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**** 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
puta
2 se ponavlja petputa pa u eksponent pišem
Koristimo pravila:
Uputa: Prebrojite koliko se puta ponavlja isti faktor i taj broj stavite u eksponent:
, , ...
1) 2 2 2 2 2
21. n
n
a a a a a a a a a a a a−
−
⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =
⋅ ⋅ ⋅ ⋅
( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
5
o 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
= →
⋅ ⋅ ⋅ ⋅ =
⋅ ⋅ =
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ =
⋅ ⋅ =
+ ⋅ + ⋅ + ⋅ + = +
⎛ ⎞ ⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
4
3 2
3
7)
8)
x xy 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 bc c c c
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠− ⋅ + ⋅ − ⋅ − ⋅ + =
= − ⋅ − ⋅ − ⋅ + ⋅ + = − ⋅ +
− − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )
( )( )
na eksponent
na
2 3
puta
2 3 4
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
22. 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=
www.mim-sraga.com 2 © M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
autor: Mladen Sraga
M-1- rješenja i kompletni postupakPOTENCIJE
2 3
puta
2
3
5
2
3
Koristimo pravila: , , ...
1 1 1 19)3 3 3 9
3
3 310) ⎛ ⎞ = ⋅⎜ ⎟3 3 27
5 5 5 5 125
2 2 2 2 2 2 3211)3 3 3 3 3 3 243
2 22 2 2 412)3 3 3 3 3 9
213)3
n
n
a a a a a a a a a a a a−
= ⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅
⎛ ⎞ = ⋅ =⎜ ⎟⎝ ⎠
⋅ =⎝ ⎠
⎛ ⎞ = ⋅ ⋅ ⋅ ⋅ =⎜ ⎟⎝ ⎠
⋅⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − ⋅ − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞− = −⎜ ⎟⎝ ⎠
22.
4
2
3
4
2 2 22 2 2 83 3 3 3 3 3 27
3 3 3 3 3 8114)4 4 4 4 4 256
4 44 4 4 1615)5 5 5 5 5 25
4 4 44 4 4 4 6416)5 5 5 5 5 5 5 125
4 417)5 5
⋅ ⋅⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ − ⋅ − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ = ⋅ ⋅ ⋅ =⎜ ⎟⎝ ⎠
⋅⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − ⋅ − = + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⋅ ⋅⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − ⋅ − ⋅ − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⋅ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞− = −⎜ ⎟⎝ ⎠
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( )
2
2
3
3
2
4 4 4 44 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⋅ − = + ⋅ =
M-1- univerzalna zbirka potpuno riješenih zadataka
( ) ( ) ( ) ( ) ( )
) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 3
puta
3
4
2 3 4 5
Prebrojimo mi
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
22. n
n
a a a a a a a a a a a a−
= ⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅
− = − ⋅ − ⋅ − = − ⋅ ⋅ =
− = − ⋅ − ⋅ − ⋅ − = + ⋅ ⋅ ⋅ =
− + − + − + − = + ⋅ − ⋅ + ⋅ −
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
nuse -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 0,00102
=
− + − = − ⋅ − ⋅ − + − ⋅ − ⋅ − ⋅ − ⋅ − − − = −
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − = ⋅ + − ⋅ − ⋅ − = − = = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎡ ⎤ ⎡ ⎤− + − = − ⋅ − + − ⋅ − ⋅ − =⎣ ⎦⎣ ⎦
= − = = 4
www.mim-sraga.com 4 © M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
M-1- rješenja i kompletni postupakPOTENCIJE
( ) ( )( ) ( )
( )
( )
je potpuno isti izraz...
ili taj isti zadatak na malo duži ali sigur
Koristimo pravila:
51) 2 3 2 3 5 5 ili
5
2) 2 4 1 2 4 7
23. 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
⋅ + ⋅ = + ⋅ ⋅ − ⋅ = − ⋅⋅ + ⋅ = + ⋅ ⋅ − ⋅ = − ⋅
⋅ ⎫⎪+ = + ⋅ = ⋅ = ⎬⎪⎭
+ + = + + ⋅ =
( )niji 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
+ + = + + = + + ⋅ = =
=
+ ( )
( )
( )
( )
ili što nije točno!!!
Jednom zauvjek treba zapamtiti pa to u zadatku treba izgledati ovako:
ili ta
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 x
a a a a a
y y y y y y y
+ = + + = + + ⋅ =
=
+ + = + + = + + ⋅ =
− = − ⋅ = ⋅ =
− + − = − + − ⋅ = ⋅ =
( )
( ) ( )
( )
( )
j 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 a
− + − = − + − = − + − ⋅ =
+ − + + − =
= − + + + − = − + ⋅ + + − ⋅ = ⋅ + ⋅ = +
+ + = + + ⋅ =
− + = − + ⋅ = − ⋅
( )
( ) ( ) ( )
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
b 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
= −
+ − = + − ⋅ = − ⋅ = −
+ − − − + − + =
= − − + − + − + =
= − − ⋅ + − ⋅ + − + ⋅ =
⎫= − ⋅ + ⋅ + ⋅ = ⎪⎬
= − + + ⎪⎭
5autor: Mladen Sraga
www.mim-sraga.com 6
M-1- univerzalna zbirka potpuno riješenih zadataka
© M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
( ) ( )( ) ( )
( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
3 3 3 3 3 3
2 2 2 2 2
ili taj isti za
Koristimo pravila:
10) 7 2 4 7 2 4 9 9
11) 2 3 2 3 1
12) 3 4 3 4 1 6
23.
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
datak 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
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:
3 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
− ⋅ − + =
= − ⋅ − + =
= − + −
+ − − + − + + − = − + ⋅ + − = − + −
+ − − + − + + − = + − − + − + + − =
= − + ⋅ + − =
= − + −
autor: Mladen Sraga
7
M-1- rješenja i kompletni postupakPOTENCIJE
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 1
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
24.n
n m n m n m n m n m n mm
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
+ − −
+
+
+
− − − + − + − − −
+ + +
+
⋅ = = = =
⋅ = =
⋅ = =
⋅ = =
⋅ = = =
⋅ ⋅ ⋅ = =
⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ = 1
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
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
x y x y x x y x x x y y x ya a a
x x x x x x x x
+ + + + + + + + + +
+ +
= = =
− ⋅ ⋅ ⋅ ⋅ ⋅ = − ⋅ ⋅ ⋅ ⋅ ⋅ = − ⋅ = −
M-1- univerzalna zbirka potpuno riješenih zadataka
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
24.n
n m n m n m n m n m n mm
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...
4 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
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 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 1
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
0 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
+ + + +
+ − + − + + − + + − −
+
+
⋅ =
⋅⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ = ⋅ =
⋅
− ⋅ − = − ⋅ − = − = −
+ ⋅ + = + = +
www.mim-sraga.com 8 © M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
M-1- rješenja i kompletni postupakPOTENCIJE
( ) ( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
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)
24.n
n m n m n m n m n m n mm
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
x x x y y y y
x yabc
+ + − − − − +
−
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
Novo MALA ŠKOLA MATEMATIKE –1 na
BESPLATNA video poduka i instrukcije
UČIMO ZAJEDNO POTENCIJE
ALGEBARSKI IZRAZIALGEBARSKI RAZLOMCI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
9autor: Mladen Sraga
M-1- univerzalna zbirka potpuno riješenih zadataka
2 2 3 2 2 3
2 2 1 3
4 4
3 2 5 3 3 5 2 3
3 5 2 3
8 5
2 3 4 2 3 4
Koristimo pravila: : :
1) 2 3 2 366
2) 5 2 5 21010
2 9 2 93)3 4 3 4
2
25.n
n m n m n m n m n m n mm
aa a a a a a a a aa
x y x y x x y yx y
x y
x y x y x x y yx y
x y
a b ab a a b b
+ −
+ +
+ +
⋅ = = = =
⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ =
=
⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ =
=
⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ =
=3
3⋅
32⋅ 2 1 3 4
2 1 7
3 7
6 2 2 6 2 2
23232
5 9 5 94)27 5 27 5
5
a a b
a b
a b
a b a b a a b b
+
+
⋅ ⋅ ⋅ =⋅
= ⋅ ⋅
=
⎛ ⎞ ⎛ ⎞− ⋅ − = − ⋅ − ⋅ ⋅ ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= +3 9⋅
9⋅
5
( ) ( )
6 2 2 1
8 3
2 3 4 3 2 5 2 3 3 2 4 5
2 3 3 2 4 5
5 5 9
4 2 2 4 2 2 1
4 2 2 1
6 3 6 3
13
1 15) 4 42 2
422
1 16) 9 93 3
13 33
3 3
a b
a b
a b c a b c a a b b c c
a b c
a b c
x y x y x x y y
x y
x y x y
+ +
+ + +
+ +
⋅ ⋅ =
=
⋅ − = ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
= − ⋅ ⋅ ⋅ =
= −
⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ ⋅ ⋅ =
= ⋅ ⋅ =
−
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M-1- rješenja i kompletni postupakPOTENCIJE
Koristimo pravila: : :
Rješenja ZADATAKA 7), 8) , 9) , 10) - šaljemo mailom
25.n
n m n m n m n m n m n mm
aa a a a a a a a aa
+ −⋅ = = = = −
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 USA OBJAŠNJENJIMA I PO
NALAZE SE Nlink: http://www.mim-sraga.
MALA Š
DODATNE U
ALGEALGEB
autor: Mladen Sraga
Z OVE ZADATKE iz POTENCIJA STUPCIMA RJEŠEVANJA ZADATAKA A NAŠOJ WEB-STRANICI com/Mala-skola-matematike--video.htm
KOLA MATEMATIKE
PUTE UZ OVE ZADATK POTENCIJEBARSKI IZRAZI
ARSKI RAZLOMCI
11
www.mim-sraga.com 12
M-1- univerzalna zbirka potpuno riješenih zadataka
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tel-01-4578-431
( )
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 mo
:
1) 2 : 2 2 2 2 2 2 2 2 162) 13 : 13 13 13 1693) 3 : 3 3 3
4) 5 : 5 5 5 5 5
26. 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
ra 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 ne,
5 : 5 5 5 5m m m m m m m− − − − − − − − −= = = →
↓
4 2 4 2 2
6 2 6 2 4
7 2 3 7 2
to se događa zbog toga što radite napamet... tj. preskaćete korake...
Preporuka: čim imamo višečlane eksponente koristite zagrade i ne preskačite korake...
5) :
6) :7) : :
x x x x
x x x xx x x x
−
−
−
= =
= =
= 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
8
8) :19) : :
10) :
11) :
12) : :13) :14) :
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 aa a
−
+ −
− −
⋅ − −−
− ⋅ −−
⋅ − ⋅ − ⋅ − −− − −
−
=
⋅ = =
= = = =
= = = =
= = = =
= = = =
= =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 3 3 2 12 112 2 2 2
ili
U slijedećim zadacima koristimo pravilo:
:115) : : : :
16) :
:
17) :
18) :
19) :
nn m n m
m
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 x xx
− −
− − −
− +
−
−
−
−−
= =
= = = =
⋅ = =
= =
= = =
= = =
= = = =
71 7 17 7 3 1 4 21 4 1743 4 34 12 12 12
13
20) :x x x x x x xx
⋅ − ⋅ −−
= = = = =
M-1- rješenja i kompletni postupakPOTENCIJE
U slijedećim zadacima koristimo pravilo: :
Rješenja ZADATAKA od 21) do 31) šaljemo u besplatnom PDF dokumentu samona zahtjev upućem mailom !
26.n
n m n mm
a a a aa
−= =
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 USA OBJAŠNJENJIMA I PO
NALAZE SE Nlink: http://www.mim-sraga.
MALA Š
DODATNE U
ALGEALGEB
autor: Mladen Sraga
Z OVE ZADATKE iz POTENCIJA STUPCIMA RJEŠEVANJA ZADATAKA A NAŠOJ WEB-STRANICI com/Mala-skola-matematike--video.htm
KOLA MATEMATIKE
PUTE UZ OVE ZADATK POTENCIJEBARSKI IZRAZI
ARSKI RAZLOMCI
13
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M-1- univerzalna zbirka potpuno riješenih zadataka
© M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
( ) ( ) ( ) ( )( )( )
Koristimo pravila:
vidi zadatke: 7),13),14),15),16),17),18),19),23),24)...
Pojavio nam se jedan problem a taj je: 22.
ogledajmo opet u kako smozadata k
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
neparni
3
Imamo neparan broj "minusa" pa će i umnožak
1 1Zaključak:
to tamo rješili:
4) 1 1 1 1
5) 1 1 1 1 1
7) 2 2 2 2 8↓
− =
−
⎫− = − ⋅ − =⎪⎪− = − ⋅ − ⋅ − = − ⎬
⎭
− = − ⋅ − ⋅ − = −
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
4 4
Imamo paran broj "minusa" pa je produkt pozitivan broj
53
53 puta bi to trebali
biti negativno tj. imati će predznak minus
8)
a se javlja ovakav problem: ...
x x x x x x
x x x x
− = − ⋅ − ⋅ − ⋅ − =
− = − ⋅ − ⋅ ⋅ −
( )( )
napisatia to je previše pisanja pa ćemo rađe gledati eksponent
53 53
54alje logički je pitanje koliko je eksponent je 54, 54 je paran b
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 neparni neparnu
roj pa je:
Zaključak:
jer je eksponent 55, a 55 je neparan broj
jer je eksponent 56, a 56 je paran broj
ilin nx x
x x
x x
x x
x x
x x
x
− =
− = −
− =
− = − =
− = =
− =
−
27.
P 22.
( ) ( ) ( ) ( ) ( )( )4 1 1
6) 1 1 1 1 1 1− =⎪
− = − ⋅ − ⋅ − ⋅ − = ⎪
Sad
53 je nepar d
( )
( )( )
1 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 br
22
n 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 I
oj
Postoji 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
−
− = − ⋅
− = − ⋅ = − ⋅ = − ⋅ = −
− = − ⋅ = − ⋅ = − ⋅ = −
− = − ⋅ = − ⋅ = ⋅ = I način
autor: Mladen Sraga
M-1- rješenja i kompletni postupakPOTENCIJE
( ) ( ) ( ) ( )( )
( )
) ( )) ( )
) ( ) ( )
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
27.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
a
⋅ ⋅ ⋅
⋅ ⋅
= = = =
=
= ⋅ = ⋅ =
= ⋅ = ⋅ =
⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠
( ) ( )
)
⎝ ⎠
15
( )
) ( ) ( )
) ( )
) ( ) ( )
3 3 32 3 2 3 2 3 3 6
3
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...
2 2 2 843 3 273
1 1 1 152 2 162
6
7
mn n n n n n mbc 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
⋅
⋅
⋅ ⋅
⋅
⋅
= =
⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= =
− = = =
) ( ) ( ) ( ) ( )) ( )
) ( )
( )
3
) ( ) ( ) ( )
) ( ) ( ) ( )
) ( ) ( )
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
43 4 4 3
I 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 n ma 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
⋅⋅
⋅
⋅
⋅ ⋅
⋅ ⋅
⎫⎪⎪⎪⎪⎪⎪⎪ =⎬
− = − ⋅ = − ⋅ = ⋅ = ⎪⎪⎪= =⎪⎪⎪= = ⎪⎭
= ⋅ ⋅ = ⋅ ⋅ =
= ⋅ ⋅ = ⋅ ⋅ =
= ⋅ ( )
) ( ) ( )) ( ) ( ) ( ) ( )
4 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
⋅ ⋅
⋅
⋅
⋅ = ⋅ ⋅ =
⎡ ⎤− = = =⎣ ⎦
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤− = − ⋅ = − ⋅ = ⋅ = = =⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
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M-1- univerzalna zbirka potpuno riješenih zadataka
© M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
27.
( )( )
( ) ( )
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
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
14 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
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
17 1 1 1
18
18 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 naj
1 1 1 1
18
x x x x
x x x x x
⋅
⋅
⋅ − ⋅ = − ⋅ = ⋅ =
⎡ ⎤ ⎡ ⎤− − = + = = =⎣ ⎦⎣ ⎦
bolji....
↓
autor: Mladen Sraga
17
M-1- rješenja i kompletni postupakPOTENCIJE
1 1
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 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 neparnoristimo 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
− =
− = −
− ⋅ = − ⋅ = ⋅ =
− = − − = − ⋅ = − ⋅ = − ⋅ = −
− = − ⋅ = − ⋅ = ⋅ =
− = − ⋅ − =
− = − ⋅ − ⋅ − = −
27.
) ( ) ( ) ( ) ( ) ( )( )
) ( ) ( )( ) ( )
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:
19
19 20
21
22 23
( )( )
( )
) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 2
3u 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 27
y y y y⋅
− = =
− = − = −
− = − = −
↑
− = − ⋅ = − ⋅ = −
↓
− = − ⋅ = − ⋅ = − ⋅ = − − = − ⋅ − ⋅ − = −
M-1- univerzalna zbirka potpuno riješenih zadataka
27. R ješenja 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
poduku , Kraljevečki ogranak 24 , Zagreb
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tel-01-4578-431
M-1- rješenja i kompletni postupakPOTENCIJE
( ) ( )
( ) ( )( ) ( )( )55 5 5
3 3 33
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) 3 2 3 5 15
3 31 1 96) 9 93 3 3
28. n nn n n n n
x xx x x x x x
a ya a a y y y
a b a b a b c abc⋅ = ⋅ =
⋅ = ⋅ = ⋅ = ⋅ =
⋅ = ⋅ = ⋅ = ⋅ =
⋅ = ⋅ =
⋅⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3
3
3
14 4 4
3 27
3 2 3 2 37)2 3 2 3
⎛ ⎞= =⎜ ⎟
⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 2
2⋅
1
3
4 44
1
5 5 55 5
5 5 5
1 1 11
1 1 48) 4 4 2 322 2 2
2 9 2 9 29)3 2 3 2
⎛ ⎞ ⎛ ⎞⎜ ⎟ = = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3
3⋅
32⋅
5
5
22 22
4 4 44
33 3 33
2 22
2 2
3 243
10)
11) 1 1
4 23 8 3 8 312) 2 84 3 4 3 4 3
2 24 413)2 2 2
a cac b b ab c b c
a b a bb a b a
x yxy xyx y x y
⎛ ⎞= =⎜ ⎟
⎝ ⎠
⋅⎛ ⎞⎛ ⎞ ⎛ ⎞⋅ = ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⋅⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ = ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⋅ ⋅⎛ ⎞ ⋅ = ⋅ = ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( )( )
2 2
2
2 2
333 23 33 2 2 3 6
2 3 2 3 62
11
možemo ostaviti u ovom obliku
ili dalje:
nakon kraćenja
2
2 4
14)
15)mm
x x y x
x x
xx xxy x xy x x x xz yz z yz z z z z zz
xy xz yz
⋅
⋅
++
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⋅ ⋅ ⎝ ⎠⎝ ⎠
= =
⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ ⋅⎛ ⎞ ⋅ = ⋅ = = = = = =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⋅⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞⎛ ⎞ ⋅⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
( )( )
( )
( )
m+1
111 2 2 12 2 2
2 1 2 22 12
nakon kraćenja
mmm m m
m mm
xy xz yz
xx x x x xz z z zzz
+++ ⋅ + +
+ +⋅ +
⎛ ⎞= ⋅ = =⎜ ⎟⎝ ⎠
⎛ ⎞⎛ ⎞⋅= = = = =⎜ ⎟⎜ ⎟⋅⎝ ⎠ ⎝ ⎠
19autor: Mladen Sraga
www.mim-sraga.com 20
M-1- univerzalna zbirka potpuno riješenih zadataka
© M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
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( ) ( )Koristimo pravilo:
Rješenja ovog ZADATKA od 16) do 18) šaljemo u besplatnom PDF dokumentu samona zahtjev upućem mailom !
28. n nn n n n na b a b a b c abc⋅ = ⋅ =
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
21autor: Mladen Sraga
M-1- rješenja i kompletni postupakPOTENCIJE
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
−
−
⎛ ⎞ =⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛⎝
29. 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 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
mješoviti broj treba prvo pretvoriti u razlomak...
primjenili smo pravilo: 75
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
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 7 410) 17 7 7 7 9 9
−− − −
−− − −
→
=
⋅ + +⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
⋅ + +⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
981
www.mim-sraga.com 22
M-1- univerzalna zbirka potpuno riješenih zadataka
© M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
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
29.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 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
xy
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
⎛ ⎞ ⎛ ⎞− ⋅ = − ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ + +⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠
M-1- rješenja i kompletni postupakPOTENCIJE
23
Koristimo pravila:
ješenja ovog ZADATKA od 18) do 27) šaljemo u besplatnom PDF dokumentu samona zahtjev upućem mailom !
ješenja ovog ZADATKA od 28) pa do 4
n n nn n
n n
a a a b bb b ab a
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0) ogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
29.
R
Rm
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
M-1- univerzalna zbirka potpuno riješenih zadataka
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 du
Koristimo pravila:1 11 , , ,
1) 2 1 1
2) 1
3) 1
3)
30. n
na a a a aa a
a
x
x yz
− −= = = =
= =
=
⎛ ⎞ ⎧= ⎨⎜ ⎟
⎩⎝ ⎠
( ) ( )
( )
( )
0 00 2 42 4
0
0 0
0
0 0
0 0 0
0
0
žim postupkom:
po pravilu:
bez obzira što je u zagradi ako je to na nultu sve
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
je 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)
www.mim-sraga.com 24 © M.I.M-Sraga centar za podukumim-sraga@zg.htnet.hr
tel-01-4578-431
autor: Mladen Sraga
M-1- rješenja i kompletni postupakPOTENCIJE
( )
( )
( )
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
− −
− −
−
−
= = = =
⋅ = ⋅ =
= =
=
=
= =
=
30.
( )( )
1 1
1
1
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 93
122)
3 10 10 10023) 0,310 3 93
24) 2
nn
nn
x y ax y ax y
aa
xx
x
− −
− −
− −
−
−−
⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
− = = =−−
= = =
=
⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−( )( )
5
5
1
152 5
125) 44
x−
−
=−
=
25
M-1- univerzalna zbirka potpuno riješenih zadataka
0 1 1Koristimo pravila:1 11 , , ,
Rješenja ovog ZADATKA od 26) do 41) šaljemo u besplatnom PDF dokumentu samona zahtjev upućem mailom !
ešenja ovog ZADATKA od 42) pa do 57) ogu se nab
nna a a a a
a a− −= = = =
aviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
30.
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
Rjm
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autor: Mladen Sraga
27
M-1- rješenja i kompletni postupakPOTENCIJE
4 32 16 2 2 2 2 2
2) 4 8 16 2 2 2 2 2 2 2
Preostala rješenja ovog ZADATKA mogu se nabaviti samo u štampanoj ( prodajnoj ) varijant
+ +
+ + ⋅
⋅ ⋅ = ⋅ ⋅ = =
⋅ ⋅ = ⋅ ⋅ = = = =
31. a 1)
( ) ( ) ( ) ( )
2 5 4 2 5 4 11
2 2 3 4 2 2 3 4 2 9 2 9 2 18
) Zapiši u obliku potencija s bazom 2:
6 5
6 5 4 3 6 4 5 3 6 4 5 3
4 3
6 4 5 3
e ove zbirke !
99 3 9 3 9 221) : : : :
32 2 2 2 2 32
3 39 22 3
32.x y
x y x y x x y y x yx y
x y
− −
− −
⎛ ⎞ ⎛ ⎞ = = ⋅ ⋅ = ⋅ ⋅ ⋅ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅= ⋅ ⋅ ⋅ =
22
⋅3
( ) ( )
( )
2 2 2 2
1 4 1 4 1 5
1 3 33 13 1
5 3 35 3 3 5 3 3
3
Preostala rješenja ovog ZADATKA mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
16 2 2 2 2 21)8 22 2
2 :2 2 2
33.x x x x
x xxx
x xx x x x
x y x y
+ + + + +
− −−−
+ − −+ − + − +
⋅ ⋅ =
⋅ ⋅= = = =
= = = = 3 5 3 2 8 8 22 2 2
Preostala rješenja ovog ZADATKA mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
x x x x− + + − + −= =
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M-1- univerzalna zbirka potpuno riješenih zadataka
( )( )
( )( )
2 2 2 24 2 2 22 4 3 2 4 2 8 4 8 4 8 12
3 2 2 3 2 2 4 6 4 6 4 102 2 3 21)
34.x xx x y x x x x x x x
y y x y y y y y y yy y
− −− ⋅ +
− ⋅ +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ = ⋅ = ⋅ = ⋅ = ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Preostala rješenja ovog ZADATKA mogu se nabaviti samo u štampanoj ( prodajnoj ) varijant
( ) ( )
( ) ( ) ( )( )
2 2 2 2 2 2 2 2 2 2 4
2 3 2 2 2 2 3 3 2
2 2 2 2 3 3 2
2 2 2 3 6
4 2 3 6
4) 4 4 16 16
6) 3 3 2 3
3 6
9 69 6
35.
ab a b a b a b
x y x x y y
x x y y
x x y yx x y y
⋅
⋅
⋅
= ⋅ ⋅ = ⋅ ⋅ =
+ = + ⋅ ⋅ + =
= ⋅ + + =
= ⋅ + + =
= + +
e ove zbirke !
Preostala rješenja ovog ZADATKA mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbirke !
DODATNE UPUTE USA OBJAŠNJENJIMA I PO
NALAZE SE Nlink: http://www.mim-sraga.
MALA Š
DODATNE U
ALGEALGEB
Z OVE ZADATKE iz POTENCIJA STUPCIMA RJEŠEVANJA ZADATAKA A NAŠOJ WEB-STRANICI com/Mala-skola-matematike--video.htm
KOLA MATEMATIKE
PUTE UZ OVE ZADATK POTENCIJEBARSKI IZRAZI
ARSKI RAZLOMCI
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tel-01-4578-431
M-1- rješenja i kompletni postupakPOTENCIJE
( ) ( ) ( ) ( ) ( ) ( )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
8 6 12 12 8 6 12 12 14 24
23)
Preostala rješenja ovog ZADATKA mogu se nabaviti samo u štampanoj ( prodajnoj ) varijante ove zbi
36.a b a b a b a b a b a b a b a b
a a b b a b a b
⋅ ⋅ ⋅ ⋅
+ +
⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅
= ⋅ ⋅ ⋅ = ⋅ =
rke !
=
Novo MALA ŠKOLA MATEMATIKE –1 na
BESPLATNA video poduka i instrukcije
UČIMO ZAJEDNO POTENCIJE
ALGEBARSKI IZRAZIALGEBARSKI RAZLOMCI
link: http://www.mim-sraga.com/Mala-skola-matematike--video.htm
KAKO NARUČITI KOMPLETNU ŠTAMPANU VARIJANTU OVE ZBIRKE ?
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29autor: Mladen Sraga
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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31
M-1- rješenja i kompletni postupakPOTENCIJE
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 3
puta
Koristeći pravila: , , ...
Izračunaj:1) 2 2 2 2 2 2)3) 4)
5) 6)
7)
21. n
n
a a a a a a a a a a a a
x x xx y x y x z x y x x z xy xy xy
x x x xx y x y x y x yy y y y
x y x y x y x y x
−
⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ ⋅ + ⋅ + ⋅ + ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− ⋅ + ⋅ − ⋅ − ⋅( )
( )( ) ( ) ( ) ( )
2 3
puta
22 3 4
3 4 3 4
2 3 5 2
8)
Koristeći pravila: , , ...
Izračunaj:
1) 5 2) 2 3) 3 4) 1
5) 1 6) 1 7) 2 8)
1 3 2 29) 10) 11) 12)3 5 3 3
213)3
n
n
a b a b a byc c c
a a a a a a a a a a a a
x
−
− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= ⋅ = ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅
−
− − − −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛−
22.
( )
( ) ( ) ( ) ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( )
3 4 2 3
422 3
3 2 3 4
2 3 4 5 20 30 45
1 2 3 4 5 2 4 5
3 5
3 4 414) 15) 16)4 5 5
417) 18) 0,2 19) 0,2 20) 0,25
21) 0,2 22) 2,5 23) 2,5 24 2,5
25) 1 1 1 1 26) 1 1 1
27) 2 2 2 2 28) 2 3 29) 3 2
130) 2 2 31)4
⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞− −⎜ ⎟⎝ ⎠
− − − −
− + − + − + − − + − + −
− + − + − + − − −
⎛ ⎞− + − ⎜ ⎟⎝ ⎠
( ) ( )
( ) ( )
2 3 22 3
2 2 2 2 2
1 32) 0,2 0,22
Koristeći pravila:
1) 2 3 2) 2 4 3) 7 24) 9 2 3 5) 2 3 5 7 26) 2 3 7) 2 48) 4 2 9 9) 3 4 5 2 8
n n n n n nc a d a c d a c a d a c d a
x x x x x a ay y y y x a x a x axy xy xy ab ab abxy xy xy y x y z x y y z
⎛ ⎞ ⎡ ⎤+ − − + −⎜ ⎟ ⎣ ⎦⎝ ⎠
⋅ + ⋅ = + ⋅ ⋅ − ⋅ = − ⋅
+ + + −− + − + − + + −+ + − +
+ − + − − − + −
( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )
2
3 3 3 2 2
2 3 2 3 2 3
2 2
3 8
10) 7 2 4 11) 2 3
12) 3 4 13) 5 2 2 7 3 3
14) 3 3 5 4 5 1 15) 3 1 6 1 1
y x y
xy xy xy x y x y
x y x y x y x y z x y z x y z
x x x x x y x y x y
+
− + + − +
+ + + − + + − + + −
− + − − + + − − + − + + −
23.
M-1- univerzalna zbirka potpuno riješenih zadataka
5 3 2 7 2 2 5 2
2 6 4 8 4 7 2 3 3 5
2 52 4 6 2 2 2 2 3 3 63 2
4 2
Koristeći pravila: : :
Izračunaj:1) 2 2 2) 5 5 3) 3 3 4) 7 75) 2 2 2 2 6) 3 2 7) 8)
9) 10) 11) 12)13) 1
24.n
n m n m n m n m n m n mm
x x m m
a a
aa a a a a a a a aa
a a x x x x x
x x x x x x x x x x a aa a a
+ − −
− −
+ −
⋅ = = = =
⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ 2 5 7 3 3 2 2 2 3
3 4 2 5 2 3 6 3 2 2
2 3 42 1 3 2 2 4 3 2
3 1 2 7 2 3 22
4) 15) 16)
17) 2 6 3 18) 2 3 4 2
1 1 119) 20) 21)2 2 2
22) 23) 24)
x x x x y x x y
m m m m
m m m n m n m n
a a a a a a a a a a
x x x x x x y x y x x x y y
x x x x
x x x x x
+ + +
+ + − +
+ − + − +
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
− ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ + ⋅ ⋅ ⋅
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⋅ ⋅ ⋅
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( )
2 3 2 2
2 3 3 5 3 1 4 7 2
2 3 43 2 2 4
1 2 2 3 4
2 2
2 3 2 1 3
32
25) 26) 2 3 27) 2 52 928) 29) 30)3 4
31) 1 1 1
32)
33)
34)
m n m n m n
m n m n m n m n x y x y
m n m n
m m m
m n m n
m m
x x xa 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
abc
− − −
+ − − + + − + +
+ −
+ + −
− −
⋅ ⋅
⋅ ⋅ ⋅
⋅ − ⋅ − + ⋅ +
+ − ⋅ + − ⋅ + −
− ⋅ +
+ ⋅ − ⋅ + ⋅ −
⎛ ⎞⎜ ⎟⎝ ⎠
2 4 3 22 2 2x y x y x y y xab ab abc c c
− − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
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M-1- rješenja i kompletni postupakPOTENCIJE
25.
P
33
( )
( )
( ) ( )
2 2 3 3 2 5 3
2 3 4 6 2 2
2 3 4 3 2 5 4 2 2
56 4 2 3 4 3 4 2
2 4 3 2 2 2 3 3
omnoži:2 3 2) 5 22 9 5 94)3 4 27 51 14 6) 92 3
7) 25 8) 2 35
19) 3 2 7 10) 22
x y x y x y x y
a b ab a b a b
a b c a b c x y x y
yxx y x y z x y z
1)
x x x x x xy x y xy x y
⋅ ⋅
⎛ ⎞⋅ − ⋅ −⎜ ⎟⎝ ⎠
⋅ − ⋅
⋅ ⋅ −
⎛ ⎞⋅ − − + − − ⋅ − +⎜ ⎟⎝ ⎠
3)
5)
9 5 7 5 5 3 3 2 2
4 2 6 2 7 2 3 7 3 4
7 1 21 12 2 29 3 52 2
4 2 8 3 2 2 3 7 2 3
3 75 8 2 4
2 3 13
Izračunaj:
1) 2 : 2 2) 13 : 13 3) 3 : 3 4) 5 : 5
5) : 6) : 7) : : 8) :
9) : 10) : 11) : 12) : :13) : 14) : : 15) : : 16) :
17) 18) 19) 20)
21)
26.x x m m
x x
m
x x x x x x x x x x
x x x x x x x x xa a a a a a a a a a a
x x x xx x x
xx
− −
+
⋅
⋅
( ) ( ) ( ) ( )
( ) ( ) ( )
1 3 7 8 7 9 111 7 10
2 2 3 5 3 2 2 3
6 42 27 4 2 3 2 3
3 3
2 3 3 4 42 2 22 4 4 2 4
3 3 3
22) 23) : 24) :
25) : 26) 2 3 : 2 3 27) :
28) 2 2 : 2 29) :
3
m m mm m
m m m m
x y x y
x x xm m m
x x ax ax x x a
a b a bx y x y x y x yc c
a b a b a bx y x y x yc c c
+ + +− +
+ + + −
+ +
− − −− − −
⎛ ⎞ ⎛ ⎞− − − − ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ ⋅ + + ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( ) ( ) ( ) ( )2 4 2 2 20) 1 1 31) 1 1x x x x x x− ⋅ + + + ⋅ − +
autor: Mladen Sraga
M-1- univerzalna zbirka potpuno riješenih zadataka
( ) ( ) ( ) ( )( )
) ( ) ) ( ) ) )
) ) ( ) ) ( ) ) ( )
) ( ) ) ( ) ) ( ) ) ( )) ( ) ) ( ) ) ( ) ) ( )
) ( ) )
2 32 2 2 3 2
42 2 32 3 2 2 2
5 2 3 42 3 4 3 4 3 4
3 3 2 32 2 3 4
25
Koristeći pravila:
Izračunaj:
3 21 2 2 3 3 44 3
15 6 7 82
9 10 2 11 2 12 2
13 14 15 16
17 18
27. mn nn n n n n n n m n m z n m zab a b abc a b c a a a a
x x x y xy
x y x x x
x x y x y x y
y y y y
x x
⋅ ⋅ ⋅= = = =
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞ −⎜ ⎟⎝ ⎠
⎡ ⎤− − − − −⎣ ⎦
− − − −( ) ) ( ) ) ( )
) ( ) ) ( ) ) ( ) ) ( )
) ) ) ( ) ) ( )
) ( ) ) ( ) ) ( ) ) ( )) ( ) ) ( ) ) ( ) ) ( )) ( )( ) ) ( )( ) ) ( )( ) ) ( )( )) ( )( ) ( ) ) ( ) ( )( ) ) ( )( ) ( )
22 2 25 5 5
3 4 2 32 2 3 3 2
2 32 32 2
4 23 2 3 2
2 3 2
4 5 43 42 3 2 3
2 2 36 5 4 3 5 72 3 3 5 2 4
19 20
21 2 22 23 2 24 3
2 225 26 27 283 3
29 30 31 32
33 2 34 3 35 2 36 2 3
37 38 39 40
41 42 43 :
m n
m xm n m n x y x y
xx x x m n
yx x
x x
ab a b y y
x x x y
x y x y a b a b
x y x y
x x y y x x
⎡ ⎤⎡ ⎤ − − −⎣ ⎦ ⎢ ⎥⎣ ⎦
− −
⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⋅ ⋅ ) ( )( ) ( )( )) ( ) ) ( ) ) ( )
) ( ) ( ) ) ( ) ) ( ) ( )) ( ) ( ) ) ( ) ( ) ) ( ) ( ) ( )
8 56 43 2
3 3 32 3 32 3 2 2 3 3 2 2 3 3 2
2 1 2 1 12 3 3 3 2
2 1 1 3 2 4 6 35 3 2 3 3 2 4
44 :
2 1 145 : 2 46 4 46 : 43 2 2
48 3 49 50
51 : 52 2 3 53 3 4
x m m mx x
m m
a a
a b ab x y x y x y x y
a b a b x x x
x x x x a a a
+ − −
− +
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⋅ ⋅
+ − +
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35
( ) ( )
3 4 4 55 5 3 5
5 5 2 2 4
Koristeći pravilo:Izračunaj:1) 2 5 2) 3 4 3) 4 6 4) 2 4
1 3 2 15) 3 2 6) 9 7) 8) 43 2 3 2
2 99) 10) 11)3 2
28. n nn n n n n
x x x x a a y y
a b a b ili a b c abc
ac b a bb c b a
⋅ = ⋅ =
⋅ ⋅ ⋅ ⋅
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝
4 3 3
2 22 3 12 3 1 2 3 6
2 4 3 2
52 3 6 2
4 3 2 2 2
3 812)4 3
413) 14) 15) 16)2
117) 18)1 2
mm
m m
xy xy x xy x x y zz yz z yzx y z x y
x y z x y xxz x y x xy y
++
⎞ ⎛ ⎞ ⎛ ⎞⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠
5 5x yx y
⎛ ⎞+⋅⎜ ⎟−⎝ ⎠
autor: Mladen Sraga
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2 2 3 4
1 1 2 3
2 2
Izračunaj:
Koristeći pravila:
2 2 1 21) 2) 3) 4)3 3 2 3
5 2 4 25) 6) 1 7) 8)7 7 5 3
1 2 19) 2 10) 1 11) 13 7 2
29.n n nn n
n n
a a a b bb b ab a
−
−
− − − −
− −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝
3 2
2 2 3 4
2 2 3 2
4 3 2 22 3 3 2 3
3 4 4 4
22 5
6
312)4
13) 14) 15) 16)
1 2 2 417) 1 18) 2 19) 2 20) 32 3 3 5
21) 22) 23) 24)
25)
x x x xy y y y
x x x x yy y y z
x yz
− −
−
⎞ ⎛ ⎞⎟ ⎜ ⎟⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
3 3 24 2 3 4 2
2 3 4 3
2 3 1 2 34 2 4 2
3 3
3 323 2 5 2
3 3
2 226) 27) 28)3 2 5
2 229) 30) 31) 32)5 5
2 3 133) 34) 35) 36)2 3
x x y x yy z z z
x y x y x x yy y xz z
x a b a yx x y c y a
− −
− −
− − −− −
−−−
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⎛ ⎞−⎛ ⎞ ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟+ −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
22 2
3 4
2 23 2
21 2 3 2 32
3
:
3 1 337) 2 2 38)2 8 4
2 1 1 3 2 339) 40)3 3 2 4 23
ya
− −− −
−− − − − −−
−
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞⋅ + ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − ⋅ ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
M-1- rješenja i kompletni postupakPOTENCIJE
( )
( ) ( ) ( ) (
( ) ( ) ( )
( )
0 1 1
02 40 0 0 0
00 0 0
0 0 0 0 0 0
2 0 00 0 0
11 1 1
Izračunaj:
1 1Koristeći pravila: 1 , , ,
1) 2 2) 3) 4)
5) 3 6) 3 7) 3 8) 3
9) 10) 2 7 11) 2 7 12) 2 7
13) 14) 15) 16) 4
17) 5 18) 19) 0,2 20)
2
37
30.
)0 0
nn
m
a a a a aa a
x yx x yz
x
x y x y x y z x y z
x x x x x
x x y
− −
−− − −
= = = =
⎛ ⎞+⎜ ⎟
⎝ ⎠
− − −
+ − − + − +
⋅
−
( )
( )
( )
( ) ( )
52 3 2
11
1 1
1 1 2 4
5 4 4 3 7 9 2 1 3 2 3 6 2 3 2 2
12 2 0 0 0 1
0
1) 3 22) 23) 0,3 24) 2 5
1 1 225) 4 26) 27) 28)34 4
2 2 3 229) 30) 31) 32)5 25
33) 3 3 3 34) 2 2 2 2 35) 5 5
136) 37) 27 23
38) 2
x x x x x x x x
m n n m
x x
ab
x y x y x y
y
−− − −
−−
− −
− − − −
− − − + − − − −
−− −
−
−
⎛ ⎞⎜ ⎟⎝ ⎠−
−
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⎛ ⎞+ ⋅ + ⋅ ⋅ ⋅ ⋅⎜ ⎟⎝ ⎠
⋅21 1 2 2 2
1 33 3 3
22 22 13 1 2 1
2 4 1 2 1 2
0 0 1 0 0 1 0 1 2
2 3 3
1 1 2 39) :2 5
2 32 2 2 340) 41) 42)2 2 2 3 2 3
43) 3 2 3 44) 3 2 3 45) 2 2 2
46) 2 2 47) 2 2
m n m n na b a b a bc c c
+ +−−
−− −− −− − − −
− − − − − −
−
− − − −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞⋅− −⎜ ⎟⎜ ⎟ ⎜ ⎟+ ⋅ +⎝ ⎠ ⎝ ⎠⎝ ⎠
⋅ − + − + −
− − 2 1 2 3
3 2 2 12
3 2 2 1
1 2 1
1 2 1
2 3 1 2 1 2 2
1 2 2 3 1 2 3 1
2 48) 3 2
2 3 2 4249) 50) 51)3 4 3 6
52) 53) 54)
55) 56) 57)
a a a bb b ca b x bc x b cc d y c d y c d
− − −
− −−
− −
− − −
− − −
− − − −
− − − − −
+ −
⋅ ⋅⋅
autor: Mladen Sraga
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M-1- univerzalna zbirka potpuno riješenih zadataka
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5 3 2 3
( ) ( )
( ) ( ) ( )
32 3 2
4 2
3 3 5
Zapiši u obliku potencija s bazom 2:
Pojednostavni i rezultati zapiši kao potenciju:
)
1) 4 32 16 2) 4 8 16 3) 2 4 8
) Zapiši u obliku potencija s bazom 3:
1) 3 9 27 2) 9 81 3 3) 3 9 27
)1) 2 2 2) 3 2
31.a
b
c
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
+ ⋅ 5 4
2 2 3 3
6 5 4 3 7 3 5 2
8 3 3 3 9 6 5 6 3 4
1 1 2
1 1
1
2 3) 6 2
4) 5 3 3 3 5) 3 5 2 5
9 3 4 21) : 2) :2 2 3 15
5 1 25 53) : 4) :7 7 49 7
16 2 4 161) 2)8 32
9 273)81
32.
33.x x x
x x
x x
x y x y x y x y
x y x y a b c a b c
+ − +
− −
−
+ ⋅
⋅ − ⋅ ⋅ + ⋅
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⋅
⋅ 1 5
1 3
25 54)125
x x
x x
+ +
+ −
⋅
42 4− ⋅
M-1- rješenja i kompletni postupakPOTENCIJE
2 2 2 22 4 2 4
3 2 3 2
3 2 3 23 2 3 2
4 3 4 3
2 2 47 2 2 2 3 6 2
3 3 2 4 2
1) 2)
3 4 3 33) 4)2 3 2 4
5 5 3 35) 6)6 3 5 5
34.x x x xy y y y
x x x xy y y y
a b a x y x yc b c z z
−− −
− −
−
− −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⋅ ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛− ⋅ − ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝
2⎞⎟⎠
)a b
a b
( ) ( ) ( )( ) ( ) (( ) ( )( )( ) ( )
( ) ( ) ( )( ) ( )( ) ( ) ( )( )
2 3 4 2 2 5 2
32 2 2 3 2 3 2
2 3 4 2 2 3 2 2 3 2
3 2 3 2 3 2 2
2 3 4 2 2 3 2
3 5 2 2 3 3 5 3 4 2
2 3 3 5 2 2 3 3 5 2 3 3 5
3 2 2 4 3
1) 2) 2 3)
4) 4 5) 2 6) 3
7) 4 8)
9) 2 10) 2 3
35. )
1) 2) 3)
4) 5) 6)
7) 8)
9)
35.x x
ab x y x y
a b c a b c a b c
a b x y xy
b
x y
a b a b a b x y
a b a b a b a b a b a b
x y x y
+
− − +
+ +
⋅ +
− − ⋅ +
+
39autor: Mladen Sraga
M-1- univerzalna zbirka potpuno riješenih zadataka
36 . I sada primjenom svih pravila za potencije riješite dopunske zadatke
( )
( )
( )
( )( )
1
1
0
POTENCIJE
1:
1: 1
n n n
n nnn m n m n n n
n
n n nmn m n m n n mn
n zmn m n m n n m z nm n
c a d a c d a
a aa a a abc a b c a ab b
a b ba a a a a ab a aa
a a a a a a a aa a
+
−− ⋅ −
− ⋅ ⋅ −
⋅ ± ⋅ = ± ⋅
⎛ ⎞⋅ = = = =⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= = = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= = = = =
( )( ) ( ) ( )( ) ( ) ( ) ( )
2 3 4 7 2 17 8 3 4 2
2 4 5 3 8 4 2 7 5 6
3 2 2 42 2 7 3 9 4 7
2 35 6 2 3 4 4 3 4 2
4 3 2 5 6 5 4 7 2
Izračunaj:1 21) 2 5 2) : 3)3 5
4) 5) : 6) :
7) : 8) : 9) 25 :5
1 110) 7 : 11) 2 3 447 7
12) 3 4 13) 2 5
36.
m
x x x x x x x x a a
a a a a a a a
a a x x x y x y
x x x a b c a b c ab c
a a a a a a x x x x
+
⋅ + ⋅ ⋅ − ⋅
⋅ ⋅
⎛ ⎞⋅ ⋅ ⋅⎜ ⎟⎝ ⎠⋅ + ⋅ + ⋅ ⋅ + ⋅ −
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )
3 6
5 3 10 2 3 5 4 2 2 3 2
3 4 1 2 1 2 3 2 3 2 2 1
4 6 3 4 6 33 2 4 3 2 4
4 6 3 43 2 4 7 2 3 3
2 3 4 37 2 3 3 2 3 2 4
3
14) 7 2 : 3 15) 3 2
16) : 17) :
18) 19) :
20) 3 2 21) :
22) : 23)
24) 1
b m m b
m n m n m n m n m n
x xx x x x
xx x
x y
x x x x x y x y z x y z
x y x y x x x
a a a a a a
a a a x y x y
x y x y a b a b
x
−
+ − + − + − − + + −
⋅
⋅ + − ⋅ ⋅
⋅
⋅ ⋅ ⋅
+ −
⋅
+ ( ) ( )
( ) ( ) ( )
( )
0 00 0 0
0 00 0 2 3 0 0
0 0
0 0
5 2 4 2 3 2
8 3 9 2
3 4 25) 2 2
26) 27) 2 3 5
3 328) 29) 30)2 2
31) 32) : 33) :
34) : 35) 12 :3
x x x
x y xy xy x y x y xy
xy x x yx y y xy
2 0 1
0
3
x x x x
x x x x
− −
− − −
+ − + ⋅ +
− ⋅ − + −
+−
⋅ ⋅x x x− −
2
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41autor: Mladen Sraga
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