ALGEBARSKI IZRAZI

M-1-Kvadrat binoma
Rješenja i kompletni postupak
34.
( x y) ( y x)
2 2
77) − 3 + = − 3 =
( )
2 2
= y −2⋅y⋅3⋅ x+ 3x
=
2 2 2
= y − 6yx+ 3 x =
= y − 6yx + 9x ili = 9x − 6 xy + y
2 2 2 2
prvom i drugom članu zamjenimo mjesta
tako da drugi bude sa predznakom
( −)
...i dalje je lako...
obadva rj. su ispravna i ista...
Taj zadatak možemo rješiti i na još dva načina:
II način:
III način:
( )
2
( ) ( )
2
− + = − ⋅ − =
2 2 2
− + = − + ⋅ − ⋅ + =
77) 3x y 1 3x y 3x y 3x 2 3x y y
2 2
( ) ( )
2 2
1 (( 3x)
2 3x y y )
2 2 2
= − + =
3 x 6xy y
= 9x − 6xy+
y
2 2
( ) ( ) ( )
( )
2
= −1 ⋅ 3x− y = 2
2
= −3 x −2⋅3⋅x⋅ y+ y =
= ⋅ − ⋅ ⋅ + =
2 3 4 2 4 2 3 2
( − ab + c ) = ( c − ab)
=
4 2 4 2 3 2 3 2
( 2c ) 2 2 c 5 a b ( 5a b )
2 ( c ) 20a b c 5 ( a ) ( b )
78) 5 2 2 5
= − ⋅ ⋅ ⋅ ⋅ ⋅ + =
2 4 2 2 3 4 2 2 2 3 2
= − + ⋅ =
= 4c
−20a b c
42 ⋅
2 3 4
22 ⋅ 32 ⋅
+ 25 =
2 3 4 2 4 2 3 2
( − xy+ z) = ( z − xy)
=
( 3z ) 2 3 z 2 x y ( 2x y )
3 ( z ) 12z x y 2 ( x ) ( y )
79) 2 3 3 2
= 9x − 6xy+
y
2 2
Uvijek dobijemo isto rješenje...pa birajte...
na koji način će te rješavat ovakve zadatke
8 2 3 4 4 6 4 6 2 3 4 8
4 2 4 2 3 2 3 2
= − ⋅ ⋅ ⋅ ⋅ ⋅ + =
2 4 2 4 2 3 2 2 2 3 2
= − + ⋅ =
42 ⋅ 2 3 4 22 ⋅ 32 ⋅
= 9 − 12 + 4 =
a
b
= 4c − 20abc + 25ab ili = 25ab − 20abc + 4c
z x y z x y
= z− xyz + xy = xy − xyz +
8 2 3 4 4 6 4 6 2 3 4 8
9 12 4 ili 4 12 9
m n 2 n m 2
( − + ) = ( − ) =
n n m m
( 2 ) 2 2 2 ( 2 )
80) 2 2 2 2
2 2
= − ⋅ ⋅ + =
n⋅2 1 n m m⋅2
= 2 −2 ⋅2 ⋅ 2 + 2 =
2n 1+ n+
m 2m
= 2 − 2 + 2
2⋅ n 1+ n+ m 2⋅m
= 2 − 2 + 2
=
=
2
n
n+ m+
1 2
( 2 ) 2 ( 2 )
= − + =
m
Rješenje možemo ostaviti u ovom obliku...
n n+ m+
1 m
= 4 − 2 + 4 ili u ovom obliku, ovisi o tome kako radite u školi...
z
M.I.M.-Sraga -2003./2013.
www.mim-sraga.com
16
© M.I.M-Sraga centar za poduku
mim-sraga@zg.htnet.hr
tel-01-4578-431