POTENCIJE

**** MLADEN SRAGA ****
2010.
UNIVERZALNA ZBIRKA
POTPUNO RIJEŠENIH ZADATAKA
PRIRUČNIK ZA SAMOSTALNO UČENJE
POTENCIJE
α

M-1- univerzalna zbirka potpuno riješenih zadataka
Rješenja svih zadataka s kompletnim postupkom i uputama
2 3
n
21. Koristimo pravila: aa ⋅ = a , aaa ⋅ ⋅ = a , aaa ⋅ ⋅ ⋅...
⋅ a=
a
2−se ponavlja pet
puta pa u eksponent
pišemo 5
n−puta
Uputa: Prebrojite koliko se puta ponavlja isti faktor i taj broj stavite u eksponent:
1) 22222 ⋅ ⋅ ⋅ ⋅
5
= 2 →
čitamo: dva na petu
1) 22222 ⋅ ⋅ ⋅ ⋅ = 2
2)
3)
4)
5)
6)
7)
8)
xxx ⋅ ⋅ = x
3
( xy) ⋅( xy) ⋅ ( xy) = ( xy)
5
xyxyxzxyxxz ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = xxxxxxyyyzz ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = x⋅y⋅ z = xyz
( x+ y) ⋅ ( x+ y) ⋅ ( x+ y) ⋅ ( x+ y) = ( x+
y)
3
grupiramo iste faktore...
⎛ x⎞ ⎛ x⎞
⎛ x⎞ ⎛ x⎞ ⎛ x⎞
⎜ ⎟⋅⎜ ⎟⋅⎜ ⎟⋅ ⎜ ⎟ = ⎜ ⎟
⎝ y⎠ ⎝ y⎠
⎝ y⎠ ⎝ y⎠ ⎝ y⎠
x− y ⋅ x+ y ⋅ x− y ⋅ x− y ⋅ x+ y =
( ) ( ) ( ) ( ) ( )
4
4
6 3 2 6 3 2
( 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 ⎠
3
3 2
= − ⋅ − ⋅ − ⋅ + ⋅ + = − ⋅ +
2 3
n
22. Koristimo pravila: a = a⋅ a , a = a⋅a⋅ a , a = a⋅a⋅a⋅...
⋅a
n−puta
2
4) ( − 1) = ( −1) ⋅( − 1)
= 1
⎫
parni
3
⎪
( )
5) ( − 1) = ( −1) ⋅( −1) ⋅( − 1)
= −1
⎬ Zaključak:
na neparni
4
⎪
( −1)
6) ( − 1) = ( −1) ⋅( −1) ⋅( −1) ⋅( − 1)
= 1 ⎪ ⎭
2 3 4
1) 5 = 5⋅ 5 = 25 2) 2 = 2⋅2⋅ 2 = 8 3) 3 = 3⋅3⋅3⋅ 3 = 81
3
( ) ( ) ( ) ( )
7) − 2 = −2 ⋅ −2 ⋅ − 2 = −8
4
( − ) = ( − ) ⋅( − ) ⋅( − ) ⋅( − )
Imamo paran broj "minusa" pa
je produkt pozitivan broj
na eksponent
− 1 = 1
eksponent
=−1
Imamo neparan broj "minusa" pa će i umnožak biti negativno tj. imati će predznak minus
8) x x x x x =
↓
4
x
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
22. Koristimo pravila:
2 3
n
a = a⋅ a , a = a⋅a⋅ a , a = a⋅a⋅a⋅...
⋅a
n−puta
9)
2
⎛1⎞
1 1 1
⎜ ⎟ = ⋅ =
⎝3⎠
3 3 9
10)
3
⎛3⎞ 3 3 3 27
⎜ ⎟ = ⋅ ⋅ =
⎝5⎠
5 5 5 125
11)
5
⎛2⎞ 2 2 2 2 2 32
⎜ ⎟ = ⋅ ⋅ ⋅ ⋅ =
⎝3⎠
3 3 3 3 3 243
12)
2
⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞
22 ⋅ 4
⎜− ⎟ = ⎜− ⎟⋅⎜− ⎟ = =
⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠
3⋅3 9
13)
3
⎛ 2 ⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞
222 ⋅ ⋅ 8
⎜− ⎟ = ⎜−
⎟⋅⎜− ⎟⋅⎜− ⎟ = − = −
⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3⎠
333 ⋅ ⋅ 27
14)
4
⎛3⎞
3 3 3 3 81
⎜ ⎟ = ⋅ ⋅ ⋅ =
⎝4⎠
4 4 4 4 256
15)
2
⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞
44 ⋅ 16
⎜− ⎟ = ⎜− ⎟⋅⎜− ⎟ = + =
⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠
5⋅5 25
16)
3
⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞
444 ⋅ ⋅ 64
⎜− ⎟ = ⎜− ⎟⋅⎜− ⎟⋅⎜− ⎟ = − = −
⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠
5⋅5⋅5 125
17)
4
⎛ 4⎞
⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞ ⎛ 4⎞
4444 ⋅ ⋅ ⋅ 256
⎜− ⎟ = ⎜−
⎟⋅⎜− ⎟⋅⎜− ⎟⋅⎜− ⎟ = + =
⎝ 5⎠
⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠ ⎝ 5⎠
5555 ⋅ ⋅ ⋅ 625
2
18) 0,2 0,2 0,2 0,04
= ⋅ =
2
( ) ( ) ( )
19) − 0,2 = −0,2 ⋅ − 0,2 = + 0,2⋅ 0,2 = 0,04
3
20) 0,2 0,2 0,2 0,2 0,04 0,2 0,008
= ⋅ ⋅ = ⋅ =
3
( ) ( ) ( ) ( ) ( )
21) − 0,2 = −0,2 ⋅ −0,2 ⋅ − 0,2 = − 0,2⋅0,2⋅ 0,2 = −0,008
( − )
autor: Mladen Sraga
2
( ) ( ) ( )
22) 2,5 = −2,
5 ⋅− 2,5 =+ 2,5⋅ 2,5 = 6, 25
3

M-1- univerzalna zbirka potpuno riješenih zadataka
22.
n
a = a⋅ a a = a⋅a⋅ a a = a⋅a⋅a⋅ ⋅a
2 3
Koristimo pravila: , , ...
3
( ) ( ) ( ) ( ) ( )
n−puta
23) − 2,5 = −2,5 ⋅ −2,5 ⋅ − 2,5 = − 2,5⋅2,5⋅ 2,5 = 15,625
4
) ( ) ( )( )( )( ) ( )
24 − 2,5 = −2,5 ⋅ −2,5 ⋅ −2,5 ⋅ − 2,5 = + 2,5⋅2,5⋅2,5⋅ 2,5 = 39,0625
2 3 4 5
( − ) + ( − ) + ( − ) + ( − ) = + ⋅( − ) ⋅ ( + ) ⋅( − )
25) 1 1 1 1 1 1 1 1 =+ 1=
1
Prebrojimo minuse -ima
ih paran broj pa će umnožak
biti pozitivan broj
20 30 45
( ) ( ) ( ) ( ) ( )
26) − 1 + − 1 + − 1 = + 1⋅ + 1 ⋅ − 1 = −1
( ) ( ) ( ) ( )
1 2 3 4
27) − 2 + − 2 + − 2 + − 2 = − 2 + 4 − 8+ 16 = 4+ 16 −2 − 8 = 10
− = ⋅ ⋅ ⋅ ⋅ − ⋅ = − =
5 2
28) 2 3 22222 33 32 9 23
− = ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ = −
4 5
29) 3 2 3333 22222 81 32
= 49
3 5
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
30) − 2 + − 2 = −2 ⋅ −2 ⋅ − 2 + −2 ⋅ −2 ⋅ −2 ⋅ −2 ⋅ −2 −8− 32 = −40
31)
2 3
⎛1⎞ ⎛ 1⎞ 1 1 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞
1 1 1−
2 1
⎜ ⎟ + ⎜− ⎟ = ⋅ + ⎜− ⎟⋅⎜− ⎟⋅⎜− ⎟ = − = = −
⎝4 ⎠ ⎝ 2 ⎠ 4 4 ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2⎠
16 8 16 16
2 2
2 3
( ) ( ) ( ) ( ) ( ) ( ) ( )
32) ⎡ − 0, 2 + − 0, 2 ⎤ = ⎡⎣
−0, 2 ⋅ − 0, 2 + −0, 2 ⋅ −0, 2 ⋅ − 0, 2 ⎤⎦
=
⎣
⎦
( )
2 2
= 0,04 − 0,008 = 0,032 = 0,001024
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
( )
ili taj isti zadatak na malo duži ali sigurniji naćin:
( ) ( )
( ) ( )
ca ⋅ + da ⋅ = c+ d⋅a ca ⋅ −da ⋅ = c−d⋅a
23. Koristimo pravila:
n n n n n n
ca ⋅ + da ⋅ = c+ d⋅a ca ⋅ −da ⋅ = c−d⋅a
5⋅ x⎫ 1) 2 3 ( 2 3)
⎪
x+ x = + ⋅ x = 5⋅ x = 5x
ili ⎬
5x
⎪⎭
2) x+ 2x+ 4x = 1+ 2 + 4 ⋅ x = 7x
( )
x+ 2x+ 4x = 1x+ 2x+ 4x = 1+ 2+ 4 ⋅ x = 7x Dakle: x = 1x
je potpuno isti izraz...
Praksa je pokazala da velika većina đaka radi istu grešku: uzimate da je:
dakle vi kada računate u glavi grešite na ovaj način:
x + 2x+ 4x = 6x ili x+ 2x+ 4x = ( 0+ 2+ 4)
⋅ x = 6x
Jednom zauvjek treba zapamtiti
x = 1x
( )
x+ 2x+ 4x = 1x+ 2x+ 4x = 1+ 2+ 4 ⋅ x = 7x
što nije točno!!!
x = 0 x
pa to u zadatku treba izgledati ovako:
što nije točno!!
( )
3) 7a− 2a = 7 −2 ⋅ a = 5⋅ a = 5a
( )
4) 9y− 2y+ 3y− y = 9 − 2 + 3 −1 ⋅ y = 9⋅ y = 9y
ili taj isti zadatak na malo duži ali sigurniji naćin:
( )
9y− 2y+ 3y− y = 9y− 2y+ 3y− 1y = 9− 2+ 3−1 ⋅ y = 9y
5) 2x+ 3a− x+ 5a+ 7x− 2a
=
( )
( )
( ) ( )
= 2x − x+ 7x+ 3a+ 5a− 2a = 2− 1+ 7 ⋅ x+ 3+ 5−2 ⋅ a = 8⋅ x+ 6⋅ a = 8x+
6a
6) 2xy + 3xy + xy = 2 + 3 + 1 ⋅ xy = 6xy
7) 2ab − 4ab + ab = 2 − 4 + 1 ⋅ ab = −1⋅ab
=−ab
( )
8) 4xy + 2xy − 9xy = 4 + 2 −9 ⋅ xy = −3⋅ xy = −3xy
2 2 2 2 2 2
2 2 2
9) y+ 3x y−4z−5x y− 2y+ 8z− 3y+ 8x y =
2 2 2
= y−2y− 3y+ 8z− 4z+ 3x y− 5x y+ 8x y =
( 1 2 3) ( 8 4) ( 3 5 8)
2
= − − ⋅ y+ − ⋅ z+ − + ⋅ x y =
2
=−4⋅ y+ 4⋅ z+ 6⋅ x y = ⎪
2
⎬
=− 4y+ 4z+ 6x y
⎪⎭
⎫
Ovo je potpuno isti izraz
autor: Mladen Sraga
5

M-1- univerzalna zbirka potpuno riješenih zadataka
( )
( ) ( )
( ) ( )
ca ⋅ + da ⋅ = c+ d⋅a ca ⋅ −da ⋅ = c−d⋅a
23. Koristimo pravila:
n n n n n n
ca ⋅ + da ⋅ = c+ d⋅a ca ⋅ −da ⋅ = c−d⋅a
10) 7xy − 2xy + 4xy = 7 − 2 + 4 ⋅ xy = 9⋅ xy = 9xy
3 3 3 3 3 3
( ) ( ) ( ) ( ) ( )
11) 2 − − 3 + = 2 −3 ⋅ + = −1⋅ + = − −
2 2 2 2 2
x y x y x y x y x y
( x+ y) + ( x+ y) − ( x+ y) = ( + − ) ⋅ ( x+ y) = ( x+
y)
12) 3 4 3 4 1 6
ili taj isti zadatak na malo duži ali sigurniji naćin:
( x y) ( x y) ( x y) ( x y) ( x y) ( x y)
= ( 3+ 4−1) ⋅ ( x+ y)
=
= 6( x+
y)
3 + + 4 + − + = 3 + + 4 + − 1 + =
2 3 2 3 2 3
13) 5 2 2 7 3 3
x y + z− x y + z+ x y − z =
2 3 2 3 2 3
5 2 3 2 7
= x y − x y + x y + z+ z =
2 3
( 5 2 3) x y ( 2 7)
= − + ⋅ + + ⋅ z =
2 3
6 9
= ⋅ x y + ⋅ z =
2 3
6 9
= x y +
2 2
( x x ) ( x x )
2
= ( 3 4) ( x 3x
5)
2
= −1⋅( x − 3x+ 5)
=
14) 3 − 3 + 5 −4 − 5 + 1 =
x
z
− ⋅ − + =
3x
5
2
=− + −
( x+ y− ) − ( x+ y− ) + ( x+ y− ) = ( − + ) ⋅ ( x+ y− ) = − ( x+ y−
)
15) 3 1 6 1 1 3 6 1 1 2 1
ili taj isti zadatak na malo duži ali sigurniji naćin:
( x y ) ( x y ) ( x y ) ( x y ) ( x y ) ( x y )
= ( 3− 6+ 1) ⋅ ( x+ y− 1)
=
=− 2( x+ y−1)
3 + −1 − 6 + − 1 + + − 1 = 3 + −1 − 6 + − 1 + 1 + − 1 =
www.mim-sraga.com 6
© M.I.M-Sraga centar za poduku
mim-sraga@zg.htnet.hr
tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
24.
n
a
Koristimo pravila: a ⋅ a = a a : a = a = a : a = a
m
a
n m n+ m n m n−m n m n−m
1) 2 ⋅ 2 = 2 = 2
5 3 5+
3 8
2) 5 ⋅ 5 = 5 = 5
2 7 2+
7 9
3) 3 ⋅ 3 = 3 = 3
x 2x x+
2x 3x
4) 7 ⋅ 7 = 7 = 7 = 7
2m−5 m−2 2m− 5+ m− 2 2m+ m−5−2 3m−7
5) 2 ⋅2 ⋅2 ⋅ 2 = 2 = 2
2 6 4 8 2+ 6+ 4+
8 20
6) 3⋅a ⋅2⋅ a = 3⋅2⋅a ⋅ a = 6⋅ a = 6a
4 7 4 7 4+
7 11
7)
x ⋅ x = x = x
2 3 2+
3 5
8)
xx ⋅ ⋅ x = x⋅x⋅ x = x = x Pazi x=
x
3 5 1 3 5 1+ 3+
5 9 1
9)
x ⋅x ⋅ x = x = x
2 4 6 2+ 4+
6 12
10)
2 5 2 5 2⋅ 2+ 2⋅ 6+ 5⋅ 3 4+ 12+
15 31
+ 2+
3 2 2 3 2 6 6 6
x ⋅x ⋅ x = x = x = x = x
11)
x ⋅x ⋅x ⋅ x = x = x = x
2 2+ a a− 2 3 2+ 2+ a+ a− 2+ 3 a+ a+ 2+ 2− 2+ 3 2a+
5
12)
13)
a ⋅ a = a = a
a
3 6 3+
6 9
⋅a
4 2
⋅ a = a ⋅a ⋅ a = a = a Pazi a = a
4 2 1 4+ 2+
1 7 1
Dosta često radite ovakve greške:
4 2 4+
2 6
a ⋅a ⋅ a = a = a ⎫⎪
što nije točno jer je:
4 2 4+ 2+
0 6⎬
a ⋅a ⋅ a = a = a ⎪⎭
a
= a
1
14)
a ⋅a ⋅a ⋅ a = a = a
2 5 7 3 2+ 5+ 7+
3 17
15)
a ⋅a ⋅ a = a = a
3x x+ 2 2x 3x+ x+ 2+ 2x 6x+
2
16)
⋅
⋅
= a = a = a
x+
y 2x
x+ 3y x+ y+ 2x+ x+ 3y x+ 2x+ x+ y+ 3y 4x+
4y
a a a
17) −2⋅x⋅6⋅x ⋅3⋅ x =−2⋅6⋅3⋅x ⋅x ⋅ x =−36⋅ x =−36
x
3 4 1 3 4 1+ 3+
4 8
autor: Mladen Sraga
7

M-1- univerzalna zbirka potpuno riješenih zadataka
24.
n
a
Koristimo pravila: a ⋅ a = a a : a = a = a : a = a
m
a
n m n+ m n m n−m n m n−m
18) 2 x x 3 x y x y 4 x x 2 x y y
2 5 2 3 6 3 2 2
⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ + ⋅ ⋅ ⋅ =
x x x y y x x x y
2+ 5 2 1 3 1 1 6 3 2+
2
= 2⋅ + 3⋅ ⋅ ⋅ ⋅ −4⋅ ⋅ + 2⋅ ⋅ =
x x y x x y
7 2+ 1 3+ 1 1+
6 3 4
= 2⋅ + 3⋅ ⋅ −4⋅ + 2⋅ ⋅ =
x x x y x y
7 7 3 4 3 4
= 2⋅ −4⋅ + 3⋅ ⋅ + 2⋅ ⋅ =
( ) ( )
= 2 −4 ⋅ x 7 + 3+ 2 ⋅x 3 ⋅ y
4 =
7 3 4
=−2⋅ x + 5⋅x ⋅ y = ⎪
7 3 4 ⎬
=− 2x
+ 5x y ⎪⎭
⎫
To je potpuno isti izraz... uobičajeni zapis je ovaj zadnji...
19)
2 3 4 2+ 3+
4 9
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞
⎜ ⎟ ⋅⎜ ⎟ ⋅ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟
⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠
20)
x ⋅ x = x = x
2m+ 1 3m+ 2 2m++ 1 3m+ 2 2m+ 3m++
1 2
=
x
5m+
3
21)
x ⋅ x = x = x = x
2m− 4 3m+ 2 2m− 4+ 3m+ 2 2m+ 3m+ 2−4 5m−2
22)
3 3 3 ⎛3
⎞
2 9
3 2 2 7
1 1 2 7 2 1 8 m
+ ⋅
m+ m+ + m− m+ m+ + + ⋅ +
⋅ m+ 9 m+
9
2 2m−7 ⎜ ⎟
2 2 ⎝2
⎠
2 2
x ⋅ x = x = x = x = x = x
23)
x ⋅x ⋅ x = x = x = x
m+ n 2m− n 3m+ 2n m+ n+ 2m− n+ 3m+ 2n m+ 2m+ 3m+ n− n+ 2n 6m+
2n
24)
x ⋅x ⋅x
= x = x = x
2m−n 3m−n 2m−2n 2m− n+ 3m− n+ 2m− 2n 2m+ 3m+ 2m−n−n−2n 7m−4n
25)
a ⋅ a = a = a = a
2m+ 3n 3m− 5n 2m+ 3n+ 3m− 5n 2m+ 3m+ 3n−5n 5m−2n
26) 2a ⋅ 3a = 2⋅3⋅ a = 6⋅ a = 6⋅ a = 6a
m− 3n+ 1 4m+ n−7 m− 3n+ 1+ 4m+ n− 7 m+ 4m+ n− 3n+ 1−7 5m−2n−6 5m−2n−6
a ⋅ a = ⋅ ⋅ a =
x+ y 2x+ y x+ y+ 2x+
y
27) 2 5 2 5 1
0⋅ a = 10a
x+ 2x+ y+ y 3x+
2y
28)
2 9 2 9 2 33 ⋅
3 3
a ⋅ a = ⋅ ⋅a ⋅ a = ⋅ ⋅ a = ⋅ a = a
3 4 3 4 3 2⋅2 2 2
3m+ 2n 2m− 4n 3m+ 2n 2m− 4n 3m+ 2n+ 2m− 4n 3m+ 2m+ 2n−4n 5m−2n
29)
30)
+
( x− y) ⋅( x− y) = ( x− y) ⋅( x− y) = ( x− y) = ( x−
y)
2 2 1 2 1 3
+
( x+ y) ⋅ ( x+ y) = ( x+ y) = ( x+
y)
3 4 3 4 7
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mim-sraga@zg.htnet.hr
tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
24.
n
a
Koristimo pravila: a ⋅ a = a a : a = a = a : a = a
m
a
n m n+ m n m n−m n m n−m
( x y ) ( x y ) ( x y ) ( x y )
m+ 1 2m+ 2 3m− 4 m+ 1+ 2m+ 2+ 3m−4
31) + −1 ⋅ + −1 ⋅ + − 1 = + − 1
=
( x y 1) ( x y 1)
m+ 2m+ 3m+ 1+ 2−4 6m−1
= + − = + −
32)
33)
2m n m 2n m 2n m 2n m 2n +
m 2n
( x −y ) ⋅ ( x + y ) = ( x + y ) ⋅ ( x + y ) = ( x + y ) = ( x + y )
1 1 1 1 2
2 3
( x+ y) ⋅( x− y)
⋅( x y) ( x y) ( x y) ( x y) ( x y) ( x y)
2m−1 m−3 2 2m−1 3 m−3
+ ⋅ − = + ⋅ + ⋅ − ⋅ − =
m
( x y) ( x y)
2+ 2 − 1 3+ m−3
= + ⋅ − =
m
( x y) ( x y)
2 + 2− 1 m+ 3−3
= + ⋅ − =
2m+
1
( x y) ( x y)
= + ⋅ −
m
34)
2
3x−2y 2
4x−y 2
x−y 2
3y−2x 2
3x− 2y+ 4x− y+ x− y+ 3y−2x
⎛ab ⎞ ⎛ab ⎞ ⎛ab ⎞ ⎛ab ⎞ ⎛ab
⎞
⎜ ⎟ ⋅⎜ ⎟ ⋅⎜ ⎟ ⋅ ⎜ ⎟ = ⎜ ⎟
=
⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠
2
⎛ab
⎞
= ⎜ ⎟
⎝ c ⎠
2
⎛ab
⎞
= ⎜ ⎟
⎝ c ⎠
3x+ 4x+ x−2x−2y−y− y+
3y
6x−
y
=
Novo MALA ŠKOLA MATEMATIKE –1 na
BESPLATNA video poduka i instrukcije
UČIMO ZAJEDNO
POTENCIJE
ALGEBARSKI IZRAZI
ALGEBARSKI RAZLOMCI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
autor: Mladen Sraga
9

M-1- univerzalna zbirka potpuno riješenih zadataka
25.
n
+ − a
Koristimo pravila: a ⋅ a = a a : a = a = a : a = a
m
a
n m n m n m n m n m n−m
2 2 3 2 2 3
1) 2xy⋅ 3xy = 2⋅3⋅x⋅x⋅yy
⋅ =
x
2+ 2 1+
3
= 6⋅ ⋅ =
= 6xy
4 4
y
3 2 5 3 3 5 2 3
2) 5xy⋅ 2xy = 5⋅2⋅x⋅x⋅y⋅ y =
x
3+ 5 2+
3
= 10⋅ ⋅ =
= 10xy
8 5
y
3)
2 2 3 9 4 2 9 2 3 4
ab⋅ ab = ⋅ ⋅a ⋅ab ⋅ ⋅ b =
3 4 3 4
2 3 ⋅3
2 1 3+
4
= ⋅ ⋅a
⋅a ⋅ b =
3 2 ⋅2
3
= ⋅a
⋅b
2
3 3 7
= ab
2
2+
1 7
4)
5 6 2 ⎛ 9 2 ⎞ 5 ⎛ 9⎞
6 2 2
− ab ⋅⎜− ab⎟ = − ⋅⎜− ⎟⋅a ⋅a ⋅b ⋅ b=
27 ⎝ 5 ⎠ 27 ⎝ 5⎠
5 9 6+ 2 2+
1
=+ ⋅ ⋅a
⋅ b =
3⋅ 9 5
1
= ab
3
8 3
1 2 3 4 3 2 5 1
2 3 3 2 4 5
5) abc ⋅− ( 4abc) = ⋅− ( 4)
⋅a ⋅a ⋅b⋅b ⋅c ⋅ c =
2 2
4 2+ 3 3+ 2 4+
5
=− ⋅a ⋅b ⋅ c =
2
5 5 9
=−2abc
4 2 1 2 1 4 2 2 1
6) 9xy⋅ xy= 9⋅ ⋅x⋅x⋅y⋅ y =
3 3
1 4+ 2 2+
1
= 33 ⋅ ⋅ ⋅x
⋅ y =
3
= 3⋅x ⋅ y = 3x y
6 3 6 3
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POTENCIJE
M-1- rješenja i kompletni postupak
25.
n
+ − a
Koristimo pravila: a ⋅ a = a a : a = a = a : a = a
m
a
n m n m n m n m n m n−m
Rješenja ZADATAKA 7), 8) , 9) , 10) - šaljemo mailom
Dakle ako trebate preostala rješenja ovog zadatka pošaljite nam poruku na:
Mail: mim-sraga@zg.htnet.hr
sa tekstom : Trebam preostala rješenja zadataka iz zbirke POTENCIJE
DODATNE UPUTE UZ OVE ZADATKE iz POTENCIJA
SA OBJAŠNJENJIMA I POSTUPCIMA RJEŠEVANJA ZADATAKA
NALAZE SE NA NAŠOJ WEB-STRANICI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
MALA ŠKOLA MATEMATIKE
DODATNE UPUTE UZ OVE ZADATK
POTENCIJE
ALGEBARSKI IZRAZI
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
11

M-1- univerzalna zbirka potpuno riješenih zadataka
n m n m
26. U slijedećim zadacima koristimo pravilo: a : a = a −
9 5 9−5 4 4
1) 2:2 = 2 = 2 = 2 = 2222 ⋅ ⋅ ⋅ = 16
7 5 7−5 2
2) 13 :13 = 13 = 13 = 169
3) 3 : 3 = 3 = 3
5x 3x 5x−3x 2x
( m−
)
3m−2 m−2 3m−2−
2 3m−2− m+ 2 3m− m+ 2−2 2m
4) 5 : 5 = 5 = 5 = 5 = 5
↓
Pazi: drugi eksponent se mora staviti u zagradu...vrlo često radite ovakvu grešku:
3m−2 m−2 5 :5
3m−2−m−2 3m−m−2−2 = 5 = 5
2m−4
= 5 →
↓
što nije točno!
Ovdje je pogreška u predznaku kod zadnjeg člana, jer prvom promjenite predznak,
a drugom ne, to se događa zbog toga što radite napamet... tj. preskaćete korake...
Preporuka: čim imamo višečlane eksponente koristite zagrade i ne preskačite korake...
5) x : x = x = x
4 2 4−2 2
6) x : x = x = x
6 2 6−2 4
7) x : x : x = x
8
14) :
7 2 3 7−2−3 2
8) x ⋅ x : x = x = x
7 3 4 7+ 3−4 6
2 1 2 1−2 −1
9) x: x = x : x = x = x =
1 1 2⋅2−1 4−1 3
2 −
2 2 2 2 2 2
7 1 7 1 7−⋅
1 3 7−3 4
−
9 3 9 3 9 9 9
2 1 2 1 2⋅2−2⋅10−1⋅5 4−20−5 21
−2
− −
5 2 2 5 2 10 10 10
4x 2x 4x−2x 2x
www.mim-sraga.com 12
1
x
10) x : x = x = x = x = x
11) x : x = x = x = x = x
12) x : x : x = x = x = x = x
13) a : a = a = a
a
a
=
x
: a = a = a
3 2 8−3−2 3
15) a: a : a a : a : a a a
= = = ili =
2 3 1 2 3 1−2−3 −4
16) a : a ⋅ a = a = a
7 2 3 7− 2+
3 8
U slijedećim zadacima koristimo pravilo:
5
x
17) = x : x = x = x
2
x
8
x
18) = x : x = x = x
3
x
5 2 5−2 3
8 3 8−3 5
3
2 3 3 3−2 1
−1
2 1 2 2 2
x
19) = x : x = x = x = x
x
7
4
1
3
a
a
7 1 7 1 7⋅3−1⋅4 21−4 17
−
4 3 4 3 12 12 12
x
20) = x : x = x = x = x = x
x
n
m
1
4
a
= a : a = a
n m n−m
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mim-sraga@zg.htnet.hr
tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
26.
U slijedećim zadacima koristimo pravilo: :
a
a
n
m
= a a = a
n m n−m
Rješenja ZADATAKA od 21) do 31) šaljemo u besplatnom PDF dokumentu samo
na zahtjev upućem mailom !
Ako trebate preostala rješenja ovog zadatka pošaljite nam poruku na:
Mail: mim-sraga@zg.htnet.hr
sa tekstom : Trebam preostala rješenja zadataka iz zbirke POTENCIJE
DODATNE UPUTE UZ OVE ZADATKE iz POTENCIJA
SA OBJAŠNJENJIMA I POSTUPCIMA RJEŠEVANJA ZADATAKA
NALAZE SE NA NAŠOJ WEB-STRANICI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
MALA ŠKOLA MATEMATIKE
DODATNE UPUTE UZ OVE ZADATK
POTENCIJE
ALGEBARSKI IZRAZI
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
13

M-1- univerzalna zbirka potpuno riješenih zadataka
27.
Koristimo pravila:
Pojavio nam se jedan problem a taj je:
m
( )
n
n
m
( ) ( ) ( ) ( )
( −x)
Pogledajmo opet u 22. zadata k kako smo to tamo rješili:
22.
2
( ) ( ) ( )
3
( ) ( ) ( ) ( )
z
n n n n n n n⋅m n n⋅m⋅z
ab = ab abc = abc a = a a = a
n
vidi zadatke: 7),13),14),15),16),17),18),19),23),24)...
4) − 1 = −1 ⋅ − 1 = 1
⎫
na parni eksponent
3
⎪
( − 1)
= 1
5) ( − 1) = ( −1) ⋅( −1) ⋅( − 1)
= −1
⎬ Zaključak:
na neparni eksponent
−
4 ⎪
( − 1)
= 1
6) ( − 1) = ( −1) ⋅( −1) ⋅( −1) ⋅( − 1)
= 1
⎭⎪
7) − 2 = −2 ⋅ −2 ⋅ − 2 = −8
8)
↓
Imamo neparan broj "minusa" pa će i umnožak biti negativno tj. imati će predznak minus
( ) ( ) ( ) ( ) ( )
4 4
− x = −x ⋅ −x ⋅ −x ⋅ − x = x
Imamo paran broj "minusa" pa
je produkt pozitivan broj
53
( − x) = ( −x) ⋅( −x) ⋅ ⋅( −x)
Sada se javlja ovakav problem: ...
53 je neparan broj pa je:
( − x)
dalje logički je pitanje koliko je
( )
( x)
55 55
56 56
53 53
= − x
( −x)
54
?
53 puta bi to trebali napisati
a to je previše pisanja
pa ćemo rađe gledati eksponent
= u eksponentu je 53 ,
eksponent je 54, 54 je paran broj pa je:
− x = − x jer je eksponent = 55, a 55 je neparan broj
− = x jer je eksponent = 56, a 56 je paran broj
( x)
54 54
− =
x
Zaključak:
( x)
( − x)
( x)
( −x)
na parni eksponent parnu
2n
2n
− = x
na neparni eksponent
= − x
neparnu
ili
− = x
2n+ 1 2n+
1
=−x
22
Sada to primjenimo u zadatcima:
( x)
( )
4 4
3 3
2n
je oznaka za parni broj
2n
+ 1 je oznaka za neparan broj
. 8) − = x zato što je eksponent = 4, a 4 je paran broj
7) − 2 = − 2 = − 8 zato što je eksponent = 3, a 3 je neparan broj
Postoji i drugi način rješavanja ovakvih zadataka:
II način
( ) ( ) ( )
( ) ( )
( − x) = ( −1⋅x)
3 3 3 3
7) − 2 = −1⋅ 2 = −1 ⋅ 2 = −1⋅ 8 = −8
( − x)
( )
3 3 3 3
− 2 = −1⋅ 2 = −1 ⋅ 2 = −1⋅ 8 = −8
Svaki negativan broj
da se zapisati u obliku:
( ) ( ) ( )
4 4 4 4 4 4
8) − x = −1⋅ x = −1 ⋅ x = 1⋅ x = x
II način
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
27.
Koristimo pravila:
( ab)
) ( )
) ( )
2 2 2 2 2
1 2x = 2 ⋅ x = 4⋅ x = 4x
2 2 2 2 2
2 3x = 3 ⋅ x = 9⋅ x = 9x
3
n n n
=
a b
⎛ ⎞ ⎛ ⎞
) ( ) ( )
m
( )
n
n
m
( ) ( ) ( ) ( )
Postupili smo prema prvom pravilu...
2 2 2
2 2
2 3 2 3 2⋅2 3⋅2 4 6
= ⋅ ⋅ = ⋅ ⋅ =
2
3 3 3 9
⎜ x y ⎟ ⎜ ⎟ x y x y x y
⎝4 ⎠ ⎝4⎠
4
16
z
n n n n n n n⋅m n n⋅m⋅z
ab = a b abc = a b c a = a a = a
m
( abc) = a b c ( a ) = a
n n n n n n ⋅ m
Postupili smo prema drugom i trećem pravilu...
4
)
3 3 3
2 3 2
3
3 2⋅3 3 6
= ⋅ ⋅ ( ) = ⋅ ⋅ =
3
⎛2 ⎞ ⎛2⎞
2 8
⎜ xy ⎟ ⎜ ⎟ x y x y x y
⎝3 ⎠ ⎝3⎠
3
27
5
4 4 4
4 4
2 3 2 3 2⋅4 3⋅4 8 12
= ⋅ ⋅ = ⋅ ⋅ =
4
1 1 1 1
⎜ x y ⎟ ⎜ ⎟ x y x y x y
⎝2 ⎠ ⎝2⎠
2
16
⎛ ⎞ ⎛ ⎞
) ( ) ( )
2
2 2⋅
2 4
⎫
6) ( x ) = x = x
⎪
⎪
2 2
⎪
2 2 2⋅2 4
7) ( − x ) = ( x ) = x = x I način
⎪
⎪
II način:
⎪
2 2 2 2
2 2 2 2⋅2 4
⎬
7) ( − x ) = ( −1⋅ x ) = ( −1) ⋅ ( x ) = 1 ⋅ x = x ⎪
⎪⎪
Prema pravilu:
n
( a )
m
= a
n⋅
m
8
9
) ( )
) ( )
3
2 2⋅3 6
x = x = x
5
2 2⋅5 10
x = x = x
⎪
⎪
⎪
⎪
⎭
) ( ) ( ) ( )
3 4
2
2 3
2
4
2
3⋅2 4⋅2 6 8
10 2xy = 2 ⋅ x ⋅ y = 4⋅x ⋅ y = 4xy
) ( ) ( ) ( )
3 4
3
3 3
3
4
3
3⋅3 4⋅3 9 12
11 2xy = 2 ⋅ x ⋅ y = 8⋅x ⋅ y = 8xy
4
) ( xy
3 4 ) = 4 ⋅ ( x
3
) ( )
12 2 2
4 4
4 3⋅4 4⋅4 12 16
⋅ y = 16⋅x ⋅ y = 16x y
13
) ⎡( ) ⎤ ( )
II način:
autor: Mladen Sraga
2
3
2
3
2⋅3 6
− y = y = y = y
⎣ ⎦
2
3
2
3
2
3 3
2 2 2 2 3 6
3
) ( ) ( ) ( ) ( )
13 ⎡ ⋅
− y ⎤ = ⎡ −1⋅ y ⎤ = ⎡ −1 ⋅ y ⎤ = ⎡1⋅ y ⎤ = y = y = y
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
15

M-1- univerzalna zbirka potpuno riješenih zadataka
27.
Koristimo pravilo:
( x)
( − x)
na parni eksponent
− = x
na neparni eksponent
parnu
= − x
neparnu
prilikom računanja načinom u
I
14), 15), 16),17), 18), 19)...
14)
( ) ( )
3 3
2 2 2⋅3 6
− y = − y = − y = − y
to je bio I način rješavanja
II način
3
) ( ) ( ) ( ) ( )
3 3 3
2 2 2 2⋅3 6 6
14 − y = −1⋅ y = −1 ⋅ y = −1⋅ y = −1⋅ y = − y
↓
2
) ( ) ( ) ( ) ( )
3
( ) Prema pravilu: ( )
− 1 = −1 − 1 = −1
2
( ) Prema pravilu: ( )
− 1 = + 1 − 1 = + 1=
1
↑
2 2 2
3 3 3 3⋅2 6 6
15 − y = −1⋅ y = −1 ⋅ y = 1⋅ y = 1⋅ y = y
neparnu
parnu
to je bio II način rješavanja
15)
2 2
( ) ( )
parnu
3 3
( − y ) = ( y )
− y = y = y = y
3 3 3⋅2 6
parnu
to je bio I način rješavanja
16
) ( ) ( )
⋅
( ) ( )
3 3
4 4 4 3 12 12
− − y = − − y = − − y = + y = y
to je bio I način rješavanja
3 3 3
) ( ) ( )
⎡( ) 3
( )
⎤
⋅
( ) ( )
16 − − y =− −1⋅ y =− −1 ⋅ y =− −1⋅ y =− − y = y
⎢⎣
⎥⎦
4 4 4 4 3 12 12
II način
17
) ( ) ( )
2 2
5 5 5⋅
2 10
− − x =− x =− x =−x
to je bio I način rješavanja
2 2 2
) ( ) ( )
⎡( ) 2
( )
⎤
⋅
( ) ( )
17 − − x =− −1⋅ x =− −1 ⋅ x =− 1⋅ x =− x =−x
⎢⎣
⎥⎦
5 5 5 5 2 10 10
II način
18
U 18) pokazat ćemo tri načina rješavanja:
2 2 2 2
5 5 5 5 5⋅
2 10
) ⎡ ( ) ⎤ ⎡ ( ) ⎤ ( ) ( )
⎣
− − x
⎦
=
⎣
+ − x
⎦
= − x = x = x = x
2 2
5 5
) ⎡ ( x ) ⎤ ⎡ ( x ) ⎤
⎣
− −
⎦
=
⎣
− ⋅ −
⎦
=
2
2 5
= ( −1
) ⋅( − x ) = = ( ) (
2
) ( )
18 1
18
2
5 5
) ( ) 2 5
2
5⋅
2 10
⎡− − x ⎤ = ⎡+ x ⎤ = ( x ) = x = x III način ...u [ ]
⎣
⎦
⎣
⎦
to je bio I način rješavanja
2 2
5 5 5⋅
2 10
1⋅−⋅ 1 x = −1 ⋅ x = 1⋅ x = x
II način
min us i minus daju plus
U ovom 18) zadatku najbrži ne III način rješavanja... dok je u 19) definitivno II način najbolji....
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
27.
19
19
2 2
) ( ) ( ) ( ) ( ) ( )
2 2 2⋅
2 4 4
⎡ ⋅
− − x ⎤ = ⎡+ − x ⎤ = − x = − x = x = x = x
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
I način u ovom zadatku pomalo zbunjuje pa je bolje to rješavati na II način izlučivanjem ( −1 ):
II način:
5 5 5 5 5 5 4 20
2
( )
2 2
5
) ( ) (( ) ) ( ) ( x )
⎣ ⎦ ⎣ ⎦
⎡ − − 2
x ⎤ = ⎡ − − 1 ⋅ 2
5 5
x ⎤ = ⎡
− − 2
1
⎢ ⎥ ⎢ ⎥ ⎢⎣
⋅
52 ⋅
( 1 x )
= ⎡− ⋅ ⎤
⎣ ⎦
=
10
2
= ⎡− ⋅ ⎤ =
⎣
1
x
⎦
2
⎤
⎥⎦
2 2
( ) ⎤ ( ) ( )
2
2
10 10 10⋅2 20
= ⎡
⎣ −1 ⋅ x ⎦ = −1 ⋅ x = 1⋅ x = x
=
I način
20
2
⋅
) ( ) ( ) ( )
− x = − x = − x = − x
5 5 2 10 10
21
) ( ) ( )
2
3
3 3 2
3
3 2⋅3 3 6
2ab = 2 ⋅a ⋅ b = 8⋅a ⋅ b = 4a b
22
23
) ( ) ( ) ( )
2 3
4
2
4
3
4
2⋅4 3⋅4 8 12
ab = a ⋅ b = a ⋅ b = ab
2
) ( ) ( ) ( )
2 2
3 3 3⋅2 6
− 2y = −2 ⋅ y = 4⋅ y = 4y
↓
2 2
( − 2) = 4 jer je: ( − 2)
= ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
2 2 2
3 3 3 3 3
4 4 4 4
−12 ⋅ = −1 ⋅ 2 = 14 ⋅ = 4
− 2 = −8 jer je: − 2 = −1⋅ 2 = −1 ⋅ 2 = −1⋅ 8 = −8
− 2 = −1⋅ 2 = −1 ⋅ 2 = 1⋅ 16 = 16 itd.
ili k
autor: Mladen Sraga
oristimo pravilo:
2
( ) ( ) ( )
ili ovako: − 2 = −2 ⋅ − 2 = 4
( )
3 3
3
) ( ) ( ) ( )
3
( ) ( ) ( ) ( )
− 2 = −2 ⋅ −2 ⋅ − 2 = −8
( x)
( − x)
na parni eksponent
− 3 = − 3 = −27
− = x
na neparni eksponent
parnu
= − x
3 3
2 2 3 2⋅3 6
24 − 3y = −3 ⋅ y = −3 ⋅ y = −27y
↓
↑
17
pa je:
( )
( )
2 2
− 2 = 2 = 4
neparnu 3 3
− 2 = − 2 = −8
3 3 3
( ) ( ) ( ) 3
3
ili ( ) ( ) ( ) ( )
− 3 = −1⋅ 3 = −1 ⋅ 3 = −1⋅ 27 = −27 − 3 = −3 ⋅ −3 ⋅ − 3 = −27

M-1- univerzalna zbirka potpuno riješenih zadataka
27.
Rješenja ovog ZADATKA od 25) do 53) šaljemo u besplatnom PDF dokumentu samo
na zahtjev upućem mailom !
Ako trebate preostala rješenja ovog zadatka pošaljite nam poruku na:
Mail: mim-sraga@zg.htnet.hr
sa tekstom : Trebam preostala rješenja zadataka iz zbirke POTENCIJE
KAKO NARUČITI KOMPLETNU ŠTAMPANU VARIJANTU OVE ZBIRKE ?
1. javite nam se mailom na: mim-sraga@zg.htnet.hr
2. nazovite nas na telefon 01-4578-431
3. nazovite nas na mobitem 098-237-534
4. pošaljite nam pismo na . M.I.M.-Sraga centar za poduku i dopisnu
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
28.
Koristimo pravilo:
n
( ) ( )
n n n n n
a ⋅ b = a⋅ b a b c = abc
n
x
( ) ( )
x x x x x x x
1) 2 ⋅ 5 = 2⋅ 5 = 10 2) 3 ⋅ 4 = 3⋅ 4 = 12
a
( ) ( )
a a a y y y y
3) 4 ⋅ 6 = 4⋅ 6 = 24 4) 2 ⋅ 4 = 2⋅ 4 = 8
( ) 5
5) 3 ⋅ 2 = 3⋅ 5 = 15
5 5 5
3 3 3
⎛1⎞ 3 ⎛1 ⎞ ⎛9⎞
⎛3⋅
3 ⎞
3
6) ⎜ ⎟ ⋅ 9 = ⎜ ⋅ 9⎟ = ⎜ ⎟ = ⎜ ⎟ = 3 = 27
⎝3⎠ ⎝3 ⎠ ⎝3⎠ ⎝ 3 ⎠
7)
4 4 4
1 1
4
4
⎛ 2 ⎞ ⎛1⎞
⎛3⎞ ⎛2⎞ ⎛3 2⎞
3
4
⎜ ⎟ ⋅ ⎜ ⎟ = ⎜ ⋅ ⎟ =
⋅
= = 1 = 1
⎝2⎠ ⎝3⎠ ⎝2 3⎠ ⎜
1 2 3 ⎟ ⎜ ⎟
⎝ 1 ⎠ ⎝1⎠
5 5 5
5 5
⎛1⎞ ⎛1 ⎞ ⎛4⎞
8) ⎜ ⎟ ⋅ 4 = ⎜ ⋅ 4⎟ = ⎜ ⎟ = 2 = 32
⎝2⎠ ⎝2 ⎠ ⎝2⎠
3
9)
5 5 5
⎛2⎞ ⎛9⎞ ⎛2 9⎞
⎛ 2 3 ⋅3
⎞
5
⎜ ⎟ ⋅ ⎜ ⎟ = ⎜ ⋅ ⎟ = ⎜ ⋅ ⎟ = 3 = 243
⎝3⎠ ⎝2⎠ ⎝3 2⎠ ⎝ 3 2 ⎠
5
10)
2 2
⎛ac ⎞ ⎛b ⎞ ⎛ac
⋅ b ⎞
⎜ ⎟ ⋅ ⎜ ⎟ = ⋅ = a
b c
⎜
b c
⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2
2
4 4 4
⎛a⎞ ⎛b⎞ ⎛a b⎞
4
11) ⋅ = ⋅ = 1 = 1
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝b⎠ ⎝a⎠ ⎝b a⎠
3 3 3
3
⎛3⎞ ⎛8⎞ ⎛3 8⎞
⎛3
42 ⋅ ⎞ 3
12) ⋅ = ⋅ = ⋅ = 2 = 8
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝4⎠ ⎝3⎠ ⎝4 3⎠ ⎝4 3 ⎠
13)
14)
2
2 2
2 2
⎛ xy ⎞ ⎛ 4 ⎞ ⎛ xy 4 ⎞ ⎛ x⋅y
22 ⋅ ⎞ ⎛2
⎞
⎜ ⎟ ⋅ ⎜ 2 ⎟ = ⎜ ⋅
2 ⎟ = ⎜ ⋅ ⎟ = ⎜ ⎟ =
⎝ 2 ⎠ ⎝ x y⎠ ⎝ 2 x y⎠
⎝ 2 xxy ⋅ ⋅ ⎠ ⎝ x⎠
ili dalje:
2
2 4
= =
x x
2 2
možemo ostaviti u ovom obliku
( x )
2
( z )
3 3 2
3
3 3 3
2 2⋅3 6
⎛ xy⎞ ⎛ x ⎞ ⎛ xy x ⎞ ⎛ xx ⋅ ⎞ ⎛ x ⎞
x x
⋅ ⎜ ⎟ = ⎜ ⋅ ⎟ = nakon kraćenja = ⎜ ⎟ = ⎜ 2 ⎟ =
3 =
2⋅3 =
6
⎜ ⎟
⎝ z ⎠ ⎝ yz⎠ ⎝ z yz⎠ ⎝ z⋅
z ⎠ ⎝ z ⎠
z z
15)
m+
1
⎛ xy ⎞ ⎛ x ⎞
⎜ ⎟ ⋅ ⎜ ⎟
⎝ z ⎠ ⎝ yz⎠
m+
1
m+1
⎛ xy x ⎞
= ⋅ = nakon kraćenja =
⎜
⎝
z
⎟
yz⎠
( x )
2
( z )
m+ 1 1 2
1
2
m+
m+
2⋅ m+ 1 2m+
2
( )
⎛ xx ⋅ ⎞ ⎛ x ⎞
x x
= ⎜ ⎟ = ⎜ ⎟ = = =
( m
z z z
)
⎝ ⋅ ⎠ ⎝ ⎠
z z
2 m+ 1 2⋅ + 1 2m+
2
autor: Mladen Sraga
19

M-1- univerzalna zbirka potpuno riješenih zadataka
28.
Koristimo pravilo:
n
( ) ( )
n n n n n
a ⋅ b = a⋅ b a b c = abc
n
Rješenja ovog ZADATKA od 16) do 18) šaljemo u besplatnom PDF dokumentu samo
na zahtjev upućem mailom !
Ako trebate preostala rješenja ovog zadatka pošaljite nam poruku na:
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sa tekstom : Trebam preostala rješenja zadataka iz zbirke POTENCIJE
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
29.
Koristimo pravila:
Izračunaj:
1)
2 2
n n
−n n
⎛a⎞ a ⎛a⎞ ⎛b⎞
b
⎜ ⎟ =
n ⎜ ⎟ = ⎜ ⎟ =
⎝b⎠ b ⎝b⎠ ⎝a⎠
a
⎛2⎞ 2 4
⎛a⎞
⎜ ⎟ 2
⎜ ⎟
⎝3⎠ 3 9
⎝b
⎠
= = primjenili smo pravilo: =
n
n
n
a
b
n
n
2)
−2 2 2
⎛2⎞ ⎛3⎞
3 9
⎜ ⎟ = ⎜ ⎟ = =
⎝ ⎠ ⎝ ⎠ 2
2
3 2 4
uputa: 2)
n n
⎛a⎞
a
po pravilu: ⎜ ⎟ =
b n
⎝ ⎠ b
−2 2 2
⎛ 2 ⎞ ⎛3⎞
3 9
⎜ ⎟ = ⎜ ⎟ = =
2
⎝
3 ⎠ ⎝2⎠
2 4
po pravilu
− n n
⎛a⎞ ⎛b⎞
⎜ ⎟ = ⎜ ⎟
⎝b⎠ ⎝a⎠
3)
3 3
⎛1⎞ 1 1
⎛a⎞
⎜ ⎟ 3
⎜ ⎟
⎝2⎠ 2 8
⎝b
⎠
= = primjenili smo pravilo: =
n
a
b
n
n
4)
4 4
⎛2⎞ 2 16
⎛a⎞
⎜ ⎟ 4
⎜ ⎟
⎝3⎠ 3 81
⎝b
⎠
= = primjenili smo pravilo: =
n
a
b
n
n
5)
−1 1
⎛5⎞ ⎛7⎞
⎜ ⎟ = ⎜ ⎟ =
⎝7⎠ ⎝5⎠
7
5
primjenili smo pravilo:
−n n n
⎛a⎞ ⎛b⎞
b
⎜ ⎟ = ⎜ ⎟ =
⎝b⎠ ⎝a⎠
a
n
−1 −1
−1
⎛ 2⎞ ⎛17 ⋅ + 2⎞
⎛9⎞
7
6) ⎜1 ⎟ = 7
⎜ = =
7
⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ 7 ⎠ 9
→
mješoviti broj treba prvo pretvoriti u razlomak...
7)
−2 2 2
⎛4⎞ ⎛5⎞
5 25
⎜ ⎟ = ⎜ ⎟ = =
2
⎝5⎠ ⎝4⎠
4 16
8)
−3 3
⎛2⎞ ⎛3⎞
3
⎜ ⎟ = ⎜ ⎟ =
⎝3⎠ ⎝2⎠
2
3
3
=
27
8
−2 −2
−2 −2 2 2
⎛ 1⎞ ⎛23 ⋅ + 1⎞
⎛6+
1⎞ ⎛7⎞ ⎛3⎞
3 9
9) ⎜2 ⎟ = 3
⎜ 3
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = =
3 3 7
2 49
→
autor: Mladen Sraga
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
mješoviti broj treba prvo pretvoriti u razlomak...
−2 −2
−2 −2 2 2
⎛ 2⎞ ⎛17 ⋅ + 2⎞
⎛7+
2⎞ ⎛9⎞ ⎛7⎞
7 4
10) ⎜1 ⎟ = 7
⎜ 7
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = =
7 7 9 9
2
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
21
7
9
81

M-1- univerzalna zbirka potpuno riješenih zadataka
29.
Koristimo pravila:
n n
−n n
⎛a⎞ a ⎛a⎞ ⎛b⎞
b
⎜ ⎟ =
n ⎜ ⎟ = ⎜ ⎟ =
⎝b⎠ b ⎝b⎠ ⎝a⎠
a
n
n
mješoviti broj treba prvo pretvoriti u razlomak...
−3 −3
−3 −3 3 3
⎛ 1⎞ ⎛12 ⋅ + 1⎞
⎛2+
1⎞ ⎛3⎞ ⎛2⎞
2 8
11) ⎜1 ⎟ = 2
⎜ 2
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = =
2 2 3
3 27
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
3
12)
−2
⎛3⎞ ⎛4⎞
⎜ ⎟ = ⎜ ⎟
⎝4⎠ ⎝3
⎠
2 2
4 16
= =
2
3 9
13)
⎛ x⎞
⎜ ⎟
⎝ y ⎠
2
=
x
y
2
2
( x)
( − x)
na parni eksponent
parnu
− = x
način oristimo pravilo: pa je:
na neparni eksponent neparnu
= − x
14) I k
2 2
2
2
⎛ x⎞ ⎛ x⎞ x ⎛ x⎞ ⎛ x⎞ ⎛ x⎞
⎛
14) ⎜− ⎟ = ⎜ ⎟ = jer je:
2
⎜− ⎟ = ⎜− ⎟ = ⎜ ⎟ = ⎜
⎝ y⎠ ⎝ y⎠ y
⎝ y⎠ ⎝ y⎠ ⎝ y⎠
⎝
parnu
parnu
x
y
⎞
⎟
⎠
2
= ...
2 2 2
2
⎛ x⎞ ⎛ x⎞ ⎛ x⎞
x x
14) ⎜− ⎟ = ⎜−1⋅ ⎟ = ( −1)
⋅ ⎜ ⎟ = 1⋅ =
⎝ y⎠ ⎝ y⎠ ⎝ y⎠
y y
2 2
2 2
II način
⎛ x⎞ ⎛ x⎞
⎜ ⎟ ⎜ ⎟
⎝ y⎠ ⎝ y⎠
II način − rastavili smo na: −1 ⋅ ...
15)
3 3
⎛ x⎞ ⎛ x⎞
x
⎜− ⎟ = − ⎜ ⎟ = −
⎝ y⎠ ⎝ y⎠
y
3 3 3
3
3
3
I način
⎛ x⎞ ⎛ x⎞ ⎛ x⎞
x
15) ⎜− ⎟ = ⎜−1⋅ ⎟ = ( −1)
⋅ ⎜ ⎟ = −1⋅
⎝ y⎠ ⎝ y⎠ ⎝ y⎠
y
x
= −
y
3 3
3 3
II način
4 4
4
( x)
( − x)
na parni eksponent
parnu
− = x
način oristimo pravilo: pa je:
na neparni eksponent neparnu
= − x
16) I k
4
⎛ x⎞ ⎛ x⎞
x
na parni eksponent
16) ⎜− ⎟ = ⎜ ⎟ = jer je:
4
( − x)
= x
⎝ y⎠ ⎝ y⎠
y
16)
II način:
4 4
⎛ x ⎞ ⎛ x⎞ 4 ⎛ x⎞
x x
⎜− ⎟ = ⎜−1⋅ ⎟ = ( −1)
⋅ ⎜ ⎟ = 1⋅ =
⎝ y ⎠ ⎝ y⎠ ⎝ y⎠
y y
4 4
4 4
parnu
2 2
2 2 2
⎛ 1⎞ ⎛12 ⋅ + 1⎞
⎛2+
1⎞ ⎛3⎞
3
17) ⎜1 ⎟ = 2
⎜ = = =
2
⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2
2
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© M.I.M-Sraga centar za poduku
mim-sraga@zg.htnet.hr
tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
29.
Koristimo pravila:
n n
−n n
⎛a⎞ a ⎛a⎞ ⎛b⎞
b
⎜ ⎟ =
n ⎜ ⎟ = ⎜ ⎟ =
⎝b⎠ b ⎝b⎠ ⎝a⎠
a
n
n
Rješenja ovog ZADATKA od 18) do 27) šaljemo u besplatnom PDF dokumentu samo
na zahtjev upućem mailom !
Rješenja ovog ZADATKA od 28) pa do 40)
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
Ako trebate preostala rješenja ovog zadatka pošaljite nam poruku na:
Mail: mim-sraga@zg.htnet.hr
sa tekstom : Trebam preostala rješenja zadataka iz zbirke POTENCIJE
autor: Mladen Sraga
23

M-1- univerzalna zbirka potpuno riješenih zadataka
1 1
= = = =
n
a
a
0 1 −1
−n
30. Koristimo pravila: a 1 , a a , a , a
0 0
1) 2 = 1 po pravilu: a = 1
0
2) x = 1
2 4
0
⎛ ⎞ ⎧
x y
3) ⎜ ⎟ = 1 ⎨
⎝ z ⎠
⎩
3)
ili taj isti zadatak dužim postupkom:
bez obzira što je u zagradi ako je eksponent
te zagrade nula sve je jednako jedan !
( x ) ⋅( y )
0
0 0
2 4
2 4
⎛ ⎞ ⋅
⎜
⎝
x y
z
⎟
⎠
11 1
= = = = 1
0
z 1 1
0 0
4) x + y = 1+ 1 = 2
5)
0
3 = 1
( )
0 0
6) 3 1 po pravilu: a 1
− = =
0 0 0
7) 3 1 zato što je: 3 1 3 1 1 1
− =− − =− ⋅ =− ⋅ =−
0
8) 3x
3 1 3
−
= − ⋅ = −
( x y)
9) + = 1
0
bez obzira što je u zagradi ako je to na nultu sve je jednako jedan !
0 ⎧
10) ( 2x− 7y)
= 1 ⎨
⎩
bez obzira što je u zagradi ako je eksponent
te zagrade nula sve je jednako jedan !
0 0 0
11) x − 2y + 7z
= 1−2⋅ 1+ 7⋅ 1 = 1− 2 + 7 = 6
( ) ( )
0
0 0
12) x − 2y + 7z
= 1− 1+ 1 = 1
0 0
13) x ⋅ x = 1⋅ 1 = 1
( x )
2
0 2
14) = 1 = 1
www.mim-sraga.com 24 © M.I.M-Sraga centar za poduku
mim-sraga@zg.htnet.hr
tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
30.
Koristimo pravila:
1 1
= 1 , = , = , =
n
a
a
0 1 −1
−n
a a a a a
x
⋅ x = ⋅ =
0 0
13) 1 1 1
( x )
2
0 2
14) = 1 = 1
m
( x )
15) = 1
0
bez obzira što je u zagradi ako je to na nuultu sve je jednako jedan !
( x)
16) 4 = 1
0
17) 5
1 1
= po pravilu: a =
5
a
−1 −1
18)
x
−1
=
1
x
19) 0,2
−1
−1 −1
⎛ 2 ⎞ ⎛1⎞
5
= ⎜ ⎟ = ⎜ ⎟ = = 5
⎝10 ⎠ ⎝5⎠
1
prvo decimalni broj pretvorimo u razlomak...
20)
−1
1 1 −n
1
x− y = = po pravilu: a =
1
n
x−
y a
( )
( x−
y)
21) 3
−2
1 1 −n
1
= = po pravilu: a =
2
n
3 9
a
22)
x
−3
=
1
3
x
23) 0,3
−2
−2 2 2
⎛ 3 ⎞ ⎛10 ⎞ 10 100
= ⎜ ⎟ = ⎜ ⎟ = =
2
⎝10 ⎠ ⎝ 3 ⎠ 3 9
( )
24) 2x
− 5
−5
=
1
( 2x
− 5)
5
25) 4
−1
=
autor: Mladen Sraga
1
4
25

M-1- univerzalna zbirka potpuno riješenih zadataka
30.
Koristimo pravila:
1 1
= 1 , = , = , =
n
a
a
0 1 −1
−n
a a a a a
Rješenja ovog ZADATKA od 26) do 41) šaljemo u besplatnom PDF dokumentu samo
na zahtjev upućem mailom !
Ako trebate preostala rješenja ovog zadatka pošaljite nam poruku na:
Mail: mim-sraga@zg.htnet.hr
sa tekstom : Trebam preostala rješenja zadataka iz zbirke POTENCIJE
Rješenja ovog ZADATKA od 42) pa do 57)
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
www.mim-sraga.com 26 © M.I.M-Sraga centar za poduku
mim-sraga@zg.htnet.hr
tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
31.
a) Zapiši u obliku potencija s bazom 2:
1)
+ +
4⋅32⋅ 16 = 2 ⋅2 ⋅ 2 = 2 = 2
2 5 4 2 5 4 11
( ) ( ) ( ) ( )
+ + ⋅
2) 4⋅8⋅ 16 = 2 ⋅2 ⋅ 2 = 2 = 2 = 2 = 2
2 2 3 4 2 2 3 4 2 9 2 9 2 18
Preostala rješenja ovog ZADATKA
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
32.
9 6 5
x y
⎛9 6 5⎞ ⎛3 4 3⎞ 2 9 3 6 4 5 3 9 2 6−4 5−3
1) ⎜ xy⎟: ⎜ xy⎟
= = : ⋅ x: x ⋅ y:
y = ⋅ ⋅ x ⋅ y =
⎝2 ⎠ ⎝2 ⎠ 3 4 3 2 2 2 3
x y
2
9 2 6−4 5−3
3⋅
3 2
= ⋅ ⋅x
⋅ y = ⋅ ⋅x ⋅ y = 3x y
2 3
2 3
2 2 2 2
Preostala rješenja ovog ZADATKA
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
33.
1)
x+ 1 4 x+ 1 4+ x+ 1 x+
5
16⋅2 2 ⋅2 2 2
= = = =
x−1 3x−3
8 2
3 x−1
3( x−1)
( 2 ) 2
( x )
x+ 5 3x− 3 x+ 5− 3 −3
x+ 5− 3x+
3
= 2 :2 = 2 = 2 = 2 = 2 = 2
x− 3x+ 5+ 3 − 2x+ 8 8−2x
Preostala rješenja ovog ZADATKA
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
autor: Mladen Sraga
27

M-1- univerzalna zbirka potpuno riješenih zadataka
34.
1)
( x )
( y )
( x )
( y )
−2 −2 4
2
3
−2 4 2
2
2
4⋅ 2
2 2
8 4 8 4+
8 12
⎛ x ⎞ ⎛ x ⎞ ⎛ y ⎞ ⎛ x ⎞ x x x x x x
⎜ ⎟ ⋅ ⎜ ⎟ = ⎜ ⎟ ⋅ = ⎜ ⎟ ⋅ = ⋅ = ⋅ = =
⎝ y ⎠ ⎝ y ⎠ ⎝ x ⎠ ⎝ y ⎠ y y y y y y
−3 2 2 2 2 3 2⋅ 2 3 2 4 6 4 6+
4 10
Preostala rješenja ovog ZADATKA
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
35.
( ) ( )
4) 4ab = 4 ⋅a ⋅ b = 16⋅a ⋅ b = 16a b
2 2 2 2 2 2 2 2⋅2 2 4
( x y ) ( x ) x y ( y )
3 ( ) 6
2 3 2 2 2 2 3 3 2
6) 3 + = 3 + 2⋅3
⋅ + =
2 2 2 2 3 3⋅
2
= ⋅ x + x y + y =
22 ⋅ 2 3 6
= 9⋅ x + 6x y + y =
= 9x + 6x y + y
4 2 3 6
Preostala rješenja ovog ZADATKA
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
DODATNE UPUTE UZ OVE ZADATKE iz POTENCIJA
SA OBJAŠNJENJIMA I POSTUPCIMA RJEŠEVANJA ZADATAKA
NALAZE SE NA NAŠOJ WEB-STRANICI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
MALA ŠKOLA MATEMATIKE
DODATNE UPUTE UZ OVE ZADATK
POTENCIJE
ALGEBARSKI IZRAZI
ALGEBARSKI RAZLOMCI
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POTENCIJE
M-1- rješenja i kompletni postupak
36.
23)
( ) ( ) ( ) ( ) ( ) ( )
ab ⋅ ab = a ⋅ b ⋅ a ⋅ b = a ⋅b ⋅a ⋅ b = a ⋅b ⋅a ⋅b
2 3 4 2 4 3 2 4 3 4 2 3 4 3 2 ⋅ 4 3 ⋅ 4 2 ⋅ 3 4 ⋅ 3 8 12 6 12
= a ⋅a ⋅b ⋅ b = a ⋅ b = a b
8 6 12 12 8+ 6 12+
12 14 24
=
Preostala rješenja ovog ZADATKA
mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
Novo MALA ŠKOLA MATEMATIKE –1 na
BESPLATNA video poduka i instrukcije
UČIMO ZAJEDNO
POTENCIJE
ALGEBARSKI IZRAZI
ALGEBARSKI RAZLOMCI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
KAKO NARUČITI KOMPLETNU ŠTAMPANU VARIJANTU OVE ZBIRKE ?
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6. nazovite nas na telefon 01-4578-431
7. nazovite nas na mobitem 098-237-534
8. pošaljite nam pismo na . M.I.M.-Sraga centar za poduku i dopisnu
poduku , Kraljevečki ogranak 24 , 10 000 Zagreb
autor: Mladen Sraga
29

M-1- univerzalna zbirka potpuno riješenih zadataka
**** MLADEN SRAGA ****
2010.
UNIVERZALNA ZBIRKA
POTPUNO RIJEŠENIH ZADATAKA
PRIRUČNIK ZA SAMOSTALNO UČENJE
POTENCIJE
SAMO ZADACI
α
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POTENCIJE
M-1- rješenja i kompletni postupak
2 3
n
21. Koristeći pravila: aa ⋅ = a , aaa ⋅ ⋅ = a , aaa ⋅ ⋅ ⋅...
⋅ a=
a
Izračunaj:
1) 22222 ⋅ ⋅ ⋅ ⋅
2)
n−puta
xxx ⋅ ⋅
( ) ( ) ( )
3) xyxyxzxyxxz ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 4) xy⋅ xy⋅
xy
5) ( x+ y) ⋅ ( x+ y) ⋅ ( x+ y) ⋅ ( x+ y)
6)
⎛ x⎞ ⎛ x⎞ ⎛ x⎞ ⎛ x⎞
⎜ ⎟⋅⎜ ⎟⋅⎜ ⎟⋅⎜ ⎟
⎝ y⎠ ⎝ y⎠ ⎝ y⎠ ⎝ y⎠
7) ( x− y) ⋅ ( x+ y) ⋅( x− y) ⋅( x− y)
⋅( x+ y)
8)
⎛a−b⎞ ⎛a−b⎞ ⎛a−b⎞
⎜ ⎟⋅⎜ ⎟⋅⎜ ⎟
⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠
22.
2 3
n
Koristeći pravila: , , ...
Izračunaj:
n−puta
( − )
2 3 4
2
1) 5 2) 2 3) 3 4) 1
( − ) ( − ) ( − ) ( −x)
3 4 3 4
5) 1 6) 1 7) 2 8)
2 3 5 2
⎛1⎞ ⎛3⎞ ⎛2⎞ ⎛ 2⎞
9) ⎜ ⎟ 10) ⎜ ⎟ 11) ⎜ ⎟ 12) ⎜−
⎟
⎝3⎠ ⎝5⎠ ⎝3⎠ ⎝ 3⎠
13)
a = a⋅ a a = a⋅a⋅ a a = a⋅a⋅a⋅ ⋅a
3 4 2 3
⎛ 2 ⎞ ⎛3⎞ ⎛ 4⎞ ⎛ 4⎞
⎜ − ⎟ 14) ⎜ ⎟ 15) ⎜−
⎟ 16) ⎜−
⎟
⎝ 3 ⎠ ⎝4⎠ ⎝ 5⎠ ⎝ 5⎠
4
⎛ 4 ⎞
17) ⎜−
⎟ 18) 0,2 19) ( −0,2)
20) 0,2
⎝ 5 ⎠
( − ) ( − ) ( − ) ) ( − )
( − ) + ( − ) + ( − ) + ( − ) ( − ) + ( − ) + ( − )
( ) ( ) ( ) ( )
2 2
3
3 2 3 4
21) 0,2 22) 2,5 23) 2,5 24 2,5
2 3 4 5 20 30 45
25) 1 1 1 1 26) 1 1 1
1 2 3 4 5 2 4 5
27) − 2 + − 2 + − 2 + −2 28) 2 −3 29) 3 −2
3 5 ⎛1
⎞
30) ( − 2) + ( −2)
31) ⎜ ⎟
⎝4
⎠
2 3
1
2 3
2
⎛ ⎞ ⎡ ⎤
( ) ( )
+ ⎜− ⎟ 32) − 0,2 + −0,2
⎝ 2 ⎠ ⎣ ⎦
23.
Koristeći pravila:
( ) ( )
ca ⋅ + da ⋅ = c+ d ⋅a ca ⋅ −da ⋅ = c−d ⋅a
n n n n n n
1) 2x+ 3x 2) x+ 2x+ 4x 3) 7a−2a
4) 9y− 2y+ 3y− y 5) 2x+ 3a− x+ 5a+ 7x−2a
6) 2xy + 3xy + xy 7) 2ab − 4ab + ab
2
xy + xy − xy y + x y − z − x y − y + z −3y+
8x y
2 2 2 2 2
8) 4 2 9 9) 3 4 5 2 8
( ) ( ) ( )
( ) ( )
10) 7 2 4 11) 2 3
3 3 3 2 2
xy − xy + xy x + y − x + y
2 3 2 3 2 3
12) 3 + + 4 + − + 13) 5 + 2 − 2 + 7 + 3 −3
x y x y x y x y z x y z x y z
2 2
( x − x+ ) − ( x − x+ ) ( x+ y− ) − ( x+ y− ) + ( x+ y−
)
14) 3 3 5 4 5 1 15) 3 1 6 1 1
autor: Mladen Sraga
31

M-1- univerzalna zbirka potpuno riješenih zadataka
Koristeći pravila: a ⋅ a = a a : a = a
n
a
m
a
= a : a = a
Izračunaj:
24.
1) 2 ⋅2 2) 5 ⋅5 3) 3 ⋅3 4) 7 ⋅7
5 3 2 7 x 2x 2m−5 m−2
5) 2 ⋅2 ⋅2 ⋅2 6) 3⋅a ⋅2⋅a 7) x ⋅x 8) x⋅x ⋅x
2 6 4 8 4 7 2 3 3 5
2 5
2 4 6 3 2 2
2 2+ a a−2 3 3 6
9) x ⋅x ⋅x 10) x ⋅x ⋅x 11) x ⋅x ⋅x ⋅x 12) a ⋅a
4 2
13) 1
a ⋅a ⋅a
n m n+ m n m n−m n m n−m
4) a ⋅a ⋅a ⋅a 15) a ⋅a ⋅a 16) a ⋅a ⋅a
2 5 7 3 3x x+ 2 2x x+ y 2x x+
3y
17) −2⋅x⋅6⋅x ⋅3⋅x 18) 2⋅x ⋅ x + 3⋅x ⋅y ⋅x⋅y−4⋅x⋅ x + 2⋅x ⋅y ⋅y
3 4 2 5 2 3 6 3 2 2
2 3 4
⎛1⎞ ⎛1⎞ ⎛1⎞
19) ⎜ ⎟ ⋅⎜ ⎟ ⋅⎜ ⎟
20) x ⋅x 21) x ⋅x
⎝2⎠ ⎝2⎠ ⎝2⎠
2m+ 1 3m+ 2 2m− 4 3m+
2
3
m+
1
2 2m− 7 m+ n 2m− n 3m+
2n
22) x ⋅x 23) x ⋅x ⋅x
24)
25) a ⋅a 26) 2a ⋅3a 27) 2a ⋅5a
( x+ y− ) ⋅ ( x+ y− ) ⋅ ( x+ y−
)
2m n m 2n
( x −y ) ⋅ ( x + y )
m−
( x+ y) ⋅( x− y) ⋅ ( x+ y) ⋅( x−
y)
2m−n 3m−n 2m−2n
2m+ 3n 3m−5n m− 3n+ 1 4m+ n− 7 x+ y 2x+
y
2 9
3 4
( ) ( ) ( ) ( )
3m+ 2n 2m−4n
2 3 4
28) a ⋅ a 29) x− y ⋅ x− y 30) x+ y ⋅ x+
y
2
m+ 1 2m+ 2 3m−4
31) 1 1 1
32)
33)
34)
⎛
⎜
⎝
ab
c
⎞
⎟
⎠
2 3 2 1 m−3
3x−2y 2
4x−y 2
x−y 2
3y−2x
⎛ab ⎞ ⎛ab ⎞ ⎛ab
⎞
⋅⎜ ⎟ ⋅⎜ ⎟ ⋅⎜ ⎟
⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠
x ⋅x ⋅x
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POTENCIJE
M-1- rješenja i kompletni postupak
25.
Pomnoži:
1)
2xy⋅3xy 2) 5xy⋅2xy
2 2 3 3 2 5 3
3)
2 2 3 9 4 5 6 2 ⎛ 9 2 ⎞
ab⋅ ab 4) − ab ⋅ ⎜ − ab ⎟
3 4 27 ⎝ 5 ⎠
5)
1 2 3 4 3 2 5 4 2 1 2
abc⋅−
( 4abc)
6) 9xy⋅
xy
2 3
7)
5
6 4 yx
2 3 4 3 4 2
25xy⋅
8) 2xyz⋅( −3xyz)
5
9) ⋅( − − 3 + 2 −7) 10) ( −2
⎛ 1
) ⋅⎜
−
⎝ 2
+
2 4 3 2 2 2 3 3
x x x x x xy x y xy x y
⎞
⎟
⎠
26.
Izračunaj:
1) 2 : 2 2) 13 : 13 3) 3 : 3 4) 5 : 5
9 5 7 5 5x 3x 3m−2 m−2
5) x : x 6) x : x 7) x : x : x 8) x ⋅ x : x
4 2 6 2 7 2 3 7 3 4
1 7 1 2 1
2 2 2 9 3 5 2 2
9) x: x 10) x : x 11) x : x 12) x : x : x
13) a : a 14) a : a : a 15) a: a : a 16) a : a ⋅ a
4x
2x
8 3 2 2 3 7 2 3
3 7
5 8 2 4
17)
x x x x
18) 19) 20)
2 3 1
x x x
3
x
21)
m+
1 3m+ 7 8m+ 7 9m+
11
x
x x m
a
22) 23) : x 24) : a
2m+ 2 3m+ 5 3m+ 2 2m−3
x x x a
− 1 7m+
10
2
x−2 2
3x−3 2
4x−4
2
6
2
4
7 4 2x+ 3y 2x+
3y
⎛ab⎞ ⎛ab⎞
25) ( x− y) :( x− y) 26) ( 2x−3 y) :( 2x−3y)
27) ⎜ :
3 ⎟ ⎜ 3 ⎟
⎝ c ⎠ ⎝ c ⎠
2m−4 4m−2 m−4
⎛ab⎞ ⎛ab⎞ ⎛ab⎞
28) ( 2x+ y) ⋅ ( 2 x+ y) :( 2x+ y)
29) ⎜ :
3 ⎟ ⋅⎜ 3 ⎟ ⎜ 3 ⎟
⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠
( ) ( ) ( ) ( )
30) x 2 −1 ⋅ x 4 + x 2 + 1 31) x 2 + 1 ⋅ x− x
2 + 1
autor: Mladen Sraga
33

M-1- univerzalna zbirka potpuno riješenih zadataka
27.
Koristeći pravila:
Izračunaj:
( )
n
n
m
( ) ( ) ( ) ( )
2 3
2 2 ⎛3 2 3⎞ ⎛2
2⎞
) ( ) ) ( ) ) )
1 2x 2 3x 3 ⎜ x y ⎟ 4 ⎜ xy ⎟
⎝4 ⎠ ⎝3
⎠
4
1
5 ⎜ ⎟ 6 7 8
⎝2
⎠
⎛ 2 3⎞
2 2 2
) x y ) ( x ) ) ( −x ) ) ( x )
2 3 4 3 4 3 4
) ( x ) ) ( x y ) ) ( x y ) ) ( x y )
2
) ⎡
2 3 4
( −y) ⎤
) ( −y ) ) ( −y ) ) −( −y
)
2
5
) −( −x
) )
2 2 3
5 2 3 4
9 10 2 11 2 12 2
3 3 2 3
13 14 15 16
⎣ ⎦
17 18
ab = a b abc = a b c a = a a = a
n n n n n n n⋅m n m z n⋅m⋅z
2
5 5 5
( ) ) ( ) ) ( x )
⎡− −x
⎤ 19 ⎡ x ⎤
⎣ ⎦
− − 20 −
⎢⎣
⎥⎦
2 2 3 3 2
) ( ab ) ) ( a b ) ) ( − y ) ) ( − y )
⎛ 2⎞ ⎛ 2⎞
m
n
) − x ) − x ) ( x ) ) ( y )
4 m
2
m n m n
) ( ) ) ( ) ) ( 3 x 2 y) ) ( 3 x 2 y
x y x y a b a b )
2 3 x
2
x x x m n
) ( ) ) ( ) ) ( ) ) ( )
3
4
4
5
x
4
x
y
2 3 2 3
( x ) ) (( y ) ) ) (( x ) ) ) (( y ) )
6
2
5 4 3
2
5
3
7
2 3 3 5 2 4
3 2
( x ) ⋅( x ) ) ( y ) ⋅ (( y ) ) ) (( x ) ) ( x ) 44 ) (( a ) ) :(( a ) )
) ( )
) ( )
2 2 2
3 4 2 3
21 2 22 23 2 24 3
2 3
2 2
25 ⎜ ⎟ 26 ⎜ ⎟ 27 28
⎝ 3 ⎠ ⎝ 3 ⎠
2 3
29 30 31 32
33 2 34 3 35 2 36 2 3
37 38 39 40
41 42 43 :
⎛ 2 3⎞ 2 ⎛ 2 3⎞ 3 2 ⎛ 2 3⎞
3 2
) ab ( ab) ) xy ⋅( xy) ) xy ( xy)
x 2 m+ 1 2m−1 m−1
2 x 3x
3 3 2
) ( ab) ⋅( ab ) ) ( x) ) ( x) ⋅( x )
m− m+
5 3 2 3 3 2 4
) ( x ) ( x ) ) ( x ) + ( x ) ) ( a ) − ( a ) + ( a )
x
8 5
6 4
3 3 3
2 3 3
2 1 1
45 ⎜ ⎟ : 2 46 ⎜ ⎟ 4 46 ⎜ ⎟ : 4
⎝3 ⎠ ⎝2 ⎠ ⎝2
⎠
48 3 49 50
2 1 1 3 2 4 6 3
51 : 52 2 3 53 3 4
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
28.
Koristeći pravilo:
Izračunaj:
n
( ) ( )
n n n n n
a ⋅ b = a⋅ b ili a b c = abc
1) 2 ⋅5 2) 3 ⋅4 3) 4 ⋅6 4) 2 ⋅4
x x x x a a y y
n
3 4 4 5
5 5 ⎛1⎞ 3 ⎛3⎞ ⎛2⎞ ⎛1⎞
5
5) 3 ⋅2 6) ⎜ ⎟ ⋅9 7) ⎜ ⎟ ⋅⎜ ⎟ 8) ⎜ ⎟ ⋅4
⎝3⎠ ⎝2⎠ ⎝3⎠ ⎝2⎠
5 5 2 2 4
4 3 3
⎛2⎞ ⎛9⎞ ⎛ac ⎞ ⎛b ⎞ ⎛a ⎞ ⎛b
⎞ ⎛3⎞ ⎛8⎞
9) ⎜ ⎟ ⋅⎜ ⎟ 10) ⎜ ⎟ ⋅⎜ ⎟ 11) ⎜ ⎟ ⋅⎜ ⎟ 12) ⎜ ⎟ ⋅⎜ ⎟
⎝3⎠ ⎝2⎠ ⎝ b ⎠ ⎝c⎠ ⎝b⎠ ⎝a⎠ ⎝4⎠ ⎝3⎠
2 2 3 3 m+
1 m+
1
2 3
2
6
2
⎛ xy ⎞ ⎛ 4 ⎞ ⎛ xy ⎞ ⎛ x ⎞ ⎛ xy ⎞ ⎛ x ⎞ ⎛ x y ⎞ ⎛ z ⎞
13) ⎜ ⎟ ⋅⎜ 14) 15) 16)
2 ⎟ ⎜ ⎟ ⋅⎜ ⎟ ⎜ ⎟ ⋅⎜ ⎟ ⎜ ⋅
4 ⎟ ⎜ 3 2 ⎟
⎝ 2 ⎠ ⎝ x y⎠ ⎝ z ⎠ ⎝ yz⎠ ⎝ z ⎠ ⎝ yz⎠
⎝ z ⎠ ⎝ x y ⎠
2 3
m
6
m
5
2
⎛ x y ⎞ ⎛ z ⎞ ⎛ x−
y⎞
⎛ x −1
⎞
17) ⎜ 18)
4 ⎟ ⋅⎜ 3 2 ⎟ ⎜ ⎟ ⋅⎜ 2 2 ⎟
⎝ z ⎠ ⎝ x y ⎠ ⎝ x + 1 ⎠ ⎝ x + 2xy+
y ⎠
5 5
⎛ x+
y⎞
⋅ ⎜ ⎟
⎝ x−
y⎠
autor: Mladen Sraga
35

M-1- univerzalna zbirka potpuno riješenih zadataka
29.
Koristeći pravila:
Izračunaj:
2 −2 3 4
⎛2⎞ ⎛2⎞ ⎛1⎞ ⎛2⎞
1) ⎜ ⎟ 2) ⎜ ⎟ 3) ⎜ ⎟ 4) ⎜ ⎟
⎝3⎠ ⎝3⎠ ⎝2⎠ ⎝3⎠
−1 −1 −2 −3
⎛5⎞ ⎛ 2⎞ ⎛4⎞ ⎛2⎞
5) ⎜ ⎟ 6) ⎜1 ⎟ 7) ⎜ ⎟ 8) ⎜ ⎟
⎝7⎠ ⎝ 7⎠ ⎝5⎠ ⎝3⎠
−2 −2
n n
−n n
⎛a⎞ a ⎛a⎞ ⎛b⎞
b
⎜ ⎟ =
n ⎜ ⎟ = ⎜ ⎟ =
⎝b⎠ b ⎝b⎠ ⎝a⎠
a
⎛ 1⎞ ⎛ 2⎞ ⎛ 1⎞ ⎛3
⎞
9) ⎜2 ⎟ 10) ⎜1 ⎟ 11) ⎜1
⎟ 12) ⎜ ⎟
⎝ 3⎠ ⎝ 7⎠ ⎝ 2⎠ ⎝4
⎠
n
n
−3 −2
2 2 3 4
⎛ x⎞ ⎛ x⎞ ⎛ x⎞ ⎛ x⎞
13) ⎜ ⎟ 14) ⎜− ⎟ 15) ⎜− ⎟ 16) ⎜−
⎟
⎝ y⎠ ⎝ y⎠ ⎝ y⎠ ⎝ y⎠
2 2 3 2
⎛ 1⎞ ⎛ 2⎞ ⎛ 2⎞ ⎛ 4⎞
17) ⎜1 ⎟ 18) ⎜−2 ⎟ 19) ⎜−2 ⎟ 20) ⎜3
⎟
⎝ 2⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 5⎠
2
4
3
3
3
2
2 3
2
⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x y ⎞
21) ⎜ 22) 23) 24)
3 ⎟ ⎜−
4 ⎟ ⎜−
4 ⎟ ⎜ 4 ⎟
⎝ y ⎠ ⎝ y ⎠ ⎝ y ⎠ ⎝ z ⎠
25)
⎛ x y
⎜ 6
⎝ z
2 5
⎞
⎟
⎠
−2
4
3
−2 3
3
4 −2
2
⎛ 2x ⎞ ⎛ x y ⎞ ⎛2x y ⎞
26) ⎜ 27) 28)
2 ⎟ ⎜ −3 −4 ⎟ ⎜ 3 ⎟
⎝3yz⎠ ⎝2 z ⎠ ⎝ 5z
⎠
4 −2 −2 4 −2
3 −1 −2 3
⎛2x y ⎞ ⎛2x y ⎞ ⎛ x⎞ ⎛ x⎞ ⎛ y⎞
⎜ ⋅
3 ⎟ ⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
29) 30) 31) 32)
⎝ 5z
⎠ ⎝ 5z
⎠
⎝ y⎠ ⎝ y⎠
⎝ x⎠
−3 −2
2 5
−3 2
3
2 2
⎛2x−
3⎞
⎛ 1 ⎞
⎛a b ⎞ ⎛a ⎞ ⎛ y ⎞ ⎛ y ⎞
33) ⎜ ⎟ 34) ⎜ ⎟ 35) ⎜ 36) ⋅
3 ⎟ ⎜ 3 ⎟ ⎜ 3 ⎟ : ⎜ 4 ⎟
⎝2x+ 3⎠ ⎝ x−
y⎠ ⎝ c ⎠ ⎝ y ⎠ ⎝a
⎠ ⎝a
⎠
2
−2 −2
−3 −2
⎛3⎞ 1 ⎛3⎞
37) ⎜ ⎟ ⋅ 2 + 2 38) ⋅⎜ ⎟
⎝2⎠ 8 ⎝4⎠
−1 −2 −3 −2 −2
−2
−3
⎛2⎞ ⎛1⎞ ⎛1⎞ ⎡⎛3⎞ 2 ⎤ ⎛3⎞
39) ⎜ ⎟ + ⎜ ⎟ −⎜ ⎟ 40) ⎢⎜ ⎟ ⋅ ⋅
−3
⎥ ⎜ ⎟
⎝3⎠ ⎝3⎠ ⎝2⎠ ⎢⎣⎝4⎠ 3 ⎥⎦
⎝2⎠
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© M.I.M-Sraga centar za poduku
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
30.
0 1 −1
Koristeći pravila: 1 , , ,
Izračunaj:
1 −n
1
a = a = a a = a =
n
a
a
2 4
0
0 0
⎛ x y ⎞
0 0
1) 2 2) x 3) ⎜ ⎟ 4) x + y
⎝ z ⎠
( )
0 0
0 0
5) 3 6) −3 7) −3 8) −3x
0 0 0 0 0 0
( + ) ( − ) − + − ( ) + (
9) x y 10) 2x 7y 11) x 2y 7z 12) x 2y 7z
m
( ) ( ) ( )
0 0 0
2 0 0
13) 14) 15) 16) 4
x ⋅ x x x x
−1 −1 −1
17) 5 18) 19) 0,2 20)
( − )
x x y
−2 −3 −2
21) 3 22) x
23) 0, 3 24) 2x−
5
−1
25) 4 26) 27) 28)
−1 −1
( −5)
( − )
−1 −1 −2 −4
−1
( )
1 1 ⎛2⎞
⎜ ⎟
4 4
⎝3
⎠
2 2 3 2a
29) 30) 31) 32)
5 2
b
33) 3 ⋅3 ⋅3 34) 2 ⋅2 ⋅2 ⋅2 35) 5 ⋅5
5x−4 4x−3 7− 9x 2x+ 1 3 2−3x x−6 2x−3 2−2x
−1
−5
)
0 0
2m−n n−2m
0 0 0 1 ⎛1
⎞
36) ( x+ y) ⋅ ( x+ y)
37) 27 ⋅x ⋅y
⋅2
⋅ ⎜ ⎟
⎝3
⎠
1 −1 2
m+ n
2
m+
2n 2
n
0 −1 ⎛1⎞ ⎛1⎞
−3
⎛ab⎞ ⎛ab⎞ ⎛ab⎞
38) y ⋅2
⋅ ⋅ ⋅2 39) ⎜ :
3 ⎟ ⎜ ⋅
3 ⎟ ⎜ 3 ⎟
⎜ ⎟ ⎜ ⎟
⎝2⎠ ⎝5⎠ ⎝ c ⎠ ⎝ c ⎠ ⎝ c ⎠
2
2 1
−2
−3 −1 −
− −
−2 −1
−2
⎛ 2 −2 ⎞ ⎛2 ⋅3
⎞
⎛2 −3
⎞
40) ⎜ 41) 42)
−2 −4 ⎟ ⎜ −1 −2 ⎟
⎜ −1 −2
⎟
⎝2 + 2 ⎠ ⎝2 ⋅ 3 ⎠
⎝2 + 3 ⎠
43) 3 ⋅2 − 3 44) 3 + 2 − 3 45) 2 + 2 −2
0 0 1 0 0 1 0 1 −2
− − 2 − 1 −
2 48) 3 2 −
+ −2
3
−2 −3 −3
−
46) 2 2 47) 2 2
−2
2
2 ⋅3 2 ⋅4
49) 50) 51)
3 4 3 ⋅6
−1
3 −2 2 −1
−3 2 2 −1
−1 −2 −1
a a a b
52) 53) 54)
−1 −2 −1
b b c
ab x bc x b c
55) 56) 57)
c d y c d y c d
2 −3 −1 2 −1 −2 2
−1 2 −2 3 −1 2 −3 −1
autor: Mladen Sraga
37

M-1- univerzalna zbirka potpuno riješenih zadataka
31.
a)
Zapiši u obliku potencija s bazom 2:
2 3 2
( ) ( )
1) 4⋅32⋅16 2) 4⋅8⋅16 3) 2 ⋅4 ⋅8
3
b) Zapiši u obliku potencija s bazom 3:
4 2
( ⋅ ⋅ ) ( ⋅ ⋅ ) ( ⋅ ⋅ )
1) 3 9 27 2) 9 81 3 3) 3 9 27
5 3 2 3
c)
Pojednostavni i rezultati zapiši kao potenciju:
1) 2 + 2 2) 3⋅2
3 3 5
4) 53 ⋅ −33 ⋅ 5) 35 ⋅ + 25 ⋅
2 2 3 3
+ 2 3) 6⋅2
5 4
−2⋅4
4
32.
⎛9 6 5⎞ ⎛3 4 3⎞ ⎛4 7 3⎞ ⎛ 2 5 2⎞
1) ⎜ x y ⎟: ⎜ x y ⎟ 2) ⎜ x y ⎟:
⎜ x y ⎟
⎝2 ⎠ ⎝2 ⎠ ⎝3 ⎠ ⎝15
⎠
⎛ 5 8 3⎞ ⎛ 1 3 3⎞ ⎛25 9 6 5⎞ ⎛5
6 3 4⎞
3) ⎜−
xy⎟: ⎜−
xy⎟ 4) ⎜ abc⎟:
⎜ abc⎟
⎝ 7 ⎠ ⎝ 7 ⎠ ⎝49 ⎠ ⎝7
⎠
33.
16⋅2 4 16
1) 2)
8 32
x+ 1 x− 1 x+
2
x−1 1−x
3)
x
9 ⋅27
x
81
x−1
4)
25 ⋅5
125
x+ 1 x+
5
+ 1 3−x
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34.
−2 −2 4
2
−2 2
4
2
⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞ ⎛ x ⎞
1) ⎜ 2)
−3 ⎟ ⋅⎜ 2 ⎟ ⎜ ⋅
−3 ⎟ ⎜ 2 ⎟
⎝ y ⎠ ⎝ y ⎠ ⎝ y ⎠ ⎝ y ⎠
3
3
2
2
3
3
2
−2
⎛ 3x ⎞ ⎛4x ⎞ ⎛ 3x ⎞ ⎛3x
⎞
3) ⎜ 4)
4 ⎟ ⋅⎜ 3 ⎟ ⎜ 4 ⎟ ⋅⎜ 3 ⎟
⎝2y ⎠ ⎝3y ⎠ ⎝2y ⎠ ⎝4y
⎠
7 2
−2 2
2
2 3
4
6 2
⎛ 5ab ⎞ ⎛ 5a ⎞ ⎛ 3xy ⎞ ⎛3xy
5) ⎜− 6)
3 ⎟ ⋅⎜ 3 2 ⎟ ⎜− 4 ⎟ ⋅⎜
2
⎝ 6c ⎠ ⎝3b c ⎠ ⎝ 5z ⎠ ⎝ 5z
⎞
⎟
⎠
−2
35.
1) ( x ) 2) ( 2x
) 3) ( a b )
2 3 4 2 2 5 2
3
( ) ( ) (
2 3 4 2 2 3 2 2 3 2
( ab − c ) ( a − bc )( a + bc )
3 2 3 2 3 2 2
( a + b ) ( x y + xy )
4) 4ab 5) 2x y 6) 3x + y
2 2 2 3 2 3 2
7) 4 8)
9) 2 10) 2 3
35. b)
1) ( x ) 2) ( y ) 3) ( a b )
2 3 4 2 2 3 2
( ab ) ab ⋅ ab ( x + y )
2 3 3 5 2 2 3 3 5 2 3 3 5
( ab −ab) ( ab −ab) ⋅ ( ab + ab)
3 2 2 4 3
( xy + xy)
4) 5) 6)
3 5 2 2 3 3 5 3 4 2
7) 8)
9)
)
autor: Mladen Sraga
39

M-1- univerzalna zbirka potpuno riješenih zadataka
36 . I sada primjenom svih pravila za potencije riješite dopunske zadatke
POTENCIJE
a
a
n
m
36.
m
(( ) )
( )
ca ⋅ ± da ⋅ = c± d⋅a
n n n
n n
n m n+
m n n n n ⎛a⎞
a
1
a ⋅ a = a ( abc)
= a b c = a = a
n
⎜ ⎟
⎝b
⎠
−n n n
m
n m n−m n n⋅m
⎛a⎞ ⎛b⎞
b
−1
:
( )
n
1
a a = a a = a ⎜ ⎟ = ⎜ ⎟ = a =
⎝b⎠ ⎝a⎠
a
a
z
n m n−m n n⋅m⋅z 0
−n
: 1
= a a = a a = a a = a =
Izračunaj:
1 2
1) 2x ⋅ x + 5x⋅x 2) x ⋅x − x : x 3) a ⋅a
3 5
b
( )
2 3 4 7 2 17 8 3 4 2
( a ⋅a ⋅a ) ( a a ) ( a a )
3 2 2 4
( ) ( ) ( ) ( )
4) 5) : 6) :
2 4 5 3 8 4 2 7 5 6
7) : 8) : 9) 25 :5
2 2 7 3 9 4 7 2
a a x x x y x y
2m+
3
5 6⎞
2 3 4 4 3 4 2
⎛ 1 1
10) ⎜ x⋅7 x: x ⎟
11) 2abc ⋅3abc⋅4abc
⎝47 7 ⎠
4 3 2 5 6 5 4 7 2
12) a ⋅ a + 3a ⋅ a + 4a⋅a 13) 2x ⋅ x + 5x ⋅x
−3x
14) 7x ⋅ x + 2 x : x −3x ⋅y 15) 3x y z ⋅2x y z
5 3 10 2 3 5 4 2b m m−2 3b
2
16) x y : x y 17) x ⋅ x : x
m+ 3 4n− 1 m+ 2 1− 2n m+ n−3 2m− 3n+ 2 m+ 2n−1
x
x
3x 2 4x 3x 2 4x
( a ) ⋅( a ) ⋅( a ) ( a ) ⋅( a ) ( a )
x
3x
2 4x
7 2 3 3
( a ) + ( a ) − ( a ) ( x y x y )
7 2 3 3 2 3 2 4
( xy ) ( xy) ( ab) ⋅( ab )
( x )
4 6 3 4 6 3
18) 19) :
4 6 3 4
20) 3 2 21) :
2 3 4 3
22) : 23)
24) 1
0 0
0 0 0
+ + 3 − 4x 25) ( x + 2) ⋅ ( x+
2)
0 0 2 3 0 0
( x y −xy ) ⋅( xy − x y ) x + y −( xy)
0 0
26) 27) 2 3 5
⋅ y
3 6
0 0
2 0 1
3xy 3+
x x y 3
0 0
0
2 − 2
28) 29) 30)
x y y xy
( )
−5 2 −4 2 −3 −2
31) ⋅
32) : 33) ⋅ :
x x x x
34) x : x 35) 12 x :3x
8 −3 −9 −2
1
n
a
x x x
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tel-01-4578-431

POTENCIJE
M-1- rješenja i kompletni postupak
Novo MALA ŠKOLA MATEMATIKE –1 na
BESPLATNA video poduka i instrukcije
UČIMO ZAJEDNO
POTENCIJE
ALGEBARSKI IZRAZI
ALGEBARSKI RAZLOMCI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
autor: Mladen Sraga
41