 1 1 2 3 1 3 2 1
1 2 2 1 1 1 2 3 1 3 2 1
3 3 2 5 4 3
12 2
2
2 2
2 2
x x y y x x y y
x y x y x y
x y x y x y
+ + + + + +
+ =
= + =
= +
z
14M.I.M-Sraga centar za poduku
tel-01-4578-431
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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15
M-1- rjeenja i kompletni postupakPOTENCIJE
4
( )
9 5 9 5 4 4
7 5 7 5 2
5 3 5 3 2
3 2 23 2 2 3 2 2 3 2 2 2
U slijedeim zadacima koristimo pravilo:
Pazi: drugi eksponent se
. :
40) 2 : 2 2 2 2 2 2 2 2 16
41) 13 : 13 13 13 169
42) 3 : 3 3 3
43) 5 : 5 5 5 5 5
n m n m
x x x x x
m mm m m m m m m
a a a
+ +
=
= = = = =
= = =
= =
= = = =
3 2 2 3 2 2 3 2 2 2 4
mora staviti u zagradu...vrlo esto radite ovakvu greku:
to nije tono!
Ovdje je pogreka u predznaku kod zadnjeg lana, jer prvom promjenite predznak,
a drugom
5 : 5 5 5 5m m m m m m m = = =
4 2 4 2 2
6 2 6 2 4
7 2
ne, to se dogaa zbog toga to radite napamet... tj. preskaete korake...
Preporuka: im imamo vielane eksponente koristite zagrade i ne preskaite korake...
44) :
45) :
46) : :
x x x x
x x x x
x x x
= =
= =3 7 2 3 2
7 3 4 7 3 4 6
2 1 2 1 2 1
1 1 2 2 1 4 1 32
2 2 2 2 2 2
7 1 7 1 7 1 3 7 3 4
9 3 9 3 9 9 9
2 2 1 2 2 2 10 1 5 4 20 5 2112
25 5 2 10 10 102
4 2 4 2 2
47) :
148) : :
49) :
50) :
51) : :
52) :x x x x
x x
x x x x x
x x x x x xx
x x x x x x
x x x x x x
x x x x x x x
a a a a
+
= =
= =
= = = =
= = = =
= = = =
= = = =
= =8 3 2 8 3 2 3
2 3 1 2 3 1 2 3 4
4
7 2 3 7 2 3 8
55 2 5 2 3
2
88 3 8 3 5
3
33 32 1
12 2
il i
U slijedeim zadacima koristimo pravilo:
53) : :
154) : : : :
55) :
:
56) :
57) :
58) :
x
nn m n m
m
a a a a a
a a a a a a a aa
a a a a a
aa a a
a
xx x x x
x
xx x x x
x
xx x x x
x
+
= =
= = = =
= =
= =
= = =
= = =
= = =3 2 1
2 2
71 7 17 7 3 1 4 21 4 1743 4 34 12 12 12
1
3
59) :
x
xx x x x x x
x
=
= = = = =
5/26/2018 Stepeni(Potencije) Operacije Sa Stepenima
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
4.
( )
( )
11 2 21 2 2 1 2 2 2 1 2 1
2 2
3 73 7 3 53 7 3 5 3 7 3 5 3 3 7 5 2
3 5
8 71 8
3 2
U slijedeim zadacima koristimo pravilo: :
60) :
61) :
62) :
nn m n m
m
mm mm m m m m m m
m
mm mm m m m m m
m
mm m
m
aa a a
a
xx x x x x x
x
xx x x x x xx
xx x
x
++ ++ + + +
+
++ ++ + + +
+
+
+
= =
= = = = =
= = = = =
= ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( )
8 7 3 2 17 3 2 1 8 7 3 2 1 8 3 7 2 1 4 6
9 119 11 2 3 7 107 10 9 11 2 3 7 10 9 11 2 3 7 10
2 3
9 2 7 11 3 10 4
7 4 7 4 3
2 3
: :
63) : : :
64) :
65) 2 3
m m mm m m m m m m m m
mm m mm m m m m m m
m
m m m
x y
x x x x x x
aa a a a a a
a
a a
x y x y x y x y
x y
+ + + + + + + + +
++ ++ + + + +
+ +
+
= = = =
= = = =
= =
= =
( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
( )
( ) ( )
2 3 2 3 2 3 2 3 2 3 0
6 4 6 4 22 2 2 2
3 3 3 3
2 4 4 2 4 2 4 4 2 4 2 4 4 2 4
2 4 4 4 2 5
: 2 3 2 3 2 3 2 3 1
66) :
67) 2 2 : 2 2 2
2 2
x y x y x y x y x y
m m m m m m m m m
m m m m
x y x y x y x y
a b a b a b a b
c c c c
x y x y x y x y x y
x y x y
+ + + +
+ + +
+ +
= = = =
= =
+ + + = + = + =
= + = +
( )
2
2 3 3 4 4 2 3 3 4 4 3 2 3 4 42 2 2 2 2
3 3 3 3 3
3 4 4 2 3 12 2 3
3 3 2
U ovom sluaju moete ostaviti i ljevo rjee
il i
68) :
x x x x x x x x x
x x x
a b a b a b a b a b
c c c c c
a b a b c
c c a b
+ + +
+ +
= = =
= = =
( ) ( )
nje...jer jo nismo obradi li pra vilo :"na minus prvu "
2 4 2 2 4 2 2 2 4 2
2 4 2 2 2 4 2
Mnoimo svaki lan prve zagrade sa svakim lanom druge zagrade...
69) 1 1 1 1 1 1 1
1
x x x x x x x x x x
x x x x x+ +
+ + = + + =
= + + =
( ) ( ) ( ) ( )
6 4 2 4 2
6 4 4 2 2
6
2 2 2 1 2 2 2 2
2 1 2 2 2 2
3 4 2 2
4 3
1
1
1
70) 1 1 1 1 1 1 1
1
1
1
x x x x x
x x x x x
x
x x x x x x x x x x
x x x x x
x x x x x
x x x
+ +
= + + =
= + + =
=
+ + = + + + + + =
= + + + =
= + + + =
= + + +
16M.I.M-Sraga centar za poduku
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17
M-1- rjeenja i kompletni postupakPOTENCIJE
( ) ( ) ( ) ( )( )( )
Koristimo pravila:
vidi zadatke: 7),13),14),15),16),17),18),19),23),24)...
vio nam se jedan problem a taj je
2.
:
ogledajmo opet u zada kako smok tt oa
zm mn nn n n n n n n m n n m z
n
ab a b abc a b c a a a a
x
= = = =
( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( )
( )
( ) ( ) ( ) ( )
na eksponent
na eksponent
2
parni
3
neparni
4
3
Imamo neparan broj "minusa" pa e i umnoak bi
1 1Zakljuak:
1 1
tamo rjeili:
. 4) 1 1 1 1
5) 1 1 1 1 1
6) 1 1 1 1 1 1
7) 2 2 2 2 8
2
=
=
= =
= =
= =
= =
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
4 4
Imamo paran broj "minusa" paje produkt p ozi tivan bro j
53
53 puta bi to trebali na
ti negativno tj. imati e predznak minus
8)
vlja ovakav problem: ...
x x x x x x
x x x x
= =
=
( )
( )
pisatia to je previe pisanjapa emo rae gledati eksponent
53 53
54je logiki je pitanje koliko je eksponent je 54, 54 je paran broj
u eksponentu je 53 ,
an broj pa je:
?
x x
x
=
=
( )
( )
( )
( )
( )
( )
na eksponent
na eksponent
54 54
55 55
56 56
2parni 2parn u
2 1neparni neparnu
pa je:
Zakljuak:
jer je eksponent 55, a 55 je neparan broj
jer je eksponent 56, a 56 je paran broj
ili
n n
n
x x
x x
x x
x x
x x
x x
x +
=
=
=
= =
= =
=
( )
( )
2 1
4 4
3 3
2 je oznaka za parn i b ro j
2 1 je oznaka za neparan broj
Sada to primjenimo u zadatcima:
.8) zato to je eksponent 4, a 4 je paran broj
7) 2 2 8 zato to je eksponent 3, a 3 je neparan broj
Po
2
n
n
n
x
x x
+
+
=
= =
= = =
( ) ( ) ( )
( ) ( )
( )
( ) ( )
( )
( ) ( ) ( )
3 3 3 3
3 3 3 3
Svaki negativan brojda se zapisati u obliku:
1
4 4 4 4 4 4II na
stoji i drugi nain rjeavanja ovakvih zadataka:
II nain
7) 2 1 2 1 2 1 8 8
2 1 2 1 2 1 8 8
8) 1 1 1
x
x x
x x x x x
=
= = = =
= = = =
= = = =
in
5.
Poja
P
Sada se ja
53 je nepar
dal
( )
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
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( ) ( ) ( ) ( )( )( )
) ( )) ( )
) ( ) ( )
2 2 2 2 2
2 2 2 2 2
2 2 22 2
2 3 2 3 2 2 3 2 4 6
2
Koristimo pravila:
Postupili smo prema prvom pravilu...
1 2 2 4 4
2 3 3 9 9
3 3 3 93
4 4 164
zm mn nn n n n n n n m n n m z
n n n
ab a b abc a b c a a a a
ab a b
x x x x
x x x x
x y x y x y x y
ab
= = = =
=
= = =
= = =
= = =
( ) ( )
) ( )
) ( ) ( )
) ( )
) ( ) ( )
3 3 3
32 3 2 3 2 3 3 63
4 4 44 4
2 3 2 3 2 4 3 4 8 12
4
22 2 2 4
2 22 2 2 2 4
Postupili smo prema drugom i treem pravilu...
I
2 2 2 843 3 273
1 1 1 15
2 2 162
6
7
mn n n n n n mc a b c a a
xy x y x y x y
x y x y x y x y
x x x
x x x x
= =
= = =
= = =
= =
= = =
) ( ) ( ) ( ) ( )
) ( )
) ( )
( )
) ( ) ( ) ( )
) ( ) ( ) ( )
) ( ) ( )
2 2 222 2 2 2 2 4
32 2 3 6
52 2 5 10
2 2 23 4 2 3 4 3 2 4 2 6 8
3 3 33 4 3 3 4 3 3 4 3 9 12
4 43 4 4 3
nain
II nain :
Prema pravilu:7 1 1 1
8
9
10 2 2 4 4
11 2 2 8 8
12 2 2
mn na a
x x x x x
x x x
x x x
x y x y x y x y
x y x y x y x y
x y x
= = = = =
= == =
= = =
= = =
= ( )
) ( ) ( )
) ( ) ( ) ( ) ( )
44 3 4 4 4 12 16
3 32 2 2 3 6
3 3 3 332 2 2 2 2 2 2 3 6II nain :
16 16
13
13 1 1 1
y x y x y
y y y y
y y y y y y y
= =
= = =
= = = = = =
5.
m
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19
M-1- rjeenja i kompletni postupakPOTENCIJE
5.
( )
( )
( ) ( )
na eksponent
na eksponent
parni par nu
neparni neparnu
3 32 2 2 3 6
Koristimo pravilo: prilikom raunanja nainom u
14), 15), 16),17), 18), 19)...
to je bio I nain rjeavanja
I
14)
II nain
1
x x
x x
y y y y
=
=
= = =
) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
) ( ) ( ) ( ) ( )
( ) ( )
3 3 332 2 2 2 3 6 6
3
2
2 2 223 3 3 3 2 6 6
2 23 3 3 2 6
Prema pravilu:
Prema pravilu:
to je bio II nain rjeavanja
to je bio I
4 1 1 1 1
1 1 1 1
1 1 1 1 1
15 1 1 1 1
15)
neparnu
par nu
y y y y y y
y y y y y y
y y y y
= = = = =
= =
= + = + =
= = = = =
= = =
( ) ( )
) ( ) ( )( ) ( )) ( ) ( ) ( ) ( ) ( ) ( )
) ( ) ( )
3 3
3 34 4 4 3 12 12
3 3 334 4 4 4 3 12 12
2 25 5 5 2 10
nain rjeavanja
to je bio I nain rjeavanja
II nain
to je bio I nain rjeavanja
16
16 1 1 1
17
1
par nu par nu
y y
y y y y y
y y y y y y
x x x x
=
= = = + =
= = = = =
= = =
) ( ) ( ) ( ) ( ) ( ) ( )
) ( ) ( ) ( ) ( )
) ( ) ( )
( ) ( )
2 2 225 5 5 5 2 10 10
2 2 2 25 5 5 5 5 2 10
2 25 5
22 5
II nain
U 18) pokazat emo tri naina rjeavanja:
to je bio I nain rjeavanja
7 1 1 1
18
18 1
1 1
x x x x x x
x x x x x x
x x
x
= = = = =
= + = = = =
= =
= = = ( ) ( ) ( )
) ( ) ( ) [ ]
2 225 5 5 2 10
2 225 5 5 5 2 10
II nain
III nain ...u min us i minus daju plus
U ovom 18) zadatku najbri ne III nain rjeavanja... dok je u 19) definitivno II nain najb
1 1 1
18
x x x x
x x x x x
= = =
= + = = =
olji....
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
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5.
) ( ) ( ) ( ) ( ) ( )
( )
) ( ) ( )( ) ( )
2 22 2 2 2 4 4
5 5 5 5 5 5 4 20
2 22 2 25 5
I nain
I nain u ovom zadatku pomalo zbunju je pa je bolje to rjeavati na II nain izluivan jem 1 :
II nain :
1 1
x x x x x x x
x x x
= + = = = = =
= =
( )( )( )
( ) ( ) ( )
) ( ) ( ) ( )
) ( ) ( )
) ( ) ( ) ( )
) ( ) ( ) ( )
( ) ( )
22
5
25 2
210
22 210 10 10 2 20
25 5 2 10 10
3 32 3 3 2 3 2 3 3 6
4 4 42 3 2 3 2 4 3 4 8 12
2 223 3 3 2 6
2 2
1
1
1 1 1
2 2 8 4
2 2 4 4
2 4 jer je: 2
x
x
x x x x
x x x x
ab a b a b a b
a b a b a b a b
y y y y
=
= =
= =
= = = =
= = =
= = =
= = =
= = =
= = ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
( )
na eksponent
na eksponent
2 2 2
3 3 3 3 3
4 4 4 4
2
3
parni parnu
neparni neparnuoristimo pravilo:
1 2 1 2 1 4 4
2 8 jer je: 2 1 2 1 2 1 8 8
2 1 2 1 2 1 16 16 itd.
ili ovako: 2 2 2 4
2 2 2 2 8
ili kx x
x x
=
=
= = =
= = = = =
= = = =
= =
= =
( )
( )
( )
) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 2
3 3
3 3
3 332 2 3 2 3 6
3 3 3 33
pa je :
il i
2 2 4
2 2 8
3 3 27
24 3 3 3 27
3 1 3 1 3 1 27 27 3 3 3 3
y y y y
= =
= =
= =
= = =
= = = = = =
19
19
20
21
22
23
27
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M-1- rjeenja i kompletni postupakPOTENCIJE
5.
25
25
25
26
26
27
) ( )
) ( )
) ( ) ( )
2 2 2 22
2 2 2 2 2 4
2
2 2 22
2 2 2 2 4
2
2 2 2 2222 2 2 2 2 4
2
25)zadatak na tri nai
2 2 2 2 4
3 3 3 93
2 2 2 4
3 3 932 2 2 2 4
1 1 13 3 3 93
x x x x x
x x x x
x x x x x
= = = =
= = + =
= = = =
) ( )
) ( )
) ( )
) ( )
) ( ) ( ) ( )
) ( ) ( ) ( )
3 3 3 33
2 2 2 2 3 6
3
3 3 33
2 2 2 3 6
3
22 2
33 3
4 4 44 4 4 4
na
I nain
II nain
2 2 2 2 8
3 3 3 273
2 2 2 8
3 3 273
28
29
30
m m m
n n n
m n m n m n m n
m m mm n m n
x x x x x
x x x x
x x x
y y y
x y x y x y x y
x y x y
= = = =
= = =
= =
= =
= = =
= =
) ( ) ( ) ( )
) ( ) ( ) ( )
) ( ) ( )
) ( ) ( )
) ( )
) ( ) ( ) ( ) ( ) ( )
2
2
2
2 2 23 2 3 2 3 2 2 2 6 4
3 2 3 2 3 2 3 2
22 2 2
33 3 3
2 2 22 2 2 2 2 2
31
32
33 2 2 2 2 4
34 3 3 3 3 27
35 2 2 2
36 2 3 2 3 2 3 2 3 2 3 4
m m n m m n m
x y x y x y x y
x x xx y x y x x y x x xy
xx x x x
xx x x x
xx x x x
m nm n m n m n m n m
x y x y
a b a b a b a b
a b a b a b a b
=
= = =
= = =
= = = =
= = = =
= =
= = = = = 9 4 9n m n =
21
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
5.
37
38
39
) ( )( ) ( )( )
) ( )( )
) ( )( )
) ( )( )
) ( )( ) ( )
) ( ) ( )( )
) ( )( ) ( )
43
2 2 3 4 24
54
3 3 4 5 60
42 2 4 8
3 3 3
26 5
2 3 2 6 2 3 5 24 15 24 15 39
24 3
3 5 3 4 5 3 2 12 30 12 30 42
35 7
2 4 2 5 3 4 7
Prema pravilu:
40
41
42
43 : :
zm
n n m z
xx x
yx
x y xy
x x x a a
y y y
x x x
y y y
x x x x x x x x
y y y y y y y y
x x x x
+
+
= = =
= =
= =
= =
= = = =
= = = =
=
) ( )( ) ( )( )
) ( ) ( ) ( ) ( )
( )
( )
30 28 30 28 2
8 56 4
3 2 3 6 8 2 4 5 144 40 144 40 104
3 32 3 3 2
2 3 2 2 3 2 2 2
32 3 3 3 2 2 2
3
6 9 2 4
6 2 9 4
6 2 9 4
:
44 : : :
2 245 : 2 : 2
3 3
2: 4
3
8: 427
8:4 :
27
2
27
x x x x
a a a a a a a a
a b ab a b a b
a b a b
a b a b
a a b b
a b
= = =
= = = =
= =
= =
= =
= =
= =
4 52
27a b=
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M-1- rjeenja i kompletni postupakPOTENCIJE
23
( ) ( ) ( ) ( )( )
) ( ) ( ) ( ) ( ) ( )
( )
( )
3 33 3 3 3 3
2 3 3 2 2 3 3 3 2
32 3 3 3 3 3 2 3
3
6 9 9 6
6 9 9 6
6 9 9 6
Koristimo pravila:
1 14 4
2 2
164
2
164
8
164
8
64
8
zm mn nn n n n n n n m n n m z ab a b abc a b c a a a a
x y x y x y x y
x y x y
x y x y
x x y y
x y
+ +
= = =
= =
= =
= =
= =
= =
=
) ( ) ( ) ( ) ( ) ( )
( )
( )
) ( )
15 15
15 15
3 33 3 3 3 3
2 3 3 2 2 3 3 3 2
32 3 3 3 3 3 2 3
3
6 9 9 6
6 9 9 6
6 9 9 6
3 3
3
3
3
3
2
8
8
1 1: 4 : 4
2 2
1: 64
2
1: 64
8
1:64 : :8
1
8 64
1
512
1 1
512
512
48 3
x y
x y
x y x y x y x y
x y x y
x y x y
x x y y
x y
x y
yx
y
x
a b
=
=
= =
= =
= =
= =
= =
= =
= =
=
( ) ( ) ( ) ( )
( )
( )
) ( ) ( )
2 2 23 2 3
2 2 3 2
2 2 6
2 2 6
2 2 6
4 7
1 3 13 3 3
3
3
3
3
3
3
49
x xx x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x
m m m
a b a b a b
a b a b
a b a b
a a b b
a b
a b
x x x
+ +
+ + +
= =
= =
= =
= =
= =
=
= =
5. =
46
47
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
5.
49
50
51
52
53
( ) ( ) ( ) ( )( )
) ( ) ( )
) ( ) ( ) ( ) ( )
) ( ) ( ) ( ) ( )
1 3 13 3 3
2 1 1 3 2 1 2 13 2
6 3 2 2
6 3 2 2
6 2 3 2
8 5
2 1 1 3 2 1 2 15 3
6 3 2 2
6
Koristimo pravila:
: :
:
zm mn nn n n n n n n m n n m z
m m m
m m m m
m m
m m
m m
m
m m m m
m m
m
ab a b abc a b c a a a a
x x x
x x x x
x x
x
x
x
x x x x
x x
x
+ + +
+
+
+
= = =
= =
= =
= =
= =
= =
=
= =
= =
= ( )
) ( ) ( )
( )
) ( ) ( ) ( )
( )
3 2 2
6 3 2 2
6 2 3 2
4 1
3 22 3 2 3 3 2
6 6
6
6
4 6 33 2 4 3 4 2 6 4 3
12 12 12
12
12
12
2 3 2 3
2 3
2 3
5
3 4 3 4
3 4
1 3 4
2
2
m
m m
m m
m
x
x
x
x x x x
x x
x
x
a a a a a a
a a a
a
a
a
+
+
=
= =
= =
=
+ = + =
= + =
= + =
=
+ = + =
= + =
= + =
= =
=
=
24M.I.M-Sraga centar za poduku
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M-1- rjeenja i kompletni postupakPOTENCIJE
6. ( )
( ) ( )
( ) ( )22 2 4 4 4
2 4
3 3
Koristimo pravilo:
1) 2 5 2 5 10 2) 3 4 3 4 12
3) 4 6 4 6 24 4) 2 4 2 4 8
5) 6) 1 1
3 8 3 87)
4 3 4 3
nn n n
x xx x x x x x
a ya a a y y y
a b c abc
a cac b b a b a bab c b c b a b a
=
= = = =
= = = =
= = = = =
=
33
3
2 2 22 2
2 2
2
2 2
3 33
moemo ostaviti u ovom obliku
ili dalje:
nakon kraenja
4 232 8
4 3
2 24 4 28)
2 2 2
2 4
9)
x yxy xy
x x y xx y x y
x x
xy x xy x
z yz z yz
= = =
= = = =
= =
= =
( )( )
( )
( )
( )
( )
333 22 2 3 6
2 3 2 3 62
1 m+11
111 2 2 12 2 2
2 1 2 22 12
22 3 6
4 3 2
nakon kraenja10)
11)
mm
mmm m m
m mm
xx x x x x
z z z z zz
xy x xy x
z yz z yz
xx x x x x
z z z zzz
x y zz x y
++
+++ + +
+ + +
= = = = =
= = =
= = = = =
( )
( )
22 22 2 1 4 22 3 6
4 3 2 4 2 1 2
22 22 2 2 4
1 2 1 2 21
2 2 1 4 22 3 6 2 3 6
4 3 2 4 3 2 4 2 1 2
nakon kraenja
nakon krae12)
mm m m
x y y z zx y zz x y z x x y
zz z z
x x xx
x y y z zx y z x y z
z x y z x y z x x y
= = = =
= = = =
= = =
( ) ( )
( )
( ) ( )
( )
( )
( )
( ) ( )
( ) ( )
1 21 2 1 2 2
1 11
5 555 2 2 2
2 2 2 2
5
2
nja
.
1 113)
1 12 2
1 1
1
1 1
1
m mmm m m m
m m m
y zy z y z y z
x xx
x y x x y x y x x y
x x y x x yx xy y x xy y
x xx y x y
x x yx y
x y x x x
x x y x y
=
= = = =
+ + = =
+ + + + + +
+ + = = + +
+=
+ + +
( )
( )
( )
( )
5
55
5
pa krat imo
11
y
x y
xx
x y x y
+= =
= =
+ +
x
25
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
7.
2 2
2
2 2 2
2
po pravilu :
pr imjeni li smo pravilo:
Koristimo pravila:
Izraunaj:
2 2 41)
3 93
2 3 3 92)
3 2 42
2uputa: 2)
3
n n
n
n n nn n
n n
n n
n
a a
b b
a a a b b
b b ab a
a a
b b
=
= = =
= = =
= = =
2 2 2
2
po pravilu
3 3
3
4 4
4
1 1
pr imjeni li smo pravilo:
pr imjeni li smo pravilo:
3 3 9
2 42
1 1 13)
2 82
2 2 164)
3 813
5 7 75)
7 5
n na b
b a
n n
n
n n
n
a a
b b
a a
b b
=
= = =
= = =
= = =
= =
11 1
2 2 2
2
3 3 3
mjeoviti broj treba prvo pretvoriti u razlomak...
pr imjeni li smo pravilo:5
1 7 22 9 76) 1
7 7 7 9
4 5 5 257)
5 4 164
2 3 38)
3 2
n n n
n
a b b
b a a
= =
+ = = =
= = =
= =
3
22 2 2 2 2
2
22 2 2 2 2
2
mjeoviti broj treba prvo pretvoriti u razlomak...
27
82
2 3 11 6 1 7 3 3 99) 2
3 3 3 3 7 497
1 7 22 7 2 9 7 7 4910) 1
7 7 7 7 9 9
=
+ + = = = = = =
+ + = = = = = =
81
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27
M-1- rjeenja i kompletni postupakPOTENCIJE
33 3 3 3 3
3
2
mjeoviti broj treba prvo pretvoriti u razlomak...
Koristimo pravila:
1 2 11 2 1 3 2 2 811) 1 2 2 2 2 3 273
3 412)
4 3
n n nn n
n n
a a a b b
b b ab a
= = =
+ + = = = = = =
=
( )
( )
na eksponent
na eksponent
2 2
2
2 2
2
parni parnu
neparni neparnu
2 2 22
2
nain oristimo pravilo: pa je:
4 16
93
13)
14) I k
14) jer je:
par nu par nu
x x
x x
x x
y y
x x x x x x x
y y y y yy
=
=
= =
=
= = = = =
( )
( )
2
2 2 2 2 22
2 2
3 3 3
3
3 3 3
3
II nain
nain rastavili smo na:
I nain
.. .
14) 1 1 1
II 1 ...
15)
15) 1 1
y
x x x x x
y y y y y
x x
y y
x x x
y y y
x x x
y y y
=
= = = =
= =
= = =
( )
( )
( )
na eksponent
na eksponent
3 3
3 3
parni parnu
neparni neparnu
4 4 4na parni eksponent parnu
4
4
II nain
nain oristimo pravilo: pa je:
II nain :
1
16) I k
16) jer je:
16)
x x
x x
x x
y y
x x xx x
y y y
x
y
=
=
=
= = =
=
( )
4 4 4 44
4 4
22 2 2 2
2
1 1 1
1 2 11 2 1 3 317) 1
2 2 2 2 2
x x x x
y y y y
= = =
+ + = = = =
7.
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
7.
22 2 2 2 2
2
22 2
I nain
II nain :
Koristimo pravila:
2 3 22 6 2 8 8 8 6418) 2
3 3 3 3 3 93
2 3 22 6 2 818) 2
3 3 3 3
n n nn n
n n
a a a b b
b b ab a
= = =
+ + = = = = = = =
+ + = = = =
( )
2 2 2 22
2
33 3 3 3 3
3
33 3 3
I nain
II nain :
8 8 8 61 1 1
3 3 93
2 3 22 6 2 8 8 8 51219) 2
3 3 3 3 3 273
2 3 22 6 2 8 819) 2 1
3 3 3 3 3
= = = =
+ + = = = = = = =
+ + = = = = =
( )
( )( )
( )
( )
3 3 33
3
22 2 2 2
2
4
4 22 2 4 8
3 4 3 4 123
33 3 3
3 3 3 3 9
4 4 3 4 3 124
3
4
I nain
8 8 51 1
3 23
3 5 44 15 4 19 19 36120) 3
5 5 5 5 255
21)
22)
22)
xx x x
y y yy
xx x x x
y y y yy
x
y
= = =
+ + = = = = =
= = =
= = = =
( )
( )
( )
( )
( )
( ) ( )
( )
33 3 3
3 3 3 93
4 3 4 3 124
22 2 33 3 3 2 6
4 4 2 4 2 84
22 2 33 3 3 2 6
2
4 4 2 4 2 84
II nain
I nain
II nain
1 1 1
23)
23) 1 1 1
xx x x
y y yy
xx x x x
y y y yy
xx x x x
y y y yy
= = = =
= = = =
= = = =
12
7
4
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M-1- rjeenja i kompletni postupakPOTENCIJE
7.
( ) ( )
( )
( )
( ) ( ) ( )
2 22 2 3 2 2 3 22 3 4 6
4 2 4 2 84
2 22 5 6
6 2 5
26
2 22 5
Koristimo pravila:
po pravilu:
po pravilu:
24)
25)
n n nn n
n n
n n n
n
n
a a a b b
b b ab a
x y x yx y x y
z z zz
x y z a b b
b az x y a
za
x y
= = =
= = =
= = = =
= =
( )
( )
6 2
2 2 5 2
12
4 10
33 3 4 4 34 12
2 3 2 3 3 6 33 2 3
3 32 3 3 3 4
3 4 2
Najk rae objanjenje:
Sve to ima minus u eksponentu
"skae" iz brojnik
2 82 826)
3 27 273
227)
2
mn ma
z
x y
z
x y
x xx x
y z y z y zy z
x y y z
z x
=
= =
=
= = =
= =
( ) ( )
( )
33 4
2
3 33 3 4
32
3 3 4 3 9 12
2 3 6
2 3
3 4
2 3 3 3 4
3 4
a u nazivnik i obrnuto...
Sada dodatna uputa: Razlomak
Ima u eksponentu "minus" pa se prvo rjeavamo tih "minusa"
8
8
512 512
2
2
2
y z
x
y z
x
y z y z
x x
x y
z
x y y z
z
= =
= =
= =
= 2 22
3 4 3 4
U brojniku je bio: i on je "skoio" u nazivnik i sada je:
U nazivniku su bili i oni su "skoili" u brojnik i sada su:
: 2 2
x xx
i z i z
29
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
7.
( )
( ) ( )
2 24 2 4
3 2 3
22 4 4 2 8
2 2 2 2 3 2 4 62 2 3
2 24 2 4
3 2 3
22 3
4
2
Koristimo pravila:
2 228)
5 5
2 4 4
25 255
2 229)
5 5
5
2
5
n n nn n
n n
a a a b b
b b ab a
x y x
z y z
x x x
y z y zy z
x y x
z y z
y z
x
= = =
= =
= = =
= =
= =
= ( ) ( )
( )
( )
( ) ( )
2 22 3
22 4
2 2 3 2
4 2
4 6
8
33 3 3 4 4 34 2 4 12
3 2 3 3 3 2 3 3 33 2 3
1 1
2 3 2 3 2 3
3 2
2
25
4
25
4
2 82 2 830)
5 5 125 1255
31)
32)
y z
x
y z
x
y z
x
x xx y x x
z y z y z yy z
x y y
y x x
x y y y y y
y x x x x
=
= =
=
= = = =
= =
= =
( )
( )
( )
( )
( ) ( )
2 12 1
3 2 2 1 1
33 3
3
2 22
33 3 3
2 5 3 3 3 9
3 2 5 3 3 2 3 5 3 6 152 5
2 32 3 2 333)
2 3 2 3 2 3
134)
1
35)
y yy x x
yx x x x y
xx x
x x x
x yx y
x y
ca b c c c
c a b a b a ba b
= = =
+ + = =
+
= =
= = = =
6 9z
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M-1- rjeenja i kompletni postupakPOTENCIJE
7.
( )
( ) ( )
( )
( )
3 23 2 2 222 2 2
3 3 4 3 2 2
3 3 4
2 3 2 2 2
3 3 3 2 4 2
6 2 4
9 6 8
6 2 4
7 2 6 8
4
7 8
7
kratimo
Koristimo pravila:
36) : :
:
:
:
1:
1
n n nn n
n n
a a a b b
b b ab a
a ya y y y
y a a y a a
a y y
y a a
a y y
y a a
a y y
y y a a
y
y a
y
= = =
= =
= =
= =
= =
= =
= 8
4
8 8 8
7 4 7 4 11
2 2 23 2
3 2 2
2 2 2
2
1 2
3 2 1 1 2 1 1 4 1 137) 2 2
2 3 8 4 9 4 2 42 2 3
1 2 1 91 1 1 1 11
9 2 4 18 4 36 36
8 21 3 1 4 1 4 1 16 1 238)
8 4 8 3 8 8 9 8 9 93
2 1 139)
3 3 2
a
y
a a a
y y y y+
=
= = =
+ = + = + = + =
+ = + = + = =
= = = = =
+
( )
3 1 2 3
2 2 22 3 2 32 3 2 3
3 2 2 3
22 2
3 3 2 3 3 3 2 59 8 1
2 1 1 2 2 2 2 2
3 2 3 4 3 2 4 27 240)
4 2 3 3 43 2 3 3
9 316 8 8 8 14 3 12
9 4 27 27 27
= + = + = + = + =
= = =
= = = =
2
8 1 8 1
27 144 27 48612 = =
31
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
8. 0 1 1
0 0
0
02 4
po pravilu:
bez obzi ra to je u zagradi ako je eksponent
te zagrade nula sve je jednako jedan !
ili taj isti zadatak du
Koristimo pravila:1 1
1 , , ,
1) 2 1 1
2) 1
3) 1
3)
n
na a a a a
a a
a
x
x y
z
= = = =
= =
=
=
( ) ( )
( )
( )
0 00 2 42 4
0
0 0
0
0 0
0 0 0
0
0
im postupkom:
po pravilu:
bez obzi ra to je u zagradi ako je to na nu ltu sve j
1 1 1 1
1 1
4) 1 1 2
3 1
6) 3 1 1
7) 3 1 zato to je: 3 1 3 1 1 1
8) 3 3 1 3
9) 1
x yx y
z z
x y
a
x
x y
= = = =
+ = + =
=
= =
= = = =
= =
+ =
( )
( ) ( )
( )
0
0 0 0
0 00
0 0
20 2
e jednako jedan !
bez obzi ra to je u zagradi ako je eksponent
te zagrade nula sve je jednako jedan !10) 2 7 1
11) 2 7 1 2 1 7 1 1 2 7 6
12) 2 7 1 1 1 1
13) 1 1 1
14) 1 1
x y
x y z
x y z
x x
x
=
+ = + = + =
+ = + =
= =
= =
5)
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33
M-1- rjeenja i kompletni postupakPOTENCIJE
8.
( )
( )
( )
0 1 1
0 0
20 2
0
0
1 1
1
1
bez obzi ra t o je u zagradi ako je to na nuultu sve je jednako jedan !
po pravilu:
Koristimo pravila:1 1
1 , , ,
13) 1 1 1
14) 1 1
15) 1
16) 4 1
1 117) 5
5
118)
19) 0,2
n
n
m
a a a a aa a
x x
x
x
x
aa
xx
= = = =
= =
= =
=
=
= =
=
=
( )( )
( )
1 1
1
1
2
2
3
3
2 2 22
2
prvo decimalni broj pretvorimo u razlomak .. .
po pravilu:
po pravilu:
2 1 5
510 5 1
1 1 120)
1 1 121) 3
3 9
122)
3 10 10 10023) 0,3
10 3 3 9
24) 2 5
n
n
n
n
x y ax y ax y
aa
xx
x
= = =
= = =
= = =
=
= = = =
( )
5
5
1
1
2 5
125) 4
4
x
=
=
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
8.
( ) ( )
( ) ( )
0 1 1
1
1 1
1
1 1
1 1
ili na drugi nain :
ili na drugi nain :
Koristimo pravila:1 1
1 , , ,
11 1 1 4 1 1 4126) 4 4
1 14 1 1 4 14 4
11 4 1 41 1 1 4 1127) 4 4
1 1 1 1 1 14 44 4
2 3 328) pr
3 2 2
n
na a a a a
a a
= = = =
= = = = = =
= = = = = = =
= =
( )
( )
1
1
11
1 1
1
1
Isti taj zadatak po pravilu ispada: vidimo da je rezultat isti!
ili krae:
ema pravilu:
11 2 1 3 1 31:
2 23 2 1 2
3 3
2 2 5 229) 2 5 10 2 5 2 5 10
5 1 5
2 5230) 2
15
n n n
n
a b b
b a a
aa
= =
= = = = =
= = = = = =
= =
( )
2
2
44
4
5 4 4 3 7 9 5 4 4 3 7 9 5 4 9 4 3 7 0
5 10
3 3 231) 3 4 12
2 1
2 232) 2
1
33) 3 3 3 3 3 3 1x x x x x x x x x
a a bab
b
+ + + +
=
= = =
= =
= = = =
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35
M-1- rjeenja i kompletni postupakPOTENCIJE
8.
( ) ( ) ( ) ( ) ( )
0 1 1
2 1 3 2 3 6 2 1 3 2 3 6 2 3 1 3 2 6 0
2 3 2 2 2 3 2 2 2 2 3 2 1
2 2 2 2 2 2 0
Koristimo pravila:1 1
1 , , ,
34) 2 2 2 2 2 2 2 1
135) 5 5 5 5 5
5
36) 1
3
n
n
x x x x x x x x x
x x x x x x
m n n m m n n m m m n n
a a a a aa a
x y x y x y x y x y
+ + + + + + + + +
+ +
+ +
= = = =
= = = =
= = = =
+ + = + = + = + =
( )
1 1
0 0 0 1
1 1 1
0 1 3
3
2 22 2 2 2 2
3 3 3 3 3
1 37) 27 2 1 1 1 2 1 2 3 6
3 1
1 1 1 1 5 1 1 1 1 538) 2 2 1 1 5
2 5 2 2 1 2 2 8 322
39) :
m n m n n m n m n n
x y
y
a b a b a b a b a b
c c c c c
+ + + + +
= = =
= = =
= =
2 02
3
22 2
23 1 3 1
2 4
2 4
2 2
2
1
1 2 41 1 1 1
2 2 8 2 82 240)1 1 1 1 1 4 12 2
4 16 162 2
1 8 7
7 168 84 1 5 8 5
16 16
m n m n n
a b
c
+ +
= =
= = = = ++ + +
= = = +
( )
2
2 2 2 22
2
2 2 2
2 22 1 2
1 2
2
7 2
5
14 5 5 5 25 251 1 1
5 14 14 196 19614
1 1 1 1 1
2 3 183 4 32 1241)1 1 1 1 1 122 3
2 2 9 183
6 3
6 2
= =
= = = = = =
= = = = =
=
2 2 2 2
2
3 2 2 4
2 3 93
= = = =
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
8. 0 1 1
22 2 2
2 22 1 2
1 2
2
Koristimo pravila:1 1
1 , , ,
3 41 1 1 1 1
4 3 18 12 3 3 4 32 12
42) 1 1 1 1 9 2 11 11 122 3
2 2 9 2 9 183
3 6
11 2
n
na a a a a
a a
= = = =
= = = = = = + + + +
=
( )
2 2 2 2 22
2
0 0 1
0 0 1
0 1 2
2
2 3
3 22 22 22 484 4841 1 1
6 22 3 3 93
1 3 11 1 3 1 243) 3 2 3 1 1 1
3 3 3 3 3
2 3 11 1 6 1 544) 3 2 3 1 1 2
3 3 3 3 3
3 4 11 1 12 1 1145) 2 2 2 1 2 3
4 4 4 42
46) 2 2
= = = = = =
= = = = =
+ = + = = = =
+ = + = = = =
= 2 3
3 2 1
3 2 1
2 3
2 3
2 2
3
3
2
3ili isti zadatak na drugi nain :
1 2 11 1 1 1 2 1 1
4 8 8 8 82 2
1 1 2 1 41 1 1 1 1 1 1 2 4 347) 2 2 2
8 4 2 8 8 82 2 2
1 1 1 1 8 9 1 148) 3 2
9 8 9 8 72 723 2
1 1
1 272 272 449)1 1 4 1 43
273
2 3
3
= = = =
+ + + = + = + = = =
= = = = =
= = = =
=3
2
3 2 2
2
3 2 3
2 2 2
vidi u 7. zadatku 27), 28), 29)
ili isti zadatak na drugi nain :
27
42
1 88
2 3 8 1 1 13 950)1616 9 16 9 2 184
1
2 3 2 8 1 1
16 9 2 9 184 4 3
=
= = = = =
= = = =
9
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37
M-1- rjeenja i kompletni postupakPOTENCIJE
8. 0 1 1
2 1
2 1
2 1 1
2 1 1
1
1
ili isti zadatak na drugi nain :
ili isti zadatak na drugi na
Koristimo pravila:1 1
1 , , ,
1 4 14
2 4 1 2 24 4 151)1 9 3 1 3 33 6
9
6 6 22 4 4 6 6 2
9 33 6 9 4
1
152)
1 1
n
na a a a a
a a
ba ba
a ab
b
= = =
= = = = =
= = =
= = =
1
1
22 2 2 22
2 2 2 2 2
2
11 11
1 1 1
1
2
222 3 23 3
1 2 2 2 2 2 22
1
in :
ili drugi nain :
ili drugi nain :
ili drugi
11
53)1 1
1
54)1 1 1
155)
1
a b
ab
ba b a ba
b a a b a
b
bb
b c b ca b bc a b bca a
a a ac c
cc
aaa ca b a cb b
c d d b d b d d
c c
=
= = = =
= = = = = =
= = = =
2 12 3 2
1 2 3 2 3 2
22
2 2 2 21 2 21
2 3 1 3 3 2 13
2 1 2 1
2 2 1 2 21 2 2
2 3 1 3 1 2 1
1 2
nain :
ili drugi nain :
1
56)1 1
57)
a ca b a c
c d b d b d
b cb c
b c y d b c y d x bc bd yx x
cxy c d c x c x c ccy d y d
b c y d b c y d x bc bd y
cxy c d c x c c x
x b
= =
= = = = =
= = =
2
22 2 3 2 32 51 2
2 3 1 2 2 2 2 2 2 22
3 1 3
3 1 2 3 21 2 2 5
2 3 1 1 2 2 2 2 2 2ili drugi nain :
1 1
1 1
cc
x b c c d c dc cx b
y c d y x b y b xy b xyy
c d c dc d c c d x b c c d
y c d x y b b xy b xy
+
+
= = = = =
= = =
d
a
=
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M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
9.
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