Algebarski razlomci PDF

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


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 !

2 www.mim-sraga.com

M.I.M.-Sraga – centar za poduku

mailto:[email protected]


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

116)

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

117)

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 y xy x x y x y yx 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⋅ −1

3y=


autor: Mladen Sraga


4 www.mim-sraga.com

( )( )( )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

−=

− +− +

↓




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

− +=

+

− − = − − = = − − − +

autor: Mladen Sraga


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

1 4 1 1 1 4

2 1 2 2

1 1 4 4

u brojniku i nazivniku podvučemo zajedničke faktore

izlučimo z.f. i

:

aa ba b

dodatna uputaa b a ba b a b

a b ab a b a ba a b a b ba 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 kvadrata a a b a b

a b a b a b

a

⋅ ⋅ ⋅ −= =

⋅ ⋅ − +

=a b⋅ ⋅ ( )2 2a b⋅ −

a b⋅ ( )2 2a b⋅ − ( )2 2

2 2

kratimo a b

aa b

=+

=+

6 www.mim-sraga.com




( )( )

( )( )( )( )( )

( )( )( )( )( )

( )( )( ) ( )( ) ( )

( )( )( )( )( )

( )

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

+ +=

⋅ − −

=

autor: Mladen Sraga


8 www.mim-sraga.com

( ) ( ) ( )( )( )

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

=−

↓ ↑

− = −



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

−=

−

− − →


autor: Mladen Sraga


10 www.mim-sraga.com

( ) ( )( )

( ) ( )

( )( )( ) ( )

( ) ( )

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

=

= + +




( )( )( )

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

autor: Mladen Sraga


( ) ( )( ) ( )

( ) ( )( )( )( )

( ) ( )( )( )( )

( ) ( )( )( )

( )( )( )( )

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

⋅ − −= =

+ +

− − − ⋅ − ⋅ − + + ⋅ −= = = =

+ + + + + +

−=

+ +

=




( )( )

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

=

+= = +

−

=


autor: Mladen Sraga



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)


222. 2 4 4 5 4 2 2 4 3 2 4

3 4 5 2 4 3 4 1 4 2 4

5 18 5 6 36 25 6 5 5

5

x y z a b c x x y y z a 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 xy y x 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= + + +


autor: Mladen Sraga


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=




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