ALGEBARSKI IZRAZI

M-1-Kvadrat binoma
Rješenja i kompletni postupak
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
M.I.M.-Sraga -2003./2013.
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15
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