4.3 - mathonline

4 – Breuken met letters
4.3 – Vermenigvuldigen
7 Bij het vermenigvuldigen van breuken doe je teller keer teller en noemer keer noemer.
a x y x ⋅ y yx x
⋅ =
=
=
y z y ⋅ z yz z
b
x
x
x⋅ x
x2
is niet te vereenvoudigen
⋅
=
=
x + 1 x + 1 ( x + 1) ⋅ ( x + 1) ( x + 1)2
c
x x − 1 x ⋅ ( x − 1) x − 1
⋅
=
=
x +1 x
( x + 1) ⋅ x x + 1
d
e
f
x⋅
x
x⋅x
x2
=
=
x +1 x +1 x +1
x⋅
x + 1 x ⋅ ( x + 1)
=
= x +1
x
x
( x + 1) ⋅
x − 1 ( x + 1) ⋅ ( x − 1) x 2 − 1
=
=
x
x
x
g a b c a ⋅ b ⋅ c abc
⋅ ⋅ =
=
x y z x ⋅ y ⋅ z xyz
h x2
⋅
yz 2
i
xy x 2 ⋅ xy x 3 y x3
= 2 = 3 = 3
z
yz ⋅ z z y z
(
)
2
2
x + xy y + 1 ( x + xy ) ⋅ ( y + 1) xy + x + xy 2 + xy x ⋅ y + 2 y + 1
y 2 + 2 y + 1 ( y + 1)
⋅
=
=
=
=
=
x
x +1
x ⋅ ( x + 1)
x ( x + 1)
x ( x + 1)
x +1
x +1
j x x +1 2
( )
yz
⋅
2
x ( x + 1) ⋅ y 2 z
2
y2 z
x −1
=
(
)
yz ⋅ x 2 − 1
=
xy 2 z ( x + 1)( x + 1)
yz ( x + 1)( x − 1)
=
xy ( x + 1)
x −1
k ab2 ax a 2b 2 x a 2 b
⋅
=
= 2
xy by bxy 2
y
l x2 + x
y
2
⋅
y
2
x −1
=
( x2 + x) y
2
2
y ( x − 1)
=
x ( x + 1) y
( x + 1)( x − 1) y
2
=
x
x
of
( x − 1) y
xy − y
m ( p − 1) 2 q 2 − 1 ( p − 1) 2 ( q 2 − 1) ( p − 1) 2 ( q − 1)(q + 1)
⋅
=
=
= ( p − 1)( q + 1)
( p − 1)( q − 1)
( p − 1)(q − 1)
q −1
p −1
n ( x + 1) 2 ( y + 1)
x2 y3
⋅
xy
xy ( x + 1)2 ( y + 1) ( x + 1) 2
=
=
y +1
x 2 y 3 ( y + 1)
xy 2
o x3 − 4 x 1 − 4 y 2 ( x3 − 4 x )(1 − 4 y 2 ) x ( x 2 − 4)(1 − 4 y 2 ) x ( x + 2)( x − 2)(1 + 2 y )(1 − 2 y )
⋅ 2
=
=
=
=
2y
2 yx( x + 2)
2 xy ( x + 2)
x + 2x
2 y ( x 2 + 2 x)
( x − 2)(1 + 2 y )(1 − 2 y )
( x − 2)(1 − 4 y 2 )
of
2y
2y
© Noordhoff Uitgevers
Uitwerkingen
1
4 – Breuken met letters
8 Bedenk dat delen door een breuk hetzelfde is als vermenigvuldigen met het omgekeerde.
a
x:
x
y x⋅ y
= x⋅ =
=y
y
x
x
b x
x x x 1 x ⋅1
x 1
:x = : = ⋅ =
=
=
y
y 1 y x y ⋅ x xy y
c
x2 :
y
x x 2 ⋅ x x3
= x2 ⋅ =
=
x
y
y
y
d x y x z x ⋅ z xz
: = ⋅ =
=
y z y y y ⋅ y y2
e
x + 1 x − 1 x + 1 x + 1 ( x + 1) ⋅ ( x + 1) ( x + 1)
:
=
⋅
=
=
x x +1
x x −1
x ⋅ ( x − 1)
x ( x − 1)
2
f a 2 bc ac a 2 bc x a 2bc ⋅ x ab
:
=
⋅
=
=
xy
x
xy ac
xy ⋅ ac
y
g
x
x + 1 ( x + 1) ⋅ ( x + 1) ( x + 1)
= ( x + 1) ⋅
=
=
( x + 1) :
x +1
x
x
x
2
h 2a + 3
2a + 3 ( 4a + 12 ) 2a + 3
1
2a + 3
2a + 3
: ( 4 ( a + 3) ) =
:
=
⋅
=
=
4a
4a
1
4a 4a + 12 4a ⋅ ( 4a + 12 ) 16a ( a + 3)
i p+q p+q p+q
( p + q) ⋅ q = q
q
:
=
⋅
=
p
q
p
p + q p ⋅ ( p + q) p
j a−2
b 2c
:
a2 + 4
a − 2 b ( a + 2 ) ( a − 2 ) ⋅ b ( a + 2 ) b ( a − 2 )( a + 2 ) ( a − 2 )( a + 2 )
= 2 ⋅
=
=
=
b ( a + 2) b c a2 + 4
b2 c ⋅ a 2 + 4
b2 c a 2 + 4
bc a 2 + 4
(
)
(
)
(
)
(
)
k x x −3 x x +2
(
) (
) x ( x − 3) 2 x − 6 x ( x − 3) ⋅ ( 2 x − 6 ) 2 x ( x − 3)( x − 3) 2 ( x − 3)2
:
=
⋅
=
=
=
( x + 2 ) 2 x − 6 ( x + 2 ) x ( x + 2 ) ( x + 2 ) ⋅ x ( x + 2 ) x ( x + 2 )( x + 2 ) ( x + 2 )2
l p
 p
  p 2q   p 2q  p + 2q p − 2 q p + 2 q
( p + 2q ) ⋅ q = p + 2q
q
=
⋅
=
:
 + 2 :  − 2 =  +
: −  =
q
q
q
p − 2 q q ⋅ ( p − 2 q ) p − 2q
q
 q
 q q  q q 
Het kan korter door direct teller en noemer met q te vermenigvuldigen.
m 2   2   x 2  x 2   x − 2   x + 2  x − 2 x
x ( x − 2) x − 2
⋅
=
=
1 −  : 1 +  =  −  :  +  = 
:
=
x 
x  x x  x x  x   x 
x x + 2 x( x + 2) x + 2

Het kan korter door direct teller en noemer met x te vermenigvuldigen.
n x2 + 2 x
xy
x 2 + 2 x ( x + 2) z x ( x + 2)( x + 2) z ( x + 2) 2
:
=
⋅
=
=
( x + 2) z
yz
yz
xy
xy 2 z
y2
o p 2 − p − 12
p 2 − p − 12 1
p 2 − p − 12
( p − 4)( p + 3) p − 4
: ( p + 3) =
⋅
=
=
=
p −3
p−3
p + 3 ( p − 3)( p + 3) ( p − 3)( p + 3) p − 3
© Noordhoff Uitgevers
Uitwerkingen
2
4 – Breuken met letters
p y2 − 2 y − 8
y−4
y 2 − 2 y − 8 x( y 2 + 6 y − 7) x ( y 2 − 2 y − 8)( y 2 + 6 y − 7)
:
=
⋅
=
=
x( y + 7) x ( y 2 + 6 y − 7)
x ( y + 7)
y−4
x( y + 7)( y − 4)
x( y + 2)( y − 4)( y + 7)( y − 1)
= ( y + 2)( y − 1)
x ( y + 7)( y − 4)
q x
 x
  x xy   x y 2  x + xy x + y 2 x + xy
y
yx (1 + y ) x (1 + y )
+
x
:
+
y
:
=
⋅
=
=
 =

 
 =  +  :  +
2
y
y
y
x+ y
y( x + y 2 ) x + y 2
y
 y
 y y  y y 
Het kan korter door direct teller en noemer met y te vermenigvuldigen.
r x
 x
  x xy   x xy  x + xy x − xy x + xy
y
y ( x + xy ) x + xy
=
⋅
=
=
=
:
 + x : − x =  +  : −  =
y
y
y
x − xy y ( x − xy ) x − xy
y
 y
 y y y y 
x (1 + y ) 1 + y
=
x (1 − y ) 1 − y
Het kan korter door direct teller en noemer met y te vermenigvuldigen.
9 a x 2 xy x
x 2 ⋅ xy x
x3 y y 2 z x3 y ⋅ y 2 z x 3 y 3 z x 2 y 2
⋅ : 2 =
=
=
=
: 2 = 2⋅
yz z y z
yz ⋅ z y z yz
x
z
yz 2 ⋅ x
xyz 2
b  2a − 4 a ( b − 1)  a
 ( 2a − 4 ) ⋅ a ( b − 1)  a
( 2a − 4 ) ⋅ a ( b − 1) ⋅ b + 1
⋅
=
=
:


:
a − 2  b + 1  ( b + 1) ⋅ ( a − 2 )  b + 1
( b + 1) ⋅ ( a − 2 ) a
 b +1
=
( 2a − 4 ) ⋅ a ( b − 1) ⋅ ( b + 1) = ( 2a − 4 )( b − 1) = 2 ( a − 2 )(b − 1) = 2 b − 1 = 2b − 2
( )
(b + 1) ⋅ ( a − 2 ) ⋅ a
( a − 2)
( a − 2)
c x 2  xy x  x 2  xy ⋅ x  x 2 x 2 y x 2 y 2 z 2 x 2 ⋅ y 2 z 2 x 2 y 2 z 2
: ⋅
:
:
=
⋅
=
= 2 2 =z
=
=
yz  z y 2 z  yz  z ⋅ y 2 z  yz y 2 z 2 yz x 2 y
yz ⋅ x 2 y
x y z
d  a −1

 a − 1 b ( a + 1) 
( a − 1) ⋅ b ( a + 1) ⋅ c
c
c
c
:
=
⋅
=

 ⋅ 2
⋅ 2
c
bc ⋅ c
a2 + 1
 a +1
 bc b ( a + 1)  a + 1  bc
( a − 1) ⋅ b ( a + 1) ⋅ c ( a − 1)( a + 1)
=
=
bc ⋅ c ⋅ a 2 + 1
c a2 + 1
(
© Noordhoff Uitgevers
)
(
)
Uitwerkingen
3

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