Kvadrat Binoma - Kvadrat Zbira i Razlike
February 9, 2017 | Author: Rastko Gajić | Category: N/A
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Rješenja i kompletni postupak
M-1-Kvadrat binoma
34. Koristeći se formulama za: kvadrat zbroja binoma i kvadrat razlike binoma ( A + B)
2
= A2 + 2 AB + B 2
( A − B)
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 )
2 3 39) x + y 3 4
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 + 7 b )
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 + 2
y
2
3 42) 4 x − 2
2
61)
( x + 1) ( 2a − b ) ( 2 x y − 3z ) (a b − c d )
65)
(
69)
(5x − 6 y )
73)
( − 2a − 3b )
77)
( − 3x + y )
49) 53) 57)
81) 85) 89)
2
2
2
2 3
) )
m
+ 3n
m
− 3n
n
+ a n +1
20)
(5x − y )
2
24)
( 2 x + 5)
28)
( 3x + 4 y )
32)
( 2 m − 5n )
2
36)
( 0, 2m + 0,3n )
2
5 40) 0, 4 x − y 3
2
2
2 1 43) x + y 3 3 2 y
2
4
2
62)
(a b
63)
66)
(
67)
(
70)
( x + 4y)
71)
( − x − 3)
74)
(−x − 4 y)
2
2
)
78) 82)
2
86) 2
90)
2
−2
3
− 5b 2
2 3
2
51)
)
− c2
2
)
55)
2
4 a 3 b − 5a 2 b 3
)
2
−2
( − 5a b + 2c ) (3 − 5 ) (2 + 2 ) (a − a ) m
m
4
3
3
2
3
3
4
2
2
79) 83)
m +1
m −1 2
87)
n −1
n +1 2
91)
1
2
52) 2
56) 60)
2
64)
9a 4 b3 − 2a 2 b 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
4
2
2
2
2
5
2
2
2 3
4
2
5
2
2 1 68) a 2 b − a 4 b3 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) xz − 2 4
2
4
2
3 2 44) x3 − y 4 4 3
2
1 75) − x − 4 y 2 2
2
2 3
2
2 47) 3 x − 0,5 y 3
59)
2
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2
( 2ab + 3c ) 2
−2
)
(8 + y )
58)
2
)
16)
2
2
2a 4 b3 − 3a 2 b7
(2 (3 (a
2
(3 − b)
2
50)
2
2
y
2
12)
2
(3 − x ) ( 2x + 3 y ) (a b + c ) ( a b − 3c )
3
54)
(x (a
2
2
1 2 46) 2 x − 1 4 3
2
3
2
2
80) 84) 88) 92)
2
2
( −2 + 2 ) (2 + 2 ) (2 − 2 ) ( 2a + 3b ) m
m
n +1 x
n
2
m −1
2
n−2 y
2
2
2
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)
2
↑ ↑
= x 2 − 2 ⋅ x ⋅1 + 12 = x 2 − 2 x + 1 ↑ ↑ ↑ = A2 − 2 ⋅ A ⋅ B + B 2
( A − 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 + 4x + x2
9)
( x − 1)
→ to možemo p isati i dalje: = x 2 − 4 x + 4
2
= x2 + 4x + 4
= 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
M I M Sraga 2003/05
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dobro je kako god napišete ...
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 )
= ( 5 x ) − 2 ⋅ 5 x ⋅ y + y 2 = 52 x 2 − 10 xy + y 2 = 25 x 2 − 10 xy + y 2
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
= 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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2
Primjeni: ( ab ) = a nb n n
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, 2 2 = 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, 72
= 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
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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
7
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 Možemo rješenje ostaviti u ovom obliku = 0,16 x 2 − xy + y = 3 9 25 2 16 4 16 2 4 = x − xy + y = = ili decimalni broj 0,16 = 3 9 100 25 100 4 2 4 25 2 x − xy + y pretvoriti u razlomak.... = 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... =
3 42) 4 x − 2
2
2
3 2 3 y = ( 4x) − 2⋅ 4x ⋅ y + y = 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 2 2 3 44) x3 − y 4 = x3 − 2 ⋅ x3 ⋅ y 4 + y 4 = 4 4 3 3 3 4 2 2 2 3 3 4 32 4 2 3 2 = 2 x − 2⋅ ⋅ ⋅ x ⋅ y + 2 y = 3 3 4 4 4 9 = x 3⋅ 2 − 1⋅ x 3 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 x− y = Odmah treba mješoviti broj pretvoriti u razlomak 46) 2 x − 1 y = 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 2003/05
64 2 40 25 2 x − xy + y 9 6 16
9
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
= 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
M I M Sraga 2003/05
10
2
( )
= x 2⋅ 2 = x 4 po pravilu: a n
m
= a n⋅m
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
=
3 2
3⋅ 2
2 3
= 9 a 2 ⋅ 2 − 6 a 2b 3 + b 6 = 9 a 4 − 6 a 2b 3 + b 6
53)
( 2a
3
− b2
) = ( 2a ) − 2 ⋅ 2 ⋅ a ⋅ b + ( b ) = 2 ( a ) − 4a b + b = 2
3 2
2
3
2
3
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 x3 y 3 + 9 y 3⋅ 2 = = 4 x 6 + 12 x3 y 3 + 9 y 6
56)
( 3x
4
+ 7 y5
) = ( 3 x ) + 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
= 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
M I M Sraga 2003/05
11
2
2
2
2
=
2
= A2 − 2 ⋅ A ⋅ B + B 2
Rješenja i kompletni postupak
M-1-Kvadrat binoma
34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A⋅ B + B 2 58)
( 2ab + 3c )
i
( A − B)
= ( 2ab ) + 2 ⋅ 2 ⋅ a ⋅ b ⋅ 3⋅ c + ( 3c ) =
2
2
2
= 22 a 2b 2 + 12abc + 32 c 2 = = 4a 2b2 + 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
) = ( 2ab ) − 2 ⋅ 2 ⋅ a ⋅ b ⋅ 3⋅ c + ( 3c ) = = 2 a ( b ) − 12ab c + 3 ( c ) = 2
2 2
2
2
2
2 2
4
2 4
4 2
2
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 2d 4
) = ( a b ) − 2⋅ a ⋅b ⋅c ⋅ d + (c d ) = = ( a ) ⋅ ( b ) − 2a b c d + ( c ) ⋅ ( d ) 2
2 3 2 2 2
2
3
3 2
2
2 3 2
4
2
4 2
4
2 2
= a 2 ⋅ 2 b 3 ⋅ 2 − 2 a 2 b 3c 2 d 4 + c 2 ⋅ 2 d 4 ⋅ 2 = = a 4 b 6 − 2 a 2b 3 c 2 d 4 + c 4 d 8 62)
(a b
2 3
− c2
) = ( a b ) − 2⋅ a ⋅b ⋅ c + ( c ) = ( a ) ⋅ ( b ) − 2a b c + c 2
2 3 2
2
2 2
3
3 2
3
=
2 2 2⋅ 2
2 3 3
=
= a 2⋅ 2b3⋅ 2 − 2a 2b3c 3 + c 4 = = a 4b6 − 2a 2b3c3 + c 4 63)
( a b − 3c ) = ( a b ) − 2 ⋅ a b ⋅ 3⋅ c + ( 3c ) = ( a ) ⋅ b − 6a bc + 3 ( c ) 3
4 2
3
2
3 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
+ 4c 5
) = ( a b ) + 2 ⋅ a ⋅ b ⋅ 4 ⋅ c + ( 4c ) = ( a ) ⋅ ( b ) + 8a b c + 4 ( c ) 2
2 3 2 2 2
2
3 2
3
5
2 3 5
2
= a 2⋅ 2b3⋅ 2 + 8a 2b3c 5 + 16c 5⋅ 2 = = a 4b6 + 8a 2b3c 5 + 16c10
M I M Sraga 2003/05
12
5 2
=
5 2
=
4 2
=
2
= A2 − 2 ⋅ A ⋅ B + B 2
Rješenja i kompletni postupak
M-1-Kvadrat binoma
34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A ⋅ B + B 2 65)
( 2a b
4 3
− 3a 2b 7
( A − B)
i
2
= A2 − 2 ⋅ A ⋅ B + B 2
) = ( 2a b ) − 2 ⋅ 2 ⋅ a ⋅b ⋅ 3⋅ a ⋅b + ( 3a b ) = = 2 ( a ) ⋅ ( b ) − 12 ⋅ a ⋅ a ⋅ b ⋅ b + 3 ( a ) ⋅ ( b ) 2
4 3 2
2
4
4 2
3
3 2
2
4
7
2
3
2 7
7
2
2
2 2
7
=
2
= 4a 4⋅ 2b3⋅ 2 − 12 ⋅ a 4+2 ⋅ b3+7 + 9a 2⋅ 2b7 ⋅ 2 = = 4a8b 6 − 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 3 2
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 4b 6 67)
( 9a b
4 3
− 2 a 2b 4
) = ( 9a b ) − 2 ⋅ 9 ⋅ a ⋅ b ⋅ 2 ⋅ a ⋅ b + ( 2 a b ) = = 9 ( a ) (b ) − 2 ⋅9⋅ 2 ⋅ a ⋅ a ⋅b ⋅b + 2 ( a ) (b ) 2
2
4 3
2
4
4
2
3
3
2
2
4
2
4
2 4
3
4
2
2
2
2
4
2
=
= 81a 4⋅ 2b3⋅ 2 − 36 ⋅ a 4+2 ⋅ b3+4 + 4a 2⋅ 2b 4⋅ 2 = = 81a8b 6 − 36a 6b 7 + 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 3 3 1 2 4 = a 2⋅ 2b 2 − ⋅ a 2+4 ⋅ b1+3 + a 4⋅ 2b3⋅ 2 = 4 3 9 1 2 4 = a 4b 2 − a 6b 4 + a 6b5 4 3 9
( )
69)
(5x − 6 y )
−2
= (5x − 6 y ) =
70)
( x + 4y)
−2
M I M Sraga 2003/05
=
((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
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
13
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)
(
2
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)
2
=
(
( − a + 7b )
2
= ( 7b − a ) = 2
= ( 7b ) − 2 ⋅ 7 ⋅ b ⋅ a + a 2 = 2
2
1 = ( − 1) ⋅ x + 4 y 2 = 2 2 1 1 = 1⋅ x + 2 ⋅ ⋅ x ⋅ 4 ⋅ y 2 + 4 y 2 2 2 2 12 = 2 x 2 + 4 xy 2 + 4 2 y 2 = 2 1 2 = x + 4 xy 2 + 16 y 4 4 2
2
) )=
2
2
1 1 75) − x − 4 y 2 = − 1⋅ x + 4 y 2 = 2 2
= 7 2 b 2 − 14ba + a 2
)
2
=
( )
M I M Sraga 2003/05
= ( − 1⋅ ( x + 4 y ) ) =
2
= ( − 1) ⋅ ( x + 4 y ) =
= ( − 1) ⋅ ( 2a + 3b ) = 2
(−x − 4 y)
14
= 49b 2 − 14ba + a 2 ili = a 2 − 14ab + 49b 2
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 )
III način:
= ( − 1⋅ ( 3 x − y ) ) = 2
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 ⋅ 3 x ⋅ 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 − 5 a b ) = = ( 2c ) − 2 ⋅ 2 ⋅ c ⋅ 5 ⋅ a ⋅ b + ( 5a b ) = 2 ( c ) − 20a b c + 5 ( a ) ⋅ ( b )
+ 2c 4
2
4
2 3 2
4 2
2
4
4 2
2
2 3 4
3
2 3 2
2
2 2
3 2
= =
= 4c 4⋅ 2 − 20a 2b3c 4 + 25a 2⋅ 2b3⋅ 2 = = 4c8 − 20a 2b3c 4 + 25a 4b6 79)
( −2x y 2
3
= 25a 4b 6 − 20a 2b3c 4 + 4c8
ili
) = ( 3z − 2 x y ) = = ( 3z ) − 2 ⋅ 3⋅ z ⋅ 2 ⋅ x ⋅ y + ( 2 x y ) = 3 ( z ) − 12 z x y + 2 ( x ) ⋅ ( y )
+ 3z 4
2
4
2
4 2
2
3 2
4
4 2
2
4 2
3
3
2
2
2 2
3 2
=
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
+ 2n
) = (2 − 2 ) = = ( 2 ) − 2⋅ 2 ⋅ 2 + ( 2 ) 2
n
n
m
2
ili
= 4 x 4 y 6 − 12 x 2 y 3 z 4 + 9 z 8
2
n
m
m
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
n
= 4n − 2n+m+1 + 4m
M I M Sraga 2003/05
( )
− 2n+m+1 + 22
m
= ili u ovom obliku, ovisi o tome kako ra dite u školi...
15
Rješenja i kompletni postupak
M-1-Kvadrat binoma
34. Kvadriramo po pravilu: ( A + B )2 = A2 + 2 ⋅ A⋅ B + B2 81)
(2
m
+ 3n ) 2 = ( 2m ) 2+ 2 ⋅ 2m ⋅ 3n + ( 3n ) 2 =
(2
m
(3
m
2
= A2 − 2 ⋅ A⋅ B + B 2
− 5m ) 2 = ( 3m ) 2− 2 ⋅ 3m ⋅ 5m + ( 5m ) 2 =
= 2m⋅ 2 + 21 ⋅ 2m ⋅ 3n + 3n ⋅ 2 =
= 3m⋅ 2 − 2 ⋅ ( 3⋅ 5) + 5m⋅ 2 =
= 22⋅ m + 21+m ⋅ 3n + 32⋅ n =
= 32⋅ m − 2 ⋅15m + 52⋅ m =
m
= ( 22 ) + 2m+1 ⋅ 3n + ( 32 ) =
= ( 32 ) − 2 ⋅15m + ( 52 ) =
= 4n + 2m+1 ⋅ 3n + 9n
= 9m − 2 ⋅15m + 25m
m
83)
82)
( A − B)
i
m
n
+ 2 n ) 2 = ( 2m ) 2 + 2 ⋅ 2 m ⋅ 2 n + ( 2 n ) 2 =
84)
=2
1+ m+ n
+2
+2
2⋅ n
m
+ 2m−1 ) 2 = ( 2m ) 2+ 2 ⋅ 2m ⋅ 2m−1 + ( 2m−1 ) 2 = = 2m ⋅ 2 + 21 ⋅ 2m ⋅ 2m−1 + 2(
= 2m⋅ 2 + 21 ⋅ 2m ⋅ 2n + 2n ⋅ 2 = 2⋅ m
(2
m
=
=2
2⋅ m
1+ m+ m−1
+2
+2
= (2
= 4m + 2m+n+1 + 4n
= 4m + ( 22 ) + 4m−1 =
n
)
2 m
+ 22 m + ( 2 m
= 2 ⋅ 4m + 4m−1
(3
m
− 3n ) 2 = ( 3m ) 2− 2 ⋅ 3m ⋅ 3n + ( 3n ) 2 = = 3m⋅ 2 − 2 ⋅ 3m+n + 3n⋅ 2 = = 32⋅ m − 2 ⋅ 3m+n + 32⋅ n = = ( 32 ) − 2 ⋅ 3m+n + ( 32 ) = m
n
= 9m − 2 ⋅ 3m+n + 9n 86)
(2
m+1
+ 2m−1 ) 2 = ( 2m+1 ) 2+ 2 ⋅ 2m+1 ⋅ 2m−1 + ( 2m−1 ) 2 = = 2( =2
m+1) ⋅ 2
2 ⋅ ( m+1)
= ( 22 )
+ 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
=
=
= 4m+1 + 22 m+1 + 4m−1 87)
(2
n
+ 21−n ) 2 = ( 2n ) 2− 2 ⋅ 2n ⋅ 21−n + ( 21−n ) 2 = = 2n ⋅ 2 − 21 ⋅ 2n ⋅ 21−n + 2(
1−n ) ⋅ 2
= 22⋅ n − 21+n+1−n + 2
2 ⋅(1−n )
= ( 22 ) − 22 + ( 22 ) n
1−n
=
=
=
= 4n − 4 + 41−n
M I M Sraga 2003/05
16
=
)
− 2 m 1
= 4m + 4m + 4m−1 =
85)
2 ⋅ ( m−1)
= ( 22 ) + 2m+n+1 + ( 22 ) = m
m−1) ⋅ 2
=
=
=
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
( )
− 21+n+1+n−2 + 2
( )
n+1
− 22 n + 22
( )
= 4n+1 − 22
n
)
⋅ 2n−2 + 2n−2
− 21 ⋅ 2 n+1 ⋅ 2 n−2 + 2(
2 ⋅ ( n+1)
= 22
(
n +1
n−2 ) ⋅ 2
2 ⋅( n−2 )
n−2
2
= =
=
=
+ 4 n −2 =
= 4n+1 − 4n + 4n−2 89)
(a
n
+ a n+1
) = ( a ) + 2⋅ a 2
n
2
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
x
2
x
2
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 2003/05
17
=
i
( A − B)
2
= A2 − 2 ⋅ A ⋅ B + B 2
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