razlika kvadratak - MIM-Sraga

( )( )
7. Koristeći formulu za razliku kvadrata: A− B A+ B = A −B
izračunaj:
2 2
( x− y)( x+ y) ( x− )( x+ ) ( x− )( x+
)
( x+ )( x− ) ( a+ b)( a−b) ( a− b)( a+
b)
( xy − z)( xy + z) ( x+ yz)( x− yz) ( ba− c)( ba+
c)
1) 2) 1 1 3) 2 2
4) 3 3 5) 3 3 6) 2 3 2 3
7) 2 5 2 5 8) 4 3 4 3 9) 3 7 3 7
⎛1 ⎞⎛1 ⎞ ⎛2 ⎞⎛2
⎞ ⎛ 3 ⎞⎛ 3
10) ⎜ x+ y⎟⎜ x− y⎟ 11) ⎜ − x⎟⎜ + x⎟ 12) ⎜5a−
b⎟⎜5a
⎝2 ⎠⎝2 ⎠ ⎝3
⎠⎝3 ⎠ ⎝ 4 ⎠⎝
+ b ⎞
4
⎟ ⎠
13) 200 − 2 200 + 2 14) 300 − 7 300 + 7 15) 100 − 9 100 + 9
( )( ) ( )( ) ( )( )
16) 48⋅52 17) 97⋅103 18) 195⋅205
19)
⎛2 ⎞⎛2 ⎞ ⎛1 1 ⎞⎛1 1 ⎞
⎜ x− 0,2⎟⎜ x+ 0,2⎟ 20) ⎜ x− y⎟⎜ x+
y⎟
⎝3 ⎠⎝3 ⎠ ⎝2 3 ⎠⎝2 3 ⎠
21)
⎛3 ⎞⎛3 ⎞
⎛2
2 4 5⎞⎛2
2 4 5⎞
⎜ xy − 2z⎟⎜ xy + 2z⎟
22) ⎜ x −5y
z ⎟⎜ x + 5y z ⎟
⎝4 ⎠⎝4 ⎠
⎝3
⎠⎝3
⎠
23)
⎛1 5 3 ⎞⎛1 5 3 ⎞
⎛4 2 5 ⎞⎛4 2 5 ⎞
⎜ x + ⎟⎜ x − ⎟
24) ⎜ x −
3 4 ⎟⎜ x +
3 4 ⎟
⎝2 yz ⎠⎝2 yz ⎠ ⎝5 7yz ⎠⎝5
7yz
⎠
25)
⎛ 2 ⎞⎛ 2 ⎞ ⎛ 1 2 ⎞⎛ 1 2 ⎞
⎜2 x− y⎟⎜2 x+ y⎟
26) ⎜1 x−2 y⎟⎜1 x−2
y⎟
⎝ 3 ⎠⎝ 3 ⎠
⎝ 3 5 ⎠⎝ 3 5 ⎠
27)
⎛ 1 2 1 ⎞⎛ 1 2 1 ⎞
( 0,3x− 5)( 0,3x+ 5)
28) ⎜1 x − y⎟⎜1
x + y⎟
⎝ 2 3 ⎠⎝ 2 3 ⎠
4 4
( x− )( x+ ) ( x − )( x + )
3 3
2 2
( 0,1x − 0,3y)( 0,1x + 0,3y) ( a− b )( a+
b )
2 2 2 3 2 3
( − a )( + a ) ( x − y )( x + y )
3 5 3 5 2 3 2 3
( x − y )( x + y ) ( x − y )( x + y )
7 2 7 2 2 3 2 3
( x − y )( x + y ) ( x − y )( x + y )
2 3 4 2 3 4 3 2 5 3 2 5
( x y − z )( x y + z ) ( a b − c )( a b + c )
3 2 3 2 5 3 5 3
( xy− z)( xy+ z) ( x − yz)( x + yz)
2 3 4 5 6 2 3 4 5 6 2 3 2 3
( x y z − a b )( x y z + a b ) ( ab + c )( ab −c
)
x y x y x y x y
( a + a )( a −a ) ( a − b )( a + b )
m n m n n m n n
( − )( + ) ( − )( + )
n m n m n+ 1 m− 2 n+ 1 m−2
( x + y )( x − y ) ( x − y )( x + y )
29) 0,1 0, 2 0,1 0, 2 30) 0,75 0,75
31) 32)
33) 3 3 34)
35) 36) 7 5 7 5
37) 5 5 38) 4 3 4 3
39) 2 3 2 3 40) 4 7 4 7
41) 3 3 3 3 42) 9 11 9 11
43) 3 4 3 4 44) 2 2
45) 46) 2 3 2 3
47) 2 2 2 2 48) 3 2 3 2
49) 50)
( x+ y− )( x+ y+ ) ( x+ y− )( x− y−
)
2 2
( − )( + ) = −
( x+ y− z)( x− y+ z) ( a+ b+ c)( a−b−c)
( x− y− )( y+ + x) ( x− y+ z)( x+ y−
z)
( x+ y−z)( x− y− z) ( x− y− z)( x+ y−
z)
51) 1 1 52) 1 1
7. Koristeći formulu za razliku kvadrata: A B A B A B
53) 3 3 54)
55) 1 1 56) 2 3 5 2 3 5
57) 2 2 58) 2 8 2 8
2 3 6 2 3 6 2 3 2 3
( x + y + z )( x + y − z ) ( x− y + z )( x+ y − z )
59) 2 3 5 2 3 5 60) 5 4 5 4

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
2 2
1)
( )( )
x− y x+ y = x − y
2 2
( )( )
2 2 2
2) 1 1 1 1
x− x+ = x − = x −
( )( )
2 2 2
3) 2 2 2 4
x− x+ = x − = x −
( )( )
2 2 2
4) 3 3 3 9
x+ x− = x − = x −
( )( ) ( )
2 2
5) 3a+ b 3a− b = 3a − b =
2 2 2
= ⋅ − =
2 2
( a b)( a b) ( a) ( b)
3
= 9a
−b
2 2
6) 2 − 3 2 + 3 = 2 − 3 =
a
2 a 3
2 2 2 2
= ⋅ − ⋅ =
4a
b
2
= −
9b
2
b
( xy z)( xy z) ( xy) ( z)
2 2
7) 2 − 5 2 + 5 = 2 − 5 =
2 x y 5 z
2 2 2 2 2
= ⋅ ⋅ − ⋅ =
= 4xy
−25z
2 2 2
( x yz)( x yz) ( x) ( yz)
2 2
8) 4 + 3 4 − 3 = 4 − 3 =
4 x 3 y z
2 2 2 2 2
= ⋅ − ⋅ ⋅ =
= 16x
−9y z
2 2 2
( ba c)( ba c) ( ba) ( c)
2 2
9) 3 − 7 3 + 7 = 3 − 7 =
3 b a 7 c
2 2 2 2 2
= ⋅ ⋅ − =
= 9ba − 49 c = ili = 9ab −49c
2 2 2 2 2 2
M I M Sraga 2005
2

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
2 2
10)
2
2
+ − = − =
⎛1 ⎞⎛1 ⎞ ⎛1
⎞
⎜ x y⎟⎜ x y⎟ ⎜ x⎟
y
⎝2 ⎠⎝2 ⎠ ⎝2
⎠
1
2
1
= x − y
4
2
2 2
= ⋅x
− y =
2
2 2
11)
2
2
− + = − =
⎛2 ⎞⎛2 ⎞ ⎛2⎞
⎜ x⎟⎜ x⎟ ⎜ ⎟ x
⎝3 ⎠⎝3 ⎠ ⎝3⎠
2
3
4
= −x
9
2
2
= − x =
2
2
3 3 2 3
⎛ ⎞⎛ ⎞ ⎛ ⎞
12) ⎜5a− b⎟⎜5a+ b⎟
= ( 5a)
−⎜
b⎟
=
⎝ 4 ⎠⎝ 4 ⎠ ⎝4
⎠
( )( )
2 2
13) 200 2 200 2 200 2 40000 4 39996
( )( )
2 2
14) 300 7 300 7 300 7 90000 49 89951
( )( )
2 2
15) 100 9 100 9 100 9 10000 81 9919
( ) ( )
2 2
16) 48 52 50 2 50 2 50 2 2500 4 2496
( ) ( )
17) 97⋅ 103 = 100−3 ⋅ 100+
3
2
2 2 3 2
= 5 ⋅a
− ⋅ b =
2
4
2 9 2
= 25a
− b
16
− + = − = − =
− + = − = − =
− + = − = − =
⋅ = − ⋅ + = − = − =
( ) ( )
2
= − = − =
2 2
100 3 10000 9 9991
⋅ = − ⋅ + = − = − =
2 2
18) 195 205 200 5 200 5 200 5 40000 25 39975
M I M Sraga 2005
3

M-1-Razlika kvadrata
Rješenja i kompletni postupak
2 2
( − )( + ) = −
7. Koristomo formulu za razliku kvadrata: A B A B A B
2
2
− + = − =
⎛2 ⎞⎛2 ⎞ ⎛2
⎞
19) ⎜ x 0.2⎟⎜ x 0.2⎟ ⎜ x⎟
0.2
⎝3 ⎠⎝3 ⎠ ⎝3
⎠
ili na drugi način:
2
3
2
2
= ⋅x
−
2
0.04
2 2 2
⎛2 ⎞⎛2 ⎞ ⎛2 ⎞ ⎛ 2 ⎞ ⎛2 ⎞ ⎛1⎞
⎜ x− 0.2⎟⎜ x+ 0.2⎟ = ⎜ x⎟ − ⎜ ⎟ = ⎜ x⎟ −⎜
⎟
⎝3 ⎠⎝3 ⎠ ⎝3 ⎠ ⎝10⎠ ⎝3 ⎠ ⎝5⎠
2 2
2 2 1
= ⋅x
− =
2 2
3 5
4 2 1
= x −
9 5
2
=
20)
2 2
⎛1 1 ⎞⎛1 1 ⎞ ⎛1 ⎞ ⎛1
⎞
⎜ x− y⎟⎜ x+ y⎟ = ⎜ x⎟ − ⎜ y⎟
=
⎝2 3 ⎠⎝2 3 ⎠ ⎝2 ⎠ ⎝3
⎠
2 2
1 2 1 2
= ⋅x
− ⋅ y =
2 2
2 3
1 2 1 2
= x − y
4 9
⎛3 ⎞⎛3 ⎞ ⎛3
⎞
2
21) ⎜ xy − 2z⎟⎜ xy + 2z⎟ = ⎜ xy⎟
− ( 2z)
=
⎝4 ⎠⎝4 ⎠ ⎝4
⎠
3
4
9
= x y −4z
16
2
2
2 2 2 2
= ⋅x ⋅y −2
⋅ z =
2
2 2 2
⎛2 2 4 5⎞⎛2
2 4 5⎞ ⎛2
2⎞
4 5 2
22) ⎜ x − 5y z ⎟⎜ x + 5y z ⎟ = ⎜ x ⎟ − ( 5y z ) =
⎝3
⎠⎝3 ⎠ ⎝3
⎠
2
3
4
9
x 25 y z
4 4 8 10
= x −25y z
9
2
2
2 2 2 4 2 5 2
= ⋅( x ) −5
⋅
2 ( y ) ⋅ ( z ) =
22 ⋅ 42 ⋅ 52 ⋅
= ⋅ − ⋅ ⋅ =
M I M Sraga 2005
4

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristomo formulu za razliku kvadrata: A− B A+ B = A −B
2 2
23)
⎛1 5 3 ⎞⎛1 5 3 ⎞ ⎛1 5⎞
⎛ 3 ⎞
⎜ x + ⎟⎜ x − ⎟ = ⎜ x ⎟ − ⎜ ⎟ =
⎝2 yz ⎠⎝2 yz ⎠ ⎝2
⎠ ⎝ yz ⎠
10
= −
2
2 2
1 5 2 3
= ⋅
2 ( x ) − =
2 2
2 y ⋅ z
1 9
x
4 yz
2 2
2
24)
2
2 2 ⎛ 2⎞
− + = −
3 4 3 4 3 4
⎛4 5 ⎞⎛4 5 ⎞ 4 ⎛ 5 ⎞
⎜ x ⎟⎜ x ⎟ ⎜ x ⎟ ⎜ ⎟
⎝5 7yz ⎠⎝5 7yz ⎠ ⎝5
⎠ ⎝7yz
⎠
2 2
4 2 2 5
= ⋅
2 ( x ) − =
2 3 2 4 2
5 7 ⋅
16 25
x
25 49⋅y
⋅z
2
( y ) ⋅( z )
22 ⋅
= ⋅ − =
32 ⋅ 42 ⋅
16 25
x
25 49yz
2
= −
6 8
=
2
2
− + = − =
⎛ 2 ⎞⎛ 2 ⎞ ⎛ 2 ⎞
25) ⎜2 x y⎟⎜2 x y⎟ ⎜2
x⎟
y
⎝ 3 ⎠⎝ 3 ⎠ ⎝ 3 ⎠
2
⎛23 ⋅ + 2 ⎞ 2
= x − y =
⎜
⎝
3
2
⎛8
⎞ 2
= x − y =
⎜
⎝3
⎟
⎠
8
3
64
= x − y
9
2
2 2
= ⋅x
− y =
2
2 2
⎟
⎠
2 2
⎛ 1 2 ⎞⎛ 1 2 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞
26) ⎜1 x−2 y⎟⎜1 x−2
y⎟
= ⎜1 x⎟ − ⎜2
y⎟
=
⎝ 3 5 ⎠⎝ 3 5 ⎠ ⎝ 3 ⎠ ⎝ 5 ⎠
2 2
⎛13 ⋅ + 1 ⎞ ⎛25 ⋅ + 2 ⎞
= ⎜ x − y =
3
⎟ ⎜
5
⎟
⎝ ⎠ ⎝ ⎠
2 2
⎛4 ⎞ ⎛12
⎞
= ⎜ x⎟ − ⎜ y⎟
=
⎝3 ⎠ ⎝ 5 ⎠
2 2
4 2 12 2
= ⋅x
− ⋅ y =
2 2
3 5
16 2 144 2
= x − y
9 25
M I M Sraga 2005
5

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( x )( x ) ( x)
2 2
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
27) 0.3 − 5 0.3 + 5 = 0.3 − 5 =
2 2
= ⋅x
− = → ⎡
⎣ = ⋅ =
2 2 2
0.3 25 0.3 0.3 0.3 0.9
2
0.09 25
= x −
⎤
⎦
2 2
2 2 2 2
⎛ 1 1 ⎞⎛ 1 1 ⎞ ⎛ 1 ⎞ ⎛1
⎞
28) ⎜1 x − y⎟⎜1 x + y⎟ = ⎜1
x ⎟ − ⎜ y ⎟ =
⎝ 2 3 ⎠⎝ 2 3 ⎠ ⎝ 2 ⎠ ⎝3
⎠
2
1⎞
2 2 1
1 ⎟ 2
( x ) ⎜ 3
( )
⎛
⎛ ⎞ 2 2
= ⎜ ⋅ − ⎟ ⋅ y =
⎝ ⎠ ⎝ ⎠
2 2
22 ⋅
22 ⋅
x y
2
⎛12 ⋅ + 1⎞
1
= ⎜ ⋅ − ⋅ =
2
⎟
⎝ ⎠ 3
2
⎛3⎞
4 1 2
= ⋅ − =
⎜ ⎟ x y
⎝2⎠
9
9 4 1 2
= x − y
4 9
2
( x )( x ) ( x)
2 2
29) 0.1 − 0.2 0.1 + 0.2 = 0.1 − 0.2 =
2 2
0.1 0.04
= ⋅x
− =
2
0.01 0.04
= x −
( x )( x ) ( x )
4 4 4 2 2
30) − 0.75 + 0.75 = − 0.75 =
x
42 ⋅
= − =
8
= −
x
0.5625
0.5625
3 3
( 0.1 − 0.3 )( 0.1 + 0. 3y) ( 0.1x ) ( 0.3y)
0.1 ( x ) 0.3
31) x y x
3 2 2
= − =
2 3 2 2 2
= ⋅ − ⋅ =
0.01 x 0.09y
32 ⋅
2
2
= ⋅ − = → ⎡0.3 = 0.3⋅ 0.3 = 0.9⎤
= 0.01x
−0.09y
6 2
y
⎣
⎦
M I M Sraga 2005
6

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
2 2
32)
( a b )( a b ) a ( b )
2 2 2 2 2
− + = − =
a
b
2 2⋅2
= − =
= a −b
2 4
( a )( a ) ( a )
33) 3 3 3
2 2 2 2 2
− + = − =
9
a
22 ⋅
= − =
= 9 −a
4
34)
( x y )( x y ) ( x ) ( y )
− + = − =
2 3 2 3 2 2 3 2
x
y
22 ⋅ 32 ⋅
= − =
= x − y
4 6
35)
( x y )( x y ) ( x ) ( y )
− + = − =
3 5 3 5 3 2 5 2
32 ⋅ 52 ⋅
= − =
x
= x − y
y
6 10
2 3 2 3 2 2 3 2
( x − y )( x + y ) = ( x ) − ( y ) =
7 ( x ) 5 ( y )
36) 7 5 7 5 7 5
2 2 2 2 3 2
= ⋅ − ⋅ =
49x
25y
22 ⋅
32 ⋅
= − =
= 49x
−25y
4 6
( x y )( x y ) ( x ) ( y )
⋅
5 ( x ) y
7 2 7 2 7 2 2 2
37) 5 − 5 + = 5 − =
2 7 2 2 2
= ⋅ − =
25
x
72 ⋅ 4
= ⋅ − =
= 25x
− y
14 4
y
2 3 2 3
( x − y )( x + y )
38) 4 3 4 3
M I M Sraga 2005
7

M-1-Razlika kvadrata
Rješenja i kompletni postupak
2 3 2 3 2 2 3 2
( x − y )( x + y ) = ( x ) − ( y ) =
4 ( x ) 3 ( y )
2 2 2 2 3 2
= ⋅ − ⋅ =
22 ⋅ 32 ⋅
= ⋅ − ⋅ =
4 6
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
38) 4 3 4 3 4 3
16 x 9 y
= 16x
−9y
2 2
2 3 4 2 3 4 2 3 2 4 2
( xy− z)( xy+ z) = ( xy) − ( z)
=
2 ( x ) ( y ) 3 ( z )
39) 2 3 2 3 2 3
2 2 2 3 2 2 4 2
= ⋅ ⋅ − ⋅ =
4 x y 9 z
22 ⋅ 32 ⋅ 42 ⋅
= ⋅ ⋅ − ⋅ =
4
= 4x
y
− 9z
6 8
3 2 5 3 2 5 3 2 2 5 2
( ab − c )( ab + c ) = ( ab ) − ( c ) =
4 ( a ) ( b ) 7 ( c )
40) 4 7 4 7 4 7
2 3 2 2 2 2 5 2
= ⋅ ⋅ − ⋅ =
16 a b 49 c
32 ⋅ 22 ⋅ 52 ⋅
= ⋅ ⋅ − ⋅ =
= 16ab
−49c
6 4 10
3 2 3 2 3 2 2 2
( xy− z)( xy+ z) = ( xy) − ( z)
=
3 ( x ) y 3 ( z )
41) 3 3 3 3 3 3
2 3 2 2 2 2 2
= ⋅ ⋅ − ⋅ =
9x y 9z
32 ⋅ 2 22 ⋅
= ⋅ − =
= 9xy
−9z
6 2 4
5 3 5 3 5 2 3 2
( x − y z)( x + y z) = ( x ) −( y z)
9 ( ) 11 ( )
42) 9 11 9 11 9 11
x y z
2 5 2 2 3 2 2
= ⋅ − ⋅ ⋅ =
81 x 121 y z
52 ⋅
32 ⋅ 2
= ⋅ − ⋅ ⋅ =
= 81x
−121y z
10 6 2
=
M I M Sraga 2005
8

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristeći formulu za razliku kvadrata: A− B A+ B = A −B
2 2
2 3 4 5 6 2 3 4 5 6 2 3 4 2 5 6 2
( x y z − a b )( x y z + a b ) = ( x y z ) − ( a b ) =
3 ( x ) ( y ) ( z ) 4 ( a ) .( b )
43) 3 4 3 4 3 4
2 2 2 3 2 4 2 2 5 2 6 2
= ⋅ ⋅ ⋅ − ⋅ =
9 x y z 16 a b
22 ⋅ 32 ⋅ 42 ⋅ 52 ⋅ 62 ⋅
= ⋅ ⋅ ⋅ − ⋅ ⋅ =
= 9x y z −16a b
4 6 8 10 12
2 3 2 3 2 2
( ab + c )( ab − c ) = ( ab ) −( c )
2 a ( b ) c
44) 2 2 2
3 2
2 2 2 2 3⋅
2
= ⋅ ⋅ − =
= 4ab
−c
2 4 6
=
45)
x y x y x y
( a a )( a a ) ( a ) ( a )
2 2
+ − = − =
a
a
x⋅2 y⋅2
= − =
= a −a
2x
2y
x y x y x 2 y 2
( a b )( a b ) ( a ) ( b )
x
y
2 ( a ) 3 ( b )
46) 2 − 3 2 + 3 = 2 − 3 =
m n m n m n
( )( ) ( ) ( )
2 2 2 2
= ⋅ − ⋅ =
x⋅2 y⋅2
= ⋅ − ⋅ =
= 4a
−9b
47) 2 2 2 2 2 2
2x
2y
2 2
− + = − =
m⋅2 n⋅2
= − =
2 2
= 2 −2
4 a 9 b
2m
2n
Možemo ostaviti u ovom obliku ↑↑ ali možemo nastaviti i dalje:
2 m 2
( 2 ) ( 2 )
n m n n n m
( )( ) ( ) ( )
2 2
− + = − =
Možemo ostaviti u ovom
n
= − =
m
= 4 −4
48) 3 2 3 2 3 2
n⋅2 m⋅2
= − =
3 2
= 3 −2
2n
2m
↑↑
obliku ali možemo nastaviti i dalje:
2 n 2
( 3 ) ( 2 )
m
= − =
n
= 9 −4
m
n
M I M Sraga 2005
9

M-1-Razlika kvadrata
Rješenja i kompletni postupak
7 . Koristimo formulu za razliku kvadrata: ( )( )
A− B A+ B = A −B
2 2
49)
n m n m n m
( x y )( x y ) ( x ) ( y )
2 2
+ − = − =
x
y
n⋅2 m⋅2
= − =
= x − y
2n
2n
50)
n+ m− n+ m− n+ m−
( x y )( x y ) ( x ) ( y )
− + = − =
1 2 1 2 1 2 2 2
( n ) ( m )
+ 1 ⋅2 −2 ⋅2
= x − y =
= x − y
2n+ 2 2m−4
( ) ( )
( x+ y− 1)( x+ y+ ) = ⎡⎣( x+ y)
− ⎤⎦ ⋅ ⎡⎣( x+
y)
2 2
= ( x+ y)
− 1 =
51)
1 1
A − B ⋅ A + B = A −B
↑ ↑ ↑ ↑
x 2xy y 1
2 2
= + + − =
2 2
x y xy
= + + 2 −1
+ 1⎤⎦
2 2
( x y )( x y ) x ( y ) x ( y )
52) + −1 − − 1 = ⎡⎣ + −1 ⎤⎦⋅⎡⎣ − + 1 ⎤⎦
=
( y 1)
2
2
= x − − =
( 2
2
1 1 )
( 2 1)
2 2
x y y
= − − ⋅ ⋅ + = →
=
2 2
x y y
− − + =
2 2
x y y
= − + 2 −1
pazi na minus ispred zagrade
( x y z)( x y z) x ( y z) 3x ( y z)
53) 3 + − 3 − + = ⎡⎣3
+ − ⎤⎦⋅⎡⎣ − − ⎤⎦
=
( 3x) ( y z)
2 2
= − − =
( )
3 x y 2yz z
2 2 2 2
= ⋅ − − + =
9x y 2yz z
2 2 2
= − + − =
2 2 2
= 9x − y − z + 2yz
M I M Sraga 2005
10

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
2 2
54)
( a b c)( a b c) a ( b c) a ( b c)
+ + − − = ⎡⎣ + + ⎤⎦⋅⎡⎣ − + ⎤⎦
=
( )
= − + =
2
a b c
2
( 2 )
2 2 2
= − + + =
a b bc c
2 2 2
= − − − =
a b 2bc c
= − − −2
2 2 2
a b c bc
( x− y− )( y+ + x) = ( x− y− ) ⋅ ( x+ y+ ) =
x ( y 1) x ( y 1)
55) 1 1 1 1
= ⎡⎣ − + ⎤⎦⋅ ⎡⎣ + + ⎤⎦
=
( y 1)
2
2
= x − + =
2 2 2
= x −( y 2 y 1 1 )
2 2
= x − y − 2y+
1
+ ⋅ ⋅ + =
( )( ) ⎡ ( ) ⎤ ⎡ (
56) 2x− 3y+ 5z 2x+ 3y− 5z = ⎣2x− 3y−5z ⎦⋅ ⎣2x+ 3y− 5z)⎤ ⎦ =
2 2
( 2x) ( 3y 5z)
2 2
2 2
2 x (( 3y) 2 3y 5z ( 5z)
)
= − − =
= ⋅ − − ⋅ ⋅ + =
( )
2 2 2
= − − + =
4x 9y 30yz 25z
4x 9y 30yz 25z
2 2 2
= − + − =
2 2 2
4 9 25 30
= x − y − z + yz
( x y z)( x y z) ( x z y) ( x z y)
= ⎡⎣( x− z)
+ 2y⎤⎦ ⎡⎣( )
2 2
= ( x−z) − ( 2y)
=
57) + 2 − −2 − = − + 2 ⋅ − − 2 =
x 2xz z 2 y
2 2 2 2
= − + − ⋅ =
= − 4 + −2
2 2 2
x y z xz
( x y z)( x y z) ( x z y) ( x z y)
( ) ( )
( x 8z) ( 2y)
⋅ x−z − 2y⎤⎦
=
58) −2 − 8 + 2 − 8 = −8 −2 ⋅ − 8 + 2 =
= ⎡⎣ x−8z −2y⎤⎦⋅⎡⎣ x− 8z + 2y⎤⎦
=
2 2
= − − =
( )
2 2 2 2
= x −2⋅x⋅ 8z+ 8z −2
⋅ y =
2 2 2 2
= − + ⋅ − =
x 16xz 8 z 4y
= − 4 + 64 −16
2 2 2
x y z xz
M I M Sraga 2005
11

M-1-Razlika kvadrata
Rješenja i kompletni postupak
( )( )
7. Koristimo formulu za razliku kvadrata: A− B A+ B = A −B
2 2
( )( ) ⎡( ) ⎤ ⎡
⎣
⎦ ⎣( )
2 3 2 6 2
= ( 2x + 3y ) − ( 5z
) =
( 2x ) 2 2x 3y ( 3y ) 5 ( z )
2 2 2 2 3 2 3 2 6⋅2
2 ( x ) 12x y 3 ( y ) 25 z
2 3 6 2 3 6 2 3 6 2 3 6
59) 2x + 3y + 5z 2x + 3y − 5z = 2x + 3y + 5z ⋅ 2x + 3y − 5z
⎤
⎦
=
Još malo upute:
2 2 2 3 3 2 2 6 2
= + ⋅ ⋅ + − ⋅ =
= ⋅ + + ⋅ − ⋅ =
4 2 3
4 12 9
= x + x y + y
2 3 6 2 3 6
( 2x + 3y + 5z )( 2x + 3y − 5z
) =
[ ( 2x 3y ) 5 z ][ ( 2x 3y ) 5 z ] ( 2x 3y ) ( 5z
)
( ) ( )
− 25z
6 12
2 3 6 2 3 6 2 3 2 6 2
= + + ⋅ + − = + − =
↓
↓
↓ ↓ ↑ ↑
A + B ⋅ A − B = A − B
2 2
I dalje kvadriramo:
( 2x ) 2 2x 3y ( 3y ) 5 ( z )
2 2 2 2 3 2 3 2 6⋅2
2 ( x ) 12x y 3 ( y ) 25 z
2 2 2 3 3 2 2 6 2
= + ⋅ ⋅ + − ⋅ =
= ⋅ + + ⋅ − ⋅
4x 12x y 9y 25z
4 2 3 6 12
= + + − =
=
2 3 2 3 2 3 2 3
( x y z )( x y z ) x ( y z ) x ( y z )
2 2 3 2
= ( 5x) −( 4y − z ) =
2 2 2 2 2 3 3 2
5 x ⎡( 4y ) 2 4y z ( z ) ⎤
⎣
⎦
2 2 2 2 2 3 3⋅2
= 25x −⎡4 ( y ) 8y z z ⎤
⎣
⋅ − +
⎦
=
2 4 2 3 6
= 25x −( 16y − 8y z + z ) =
60) 5 − 4 + 5 + 4 − = ⎡5 − 4 − ⎤⋅ ⎡5 + 4 − ⎤
⎣ ⎦ ⎣ ⎦
=
= ⋅ − − ⋅ ⋅ + =
= 25x −16y −8y z −z
2 4 2 3 6
M I M Sraga 2005
12