ALGEBARSKI RAZLOMCI

Rješenja
ALGEBARSKI RAZLOMCI
**** MLADEN SRAGA ****
2011.
UNIVERZALNA ZBIRKA
POTPUNO RIJEŠENIH ZADATAKA
PRIRUČNIK ZA SAMOSTALNO UČENJE
SKUP REALNIH BROJEVA
ALGEBARSKI RAZLOMCI
α
Skup realnih brojeva
1
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
Autor:
MLADEN SRAGA
Grafički urednik:
Mladen Sraga
BESPLATNA - WEB-VARIJANTA
Tisak:
M.I.M.-SRAGA d.o.o.
CIP-Katalogizacija u publikaciji Nacionalna i sveučilišna knjižnica, Zagreb
© M.I.M-Sraga d.o.o. 1992./2011.
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i tražite dodatne upute i objašnjenja ...
Dodatne upute i objašnjenja možete zatražiti i na mail: mim-sraga@zg.htnet.hr
Ovo je jako skraćena varijanta naše zbirke … samo oglednih 40-ak zadataka ….
M.I.M.-SRAGA d.o.o. zadržava sva prava na reproduciranje , umnažanje , prodaju ove
zbirke potpuno riješenih zadataka isključivo u okviru svog programa poduke i dopisne
poduke.
Nikakva komercijalna upotreba ove zbirke nije dozvoljena bez pismene dozvole
nakladnika !
2
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M.I.M.-Sraga – centar za poduku

Rješenja
ALGEBARSKI RAZLOMCI
Ovo nisu svi zadaci iz ove zbirke ,
Ovo je samo manji dio od oko 12% zadataka iz kompletne zbirke …
I ovdje su postavljeni samo kao ogledni primjerci ….
Ali vam mogu poslužiti kao solidna vježba pred testove ili ispitivanja u školi …
201.
16)
2 2
+ +
1. ( a + b )
− =
( − =
)( + ) ( a 2 − b 2) ( a 2 + b
2)
2
a
2
b
2
a
2
b
a b a b a b
4 4 2 2 2 2
=
a
1
− b
2 2
17)
↓
( ) ( ) ( )( )
a − b = a − b = a − b a + b
4 4 2 2 2 2 2 2 2 2
↑
razlika kvadrata !
2 2 2 2
a −b a −b
= =
4 4 2 2 2 2
a − b a − b a + b
( )( )
2 2
1. ( a − b )
( 2 2 2 2
a − b ) ( a + b )
=
a
1
+ b
2 2
19)
x− y x− y x−
y 1⋅( x−
y)
= = =
2 2
( y− x) ( x− y) ( x− y) ⋅( x−
y)
( x− y)
⋅( x−
y)
↓ ↑
( a− b) = ( b−a)
2 2
1
=
x − y
sjetimo se ovog pravila ( pogledaj u naše formule !! )
20)
ili kraće :
2 2
x y−
xy xxy ⋅ ⋅ −xyy
⋅ ⋅
= =
2 2 3 2 2 1
3x y −3xy 3⋅x⋅x⋅y −3⋅x⋅y ⋅y
xxy ⋅ ⋅ − xyy ⋅ ⋅
= =
2 2 1
3⋅xxy ⋅ ⋅ −3⋅xy ⋅ ⋅y
( )
2
( )
x⋅y⋅ x−
y
= =
3⋅x⋅y ⋅ x−
y
1
=
3y
( )
( )
2 2
x y−
xy x⋅y⋅ x−
y
= =
3 3 3
2 2 3 2
x y − xy ⋅x⋅y ⋅ x−
y
1⋅
x ⋅ y ⋅( x−
y)
3⋅
x ⋅y⋅ y ⋅( x−
y)
1⋅
x ⋅ y ⋅( x−
y)
3⋅
x ⋅y⋅ y ⋅( x−
y)
=
1
=
3y
Skup realnih brojeva
3
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
202.
3)
1
+ +
( a+
b)
= =
+ + − + ( a b)
a b a b
a b a b a ab b
( )( )
3 3 2 2
2 2
+ ( a − ab+
b )
=
1
a − ab+
b
2 2
↓
treba prepoznati formulu za zbroj kubova
4)
x
y
2
2
− xy
− xy
xx ⋅ − xy ⋅
= =
y⋅y−x⋅y
⋅( − ) ⋅( − + ) ⋅( −1) ⋅( − ) x⋅− ( 1) ⋅( y−x)
⋅( − ) ⋅( − ) ⋅( − ) y⋅( y−
x)
x x y x y x x y x
= = = =
y y x y y x y y x
( 1)
= x⋅ − = − x
6)
2
( x− y) ( x− y)( x−
y)
2 2
x − y ( x− y)( x+
y)
= =
( x−
y)
( x−
y)
( x− y)
( x+
y)
=
x−
y
x+
y
9)
− y ( x− y)( x+
y)
( x− y) ( x+
y)
= =
+ + − + ( x y)
− +
2 2
x
x y x y x xy y
( )( )
3 3 2 2
2 2
+ ( x xy y )
=
x−
y
x − xy+
y
2 2
↓
zbroj kubova
4
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M.I.M.-Sraga – centar za poduku

Rješenja
203.
6.)
x
x
− y
− y
6 6
2 2
( ) ( )
( )( )
( )( )
3 2 3 2 3 3 3 3
x − y x − y x + y
= = =
x− y x+ y x− y x+
y
( )( )
ALGEBARSKI RAZLOMCI
2 2 2 2
( x− y)( x + xy+ y )( x+ y)( x − xy+
y )
= =
( x− y)( x+
y)
=
( x−
y)
( x 2 xy y 2) ( x y 2 2
+ + + ) ( x − xy+
y )
( x− y)
( x+
y)
2 2 2 2
( x xy y )( x xy y )
= + + − + =
2 2 2 2
( x y xy)( x y xy)
= + + + − =
( x y ) ( xy)
2 2 2 2
= + − =
( ) 2 ( )
x x y y x y
2 2 2 2 2 2 2 2
= + ⋅ ⋅ + − =
=
= x + 2x y + y − x y = x + 2x y −1⋅ x y + y = x + x y + y
4 2 2 4 2 2 4 2 2 2 2 4 4 2 2 4
8.)
2 2 2
( ) ( ) ( )( )
2 2 2 2
− − − − − − ( ) ( ) ( )
( x )
= − y − 1 ( x− y+
1 ) x− y+
1
=
( x+ y) ⋅( x− y−1)
x+
y
( )( )
( ) ( )
x− y −1 x− y −1 x− y−1 x− y+ 1 x− y−1 x− y+
1
= = = =
x x y y x y x y x− y ⋅ x+ y −1⋅ x+ y x+ y ⋅ x− y−1
2 2 2
( − ) − = ( − ) − = ( to je razlika kvadrata ) = ( − − )( − + )
uputa : x y 1 x y 1 .... x y 1 x y 1
Skup realnih brojeva
5
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
204.
1)
x ⋅y x ⋅y x
2
2 2 2
x y
= = =
2 2 2 2 2
x − x y x ⋅1− x ⋅y x ⋅ 1− y
2
( )
x
⋅
⋅ y
( 1−
y)
y
=
1 − y
ALGEBARSKI RAZLOMCI
5)
4x y−
4xy
2x y−
2xy
2 2
2 2
( )
( )
4⋅xxy
⋅ ⋅ −4⋅xyy
⋅ ⋅ 4⋅x⋅y⋅ x−
y
= = =
2⋅xxy ⋅ ⋅ −2⋅xyy ⋅ ⋅ 2⋅xy ⋅ ⋅ x−
y
2⋅
2⋅
x ⋅ y ⋅( x−
y)
2 ⋅ x ⋅ y ⋅( x−
y)
2
= = 2
1
ili kraće:
4x y−
4xy
2x y−
2xy
2 2
2 2
( )
( )
4⋅x⋅y⋅ x−
y
= =
2⋅x⋅y⋅ x−
y
2⋅
2 ⋅ x ⋅ y ⋅( x−
y)
2 ⋅ x ⋅ y ⋅( x−
y)
2
= = 2
1
205.
7)
4 2 3 2 2 1 2 1 2
ab−
ab a ⋅a ⋅b −a ⋅b ⋅b
= =
5 5 1 4 1 1 1 4
ab−ab a⋅a ⋅b −a⋅b⋅b
( )
( )
( )
( )( )
2 1 2 2 2 2
a ⋅b ⋅ a −b a⋅a⋅b⋅ a −b
= = =
1 1 4 4 2 2 2 2
a ⋅b ⋅ a −b a⋅b⋅ a − b a + b
a ⋅a⋅ b ⋅( a
2 −b
2
)
a ⋅ b ( a
2 b
2
) a + b
2 2
⋅ − ( )
=
a
a
+ b
2 2
dodatna uputa :
4 2 3 2+ 2 1 2 1+
1
ab−
ab a ⋅b −a ⋅b
= =
5 5 1+ 4 1 1+
4
ab−ab a ⋅b −ab
⋅
2 2 1 2 1 2
a ⋅a ⋅b −a ⋅b ⋅b
= =
1 4 1 1 1 4
a ⋅a ⋅b −a ⋅b ⋅b
⋅ ⋅( − )
⋅ ⋅( − )
2 2
aab ⋅ ⋅ ⋅( a −b)
(
2 2)( 2 2)
a ⋅a⋅ b ⋅( a
2 −b
2
)
⋅ b ( 2 2
⋅ a − b ) ( a + b )
2 1 2 2
a b a b
= =
1 1 4 4
a b a b
= =
ab ⋅ ⋅ a − b a + b
=
a
=
u brojniku i nazivniku podvučemo zajedničke faktore
izlučimo z.f. i u brojniku i u nazivniku
u nazivniku treba prepoznati razliku kvadrata
kratimo
=
a
a
+ b
2 2
6
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M.I.M.-Sraga – centar za poduku

Rješenja
ALGEBARSKI RAZLOMCI
206.
10)
( )
( 3 − x)
2
( )( )
( )( )( )
( x−3)( x−3)
( x−3)( x−3)( 3−
x)
( )( )
( )( )( )
1⋅( x −3)
( x − 3)
( x − 3)
( x − 3)
( 3 − x)
( )( )
( ) ( )( ) ( )
x−3 x−3 x−3 x−3 x−3 x−3 x−3
= = =
3
3−x 3−x 3− x − x+ 3 − x+ 3 3− x −1⋅ x−3 ⋅ −1 x−3 ⋅ 3−
x
= =
1
=
3 − x
=
3 3
x + y
11) = ?
( x+
y)
3
12)
( − )
2 2
( x− y)( x + xy+
y )
3 3
x − y
= =
3 1+
2
x y x y
( − )
2 2
( x− y)( x + xy+
y )
= =
( x− y) ⋅( x−
y)
1 2
( )
2 2
( x−
y)
x + xy+
y + +
=
( x− y)
⋅( x− y) ( x−
y)
x xy y
2 2
2 2
Skup realnih brojeva
7
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
208.
9)
1
2 2
− − −
( x − y )
= = =
− ( ) −( ) ( − )( + ) ( 2 2 2 2
x − y ) ( x + y )
2 2 2 2 2 2
x y x y x y
x y x y x y x y
4 4 2 2 2 2 2 2 2 2
=
x
1
+ y
2 2
12)
2 2 2
1−x
1 −x
= =
2 2 2
x − 2x+ 1 x − 2x+
1
( )( )
( x−
)
( )( )
( −x)
( )( )
( 1−x)( 1−x)
1− x 1+ x 1− x x+ 1 1− x x+
1
= = = =
2 2
1 1
( 1−
x)
( x + 1)
( 1− x)
( 1−
x)
x + 1
=
1 − x
pravilo kaže :
↓
↑
( a− b) = ( b−a)
2 2
15)
pravilo :
2
x − x xx ⋅ −1⋅x
x⋅( x−1)
x⋅( x−1)
= = =
2 2
( 1−x) ( x− 1)
( x−1)( x−1)
( x−1) ( x−1)
↓ ↑
( a− b) = ( b−a)
2 2
=
x
x −1
8
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ALGEBARSKI RAZLOMCI
209.
8)
3 2
1 2 1 1 1
xy+ 2xy+
xy x x y x x y x y
3 1 2 1
xy− xy x⋅x⋅y− x⋅y⋅1
⋅ ⋅ + 2⋅ ⋅ ⋅ + ⋅ ⋅1
= =
( 2 1)
⋅ ⋅( −1)
1 2
x ⋅y⋅ x + ⋅ x+
x
= =
1 2
x y x
( x+ 1)( x+
1)
( x+ 1) ( x+
1)
= =
⋅ − ( x− 1)( x+
1)
( x− 1) ( x+
1)
⋅ y ⋅ ( x + 1) 2
2 2
x ⋅ y ( x 1 )
=
x + 1
x −1
210.
( )
2
2 2
3 12 2 3 3 12
x− + x x − ⋅x⋅ + + x
2 2 2
x −9 x −3
2 2
x − 6x+ 12x+
3
= =
1) = = → pogledati u 57. zadatak kako se to radi ...
( x− 3)( x+
3)
2 2
x + 6x+
3
2
( x + 3)
( x+ 3) ( x+
3)
( x− 3)( x+ 3)
( x− 3)( x+ 3)
( x− 3) ( x+
3)
= = =
=
x + 3
x − 3
4)
( )
2
2 2
2 8 4 4 8 4 8 4
x+ − x x + x+ − x x + x− x+
= = =
−5 −6 −3 −2 −6
xx ⋅ −3⋅x−2⋅x−2⋅3
2 2
x x x x x
2
2
x − 4x+
4 ( x − 2)
( x−2) ( x−2)
⋅( −3) −2⋅( −3)
( −3)( − 2)
( x−3) ( x−2)
= = =
x x x x x
=
x − 2
x − 3
uputa uz ovaj zadatak :
2
x −5x−6
→
je kvadratni trinom ...
tehnika kako se taj izraz rješava na faktore objašnjena je u 66. zadatku ove zbirke !
Skup realnih brojeva
9
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
211.
9)
2
x xz xy y x z
( ) ( )
2 2
( x+ y)
− z
( ) 1 ( )
2 2 2 2 2 2
x + y − z + 2xy x + 2xy+ y − z
= =
+ + − − − x⋅ x+ z+ y − y+ x+ z x⋅ x+ y+ z − ⋅ y+ x+
z
( x+ y− z)( x+ y+
z)
( x+ y− z) ( x+ y+
z)
+ −
=
( x+ y+ z) ⋅( x− 1)
( x+ y+ z)
⋅( x −1) x −1
= =
x y z
=
212.
1)
( x ) − ( y )
4 2 4 2
8 8
x − y
= =
3 2 2 3 1 2 2 2 1 2
x + xy −x y− y x ⋅ x + x⋅y − y⋅x − y ⋅y
4 4 4 4
( x − y )( x + y )
2 2 2 2
( ) ( )
= =
x⋅ x + y − y⋅ x + y
=
⎡
⎣
2 2 2 2 4 4
( x ) − ( y ) ⎤( x + y )
⎦
2 2
( x + y )( x−
y)
2 2 2 2 4 4
( x − y )( x + y )( x + y )
= =
2 2
( x + y )( x−
y)
2 2 2 2 4 4
( x − y ) ( x + y ) ( x + y )
2 2
( x + y ) ( x−
y)
=
2 2 4 4
( x − y )( x + y )
= =
( x−
y)
4 4
( x y)( x y )
= + +
( x−
y)
4 4
( x+ y)( x + y )
( x−
y)
=
10
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M.I.M.-Sraga – centar za poduku

Rješenja
213.
U ovom zadatku se ponovo susrečemo sa KVADRATNIM TRINOMOM
u zadatku br. 66. smo dali detaljnu uputu kako se kvadratni trinom rastavlja na faktore ....
ALGEBARSKI RAZLOMCI
1)
2 2 2
x −4 x −2
= =
− −2 − 2 + 1 −2
2 2
x x x x x
( x− 2)( x+
2)
( x− 2)( x+
2)
( 2)
1 ( 2 ) ( x− 2)( x+
1)
= x ⋅ x − + ⋅ x −
uputa kako rastaviti
kvadratni trinom iz brojnika:
= =
( x − 2)
( x + 2)
( x − 2)
( x + 1)
=
x + 2
x + 1
a = 1
m+ n = b⎫
m+ n = −1 ⎫ m+ n = −1⎫
− − = = − ⇒ ⎬⇒ ⎬⇒ ⎬⇒ = − = 1
c =−2
mn ⋅ = ac ⋅ ⎭ mn ⋅ = 1⋅( −2)
⎭ mn ⋅ = −2
⎭
2
x x 2 b 1 m 2, n
m =− 2 , n = 1
− − 2 = − 2 + 1 − 2 =
= − − − =
= − − − =
= − −
2 2
x x x x x
x( x 2) 1( x 2)
x( x 2) 1( x 2)
( x 2)( x 1)
2)
2 2
( ) ( )
( ) ( )
2
2 2 2
2
4x
− 4x+ 1 2 ⋅x
−2⋅2x⋅ 1+
1 2x −2⋅2x⋅ 1+ 1 2x−1
2 =
2
= =
2x − 5x+ 2 2x −4x− 1x+
2 2⋅xx ⋅ −2⋅2⋅x−1⋅ x+ 2 2x⋅ x−2 −1⋅ x−2
=
uputa kako rastaviti
kvadratni trinom iz brojnika:
( 2x−1)( 2x−1)
( 2x−1) ( 2x−1)
( x−2)( 2x− 1)
( x−2) ( 2x−1)
= =
a = 2
m+ n = b⎫ m+ n = − 5⎫ m+ n = −5⎫
− + = =− ⇒ ⎬⇒ ⎬⇒ ⎬⇒ =− =−
c = 2
mn ⋅ = ac ⋅ ⎭ mn ⋅ = 2 ⋅ 2⎭ mn ⋅ = 4 ⎭
2
2x 5x 2 b 5 m 4, n 1
m =− 4 , n =−1
− + = − − + =
= − − − =
= − − − =
= − −
2 2
2x 5x 2 2x 4x 1x
2
2x( x 2) 1( x 2)
2x( x 2) 1( x 2)
( x 2)( 2x
1)
2x
−1
=
x − 2
Skup realnih brojeva
11
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
214.
4)
3 2 2 3 2 1 2 1 2 1 2
x − x y+ xy − x x ⋅x −x ⋅ y+ x ⋅y − x ⋅x
= =
3 2 2 3 2 1 2 2 2 1
x + x y+ xy + y x ⋅ x + x ⋅ y+ x⋅ y + y ⋅y
( ) ( )
2 2 2
x ⋅ x− y + x⋅ y − x
= =
⋅ + + ⋅ +
( ) ( )
2 2
x x y y x y
( ) ( )( )
( )( )
( ) ⎡ ( )
2 2
( x+ y)( x + y )
( ) ( )( )
x⋅x⋅ x− y + x⋅ y− x y+ x x⋅x⋅ x− y − x⋅ x− y y+
x
= =
2 2 2 2
x+ y x + y x+ y x + y
x⋅ x− y ⎣x− y+
x ⎤
=
⎦
=
( )( )
=
=
( )( )
x⋅ x− y x− y−x
2 2
( x+ y)( x + y )
( )( )
x⋅ x− y −y
= =
2 2
( x+ y)( x + y )
( )
( )( )
( 1) ( )
( )( )
( )
−xy x − y −xy⋅ − ⋅ − x + y + xy⋅ y − x
= = = =
2 2 2 2 2 2
x+ y x + y x+ y x + y x+ y x + y
( )( )
=
( − x)
xy y
2 2
( x+ y)( x + y )
12
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Rješenja
ALGEBARSKI RAZLOMCI
215.
1)
( 1 )
( )
x x+
1 x x 1 x
a + a a ⋅ 1+ a ⋅a
a ⋅ + a a
= = =
x x+
1 x x 1 x
a −a a ⋅1−a ⋅a a ⋅ 1−a
a
x
x
( 1 a)
( 1−a)
⋅ +
⋅
1+
a
=
1 − a
2)
( 1 a)
( )
x x+
1 x x 1 x
a + a a ⋅ 1+ a ⋅ a 1+
a
= = =
x x+
1 x x 1 x
a −a a ⋅1−a ⋅a a ⋅ 1−a 1−a
a
⋅ +
3)
x+
2 x x 2 x
a −a a ⋅a −1⋅a
= =
x+
1 x x 1 x
a + a a ⋅ a + 1⋅a
2
( )
( )
( )( )
x
( )
x
x
a ⋅ a −1 a ⋅ a− 1 a+
1
= = =
x
a ⋅ a+ 1 a ⋅ a+
1
a
x
⋅( a− 1) ( a+
1)
x
a ⋅ ( a + 1)
a −1
= = a −1
1
4)
x 4
a ⋅( a −1)
( )
x+
4 x x 4 x
a −a a ⋅a −a
⋅1
= =
=
− + − ⋅ − ⋅ + ⋅ − ⋅1 ⋅ − + −1
x+ 3 x+ 2 x+
1 x x 3 x 2 x 1 x x 3 2
a a a a a a a a a a a a a a a
=
a
x
x 4
a ⋅( a −1)
3 2
( a a a 1)
⋅ − + −
=
2 2
( a − 1)( a + 1)
( 1) 1 ( 1)
2
a a a
( a )
4
2 2 2
a −1
−1
= = =
3 2 2 1 2
a − a + a−1 a ⋅a −a ⋅ 1+ 1⋅ a−1
= =
⋅ − + ⋅ −
( )
2
( a− 1)( a+ 1)( a + 1)
2
( a− 1)( a + 1)
= =
( a −1)
2
( a+ 1) ( a + 1)
2
a − ( a +1)
( 1)
=
a + 1
= = a + 1
1
Skup realnih brojeva
13
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
221.
ALGEBARSKI RAZLOMCI
4)
( 1) ( 1 )
2 2
x−1 1− y x−1 1− y x ⋅ x− + y ⋅ − y x − x+ y−
y
+ = + = = =
2 2 2 2
xy xy x ⋅ y ⋅ y x ⋅ x ⋅ y x ⋅ x ⋅ y ⋅ y
xy
( x− y)( x+ y) −1⋅( x− y) ( x− y)( x+ y−1)
2 2
x − y − x+
y
= = =
xy xy xy
2 2 2 2 2 2
5
7)
2 4 3x 2 4 3x 2 4 3x
− + = − + = − + =
2 2
x− 3 x+ 3 x −9 x− 3 x+ 3 9 x− 3 x+ 3 x− 3 x+
3
x −
( )( )
( x+ ) − ( x− ) + x x+ − x+
2+
3x
=
( x− 3)( x+ 3)
( x− 3)( x+
3)
2 3 4 3 3 2 6 4 1
= =
2x− 4x+ 3x+ 6 + 12 x+
18
= =
2 2 2
x −3 x −9
14
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Rješenja
ALGEBARSKI RAZLOMCI
222.
2)
2 4 4 5 4
2 2 4 3 2 4
5x y z 18a b c 5 ⋅ x ⋅ x ⋅ y ⋅ y ⋅ z 6 ⋅3⋅ ⋅ a ⋅ a ⋅ b ⋅ c
⋅ = ⋅ =
3 4 5 2 4 3 4 1 4 2 4
6abc 25xy z 6 ⋅ a ⋅ b ⋅ c ⋅ c 5⋅5
⋅ x⋅ y ⋅ z
=
5
⋅
6
x
⋅
⋅ x ⋅
3
a
⋅
y
4
b
2
⋅
⋅ y
1
c
⋅
z
2 4
⋅ c
xy ⋅3 ⋅ a 3xy a
= =
4 4
c ⋅ 5 5c
2 2 2 2
4
⋅
6
⋅3 ⋅ a
3
5 ⋅5 ⋅ x
2 4
⋅ a ⋅ b ⋅ c
⋅
2
y
⋅
4
z
=
5)
( )( )
( − )
( )( )
3 3 3 3
2 2 2 2
x + y x − y x+ y x − xy+ y x− y x + xy+
y
⋅ = ⋅ =
2 2 2 2
x−
y x − xy+ y x y
x − xy+
y
=
2 2
( x+ y) ( x − xy+
y )
( x−
y)
⋅
2 2
( x + xy+
y )
2 2
( x − xy+
y )
( x−
y)
2 2
( x y)( x xy y )
= + + +
Skup realnih brojeva
15
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
225.
ALGEBARSKI RAZLOMCI
4)
( 2)
( )( )
( )
( 2)( 2)
2 2
2 2 2 2 2 2
x + 4x+ 4 x −4 x + 2 ⋅ 2 ⋅ x+
2 x − 2xy+
y x+ x−
y
:
= ⋅ = ⋅ =
2 2 2 2 2 2
x − y x − 2xy+ y x− y x+ y x −2
x− y x+ y x− x+
=
( )( )
( x+ 2) ( x+
2)
( x−
y)
( x−
y)
⋅
( x− y)
( x+
y)
( x− 2) ( x+
2)
=
( x+ 2)( x−
y)
( x+ y)( x−2)
5)
( )
( x 1)
( )( )
( x+ 1− y)( x+ 1+
y)
( x−1− y)( x− 1+ y)
( x+ 1− y)( x+ 1+ y)
x ⋅ ( x+ 1−
y)
x ⋅ ( x−1−
y)
( x−1−
y)
( x− 1+
y)
( x+ 1−
y)
⋅
( x+ 1− y)
( x+ 1+ y)
( x−1−
y)
2 2 2 2
x−1 − y x −x − xy x−1− y x− 1+
y x + x − xy
:
= ⋅ =
2 2 2 2
+ − y x + x−xy x − x−
xy
= ⋅ =
=
x− 1+ y x+ y−1
= =
x+ 1+ y x+ y+
1
=
8)
( x− y)( x− y) + ( x+ y)( x+
y)
( ) ( )
3 2 3 2
⎛ x− y x+ y⎞
x + xy x − xy
⎜ + ⎟:
= ⋅ =
3 2 3 2
⎝ x+ y x+ y⎠
x − xy x+ y ⋅ x−
y x + xy
2 2 2 2
( x− y) + ( x+
y) x( x − y )
2 2 2 2
x − y x( x + y )
2 2 2 2
2 2
x − 2xy+ y + x + 2xy+
y ( x − y )
2 2 2 2
x − y ( x + y )
2 2
2 2 2 2
2x + 2y x − y
2 ( x + y ) ( x
2 − y
2
)
⋅
2 2 2 2
x − y x + y ( x
2 − y
2
) ( x
2 + y
2
)
= ⋅ =
= ⋅ =
= ⋅ =
= 2
16
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ALGEBARSKI RAZLOMCI
Ovo su ogledni primjeri stranica iz
ZBIRKE POTPUNO RIJEŠENIH ZADATAKA
ALGEBARSKI IZRAZI
PRIRUČNIK ZA SAMOSTALNO UČENJE
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izdavač: M.I.M.-Sraga
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Skup realnih brojeva
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ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
Cijena kompletne zbirke ALGEBARSKI RAZLOMCI za PRVI razred srednje
škole je 120 kn sa popustom = 60kn
Sve dodatne informacije i narudžbe na:
01-4578-431
ili
098-237-534
ili na mail: mim-sraga@zg.htnet.hr
iz naše ponude izdvajamo:
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18
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Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
Rješenja
ALGEBARSKI RAZLOMCI
79.
Tehnika rješavanja ovog zadatka je sljedeća: brojnik i nazivnik rastavimo na faktore i kratimo...
1.)
xy x⋅
y
= =
x− xy x⋅ − y
( 1 )
x ⋅ y
( )
x ⋅ 1 − y
y
=
1 − y
-rastavimo na faktore, kratimo x u brojniku i nazivniku
Ovi prvi zadaci su jednostavni, samorastavimo na faktore i kratimo
2.)
3
ab
ab−
ab
2 2
2 2
abb ⋅ ⋅ abb ⋅ ⋅
a
= = = kratimo:
aab ⋅ ⋅ −abb ⋅ ⋅ ab ⋅ ⋅ a−b
a
( )
2
⋅ b ⋅b
⋅ b ⋅( a−b)
=
2
b
a−
b
3.)
( − )
( )
( )
( )
ax − bx x⋅
a b x ⋅ a−b
= =
ax + bx x⋅ a + b x ⋅ a+
b
=
a−
b
a+
b
kratimo
4.)
xz − yz
2
z + 3z
( − )
( z 3)
( )
( )
z⋅
x y z ⋅ x−
y
= =
z⋅ + z ⋅ z + 3
=
x−
y
z + 3
kratimo
5.)
( )
( 1)
( )
( )
( )
2
a + a aa ⋅ + a a⋅ a+
a ⋅ a + 1
= = =
ax −ay a⋅ x − y a⋅ x − y a ⋅ x−
y
=
a + 1
x−
y
6.)
2
a − 2ab
2
ab − 2b
( −2
)
( )
aa ⋅ −2⋅ab
⋅ a⋅
a b
= = =
ab ⋅ −2⋅bb ⋅ b⋅ a−2b
( −2
)
( −2
)
a⋅
a b
b⋅
a b
=
a
b
7.)
( )
R.
KVADRATA
( )
a
=
( 3 − 4 )( 3 + 4 ) ( 3 −4b)
2
3a
+ 4ab
a⋅ 3a+ 4b a 3a+
4b
= =
2 3 2 2
9ab−16b b( 9a −16b
) b a b a b b a
8.)
16x − 36xy
2
6xy
− 9y
3 2
R.
KVADRATA
2 2
( )
( − )
( )( )
( − )
4x 4x − 9y 4x 2x− 3y 2x+
3y
= =
3y 2x 3y 3y 2x 3y
kratimo
=
( + )
4x 2x 3y
3y
Skup realnih brojeva
19
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
ALGEBARSKI RAZLOMCI
10.
5 3 2 3 2 2
12a −27a b 3 a (4 a −b ) 3 a(2a− 3 b)(2a+ 3 b) 3 a(2a+
3 b)
= = =
3 2 2 2
8 12 4 (2 3 ) 4(2 b a−
3) b 4b
ab−
ab ab a−
b
kratimo
a
sa a
2 3
kratimo
još jednom isti zadatak:
5 3 2 3 2 2
12a −27a b 3 a (4 a −b
)
3 2 2 2
8 12 4 (2 3 )
= =
ab−
ab ab a−
b
( )
( )
2
3⋅a
⋅a⋅
2 2 2
2 2
4a −b 3⋅
a ⋅a⋅
4a −b
=
=
2
4 ⋅a ⋅b⋅(2a−3 b)
2
4⋅ a ⋅b⋅(2a−3 b)
3 a(2a− 3 b)(2a+
3 b)
3 a(2a−
3 b)
(2a+
3 b)
= =
4(2 b a−
3) b
4 b(2a−
3 b)
3 a(2a+
3 b)
=
4b
=
=
11.
4 3 2 2 2 2 2
2a − 8a b+ 8a b 2 a ( a − 4ab+
4 b )
4 3 3
a 2 a b a ( a 2 b)
−
= =
−
2
2⋅
a ⋅( a−2 b)
=
2
a ⋅a⋅( a−2 b)
( a b) ( a b)
a⋅( a−2b)
2⋅ −2 ⋅ −2
=
2( a−
2 b)
=
a
2
treba prepoznati kvadrat razlike
( ) ( − )
= a− b = a− b ⋅ a b
=
2
rastavimo: ( 2 ) 2 2
12.
2 2
a − 6a+ 9 ( a−3)
= =
2
a − 9 ( a− 3)( a+
3)
( a − 3) ⋅( a −3)
( a − 3) ⋅ ( a + 3)
=
( a − 3)
( a + 3)
13.
a
2
a a a
2
−4 ( − 2)( + 2)
= =
+ a−6
( a− 2)( a+
3)
↓
( a − 2) ( a + 2)
=
( a − 2) ( a + 3)
a + 2
a + 3
+ − 6= + 3 −2 −6 − ( + 3) − 2( + 3) = ( + 3)( −2)
2 2 2
a a a a a a a a a a
14.
− ( − )( + ) ( a− b)( a+
b)
= =
3 3 2 2
a + b ( a+ b)
a − ab+ b ( a b)
a − ab+
b
2 2
a b a b a b
( )
2 2
+ ( )
=
a−
b
a − ab+
b
2 2
20
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Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
ALGEBARSKI RAZLOMCI
17.)
18.)
( a− b)( a+
b)
2 2
a − b
= =
3 2 2 3 3 2 2 3
a + ab −a b−b a − a b+ ab −b
( a− b)( a+
b)
2( − ) +
2( − )
( )( )
= =
a a b b a b
2 2
( a− b)( a + b )
( a− b)( a+
b)
2 2 2 2
a a b b a b a b
a− b a+ b a+
b
= =
a + b
2 2
a − b
= =
− − − − − −
( a− b)( a+
b)
( a− b)( a+ b) − ( a+
b)
( a− b)( a+
b)
( a+ b)( a−b−1)
= =
= =
a−
b
=
a−b−1
2 2
19.)
KV . ZBROJA
RKV . .
2 2 2
2 2
a + 2ab+ b −c
( a+ b)
−c
= =
a+ b+
c a+ a+ b+ c c a+ b+ c a+
c
( ) ( )
( )( )
( a+ b− c)( a+ b+
c)
( a+ b+ c)( a+
c)
= =
a+ b−c
=
a+
c
kratimo
20.)
2 2 2 2 2 2
a + b − c + 2ab a + 2ab+ b −c
= =
2 2 2 2 2 2
a − b + c + 2ac a + 2ac+ c −b
2 2
( a+ b)
−c
2 2
( a+ c)
−b
( a+ b− c)( a+ b+
c)
( a+ c− b)( a+ b+
c)
= =
= =
( a+ b− c)( a+ b+
c)
( a− b+ c)( a+ b+
c)
= =
( a+ b− c) ( a+ b+
c)
( a− b+ c) ( a+ b+
c)
=
=
a+ b−c
a− b+
c
Skup realnih brojeva
21
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
ALGEBARSKI RAZLOMCI
21.)
2 2
a + 6a+ 5 a + a+ 5a+
5
= =
3 2 3 2
a + 5a −a−5 a − a+ 5a
−5
a( a+ 1) + 5( a+
1)
= =
a a
2 2
( − 1) + 5( a −1)
( a+ 1)( a+
5)
2
( a − 1)( a+
5)
= =
( a+ 1)( a+
5)
( a+ 1)( a− 1)( a+
5)
= =
( a + 1)
( a + 5)
( a + 1)
( a− 1) ( a+
5)
=
=
1
a −1
22.)
+ 2 + 2 + 2 + 2
= =
4 2 2
x + 1 − 1 x+ 1 − 1 x+ 1 + 1
2 2
x x x x
( ) ( )
( )(( ) )
2
x + 2x+
2
= =
2 2
( x + 2x+ 1− 1)( x + 2x+ 1+
1)
2
x + 2x+
2
= =
2 2
( x + 2x)( x + 2x+
2)
2
1⋅ ( x + 2x+
2)
=
( x 2 + 2x ) ( x 2 + 2x+
2 )
=
1
= =
2
x + 2x
1
=
x x
( + 2)
23.)
( )
2
2
( 2a)( a−1) −4( 2a−3)
( 2a−3) ( a−1)
−4
= =
2 2
( a+ 1) ( a− 3)
( a+ 1) ( a−3)
( 2a−3)( a−1−2)( a− 1+
2)
( a+ 1)( a+ 1)( a−3)
( 2a−3)( a− 3)( a+
1)
= =
( 2+ 1)( a+ 1)( a−3)
= =
2a
− 3
=
a + 1
22
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Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
ALGEBARSKI RAZLOMCI
24.)
( )( )
2 2
( ) ⎡ ( )
2
( ) ( )
2 2 2
4 4 1 2 3 2 1 3 3
a − a+ a − a− a− a + a− a−
= =
a − 6a+ 9 ⎣a − 1+ a a+ 1 ⎤
⎦ a−3 ⎣ a− 1 a+ 1 + a a+
1 ⎤⎦
(
2
) ⎡( )( ) ( )
2
( 2a− 1) ( a( a+ 1) − 3( a+
1)
)
2
( a− 3) ( a+ 1)( a− 1+
a)
2
( 2a+ 1) ( a+ 1)( a−3)
2
( a− 3) ( a+ 1)( 2a−1)
2a
−1
a − 3
= =
= =
25.)
2
( x+ 2y)
− 4 ( x+ 2y− 2)( x+ 2y+
2)
= =
− ( − ) ( x− 2y)( x+ 2y) −2( x−2y)
( x−2y)
( x+ 2y
−2)
2 2
x + 4xy+ 4y
−4
4 2 2
2 2
x y x y
x+ 2y+
2
=
x−
2y
=
26.)
2 2 2
( a b c 2bc)( a b c)
( )( )
2 2 2
a − ( b + 2bc+ c ) ⎤( a+ b−c)
( )( )
− − − + − ⎡
=
⎣ ⎦
=
2 2 2 2 2 2
a+ b+ c a − b + c − 2ac a+ b+ c a − 2ac+ c −b
2
( ) ⎤( )
⎡
2
a − b+ c a+ b−c
=
⎣
⎦
=
2 2
( a+ b+ c) ⎡( a−c)
−b
⎤
⎣
⎦
⎡⎣a− ( b+ c) ⎤⎦( a+ b+ c)( a+ b−c)
( a+ b+ c)( a−c−b)( a− c+
b)
( a−b− c)( a+ b−c)
1
( a−b− c)( a+ b−c)
= =
= =
27.)
2 2 2
a b c + 2ab + 2bc + 2ac
a+ b+ c a+ b+ c a+ b+ c a+ b+
c
= =
2 2 2 2 2 2 2
2
a −b −c − 2bc a b + 2bc+ c a − b+
c
+ + ( )( )
( )
( )( )
a− ( b+ c) ⎤( a+ b+
c)
a+ b+ c a+ b+ c a+ b+
c
= =
⎡⎣
⎦
a−b−c
( )( )
( )
=
28.)
2 2
x xy xz y yz
( )
( )
2 2
x xy xy y z x y
− 3 + + 2 −2
2 − + 2 + −2
= =
2 2 2 2 2 2
x − y + 2yz−z x − y − 2yz+
z
( −2 ) − ( − 2 ) + ( −2
)
2
2
x −( y−z)
( x−2y)
( x−
y ) ( )( )
x−( y− z) ⎤( x+ y−
z)
( )( )
x x y y x y z x y
= =
=
⎡⎣
⎦
+ z x−2y x− y+ z x − 2y
= =
x − y+ z x+ y− z x+ y−
z
Skup realnih brojeva
23
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
Rješenja
ALGEBARSKI RAZLOMCI
29.)
2 2 2
a − ab + ac + bc a − ab − ab + b + ac − bc
( )
( )( )
−( − ) ⎤( + − )
( −2 ) − ( − 2 ) + ( −2
)
( )
3 2 2 2 2 a a b b a b c a b
= = =
2 2 2 2 2 2 2
2
a − b + 2bc−c a b − 2bc+ c a − b−c
( )( )
( )( )
a−2b a− b+ c a−2b a− b+ c a−
2b
= = =
⎡⎣a b c ⎦ a b c a− b+ c a+ b− c a+ b−c
30.)
( )
2 2 2 2
xy⋅ a − b + abx −aby 2 2 2 2 2 2 2 2
ax− y− bxy+ abx −aby a xy+ abx −aby −b xy
=
= =
2 2 2 2
2 2 2 2 2 2 2 2
abx + aby + xy⋅ a + b abx + aby + a xy + b xy aby + b xy + abx + a xy
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
ax⋅ ay + bx −by⋅ ay + bx ay + bx ⋅ ax −by ax − by
= =
=
by⋅ ay + bx + ax⋅ bx + ay ay + bx ⋅ by + ax ax + by
31.)
a x + b − abx + a x − ab a x − abx + b + a x − ab
=
ax⋅ ax −b ⋅ ax + b ⋅ ax − b + 3a
2 2 2 2 2 2 2 2
3 3 6 9 9 3 6 3 9 9
3 3 2
( ax abx) ( ax b 3a)
− ⋅ − + ( ) ( ) ( )
=
2 2 2
( ax abx b) a( ax b)
3⋅ − 2 + + 9 ⋅ −
= =
ax⋅ ax −b ⋅ ax + b ⋅ ax − b + a
( ) ( ) ( 3 )
2
3⋅( ax − b) + 9a⋅( ax −b)
( ) ( ) ( 3 )
= =
ax⋅ ax −b ⋅ ax + b ⋅ ax − b + a
( ax b) ( ax b) a ( ax b)
⋅( − ) ⋅ ( + ) ⋅( − + 3 )
3⋅ − ⋅ − + 9 ⋅ −
= =
ax ax b ax b ax b a
( ax −b) ⋅( 3. ( ax − b)
+ 9a)
( ) ( ) ( 3 )
= =
ax⋅ ax −b ⋅ ax + b ⋅ ax − b + a
=
3ax − 3b + 9a
ax⋅ ax + b ⋅ ax − b +
( ) ( 3a)
=
3⋅( ax − b + 3a)
3
( ) ( 3 ) ( )
= =
ax⋅ ax + b ⋅ ax − b + a ax⋅ ax + b
24
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Rješenja
32.)
Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
3 2 3 2
2 + 3 −2 −3 2 − 2 + 3 −3
2
2 2 3 1
x x y x y x x x y y
= =
x − x+ y⋅ x−
( ) 2x⋅( x− 1) + 3y⋅( x−1)
2 2
( ) ( )
2x⋅ x − 1 + 3y⋅ x −1
= =
( x−1) ⋅ ( 2x+
3y)
2
( x −1) ⋅ ( 2x+
3y)
= =
( x−1) ⋅ ( 2x+
3y)
( x−1) ⋅ ( x+
1)
( x −1)
= = x + 1
ALGEBARSKI RAZLOMCI
33.)
3 2 2 3 3 2 2 3
3x + xy −6x y−2y 3x − 6x y+ xy −2y
= =
5 4 4 5 5 4 4 5
9x − xy − 18x y+ 2y 9x −18x y− xy + 2y
2
2
3x
⋅( x− 2y) + y ⋅( x−2y)
=
4 4
9x ⋅ x−2y − y ⋅ x−2y
( ) ( )
2 2
( x−2y) ⋅ ( 3x + y )
4 4
( x−2y) ⋅( 9x − y )
2 2
( 3x
+ y )
2 2 2 2
( 3x − y ) ⋅ ( 3x + y )
= =
= =
=
3x
1
− y
2 2
=
34.)
( ) ( ) ( ) ( )
ac + ad + bc + bd −a⋅ c + d a⋅ c + d + b⋅ c + d −a⋅ c + d
= =
2 2 2 2 2 2 2 2
ac − ad + bc − bd ac + bc −ad −bd
( c+ d) ⋅ ( a+ b−a)
2
⋅ ( + ) − ⋅ ( + )
( c+ d)
⋅b
= =
2
c a b d a b
= =
=
=
2 2
( a+ b) ⋅( c −d
)
( c+ d)
⋅b
( a+ b)
⋅( c−d) ⋅ ( c+
d)
b
( a+ b) ⋅( c−d)
=
Skup realnih brojeva
25
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
ALGEBARSKI RAZLOMCI
Univerzalna zbirka potpuno korak po korak riješenih zadataka
za prvi razred svih srednjih škola
Priručnik za samostalno učenje MATEMATIKA-1-
SKUPU REALNIH BROJEVA:
Više informacija o toj zbirci : www.mim-sraga.com
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video instrukcije ili video poduka na:
+
MATEMATIČKE FORMULE I NJIHOVA PRIMJENA - MEMENTO iz MATEMATIKE
26
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M.I.M.-Sraga – centar za poduku

Rješenja
ALGEBARSKI RAZLOMCI
Priručnici za samostalno učenje MATEMATIKA 2
Skup realnih brojeva
27
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga

Rješenja
Matematika −1 − riješeni zadaci po Pavkovi ć − Veljan
ALGEBARSKI RAZLOMCI
79.
Skrati razlomke:
3
xy ab ax bx
= =
2 2
x− xy ax+
bx
1.) 2.) 3.)
ab−
ab
−
=
− + −
4.) = 5.) =
6.)
z + 3z ax − ay
ab −2b
2 2
xz yz a a a 2ab
2 2
=
2 3 2
3a + 4ab 16x −36xy
7.) = 8.)
=
2 3 2
9ab−16b 6xy−9y
5 3 2 4 3 2 2
12a −27a b 2a − 8a b+
8a b
10. = 11.
=
3 2 2 4 3
8ab−12ab a −2ab
2
a − 6a+ 9 a −4
12.
= 13. = 14.
2 2
a − 9 a + a−6
2 2 2 2
a −b a −b
17.) = 18.)
=
3 2 2 3 2 2
a + ab −a b−b a −a−b−b
a
a
2 2
−b
+ b
2
3 3
=
2 2 2 2 2 2
a + 2ab+ b − c a + b − c + 2ab
19.) = 20.)
=
2 2 2
a+ b+ c a+ a+ b+ c c a − b + c + 2ac
( ) ( )
2 2
a + 6a+ 5 x + 2x+
2
21.) = 22.)
=
3 2 4
a + 5a −a− 5 x + 1 −1
( )
23.)
2
( 2a)( a−1) −4( 2a−3)
2
( a+ 1) ( a−3)
2 2
( 4a − 4a+ 1)( a −2a−3)
2 2
( a − 6a+ 9) ⎡a − 1+ a( a+
1)
= 24.)
=
⎣
⎤
⎦
( )
2 2 2
( a −b −c − 2bc)( a+ b−c)
2 2
x + 4xy+ 4y
−4
25.) = 26.)
=
− − + + − + −
( )( )
2 2 2 2 2
x 4y 2 x 2y a b c a b c 2ac
2 2 2 2 2
a + b + c + 2ab + 2bc + 2ac x − 3xy + xz + 2y −2yz
27.) = 28.)
=
2 2 2 2 2 2
a −b −c −2bc x − y + yz−
z
29.)
2
a 3ab ac 2bc
a
2 2 2 2
− + +
xy⋅ a − b + abx −aby
= 30.)
=
2 2 2 2 2
− b + 2bc − c abx + aby + xy ⋅ a + b
2 2
( )
( )
28
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