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\(a)\)\(3x^2-6xy+3y^2-12\)
\(=\)\(3\left(x^2-2xy+y^2\right)-12\)
\(=\)\(3\left(x-y\right)^2-12\)
\(=\)\(3\left[\left(x-y\right)^2-4\right]\)
\(=\)\(3\left(x-y-4\right)\left(x-y+4\right)\)
\(b)\)\(x^2+5x+6\)
\(=\)\(\left(x^2+2x\right)+\left(3x+6\right)\)
\(=\)\(x\left(x+2\right)+3\left(x+2\right)\)
\(=\)\(\left(x+2\right)\left(x+3\right)\)
Chúc bạn học tốt ~
a) 3x2-6xy+3y2-12=3(x2-2xy+y2)-12=3(x-y)2-12=3[(x-y)2-4]=3(x-y-2)(x-y+2)
b)x2+3x+2x+6=x(x+3)+2(x+3)=(x+3)(x+2)
a: \(=\left(x+2-y\right)\left(x+2+y\right)\)
c: \(=\left(x-y\right)^2\)
a , 3x2 + 3y2 - 6xy - 12
= 3 ( x2 + y2 - 2xy - 4 )
= 3 ( x - y )2 - 22
= 3 ( x - y + 2 ) ( x - y - 2 )
3x^2 +3y^2 -6xy -12
=3(x^2 - 2xy +y^2 - 2^2 )
=3 (x-y)^2 - 2^2
=3(x-y-2)(x-y+2)
3(x+y) -(x^2+2xy+y^2)
=3(x+y) -(x+y)^2
(x+y)(3-x-y)
a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)
b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)
\(=\left(2x+y\right)\left(4x^2-8xy+y^2\right)\)
c) \(\left(3x+2\right)^2-2\left(x-1\right)\left(3x+2\right)+\left(x-1\right)^2\)
\(=[\left(3x+2\right)-\left(x-1\right)]^2=\left(3x+2-x+1\right)^2=\left(2x+3\right)^2=\left(2x+3\right)\left(2x+3\right)\)
k) = x( 2x - 1 ) - 3y( 2x - 1 ) = ( 2x - 1 )( x - 3y )
l) = x( x - y ) + 5( x - y ) = ( x - y )( x + 5 )
m) = ( a2 - 4a + 4 )( a2 + 4a + 4 ) = ( a - 2 )2( a + 2 )2
n) = y2( x2 - 1 ) - ( x2 - 1 ) = ( x - 1 )( x + 1 )( y - 1 )( y + 1 )
q) = 3[ ( x - y )2 - 4z2 ] = 3( x - y - 2z )( x - y + 2z )
a.\(xz+yz-5\left(x+y\right)\)
\(=z\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(z-5\right)\)
b.\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?
d.\(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)'
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
Trả lời:
a, xz + yz - 5 ( x + y )
= ( xz + yz ) - 5 ( x + y )
= z ( x + y ) - 5 ( x + y )
= ( x + y ) ( z - 5 )
b, 3x2 - 3xy - 5x + 5y
= ( 3x2 - 3xy ) - ( 5x - 5y )
= 3x ( x - y ) - 5 ( x - y )
= ( x - y ) ( 3x - 5 )
c, x2 + 6x - y2 - 3z2
= - ( 3x2 - x2 + y2 - 6x )
d, 3x2 + 6xy + 3y2 - 3z2
= 3 ( x2 + 2xy + y2 - x2 )
= 3 [ ( x2 + 2xy + y2 ) - z2 ]
= 3 [ ( x + y )2 - z2 ]
= 3 ( x + y - z ) ( x + y + z )
6xy + 48x2y2 - 3y2 - 3x2
=3(16x2y2-x2+2xy-y2)
=3[(4xy)2-(x-y)2]
=3[(4xy+(x-y)]*[(4xy)-(x-y)]
=3(4xy+x-y)(4xy-x+y)
a) x2 +x -y2 + y = ( x2 -y2 ) +(x+y)
= (x-y)(x+y) +(x+y)
=(x+y)( x-y+1)
b) 3x2 +3y2 -6xy -12 = 3(x2 +y2 - 2xy) -12
=3 [ (x-y)2 -4]
= 3( x-y-2)(x-y+2)
a) x2 + x - y2 + y
= (x2 - y2) + (x + y)
= (x + y) (x - y) + (x + y)
= x + y
b) 3x2 + 3y2 - 6xy - 12
= 3 (x2 + y2 - 2xy - 4)
= 3 [(x2 - 2xy + y2) - 4]
= 3 [(x - y)2 - 22]
= 3 (x - y + 2) (x - y - 2)
(sai thì thôi)