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Rješenja ALGEBARSKI RAZLOMCI
**** MLADEN SRAGA ****
2011.
UNIVERZALNA ZBIRKA
POTPUNO RIJEŠENIH ZADATAKA PRIRUČNIK ZA SAMOSTALNO UČENJE
SKUP REALNIH BROJEVA
ALGEBARSKI RAZLOMCI
α
1 Skup realnih brojeva 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.
Potpunu garanciju na kompletnu zbirku daje: centar za dopisnu poduku M.I.M.-SRAGA -dakle sve što vam se čini nejasno krivo ili sumnjivo - zovite 01-4578-431 ili 01-4579-130 i tražite dodatne upute i objašnjenja ... Dodatne upute i objašnjenja možete zatražiti i na mail: [email protected]
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 !
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3 Skup realnih brojeva
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
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 …
( )( )( )2 2
2 2 2 2
4 4 2 2 2 2
1.16)
201.a ba b a b
a b a b a b
++ += =
− − + ( ) ( )2 2 2 2a b a b− +
( ) ( ) ( )( )
( )( )( )
2 2
4 4 2 2 2 2 2 2 2 2
razlika kvadrata !
2 22 2 2 2
4 4 2 2 2 2
1
1.17)
a b
a b a b a b a b
a ba b a ba b a b a b
=−
↓
− = − = − +
↑
−− −= =
− − + ( )2 2a b− ( )
( ) ( ) ( ) ( )( )
2 22 2
2 2
1
119)
a ba b
x yx y x y x yx y x yy x x y
=++
⋅ −− − −= = =
− ⋅ −− − ( )x y− ( )
( ) ( )
( )( )
2 2
2 2
2 2 3 2 2 1
2 2 1
2
sjetimo se ovog pravila ( pogledaj u naše formule !! )
1
20)3 3 3 3
3 3
1
3
x yx y
a b b a
x x y x y yx y xyx y xy x x y x y y
x x y x y yx x y x y y
xx y x yx y x y
=−⋅ −
↓ ↑
− = −
⋅ ⋅ − ⋅ ⋅−= =
− ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅⋅ ⋅ − ⋅ ⋅
= =⋅ ⋅ ⋅ − ⋅ ⋅ ⋅
⋅⋅ ⋅ −= =
⋅ ⋅ ⋅ −
y⋅ ( )x y⋅ −
3 x⋅ y y⋅ ⋅ ( )x y⋅ −
( )( )
2 2
2 2 3 2
:
13
1
3 3 3
ili kraće
y
xx y x yx y xyx y xy x y x y
=
=
⋅⋅ ⋅ −−= =
− ⋅ ⋅ ⋅ −
y⋅ ( )x y⋅ −
3 x⋅ y y⋅ ⋅ ( )x y⋅ −
13y
=
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( )( )( )1
3 3 2 23)
202.
a ba b a ba b a b a ab b
++ += =
+ + − + ( )a b+ ( )
( )( )
( )( )
( ) ( )( )
( ) ( )
2 22 2
2
2
treba prepoznati formulu za zbroj kubova
1
4)
11
a ab ba ab b
x x x yx xyy y x yy xy
x y xx x y x y x x y xy y x y y x y y x
=− +− +
↓
⋅ − ⋅−= =
⋅ − ⋅−
⋅ − ⋅ −⋅ − ⋅ − + ⋅ − ⋅ −= = = =
⋅ − ⋅ − ⋅ − ( )y y x⋅ −( )
( ) ( )( )( )( )
( )2
2 2
1
6)
x x
x yx y x y x yx y x yx y
= ⋅ − = −
−− − −= =
− +−
( )( )
x y
x y
−
− ( )
( )( )( )( )
( ) ( )2 2
3 3 2 29)
x yx yx y
x y x yx y x yx yx y x y x xy y
−=
++
− +− +−= =
+ + − + ( )x y+ ( ) 2 22 2
zbroj kubova
x yx xy yx xy y
−=
− +− +
↓
Rješenja ALGEBARSKI RAZLOMCI
5
Skup realnih brojeva ALGEBARSKI RAZLOMCI autor:
203.( ) ( )( )( )
( )( )( )( )
( )( )( )( )( )( )
( )
3 2 3 2 3 3 3 36 6
2 2
2 2 2 2
6.)x y x y x yx yx y x y x y x yx y
x y x xy y x y x xy y
x y x y
x y
− − +−= = =
− + − +−
− + + + − += =
− +
−=
( ) ( )2 2x xy y x y+ + + ( )( )
2 2x xy y
x y
− +
− ( )x y+
( )( )
( )( )
( ) ( )
( ) ( )
( ) ( ) ( )( )( ) ( ) ( )
( )( )( ) ( )
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2 2 2 2 2
4 2 2 4 2 2 4 2 2 2 2 4 4 2 2 4
2 2 2
2 2 2 2
2
2 2 1
1 1 1 1 1 18.)
1 1
x xy y x xy y
x y xy x y xy
x y xy
x x y y x y
x x y y x y x x y x y y x x y y
x y x y x y x y x y x yx y x y x y x y x yx x y y x y x y
x
=
= + + − + =
= + + + − =
= + − =
= + ⋅ ⋅ + − =
= + + − = + − ⋅ + = + +
− − − − − − − + − − − += = = =
− ⋅ + − ⋅ + + ⋅ − −− − − − − −
=( )1y− − ( )( ) ( )
1
1
x y
x y x y
− +
+ ⋅ − −
( ) ( ) ( ) ( )( )2 2 2 to je razlika kvadrata
1
: 1 1 .... 1 1
x yx y
uputa x y x y x y x y
− +=
+
− − = − − = = − − − +
Mladen Sraga
Rješenja ALGEBARSKI RAZLOMCI
204.
( )2 2 22
2 2 2 2 21)1 1x y x y xx y
x x y x x y x y⋅ ⋅
= = =− ⋅ − ⋅ ⋅ − 2
yx
⋅
( )
( )( )
2 2
2 2
11
2 244 44 45)2 2 22 2
yyy
x y x yx x y x y yx y xyx x y x y y x y x yx y xy
=−⋅ −
⋅⋅ ⋅ ⋅ −⋅ ⋅ ⋅ − ⋅ ⋅ ⋅−= = =
⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ −−
x⋅ y⋅ ( )x y⋅ −
2 x⋅ y⋅ ( )x y⋅ −
( )( )
2 2
2 2
2 21
ili kraće:
2 244 422 2x y x yx y xyx y x yx y xy
= =
⋅⋅ ⋅ ⋅ −−= =
⋅ ⋅ ⋅ −−
x⋅ y⋅ ( )x y⋅ −
2 x⋅ y⋅ ( )x y⋅ −
2 21
= =
( )( )
( )( )( )
2 2 1 2 1 24 2 3
5 5 1 4 1 1 1 4
2 1 2 2 2 2
1 1 4 4 2 2 2 2
7)
205.
a a b a b ba b a ba b ab a a b a b b
aa b a b a a b a b
a b a b a b a b a b
⋅ ⋅ − ⋅ ⋅−= =
− ⋅ ⋅ − ⋅ ⋅
⋅ ⋅ − ⋅ ⋅ ⋅ −= = =
⋅ ⋅ − ⋅ ⋅ − +
a b⋅ ⋅ ( )2 2a b⋅ −
a b⋅ ( )2 2a b⋅ − ( )
( )( )
2 22 2
2 2 1 2 1 14 2 3
5 5 1 4 1 1 4
2 2 1 2 1 2
u brojniku i nazivniku podvučemo zajedničke faktore
:
aa ba b
dodatna uputaa b a ba b a b
a b ab a b a ba a b a b b
+ +
+ +
=
1 4 1 1 1 4
2 1 2 2
1 1 4 4izlučimo z.f. i
a a b a b b
a b a b
a b a b
++
⋅ − ⋅−= =
− ⋅ − ⋅
⋅ ⋅ − ⋅ ⋅= =
⋅ ⋅ − ⋅ ⋅
⋅ ⋅ −= =
⋅ ⋅ −
( )( )( )
2 2
2 2 2 2
u brojniku i u nazivniku
u nazivniku treba prepoznati razliku kvadrat a a b a b
a b a b a b
a
⋅ ⋅ ⋅ −= =
⋅ ⋅ − +
=
a
a b⋅ ⋅ ( )2 2a b⋅ −
a b⋅ ( )2 2a b⋅ − ( )2 2
2 2
kratimo a b
aa b
=+
=+
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7
Skup realnih brojeva ALGEBARSKI RAZLOMCI autor: Mlade
( )( )
( )( )( )( )( )
( )( )( )( )( )
( )( )( ) ( )( ) ( )
( )( )( )( )( )
( )
2
3
3 3 3 3 3 3 310)
3 3 3 3 3 3 1 3 1 3 33
1 33 33 3 3
206.
x x x x x x xx x x x x x x x xx
xx xx x x
− − − − − − −= = =
− − − − + − + − − ⋅ − ⋅ − − ⋅ −−
⋅ −− −= =
− − −
( )3x −
( )3x − ( )3x − ( )
( )
( )( )( )
( )( )( )( ) ( )
( )
3 3
3
2 23 3
3 1 2
2 2
1 2
133
11) ?
12)
xx
x yx y
x y x xy yx yx y x y
x yx y x xy y
x y x y
+
=−−
+=
+
− + +−= =
− −
−− + += =
− ⋅ −
( )( )
2 2x xy y
x y
+ +
− ( ) ( )
2 2
2 2
x xy yx y x y
+ +=
⋅ − −
=
n Sraga
Rješenja ALGEBARSKI RAZLOMCI
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( ) ( ) ( )( )( )
12 2
2 2 2 2 2 2
4 4 2 2 2 2 2 2 2 29)
208.x yx y x y x y
x y x y x y x y
−− − −= = =
− − − + ( )2 2x y− ( )
( )( )( )
( )( )( )
( )( )( )( )
( )
2 22 2
2 2 2
2 2 2
2 2
1
1 112)2 1 2 1
11 1 1 1 1 11 11 1
x yx y
x xx x x x
xx x x x x xx xx x
=++
− −= =
− + − +−− + − + − +
= = = =− −− −
( )( )
1
1
x
x
+
− ( )
( ) ( )
( ) ( )( )
( )( )( )
2 2
2
2 2
:
111
11115)1 11 1
pravilo kaže
xxx
a b b a
x xx xx x xx xx xx x
+=
−−
↓ ↑
− = −
⋅ −⋅ −⋅ − ⋅−= = =
− −− − ( ) ( )1 1x x− −
( ) ( )2 2:
1
pravilo
xx
a b b a
=−
↓ ↑
− = −
Rješenja ALGEBARSKI RAZLOMCI
9 Skup realnih brojeva
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
209.
( )( )
1 2 1 1 13 2
3 1 2 1
1 2
1 2
2 128)1
2 1
1
x x y x x y x yx y x y xyx y xy x x y x y
x y x x x
x y x
⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅+ += =
− ⋅ ⋅ − ⋅ ⋅
⋅ ⋅ + ⋅ += =
⋅ ⋅ −
y⋅ ( )21x
x
⋅ +
y⋅ ( )( )( )( )( )
( ) ( )2 2
1 11 11 11
x xx xx xx
+ ++ += =
− +⋅ − ( ) ( )1 1x x− +
11
xx+
=−
( )
( )( )
( )( )( )
( )( )( ) ( )
2 2 2
2 2 2
2 2
22 2
pogledati u 57. zadatak kako se to radi 3 12 2 3 3 121) ...
9 36 12 3
3 3
3 336 33 3 3 3
210.x x x x xx x
x x xx x
x xxx xx x x x
→− + − ⋅ ⋅ + +
= =− −
− + += =
− +
+ +++ += = =
− + − + ( ) ( )3 3x x− +
( )
( ) ( )( )
( )( )( ) ( )
2 2 2
2 2
22
33
2 8 4 4 8 4 8 44)3 2 2 35 6 3 2 6
2 224 43 2 3 3 2
xx
x x x x x x x xx x x xx x x x x
x xxx xx x x x x
+=
−
+ − + + − + − += = =
⋅ − ⋅ − ⋅ − ⋅− − − − −
− −−− += = =
⋅ − − ⋅ − − − ( ) ( )3 2x x− −
2
:
je kvadratni trinom ...
tehnika kako se taj izraz rješava na faktore objašnjena je u 66. zadatku ove zbirke !
23
5 6 uputa uz ovaj zadatak
xx
x x
−=
−
− − →
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( ) ( )( )
( ) ( )
( )( )( ) ( )
( ) ( )
2 22 2 2 2 2 2
2
2 29)1
1
211.x y zx y z xy x xy y z
x x z y y x z x x y z y x zx xz xy y x z
x y z x y zx y z x y zx y z x
+ −+ − + + + −= =
⋅ + + − + + ⋅ + + − ⋅ + ++ + − − −
+ − + ++ − + += =
+ + ⋅ − ( )x y z+ + ( ) 11x y zxx+ −
=−⋅ −
=
( ) ( )
( )( )( ) ( )
( ) ( ) ( )( )( )
( )( )( )( )( )
( ) ( )
4 2 4 28 8
3 2 2 3 1 2 2 2 1 2
4 4 4 4
2 2 2 2
2 2 2 2 4 4
2 2
2 2 2 22 2 2 2 4 4
2 2
1)
212.
x yx yx xy x y y x x x y y x y y
x y x y
x x y y x y
x y x y
x y x y
x y x yx y x y x y
x y x y
−−= =
+ − − ⋅ + ⋅ − ⋅ − ⋅
− += =
⋅ + − ⋅ +
⎡ ⎤− +⎣ ⎦=+ −
− +− + += =
+ −
( )( )
4 4
2 2
x y
x y
+
+ ( )
( )( )( )
( )2 2 4 4
x y
x yx y x y
x y
=−
−− += =
−
( )( )( )
4 4x y x y
x y
+ +
−
( )( )4 4x y x y
=
= + +
Rješenja ALGEBARSKI RAZLOMCI
ALGEBARSKI RAZLOMCI 11
Skup realnih brojeva autor: Mladen Sraga
( )( )( )
2 2 2
2 2
U ovom zadatku se ponovo susrečemo sa KVADRATNIM TRINOMOMu zadatku br. 66. smo dali detaljnu uputu kako se kvadratni trinom rastavlja na faktore ....
4 21)2 2 1 2
2 22 1
213.
x xx x x x x
x xx x x
− −= =
− − − + −− +
=⋅ − + ⋅ −( )
( )( )( )( )
( )22 22 2 1
xx xx x
−− += =
− +
( )( )
2
2
x
x
+
− ( )
( )
( ) ( )( ) ( )
( )( )
2
2 2
2
uputa kako rastavitikvadratni trinom iz brojnika:
2 , 1
211
1 1 12 1 2,
1 2 22
2 2 1 22 1 22 1 2
2 1
4 42)
m n
xxx
a m nm n b m nx x b m n
m nm n a c m nc
x x x x xx x xx x xx x
x x
= − =
+=
++
= + = −+ = + = −⎫⎫ ⎫− − = = − ⇒ ⇒ ⇒ ⇒ = − =⎬ ⎬ ⎬⋅ = ⋅ −⋅ = ⋅ ⋅ = −⎭ ⎭⎭= −
− − = − + − == − − − == − − − == − −
− ( ) ( )( ) ( )
( )( )( )( )
( ) ( )
2 222 2 2
2 2
2 2 2 1 1 2 12 2 2 1 112 2 2 1 2 2 2 1 22 5 2 2 4 1 2
2 1 2 12 1 2 12 2 1
x x xx xx x x x x x xx x x x x
x xx xx x
− ⋅ ⋅ + −⋅ − ⋅ ⋅ ++= = =
⋅ ⋅ − ⋅ ⋅ − ⋅ + ⋅ − − ⋅ −− + − − +
− −− −= =
− − ( ) ( )2 2 1x x− −
( ) ( )( ) ( )
( )( )
2
2 2
uputa kako rastavitikvadratni trinom iz brojnika:
4 , 1
2 12
2 5 52 5 2 5 4, 1
2 2 42
2 5 2 2 4 1 22 2 1 22 2 1 2
2 2 1
m n
xx
a m n b m n m nx x b m n
m n a c m n m nc
x x x x xx x xx x xx x
= − = −
−=
−
= + = + = − + = −⎫ ⎫ ⎫− + = = − ⇒ ⇒ ⇒ ⇒ = − = −⎬ ⎬ ⎬⋅ = ⋅ ⋅ = ⋅ ⋅ =⎭ ⎭ ⎭=
− + = − − + == − − − == − − − == − −
=
1
Rješenja ALGEBARSKI RAZLOMCI
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( ) ( )( ) ( )
( ) ( )( )( )( )
( ) ( )( )( )( )
( ) ( )( )( )
( )( )( )( )
2 1 2 1 2 1 23 2 2 3
3 2 2 3 2 1 2 2 2 1
2 2 2
2 2
2 2 2 2
2 2
2 2
4)
214.
x x x y x y x xx x y xy xx x y xy y x x x y x y y y
x x y x y x
x x y y x y
x x x y x y x y x x x x y x x y y xx y x y x y x y
x x y x y x
x y x y
x x y x y xx y x y
⋅ − ⋅ + ⋅ − ⋅− + −= =
+ + + ⋅ + ⋅ + ⋅ + ⋅
⋅ − + ⋅ −= =
⋅ + + ⋅ +
⋅ ⋅ − + ⋅ − + ⋅ ⋅ − − ⋅ − += =
+ + + +
⋅ − − +⎡ ⎤⎣ ⎦= =+ +
⋅ − − −=
+ +
( )( )( )( )
( )( )( )
( ) ( )( )( )
( )( )( )
( )( )( )
2 2
2 2 2 2 2 2
2 2
1
x x y yx y x y
xy x y xy x y xy y xx y x y x y x y x y x y
xy y xx y x y
⋅ − −= =
+ +
− − − ⋅ − ⋅ − + + ⋅ −= = = =
+ + + + + +
−=
+ +
=
Rješenja ALGEBARSKI RAZLOMCI
13 Skup realnih brojeva
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
( )( )
1 1
1 1
111)1 1
215.x xx x x x
x x x x x
a a aa a a a aa a a a a a a
+
+
⋅ ++ ⋅ + ⋅= = =
− ⋅ − ⋅ ⋅ −( )1
x
a
a
⋅ +
( )
( )( )
( )( )
( )( )( )
1 1
1 1
2 2
1 1
2
111
11 12)1 1 1
13)1
1 1 11 1
xx x x x
x x x x x
x x x x
x x x x
xx x
x x
aaa
a aa a a a a aa a a a a a a a
a a a a aa a a a a
aa a a a aa a a a
+
+
+
+
+=
−⋅ −
⋅ ++ ⋅ + ⋅ += = =
− ⋅ − ⋅ ⋅ − −
− ⋅ − ⋅= =
+ ⋅ + ⋅
⋅ − ⋅ − += = =
⋅ + ⋅ +
( ) ( )1 1a a⋅ − +xa ( )1a⋅ +
( )( )
44 4
3 2 1 3 2 1 3 2
1 11
114)1 1
xx x x x
x x x x x x x x x
x
a a
a aa a a a aa a a a a a a a a a a a a a a
a
+
+ + +
−= =
⋅ −− ⋅ − ⋅= =
− + − ⋅ − ⋅ + ⋅ − ⋅ ⋅ − + −
=( )4 1
x
a
a
⋅ −
( )
( )( )
( )( )( ) ( )
( )( )( )( )( )
( )
3 2
2 2 24
3 2 2 1 2
2 2
2
2
2
1
111 1 1 1
1 11 1 1
11 1 1
1 1
a a a
aaa a a a a a a
a aa a a
aa a a
a a
=⋅ − + −
−−= = =
− + − ⋅ − ⋅ + ⋅ −
− += =
⋅ − + ⋅ −
−− + += =
− +
( ) ( )21 1a a+ +
( )1a − ( )2 1a +
1 11a a
=
+= = +
−
=
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221. ( ) ( )
( )( ) ( ) ( )(
)
( )( )( ) ( )
( )( )
2 2
2 2 2 2
2 2
2 2 2 2 2 2
2 2
1 11 1 1 1
1 1
2 4 3 2 4 3 2 4 33 3 3 3 3 3 3 39 9
2 3 4 3 3 2 6 4 13 3
x x y yx y x y x x y yx y y x x y x x y yxy x y x y
x y x y x y x y x yx y x yx y x y x y
x x xx x x x x x x xx x
x x x x xx x
⋅ − + ⋅ −− − − − − + −+ = + = = =
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅4)
− + − ⋅ − − + −− − += = =
− + = − + = − + =− + − + − + − +− −
+ − − + + − += =
− + ( )( )
2 2 2
2 33 3
2 4 3 6 12 183 9
xx x
x x x xx x
+=
− +
− + + + += =
− −
5
7)
Rješenja ALGEBARSKI RAZLOMCI
15 Skup realnih brojeva ALGEBARSKI RAZLOMCI
autor:
222. 2 2 4 3 2 42 4 4 5 4
3 4 5 2 4 3 4 1 4 2 4
5 6 35 186 25 6 5 5
5
x x y y z a a b cx y z a b ca b c xy z a b c c x y z
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ = ⋅ =
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
=
2)
x⋅ 2x y⋅ ⋅ 2 4y z⋅ ⋅
6 3a⋅ 4b⋅ 1c⋅ 4
6c
⋅⋅
33 a⋅ ⋅ 2 4a b⋅ ⋅ c⋅5 5 x⋅ ⋅ 2y⋅ 4z⋅
( ) ( )( )
( )( )
( ) ( )
2 2 2 2
4 4
2 2 2 23 3 3 3
2 2 2 2
2 2
3 35 5
xy a xy ac c
x y x xy y x y x xy yx y x yx y x yx xy y x xy y
x y x xy y
=
⋅ ⋅= =
⋅
+ − + − + ++ −⋅ = ⋅ =
− −− + − +
+ − +=
5)
( )x y−
( )x y−⋅
( )( )
2 2
2 2
x xy y
x xy y
+ +
− +( )( )2 2x y x xy y= + + +
Mladen Sraga
Rješenja ALGEBARSKI RAZLOMCI
225.
( )( )( )
( )( )( )
( )( )( ) ( )
2 22 22 2 2 2
2 2 2 2 2 2
22 2 24 4 4 2:2 22 2
2 2
x x yx xx x x x xy yx y x y x y x y x xx y x xy y x
x x
+ −+ ⋅ ⋅ ++ + − − += ⋅ = ⋅ =
− + − + − +− − + −
+ +=
4)
( )x y− ( )( )x y
x y
−⋅
+
( )( ) ( )2 2
x y
x x
−
− +
( )( )( )( )
( )( )
( )( )( )( )( )( )( )( )
( )( )
( )
2 2 2 2
2 2 22
22
1 1 1:
1 11
1 1 11 1 1
1
x x yx y x
x y x y x yx x xy x x xyx y x yx x xy x x xyx y
x y x y x x yx y x y x x y
x y
+ −=
+ −
− − − − − +− − + −= ⋅ =
+ − + ++ − − −+ −
− − − + ⋅ + −= ⋅ =
+ − + + ⋅ − −
− −=
5)
( )
( )1
1
x y
x y
− +
+ − ( )( )1
1
x y
x y
+ −⋅
+ + ( )1x y− −
( )( ) ( )( )( ) ( )
( ) ( ) ( )( )
( )( )
( )
3 2 3 2
3 2 3 2
2 2 2 2
2 2 2 2
2 22 2 2 2
2 2 2 2
2 22 2 2 2
2 2 2 2
1 11 1
:
2 2
22 2
x y x yx y x y
x y x y x y x yx y x y x xy x xyx y x y x y x yx xy x xy
x x yx y x yx y x x y
x yx xy y x xy yx y x y
x yx y x yx y x y
=
− + + −= =
+ + + +
− − + + +⎛ ⎞− + + −+ = ⋅ =⎜ ⎟+ + + ⋅ −− +⎝ ⎠
−− + += ⋅ =
− +
−− + + + += ⋅ =
− +
++ −= ⋅ =
− +
8)
( )2 2x y−
( )2 2x y−⋅( )2 2x y+
2=
16 www.mim-sraga.com
M.I.M.-Sraga – centar za poduku
Rješenja ALGEBARSKI RAZLOMCI
Ovo su ogledni primjeri stranica iz ZBIRKE POTPUNO RIJEŠENIH ZADATAKA ALGEBARSKI IZRAZI PRIRUČNIK ZA SAMOSTALNO UČENJE Autor: Mladen Sraga izdavač: M.I.M.-Sraga kompletnu zbirku možete kupiti preko: www.mim-sraga.com ili narudžbom na 01-4578-431 ili na mail: [email protected] Puna cijena kompletne zbirke ALGEBARSKI RAZLOMCI za PRVI razred srednje škole je 120 kn trenutno sa popustom od 50% prodaje se za 60kn
17 Skup realnih brojeva ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
Rješenja ALGEBARSKI RAZLOMCI
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Rješenja ALGEBARSKI RAZLOMCI
ALGEBARSKI RAZLOMCI 19
Skup realnih brojeva autor: Mladen Sraga
( )
79.Tehnika rješavanja ovog zadatka je sljedeća: brojnik i nazivnik rastavimo na faktore i kratimo...
1x y xxy
x xy x y⋅
= =− ⋅ −
yx
⋅
( )
( )2 23
2 2
-rastavimo na faktore, kratimo u brojniku i nazivniku
Ovi prvi zadaci su jednostavni, samorastavimo na faktore i kratimo
kratimo:
11x
yyy
a b b a b b aaba a b a b b a b a ba b ab
=−⋅ −
⋅ ⋅ ⋅ ⋅= = =
⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ −−
1.)
b⋅ 2ba
⋅b⋅ ( )
2.)
( )( )
2ba ba b
x a b xax bxax bx x a b
=−⋅ −
⋅ −−= =
+ ⋅ +( )a b
x⋅ −
( )3.)
( )( )2
kratimo
33
a ba ba b
z x y zxz yzz zz z
−=
+⋅ +
⋅ −−= =
⋅ ++( )x y
z⋅ −
( )
( )( )( )
2
kratimo
33
1
x yzz
a a aa a aa aax ay a x y a x y
−=
+⋅ +
⋅ +⋅ ++= = =
− ⋅ − ⋅ −
4.)
( )1aa⋅ +
( )
( )( )
( )2
2
1
22222 22
ax yx y
a a ba a ba a a ba aba b b b b a bab b
+=
−⋅ −
⋅ −⋅ −⋅ − ⋅ ⋅−= = =
⋅ − ⋅ ⋅ ⋅ −−
5.)
( )2b a b⋅ −
ab
=6.)
( )( )
( )( )( ) ( )
( )( )
( )( )( )
( )
2
2 3 2 2
.
.
2 23 2
2
3 4 3 43 47.)3 4 3 49 16 9 16
4 4 9 4 2 3 2 316x 36
3 4
4 2 33
kratimo
ab
3 2 3 3 2 36 9
R KVADRATA
R KVADRATA
a a b a a ba abb a b a b b aa b b b a b
x x y x x y x yxyy x y y x yxy y
⋅ + ++= =
− +− −
− − +−= =
− −−
x x yy
=−
+=
ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
8.)
Rješenja ALGEBARSKI RAZLOMCI
www.mim-sraga.com
ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
20
M.I.M.-Sraga – centar za poduku
5 3 2 3 2 2
3 2 2 2
2 3
5 3 2 3 2 2
3 2 2 2
2
kratimo kratimo
još jednom isti zadatak:
12 27 3 (4 ) 3 (2 3 )(2 3 ) 3 (2 3 )10.4 (2 3 ) 48 12 4 (2 3 )
sa
12 27 3 (4 )8 12 4 (2 3 )
3
a a b a a b a a b a b a a bb a b ba b a b a b a b
a a
a a b a a ba b a b a b a b
a a
− − − + += = =
−− −
− −= =
− −
⋅ ⋅ ⋅=
( )2 2 2
2
4 3
4 (2 3 )
a b a
a b a b
− ⋅=
⋅ ⋅ ⋅ −
( )2 2
2
4
4
a a b
a
⋅ ⋅ −
⋅ (2 3 )
3 (2 3 )3 (2 3 )(2 3 )4 (2 3 )
b a b
a a ba a b a bb a b
=⋅ ⋅ −
−− += =
−
(2 3 )4 (2 3 )
a bb a b
+
−
4 3 2 2 2 2 2
4 3 3
2
treba prepoznati kvadrat razlike
3 (2 3 )4
2 8 8 2 ( 4 4 )11. 2 ( 2 )
2
a a bb
a a b a b a a ab ba a b a a b
a
=
+=
− + − += =
− −
⋅=
2
2
( 2 )a ba
⋅ − ( ) ( )
( ) ( )
2rastavimo: ( 2 ) 2 2( 2 )
2 2 2
a b a b a ba a b
a b a b
= − = − ⋅⋅ ⋅ −
⋅ − ⋅ −=
−
( )2a a b⋅ −
2 2
2
2( 2 )
( 3)6 9 ( 3)12.( 3)( 3)9
a ba
aa a aa aa
=
−=
−− + −= =
− +−
( 3)( 3)
aa
⋅ −
−
2
2
( 3)( 3)( 3)
( 2)4 ( 2)( 2)13.( 2)( 3)6
aaa
aa a aa aa a
−=
+⋅ +
−− − += =
− ++ −
( 2)( 2)
aa
+
−
( )
2 2 2
2 2
3 3 2 2
23( 3)
6 3 2 6 ( 3) 2( 3) ( 3)( 2)
( ) ( )( )( )14.( )
aaa
a a a a a a a a a a
a b a ba b a b a ba b a b a ab b
+=
++
↓
+ − = + − − − + − + = + −
− +− − += =
+ + − + ( )a b+ ( ) 2 22 2
a ba ab ba ab b
−=
− +− +
Rješenja ALGEBARSKI RAZLOMCI
ALGEBARSKI RAZLOMCI
ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
21 Skup realnih brojeva
autor: Mladen Sraga
( )( )
( )( )( ) ( )
( )( )( )( )
( )( )
( )( )( )( ) ( )( )( )
( )( )
2 2
3 2 2 3 3 2 2 3
2 2
2 22 2
2 2
2 2 2 2
.
2 2 2
17.)
18.)
1
1
19.)
2KV ZBROJA
a b a ba ba ab a b b a a b ab b
a b a ba a b b a b
a b a b a ba ba b a b
a b a ba ba a b b a b a b
a b a ba b a b a b
a b a ba b a ba ba b
a ab b ca b
− +−= =
+ − − − + −− +
= =− + −
− + += =
+− +
− +−= =
− − − − − −− +
= =− + − +
− += =
+ − −
−=
− −
+ + −+ +( ) ( )
( )( )( )( )( )( )( )
( )( )( )( )( )( )
( )( )( )( )
( ) ( )
. .
2 2
2 2 2 2 2 2
2 2 2 2 2 2
2 2
2 2
kratimo
2 220.)2 2
R KV
a b cc a a b c c a b c a c
a b c a b ca b c a c
a b ca c
a b c ab a ab b ca b c ac a ac c b
a b c
a c b
a b c a b ca c b a b c
a b c a b ca b c a b ca b c a b c
+ −= =
+ + + + + +
+ − + += =
+ + +
+ −=
+
+ − + + + −= =
− + + + + −
+ −= =
+ −
+ − + += =
+ − + +
+ − + ++ − + += =
− + + + ( ) ( )a b c a b c− + + +
a b ca b c
=
+ −=
− +
Rješenja ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
ALGEBARSKI RAZLOMCI
www.mim-sraga.com22
M.I.M.-Sraga – centar za poduku
( ) ( )( ) ( )( )( )( )( )
( )( )( )( )( )
( )
2 2
3 2 3 2
2 2
2
6 5 5 521.)5 5 5 5
1 5 11 5 1
1 51 5
11 51 1 5
a a a a aa a a a a a
a a aa a a
a aa a
aa aa a a
+ + + + += =
+ − − − + −+ + +
= =− + −
+ += =
− +
++ += =
+ − +
( )5a +
( )1a + ( ) ( )1 5a a− +
( ) ( )( ) ( )( )
( )( )
( )( )( )
2 2
4 2 2
2
2 2
2
2 2
2
11
2 2 2 222.)1 1 1 1 1 1
2 22 1 1 2 1 1
2 22 2 2
1 2 2
a
x x x xx x x
x xx x x x
x xx x x x
x x
=
=−
+ + + += =
+ − + − + +
+ += =
+ + − + + +
+ += =
+ + +
⋅ + +=( ) ( )2 22 2 2x x x x+ + +
( )
( )( ) ( )( ) ( )
( ) ( )( )( ) ( )
( )( )( )( )( )( )
( )( )( )( )( )( )
2
22
2 2
12
12
2 3 1 42 1 4 2 323.)
1 3 1 3
2 3 1 2 1 21 1 3
2 3 3 12 1 1 3
2 31
x x
x x
a aa a a
a a a a
a a aa a a
a a aa a
aa
=
= =+
=+
− − −− − −= =
+ − + −
− − − − += =
+ + −
− − += =
+ + −
−=
+
Rješenja ALGEBARSKI RAZLOMCI
ALGEBARSKI RAZLOMCI
ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
23 Skup realnih brojeva
autor: Mladen Sraga
( )( )
( ) ( )( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )( )
( ) ( )( )( ) ( )( )
( )( )
( ) ( ) ( )( )( )( )
22 2 2
22 2
2
2
2
2
22 2
2 2
4 4 1 2 3 2 1 3 324.)
6 9 1 1 3 1 1 1
2 1 1 3 1
3 1 1
2 1 1 3 2 133 1 2 1
2 4 2 2 2 24 4 425.)2 2 2 2 24 2 2
a a a a a a a a
a a a a a a a a a a
a a a a
a a a a
a a a aaa a a
x y x y x yx xy yx y x y x y x y xx y x y
− + − − − + − −= =
⎡ ⎤− + − + + − − + + +⎡ ⎤⎣ ⎦ ⎣ ⎦
− + − += =
− + − +
+ + − −= =
−− + −
+ − + − + ++ + −= =
− + − − −− − ( )
( )( )( ) ( )
( ) ( )( ) ( )
( ) ( )
( ) ( )
( ) ( )( )( )( ) ( )( ) ( )( ) ( )
2 2 22 2 2
2 2 2 2 2 2
22
2 2
2 2 2
2 22 2
2
2226.)
2 2
1
27.)
yx yx y
a b bc c a b ca b c bc a b c
a b c a b c ac a b c a ac c b
a b c a b c
a b c a c b
a b c a b c a b ca b c a c b a c b
a b c a b ca b c a b c
a b c
=+ −
+ +=
−
⎡ ⎤− + + + −− − − + − ⎣ ⎦= =+ + − + − + + − + −
⎡ ⎤− + + −⎣ ⎦= =⎡ ⎤+ + − −⎣ ⎦
− + + + + −⎡ ⎤⎣ ⎦= =+ + − − − +
− − + −= =
− − + −
+ + ( )( )( )
( )( )( )
( )( )( ) ( )
( )( )
( ) ( ) ( )( )
( )
2 2 2 22 2 2 2
2 22 2
2 2 2 2 2 2
22
2 2 22 2
2 2 23 2 228.)2 2
2 2 2
2
a b c a b c a b c a b cab bc aca b c bc a b bc c a b c
a b c a b c a b ca b ca b c a b c
x xy xy y z x yx xy xz y yzx y yz z x y yz z
x x y y x y z x y
x y z
x y x y
+ + + + + + + ++ + += =
− − − + + − +
+ + + + + += =
− −− + + +⎡ ⎤⎣ ⎦
− + + −− + + −= =
− + − − − +
− − − + −= =
− −
− −=
( )( ) ( )
( )( )( )( )
2 2z x y x y z x yx y z x y z x y zx y z x y z
+ − − + −= =
− + + − + −− − + −⎡ ⎤⎣ ⎦
=
Rješenja ALGEBARSKI RAZLOMCI
-sraga.com
ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
( )( ) ( ) ( )
( )
( )( )( ) ( )
( ) ( )( )( )
( )( )
2 2 2
2 2 2 22 2 2 2
2 2 2 2 2 2 2
2 2 2 2
2 2 23 2 2 2 229.)2 2
2 2 2
30.)
a a b b a b c a ba ab ac bc a ab ab b ac bca b bc c a b bc c a b c
a b a b c a b a b c a ba b c a b c a b ca b c a b c
xy a b abx aby a x y b xy abx ababx aby xy a b
− − − + −− + + − − + + −= = =
− + − − + − −
− − + − − + −= = =
− + + − + −− − + −⎡ ⎤⎣ ⎦
⋅ − + − − − + −=
+ + ⋅ +
( ) ( )( ) ( )
( ) ( )( ) ( )
( ) ( )
2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
3 3 2
3 3 6 9 9 3 6 3 9 931.)3
y a xy abx aby b xyabx aby a xy b xy aby b xy abx a xy
ax ay bx by ay bx ay bx ax by ax byby ay bx ax bx ay ay bx by ax ax by
a x b abx a x ab a x abx b a x abaxa x ab x ax b a
+ − −= =
+ + + + + +
⋅ + − ⋅ + + ⋅ − −= =
⋅ + + ⋅ + + ⋅ + +
+ − + − − + + −=
− ⋅ − + ( ) ( ) ( )
=
( ) ( )( ) ( ) ( )
( ) ( )( ) ( ) ( )
( ) ( ) ( )( ) ( ) ( )
( ) ( )( )( ) ( ) ( )
( )
2 2 2
2
3
3 2 9
3
3 93
3 93
3. 93
3 3 9
ax b ax b ax b a
a x abx b a ax b
ax ax b ax b ax b a
ax b a ax bax ax b ax b ax b a
ax b ax b a ax bax ax b ax b ax b a
ax b ax b aax ax b ax b ax b a
ax b aax ax b ax b
=⋅ − ⋅ + ⋅ − +
⋅ − + + ⋅ −= =
⋅ − ⋅ + ⋅ − +
⋅ − + ⋅ −= =
⋅ − ⋅ + ⋅ − +
⋅ − ⋅ − + ⋅ −= =
⋅ − ⋅ + ⋅ − +
− ⋅ − += =
⋅ − ⋅ + ⋅ − +
− +=
⋅ + ⋅ − +( )
( )( ) ( ) ( )
3
3 3 33
a
ax b aax ax b ax b a ax ax b
=
⋅ − += =
⋅ + ⋅ − + ⋅ +
24 www.mim
M.I.M.-Sraga – centar za poduku
Rješenja ALGEBARSKI RAZLOMCI ć− − −Matematika 1 riješeni zadaci po Pavkovi Veljan
25 Skup realnih brojeva
ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
( ) ( ) ( )( ) ( )( ) ( )
( ) ( )( ) ( )( ) ( )
( )
3 2 3 2
2
2 2
2
3 2 2 3 3 2 2 3
5 4 4 5 5 4 4 5
2
2 3 2 3 2 2 3 332.)2 1 3 12 2 3 1
2 1 3 1
1 2 3
1 2 3
1 2 3
1 11
1
3 6 2 3 6 233.)9 18 2 9 18 2
3
x x y x y x x x y yx x y xx x y x
x x y x
x x y
x x y
x x y
x xx
x
x xy x y y x x y xy yx xy x y y x x y xy y
x x
+ − − − + −= =
⋅ − + ⋅ −− + ⋅ −
⋅ − + ⋅ −= =
− ⋅ +
− ⋅ += =
− ⋅ +
− ⋅ += = +
−
+ − − − + −= =
− − + − − +
⋅=
( ) ( )( ) ( )
( ) ( )( ) ( )
( )( ) ( )
( ) ( ) ( ) ( )
( ) ( )( ) ( )( )
( ) ( )( )
( )
2
4 4
2 2
4 4
2 2
2 2 2 2
2 2
2 2 2 2 2 2 2 2
2
2 2
2 29 2 2
2 3
2 9
3
3 3
13
34.)
y y x yx x y y x y
x y x y
x y x y
x y
x y x y
x y
ac ad bc bd a c d a c d b c d a c dac ad bc bd ac bc ad bd
c d a b ac a b d a b
c d ba b c d
c d ba b
2
− + ⋅ −=
⋅ − − ⋅ −
− ⋅ += =
− ⋅ −
+= =
− ⋅ +
=−
+ + + − ⋅ + ⋅ + + ⋅ + − ⋅ += =
− + − + − −
+ ⋅ + −= =
⋅ + − ⋅ +
+ ⋅= =
+ ⋅ −
+ ⋅=
+ ( ) ( )
( ) ( )
c d c db
a b c d
=⋅ − ⋅ +
=+ ⋅ −
Rješenja ALGEBARSKI RAZLOMCI
Univerzalna zbirka potpuno korak po korak riješenih zadataka za prvi razred svih srednjih škola
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MATEMATIČKE FORMULE I NJIHOVA PRIMJENA - MEMENTO iz MATEMATIKE
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Rješenja ALGEBARSKI RAZLOMCI
Priručnici za samostalno učenje MATEMATIKA 2
27 Skup realnih brojeva ALGEBARSKI RAZLOMCI
autor: Mladen Sraga
Rješenja ALGEBARSKI RAZLOMCI
www.mim-sraga.com
š ć− − −Matematika 1 rije eni zadaci po Pavkovi Veljan
28
M.I.M.-Sraga – centar za poduku
3
2 2
2 2
2 2
2 3 2
2 3 2
5 3 2 4 3 2 2
3 2 2 4 3
2
79.Skrati razlomke:
1.) 2.) 3.)
24.) 5.) 6.)3 2
3 4 16x 367.) 8.)9 16 6 9
12 27 2 8 810. 11.8 12 2
612.
xy ab ax bxx xy ax bxa b ab
xz yz a a a abax ayz z ab b
a ab xya b b xy y
a a b a a b a ba b a b a a b
a
−= =
− +−
− + −= =
−+ −
+ −= =
− −
− − += =
− −
−
=
=
( ) ( )
( )
( )( ) ( )( ) ( )
2 2
2 2
2 2 2 2
3 2 2 3 2 2
2 2 2 2 2 2
2 2 2
2 2
3 2 4
2
2
9 413. 14.9 6
17.) 18.)
2 219.) 20.)2
6 5 2 221.) 22.)5 5 1 1
2 1 4 2 323.)
1 3
a aa a aa b a b
a ab a b b a a b b
a ab b c a b c aba b c a a b c c a b c ac
a a x xa a a x
a a a
a a
+ −= =
− + −− −
= =+ − − − − −
+ + − + − += =
+ + + + + − + +
+ + + += =
+ − − + −
− − −
+ −
( )( )( ) ( )
2
3 3
a ba b
−=
+
( )( )( )( )( )
2 2
2 2
2 2 22 2
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2 2
2
2 2
4 4 1 2 324.)
6 9 1 1
24 4 425.) 26.)4 2 2 2
2 2 2 3 2 227.) 28.)2
3 229.)
a a a a
a a a a a
a b c bc a b cx xy yx y x y a b c a b c ac
a b c ab bc ac x xy xz y yza b c bc x y yz z
a ab ac bca b
− + − −= =
⎡ ⎤− + − + +⎣ ⎦
− − − + −+ + −= =
− − + + − + −
+ + + + + − + + −= =
− − − − + −
− + +−
( )( )
2 2 2 2
2 2 2 2 230.)
2
xy a b abx aby
bc c abx aby xy a b
⋅ − + −= =
+ − + + ⋅ +