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