2 i 3 ISPIT ZNANJA

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2011.
POTPUNO RIJEŠENI ZADACI
PRIRUČNIK ZA SAMOSTALNO UČENJE
KOMPLETNA
RJEŠENJA
2. i 3. ISPIT ZNANJA
α

ALGEBARSKI IZRAZI – test br.2. i 3.
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MATEMATIKA –1
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ALGEBARSKI IZRAZI – test br.2. i 3.
2. i 3. test – ALGEBARSKI IZRAZI
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1.
( ) ( ) ( ) ( )
43 4 25 2( 2)
a ) 4a⋅ 3a−b −2b⋅ 5a− 2b = 4a⋅ 3a+ 4a⋅ −b −2b⋅5a−2b⋅ − 2b
=
1 1 1 1
= ⋅ ⋅a ⋅a − ⋅a⋅b− ⋅ ⋅b⋅a− ⋅ − ⋅b ⋅ b =
12 a 4ab 10ab 4 b
2 1+
1
= ⋅ − − + ⋅ =
12a 14ab 4b
2 2
= − + ↓
vidi objašnjenje
( )
−4ab − 10ab = −4− 10 ab = −14ab
2.
( − ) ⋅ ( + ) −( − ) ⋅ ( + ) = ⋅ ( + ) − ⋅ ( + ) −⎣
⋅ ( + ) − ⋅ ( + ) ⎦
xx x2 1 x 1 2 ( xx x1 3 x 3 1)
b) x 1 x 2 x 3 x 1 x x 2 1 x 2 ⎡x x 1 3 x 1 ⎤ =
Kvadriramo po pravilu:
To pravilo možemo pisati i ovako:
= ⋅ + ⋅ − ⋅ − ⋅ − ⋅ + ⋅ − ⋅ − ⋅ =
= 0
( )
( )
2 2
= x + 2x−x−2− x + 1x−3x− 3 =
2 2
= x + x−2− x −2x− 3 =
2 2
= x + x−2− x + 2x+ 3 =
2 2
= x − x + x+ 2x− 2+ 3 =
= 3x
+ 1
( ) ( ) ( )
( ) ( )
2 2 2 2 2 2
A+ B = A + 2⋅A⋅ B+ B i A− B = A −2⋅A⋅ B+
B
2 2 2
a) 3x− 4y = 3x −2⋅3x⋅ 4y+ 4y
=
2 2 2 2
= 3 x − 24xy+ 4 y =
= 9x − 24xy+
16y
2 2
( ) i ( )
2 2 2 2 2
I + II = I + 2⋅I ⋅ II + II I − II = I −2⋅I ⋅ II + II
2
2 2 2
3 4 3 3 4 4
⎛2 3 ⎞ ⎛2 ⎞ 2 3 ⎛3
⎞
b ) ⎜ x − y ⎟ = ⎜ x ⎟ −2⋅ x ⋅ y + ⎜ y ⎟ =
⎝3 4 ⎠ ⎝3 ⎠ 3 4 ⎝4
⎠
2 2
2 3 2 2 3 3 4 3 4 2
2 3
=
2 ( x ) −2⋅ ⋅ ⋅x ⋅ y +
2 ( y ) = →2⋅ ⋅ =
3 3 4 4
3 4
4 9
x 1 x y y
9 16
4 6 3 4 9 8
= x − x y + y
9 16
32 ⋅ 3 4 42 ⋅
= − ⋅ + =
2⋅23
⋅
34 ⋅
= 1
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2.
( ) ( ) ( )
2 4 2 4 5 2 5 2
3 ( x ) 42x y 7 ( y )
4 5 2 4 2 4 5 5 2
c) 3x + 7y = 3x + 2⋅3⋅x ⋅7⋅ y + 7y
=
= + + =
42 ⋅
4 5 52 ⋅
= 9x + 42x y + 49y
=
= 9x + 42x y + 49y
8 4 5 10
3.
( )
2 2 2
( ) ( ) ( ) ( )
2 2
2 ( x 2x 1) 3 ( 4x 12x
9)
d) 2 x 1 3 2x 3 2 x 2 x 1 1 3 2x 2 2x
3 3
2 2 2
− − + = ⋅ − ⋅ ⋅ + − ⋅ + ⋅ ⋅ + =
= ⋅ − + − ⋅ + + =
2 2
= 2⋅x −2⋅ 2x+ 2⋅1−3⋅4x −3⋅12x−3⋅ 9 =
2
= 2x
− 4x+
2
2 12x
36x
27
2 2
= 2x −12x −4x− 36x+ 2− 27 =
=− − −
2
10x
40x
25
− − − =
2
2 4 5 2 4 5 2 4 5 2
⎛2 ⎞⎛2 ⎞ ⎛2
⎞
a) ⎜ 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 4 2 5 2
= ⋅( x ) −5
⋅
2
( y ) ⋅ ( z ) =
22 ⋅ 42 ⋅ 52 ⋅
= ⋅ − ⋅ ⋅ =
( )( ) ( ) ( )
b) 3x+ y−z 3x− y+ z = ⎡⎣3x+ y− z ⎤⎦⋅⎡⎣3x− y− z ⎤⎦
=
2 2
= ( 3x) −( y− z)
=
( )
2 2 2 2
= 3 ⋅x − y − 2yz+ z =
2 2 2
= 9x − y + 2yz− z =
2 2 2
= 9x − y − z + 2yz
2 2 2 2 2
( )( ) ( ) ( )
c) x− 1 x+ 1 − x+ 2 = x −1 − x + 2⋅x⋅ 2+ 2 =
2 2
= x −1−x −4x− 4 =
2 2
= x − x −4x−1− 4 =
=−4x
−5
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4.
Koristimo se formulama za:
kub zbroja i kub razlike
( ) ( )
3 3 2 2 3 3 3 2 2
A+ B = A + 3A B+ 3AB + B A− B = A − 3A B+ 3AB −B
( )
3 3 2 2 3
A − B = A − 3⋅A ⋅ B + 3⋅ A⋅ B − B
↑ ↑ ↓ ↓ ↓ ↓
( ) ( ) ( ) ( ) ( )
3 3 2 2 3
a) 3x− 5y = 3x −3⋅ 3x ⋅ 5y+ 3⋅3x⋅ 5y − 5y
=
3 3 2 2 2 2 3 3
= 3 ⋅x
−3⋅3 ⋅x ⋅5⋅ y+ 9⋅x⋅5 ⋅y −5
⋅ y =
x x y x y y
3 2 2 3
= 27 −3⋅9⋅5⋅ ⋅ + 9⋅25⋅ ⋅ − 125 =
= 27x − 135x y+ 225xy −125y
3 2 2 3
3
( ) ( ) ( ) ( ) ( )
( ) ( ) 3 ( ) ( ) 3
4 2 3 3 4 2 3 4 2 2 3 4 2 3 2 3 3
b) x y − z = x y −3⋅ x y ⋅ z + 3⋅x y ⋅ z − z =
x y x y z x y z z
4 3 2 3 4 2 2 2 3 4 2 3⋅2 3⋅3
= ⋅ − ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − =
= x ⋅y −3⋅x ⋅y ⋅ z + 3x y z
43 ⋅ 23 ⋅ 42 ⋅ 22 ⋅ 3 4 2 6
= x y − 3x y z + 3x y z − z
12 6 8 4 3 4 2 6 9
9
− z =
5.
2 2
( ) ( ) ( )( )
2 2 2 2 2 2
a) 25x 49y 5 x 7 y 5x 7y 5x 7y 5x 7y
− = ⋅ − ⋅ = − = − +
ili ovako:
2
2
2 36 2 2 2 6 2
2 ⎛ 6 ⎞ ⎛ 6 ⎞⎛ 6
− = ⋅ − ⋅ =
2
( ) − = − +
b) 1.44x y 1.2 x y 1.2x ⎜ y⎟ ⎜1.2x y⎟⎜1.2x
y ⎞ ⎟
169 13
⎝13 ⎠ ⎝ 13 ⎠⎝ 13 ⎠
2
2 2
2 36 2 2 2 6 2
2 ⎛ 6 ⎞ ⎛12 ⎞ ⎛ 6
− = ⋅ − ⋅ =
2
( ) − = −
1.44x y 1.2 x y 1.2x ⎜ y⎟ ⎜ x⎟
169 13
⎝13 ⎠ ⎝10
⎠
2 2
⎛6 ⎞ ⎛ 6 ⎞ ⎛6 6 ⎞⎛6 6 ⎞
= ⎜ x⎟ − ⎜ y⎟ = ⎜ x− y⎟⎜ x+
y⎟
⎝5 ⎠ ⎝13 ⎠ ⎝5 13 ⎠⎝5 13 ⎠
⎞
⎜ y ⎟
⎝13
⎠
2
=
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6.
( x ) 2 x y ( y )
( x y )
( ) ( ) ( )
( ) ( )
a) x − 2x y + y = x −2 ⋅x ⋅ y + y = → x = x = x , y = y = y
6 3 2 4 32 ⋅ 3 2 22 ⋅ 6 32 ⋅ 3 2 4 22 ⋅ 2 2
= −
= −
3 2 3 2 2 2
3 2 2
Još jednom isti zadatak :
⋅ ⋅ + =
x − 2x y + y = x −2⋅x ⋅ y + y = x −2⋅x ⋅ y + y = x − y
6 3 2 4 3⋅2 3 2 2⋅2 3 2 3 2 2 2 3 2 2
↓ ↓ ↓ ↑ ↑
( )
− 2⋅ ⋅ + = −
2 2
A A B B A B
2
2 2
b) 20xy−4x − 25y = −4x
+ 20xy
− 25y
=
2 2
2 2
( x xy y )
2 2 2 2
( x x y y )
=−1⋅ 4 − 20 + 25 =
=−1⋅ 2 ⋅ −2⋅2⋅5⋅ ⋅ + 5 ⋅ =
2 2
(( x) x y ( y)
)
=−1⋅ 2 −2⋅2 ⋅ 5 + 5 =
( x y)
=−1⋅ 2 −5
2
7.
7 4 5 3 4 2 4 3 2 2 4 1 2 1 4 2
a) a b + a b − a b = a ⋅a ⋅b ⋅ b + a ⋅a ⋅b ⋅b −a ⋅b
⋅ 1=
a a b b a a b b a b
4 3 2 2 4 1 2 1 4 2
= ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ ⋅ =
( 1)
( 1)
a b a b a b
4 2 3 2 1 1
= ⋅ ⋅ ⋅ + ⋅ − =
4 2 3 2
ab ab ab
= + −
1
3 2 2 3 2 2
b) x y− 6x y + 9xy = x⋅x ⋅y−6⋅x⋅x⋅y⋅ y+ 9⋅x⋅y⋅ y =
(
2
6 9
2)
(
2
6 9
2)
2
( x 2 x 3y 3y
2
)
= x⋅y⋅ x − xy+ y =
= x⋅y⋅ x − xy+ y =
= xy⋅ − ⋅ ⋅ + ( ) =
( 3 )
= xy⋅ x − y
2
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8.
( − ) − + = ⋅( − ) − ⋅( − ) =
= ( y−1)( x−1)
a) x y 1 y 1 x y 1 1 y 1
9.
( ) ( ) ( ) ( )
3 2 2 2 1 2
b)
a b c d a b d c a a b b c d a b c d
− + − = ⋅ ⋅ ⋅ − + ⋅ ⋅ − + =
( ) ( )
2 2 2
a) x 3y 12xy x 2 x 3y 3y 12xy
+ − = + ⋅ ⋅ + − =
2 2
= x + 6xy− 12xy+ 9y
=
x 6xy 9y
2 2
= − + =
( x 3y)
( c d)
=−1
⋅ −
( ) ( 1)( )
2 2
= a ⋅a⋅b⋅b c− d + a ⋅b − c− d =
( )( 1)
2
= abc−d ab−
kvadriramo prvu zagradu
grupiramo
( ) treba prepoznati kvadrat razlike ( )
2 2
x 2 x 3y 3y a b
= − ⋅ ⋅ + = −
= −
2
2
10.
a) ( 1 x)( 5x 2) ( x 1)( 2x 3) ( 1 x)( 5x 2) ( 1 x)( 2x
3)
= ( 1− x)( 5x+ 2) −1( 1− x)( 2x+ 3)
=
= ( 1− x)( 5x+ 2) −11 ( − x)( 2x+ 3)
=
− + + − + = − + + − + + =
( 1 x) ⎡⎣5x 2 1 ( 2x
3)
( 1 x)( 5x 2 2x
3)
( 1 x)( 3x
1)
= − + − ⋅ + ⎤⎦
=
= − + − − =
= − −
podcrtamo zajednički faktor
izlučimo Z.F.
11.
a) ( 1−2x)( 3− x) − ( 2x+ 3)( 2x−1)
− 1+ 2x
=
( 1 2x)( 3 x) ( 2x 3) ( 1) ( 1 2x) 1 ( 1 2x)
= ( 1−2x)( 3− x) + ( 2x+ 3)( 1−2x) −11 ( − 2x)
=
= ( 1−2x) ⋅( 3− x+ 2x+ 3− 1)
=
= ( 1−2x) ⋅( − x+ 2x+ 3+ 3− 1)
=
= ( 1− 2x)( x+
5)
= − − − + ⋅ − ⋅ − − ⋅ − =
podcrtamo zajednički faktor
izlučimo Z.F.
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12.
( ) ( )
( ) 1( )
a) ac−bc− a+ b = c a−b −1
a− b =
= ca−b − a− b =
( a b)( c 1)
= − −
podcrtamo zajednički faktor, izlučimo Z.F.
13.
3 2 2 2 2 2 2 2
a) x − 2x + 3xy − 6y = x⋅x −2⋅ x + 3⋅x⋅y −2⋅ 3y
=
( 2) 3 ( 2)
( 2) 3 ( 2)
= ⋅ − + ⋅ − =
2 2
x x y x
= ⋅ − + ⋅ − =
2 2
x x y x
2 2
( x 2) ( x 3y
)
= − ⋅ +
podcrtamo zajednički faktor
izlučimo Z.F.
14.
2 3 2 2 3 2 1 2 2 2 1
a)
x y −xy z− z xy− z = x⋅x⋅y ⋅y − x⋅y ⋅z− z ⋅x⋅y− z ⋅ z =
( ) ( )
xy xy z z xy z
2 1 2 1
= − − ⋅ − =
( ) ( )
2 2
xy xy z z xy z
= − − ⋅ − =
2 2
( xy z)( xy z )
= − −
15.
( ) ( ) ( ) ( )
4 2 4 2
a) x 3 y y 3 y x 3 y y 3 y
− − − = − − − =
4 2
( 3 y)( x y )
( 3 )(( ) )
2 2
( 3 y)( x y)( x y)
= − − =
= − y x − y = prepoznaj: a −b
= − − +
2 2 2 2 2
16.
3 2 2 3 2 2 2 2
a)
a a
−
b− ab + b = a ⋅a−a ⋅b−a⋅ b + b ⋅ b =
( ) ( )
= − − − =
2 2
a a b b a b
2 2
( a b)( a b )
= − − =
( a b)( a b)( a b)
= − − + =
2
( a b) ( a b)
= − ⋅ +
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17.
( ) ( ) ( ) ⎡( ) ( )
3 3 3 3 3 3
3 3 2 2
a) 27x − 125y = 3 x − 5 y = 3x − 5y = 3x− 5y 3x + 3x⋅ 5y+ 5y
⎤ =
⎣
⎦
2 2
( 3x 5y)( 9x 15xy 25y
)
= − + +
( ) ( ) ⎡ ( ) ⎤ ⎡( ) ( )( ) ( )
3 3 2 2
b) x+ 3 − x− 1 = x+ 3− x−1 ⋅ x+ 3 + x+ 3 x− 1 + x− 1 ⎤
⎣ ⎦ =
⎣ ⎦
2 2 2
( x 3 x 1)( x 6x 9 x x 3x 3 x 2x
1)
2 2 2
4 ( x x x 6x x 3x 2x
9 3 1)
2
4 ( 3x
6x
7)
= + − + + + + − + − + − + =
= ⋅ + + + − + − + − + =
= ⋅ + +
18) a)
a = 2
2
m+ n = b⎫ m+ n = − 5⎫ m+ n = −5⎫
2x − 5x+ 2 = b =−5 ⇒ ⎬⇒ ⎬⇒ ⎬⇒ m =− 4, n =−1
mn ⋅ = ac ⋅ ⎭ mn ⋅ = 2 ⋅ 2⎭ mn ⋅ = 4 ⎭
c = 2
m =− 4 , n =−1
2 2
2x 5x 2 2x 4x 1x
2
− + = − − + =
2x( x 2) 1( x 2)
2x( x 2) 1( x 2)
( x 2)( 2x
1)
= − − − =
= − − − =
= − −
18.
a = 1
2
m+ n = b⎫
m+ n = 13 ⎫ m+ n = 13⎫ b) x + 13x+ 12 = b = 13 ⇒ ⎬⇒ ⎬⇒
⎬ ⇒ m = 12, n = 1
mn ⋅ = ac ⋅ ⎭ mn ⋅ = 1⋅12⎭
mn ⋅ = 12
c = 12
⎭
m = 12 , n = 1
+ 13 + 12 = + 12 + 1 + 12 =
2 2
x x x x x
x( x 12) 1( x 12)
x( x 12) 1( x 12)
( x 12)( x 1)
= + + + =
= + + + =
= + +
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19.
3 2 3 2 3
a) a − 9a + 27a− 27 = a −3⋅3⋅ a + 3⋅9⋅a− 3 =
( )
3 2 2 3
= a − ⋅a ⋅ a+ ⋅a⋅ − = a−
3 3 3 3 3
↓ ↓↓ ↓ ↓↓↓ ↓ ↑ ↑
( )
3 2 2 3
a −3⋅a ⋅ b+ 3⋅a⋅b − b = a−b
ili drugi način:
3 2 3 2 3
a − 9a + 27a− 27 = a − 9a + 27a− 3 =
3 3
= a −3 −9⋅a⋅ a+ 3⋅9⋅ a =
2 2
( a 3) ( a a 3 3 ) 9a ( a 3)
2
= ( a−3) ⋅ ( a + 3a+ 9− 9a)
=
2
= ( a −3) ⋅ ( a + 3a− 9a+ 9)
=
2
= ( a−3) ⋅( a − 6a+ 9)
=
= − ⋅ + ⋅ + − ⋅ − =
( a 3) ( a 3) ( a 3)
( a 3)
1 2 1+
2
= − ⋅ − = − =
= −
3
3
3
3 3 2 2 2 3 3 3 3 2 2 2 3 3
b) 27ab − 54abc+ 36abc − 8c = 3⋅a ⋅b −329 ⋅ ⋅ ⋅a ⋅b ⋅ c+ 334 ⋅ ⋅ ⋅abc ⋅ ⋅ −2⋅ c =
ili drugi način:
2 2 2 2 2
( 3 ab) 3 3 a b 2 c 3 3 ab2 c ( 2 c)
( 3ab) 3 ( 3ab) 2c 3 3ab ( 2c) ( 2c)
( 3ab
2c)
3 3
= ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ =
3 2 2 3
= − ⋅ ⋅ + ⋅ ⋅ − =
= −
ab − abc+ abc − c = ⋅a ⋅b − ⋅c − ⋅ ⋅aabbc ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅abcc
⋅ ⋅ ⋅ =
3 3 2 2 2 3 3 3 3 3 3
27 54 36 8 3 2 3 18 2 18
3
3 3
( 3ab) ( 2c) 18 a b c ( 3 a b 2 c)
= − − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ =
2 2
( ) ( )
( ) ( ) ( )
= 3ab −2c ⋅ 3ab + 3ab⋅ 2c + 2c −18abc⋅ 3ab −2c
2 2 2
( 3ab 2c) ( 9a b 6abc 4c 18abc)
= − ⋅ + + − =
(
2
) =
(
2 2
)
2
= ( 3ab −2c) ⋅ ( 3ab)
− 12abc
+( 2c)
( 3ab 2c) ( 3ab) 2 3ab 2c ( 2c)
= − ⋅ − ⋅ ⋅ + =
( 3ab 2c) ( 3ab 2c)
1 2
= − ⋅ − =
+
( 3ab 2c) ( 3ab 2c)
1 2 3
= − = −
=
Drugi i treći ispit znanja iz matematike za prvi razred gimnazije , tehničkih škola i svih ostalih škola …
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ALGEBARSKI IZRAZI – test br.2. i 3.
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Drugi i treći ispit znanja iz matematike za prvi razred gimnazije , tehničkih škola i svih ostalih škola …
Autori: Ivana i Mladen Sraga
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ALGEBARSKI IZRAZI – test br.2. i 3.
www.mim-sraga.com
Novo MALA ŠKOLA MATEMATIKE 1 na
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UČIMO ZAJEDNO
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Drugi i treći ispit znanja iz matematike za prvi razred gimnazije , tehničkih škola i svih ostalih škola …
Autori: Ivana i Mladen Sraga
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