algebarski izrazi zadaci za vjezbu

**** MLADEN SRAGA **** © - 2003.-2013.

UNIVERZALNA ZBIRKA POTPUNO RIJEŠENIH ZADATAKA PRIRUČNIK ZA SAMOSTALNO UČENJE

ALGEBARSKI IZRAZI KVADRAT ZBROJA KVADRAT RAZLIKE

α

Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Koristeći se formulama za:

kvadrat zbroja

( A + B)

2

i

kvadrat razlike

( A − B)

= A2 + 2 AB + B 2

2

= A2 − 2 AB + B 2

izračunaj: 1)

( x + y)

2

2)

( y + x)

2

3)

( x − y)

2

4)

( y − x)

5)

( x + 3)

2

6)

( x − 5)

2

7)

(2 − x)

2

8)

(2 + x)

9)

( x − 1)

2

10)

( x − 4)

2

11)

( a − 2)

2

15)

(7 − x)

2

19)

( x + 5y)

23)

( 3x + 2 )

2

27)

(5x + 2 y )

2

31)

( 3m + n )

35)

( 0,5 x − 2 y )

13)

( y − 5)

2

14)

( x + 6)

17)

(2 + z)

2

18)

( z − 2)

21)

(7x + y )

22)

( x − 3y)

25)

( 2a − 5b )

26)

( 2x + 3y )

29)

( 3x − 4 y )

30)

( 5a + 7b )

33)

( 0, 2 x + 3)

34)

(1, 2 x + 0,3)

1⎞ ⎛ 37) ⎜ x + ⎟ 2⎠ ⎝

2 2 2 2

2

⎛3 ⎞ 38) ⎜ x + 1⎟ ⎝2 ⎠

⎛5 ⎞ 41) ⎜ x + 0, 2 y ⎟ ⎝2 ⎠ ⎛ 1 ⎞ 45) ⎜1 x + y ⎟ ⎝ 2 ⎠

2

61) 65)

(

69)

(5x − 6 y )

73)

( − 2a − 3b )

77)

( − 3x + y )

53) 57)

81) 85) 89)

3 ⎛ 42) ⎜ 4 x − 2 ⎝

2

( x + 1) ( 2a − b ) ( 2 x y − 3z ) (a b − c d )

49)

3

2

2

2

3

2

2 3

2

4

2

2

2a 4b3 − 3a 2b7

(2 (3 (a

m

m

n

) −3 )

−2

2

2

)

2

20)

(5x − y )

2

24)

( 2 x + 5)

28)

( 3x + 4 y )

32)

( 2m − 5n )

2

36)

( 0, 2m + 0,3n )

2

5 ⎞ ⎛ 40) ⎜ 0, 4 x − y ⎟ 3 ⎠ ⎝

2

2

2

⎞ y⎟ ⎠

2 ⎞ ⎛1 43) ⎜ x + y 3 ⎟ 3 ⎠ ⎝2 ⎞ y⎟ ⎠

2

59)

62)

(a b

63)

66)

(

67)

(

70)

( x + 4y)

71)

( − x − 3)

74)

(−x − 4 y)

n

2

86) 90)

2

−2

3

− 5b 2

2 3

2

51)

)

2

)

− c2

55)

2

4a 3b − 5a 2b3

)

2

−2

( − 5a b + 2 c ) (3 − 5 ) (2 + 2 ) (a − a ) m

m

4

3

4

2

2

79) 83)

m +1

m −1 2

87)

n −1

n +1 2

91)

2

2

3 2

3

4

52) 56) 60)

2

64)

9a 4b 3 − 2a 2b 4

)

3

m

n

n

1− n

x

2

2

4

2

2

x

2

2

2 2 2

2

2

⎞ y⎟ ⎠

( 3a − b ) ( 3x + 7 y ) ( 2ab − 3c ) ( a b + 4c ) 2

3 2

4

5

2

2

2

2

4

2 3

2

5

2

2

2 ⎛1 ⎞ 68) ⎜ a 2b − a 4b3 ⎟ 3 ⎝2 ⎠ 72)

(−x − 2 y)

76)

( − a + 7b )

2

( − 2 x y + 3z ) (2 + 2 ) (2 + 2 ) ( 2a − 3b ) 2

2

3 ⎛1 48) ⎜ x z − 2 ⎝4

3 2

2

2

3 ⎞ ⎛2 44) ⎜ x 3 − y 4 ⎟ 4 ⎠ ⎝3

2

⎛ 1 ⎞ 75) ⎜ − x − 4 y 2 ⎟ ⎝ 2 ⎠

2

2 3

2

⎛ 2 ⎞ 47) ⎜ 3 x − 0,5 y ⎟ ⎝ 3 ⎠

( 2ab + 3c ) 2

82)

M.I.M.-Sraga -2003./2013. www.mim-sraga.com

2

58)

78)

2

(8 + y )

54)

2

)

16)

2

(3 − x ) ( 2x + 3 y ) (a b + c ) ( a b − 3c )

+ 3n

+ a n +1

(3 − b)

2 ⎛3 39) ⎜ x + 3 ⎝4

⎞ y⎟ ⎠

2

12)

2

(x (a

50)

)

2

2

1 ⎛ 2 46) ⎜ 2 x − 1 4 ⎝ 3

2

2

2

2

80) 84) 88) 92)

2

2

( −2 + 2 ) (2 + 2 ) (2 − 2 ) ( 2a + 3b ) m

2

2

n

m −1 2

m

n +1

x

n−2

y

2

2

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 i ( A − B )2 = A2 − 2 ⋅ A⋅ B + B 2 2 2 2 2 2 2 To pravilo možemo pisati i ovako: ( I + II ) = I + 2 ⋅ I ⋅ II + II i ( I − II ) = I − 2 ⋅ I ⋅ II + II Sve o kvadratu binoma morali bi znati već iz 8. razreda ali ponovim o to još jednom.

Pogledajmo kako to radimo na dva primjera:

a)

( x + 1)

2

a)

( x + 1)

2

( x − 1)

b)

2

= x 2 + 2 ⋅ x ⋅1 + 12 = x 2 + 2 x + 1

objašnjenje:

( x + 1)

2

= x 2 + 2 ⋅ x ⋅1 + 12 = x 2 + 2 x + 1

↑ ↑ ↑ ↑ ↑ 2 2 ( A + B ) = A + 2 ⋅ A⋅ B + B2

( x + 1)

ili

↑ ↑

( I + II )

2

2

= x 2 + 2 ⋅ x ⋅1 + 12 = x 2 + 2 x + 1 ↑

↑

↑

= I 2 + 2 ⋅ I ⋅ II + II 2

↓ ovo pravilo čitamo ovako: Prvi plus drugi na kvadrat jednako je prvi na kvadrat, plus dvostruki prvi puta drugi, plus drugi na kvadrat.

b)

( x − 1)

2

= x 2 − 2 ⋅ x ⋅1 + 12 = x 2 − 2 x + 1

objašnjenje:

( x − 1) ↑ ↑

( A − B)

2

2

= x 2 − 2 ⋅ x ⋅1 + 12 = x 2 − 2 x + 1 ↑ ↑ ↑ = A2 − 2 ⋅ A ⋅ B + B 2

( x − 1)

ili

2

= x 2 − 2 ⋅ x ⋅1 + 12 = x 2 − 2 x + 1

↑ ↑ ↑ ↑ ↑ 2 2 ( I − II ) = I − 2 ⋅ I ⋅ II + II 2

1)

( x + y)

= x 2 + 2 xy + y 2

2)

( y + x)

2

= y 2 + 2 ⋅ y ⋅ x + x 2 = x 2 + 2 xy + y 2

3)

( x − y)

2

= x 2 − 2 xy + y 2

4)

( y − x)

2

= y 2 − 2 ⋅ y ⋅ x + x 2 = x 2 + 2 xy + y 2

5)

( x + 3)

2

= x 2 + 2 ⋅ x ⋅ 3 + 32 =

2

= x2 + 6x + 9

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 To pravilo možemo pisati i ovako:

6)

( x − 5)

2

= A2 + 2 ⋅ A ⋅ B + B 2 i

( I + II )

2

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

= I 2 + 2 ⋅ I ⋅ II + II 2 i ( I − II ) = I 2 − 2 ⋅ I ⋅ II + II 2 2

= x 2 − 2 ⋅ x ⋅ 5 + 52 = = x 2 − 10 x + 25

7)

(2 − x)

2

= 22 − 2 ⋅ 2 ⋅ x + x 2 = = 4 − 4x + x2

8)

(2 + x)

2

= 22 + 2 ⋅ 2 ⋅ x + x 2 = = 4 + 4 x + x2

9)

( x − 1)

→ to možemo p isati i dalje: = x 2 − 4 x + 4

2

= x2 + 4x + 4

dobro je kako god napišete ...

= x 2 − 2 ⋅ x ⋅1 + 12 = = x2 − 2x + 1

10)

( x − 4)

2

= x 2 − 2 ⋅ x ⋅ 4 + 42 = = x 2 − 8 x + 16

11)

( a − 2)

2

= a 2 − 2 ⋅ a ⋅ 2 + 22 = = a 2 − 4a + 4

12)

(3 − b )

2

= 32 − 2 ⋅ 3 ⋅ b + b 2 = = 9 − 6b + b 2

13)

( y − 5)

2

= y 2 − 2 ⋅ y ⋅ 5 + 52 = = y 2 − 10 y + 25

14)

( x + 6)

2

= x 2 + 2 ⋅ x ⋅ 6 + 62 = = x 2 + 12 x + 36

15)

(7 − x)

2

= 72 − 2 ⋅ 7 ⋅ x + x2 = = 49 − 14 x + x 2

16)

(8 + y )

2

= 82 + 2 ⋅ 8 ⋅ y + y 2 = = 64 + 16 y + y 2

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2

= A2 + 2 ⋅ A ⋅ B + B 2 i

To pravilo možemo pisati i ovako:

17)

(2 + z)

2

( I + II )

2

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

= I 2 + 2 ⋅ I ⋅ II + II 2 i ( I − II ) = I 2 − 2 ⋅ I ⋅ II + II 2 2

= 22 + 2 ⋅ 2 ⋅ z + z 2 = = 4 + 4z + z2

18)

( z − 2)

2

= z 2 − 2 ⋅ z ⋅ 2 + 22 = = z2 − 4z + 4

19)

( x + 5y)

= x2 + 2 ⋅ x ⋅ 5 y + (5 y ) = 2

2

= x 2 + 10 xy + 52 y 2 = = x 2 + 10 xy + 25 y 2 Još jednom isti zadatak:

19)

( x + 5y)

= x 2 + 2 ⋅ x ⋅ 5 y + ( 5 y ) = x 2 + 10 xy + 52 y 2 = x 2 + 10 xy + 25 y 2

2

2

7 Primjeni pravilo: ( ab ) = a n b n n

u našem slučaju:

(5 y )

2

= 52 y 2 = 25 y 2

7 i ovdje

20)

(5x − y )

2

= ( 5 x ) − 2 ⋅ 5 x ⋅ y + y 2 = 52 x 2 − 10 xy + y 2 = 25 x 2 − 10 xy + y 2 2

ili možemo pisati i ovako:

(5x − y )

= (5x ) − 2 ⋅ 5x ⋅ y + y 2 =

2

2

= 52 x 2 − 10 xy + y 2 = = 25 x 2 − 10 xy + y 2

21)

(7x + y )

2

= (7 x) + 2⋅ 7 x ⋅ y + y2 = 2

= 7 2 x 2 + 14 xy + y 2 = = 49 x 2 + 14 xy + y 2

22)

( x − 3y)

2

= x2 − 2 ⋅ x ⋅3 y + (3 y ) = 2

= x 2 − 6 xy + 32 y 2 = = x 2 − 6 xy + 9 y 2

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 i ( A − B )2 = A2 − 2 ⋅ A⋅ B + B 2 2 2 2 2 2 2 To pravilo možemo pisati i ovako: ( I + II ) = I + 2 ⋅ I ⋅ II + II i ( I − II ) = I − 2 ⋅ I ⋅ II + II 23)

( 3x + 2 )

= ( 3 x ) + 2 ⋅ 3 x ⋅ 2 + 22 =

2

2

= 32 x 2 + 12 x + 4 = = 9 x 2 + 12 x + 4

24)

( 2 x + 5)

= ( 2 x ) + 2 ⋅ 2 x ⋅ 5 + 52 =

2

2

= 22 x 2 + 20 x + 25 = = 4 x 2 + 20 x + 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.

25)

( 2a − 5b )

2

= ( 2a ) − 2 ⋅ 2a ⋅ 5b + ( 5b ) = 22 a 2 − 20ab + 52 b 2 = 4a 2 − 20ab + 25b 2 2

2

Još jednom isti zadatak: ( 2a − 5b ) = ( 2a ) − 2 ⋅ 2a ⋅ 5b + ( 5b ) = 2

2

2

Primjeni: ( ab ) = a nb n n

= 22 a 2 − 20ab + 52 b 2 = = 4a 2 − 20ab + 25b 2

26)

(2x + 3y )

2

= ( 2 x ) + 2 ⋅ 2 x ⋅3 y + (3 y ) = 2

2

= 22 x 2 + 12 xy + 32 y 2 = = 4 x 2 + 12 xy + 9 y 2

27)

(5x + 2 y )

2

= (5x ) + 2⋅5x ⋅ 2 y + ( 2 y ) = 2

2

= 52 x 2 + 20 xy + 22 y 2 = = 25 x 2 + 20 xy + 4 y 2

28)

( 3x + 4 y )

2

= ( 3x ) + 2 ⋅ 3x ⋅ 4 y + ( 4 y ) = 2

2

= 32 x 2 + 24 xy + 42 y 2 = = 9 x 2 + 24 xy + 16 y 2

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 To pravilo možemo pisati i ovako:

29)

( 3x − 4 y )

= A2 + 2 ⋅ A ⋅ B + B 2 i

( I + II )

2

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

= I 2 + 2 ⋅ I ⋅ II + II 2 i ( I − II ) = I 2 − 2 ⋅ I ⋅ II + II 2 2

= ( 3x ) − 2 ⋅ 3x ⋅ 4 y + ( 4 y ) =

2

2

2

= 32 x 2 − 24 xy + 42 y 2 = = 9 x 2 − 24 xy + 16 y 2

30)

( 5a + 7b )

= ( 5a ) + 2 ⋅ 5a ⋅ 7b + ( 7b ) =

2

2

2

= 52 a 2 + 70ab + 7 2 b 2 = = 25a 2 + 70ab + 49b 2

31)

( 3m + n )

= ( 3m ) + 2 ⋅ 3m ⋅ n + n 2 =

2

2

= 32 m 2 + 6mn + n 2 = = 9m 2 + 6mn + n 2

32)

( 2m − 5n )

2

= ( 2m ) − 2 ⋅ 2m ⋅ 5n + ( 5n ) = 2

2

= 22 m 2 − 20mn + 52 n 2 = = 4m 2 − 20mn + 25n 2

33)

( 0, 2 x + 3)

2

= ( 0, 2 x ) + 2 ⋅ 0, 2 ⋅ x ⋅ 3 + 32 = 0, 22 x 2 + 1, 2 x + 9 = 0, 04 x 2 + 1, 2 x + 9 2

↓

↓

2 ⋅ 0, 2 ⋅ 3 = 1, 2

0, 22 = 0, 2 ⋅ 0, 2 = 0, 04

Još jednom taj isti zadatak:

33)

( 0, 2 x + 3)

2

= ( 0, 2 x ) + 2 ⋅ 0, 2 ⋅ x ⋅ 3 + 32 = 2

= 0, 22 x 2 + 1, 2 x + 9 =

Uputa:

→

2 ⋅ 0, 2 ⋅ 3 = 1, 2

→

0, 22 = 0, 2 ⋅ 0, 2 = 0, 04

= 0, 04 x 2 + 1, 2 x + 9

34)

(1, 2 x + 0,3)

2

= (1, 2 x ) + 2 ⋅1, 2 x ⋅ 0,3 + 0,32 =

→

2 ⋅1, 2 ⋅ 0,3 = 0, 7 2

= 1, 22 x 2 + 2 ⋅1, 2 ⋅ 0,3 ⋅ x + 0, 09 =

→

1, 22 = 1, 2 ⋅1, 2 = 1, 44

2

= 1, 44 x 2 + 0, 72 x + 0, 09

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2

( I + II )

To pravilo možemo pisati i ovako:

35)

( 0,5 x − 2 y )

2

( A − B)

= A2 + 2 ⋅ A ⋅ B + B 2 i 2

2

= A2 − 2 ⋅ A ⋅ B + B 2

= I 2 + 2 ⋅ I ⋅ II + II 2 i ( I − II ) = I 2 − 2 ⋅ I ⋅ II + II 2 2

= ( 0,5 x ) − 2 ⋅ 0,5 ⋅ x ⋅ 2 ⋅ y + ( 2 y ) = 2

2

= 0,52 x 2 − 2 x + 22 y 2 = = 0, 25 x 2 − 2 x + 4 y 2

36)

( 0, 2m + 0,3n )

2

= ( 0, 2m ) + 2 ⋅ 0, 2 ⋅ m ⋅ 0,3 ⋅ n + ( 0,3n ) = 2

2

= 0, 22 m 2 + 0,12mn + 0,32 n 2 = = 0, 04m 2 + 0,12mn + 0, 09n 2 2

2

1⎞ 1 ⎛1⎞ ⎛ 37) ⎜ x + ⎟ = x 2 + 2 ⋅ x ⋅ + ⎜ ⎟ = 2⎠ 2 ⎝2⎠ ⎝ 1 12 = x2 + 2 ⋅ ⋅ x + 2 = 2 2 1 = x2 + x + 4

2

2

3 ⎛3 ⎞ ⎛3 ⎞ 38) ⎜ x + 1⎟ = ⎜ x ⎟ + 2 ⋅ ⋅ x ⋅1 + 12 = 2 ⎝2 ⎠ ⎝2 ⎠ 32 = 2 x 2 + 3x + 1 = 2 9 = x 2 + 3x + 1 4

2 ⎛3 39) ⎜ x + 3 ⎝4

2

2

2

3 2 ⎞ ⎛3 ⎞ ⎛2 ⎞ y ⎟ = ⎜ x ⎟ + 2⋅ ⋅ x⋅ ⋅ y + ⎜ y ⎟ = 4 3 ⎠ ⎝4 ⎠ ⎝3 ⎠ 2 ⋅ 3⋅ 2 32 3 2 22 3 2 = 2 x 2 + 2 ⋅ ⋅ ⋅ x ⋅ y + 2 y 2 = → drugi član ⋅ 2 ⋅ ⋅ xy = xy = 1xy 4 3 4 3 4⋅3 4 3 9 2 4 = x + xy + y 2 16 9

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2

= A2 + 2 ⋅ A ⋅ B + B 2 i

2

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

2

5 ⎞ 5 2 ⎛ ⎛5 ⎞ 40) ⎜ 0, 4 x − y ⎟ = ( 0, 4 x ) − 2 ⋅ 0, 4 ⋅ x ⋅ ⋅ y + ⎜ y ⎟ = 3 ⎠ 3 ⎝ ⎝3 ⎠ 2 4 5 5 = 0, 42 x 2 − 2 ⋅ ⋅ ⋅ x ⋅ y + 2 y 2 = 10 3 3 4 25 2 = 0,16 x 2 − xy + Možemo rješenje ostaviti u ovom obliku y = 3 9 16 2 4 25 2 16 4 = y = ili decimalni broj 0,16 = x − xy + = 100 3 9 100 25 4 2 4 25 2 pretvoriti u razlomak.... x − xy + y = 25 3 9

2

2

5 2 ⎛5 ⎞ ⎛5 ⎞ 41) ⎜ x + 0, 2 y ⎟ = ⎜ x ⎟ + 2 ⋅ ⋅ x ⋅ 0, 2 ⋅ y + = ( 0, 2 y ) = 2 ⎝2 ⎠ ⎝2 ⎠ 52 = 2 x 2 + 1⋅ xy + 0, 22 y 2 = 2 25 2 x + xy + 0, 04 y 2 = 4 2

5 → 2 ⋅ ⋅ 0, 2 = 5 ⋅ 0, 2 = 1 2

2

2

5 2 ⎛5 ⎞ ⎛5 ⎞ ⎛ 2 ⎞ ili taj z adatak na drugi način: ⎜ x + 0, 2 y ⎟ = ⎜ x ⎟ + 2 ⋅ ⋅ x ⋅ ⋅ y + = ⎜ y⎟ = 2 10 ⎝2 ⎠ ⎝2 ⎠ ⎝ 10 ⎠ 2

52 2 5 1 ⎛1 ⎞ x + 2⋅ ⋅ ⋅ x ⋅ y + ⎜ y ⎟ = 2 2 5 2 ⎝5 ⎠ 25 2 1 2 = x + xy + y 4 25 4 1 Vidimo da su obadva rješenja ista jer je: 0, 04 = pa sada kako je kome lakše = 100 25 računat obadva načina su ispravna... =

2

2

3 ⎞ 3 2 ⎛ ⎛3 ⎞ 42) ⎜ 4 x − y ⎟ = ( 4 x ) − 2 ⋅ 4 x ⋅ y + ⎜ y ⎟ = 2 ⎠ 2 ⎝ ⎝2 ⎠ 3 32 = 42 x 2 − 2 ⋅ 4 ⋅ ⋅ x ⋅ y + 2 y 2 = 2 2 2 3 = 16 x 2 − 12 xy + 2 y 2 = 2 9 = 16 x 2 − 12 xy + y 2 4

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 2

2

i

2

= A2 − 2 ⋅ A ⋅ B + B 2

2

2 ⎞ 1 2 ⎛1 ⎛1 ⎞ ⎛2 ⎞ 43) ⎜ x + y 3 ⎟ = ⎜ x ⎟ + 2 ⋅ ⋅ x ⋅ ⋅ y 3 + ⎜ y 3 ⎟ = 3 ⎠ 2 3 ⎝2 ⎝2 ⎠ ⎝3 ⎠ 2 2 1 2 2 = 2 x 2 + xy 3 + 2 y 3 2 = 2 3 3 1 2 4 = x 2 + xy 3 + y 6 4 3 9

( )

2

( A − B)

2

→

(y )

3 2

= y 3⋅ 2 = y 6

2

3 3 ⎞ 2 ⎛3 ⎞ ⎛2 ⎛2 ⎞ 44) ⎜ x3 − y 4 ⎟ = ⎜ x 3 ⎟ − 2 ⋅ x3 ⋅ y 4 + ⎜ y 4 ⎟ = 4 4 ⎠ 3 ⎝4 ⎠ ⎝3 ⎝3 ⎠ 2 2 3 2 2 3 3 4 32 4 2 = 2 x − 2⋅ ⋅ ⋅ x ⋅ y + 2 y = 3 3 4 4 4 9 = x 3⋅ 2 − 1⋅ x3 y 4 + y 4⋅ 2 = 9 16 4 6 3 4 9 8 = x −x y + y 9 16

( )

2 3 2 ⋅ 2 ⋅3 → 2⋅ ⋅ = =1 3 4 3⋅ 4

( )

2

1 ⎛ 1 ⎞ 45) ⎜1 x + y ⎟ = Odmah treba mješoviti broj: 1 2 ⎝ 2 ⎠ 2

2

pretvoriti u razlomak pa tek tada kvadri rati:

2

3 ⎛ 1 ⎞ ⎛3 ⎞ ⎛3 ⎞ 2 ⎜1 x + y ⎟ = ⎜ x + y ⎟ = ⎜ x ⎟ + 2 ⋅ x ⋅ y + y = 2 ⎝ 2 ⎠ ⎝2 ⎠ ⎝2 ⎠ 2 3 = 2 x 2 + xy + y 2 2

2

1 1.2 + 1 3 →1 = = 2 2 2

2

1 ⎞ 1⋅ 4 + 1 ⎞ ⎛ 2 ⎛ 2⋅3 + 2 46) ⎜ 2 x − 1 y ⎟ = ⎜ x− y ⎟ = Odmah treba mješoviti broj pretvoriti u razlomak 4 ⎠ 4 ⎝ 3 ⎝ 3 ⎠ 2

5 ⎞ ⎛8 = ⎜ x− y⎟ = 4 ⎠ ⎝3

→ pa tek tad a kvadrirati

2

2

8 5 ⎛8 ⎞ ⎛5 ⎞ = ⎜ x ⎟ − 2⋅ x ⋅ y + ⎜ y ⎟ = 3 4 ⎝3 ⎠ ⎝4 ⎠ 2 2 8 8⋅5 5 = 2 x2 − ⋅ x⋅ y + 2 y2 = 3 3⋅ 2 4 =

M.I.M.-Sraga -2003./2013. www.mim-sraga.com

64 2 40 25 2 x − xy + y 9 6 16

10

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2

2

i

2

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

2

5 ⎞ 1 ⎞ ⎛ 2 ⎞ ⎛ 3⋅ 3 + 2 ⎛ 11 47) ⎜ 3 x − 0,5 y ⎟ = ⎜ x − y⎟ = ⎜ x− y⎟ = 10 ⎠ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎝ 3 2

2

11 1 ⎛ 11 ⎞ ⎛1 ⎞ = ⎜ x ⎟ − 2⋅ ⋅ x ⋅ ⋅ y + ⎜ y ⎟ = 3 2 ⎝ 3 ⎠ ⎝2 ⎠ 2 2 11 11 1 = 2 x 2 − xy + 2 y 2 = 3 3 2 121 2 11 1 x − xy + y 2 = 9 3 4

2

2

2

3 ⎞ 1 3 ⎛1 ⎛1 ⎞ ⎛3 ⎞ 48) ⎜ xz − y ⎟ = ⎜ xz ⎟ − 2 ⋅ ⋅ x ⋅ z ⋅ ⋅ y + ⎜ y ⎟ = 2 ⎠ 4 2 ⎝4 ⎝4 ⎠ ⎝2 ⎠ 2 1 1 3 32 = 2 x2 y 2 − 2⋅ ⋅ ⋅ x ⋅ z ⋅ y + 2 y 2 = 4 4 2 2 1 2 2 3 9 2 = x y − xzy + y 16 4 4

49)

(x

2

) = ( x ) + 2⋅ x

+1

2

2

2

2

⋅1 + 12 =

( )

→ x2

2

( )

= x 2⋅ 2 = x 4 po pravilu: a n

m

= a n⋅m

= x 2⋅2 + 2 x 2 + 1 = = x4 + 2 x2 + 1

50)

(x

2

−2

) = ( x ) − 2⋅ x 2

2

2

2

⋅ 2 + 22 =

= x 2⋅2 − 4 x 2 + 4 = = x4 − 4 x2 + 4

51)

(3 − x ) 4

2

( )

= 32 − 2 ⋅ 3 ⋅ x 4 + x 4

2

=

= 9 − 6 x 4 + x 4⋅ 2 = = 9 − 6 x 4 + x8

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 52)

( 3a

2

− b3

) = ( 3a ) − 2 ⋅3⋅ a ⋅ b + ( b ) = 3 ( a ) − 6a b + b = 2

2

2

2

2

2

2

3

( A − B)

i

2

= A2 − 2 ⋅ A ⋅ B + B 2

=

3 2

3⋅ 2

2 3

= 9a 2⋅ 2 − 6a 2b3 + b 6 = 9a 4 − 6a 2b3 + b6

53)

( 2a

3

− b2

) = ( 2a ) − 2 ⋅ 2 ⋅ a ⋅ b + ( b ) = 2 ( a ) − 4a b + b = 2

3 2

3

3 2

2

2

2

=

2

=

2

2⋅ 2

3 2

= 4a 3⋅ 2 − 4a 3b 2 + b 4 = = 4a 6 − 4a 3b 2 + b 4

54)

(a

3

− 5b 2

) = ( a ) − 2 ⋅ a ⋅ 5⋅ b + ( 5b ) = a − 10a b + 5 ( b ) = 2

3 2

3⋅ 2

3

2

3 2

2

2

2 2

= a 6 − 10a 3b 2 + 25b 4

55)

( 2x

3

+ 3 y3

) = ( 2 x ) + 2 ⋅ 2 ⋅ x ⋅ 3⋅ y + ( 3 y ) = 2 ( x ) + 12 x y + 3 ( y ) = 2

3 2

2

3

3 2

3

3

3

=

3 2

2

3 2

= 4 x3⋅ 2 + 12 x 3 y 3 + 9 y 3⋅ 2 = = 4 x 6 + 12 x3 y 3 + 9 y 6

56)

( 3x

4

+ 7 y5

) = ( 3x ) + 2 ⋅ 3⋅ x ⋅ 7 ⋅ y + ( 7 y ) = = 3 ( x ) + 42 x y + 7 ( y ) = 2

4

2

2

4

4

2

4

5

5

5 2

2

5 2

= 9 x 4⋅ 2 + 42 x 4 y 5 + 49 y 5⋅ 2 = = 9 x8 + 42 x 4 y 5 + 49 y10

57)

(2x y 2

3

− 3z 2

) = ( 2 x y ) − 2 ⋅ 2 ⋅ x ⋅ y ⋅ 3⋅ z + ( 3z ) = = 2 ( x ) ( y ) − 4 ⋅ 3⋅ x ⋅ y ⋅ z + 3 ( z ) 2

2

2

3 2

2 2

2

3 2

3

2

2

3

2

2

2

2

2 2

=

= 4 x 2⋅ 2 y 3⋅ 2 − 12 x 2 y 3 z 2 + 9 z 2⋅ 2 = = 4 x 4 y 6 − 12 x 2 y 3 z 2 + 9 z 4

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 58)

( 2ab + 3c )

2

i

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

= ( 2ab ) + 2 ⋅ 2 ⋅ a ⋅ b ⋅ 3⋅ c + ( 3c ) = 2

2

= 22 a 2b 2 + 12abc + 32 c 2 = = 4a 2b 2 + 12abc + 9c 2 59)

( a b + c ) = ( a b ) + 2 ⋅ a ⋅b ⋅ c + ( c ) = ( a ) ⋅ b + 2a bc + c = 2

3 2

2

2

2

2

2

3

2

2

=

3 2

3

2

= a 4b 2 + 2a 2bc 3 + c 2 60)

( 2ab

2

− 3c 4 ) 2 = ( 2ab 2 ) 2 − 2 ⋅ 2 ⋅ a ⋅ b 2 ⋅ 3⋅ c 4 + ( 3c 4 ) 2 = = 22 a 2 ( b 2 ) 2 − 12ab 2c 4 + 32 ( c 4 ) 2 = = 4a 2b 2⋅ 2 − 12ab 2c 4 + 9c 4⋅ 2 = = 4a 2b 4 − 12ab 2c 4 + 9c8

61)

(a b

2 3

− c 2 d 4 ) 2 = ( a 2b3 ) 2 − 2 ⋅ a 2 ⋅ b3 ⋅ c 2 ⋅ d 4 + ( c 2 d 4 ) 2 = = ( a 2 ) 2 ⋅ ( b3 ) 2 − 2a 2b3c 2 d 4 + ( c 2 ) 2 ⋅ ( d 4 ) 2 = = a 2⋅ 2b3⋅ 2 − 2a 2b3c 2 d 4 + c 2⋅ 2 d 4⋅ 2 = = a 4b6 − 2a 2b3c 2 d 4 + c 4 d 8

62)

(a b

2 3

− c 2 ) 2 = ( a 2b 3 ) 2 − 2 ⋅ a 2 ⋅ b 3 ⋅ c 3 + ( c 2 ) 2 = = ( a 2 ) 2 ⋅ ( b 3 ) 2 − 2 a 2b 3c 3 + c 2 ⋅ 2 = = a 2⋅ 2b3⋅ 2 − 2a 2b3c 3 + c 4 = = a 4b6 − 2a 2b3c 3 + c 4

63)

( a b − 3c ) = ( a b ) − 2 ⋅ a b ⋅3⋅ c + ( 3c ) = ( a ) ⋅ b − 6a bc + 3 ( c ) 3

4

2

3

3 2

2

3

2

4

3

4

2

4

2

=

4

2

=

= a 3⋅ 2b 2 − 6a 3bc 4 + 9c 4⋅ 2 = = a 6b 2 − 6a 3bc 4 + 9c8 64)

(a b

2 3

+ 4c5 ) 2 = ( a 2b3 ) 2 + 2 ⋅ a 2 ⋅ b3 ⋅ 4 ⋅ c 5 + ( 4c5 ) 2 = = ( a 2 ) 2 ⋅ ( b3 ) 2 + 8a 2b3c 5 + 42 ( c5 ) 2 = = a 2⋅ 2b3⋅ 2 + 8a 2b3c5 + 16c 5⋅ 2 = = a 4b6 + 8a 2b3c 5 + 16c10

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M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 65)

( 2a b

4 3

( A − B)

i

2

= A2 − 2 ⋅ A ⋅ B + B 2

− 3a 2b7 ) 2 = ( 2a 4b3 ) 2 − 2 ⋅ 2 ⋅ a 4 ⋅ b3 ⋅ 3⋅ a 2 ⋅ b7 + ( 3a 2b7 ) 2 = = 22 ( a 4 ) 2 ⋅ ( b3 ) 2 − 12 ⋅ a 4 ⋅ a 2 ⋅ b3 ⋅ b7 + 32 ( a 2 ) 2 ⋅ ( b7 ) 2 = = 4a 4⋅ 2b3⋅ 2 − 12 ⋅ a 4+ 2 ⋅ b3+7 + 9a 2⋅ 2b7 ⋅ 2 = = 4a8b6 − 12a 6b10 + 9a 4b9

66)

( 4a b − 5a b ) = ( 4a b ) − 2 ⋅ 4 ⋅ a ⋅b ⋅5⋅ a ⋅ b + ( 5a b ) = = 4 ( a ) ⋅b − 2 ⋅ 4 ⋅5⋅ a ⋅ a ⋅b ⋅b + 5 ( a ) ⋅(b ) 3

2 3 2

3

2

2

3

3 2

2

2

3

3

2

2 3 2

1

3

2

2

2

3 2

=

= 16a 3⋅ 2b 2 − 40a 3+2 ⋅ b1+3 + 25a 2⋅ 2b3⋅ 2 = = 16a 6b 2 − 40a 5b 4 + 25a 4b6 67)

( 9a b

4 3

− 2 a 2 b 4 ) = ( 9 a 4b 3 ) − 2 ⋅ 9 ⋅ a 4 ⋅ b 3 ⋅ 2 ⋅ a 2 ⋅ b 4 + ( 2 a 2b 4 ) = 2

2

2

= 92 ( a 4 ) ( b3 ) − 2 ⋅ 9 ⋅ 2 ⋅ a 4 ⋅ a 2 ⋅ b3 ⋅ b 4 + 22 ( a 2 ) ( b 4 ) = 2

2

2

2

= 81a 4⋅ 2b3⋅ 2 − 36 ⋅ a 4+ 2 ⋅ b3+ 4 + 4a 2⋅ 2b 4⋅ 2 = = 81a8b6 − 36a 6b7 + 4a 4b8 2

2

2

2 1 2 ⎛1 ⎞ ⎛1 ⎞ ⎛2 ⎞ 68) ⎜ a 2b − a 4b3 ⎟ = ⎜ a 2b ⎟ − 2 ⋅ ⋅ a 2 ⋅ b ⋅ ⋅ a 4 ⋅ b3 + ⎜ a 4b3 ⎟ = 2 3 2 2 3 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2 2 1 1 2 2 = 2 ⋅ ( a 2 ) 2 ⋅ b 2 − 2 ⋅ ⋅ ⋅ a 2 ⋅ a 4 ⋅ b1 ⋅ b3 + 2 ( a 4 ) 2 ⋅ ( b3 ) 2 = 2 2 3 3 4 1 2 = a 2⋅ 2b 2 − ⋅ a 2+ 4 ⋅ b1+3 + a 4⋅ 2b3⋅ 2 = 9 4 3 1 2 4 = a 4b 2 − a 6b 4 + a 6b 5 4 3 9 69)

(5x − 6 y )

−2

= (5x − 6 y ) =

70)

( x + 4y)

−2

M.I.M.-Sraga -2003./2013. www.mim-sraga.com

=

((5x − 6 y ) )

2 −1

1

(5x )

2

− 2⋅5⋅ x ⋅ 6 ⋅ y + ( 6 y )

= ( x + 4y) =

2 ⋅ ( − 1)

2 ⋅ ( − 1)

=

(( x + 4 y )

1 x + 2⋅ x ⋅4⋅ y + ( 4 y ) 2

2

=

2

)

=

2 −1

=

1

(5x − 6 y )

2

=

1 1 = 2 2 2 5 x − 30 xy + 6 y 25 x − 30 xy + 36 y 2 2

=

2

1

( x + 4y)

2

=

1 1 = 2 2 2 x + 8 xy + 4 y x + 8 xy + 16 y 2 2

14

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34.

71)

( −a − b )

U formulama piše:

Pokažimo zašto je to tako:

( −a − b )

Objašnjenje:

2

2

= (a + b)

( −a − b )

2

2

= ( − 1⋅ ( a + b ) ) = ( − 1) ⋅ ( a + b ) = 1⋅ ( a + b ) = ( a + b ) 2

2

2

2

= ( − 1⋅ ( a + b ) ) =

Izlučili smo zajednički faktor

= ( − 1) ⋅ ( a + b ) =

pa smo primjenili pravilo:

= 1⋅ ( a + b ) =

( − 1)

2

2

2

2

= (a + b)

2

2

( − 1) ,

( a ⋅b )

n

= a n ⋅bn

=1

2

Sada taj postupak primjenimo u: 71. , 72. , 73. , 74. ,

( − x − 3)

72)

(−x − 2 y)

2

(

)

= ( − 1⋅ ( x + 3) ) = ( − 1) ⋅ ( x + 3) = 1⋅ x 2 + 2 ⋅ x ⋅ 3 + 32 = x 2 + 6 x + 9

71)

2

2

2

2

= ( − 1⋅ ( x + 2 y ) ) =

Izlučimo zajednički faktor

= ( − 1) ⋅ ( x + 2 y ) =

pa primje nimo pravilo:

2

2

2

(

= 1⋅ x 2 + 2 ⋅ x ⋅ 2 ⋅ y + ( 2 y )

(

2

)=

( − 1) ,

( a ⋅b )

n

= a n ⋅bn

)

= 1⋅ x 2 + 4 xy + 2 2 y 2 = = x 2 + 4 xy + 4 y 2

73)

( − 2a − 3b )

= ( − 1⋅ ( 2a + 3b ) ) = 2

2

74)

(

= ( − 1⋅ ( x + 4 y ) ) = 2

2

= ( − 1) ⋅ ( x + 4 y ) =

= ( − 1) ⋅ ( 2a + 3b ) = 2

(−x − 4 y)

2

2

= 1⋅ ( 2a ) + 2 ⋅ 2 ⋅ a ⋅ 3 ⋅ b + ( 3b ) 2

(

2

2

(

)=

(

= 1⋅ x 2 + 2 ⋅ x ⋅ 4 ⋅ y + 4 y 2

)

( )

= 1⋅ 2 2 ⋅ a 2 + 12ab + 32 b 2 =

= x 2 + 8 xy + 4 2 y 2

= 4a + 12ab + 9b

= x 2 + 8 xy + 16 y 4

2

2

76)

=

( − a + 7b )

2

= ( 7b − a ) = 2

= ( 7b ) − 2 ⋅ 7 ⋅ b ⋅ a + a 2 = 2

2

⎛1 ⎞ = ( − 1) ⋅ ⎜ x + 4 y 2 ⎟ = ⎝2 ⎠

= 7 2 b 2 − 14ba + a 2

⎛ ⎛ 1 ⎞2 1 = 1⋅ ⎜ ⎜ x ⎟ + 2 ⋅ ⋅ x ⋅ 4 ⋅ y 2 + 4 y 2 ⎜⎝ 2 ⎠ 2 ⎝ 2 12 = 2 x 2 + 4 xy 2 + 42 y 2 = 2 1 2 = x + 4 xy 2 + 16 y 4 4

(

)

2

⎞ ⎟⎟ = ⎠

( )

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2

2

2

⎛ ⎛ 1 ⎞ ⎛1 ⎞⎞ 75) ⎜ − x − 4 y 2 ⎟ = ⎜ − 1⋅ ⎜ x + 4 y 2 ⎟ ⎟ = ⎝ 2 ⎠ ⎝2 ⎠⎠ ⎝ 2

2

) )=

15

= 49b 2 − 14ba + a 2 ili = a 2 − 14ab + 49b 2

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

prvom i drugom članu zamjenimo mjesta

34. 77) ( − 3x + y )2 = ( y − 3x )2 =

tako da drugi bude sa predznakom ( − ) ...i dalje je lako...

= y 2 − 2 ⋅ y ⋅ 3⋅ x + ( 3x 2 ) = = y 2 − 6 yx + 32 x 2 = = y 2 − 6 yx + 9 x 2

= 9 x 2 − 6 xy + y 2

ili

obad va rj. su ispravna i ista...

Taj zadatak možemo rješiti i na još dva načina: II način:

77)

( − 3x + y )

2

III način:

= ( − 1⋅ ( 3x − y ) ) = 2

( − 3x + y )

= ( − 1) ⋅ ( 3x − y ) = 2

2

(

2

= ( − 3x ) + 2 ⋅ ( − 3x ) ⋅ y + y 2 = 2

= ( − 3) x 2 − 2 ⋅ 3 ⋅ x ⋅ y + y 2 = 2

)

= 1⋅ ( 3x ) − 2 ⋅ 3x ⋅ y + y 2 = 2

= 9 x 2 − 6 xy + y 2

= 32 x 2 − 6 xy + y 2 = Uvijek dobijemo isto rješenje...pa birajte...

= 9 x 2 − 6 xy + y 2 78)

( −5a b

2 3

na koji način će te rješavat ovakve zadatke

+ 2c 4 ) 2 = ( 2c 4 − 5a 2b3 ) 2 = = ( 2c 4 ) 2 − 2 ⋅ 2 ⋅ c 4 ⋅ 5 ⋅ a 2 ⋅ b3 + ( 5a 2b3 ) 2 = = 22 ( c 4 ) 2 − 20a 2b3c 4 + 52 ( a 2 ) 2 ⋅ ( b3 ) 2 = = 4c 4⋅ 2 − 20a 2b3c 4 + 25a 2⋅ 2b3⋅ 2 = = 4c8 − 20a 2b3c 4 + 25a 4b6

79)

( −2 x y 2

3

ili

= 25a 4b6 − 20a 2b3c 4 + 4c8

+ 3z 4 ) 2 = ( 3z 4 − 2 x 2 y 3 ) 2 = = ( 3z 4 ) 2− 2 ⋅ 3⋅ z 4 ⋅ 2 ⋅ x 2 ⋅ y 3 + ( 2 x 2 y 3 ) 2 = = 32 ( z 4 ) 2 − 12 z 4 x 2 y 3 + 22 ( x 2 ) 2 ⋅ ( y 3 ) 2 = = 9 z 4⋅ 2 − 12 x 2 y 3 z 4 + 4 x 2⋅ 2 y 3⋅ 2 = = 9 z 8 − 12 x 2 y 3 z 4 + 4 x 4 y 6

80)

( −2

m

ili

= 4 x 4 y 6 − 12 x 2 y 3 z 4 + 9 z 8

+ 2n ) 2 = ( 2n − 2m ) 2 = = ( 2n ) 2 − 2 ⋅ 2n ⋅ 2m + ( 2m ) 2 = = 2n ⋅ 2 − 21 ⋅ 2n ⋅ 2m + 2m ⋅ 2 = = 22 n − 21+ n+ m + 22 m =

Rješenje možemo ostaviti u ovom obliku...

= 22⋅ n − 21+n+m + 22⋅ m = = ( 22 ) − 2n+m+1 + ( 22 ) = n

= 4n − 2n+ m+1 + 4m

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m

ili u ovom obliku, ovisi o tome kako ra dite u školi...

16

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Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A⋅ B + B 2 81)

(2

m

) = ( 2 ) + 2⋅ 2

+ 3n

2

m 2

( )

⋅ 3n + 3n

m

82)

=

2

( A − B)

i

(3

m

− 5m

) = (3 ) − 2⋅3 2

+ 2n

m

( )

⋅ 5m + 5m

2

= 22⋅ m + 21+m ⋅ 3n + 32⋅ n =

= 32⋅ m − 2 ⋅15m + 52⋅ m =

m

( )

+ 2m+1 ⋅ 3n + 32

n

) = ( 2 ) + 2⋅ 2 2

m 2

m

( )

= 32

=

=2

( )

= 22

1+ m+ n

+2 m

( )

⋅ 2 n + 2n +2

( )

m

− 2 ⋅15m + 52

m

=

= 9m − 2 ⋅15m + 25m 2

84)

=

2⋅ n

=

( )

+ 2m+n+1 + 22

(2

m

+ 2m−1

) = ( 2 ) + 2⋅ 2 2

=2 n

m 2

m

(

⋅ 2m−1 + 2m−1

= 2m ⋅ 2 + 21 ⋅ 2m ⋅ 2m−1 + 2(

= 2m⋅ 2 + 21 ⋅ 2m ⋅ 2n + 2n ⋅ 2 = 2⋅ m

=

m

= 4n + 2m+1 ⋅ 3n + 9n m

m 2

= 3m⋅ 2 − 2 ⋅ ( 3⋅ 5) + 5m⋅ 2 =

( )

(2

= A2 − 2 ⋅ A⋅ B + B 2

= 2m⋅ 2 + 21 ⋅ 2m ⋅ 3n + 3n ⋅ 2 = = 22

83)

2

2⋅ m

( )

= 22

=

1+ m+ m−1

+2 m

= 4m + 22

2 ⋅ ( m−1)

( )

+ 22 m + 2 2

( )

= 4m + 2m+n+1 + 4n

+2

m

m−1) ⋅ 2

m−1

)

2

=

=

=

=

+ 4m−1 =

= 4m + 4m + 4m−1 = = 2 ⋅ 4m + 4m−1 85)

(3

m

− 3n

) = (3 ) − 2⋅3 2

m 2

m

( )

⋅ 3n + 3n

=

2

= 3m⋅ 2 − 2 ⋅ 3m+n + 3n ⋅ 2 = = 32⋅ m − 2 ⋅ 3m+n + 32⋅ n =

( )

= 32

( )

m

− 2 ⋅ 3m+n + 32

n

=

= 9m − 2 ⋅ 3m+n + 9n 86)

(2

m+1

) = ( 2 ) + 2⋅ 2

+ 2m−1

m+1 2

2

= 2( =2

m+1) ⋅ 2

2 ⋅ ( m+1)

( )

= 22

m+1

(

⋅ 2m−1 + 2m−1

+ 21 ⋅ 2m+1 ⋅ 2m−1 + 2( + 21+m+1+m−1 + 2

m+1

( )

+ 22 m+1 + 22

m−1) ⋅ 2

2 ⋅ ( m−1)

m−1

)

2

=

=

=

=

= 4m+1 + 22 m+1 + 4m−1 87)

(2

n

+ 21−n

) = ( 2 ) − 2⋅ 2 ⋅ 2 2

n 2

n

1−n

( )

+ 21−n

= 2n ⋅ 2 − 21 ⋅ 2n ⋅ 21−n + 2(

1−n ) ⋅ 2

= 22⋅ n − 21+n+1−n + 2

( )

= 22

n

2 ⋅(1−n )

( )

− 22 + 2 2

1−n

2

=

=

=

=

= 4n − 4 + 41−n

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17

© M.I.M-Sraga centar za poduku mim-sraga@zg.htnet.hr tel-01-4578-431

Rješenja i kompletni postupak

M-1-Kvadrat binoma

34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 88)

(2

n +1

− 2n −2

) = ( 2 ) − 2⋅2 n+1 2

2

= 2( =2

n +1) ⋅ 2

2 ⋅ ( n +1)

( )

= 22

(

n+1

− 21 ⋅ 2n+1 ⋅ 2n−2 + 2( − 21+n+1+ n−2 + 2

( )

n +1

− 22 n + 22

( )

= 4n+1 − 22

n

)

⋅ 2 n− 2 + 2 n− 2

n−2 ) ⋅ 2

2 ⋅( n−2 )

n−2

2

i

( A − B)

2

= A2 − 2 ⋅ A ⋅ B + B 2

= =

=

=

+ 4 n−2 =

= 4n+1 − 4n + 4n−2 89)

(a

n

+ a n+1

) = ( a ) + 2⋅ a 2

2

n

n

(

⋅ a n+1 + a n+1

= a n ⋅ 2 + 2 ⋅ a n+n+1 + a (

n +1) ⋅ 2

)

=

2

=

= a 2 n + 2a 2 n+1 + a 2 n+1 90)

(a

n −1

− a n+1

) = ( a ) − 2⋅a n−1 2

2

= a( =a 91)

( 2a

x

− 3b x

n −1) ⋅ 2

2 n −2

n −1

− 2 ⋅ a n−1+ n+1 + a (

− 2a + a 2n

(

)

n +1) ⋅ 2

=

⋅ a n+1 + a n+1

=

2

2 n+2

) = ( 2a ) − 2 ⋅ 2 ⋅ a ⋅3⋅b + ( 3b ) = = 2 ( a ) − 12 ⋅ a ⋅ b + 3 ( b ) = 2

x

2

2

x

2

x

x

x

2

x

2

x

2

x

= 4a x ⋅ 2 − 12a xb x + 9b x ⋅ 2 = = 4a 2 x − 12a xb x 9b 2 x 92)

( 2a

x

+ 3b y

) = ( 2a ) + 2 ⋅ 2a = 2 ( a ) + 12a b 2

2

x

2

x

2

x

x y

( ) + 3 (b ) =

⋅ 3b y + 3b y 2

y

2

=

2

= 4a x ⋅ 2 + 12a xb y + 9b y ⋅ 2 = = 4a 2 x + 12a xb y + 9b 2 y

M.I.M.-Sraga -2003./2013. www.mim-sraga.com

18

© M.I.M-Sraga centar za poduku mim-sraga@zg.htnet.hr tel-01-4578-431