ALGEBARSKI IZRAZI

**** MLADEN SRAGA ****
© - 2003.-2013.
UNIVERZALNA ZBIRKA
POTPUNO RIJEŠENIH ZADATAKA
PRIRUČNIK ZA SAMOSTALNO UČENJE
ALGEBARSKI IZRAZI
KVADRAT ZBROJA
KVADRAT RAZLIKE
α

M-1-Kvadrat binoma
Rješenja i kompletni postupak
34.
Koristeći se formulama za:
izračunaj:
( ) ( )
2 2 2 2 2
A+ B = A + 2AB+ B A− B = A − 2AB+
B
( x+ y) ( y+ x) ( x−
y) ( y−x)
2 2 2 2
1) 2) 3) 4)
( x+ ) ( x− ) ( −x) ( + x)
2 2 2 2
5) 3 6) 5 7) 2 8) 2
2 2 2
( x− ) ( x− ) ( a−
) ( −b)
9) 1 10) 4 11) 2 12) 3
2 2
( y− ) ( x+ ) ( − x)
16) ( 8+
y)
13) 5 14) 6 15) 7
kvadrat zbroja i kvadrat razlike
2 2
2 2 2
( + z) ( z− ) ( x+
y) (
17) 2 18) 2 19) 5 20) 5x−
y
2 2 2
( + ) ( − ) ( + ) (
21) 7x y 22) x 3y 23) 3x 2 24) 2x+
5
2 2 2
( − ) ( + ) ( + ) (
25) 2a 5b 26) 2x 3y 27) 5x 2y 28) 3x+
4y
2 2 2
( − ) ( + ) ( + ) (
29) 3x 4y 30) 5a 7b 31) 3m n 32) 2m−5n
2 2 2
( x+ ) ( x+ ) ( x−
y) (
33) 0,2 3 34) 1,2 0,3 35) 0,5 2 36) 0,2m+
0,3n
2 2 2
⎛ 1 ⎞
⎛3 ⎞ ⎛3 2 ⎞ ⎛ 5 ⎞
37) ⎜x
+ ⎟ 38) ⎜ x+ 1⎟ 39) ⎜ x+
y⎟ 40) ⎜0,4x−
y⎟ ⎝ 2 ⎠
⎝2 ⎠ ⎝4 3 ⎠ ⎝ 3 ⎠
2 2 2
⎛5 ⎞ ⎛ 3 ⎞ ⎛1 2 3⎞ ⎛2 3 3 4⎞
41) ⎜ x+ 0,2y⎟ 42) ⎜4x− y⎟ 43) ⎜ x+
y ⎟ 44) ⎜ x − y ⎟
⎝2 ⎠ ⎝ 2 ⎠ ⎝2 3 ⎠ ⎝3 4 ⎠
2 2 2
⎛ 1 ⎞ ⎛ 2 1 ⎞ ⎛ 2 ⎞ ⎛1 3 ⎞
45) ⎜1 x+ y⎟ 46) ⎜2 x−1 y⎟ 47) ⎜3 x−0,5y⎟ 48) ⎜ xz−
y⎟ ⎝ 2 ⎠ ⎝ 3 4 ⎠ ⎝ 3 ⎠ ⎝4 2 ⎠
49)
2 2 2 2 4 2
( x + ) ( x − ) ( −x
) (
1 50) 2 51) 3 52) 3a
( − ) ( − ) ( + ) (
( − ) ( + ) ( + ) (
( − ) ( − ) ( − ) (
53) 2a b 54) a 5b 55) 2x 3y 56) 3x
3 2 2 3 2 2 3 3 2 4
57) 2xy 3z 58) 2ab 3c 59) ab c 60) 2ab
2 3 2 2 2 2 3 2 2
61) ab cd 62) ab c 63) ab 3c 64) ab
2 3 2 4 2 2 3 2 2 3 4 2 2 3
( ) ( )
( )
4 3 2 7 2 3 2 3 2 4 3
65) 2ab −3ab 66) 4ab−5ab 67) 9ab
−2ab
68)
2 4 2 2
−2 −2 2
( − ) ( + ) ( − − ) (
2
)
)
−b
2
2
)
)
2
2
)
5 2
)
c)
c)
2 3 2
+ 7y
69) 5x 6y 70) x 4y 71) x 3 72) −x−2y
2 2 ⎛ 1 2 ⎞
73) ( −2a−3b) 74) ( −x−4y)
75) ⎜− x−4y ⎟ 76) −a
⎝ 2 ⎠
−3
+ 4
)
( + 7b)
2 2 3 4 2 2 3 4 2
m n 2
( − x+ y) ( − a b + c ) ( − x y + z ) ( − + 2 )
m n 2 m m 2 m n 2
m m−1 2
( + ) ( − ) ( + ) ( + 2 )
m n 2 m+ 1 m−1 2 n 1− n 2
n+
n−
( − 3 ) 86) ( 2 + 2 ) 87) ( 2 + 2 ) 88) ( 2 −2
)
n n+ 1 2 n− 1 n+
1 2 x x 2
x y
( a + a ) ( a −a ) ( a − b ) ( a + 3b
)
77) 3 78) 5 2 79) 2 3 80) 2
81) 2 3 82) 3 5 83) 2 2 84) 2
85) 3
89) 90) 91) 2 3 92) 2
2
2
2
)
2
2
4 2
5 2
⎛1 2 4 3⎞
⎜ ab−
ab ⎟
⎝2 3 ⎠
1 2 2
2
2
2
2
2
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
2 2
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
To pravilo možemo pisati i ovako:
( ) ( )
2 2 2 2
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
Sve o kvadratu binoma morali bi znati već iz 8. razreda ali ponovimo to još jednom.
Pogledajmo kako to radimo na dva primjera:
( + ) ( − )
2 2
a) x 1 b) x 1
ili
( )
objašnjenje:
( )
( )
2 2 2 2
a) x+ 1 = x + 2⋅x⋅ 1+ 1 = x + 2x+
1
2 2 2 2
x+ 1 = x + 2⋅x⋅ 1+ 1 = x + 2x+
1
↑ ↑ ↑ ↑ ↑
A+ B = A + 2 ⋅A⋅ B+
B
( )
( )
2 2 2
( )
( )
2 2 2 2
x+ 1 = x + 2⋅x⋅ 1+ 1 = x + 2x+
1
↑ ↑ ↑ ↑ ↑
2 2
I + II = I + 2⋅I⋅II
+ II
↓
ovo pravilo čitamo ovako:
2 2 2 2
2 2 2 2
2
Prvi plus drugi na kvadrat
jednako je prvi na kvadrat, plus dvostruki prvi puta drugi, plus drugi na kvadrat.
b) x− 1 = x −2⋅x⋅ 1+ 1 = x − 2x+
1
objašnjenje:
x− 1 = x −2⋅x⋅ 1+ 1 = x − 2x+
1
↑ ↑ ↑ ↑ ↑
A− B = A −2⋅A⋅ B+
B
( )
2 2 2
ili
( )
( )
2 2 2 2
x− 1 = x −2⋅x⋅ 1+ 1 = x − 2x+
1
↑ ↑ ↑ ↑ ↑
I − II = I −2⋅I⋅ II + II
2 2 2
( ) ( )
2 2 2 2 2 2
1) x + y = x + 2xy + y 2) y + x = y + 2⋅y⋅ x + x = x + 2xy + y
( ) ( )
2 2 2 2 2 2
3) x − y = x − 2xy+ y 4) y− x = y −2⋅y⋅ x+ x = x
2 2
+ 2xy+
y
2 2
( )
2 2 2
5) x+ 3 = x + 2⋅x⋅ 3 + 3 =
6x
9
2
= + +
x
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tel-01-4578-431

M-1-Kvadrat binoma
Rješenja i kompletni postupak
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
To pravilo možemo pisati i ovako:
( )
2 2 2
6) x− 5 = x −2⋅x⋅ 5 + 5 =
x
10x
25
2
= − +
( ) ( )
2 2 2 2 2 2
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
( )
2 2 2
7) 2 − x = 2 −2⋅2⋅ x+ x =
= 4− 4 +
2
x x →
to možemo pisati i dalje:
2
= x − x+
4 4
( )
2 2 2
8) 2 + x = 2 + 2⋅2⋅ x+ x =
= + + = + +
2 2
4 4x x x 4x
4
dobro je kako god napišete ...
( )
( )
( )
( )
2 2 2
9) x− 1 = x −2⋅x⋅ 1+ 1 =
2
= x − x+
2 1
2 2 2
10) x− 4 = x −2⋅x⋅ 4+ 4 =
2
= x − x+
8 16
2 2 2
11) a− 2 = a −2⋅a⋅ 2+ 2 =
2
= a − a+
4 4
2 2 2
12) 3− b = 3 −2⋅3⋅ b+ b =
( )
( )
( )
( )
= 9− 6b+
b
2 2 2
13) y− 5 = y −2⋅y⋅ 5+ 5 =
2
2
= − +
y
10y
25
2 2 2
14) x+ 6 = x + 2⋅x⋅ 6+ 6 =
2
= + +
x
12x
36
2 2 2
15) 7− x = 7 −2⋅7⋅ x+ x =
= 49− 14x+
x
2 2 2
16) 8+ y = 8 + 2⋅8⋅ y+ y =
= 64+ 16y+
y
2
2
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
To pravilo možemo pisati i ovako:
( ) ( )
2 2 2 2 2 2
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
( )
( )
2 2 2
17) 2 + z = 2 + 2⋅2⋅ z+ z =
= 4+ 4z+
z
2 2 2
18) z− 2 = z −2⋅z⋅ 2 + 2 =
2
= − +
2 2
( x+ y)
= x + 2 x 5y ( 5y)
19) 5
Još jednom isti zadatak:
z
2
4z
4
( ) ( )
( ) ( )
2 2 2
= x + xy+ y =
2 2
2
2 2 2 2 2 2 2 2
Primjeni pravilo:
2 2
⋅ ⋅ + =
10 5
= x + 10xy+
25y
19) x+ 5y = x + 2⋅x⋅ 5y+ 5y = x + 10xy+ 5 y = x + 10xy+
25y
i ovdje
n n n
( ) u našem slučaju: ( )
2 2 2 2
ab = a b 5y = 5 y = 25y
20) 5x− y = 5x −2⋅5x⋅ y+
y = 5 x − 10xy+ y = 25x − 10xy+
y
2 2 2 2 2 2
ili možemo pisati i ovako:
( ) ( )
2 2 2
5x− y = 5x −2⋅5x⋅ y+ y =
5 10
2 2 2
= x − xy+ y =
= 25x − 10xy+
y
2 2
( ) ( )
2 2 2
21) 7x+ y = 7x + 2⋅7x⋅ y+ y =
2 2 2
= x + xy+ y =
7 14
= 49x + 14xy+
y
2 2
2
( x y) x x y ( y)
2 2
22) − 3 = −2⋅ ⋅ 3 + 3 =
2 2 2
= x − xy+ y =
6 3
= x − 6xy+
9y
2 2
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
To pravilo možemo pisati i ovako:
( ) ( )
2 2 2
23) 3x+ 2 = 3x + 2⋅3x⋅ 2+ 2 =
2 2
3 x 12x
4
= + + =
= x + x+
2
9 12 4
( ) ( )
2 2 2 2 2 2
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
( ) ( )
2 2 2
24) 2x+ 5 = 2x + 2⋅2x⋅ 5+ 5 =
2 2
2 x 20x
25
= + + =
= x + x+
2
4 20 25
Sve ove zadatke možemo rješavati u jednom redu ali tada je teško pisati upute,
kada svaki korak pišem u novi red tada desno od zadatka ima mjesta za uputu.
2 2
( a− b) = ( a)
− ⋅ a⋅ b+( )
25) 2 5 2 2 2 5
2 2 2 2 2 2 2
5b = 2 a − 20ab+ 5 b = 4a − 20ab+
25b
Još jednom isti zadatak:
2 2 2
( 2 5 ) ( 2 ) 2 2 5 ( 5 ) Primjeni: ( )
2 2 2 2
= − + =
2 a 20ab 5 b
= 4a − 20ab+
25b
2 2
n n n
a− b = a − ⋅ a⋅ b+ b = ab = a b
( x y) ( x) x y ( y)
2 2 2
26) 2 + 3 = 2 + 2⋅2 ⋅ 3 + 3 =
2 2 2 2
= + + =
2 x 12xy 3 y
= 4x + 12xy+
9y
2 2
2 2
( x+ y) = ( x)
+ ⋅ x 2y ( 2y)
27) 5 2 5 2 5
2 2 2 2
= x + xy+ y =
5 20 2
= 25x + 20xy+
4y
2 2
2
⋅ + =
( x y) ( x) x y ( y)
2 2 2
28) 3 + 4 = 3 + 2⋅3 ⋅ 4 + 4 =
2 2 2 2
= x + xy+ y =
3 24 4
= 9x + 24xy+
16y
2 2
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
34.
Kvadriramo po pravilu:
To pravilo možemo pisati i ovako:
( x y) ( x) x y ( y)
( ) ( )
2 2 2
29) 3 − 4 = 3 −2⋅3 ⋅ 4 + 4 =
2 2 2 2
= x − xy+ y =
2 2
2 2 2 2 2 2
A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
3 24 4
= 9x − 24xy+
16y
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
2
( a+ b) = ( a)
25a 7b ( 7b)
30) 5 7 5
2 2
+ ⋅ ⋅ + =
2 2 2 2
= + + =
5 a 70ab 7 b
= 25a + 70ab+
49b
2 2
( ) ( )
2 2 2
31) 3m+ n = 3m + 2⋅3m⋅ n+ n =
2 2 2
= + + =
3 m 6mn n
= 9m + 6mn+
n
2 2
( m n) ( m) m n ( n)
2 2 2
32) 2 − 5 = 2 −2⋅2 ⋅ 5 + 5 =
2 2 2 2
= − + =
2 m 20mn 5 n
= 4m − 20mn+
25n
2 2
( ) ( )
2 2 2 2 2 2
33) 0,2x+ 3 = 0,2x + 2⋅0,2⋅x⋅ 3+ 3 = 0,2 x + 1,2x+ 9 = 0,04x
+ 1,2x
+ 9
Još jednom taj isti zadatak:
( ) ( )
2
2 0,2 3 1,2 0,2 0,2 0,2 0,04
2 2 2
2 2 2
0,2 1,2 9 0,2 0,2 0,2 0,04
2
0,04 1, 2 9
Uputa:
33) 0,2x+ 3 = 0,2x + 2⋅0,2⋅x⋅ 3+ 3 = → 2⋅0,2⋅ 3 = 1,2
= x + x+ = → = ⋅ =
= x + x+
↓
↓
⋅ ⋅ = = ⋅ =
( ) ( )
2 2 2
34) 1,2x+ 0,3 = 1,2x + 2⋅1,2x⋅ 0,3+ 0,3 = → 2⋅1,2⋅0,3 = 0,72
= x + ⋅ ⋅ ⋅ x+ = → = ⋅
2 2 2
1, 2 2 1, 2 0, 3 0, 09 1, 2 1, 2 1, 2 1,
= x + x+
2
1,44 0,72 0,09
=
44
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
To pravilo možemo pisati i ovako:
( ) ( )
2 2 2 2 2 2
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
( x y) ( x) x y ( y)
2 2 2
35) 0,5 − 2 = 0,5 −2⋅0,5⋅ ⋅2⋅ + 2 =
2 2 2 2
= x − x+ y =
0,5 2 2
= 0, 25x − 2x+
4y
2 2
( ) ( ) ( )
2 2 2
36) 0,2m+ 0,3n = 0,2m + 2⋅0,2⋅m⋅0,3⋅ n+ 0,3n
=
0, 2 m 0,12mn 0,3 n
2 2 2 2
= + + =
= 0,04m + 0,12mn+
0,09n
2 2
2 2
2
⎛ 1⎞ 1 ⎛1⎞
37) ⎜x+ ⎟ = x + 2⋅x⋅ + ⎜ ⎟ =
⎝ 2⎠ 2 ⎝2⎠
2
2
= x + ⋅ ⋅ x+ =
2
2
= x + x+
1 1
2 2 2
1
4
2 2
2
+ = + ⋅ ⋅ ⋅ + =
⎛3 ⎞ ⎛3 ⎞ 3
38) ⎜ x 1⎟ ⎜ x⎟
2 x 1 1
⎝2 ⎠ ⎝2 ⎠ 2
3
2
9
x
4
2
2
= x + 3x+ 1=
2
2
= + +
3x
1
2 2 2
⎛3 2 ⎞ ⎛3 ⎞ 3 2 ⎛2
⎞
39) ⎜ x+ y⎟ = ⎜ x⎟ + 2⋅ ⋅x⋅ ⋅ y+ ⎜ y⎟
=
⎝4 3 ⎠ ⎝4 ⎠ 4 3 ⎝3
⎠
2 2
3 2 3 2 2 2
3 2 232 ⋅ ⋅
= x + 2 ⋅ ⋅ ⋅x⋅ y+ y = → drugi član⋅ 2⋅ ⋅ xy = xy = 1xy
2 2
4 4 3 3
4 3 4⋅3
9 4
= x + xy+
y
16 9
2 2
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
( ) ( )
2 2 2 2 2 2
2 2 2
3 3 3
⎛1 2 ⎞ ⎛1 ⎞ 1 2 ⎛2
⎞
43) ⎜ x+ y ⎟ = ⎜ x⎟ + 2⋅ ⋅x⋅ ⋅ y + ⎜ y ⎟ =
⎝2 3 ⎠ ⎝2 ⎠ 2 3 ⎝3
⎠
1 2 2
= + + ( ) = → ( ) = =
2 3 3
1 2 2 3 4 6
= x + xy + y
4 3 9
2 2
2 3 3 2 3 2 3⋅2 6
x xy y y y y
2 2
2 2
3 4 3
⎛2 3 ⎞ ⎛2 ⎞ 2 3 3 4 ⎛3
4⎞
44) ⎜ x − y ⎟ = ⎜ x ⎟ −2⋅
x ⋅ y + ⎜ y ⎟ =
⎝3 4 ⎠ ⎝3 ⎠ 3 4 ⎝4
⎠
32 ⋅ 3 4 42 ⋅
= − ⋅ + =
2
2 2
2 3 2 2 3 3 4 3 4 2
2 3 2⋅2⋅3
=
2 ( x ) −2⋅ ⋅ ⋅x ⋅ y +
2 ( y ) = →2⋅ ⋅ = = 1
3 3 4 4 3 4 3⋅4
4 9
x 1 x y y
9 16
4 6 3 4 9 8
= x − x y + y
9 16
⎛
⎜
⎝
2
1 ⎞
1
x+ y⎟
=
2 ⎠
2
45) 1 Odmah treba mješoviti broj: 1
2 2 2
2
1 x y x y x 2 x y y
1
2 2
pretvoriti u razlomak pa tek tada kvadrirati:
⎛ 1 ⎞ ⎛3 ⎞ ⎛3 ⎞ 3 1 1.2+
1 3
⎜ + ⎟ = ⎜ + ⎟ = ⎜ ⎟ + ⋅ ⋅ + = → = =
⎝ 2 ⎠ ⎝2 ⎠ ⎝2 ⎠ 2 2 2 2
2
3
= x + xy+
y
2
2
2 2
⎛ 2 1 ⎞ ⎛2⋅ 3+ 2 1⋅ 4+
1 ⎞
46) ⎜2 x− 1 y⎟ = ⎜ x− y⎟
=
⎝ 3 4 ⎠ ⎝ 3 4 ⎠
2
⎛8 5 ⎞
= ⎜ x− y⎟
= →
⎝3 4 ⎠
2 2
⎛8 ⎞ 8 5 ⎛5
⎞
= ⎜ x⎟ −2⋅ x⋅ y+ ⎜ y⎟
=
⎝3 ⎠ 3 4 ⎝4
⎠
2 2
8 2 8⋅5 5 2
= x − ⋅x⋅ y+ y =
2 2
3 3⋅2 4
64 40 25
= x − xy+
y
9 6 16
2 2
Odmah treba mješoviti broj pretvoriti u razlomak
pa tek tada kvadrirati
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M-1-Kvadrat binoma
Rješenja i kompletni postupak
34. Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
( ) ( )
2 2 2 2 2 2
2 2 2
⎛ 2 ⎞ ⎛3⋅ 3+
2 5 ⎞ ⎛11 1 ⎞
47) ⎜3 x− 0,5y⎟ = ⎜ x− y⎟ = ⎜ x− y⎟
=
⎝ 3 ⎠ ⎝ 3 10 ⎠ ⎝ 3 2 ⎠
2 2
⎛11 ⎞ 11 1 ⎛1
⎞
= ⎜ x⎟ −2⋅ ⋅x⋅ ⋅ y+ ⎜ y⎟
=
⎝ 3 ⎠ 3 2 ⎝2
⎠
2 2
11 2 11 1 2
= x − xy+ y =
2 2
3 3 2
121 2 11 1 2
= x − xy+
y
9 3 4
48)
2 2 2
⎛1
3 ⎞ ⎛1 ⎞ 1 3 ⎛3
⎞
⎜ xz − y⎟ = ⎜ xz⎟ −2⋅ ⋅x⋅z⋅ ⋅ y+ ⎜ y⎟
=
⎝4
2 ⎠ ⎝4 ⎠ 4 2 ⎝2
⎠
2 2
1 2 2 1 3 3 2
= xy −2⋅ ⋅ ⋅xzy ⋅ ⋅ + y =
2 2
4 4 2 2
1 2 2 3 9 2
= xy − xzy+
y
16 4 4
( ) ( ) ( )
x + = x + ⋅x ⋅ + = → x = x = x
2 2 2 2 2 2 2 2 2⋅2 4
49) 1 2 1 1 po pravilu:
22 ⋅ 2
= + + =
x
2x
1
4 2
= + +
x
2x
1
( a )
= a
n m n⋅m
2
( ) ( )
2 2 2 2 2
50) x − 2 = x −2⋅x
⋅ 2+ 2 =
= x
x
− 4x
+ 4 =
22 ⋅ 2
4x
4
4 2
= − +
2
( x ) x ( x )
51) 3 3 2 3
4 2 4 4 2
− = − ⋅ ⋅ + =
x
x
4 4⋅2
= 9− 6 + =
= 9− 6x
+ x
4 8
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34. ( ) ( )
2
( a b ) ( a ) a b ( b )
2
2 2 2 3 3⋅2
= 3 ( a ) − 6a b + b =
2 2 2 2 2
Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
2 3 2 2 2 3 3 2
52) 3 − = 3 −2⋅3⋅ ⋅ + =
= 9a − 6a b + b = 9a − 6a b + b
22 ⋅ 2 3 6 4 2 3 6
2
2
( a b ) ( a ) a b ( b )
2 3
2
3 2 2⋅2
= 2 ( a ) − 4a b + b =
3 2 3 2 3 2 2 2
53) 2 − = 2 −2⋅2⋅ ⋅ + =
32 ⋅ 3 2 4
= 4a − 4a b + b =
= 4a − 4a b + b
6 3 2 4
( a b ) ( a ) a b ( b )
32 ⋅ 3 2 2 2 2
= a − 10a b + 5 ( b ) =
3 2 2 3 2 3 2 2 2
54) − 5 = −2⋅ ⋅5⋅ + 5 =
= a − 10a b + 25b
6 3 2 4
( x y ) ( x ) x y ( y )
2 3 2 3 3 2 3 2
2 ( x ) 12x y 3 ( y )
3 3 2 3 2 3 3 3 2
55) 2 + 3 = 2 + 2⋅2⋅ ⋅3⋅ + 3 =
= + + =
x x y y
32 ⋅ 3 3 32 ⋅
= 4 + 12 + 9 =
= 4x + 12x y + 9y
6 3 3 6
( x y ) ( x ) x y ( y )
2 4 2 4 5 2 5 2
= 3 ( x ) 42x y 7 ( y )
4 5 2 4 2 4 5 5 2
56) 3 + 7 = 3 + 2⋅3⋅ ⋅7⋅ + 7 =
+ + =
x x y y
42 ⋅
4 5 52 ⋅
= 9 + 42 + 49 =
= 9x + 42x y + 49y
8 4 5 10
2 3 2 2 2 3 2 2 3 2 2 2
( xy− z) = ( xy) − ⋅ ⋅x⋅y⋅ ⋅ z + ( z)
=
2 ( x ) ( y ) 4 3 x y z 3 ( z )
57) 2 3 2 2 2 3 3
2 2 2 3 2 2 3 2 2 2 2
= − ⋅ ⋅ ⋅ ⋅ + =
x y x y z z
22 ⋅ 32 ⋅ 2 3 2 22 ⋅
= 4 − 12 + 9 =
= 4xy − 12xyz + 9z
4 6 2 3 2 4
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34. ( ) ( )
2 2
2 2
Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
2 2
( ) ( ) ( )
58) 2ab+ 3c = 2ab + 2⋅2⋅a⋅b⋅3⋅ c + 3c
=
2 2 2 2 2
= 2 ab + 12abc+ 3 c =
= 4a b + 12abc + 9c
2 3 2 2 2 2 3 3 2
59) ab+ c = ab + 2⋅a ⋅bc ⋅ + c =
2 4
60) ( 2ab
3c
)
2 2 2
2 2 2
( ) ( ) ( )
( )
2 2 2 2 3 2
= a ⋅ b + 2a bc + c =
= ab + 2abc + c
4 2 2 3 2
− ( 2ab ) 2 2 a b 3 c ( 3c
)
2 2 2 2 2 4 2 4 2
2 a ( b ) 12ab c 3 ( c )
2 2 2 2 4 4 2
2 2⋅2 2 4 4⋅2
= 4ab − 12abc + 9c
=
= 4ab − 12abc + 9c
2 4 2 4 8
2 3 2 4 2 2 3 2 2 3 2 4 2 4 2
( ab − cd ) = ( ab) − ⋅a ⋅b⋅c ⋅ d + ( cd ) =
( a ) ( b ) 2a b c d ( c ) ( d )
61) 2
= − ⋅ ⋅ ⋅ ⋅ ⋅ + =
= − + =
2 2 3 2 2 3 2 4 2 2 4 2
= ⋅ − + ⋅ =
22 ⋅ 32 ⋅ 2 3 2 4 22 ⋅ 42 ⋅
= a b − 2a b c d + c d =
4 6 2 3 2
= ab −2abcd
4 4 8
( ab c ) ( ab) a b c ( c )
2 2 3 2 2 3 3 2⋅2
( a ) ( b ) 2a b c c
2 3 2 2 2 3 2 2 3 3 2 2
62) − = −2⋅ ⋅ ⋅ + =
a b 2a b c c
22 ⋅ 32 ⋅ 2 3 3 4
= − + =
= ab − 2abc + c
4 6 2 3 3 4
( ab c ) ( ab) ab c ( c )
( a ) b 6a bc 3 ( c )
3 4 2 3 2 3 4 4 2
63) − 3 = −2⋅ ⋅3⋅ + 3 =
3 2 2 3 4 2 4 2
= ⋅ − + =
a b 6a bc 9c
32 ⋅ 2 3 4 42 ⋅
= − + =
= ab − 6abc + 9c
6 2 3 4 8
+ c d
= ⋅ − + =
5 2 2 3 2 2 3 5 5 2
( ab + c) = ( ab) + 2⋅a ⋅b⋅4⋅ c + ( 4c)
=
2 3
64) 4
( a ) ( b ) 8a b c 4 ( c )
2 2 3 2 2 3 5 2 5 2
= ⋅ + + =
22 ⋅ 32 ⋅ 2 3 5 52 ⋅
= a b + 8a b c + 16c
=
= ab + 8abc + 16c
4 6 2 3 5 10
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34. ( ) ( )
2 2 2 2 2 2
Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
4 3 2 7 2 4 3 2 4 3 2 7 2 7 2
( ab − ab ) = ( ab) − ⋅ ⋅a ⋅b⋅ ⋅a ⋅ b + ( ab ) =
2 ( a ) ( b ) 12 a a b b 3 ( a ) ( b )
65) 2 3 2 2 2 3 3
2 4 2 3 2 4 2 3 7 2 2 2 7 2
= ⋅ − ⋅ ⋅ ⋅ ⋅ + ⋅ =
42 ⋅ 32 ⋅ 4+ 2 3+ 7 22 ⋅ 72 ⋅
= 4a b −12⋅a ⋅ b + 9a b =
= 4ab − 12ab + 9ab
8 6 6 10 4 9
66)
3 2 3 2 3 2 3 2 3 2 3 2
( 4ab− 5ab) = ( 4ab) −2⋅4⋅a⋅b⋅5⋅a ⋅ b + ( 5ab)
=
4 ( a ) b 2 4 5 a a b b 5 ( a ) ( b )
2 3 2 2 3 2 1 3 2 2 2 3 2
= ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ =
3⋅ 2 2 3+ 2 1+ 3 2⋅2 3⋅2
= 16 −40 ⋅ + 25 =
a b a b a b
= 16ab − 40ab + 25ab
6 2 5 4 4 6
2 2 2
4 3 2 4 4 3 4 3 2 4 2 4
( ab − ab ) = ( ab) − ⋅ ⋅a ⋅b⋅ ⋅a ⋅ b + ( ab ) =
2 2
2 4 3 4 2 3 4 2 2 4
= 9 ( a ) ( b ) −2⋅9⋅2⋅ a a b b 2 ( a ) ( b )
67) 9 2 9 2 9 2 2
a b a b a b
42 ⋅ 32 ⋅ 4+ 2 3+ 4 22 ⋅ 42 ⋅
= 81 −36⋅ ⋅ + 4 =
= 81ab − 36ab + 4ab
8 6 6 7 4 8
2 2 2
2 4 3 2 2 4 3 4 3
⎛1 2 ⎞ ⎛1 ⎞ 1 2 ⎛2
⎞
68) ⎜ ab− ab ⎟ = ⎜ ab⎟ −2⋅ ⋅a ⋅b⋅ ⋅a ⋅ b + ⎜ ab ⎟ =
⎝2 3 ⎠ ⎝2 ⎠ 2 3 ⎝3
⎠
2 2
⋅ ⋅ ⋅ + =
2 2
1 2 2 2 1 2 2 4 1 3 2 4 2 3 2
= ⋅
2 ( a ) ⋅b −2⋅ ⋅ ⋅a ⋅a ⋅b ⋅ b +
2 ( a ) ⋅ ( b ) =
2 2 3 3
1 22 ⋅ 2 2 2+ 4 1+ 3 4 4⋅2 3⋅2
= a b − ⋅a ⋅ b + a b =
4 3 9
1 4 2 2 6 4 4 6 5
= ab − ab + ab
4 3 9
( x y) ( x y) ( ) ( x y)
( 5x) −2⋅5⋅x⋅6⋅ y+
( 6y)
−
⋅
( ) ( ) ( −
x y x y ) ( x y)
1
( )
( 5x−
6y)
−2 2⋅ −1 2 −1
69) 5 − 6 = 5 − 6 = 5 − 6 = =
1 1 1
= = =
5 x − 30xy + 6 y 25x − 30xy + 36y
2 2 2 2 2 2 2 2
2 2 1 2
−1
1
( )
( x+
4y)
70) + 4 = + 4 = + 4 = =
1 1 1
= = =
2
2 2 2 2 2
x + 2⋅x⋅4⋅ y+
4y
x + 8xy+ 4 y x + 8xy
+ 16y
( )
2
2
2
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34.
71)
U formulama piše:
Pokažimo zašto je to tako:
Objašnjenje:
( −a− b) = ( a+
b)
2 2
2 2 2
( −a− b) = ( −1⋅ ( a+ b)
) = ( −1) ⋅ ( a+ b) = 1⋅ ( a+ b) = ( a+
b)
2
( ) ( ),
2
( ) ( )
−a− b = −1⋅ a+ b = Izlučili smo zajednički faktor −1
2 2
( ) ( ) ( )
= −1
⋅ a+ b = pa smo primjenili pravilo: a⋅b
( a b) ( )
( a b)
2
2 2
= 1⋅ + = − 1 = 1
= +
2 2
n n n
= a ⋅b
2
Sada taj postupak primjenimo u: 71. , 72. , 73. , 74. ,
2
( ) ( ) ( ) ( )
( ) ( )
2 2 2 2 2 2
71) −x− 3 = −1⋅ x+ 3 = −1 ⋅ x+ 3 = 1⋅ x + 2⋅x⋅ 3+ 3 = x + 6x+
9
2
( ) Izlučimo zajednički faktor ( )
2
( x y) ( x y)
72) − − 2 = −1⋅ + 2 = −1 ,
2 2
= ( −1) ⋅ ( x+ 2y)
= pa primjenimo pravilo: ( )
2
2
= 1⋅ ( x + 2⋅x⋅2⋅ y+ ( 2y)
) =
2 2 2
( x xy y )
= 1⋅ + 4 + 2 =
= x + 4xy+
4y
2 2
n n n
ab ⋅ = a ⋅b
( )
2
( a b) ( a b)
73) −2 − 3 = −1⋅ 2 + 3 =
2 2
( 1) ( 2a
3b)
(( ) ( ) )
= − ⋅ + =
2 2 2 2
( a ab b )
2
( )
2
( x y) ( x y)
2 2 2
2 2
74) − − 4 = −1⋅ + 4 =
( 1) ( x 4y)
2 2
= 1⋅ 2a + 2⋅2⋅a⋅3⋅ b+ 3b = = 1⋅ x + 2⋅x⋅4⋅ y+
4y
= 1⋅ 2 ⋅ + 12 + 3 =
= 4a + 12ab+
9b
= − ⋅ + =
2
2
( ( ) )
2
2 2 2
8 4 ( )
= x + xy+ y =
= x + 8xy+
16y
2 4
2
=
2
2
1 2 ⎛ 1 2 ⎞
2 2
− − = − ⋅ + = − + = − =
⎛ ⎞ ⎛ ⎞
75) ⎜ x 4y ⎟ 1 4 76) ( 7 ) ( 7 )
2
⎜ ⎜ x y ⎟
2
⎟
a b b a
⎝ ⎠ ⎝ ⎝ ⎠⎠
2 2
2
= ( 7b)
−2⋅7⋅b⋅ a+ a =
2 ⎛1
2 ⎞
= ( −1)
⋅ ⎜ x+ 4y
⎟ =
2 2 2
⎝2
⎠
= 7 b − 14ba+
a
2
2 2
⎛⎛1 ⎞ 1
2 ⎞
= 49b − 14ba+
a
2 2
= 1⋅ x + 2⋅ ⋅x⋅4⋅ y + ( 4y
) =
⎜⎜
⎟
⎝2 ⎠ 2
⎟
ili
⎝
⎠
2 2
2
= a − 14ab+
49b
1
2
2 2 2 2
= x + 4xy + 4
2
( y ) =
2
1 2 2 4
= x + 4xy + 16y
4
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34.
( x y) ( y x)
2 2
77) − 3 + = − 3 =
( )
2 2
= y −2⋅y⋅3⋅ x+ 3x
=
2 2 2
= y − 6yx+ 3 x =
= y − 6yx + 9x ili = 9x − 6 xy + y
2 2 2 2
prvom i drugom članu zamjenimo mjesta
tako da drugi bude sa predznakom
( −)
...i dalje je lako...
obadva rj. su ispravna i ista...
Taj zadatak možemo rješiti i na još dva načina:
II način:
III način:
( )
2
( ) ( )
2
− + = − ⋅ − =
2 2 2
− + = − + ⋅ − ⋅ + =
77) 3x y 1 3x y 3x y 3x 2 3x y y
2 2
( ) ( )
2 2
1 (( 3x)
2 3x y y )
2 2 2
= − + =
3 x 6xy y
= 9x − 6xy+
y
2 2
( ) ( ) ( )
( )
2
= −1 ⋅ 3x− y = 2
2
= −3 x −2⋅3⋅x⋅ y+ y =
= ⋅ − ⋅ ⋅ + =
2 3 4 2 4 2 3 2
( − ab + c ) = ( c − ab)
=
4 2 4 2 3 2 3 2
( 2c ) 2 2 c 5 a b ( 5a b )
2 ( c ) 20a b c 5 ( a ) ( b )
78) 5 2 2 5
= − ⋅ ⋅ ⋅ ⋅ ⋅ + =
2 4 2 2 3 4 2 2 2 3 2
= − + ⋅ =
= 4c
−20a b c
42 ⋅
2 3 4
22 ⋅ 32 ⋅
+ 25 =
2 3 4 2 4 2 3 2
( − xy+ z) = ( z − xy)
=
( 3z ) 2 3 z 2 x y ( 2x y )
3 ( z ) 12z x y 2 ( x ) ( y )
79) 2 3 3 2
= 9x − 6xy+
y
2 2
Uvijek dobijemo isto rješenje...pa birajte...
na koji način će te rješavat ovakve zadatke
8 2 3 4 4 6 4 6 2 3 4 8
4 2 4 2 3 2 3 2
= − ⋅ ⋅ ⋅ ⋅ ⋅ + =
2 4 2 4 2 3 2 2 2 3 2
= − + ⋅ =
42 ⋅ 2 3 4 22 ⋅ 32 ⋅
= 9 − 12 + 4 =
a
b
= 4c − 20abc + 25ab ili = 25ab − 20abc + 4c
z x y z x y
= z− xyz + xy = xy − xyz +
8 2 3 4 4 6 4 6 2 3 4 8
9 12 4 ili 4 12 9
m n 2 n m 2
( − + ) = ( − ) =
n n m m
( 2 ) 2 2 2 ( 2 )
80) 2 2 2 2
2 2
= − ⋅ ⋅ + =
n⋅2 1 n m m⋅2
= 2 −2 ⋅2 ⋅ 2 + 2 =
2n 1+ n+
m 2m
= 2 − 2 + 2
2⋅ n 1+ n+ m 2⋅m
= 2 − 2 + 2
=
=
2
n
n+ m+
1 2
( 2 ) 2 ( 2 )
= − + =
m
Rješenje možemo ostaviti u ovom obliku...
n n+ m+
1 m
= 4 − 2 + 4 ili u ovom obliku, ovisi o tome kako radite u školi...
z
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34.
m n m m n n
( ) ( ) ( )
( ) ( )
2 2 2
+ = + ⋅ ⋅ + =
m⋅2 1 m n n⋅2
= 2 + 2 ⋅2 ⋅ 3 + 3 =
m
2 m+
1 n 2
( 2 ) 2 3 ( 3 )
2 2 2 2 2 2
Kvadriramo po pravilu: A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
81) 2 3 2 2 2 3 3
2⋅ m 1+ m n 2⋅n
= 2 + 2 ⋅ 3 + 3 = =
= + ⋅ + =
= + ⋅ +
n m+
1 n n
4 2 3 9
m n m m n n
( ) ( ) ( )
83) 2 2 2 2 2 2 2
2 2 2
+ = + ⋅ ⋅ + =
2 2 2 2 2
m⋅2 1 m n n⋅2
= + ⋅ ⋅ + =
m
2 m+ n+
1 2
( 2 ) 2 ( 2 )
n
m m m m m m
( ) ( ) ( )
82) 3 5 3 2 3 5 5
2 2 2
− = − ⋅ ⋅ + =
( )
m⋅2 m m⋅2
= 3 −2⋅ 3⋅ 5 + 5 =
m
2 m 2
( 3 ) 2 15 ( 5 )
m m− m m m− m−
( ) ( ) ( )
1 2 2 1 1 2
+ = + ⋅ ⋅ + =
m⋅2 1 m m−1
2 2 2 2 2
2⋅ m 1+ m+ n 2⋅n 2⋅m
1+ m+ m−1
2 2 2 2 2
m m+ n+
1 n
4 2 4
n
2⋅m m 2⋅m
3 −2⋅ 15 + 5 =
= − ⋅ + =
= 9 −2⋅ 15 + 25
m m m
84) 2 2 2 2 2 2 2
= + + = = +
= + + =
= + +
m n m m n n
( ) ( ) ( )
85) 3 3 3 2 3 3 3
2 2 2
− = − ⋅ ⋅ + =
m⋅ 2 m+ n n⋅2
= 3 −2⋅ 3 + 3 =
2⋅ m m+ n 2⋅n
= 3 −2⋅ 3 + 3 =
m
2 m+
n 2
( 3 ) 2 3 ( 3 )
= − ⋅ + =
+
= 9 −2⋅ 3 + 9
m m n n
m+ m− m+ m+ m− m−
( ) ( ) ( )
1 1 2 1 2 1 1 1 2
86) 2 + 2 = 2 + 2⋅2 ⋅ 2 + 2 =
= 2
( m 1)
+ ⋅2 1 m+ 1 m−1
m−1 ⋅2
n
( )
⋅ ( m+ ) 1+ m+ 1+ m−1
⋅( m−
)
2 1 2 1
m
2 2m+
1 2
( 2 ) 2 ( 2 )
+ 1 m−1
m+ 1 2m+ 1 m−1
n −n n n −n −n
( ) ( ) ( )
1 2 2 1 1 2
+ = − ⋅ ⋅ + =
n
2 2 2
( 2 ) 2 ( 2 )
+ 2 ⋅2 ⋅ 2 + 2 =
= 2 + 2 + 2 =
= + + =
= 4 + 2 + 4
87) 2 2 2 2 2 2 2
( −n)
n⋅2 1 n 1−n
1 ⋅2
= 2 −2 ⋅2 ⋅ 2 + 2 =
⋅( −n)
2⋅
n 1+ n+ 1−n
2 1
2 2 2
= − + =
n
= 4 − 4+
4
1−n
1−n
= − + =
m
( m−
)
1 ⋅2
= + ⋅ ⋅ + =
m
2 2m
2
( 2 ) 2 ( 2 )
m
m 2 m−1
4 ( 2 ) 4
m m m−1
= 4 + 4 + 4 =
m
= 24 ⋅ + 4
m−1
⋅( m−
)
2 1
+ 2 =
m−1
= + + =
= + + =
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2 2 2 2 2
2
34. Kvadriramo po pravilu: ( A+ B) = A + 2⋅A⋅ B+ B i ( A− B)
= A −2⋅A⋅ B+
B
n+ n− n+ n+ n− n−
( ) ( ) ( )
1 2 2 1 2 1 2 2 2
88) 2 − 2 = 2 −2⋅2 ⋅ 2 + 2 =
( n ) 1 n+ 1 n−2
( n )
+ 1 ⋅2 −2 ⋅2
= 2 −2 ⋅2 ⋅ 2 + 2 =
( n ) 1+ n+ 1+ n−2
( n )
2⋅ + 1 2⋅ −2
= 2 − 2 + 2 =
n
2
( 2 )
2n
2
2
( 2 )
n+ 1
4
n
2
( 2 )
n−2
4
+ 1 n−2
= − + =
= − + =
= 4 − 4 + 4
n+ 1 n n−2
89)
n n+ 1 2 n 2 n n+ 1 n+
1 2
( a + a ) = ( a ) + 2⋅a ⋅ a + ( a ) =
( n )
n⋅ 2 n+ n+
1 + 1 ⋅2
= a + 2⋅ a + a =
= a + 2a + a
2n 2n+ 1 2n+
1
n− n+ n− n− n+ n+
( a a ) ( a ) a a ( a )
1 1 2 1 2 1 1 1 2
90) − = −2⋅ ⋅ + =
( n ) n−+ 1 n+
1 ( n )
−1 ⋅ 2 + 1 ⋅2
= a −2⋅ a + a =
= a − 2a + a
2n− 2 2n 2n+
2
x x x x x x
( a b ) ( a ) a b ( b )
2
2
2 x x x 2 x
= 2 ( a ) −12⋅a ⋅ b + 3 ( b )
2 2 2
91) 2 − 3 = 2 −2⋅2⋅ ⋅3⋅ + 3 =
x⋅2 x x x⋅2
= a − a b + b =
4 12 9
= 4a −12a b 9b
2x x x 2x
x y x x y y
( a b ) ( a ) a b ( b )
2 x 2 x y 2 y 2
= 2 ( a ) + 12a b + 3 ( b ) =
2 2 2
+ = + ⋅ ⋅ + =
92) 2 3 2 2 2 3 3
x⋅2 x y y⋅2
= a + a b + b =
4 12 9
= 4a + 12a b + 9b
2x x y 2y
=
M.I.M.-Sraga -2003./2013.
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