Download - Kompletni postupak i rješenja
M-1- rješenja i kompletni postupakPOTENCIJE
Rješenja svih zadataka s kompletnim postupkom i uputama 1. 2 3
puta
2 se ponavlja petputa pa u eksponent pišemo
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...
čitamo: dva na petu2
1) 2 2 2 2 2 22)3)
4)
5)
6)
x x x xx 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 xy y y
= →
⋅ ⋅ ⋅ ⋅ =
⋅ ⋅ =
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ =
⋅ ⋅ =
+ ⋅ + ⋅ + ⋅ + = +
⋅ ⋅
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
4
3 2
3
7)
8)
x xy 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 bc c c c
⋅ =
− ⋅ + ⋅ − ⋅ − ⋅ + =
= − ⋅ − ⋅ − ⋅ + ⋅ + = − ⋅ +
− − − − ⋅ ⋅ =
2.
( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( )( )
na eksponent
na
2 3
puta
2 3
2
parni3
neparni4
Zaključak:
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 1
5) 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" pa je produkt pozitivan broj
Imamo neparan broj "minusa" pa će i umnožak biti negativno tj. imati će predznak minus
1
7) 2 2 2 2 8
8) x x x x x
↓
= −
− = − ⋅ − ⋅ − = −
− = − ⋅ − ⋅ − ⋅ − 4x=
4
7
8
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za [email protected]
tel-01-4578-431
2 3
puta
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 83 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 2565 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.
M-1- rješenja i kompletni postupakPOTENCIJE
2.
( ) ( ) ( ) ( ) ( )
) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 3
puta
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 će umnožakbiti pozitivan broj
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
10
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za [email protected]
tel-01-4578-431
3. ( ) ( )
( ) ( )
( )
( )
je potpuno isti izraz...
ili taj isti zadatak na malo duži ali sigurn
Koristimo pravila:
51) 2 3 2 3 5 5 ili
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 ac a d a c d a c a d a c d a
xx x x x x
x
x x x x x
⋅ + ⋅ = + ⋅ ⋅ − ⋅ = − ⋅⋅ + ⋅ = + ⋅ ⋅ − ⋅ = − ⋅
⋅ + = + ⋅ = ⋅ =
+ + = + + ⋅ =
( )iji naćin:
Dakle:
Praksa je pokazala da velika većina đaka radi istu grešku: uzimate da je: što nije točno!!
dakle vi kada računate u glavi grešite na ovaj način:
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 što nije točno!!!
Jednom zauvjek treba zapamtiti pa to u zadatku treba izgledati 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 duži ali sigurniji naćin:
9 2 3 9 2 3 1 9 2 3 1 9
5) 2 3 5 7 22 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 ax 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
2Ovo je potpuno isti izraz
8) 4 2 9 4 2 9 3 3
9) 3 4 5 2 8 3 82 3 8 4 3 5 8
1 2 3 8 4 3 5 8
4 4 64 4 6
ab
xy xy xy xy xy xy
y x y z x y y z y x yy y y z z x y x y x y
y z x y
y z x yy z x y
= −
+ − = + − ⋅ = − ⋅ = −
+ − − − + − + =
= − − + − + − + =
= − − ⋅ + − ⋅ + − + ⋅ =
= − ⋅ + ⋅ + ⋅ =
= − + +
11
M-1- rješenja 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 ac 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 duži ali sigurniji naćin:
3 4 3 4 1
3 4 1
6
13) 5 2 2 7 3 35 2 3 2 75 2 3 2 7
6 96 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 zx y x y x y z z
x y z
x y zx y z
x x x x
+ + + − + = + + + − + =
= + − ⋅ + =
= +
+ − + + − =
= − + + + =
= − + ⋅ + + ⋅ =
= ⋅ + ⋅ =
= +
− + − − + =
= ( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )( ) ( )
( )
2
2
2
ili taj isti zadatak na malo duži ali sigurniji naćin:
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.
12
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za [email protected]
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 aa
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 15 315 223 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 često radite ovakve greške:
što nije točno 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.
−
M-1- rješenja 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 22 3 4 22 3 4 22 4 3 22 4
nn m n m n m n m n m n m
m
aa a a a a a a a aa
x x x y x y x x x y yx x x y y x x x yx x y x x yx 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... uobičajeni zapis je ovaj zadnji...
3 2
2 52 5
1 1 1 1 119)2 2 2 2 2
20) m m m m m m
x x y
x x yx 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 92 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
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 aa
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 abc c c c c
abc
− − − −
+ − + −
+ − + −
+
− − − − − + − + − + −
+
⋅ − = + ⋅ + ⋅ − ⋅ − =
= + ⋅ − =
= + ⋅ − =
= + ⋅ −
⋅ ⋅ ⋅ = =
=
( ) ( )
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 33 3 3
136) 25 5 5 5 55 5
37) 2 3 2 3 6
x x x y y y y
x yabc
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 73 2 7
1 139) 2 2 2 22 2
2
z x y
x x x x x x x x x x x x x x
x x x x xx 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 22
2 22 2
x x y y x x y y
x y x y x yx y x y x y
+ + + + + +
+ ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ =
= − ⋅ ⋅ + ⋅ − ⋅ ⋅ =
= − + −
z
14 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
15
M-1- rješenja 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 slijedećim zadacima koristimo pravilo:
Pazi: drugi eksponent se
. :
40) 2 : 2 2 2 2 2 2 2 2 1641) 13 : 13 13 13 16942) 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 često radite ovakvu grešku:
što nije točno!
Ovdje je pogreška u predznaku kod zadnjeg člana, 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 događa zbog toga što radite napamet... tj. preskaćete korake...
Preporuka: čim imamo višečlane eksponente koristite zagrade i ne preskačite korake...
44) :
45) :46) : :
x x x x
x x x xx 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 1 322 2 2 2 2 2
7 1 7 1 7 1 3 7 3 49 3 9 3 9 9 9
2 2 1 2 2 2 10 1 5 4 20 5 211 225 5 2 10 10 102
4 2 4 2 2
47) :148) : :
49) :
50) :
51) : :52) :x x x x
x xx 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 xa a a a
− −
+ −
− −
⋅ − −−
− ⋅ −−
⋅ − ⋅ − ⋅ − −− − −
−
= =
⋅ = =
= = = =
= = = =
= = = =
= = = =
= =8 3 2 8 3 2 3
2 3 1 2 3 1 2 3 44
7 2 3 7 2 3 8
55 2 5 2 3
2
88 3 8 3 5
3
33 32 112 2
ili
U slijedećim 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
a a a aa
x x x x xxx x x x xx
x x x x xx
− −
− − −
− +
−
−
−
−
= =
= = = =
⋅ = =
= =
= = =
= = =
= = =3 2 1
2 2
71 7 17 7 3 1 4 21 4 1743 4 34 12 12 12
13
59) :
x
x x x x x x xx
−
⋅ − ⋅ −−
=
= = = = =
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 slijedećim 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
a a a aa
x x x x x x xxx x x x x x xx
x x xx
−
++ − ++ + + − − − + − − −
+
++ − ++ + + − − − + −
+
+−
+
= =
= = = = =
= = = = =
= ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( )
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
a a a a a a aa
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 bc 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 slučaju možete ostaviti i ljevo rješe
ili
68) :x x x x x x x x x
x x x
a b a b a b a b a bc c c c c
a b a b cc c a b
−
− − − − + − − − + − − − +
+ − + − − −
⋅ = = =
= = =
( ) ( )
nje...jer još nismo obradili pravilo:"na minus prvu"
2 4 2 2 4 2 2 2 4 2
2 4 2 2 2 4 2
Množimo svaki član prve zagrade sa svakim članom 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
11
1
70) 1 1 1 1 1 1 1
11
1
x x x x xx x x x xx
x x x x x x x x x x
x x x x xx x x x xx x x
+ +
= + + − − − =
= + − + − − =
= −
+ ⋅ − + = ⋅ + ⋅ − + ⋅ + ⋅ + ⋅ − + ⋅ =
= − + + − + =
= − + − + + =
= − + + +
16 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
17
M-1- rješenja 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 smo k 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
parni3
neparni4
3
Imamo neparan broj "minusa" pa će i umnožak bi
1 1Zaključak:
1 1
tamo rješili:
. 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" pa je produkt pozitivan broj
53
53 puta bi to trebali na
ti negativno tj. imati će predznak minus
8)
vlja ovakav problem: ...
x x x x x x
x x x x
− = − ⋅ − ⋅ − ⋅ − =
− = − ⋅ − ⋅ ⋅ −
( )( )
pisatia to je previše pisanja pa ćemo rađe gledati eksponent
53 53
54je logički 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
2 parni 2parnu
2 1 neparni neparnu
pa je:
Zaključak:
jer je eksponent 55, a 55 je neparan broj
jer je eksponent 56, a 56 je paran broj
ilin 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 parni broj
2 1 je oznaka za neparan broj
Sada to primjenimo u zadatcima:
.8) zato što je eksponent 4, a 4 je paran broj
7) 2 2 8 zato što 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 broj da se zapisati u obliku:
1
4 4 4 4 4 4 II na
stoji i drugi način rješavanja ovakvih zadataka:II način
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
−
− = − ⋅
− = − ⋅ = − ⋅ = − ⋅ = −
− = − ⋅ = − ⋅ = − ⋅ = −
− = − ⋅ = − ⋅ = ⋅ = čin
5.
⋅
Poja
P
Sada se ja
53 je nepar
dal
( )
18
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za [email protected]
tel-01-4578-431
( ) ( ) ( ) ( )( )( )
) ( )) ( )
) ( ) ( )
2 2 2 2 2
2 2 2 2 2
2 2 22 22 3 2 3 2 2 3 2 4 62
Koristimo pravila:
Postupili smo prema prvom pravilu...
1 2 2 4 4
2 3 3 9 9
3 3 3 934 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 3 332 3 2 3 2 3 3 63
4 4 44 42 3 2 3 2 4 3 4 8 124
22 2 2 4
2 22 2 2 2 4
Postupili smo prema drugom i trećem pravilu...
I
2 2 2 843 3 273
1 1 1 152 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
način
II način:Prema pravilu:
7 1 1 1
8
9
10 2 2 4 4
11 2 2 8 8
12 2 2
mn na ax 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 način:
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
19
M-1- rješenja i kompletni postupakPOTENCIJE
5.
( )( )
( ) ( )
na eksponent
na eksponent
parni parnu
neparni neparnu
3 32 2 2 3 6
Koristimo pravilo: prilikom računanja načinom u
14), 15), 16),17), 18), 19)...
to je bio I način rješavanja
I
14)
II način
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 način rješavanja
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
parnu
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
način rješavanja
to je bio I način rješavanja
II način
to je bio I način rješavanja
16
16 1 1 1
17
1
parnu parnuy 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 način
U 18) pokazat ćemo tri načina rješavanja:
to je bio I način rješavanja
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 način
III način ...u min us i minus daju plus
U ovom 18) zadatku najbrži ne III način rješavanja... dok je u 19) definitivno II način najb
1 1 1
18
x x x x
x x x x x
⋅
⋅
⋅ − ⋅ = − ⋅ = ⋅ =
− − = + = = =
olji....
20
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
M.I.M-Sraga centar za [email protected]
tel-01-4578-431
5.
) ( ) ( ) ( ) ( ) ( )( )
) ( ) ( )( ) ( )
2 22 2 2 2 4 45 5 5 5 5 5 4 20
2 22 2 25 5
I način
I način u ovom zadatku pomalo zbunjuje pa je bolje to rješavati na II način izlučivanjem 1 :
II način:
1 1
x x x x x x x
x x x
⋅ ⋅
−
− − = + − = − = − = = =
− − = − − ⋅ = − − ⋅ ( )( )( )
( ) ( ) ( )
) ( ) ( ) ( )
) ( ) ( )
) ( ) ( ) ( )
) ( ) ( ) ( )
( ) ( )
225
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:
ili
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
M-1- rješenja i kompletni postupakPOTENCIJE
5. 25 25 25 26 26 27
) ( )
) ( )
) ( ) ( )
2 2 2 222 2 2 2 2 42
2 2 222 2 2 2 42
2 2 2 2222 2 2 2 2 42
25) zadatak na tri nači
2 2 2 2 43 3 3 93
2 2 2 43 3 93
2 2 2 2 41 1 13 3 3 93
x x x x x
x x x x
x x x x x
⋅
⋅
⋅
− = = ⋅ = ⋅ =
− = − ⋅ = + ⋅ =
− = − ⋅ ⋅ = − ⋅ ⋅ = ⋅ ⋅ =
) ( )
) ( )
) ( )
) ( )
) ( ) ( ) ( )
) ( ) ( ) ( )
3 3 3 332 2 2 2 3 63
3 3 332 2 2 3 63
2 2 2
3 3 3
4 4 4 4 4 4 4
na
I način
II način
2 2 2 2 83 3 3 273
2 2 2 83 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
2 2 2 2
3 3 3 3
2 2 2 2 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
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
5. 37 38 39
) ( )( ) ( )( )
) ( )( )
) ( )( )
) ( )( )
) ( )( ) ( )
) ( ) ( )( )
) ( )( ) ( )
432 2 3 4 24
543 3 4 5 60
42 2 4 8
3 3 3
26 52 3 2 6 2 3 5 24 15 24 15 39
24 33 5 3 4 5 3 2 12 30 12 30 42
35 72 4 2 5 3 4 7
Prema pravilu:
40
41
42
43 : :
zmn n m z
x x 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 43 2 3 6 8 2 4 5 144 40 144 40 104
3 32 3 3 22 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 : 23 3
2 : 438 : 4
278 :4 :
272
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 5227a b=
22 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
M-1- rješenja i kompletni postupakPOTENCIJE
23
( ) ( ) ( ) ( )( )
) ( ) ( ) ( ) ( ) ( )
( )
( )
3 33 3 3 3 32 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 42 2
1 6421 6481 648648
zm mn nn n n n n n n m n n m zab 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 32 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
33
3
3
2
88
1 1: 4 : 42 2
1 : 6421 : 6481:64 : :8
18 64
1512
1 1512
512
48 3
x yx 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
yx
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
333
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 ba ba b
x x x
⋅ ⋅ ⋅
+ +
+ ⋅ + +
⋅ = ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ =
=
= =
5. ⋅= 46 47
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 xxxx
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 32 3
5
3 4 3 4
3 41 3 4
22
m
m m
m m
m
xxx
x x x x
x xx
x
a a a a a a
a a aa
aa
− −
− − +
− − +
−
⋅ ⋅
⋅ ⋅ ⋅
=
= =
= =
=
+ = ⋅ + ⋅ =
= ⋅ + ⋅ =
= + ⋅ =
=
− + = − ⋅ + ⋅ =
= − ⋅ + ⋅ =
= − + ⋅ =
= ⋅ =
=
⋅=
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M-1- rješenja i kompletni postupakPOTENCIJE
6. ( )
( ) ( )( ) ( )
22 2 4 4 42 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
=
⋅ = ⋅ = ⋅ = ⋅ =
⋅ = ⋅ = ⋅ = ⋅ =
⋅ ⋅ = ⋅ = ⋅ = ⋅ = =
⋅ = ⋅
333
2 2 22 2
2 2
2
2 2
3 33
možemo ostaviti u ovom obliku
ili dalje:
nakon kraćenja
4 23 2 84 3
2 24 4 28)2 2 2
2 4
9)
x yxy xyx x y xx y x y
x x
xy x xy xz 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 kraćenja10)
11)
mm
mmm m m
m mm
xx x x x xz z z z zz
xy x xy xz yz z yz
xx x x x xz z z zzz
x y zz x y
⋅
⋅
++
+++ ⋅ + +
+ +⋅ +
⋅= = = = = ⋅
⋅ = ⋅ = =
⋅= = = = = ⋅
⋅
( )( )
22 2 2 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 kraćenja
nakon kraće12)mm m m
x y y z zx y zz x y z x x y
zz z zx x xx
x y y z zx y z x y zz 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 11
1 11
m mm m m m m
m m m
y zy z y z y zx xx
x y x x y x y x x yx x y x x yx xy y x xy y
x xx y x yx x yx y
x y x x xx x y x y
⋅ ⋅
⋅
=
⋅ ⋅= = = =
− − + − − + ⋅ ⋅ = ⋅ ⋅ = + − + −+ + + +
− ⋅ +− + = ⋅ ⋅ = + −+
− − ⋅ += ⋅ ⋅
+ + ⋅ +( )( )
( )( )
5
55
5
pa kratimo
11
yx y
xxx y x y
+= = −
− −= = + +
x
25
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
7.
2 2
2
2 2 2
2
po pravilu:
primjenili smo pravilo:
Koristimo pravila:
Izračunaj:
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 ab b
a a a b bb b ab a
a ab b
−
−
=
= = =
= = =
= = =
2 2 2
2
po pravilu
3 3
3
4 4
4
1 1
primjenili smo pravilo:
primjenili smo pravilo:
3 3 92 42
1 1 13) 2 82
2 2 164) 3 813
5 7 75)7 5
n na bb a
n n
n
n n
n
a ab b
a ab b
−
−
=
−
= = =
= = =
= = =
= =
11 1
2 2 2
2
3 3 3
mješoviti broj treba prvo pretvoriti u razlomak...
primjenili smo pravilo: 5
1 7 22 9 76) 17 7 7 9
4 5 5 257)5 4 164
2 3 38)3 2
n n n
n
a b bb a a
−
−− −
−
−
→
= =
⋅ + = = =
= = =
= = 3
22 2 2 2 2
2
22 2 2 2 2
2
mješoviti broj treba prvo pretvoriti u razlomak...
2782
2 3 11 6 1 7 3 3 99) 23 3 3 3 7 497
1 7 22 7 2 9 7 7 4910) 17 7 7 7 9 9
−− − −
−− − −
→
=
⋅ + + = = = = = =
⋅ + + = = = = = = 81
26 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
27
M-1- rješenja i kompletni postupakPOTENCIJE
33 3 3 3 3
3
2
mješoviti broj treba prvo pretvoriti u razlomak...
Koristimo pravila:
1 2 11 2 1 3 2 2 811) 12 2 2 2 3 273
3 412)4 3
n n nn n
n n
a a a b bb b ab a
−
−− − −
−
= = =
⋅ + + = = = = = =
=
( )( )
na eksponent
na eksponent
2 2
2
2 2
2
parni parnu
neparni neparnu
2 2 22
2
način oristimo pravilo: pa je:
4 1693
13)
14) I k
14) jer je:parnu parnu
x x
x x
x xy y
x x x x x x xy y y y yy
− =
− = −
= =
=
− = = − = − = =
( )
( )
2
2 2 2 2 22
2 2
3 3 3
3
3 3 33
II način
način rastavili smo na:
I način
...
14) 1 1 1
II 1 ...
15)
15) 1 1
y
x x x x xy y y y y
x xy y
x x xy y y
x x xy y y
=
− = − ⋅ = − ⋅ = ⋅ =
− − ⋅
− = − = −
− = − ⋅ = − ⋅ = −
( )( )
( )
na eksponent
na eksponent
3 3
3 3
parni parnu
neparni neparnu
4 4 4na parni eksponent parnu
4
4
II način
način oristimo pravilo: pa je:
II način:
1
16) I k
16) jer je:
16)
x x
x x
x xy y
x x x x xy y y
xy
− =
− = −
⋅ = −
− = = − =
− = −
( )4 4 4 4
4
4 4
22 2 2 2
2
1 1 1
1 2 11 2 1 3 317) 12 2 2 2 2
x x x xy y y y
⋅ = − ⋅ = ⋅ =
⋅ + + = = = =
7.
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
7.
22 2 2 2 2
2
22 2
I način
II način:
Koristimo pravila:
2 3 22 6 2 8 8 8 6418) 23 3 3 3 3 93
2 3 22 6 2 818) 23 3 3 3
n n nn n
n n
a a a b bb b ab a
− = = =
⋅ + + − = − = = − = − = = =
⋅ + + − = − = = − = −
( )2 2 2 2
2
2
33 3 3 3 3
3
33 3 3
I način
II način:
8 8 8 61 1 13 3 93
2 3 22 6 2 8 8 8 51219) 23 3 3 3 3 273
2 3 22 6 2 8 819) 2 13 3 3 3 3
= − ⋅ = − ⋅ = ⋅ =
⋅ + + − = − = = − = − = − = − = −
⋅ + + − = − = = − = − = − ⋅
( )
( )( )
( )( )
3 3 33
3
22 2 2 2
2
44 22 2 4 8
3 4 3 4 123
33 3 33 3 3 3 9
4 4 3 4 3 124
3
4
I način
8 8 51 13 23
3 5 44 15 4 19 19 36120) 35 5 5 5 255
21)
22)
22)
xx x xy y yy
xx x x xy y y yy
xy
⋅
⋅
⋅
⋅
= − ⋅ = − ⋅ = −
⋅ + + = = = = =
= = =
− = − = − = − = −
−
( )( )( )
( )( )
( )( )( )
33 3 33 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 62
4 4 2 4 2 84
II način
I način
II način
1 1 1
23)
23) 1 1 1
xx x xy y yy
xx x x xy y y yy
xx x x xy y y yy
⋅
⋅
⋅
⋅
⋅
⋅
= − ⋅ = − ⋅ = − ⋅ = −
− = = = =
− = − ⋅ = − ⋅ = ⋅ =
127
4
28 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
M-1- rješenja 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 bb b ab a
x y x yx y x yz z zz
x y z a b bb 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
Najkrače objašnjenje:
Sve što ima minus u eksponentu
"skaće" iz brojnik
2 82 826)3 27 273
227)2
m n ma
zx yzx y
x xx xy z y z y zy z
x y y zz 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 rješavamo tih "minusa"
8
8
512 512
2
22
y zx
y z
x
y z y zx x
x yz
x y y zz
⋅ ⋅
⋅
−
− −
−
− −
= =
⋅ ⋅= =
⋅ ⋅= =
= 2 22
3 4 3 4
U brojniku je bio: i on je "skočio" u nazivnik i sada je:
U nazivniku su bili i oni su "skočili" u brojnik i sada su:
: 2 2
x xx
i z i z
−
− −
29
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 4 425 255
2 229)5 5
52
5
n n nn n
n n
a a a b bb b ab a
x y xz y z
x x xy z y zy z
x y xz y z
y zx
−
−
⋅
⋅ ⋅
− −−
= = =
= =
⋅ ⋅= = =
⋅ ⋅⋅ ⋅
= =
= =
⋅=
( ) ( )( )
( )( ) ( )
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
254
254
2 82 2 830)5 5 125 1255
31)
32)
y z
x
y zx
y zx
x xx y x xz y z y z yy z
x y yy x x
x y y y y yy x x x x
⋅ ⋅
⋅
⋅−
⋅ ⋅
−
−
⋅=
⋅
⋅ ⋅= =
⋅
=
⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅
= =
⋅ = ⋅ = ⋅
( )( )
( )
( )( ) ( )
2 12 1
3 2 2 1 1
33 3
3
2 22
33 3 32 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 xyx x x x y
xx xx x x
x y x yx y
ca b c c cc a b a b a ba b
−
−
− ⋅
⋅ ⋅
⋅= ⋅ = =
⋅
+− + = = + − −
− = = − −
= = = = ⋅ ⋅
6 9z
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tel-01-4578-431
M-1- rješenja i kompletni postupakPOTENCIJE
7.
( )( ) ( )
( )( )
3 23 2 2 222 2 2
3 3 4 3 2 23 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 bb b ab a
a ya y y yy a a y a a
a y yy a aa y yy a aa y yy y a a
yy 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 22 3 8 4 9 4 2 42 2 3
1 2 1 91 1 1 1 119 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
ay
a a ay 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 12 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 3 129 4 27 27 27
−
− − −− −−
−
−− −
= + − = + − = + = + =
⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ =
⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ = 2
8 1 8 127 144 27 48612⋅ = ⋅ =
31
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
8. 0 1 1
0 0
0
02 4
po pravilu:
bez obzira što je u zagradi ako je eksponent
te zagrade nula sve je jednako jedan !
ili taj isti zadatak duž
Koristimo pravila:1 11 , , ,
1) 2 1 1
2) 1
3) 1
3)
nna a a a a
a a
a
x
x yz
− −= = = =
= =
=
=
( ) ( )
( )
( )
0 00 2 42 4
0
0 0
0
0 0
0 0 0
0
0
im postupkom:
po pravilu:
bez obzira što je u zagradi ako je to na nultu sve j
1 1 1 11 1
4) 1 1 2
3 1
6) 3 1 1
7) 3 1 zato što je: 3 1 3 1 1 1
8) 3 3 1 3
9) 1
x yx yz z
x y
a
x
x y
⋅ ⋅= = = =
+ = + =
=
− = =
− = − − = − ⋅ = − ⋅ = −
− = − ⋅ = −
+ =
( )
( ) ( )
( )
0
0 0 0
0 00
0 0
20 2
e jednako jedan !
bez obzira što 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- rješenja i kompletni postupakPOTENCIJE
8.
( )
( )
( )
0 1 1
0 0
20 2
0
0
1 1
1
1
bez obzira što je u zagradi ako je to na nuultu sve je jednako jedan !
po pravilu:
Koristimo pravila:1 11 , , ,
13) 1 1 1
14) 1 1
15) 1
16) 4 1
1 117) 5 5
118)
19) 0, 2
nn
m
a a a a aa a
x x
x
x
x
aa
xx
− −
− −
−
−
= = = =
⋅ = ⋅ =
= =
=
=
= =
=
=
( )( )
( )
1 1
11
22
33
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) 0,310 3 3 9
24) 2 5
nn
nn
x y ax y ax y
aa
xx
x
− −
− −
− −
−
−−
= = =
− = = =−−
= = =
=
= = = =
−( )
55
1
12 5
125) 44
x−
−
=−
=
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 način:
ili na drugi način:
Koristimo pravila:1 11 , , ,
11 1 1 4 1 1 4126) 4 41 14 1 1 4 1
4 4
11 4 1 41 1 1 4 1127) 4 41 1 1 1 1 14 4
4 4
2 3 328) pr3 2 2
nna a a a a
a a− −
− −
− −
−
= = = =
⋅ ⋅= = = = = =
⋅
⋅ − ⋅ −⋅= = = − = − = = = −
⋅− −−−
= =
( )( )
11
11
1 1
1
1
Isti taj zadatak po pravilu ispada: vidimo da je rezultat isti!
ili kraće:
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 5 105 1 5
2 5230) 215
n n n
na b bb 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 4 122 1
2 232) 21
33) 3 3 3 3 3 3 1x x x x x x x x x
a a b abb
−
−
− − − − + − + − + − − − +
⋅ − = −
⋅= = ⋅ =
⋅= =
⋅ ⋅ = = = =
34 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
35
M-1- rješenja 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 11 , , ,
34) 2 2 2 2 2 2 2 1
135) 5 5 5 5 55
36) 1
3
nn
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 10 0 0 1
1 1 10 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 63 1
1 1 1 1 5 1 1 1 1 538) 2 2 1 1 52 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 bc 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 12 2 8 2 82 240)
1 1 1 1 1 4 12 24 16 162 2
1 8 77 168 8
4 1 5 8 516 16
m n m n na bc
+ − − +
−− −
−− −
− −
− −
−
= =
− ⋅ − − −= = = = ⋅ ++ + +
− − ⋅= = = − + ⋅
( )
2
2 2 2 22
2
2 2 2
2 22 1 2
1 2
2
7 25
14 5 5 5 25 251 1 15 14 14 196 19614
1 1 1 1 12 3 183 4 32 1241)
1 1 1 1 1 122 32 2 9 183
6 36 2
−
−
− − −
− −− −
− −
⋅ = − =
= − = − = − ⋅ = − ⋅ = ⋅ =
⋅ ⋅ ⋅ = = = = = ⋅ ⋅ ⋅
⋅= ⋅
2 2 2 2
2
3 2 2 42 3 93
− − = = = =
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 11 , , ,
3 41 1 1 1 14 3 18 12 3 3 4 32 1242)
1 1 1 1 9 2 11 11 122 32 2 9 2 9 183
3 611 2
nna a a a a
a a− −
−− − −
− −− −
− −
= = = =
− − − − ⋅ ⋅− = = = = = − = + ⋅+ + + ⋅
⋅= −
⋅( )
2 2 2 2 22
2
0 0 1
0 0 1
0 1 22
2 3
3 22 22 22 484 4841 1 16 22 3 3 93
1 3 11 1 3 1 243) 3 2 3 1 1 13 3 3 3 3
2 3 11 1 6 1 544) 3 2 3 1 1 23 3 3 3 3
3 4 11 1 12 1 1145) 2 2 2 1 2 34 4 4 42
46) 2 2
− −
−
− −
= − = − = − ⋅ = − ⋅ = ⋅ = ⋅
⋅ − −⋅ − = ⋅ − = − = = =
⋅ − −+ − = + − = − = = =
⋅ − −+ − = + − = − = = =
− = 2 3
3 2 13 2 1
2 32 3
2 2
3
3
2
3ili isti zadatak na drugi način:
1 2 11 1 1 1 2 1 14 8 8 8 82 2
1 1 2 1 41 1 1 1 1 1 1 2 4 347) 2 2 28 4 2 8 8 82 2 2
1 1 1 1 8 9 1 148) 3 29 8 9 8 72 723 2
1 11 272 272 449)
1 1 4 1 43273
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 način:
2742
1 882 3 8 1 1 13 950)1616 9 16 9 2 1841
2 3 2 8 1 116 9 2 9 184 4 3
−
−
=
⋅⋅ ⋅= = = = =
⋅ ⋅
⋅= = = =
⋅ ⋅⋅
9
36 M.I.M-Sraga centar za [email protected]
tel-01-4578-431
37
M-1- rješenja i kompletni postupakPOTENCIJE
8. 0 1 1
2 1
2 1
2 1 1
2 1 1
1
1
ili isti zadatak na drugi način:
ili isti zadatak na drugi nač
Koristimo pravila:1 11 , , ,
1 4 142 4 1 2 24 4 151)1 9 3 1 3 33 6 96 6 2
2 4 4 6 6 29 33 6 9 4
1152)
1 1
nna a a a a
a a
ba baa ab
b
− −
−
−
−
−
−
−
= = =
⋅⋅ ⋅= = = = =
⋅⋅ ⋅
⋅ ⋅= = =
⋅ ⋅
⋅= = =
⋅
1
1
22 2 2 22
2 2 2 2 2
2
11 11
1 1 1
1
22
22 3 23 3
1 2 2 2 2 2 22
1
in:
ili drugi način:
ili drugi način:
ili drugi
1153)
1 1
1
54)1 1 1
1
55)1
a bab
ba b a bab a a b a
b
bb b c b ca b bc a b bca aa a ac c
cc
aa a ca b a cb bc d d b d b dd
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
način:
ili drugi način:
1
56)1 1
57)
a ca b a cc d b d b d
b cb c b c y d b c y dx bc bdyx xcxy c d c x c x c cc
y d y db c y d b c y dx bc bdy
cxy c d c x c c x
x b
−
−
−
− −
−
− −
− −
⋅= =
⋅
⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
= = = = =⋅ ⋅ ⋅⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = =
⋅ ⋅ ⋅
22
2 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 način:
1 1
1 1
cc x b c c d c dc cx b
y c d y x b y b xy b xyyc d c d
c d c c dx b c c dy c d x y b b xy b xy
+
− −
+− −
− −
⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = = = =
⋅ ⋅⋅ ⋅⋅
⋅ ⋅ ⋅= = =
⋅ ⋅
d
a
=
M-1- univerzalna zbirka zadataka M-1- univerzalna zbirka zadataka
9.
38 M.I.M-Sraga centar za [email protected]
tel-01-4578-431