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3x2 + 6xy + 3y2 – 3z2
= 3.(x2 + 2xy + y2 – z2)
(Nhận thấy xuất hiện x2 + 2xy + y2 là hằng đẳng thức nên ta nhóm với nhau)
= 3[(x2 + 2xy + y2) – z2]
= 3[(x + y)2 – z2]
= 3(x + y – z)(x + y + z)
b, (\(x^2\) - \(xy\) ) + (\(x-y\))
= (\(x-y\)).\(x\) + (\(x-y\))
= (\(x-y\)).(\(x\) + 1)
c, \(x^2\) - 2\(x\) - 15
= (\(x^2\) - 2\(x\) + 1) - 16
= (\(x\) - 1)2 - 42
= (\(x-1-4\)).(\(x-1+4\))
= (\(x-5\)).(\(x+3\))
a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)
a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)
b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)
c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)
a: \(3x^2-6xy+3y^2-12x^2\)
\(=3\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\left[\left(x-y\right)^2-4x^2\right]\)
\(=3\left(x-y-2x\right)\left(x-y+2x\right)\)
\(=3\left(-x-y\right)\left(3x-y\right)\)
b: \(3x^2y^2-6x^2y^3+12x^2y^2\)
\(=3x^2y^2\left(1-2y+4\right)\)
\(=3x^2y^2\left(-2y+5\right)\)
c: Ta có: \(3x^2-3y^2+12x-12y\)
\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
a,3x2-6xy+3y2
= 3(x2- 2xy+ y2)
= 3(x- y)2
b,xy-9x+y-9
= (xy+ y)- (9x+ 9)
= y(x+ 1)- 9(x+ 1)
= (x+1)(y- 9)
Chúc bạn học tốt
a,\(3x^2-6xy+3y^2\)
=\(3\left(x^2-2xy+y^2\right)\)
=\(3\left(x-y\right)^2\)
b,xy-9x+y-9
=\(\left(xy+y\right)-\left(9x+9\right)\)
=\(y\left(x+1\right)-9\left(x+1\right)\)
=\(\left(x+1\right)\left(y-9\right)\)
a)\(=3x\left(x+2y\right)\)
c)\(=\left(x-7\right)\left(x-1\right)\)
b)\(=x\left(x-2y\right)+3\left(x-2y\right)=\left(x+3\right)\left(x-2y\right)\)
d)\(=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(a,3x^2+6xy=3x\left(x+2y\right)\\ c,x^2-8x+7=\left(x^2-x\right)-\left(7x-7\right)=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\\ b,x^2-2xy+3x-6y=\left(x^2+3x\right)-\left(2xy+6y\right)=x\left(x+3\right)-2y\left(x+3\right)=\left(x+3\right)\left(x-2y\right)\\ d,4x^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
sai rồi vì (x2+6xy+9) ko có y2 nên ko thể có hằng đẳng thức được
Trả lời:
a, 3x2y - 6xy = 3xy ( x - 2 )
b, x2 - y2 - 9x + 9y
= ( x2 - y2 ) - ( 9x - 9y )
= ( x - y )( x + y ) - 9 ( x - y )
= ( x - y )( x + y - 9 )
c, x3 - 6x2 - y2x + 9x
= x ( x2 - 6x - y2 + 9 )
= x [ ( x2 - 6x + 9 ) - y2 ]
= x [ ( x - 3 )2 - y2 ]
= x ( x - 3 - y )( x - 3 + y )
3x2y - 6xy = 3xy( x - 2 )
x2 - y2 - 9x + 9y = ( x - y )( x + y ) - 9( x - y ) = ( x - y )( x + y - 9 )
x3 - 6x2 - y2x + 9x = x( x2 - 6x - y2 + 9 ) = x[ ( x - 3 )2 - y2 ] = x( x - y - 3 )( x + y - 3 )