ZBIRKA POTPUNO RIJEŠENIH ZADATAKA

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
1. Koristimo pravilo koje kaže da je: a = a⋅a i − a = a
( ) 2
2 2
2 2 2
1) 3 = 3⋅ 3 = 9 2) 4 = 4⋅ 4 = 16 3) 5 = 5⋅ 5 = 25
↓
2
a
↑ ↑
= a⋅a
2 2 2
4) 7 7 7 49 5) 11 11 11 121 6) 12 12 12 144
= ⋅ = = ⋅ = = ⋅ =
2 2 2
7) 14 14 14 196 8) 15 15 15 225 9) 20 20 20 400
= ⋅ = = ⋅ = = ⋅ =
2
2 2
10) 21 21 21 441 11) 22
= ⋅ =
2 2
13) 5 1 5 1 5 5 1 25 25
− = − ⋅ = − ⋅ ⋅ = − ⋅ = −
( )
= 22⋅ 22 = 484 12) 25 = 25⋅ 25 = 625
2 2
( −n
) ( − ⋅n
)
Objašnjenje svaki negativan broj da se zapisati u obliku 1
2 2
Dakle: 1 gdje je sada -pozitivan broj, a njega znamo kvadrirati
− n = − ⋅n n
2 2
Pa u ovom zadatku imamo: − 5 =−1⋅ 5 =−1⋅5⋅ 5 =−1⋅ 25 =−25
↓ ↓
− n = − ⋅ n n =
1 , u ovom zadatku: 5
2 2 2 2
2 2
14) − 10 =−1⋅ 10 =−1⋅10⋅ 10 =−1⋅ 100 =−100
2 2
15) − 20 =−1⋅ 20 =−1⋅20⋅ 20 =−1⋅ 400 =−400
2 2
16) − 25 =−1⋅ 25 =−1⋅25⋅ 25 =−1⋅ 625 =−625
2
( ) ( ) ( )
17) − 5 = −5 ⋅ − 5 = + 25 = 25
( ) ( )
II način po pravilu: a a
2 2 2 2
− = − 5 = 5 = 25
vidi 2. zadatak
III
način:
( ) ( ) ( )
2 2 2 2
− 5 = −1⋅ 5 = −1 ⋅ 5 = + 1⋅5⋅ 5 = 25
više o
III
načinu u poglavlju potncije.
2
( ) ( ) ( )
18) − 10 = −10 ⋅ − 10 = + 10⋅ 10 = 100
II
III
način
način
( )
2 2
− 10 = 10 = 10⋅ 10 = 100
2
( − 10)
( ) ( )
2
( ) ( ) ( ) ( )
2 2 2
= −1⋅ 10 = −1 ⋅ 10 = −1 ⋅ 10 = −1 ⋅ −1 ⋅10⋅ 10 = 100
2