ZBIRKA POTPUNO RIJEŠENIH ZADATAKA

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

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

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
2. Koristimo pravila: I način: a = a⋅a , − ⋅ − = + − a = a
I način:
( ) ( ) ( )
( 1)
2
a = a ⋅ a u ovom zadatku: a = −
↑ ↓ ↓
( ) ( ) ili II način: ( )
2 2 2
2 2
( ) ( ) ( )
1) − 1 = −1 ⋅ − 1 = + 1 = 1
ili duže − 1 = −1 ⋅ − 1 = + 1⋅ 1 = 1
( −⋅− ) ( ) =+ ( −⋅− ) ( ) =+
II način po pravilu:
( )
2 1
− 1 = 1 = 11 ⋅ = 1
( a)
2 2
− = a →suprotni brojevi imaju jednake kvadrate
I način:
2
a = a ⋅ a u ovom zadatku: a = −
↑ ↓ ↓
( ) ( ) ( )
( )
( a)
( 2)
2 2
2 2
2 2
( ) ( ) ( )
2) − 2 = −2 ⋅ − 2 = 4
ili duže − 2 = −2 ⋅ − 2 = + 2⋅ 2 = 4
( −⋅− ) ( ) =+ ( −⋅− ) ( ) =+
II način po pravilu:
− 2 = 2 = 2⋅ 2 = 4
− = a
I način:
2
( ) ( ) ( )
3) − 3 = −3 ⋅ − 3 = + 9 = 9 ili duže − 3 = −3 ⋅ − 3 = + 3⋅ 3 = 9
( ) ( )
−⋅− =+
2
( ) ( ) ( )
( ) ( )
−⋅− =+
II način po pravilu:
( )
2 2
− 3 = 3 = 3⋅ 3 = 9
( a)
2 2
− = a
po pravilu:
= ⋅
2
a a a
↑ ↓ ↓
2
( ) ( ) ( )
4) − 5 = −5 ⋅ − 5 = + 5⋅ 5 = 25
( )
2 2
( ) ( )
−⋅− =+
II način po pravilu:
( a)
− 5 = 5 = 5⋅ 5 = 25
2 2
− = a
M-8-rješenja--radna varijanta ( nedovršena)
3
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
2. Koristimo pravila: I način: a = a⋅a , − ⋅ − = + −a = a
I način po pravilu:
2
a = a ⋅ a
↑ ↓ ↓
2
( ) ( ) ( )
5) − 11 = −11 ⋅ − 11 = + 11⋅ 11 = 121
( ) ( )
−⋅− =+
2
( ) ( ) ( )
( ) ( )
( )
6) − 13 = −13 ⋅ − 13 = + 13⋅ 13 = 169
−⋅− =+
2
( ) ( ) ( )
2
( ) ( ) ( )
2
( ) ( ) ( )
( ) ( ) ili II način: (
2 2 2
II način po pravilu:
2 2
( a)
− 11 = 11 = 11⋅ 11 = 121
II način po pravilu:
( )
7) − 15 = −15 ⋅ − 15 = + 15⋅ 15 = 225
( )
8) − 17 = −17 ⋅ − 17 = + 17⋅ 17 = 289
( )
2 2
II način po pravilu:
II način
9) − 20 = −20 ⋅ − 20 = + 20⋅ 20 = 400
2
( ) ( ) ( )
2
( ) ( ) ( )
II način
2 2
2 2
2 2
− = a
( a)
( a)
2 2
− = a
− 13 = 13 = 13⋅ 13 = 169
2 2
− = a
− 15 = 15 = 15⋅ 15 = 225
− 17 = 17 = 17⋅ 17 = 289
( )
10) − 22 = −22 ⋅ − 22 = + 22⋅ 22 = 484
II način
( )
11) − 25 = −25 ⋅ − 25 = + 25⋅ 25 = 625
II način
( )
2 2
− 20 = 20 = 20⋅ 20 = 400
2 2
− 22 = 22 = 22⋅ 22 = 484
2 2
− 25 = 25 = 25⋅ 25 = 625
)
4

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
2. Koristimo pravila: I način: a = a⋅a , − ⋅ − = + − a = a
2
( ) ( ) ( )
12) − 30 = −30 ⋅ − 30 = + 30⋅ 30 = 900
II način
( )
2 2
− 30 = 30 = 30⋅ 30 = 900
po pravilu:
( ) ( )
II način
2
a a a
( ) ( ) ( )
2 2
( ) ( )
2 2
( ) ( )
( ) ( ) ili II način: ( )
2 2 2
2
( 1) ( 1) ( 1)
= ⋅ dakle: − = − ⋅ −
↑ ↓ ↓
13) − − 1 =− − 1 =− −1 ⋅ − 1 =− + 1 =−1
2
− − 1 =−1⋅ − 1 =−1⋅
1 = −11 ⋅ = −1
( ) ( )
II način
( ) ( ) ( )
2 2
( ) ( )
po pravilu:
( a)
( ) ( ) ( )
14) − − 5 =− − 5 =− −5 ⋅ − 5 =− + 5⋅ 5 =− + 25 =−25
2 2 2
− − 5 =−1⋅ − 5 =−1⋅ 5 =−1⋅5⋅ 5 =−2
↓
− = a
2 2
( ) ( )
ili kraće:
( ) ( ) ( )
2 2
( ) ( )
2 2
( ) ( )
( ) ( ) ( )
15) − − 10 =− − 10 =− −10 ⋅ − 10 =− + 10⋅ 10 =− + 100 =−100
− − 10 = − − 10 = − + 100 = −100
II način
( ) ( )
2 2
2
− − 10 =−1⋅ − 10 =−1⋅ 10 = −1⋅10⋅ 10 = −100
po pravilu:
↓
− = a
( a)
2 2
( ) ( )
ili kraće:
( ) ( ) ( )
2 2
( ) (( ) ) ( )
2 2
( ) ( ) ( )
16) − − 18 =− − 18 =− −18 ⋅ − 18 =− + 18⋅ 18 =− + 324 =−324
− − 18 = − − 18 = − + 324 = −324
II način
( ) ( )
2 2 2
− − 18 =−1⋅ − 18 =−1⋅ 18 =−1⋅ 324 =−324
po pravilu:
↓
− =
( a)
↑
a
2 2
M-8-rješenja--radna varijanta ( nedovršena)
5
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
3.
Ovaj zadatak je isti kao 1. samo sada kvadriramo decimalne brojeve, neki pozitivne decimalne
brojeve prilikom kvadriranja stavljaju u zagradu a neki ne, ja ću pokazati obadvije varijante vidi 3.A
( )
1) 0. 3 = 0.3⋅ 0.3 = 0.09
( )
( )
2
2
2
0.3 = 0.3⋅ 0.3 = 0.09
decimalne brojeve bi trebali znati množiti iz 6 razreda:
II način
1) 0.3
( )
( )
2
II način
2
II način
( )
( )
↓
Još jednom isti zadatak s time da decimalni broj pretvorimo u razlomak pa kvadriramo...
2 2
⎛ 3 ⎞ 3 33 ⋅ 9
= ⎜ ⎟ = = = = 9:100 = 0.09
2
⎝10 ⎠ 10 10⋅10 100
2) 0.1 = 0.1⋅ 0.1 = 0.01
2
2 2
2 ⎛ 1 ⎞ 1 1
2) ( 0.1)
= ⎜ ⎟ = = = 1:100 = 0.01
2
⎝10 ⎠ 10 100
3) 0.5 = 0.5⋅ 0.5 = 0.25
2 2
2 ⎛ 5 ⎞ 5 55 ⋅ 25 25 1
0.5 = ⎜ ⎟ = = = = = = 1:4 = 0.25
2
⎝10 ⎠ 10 10⋅10 100 25⋅4 4
4) 0.01 = 0.01⋅0.01 = 0.0001
II način
( )
2 2
2 ⎛ 1 ⎞ 1 1 1
0.01 = ⎜ ⎟ = = = = 1:10000 = 0.0001
2
⎝100 ⎠ 100 100⋅100 10000
0.3⋅ 0.3 = 0.09
00
09
0,0 9
0.01⋅ 0.01 = 0.0001
000
000
001
0.0001
( )
2
5) 0.04 = 0.04⋅ 0.04 = 0.0016
II način
( )
( )
II način
2
2 2
2 ⎛ 4 ⎞ 4 44 ⋅ 16
0.04 = ⎜ ⎟ = = = = 16:10000 = 0.0016
2
⎝100 ⎠ 100 100⋅100 10000
6) 0.003 = 0.003⋅ 0.003 = 0.000009
( )
2 2
2 ⎛ 3 ⎞ 3 9 9
0.003 = ⎜ ⎟ = = = = 9:1000000 = 0.000009
2
⎝1000 ⎠ 1000 1000⋅1000 1000000
6

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
2
3. = ⋅ =
( )
7) 1.3 1.3 1.3 1.69
II način decimalni broj pretvorimo u decimalni razlomak pa kvadriramo...
( )
2 2
2 ⎛13⎞
13 13⋅13
169
1.3 = ⎜ ⎟ = = = = 169:100 = 1.69
2
⎝10 ⎠ 10 10⋅10 100
( )
2
8) 2.5 = 2.5⋅ 2.5 = 6.25
II način
( 2.5)
2
2 2
⎛25⎞
25
= ⎜ ⎟ = =
2
⎝10 ⎠ 10
25⋅25 625
= = 625:100 = 6.25
10⋅10 100
( )
2
9) 1.1 = 1.1⋅ 1.1 = 1.21
II način
( )
2 2
2 ⎛11⎞
11 11⋅11
121
1.1 = ⎜ ⎟ = = = = 121:100 = 1.21
2
⎝10 ⎠ 10 10⋅10 100
( )
II način
( )
2
10) 0.11 = 0.11⋅ 0.11 = 0.0121
2 2
2 ⎛ 11 ⎞ 11 11⋅11
121
0.11 = ⎜ ⎟ = = = = 121:10000 = 0.0121
2
⎝100 ⎠ 100 100⋅100 10000
( )
II način
( )
2
11) 0.009 = 0.009⋅ 0.009 = 0.000081
2 2
2 ⎛ 9 ⎞ 9 99 ⋅ 81
0.009 = ⎜ ⎟ = = = = 81:1000000 = 0.000081
2
⎝1000 ⎠ 1000 1000⋅1000 1000000
( )
II način
( )
2
12) 0.14 = 0.14⋅ 0.14 = 0.0196
2 2
2 ⎛ 14 ⎞ 14 14⋅14
196
0.14 = ⎜ ⎟ = = = = 196: 10000 = 0.0196
2
⎝100 ⎠ 100 100⋅100 10000
M-8-rješenja--radna varijanta ( nedovršena)
7
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
3.
( − )
2 2
( ) ( ) ( ) ( )
13) − 0.2 =−1⋅ 0.2 =−1⋅ 0.2⋅ 0.2 =−1⋅ 0.04 =−0.04
1 možemo i ne moramo pisati :
još jednom isti zadatak
2 2
( ) ( ) ( ) ( )
: − 0.2 = − 0.2 = − 0.2⋅ 0.2 = − 0.04 = −0.04
II način
decimalni broj pretvorimo u decimalni razlomak pa kvadriramo.. .
ili
2 2
2 ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 4 ⎞
− ( 0.2) =−1⋅ ⎜ ⎟ =−1⋅ ⎜ 1 1
2 ⎟ =− ⋅ ⎜ ⎟ =− ⋅ ( 4:100) =−1⋅ ( 0.04)
=−0.04
⎝10 ⎠ ⎝10
⎠ ⎝100
⎠
2 2
2 2
( ) ( ) ( ) ( )
II način
2 ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 4 ⎞
− ( 0.2) = − ⎜ ⎟ = − ⎜ 2 ⎟ = − ⎜ ⎟ = − ( 4:100) = − ( 0.04)
= −0.04
⎝10 ⎠ ⎝10
⎠ ⎝100
⎠
14) − 1.2 = − 1.2 = − 1.2⋅ 1.2 = − 1.44 = −1.44
2
2
2 ⎛12 ⎞ ⎛12 ⎞ ⎛144
⎞
− ( 1.2)
= − ⎜ ⎟ = −⎜ 2 ⎟ =− ⎜ ⎟ =− ( 144:100) =− ( 1.44)
=−1.44
⎝10
⎠ ⎝10
⎠ ⎝100
⎠
2
( ) ( ) ( )
15) − 0.06 = − 0.06⋅ 0.06 = − 0.0036 = −0.0036
II način
2 2
2 ⎛ 6 ⎞ ⎛ 6 ⎞ ⎛ 36 ⎞
− ( 0.06) = − ⎜ ⎟ = − ⎜ 2 ⎟ = − ⎜ ⎟ = − ( 36:10000)
= −0.0036
⎝100 ⎠ ⎝100
⎠ ⎝10000
⎠
2
( ) ( ) ( )
16) − 0.13 =− 0.13⋅ 0.13 =− 0.0169 =−0.0169
II način
2 2
2 ⎛ 13 ⎞ ⎛ 13 ⎞ ⎛ 169 ⎞
− ( 0.13) = − ⎜ ⎟ = − ⎜ 2 ⎟ = − ⎜ ⎟ = − ( 169:10000)
= −0.0169
⎝100 ⎠ ⎝100
⎠ ⎝10000
⎠
8

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
3.A
Ovaj zadatak je isti kao 3. samo sada decimalne brojeve zapisujemo bez zagrade:
2
1) 0.3 = 0.3⋅ 0.3 = 0.09
2
0.3 = 0.3⋅ 0.3 = 0.09
↓
decimalne brojeve bi trebali znati množiti iz 6 razreda:
II način Još jednom isti zadatak s time da decimalni broj pretvorimo u razlomak pa kvadriramo...
2 2
2 ⎛ 3 ⎞ 3 33 ⋅ 9
1) 0.3 = = = = = 9:100 = 0.09
2
⎜ ⎟
⎝10 ⎠ 10 10⋅10 100
2
2) 0.1 = 0.1⋅ 0.1 = 0.01
II način
2 2
2 ⎛ 1 ⎞ 1 1
2) 0.1 = = = = 1:100 = 0.01
2
⎜ ⎟
⎝10 ⎠ 10 100
2
3) 0.5 = 0.
5⋅ 0.5 = 0.25
II način
2 2
2 ⎛ 5 ⎞ 5 55 ⋅ 25 25 1
0.5 = = = = = = = 1:4 = 0.25
2
⎜ ⎟
⎝10 ⎠ 10 10⋅10 100 25⋅4 4
2
4) 0.01 = 0.01⋅ 0.01 = 0.0001
II način
2 2
2 ⎛ 1 ⎞ 1 1 1
0.01 = = = = = 1:10000 = 0.0001
2
⎜ ⎟
⎝100 ⎠ 100 100⋅100 10000
2
5) 0.04 = 0.04⋅ 0.04 = 0.0016
II način
0.04
2
2 2
⎛ 4 ⎞ 4 44 ⋅ 16
= ⎜ ⎟ = = = = 16:10000 = 0.0016
2
⎝100 ⎠ 100 100⋅100 10000
2
6) 0.003 = 0.003⋅ 0.003 = 0.000009
II način
2 2
2 ⎛ 3 ⎞ 3 9 9
0.003 = = = = = 9:1000000 = 0.000009
2
⎜ ⎟
⎝1000 ⎠ 1000 1000⋅1000 1000000
M-8-rješenja--radna varijanta ( nedovršena)
9
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
3.A = ⋅ =
2
7) 1.3 1.3 1.3 1.69
II način
decimalni broj pretvorimo u decimalni razlomak pa kvadriramo...
2 2
2 ⎛13⎞
13 13⋅13
169
1.3 = = = = = 169:100 = 1.69
2
⎜ ⎟
⎝10 ⎠ 10 10⋅10 100
2
8) 2.5 = 2.5⋅ 2.5 = 6.25
II način
2.5
2
2 2
⎛25⎞
25
= ⎜ ⎟ = =
2
⎝10 ⎠ 10
25⋅25 625
= = 625:100 = 6.25
10⋅10 100
= ⋅ =
2
9) 1.1 1.1 1.1 1.21
II način
2 2
2 ⎛11⎞
11 11⋅11
121
1.1 = = = = = 121:100 = 1.21
2
⎜ ⎟
⎝10 ⎠ 10 10⋅10 100
2
10) 0.11 0.11 0.11 0.0121
II način
= ⋅ =
2 2
2 ⎛ 11 ⎞ 11 11⋅11
121
0.11 = = = = = 121:10000
2
⎜ ⎟
⎝100 ⎠ 100 100⋅100 10000
= 0.0121
= ⋅ =
2
11) 0.009 0.009 0.009 0.000081
2 2
2 ⎛ 9 ⎞ 9 99 ⋅ 81
0.009 = = = = = 81:1000000 = 0.000081
2
⎜ ⎟
⎝1000 ⎠ 1000 1000⋅1000 1000000
= ⋅ =
2
12) 0.14 0.14 0.14 0.0196
2 2
2 ⎛ 14 ⎞ 14 14⋅14
196
0.14 = = = = = 196:10000 = 0.0196
2
⎜ ⎟
⎝100 ⎠ 100 100⋅100 10000
13)
− =− ⋅ =− ⋅ ⋅ =− ⋅ =−
2 2
0.2 1 0.2 1 0.2 0.2 1 0.04 0.04
II način
ili
2 2
2 ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 4 ⎞
0.2 1 1 ⎜ 1 1
2 ⎟
( 4:100) 1 ( 0.04)
0.04
− =− ⋅ ⎜ ⎟ =− ⋅ =− ⋅ ⎜ ⎟ =− ⋅ =− ⋅ =−
⎝10 ⎠ ⎝10
⎠ ⎝100
⎠
2 2
2 ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 4 ⎞
0.2 ⎜ 2 ⎟
( 4:100) ( 0.04)
0.04
− = − ⎜ ⎟ = − = − ⎜ ⎟ = − = − = −
⎝10 ⎠ ⎝10
⎠ ⎝100
⎠
10

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
3.A
2 2
14) − 1.2 =−1⋅ 1.2 =−1⋅1.2⋅ 1.2 =−1⋅ 1.44 =−1.44
II način
decimalni broj pretvorimo u decimalni razlomak pa kvadriramo...
2 2
2 2 ⎛12 ⎞ ⎛12 ⎞ ⎛144
⎞
1.2 1 1.2 1 ⎜ 2 ⎟
( 144:100) ( 1.44)
1.44
− = − ⋅ = − ⋅ ⎜ ⎟ = − = − ⎜ ⎟ = − = − = −
⎝10 ⎠ ⎝10
⎠ ⎝100
⎠
15) − 0.06
II način
= −1⋅ 0.06 = −1⋅0.06⋅ 0.06 = −1⋅ 0.0036 = −0.0036
2 2
2 2
2 2 ⎛ 6 ⎞ ⎛ 6 ⎞ ⎛ 36 ⎞
0.06 1 0.06 1 ⎜ 2 ⎟
( 36:10000)
0.0036
− = − ⋅ = − ⋅ ⎜ ⎟ = − = − ⎜ ⎟ = − = −
⎝100 ⎠ ⎝100
⎠ ⎝10000
⎠
2 2
16) − 0.13 = −1⋅ 0.13 = .1⋅0.13⋅ 0.13 = −1⋅ 0.0169 = −0.0169
II način
0.13 1 0.13
2 2
− = − ⋅ = −
2 2
⎛ 13 ⎞ ⎛ 13 ⎞ ⎛ 169 ⎞
1⋅ ⎜ ⎟ = −1⋅ ⎜ 1
2 ⎟ = − ⋅ ⎜ ⎟ =
⎝100 ⎠ ⎝100
⎠ ⎝10000
⎠
=−1⋅ 169:10000 =−1⋅ 0.0169 =−0.0169
( )
M-8-rješenja--radna varijanta ( nedovršena)
11
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
4.
Koristimo sve što smo naučili u prva tri zadatka:
I način po pravilu:
2
a = a ⋅ a
↑ ↓ ↓
2
( ) ( ) ( )
1) − 0.1 = −0.1 ⋅ − 0.1 = + 0.1⋅ 0.1 = 0.01
II način po pravilu:
( )
2 2
( a)
2 2
− = a
− 0.1 = 0.1 = 0.1⋅ 0.1 = 0.01
III način decimalni broj pretvorimo u decimalni razlomak....
( )
2
( ) ( ) ( )
II način po pravilu:
( )
2 2
( a)
2 2
2 2 ⎛ 1 ⎞ 1 1
− 0.1 = 0.1 = ⎜ ⎟ = = = 1:100 = 0.01
2
⎝10 ⎠ 10 100
2) − 0.2 = −0.2 ⋅ − 0.2 = + 0.2⋅ 0.2 = 0.04
2 2
− = a
− 0.2 = 0.2 = 0.2⋅ 0.2 = 0.04
III način decimalni broj pretvorimo u decimalni razlomak.. ..
( )
2
( ) ( ) ( )
II način po pravilu:
( )
( )
2 2
( a)
2 2
2 2 ⎛ 2 ⎞ 2 4
− 0.2 = 0.2 = ⎜ ⎟ = = = 4:100 = 0.04
2
⎝10 ⎠ 10 100
3) − 2.1 = −2.1 ⋅ − 2.1 = + 2.1⋅ 2.1 = 4.41
2 2
− 2.1 = 2.1 = 2.1⋅ 2.1 = 4.41
III način decimalni broj pretvorimo u decimalni razlomak....
2 2 ⎛ 21⎞
− 2.1 = 2.1 = ⎜ ⎟
⎝10
⎠
− = a
2
( ) ( ) ( )
II način po pravilu:
( )
( )
2 2
( a)
2 2
21 441
= = = 441:100 = 4.41
2
10 100
4) − 0.04 = −0.04 ⋅ − 0.04 = + 0.04⋅ 0.04 = 0.0016
2 2
− = a
− 0.04 = 0.04 = 0.04⋅ 0.04 = 0.0016
III način decimalni broj pretvorimo u decimalni razlomak....
2 2
2 2 ⎛ 4 ⎞ 4 16
− 0.04 = 0.04 = ⎜ ⎟ = = = 16:10000 = 0.0016
2
⎝100 ⎠ 100 10000
12

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
4.
Koristimo sve što smo naučili u prva tri zadatka:
2
( ) ( ) ( )
5) − 2.5 = −2.5 ⋅ − 2.5 = + 2.5⋅ 2.5 = 6.25
II način po pravilu:
( )
( )
2 2
2 2
( a)
M-8-rješenja--radna varijanta ( nedovršena)
2 2
− 2.5 = 2.5 = 2.5⋅ 2.5 = 6.25
III način decimalni broj pretvorimo u decimalni razlomak....
− 2.5 = 2.5
− = a
2
( ) ( ) ( )
II način:
III način
( )
( )
2 2
⎛25⎞
25 625
= ⎜ ⎟ = = = 625:100 = 6.25
2
⎝10 ⎠ 10 100
6) − 0.001 = −0.001 ⋅ − 0.001 = + 0.001⋅ 0.0001 = 0.000001
2 2
− 0.001 = 0.001 = 0.001⋅ 0.0001 = 0.000001
2 2
2 2 ⎛ 1 ⎞ 1 1
− 0.001 = 0.001 = ⎜ ⎟ = = = 1:1000000 = 0.000001
2
⎝1000 ⎠ 1000 1000000
2
( ) ( ) ( )
7) − 0.012 = −0.012 ⋅ − 0.012 = + 0.012⋅ 0.012 = 0.000144
II način:
III način
( )
( )
2 2
− 0.012 = 0.012 = 0.012⋅ 0.012 = 0.000144
2
( ) ( ) ( )
2 2
2 2 ⎛ 12 ⎞ 12 144
− 0.012 = 0.012 = ⎜ ⎟ = = = 144:1000000 = 0.000144
2
⎝1000 ⎠ 1000 1000000
8) − 0.025 = −0.025 ⋅ − 0.025 = + 0.025⋅0.025 = 0.000625
II način:
III način
( )
( )
2 2
− 0.025 = 0.025 = 0.025⋅ 0.025 = 0.000625
2 2
2 2 ⎛ 25 ⎞ 25 625
− 0.025 = 0.025 = ⎜ ⎟ = = = 625:1000000 = 0.000625
2
⎝1000 ⎠ 1000 1000000
2 2
( ) ( ) ( ) ( )
9) − − 0.3 =− 0.3 =− 0.3⋅ 0.3 =− 0.09 =−0.09
↓
u ovom tipu zadataka raditi ćemo samo po
( a)
2 2
− = a
II načinu po pravilu: u ovom zadatku:
2 2
( ) ( )
( ) ( ) ( )
− − 0.3 =− 0.3 =− 0.3⋅ 0.3 =− 0.09 =−0.09
minus ispred zagrade ostaje do zadnjeg koraka kada konačnom rješenju daje negativan predznak
ili minus ispred zagrade možemo pisati i kao:
2 2 2
( ) ( ) ( )
13
( −1) ⋅( zagrada)
( ) ( )
− − 0.5 =−1⋅ − 0.5 =−1⋅ 0.5 =−1⋅ 0.5⋅ 0.5 =−1⋅ 0.25 =−0.25
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
4.
Koristimo sve što smo naučili u prva tri zadatka:
2 2
( ) ( ) ( ) ( )
10) − − 0.5 =− 0.5 =− 0.5⋅ 0.5 =− 0.25 =−0.25
↓
u ovom tipu zadataka raditi ćemo samo po
−
( a)
2 2
( ) ( )
2 2
− = a
II načinu po pravilu: u ovom zadatku:
( ) ( )
− − 0.5 = − 0.5 = − 0.5⋅ 0.5 = −( 0.25)
=−0.25
minus ispred zagrade ostaje do zadnjeg koraka kada konačnom rješenju daje negativan predznak
2 2 2
( ) ( ) ( )
( −1) ⋅( )
ili minus ispred zagrade možemo pisati i kao: zagrada
( ) ( )
− 0.5 =−1⋅ − 0.5 =−1⋅ 0.5 =−1⋅ 0.5⋅ 0.5 =−1⋅ 0.25 =−0.
25
ili
2 2
( ) ( ) ( ) ( )
2 2
−( − ) =− ( ) =−
11) − − 0.04 =− 0.04 =− 0.04⋅ 0.04 =− 0.0016 =−0.0016
ili nešto kraći postupak:
0.04 0.04 0.0016
drugi i treći korak napravili smo "u glavi"
2 2 2
( ) ( ) ( ) ( )
− − 0.04 = −1⋅ − 0.04 = −1⋅ 0.04 = −1⋅ 0.04⋅ 0.04 = −1⋅ 0.0016 = −0.0016
12
ili
ili
2 2
( ) ( ) ( ) ( )
− −0.12 =− 0.12 =− 0.12⋅ 0.12 =− 0.144 =−0.0144
2 2 2
( ) ( ) ( ) ( )
− − 0.12 = −1⋅ − 0.12 = −1⋅ 0.12 = −1⋅ 0.12⋅ 0.12 = −1⋅ 0.144 = −0.0144
2 2
( ) ( ) ( ) ( )
13) − − 1.1 =− 1.1 =− 1.1⋅ 1.1 =− 1.21 =−1.21
2 2 2
( ) ( ) ( ) ( )
− − 1.1 = −1⋅ − 1.1 = −1⋅ 1.1 = −1⋅ 1.11.1 ⋅ = −11.21 ⋅ = −1.21
2 2
( ) ( )
14) 2.5 2.5
ili
ili
− − = − = ( ) ( )
− 2.5⋅ 2.5 =− 6.25 =−6.25
2 2 2
( ) ( ) ( ) ( )
− − 2.5 =−1⋅ − 2.5 =−1⋅ 2.5 =−1⋅ 2.5⋅ 2.5 =−1⋅ 6.25 =−6.25
2 2
( ) ( ) ( ) ( )
15) − − 1.4 =− 1.4 =− 1.4⋅ 1.4 =− 1.96 =−1.96
2 2 2
( ) ( ) ( ) ( )
− − 1.4 =−1⋅ − 1.4 =−1⋅ 1.4 =−1⋅ 1.4⋅ 1.4 =−1⋅ 1.96 =−1.96
2 2
( ) ( ) ( ) ( )
16) − − 0.009 = − 0.009 = − 0.009⋅ 0.009 = − 0.000081
=−0.000081
ili
( 0.009) 1 ( 0.009)
2 2
− − = − ⋅ − =
2
( ) ( ) ( )
=−1⋅ 0.009 =− 0.009⋅ 0.009 =− 0.000081 =−0.000081
14

M-8-Kvadriranje i korjenovanje
Rješenja i kompletni postupak
5.
1)
Koristimo pravila:
Inačin:
Inačin:
2 2
a
2
⎛a⎞
⎜ ⎟
⎝b
⎠
2 2
= ⋅ II način: =
a a
⎛1⎞ 1 1 11 ⋅ 1 ⎛1⎞
1 1 11 ⋅ 1
⎜ ⎟ = ⋅ = = ili kraće: ⎜ ⎟ = ⋅ = =
⎝2⎠ 2 2 2⋅2 4 ⎝2⎠
2 2 2⋅2 4
a
b
2
II način: postoji pravilo koje kaže:
2 2
⎛1⎞
1 1
⎜ ⎟ = =
⎝ ⎠ 2
2
2 4
⎛a⎞
⎜ ⎟
⎝b
⎠
2 2
a
=
2
b
pa po njemu možemo rješiti ovaj zadatak:
2)
Inačin:
2 2 2
⎛1⎞ 1 1 1 ⎛1⎞
1 1
⎜ ⎟ = ⋅ = II način: ⎜ ⎟ = =
⎝ ⎠ ⎝ ⎠ 4
2
4 4 4 16 4 16
3)
Inačin:
II način
2 2
2 2
⎛2⎞ 2 2 22 ⋅ 4 ⎛2⎞
2 2 4
⎜ ⎟ = ⋅ = = ili kraće: ⎜ ⎟ = ⋅ =
⎝3⎠ 3 3 3⋅3 9 ⎝3⎠
3 3 9
⎛2⎞
2 4
⎜ ⎟ = =
2
⎝3⎠
3 9
4)
Inačin:
2
2 2
⎛4⎞
4 4 16
⎛4⎞
4 44 ⋅ 16
⎜ ⎟ = ⋅ =
II način: ⎜ ⎟ = = =
2
⎝5⎠
5 5 25 ⎝5 ⎠ 5 5⋅5 25
2 2
2
2 2 2 4
2
5) Inačin: rješenje možemo zapisati i ovako: 2
II način
⎛3⎞
3 3 9 9 9 1
⎜ ⎟ = ⋅ = =
⎝2⎠
2 2 4 4 4 4
⎛3⎞
3 33 ⋅ 9
⎜ ⎟ = = =
⎝ ⎠ 2 ⋅
6)
Inačin:
2 2 2
⎛5⎞ 5 5 25 ⎛5⎞
5 25
⎜ ⎟ = ⋅ = II način: ⎜ ⎟ = =
2
⎝4⎠ 4 4 16 ⎝4⎠
4 16
7)
2 2 2
⎛ 9 ⎞ 9 9 81 ⎛9⎞
9 99 ⋅ 81
⎜ ⎟ = ⋅ = II način: ⎜ ⎟ = = =
2
⎝7
⎠ 7 7 49 ⎝7⎠
7 7⋅7 49
8)
2 2 2
⎛11⎞ 11 11 121 ⎛11⎞
11 11⋅11
121
⎜ ⎟ = ⋅ = II način: ⎜ ⎟ = = =
2
⎝ 8 ⎠ 8 8 64 ⎝ 8 ⎠ 8 8⋅8 64
M-8-rješenja--radna varijanta ( nedovršena)
15
M.I.M.-Sraga --centar za poduku--01-4578-431

Rješenja i kompletni postupak
M-8-Kvadriranje i korjenovanje
5.
Koristimo pravilo:
Inačin:
2 2
2 2
2
2
2 2
= ⋅ II način: =
⎛ 1⎞ ⎛12 ⋅ + 1⎞
⎛2+
1⎞ ⎛3⎞
3 3 9
9) ⎜1 ⎟ = 2
⎜ 2
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⋅ =
⎝ ⎠ ⎝ ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 4
a
a a
↓
↑
⎛a⎞
⎜ ⎟
⎝b
⎠
mješoviti prvo pretvorimo u razlomak pa tek tada kvadriramo
a
= a⋅a
a
b
2
II način
⎛ 1 ⎞
⎜1 ⎟
⎝ 2 ⎠
2
2 2 2 2
⎛1⋅ 2+ 1⎞ ⎛2+
1⎞ ⎛3⎞
3 9
= ⎜
2
2
⎟ = ⎜ ⎟ = ⎜ ⎟ = =
⎝ ⎠ ⎝ 2 ⎠ ⎝2⎠
2 4
Inačin
II način
2 2
2 2
⎛ 2⎞ ⎛23 ⋅ + 2⎞
⎛6+
2⎞ ⎛8⎞
8 8 64
10) ⎜2 ⎟ = 3
⎜ 3
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⋅ =
⎝ ⎠ ⎝ ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ 3 3 9
2 2
2 2 2
⎛23 ⋅ + 2⎞
+
2 = 3
⎜ 3
⎟ = = = =
3 3
2 9
⎛ 2⎞ ⎛6 2⎞ ⎛8⎞
8 64
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3
Inačin
2
2 2 2
⎛ 1 ⎞ ⎛35 ⋅ + 1⎞
⎛15+
1⎞ ⎛16 ⎞ 16 16 256
11) ⎜3 ⎟ = 5
⎜
⎝ ⎠ ⎝ 5
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⋅ =
⎠ ⎝ 5 ⎠ ⎝ 5 ⎠ 5 5 25
II način
2 2
2 2 2
⎛ 1 ⎞ ⎛35 ⋅ + 1⎞
⎛15+
1⎞ ⎛16 ⎞ 16 1616 ⋅ 256
⎜3 ⎟ = 5
⎜ 5
⎟ = ⎜ ⎟ = ⎜ ⎟ = = =
5 5 5
2
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 5 ⋅ 5 25
Inačin
2 2
2 2
⎛ 5 ⎞ ⎛16 ⋅ + 5⎞
⎛6 + 5⎞ ⎛11⎞
11 11 121
12) ⎜1 ⎟ = 6
⎜ 6
⎟ = ⎜ ⎟ = ⎜ ⎟ = ⋅ =
⎝ ⎠ ⎝ ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ 6 6 36
II način
2 2
2 2 2
⎛ 5 ⎞ ⎛16 ⋅ + 5⎞ ⎛6 + 5⎞ ⎛11⎞
11 1111 ⋅ 121
⎜1 ⎟ =
⎝ 6
⎜
2
6
⎟ = ⎜ ⎟ = ⎜ ⎟ = = =
⎠ ⎝ ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ 6 6⋅6
36
16