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a) x3 + x2y - x2z - xyz
= ( x3 + x2y ) - ( x2z + xyz )
= x2( x + y ) + xz( x + y )
= ( x + y )( x2 + xz )
= x( x + y )( x + z )
b) x2 - y2 + 6x + 9
= ( x2 + 6x + 9 ) - y2
= ( x + 3 )2 - y2
= ( x - y + 3 )( x + y + 3 )
c) x2 - 4xy - x + 2y + 4y2
= ( x2 - 4xy + 4y2 ) - ( x - 2y )
= ( x - 2y )2 - ( x - 2y )
= ( x - 2y )( x - 2y - 1 )
d) 18x3 - 12x2 + 3x - 2
= ( 18x3 - 12x2 ) + ( 3x - 2 )
= 6x2( 3x - 2 ) + ( 3x - 2 )
= ( 3x - 2 )( 6x2 + 1 )
e) a2 + 2ab + b2 - c2 + 2cd - d2
= ( a2 + 2ab + b2 ) - ( c2 - 2cd + d2 )
= ( a + b )2 - ( c - d )2
= ( a + b - c + d )( a + b + c - d )
f) xz - yz - x2 + 2xy - y2
= z( x - y ) - ( x2 - 2xy + y2 )
= z( x - y ) - ( x - y )2
= ( x - y )( z - x + y )
a) x3 + x2y - x2z - xyz
= ( x3 + x2y ) - ( x2z + xyz )
= x2( x + y ) + xz( x + y )
= ( x + y )( x2 + xz )
= x( x + y )( x + z )
b) x2 - y2 + 6x + 9
= ( x2 + 6x + 9 ) - y2
= ( x + 3 )2 - y2
= ( x - y + 3 )( x + y + 3 )
c) x2 - 4xy - x + 2y + 4y2
= ( x2 - 4xy + 4y2 ) - ( x - 2y )
= ( x - 2y )2 - ( x - 2y )
= ( x - 2y )( x - 2y - 1 )
d) 18x3 - 12x2 + 3x - 2
= ( 18x3 - 12x2 ) + ( 3x - 2 )
= 6x2( 3x - 2 ) + ( 3x - 2 )
= ( 3x - 2 )( 6x2 + 1 )
e) a2 + 2ab + b2 - c2 + 2cd - d2
= ( a2 + 2ab + b2 ) - ( c2 - 2cd + d2 )
= ( a + b )2 - ( c - d )2
= ( a + b - c + d )( a + b + c - d )
f) xz - yz - x2 + 2xy - y2
= z( x - y ) - ( x2 - 2xy + y2 )
= z( x - y ) - ( x - y )2
= ( x - y )( z - x + y )
a, \(x^3-6x^2+9x\)
\(=x\left(x^2-6x+9\right)\)
\(=x \left(x-3\right)\)
Câu b, c cũng tượng tự nha bn , dễ mà
#hoc_tot#
b) \(x^2-2xy+3x-6y=x\left(x-2y\right)+3\left(x-2y\right)=\left(x-2y\right)\left(x+3\right)\)
c)\(x^2-8x+7=x^2-x-7x+7=x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(x-7\right)\)
a)\(x^3-6x^2+9x=x\left(x^2-2\cdot x\cdot3+3^2\right)=x\left(x-3\right)^2\)
~ Chúc bạn học tốt ~
a) xy – 3x + 2y – 6
= (xy - 3x) + (2y - 6)
= x(y - 3) + 2(y - 3)
= (y - 3)(x + 2)
b) x2y + 4xy + 4y – y3
= y(x2 + 4x + 4 - y2)
= y[(x2 + 4x + 4) - y2]
= y[(x + 2)2 - y2]
= y(x + 2 + y)(x + 2 - y)
c) x2 + y2 + xz + yz + 2xy
= (x2 + 2xy + y2) + (xz + yz)
= (x + y)2 + z(x + y)
= (x + y)(x + y + z)
d) x3 + 3x2 – 3x – 1
= (x3 - 1) + (3x2 - 3x)
= (x - 1)(x2 + x + z) + 3x(x - 1)
= (x - 1)(x2 + 4x + 1)
a )
\(xy-3x+2y-6\)
\(=\left(xy+2y\right)-3x-6\)
\(=y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(y-3\right)\left(x+2\right)\)
b )
\(x^2y+4xy+4y-y^3\)
\(=y\left(x^2+4x+4-y^2\right)\)
\(=y\left[\left(x+2\right)^2-y^2\right]\)
\(=y\left(x+2-y\right)\left(x+2+y\right)\)
c )
\(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
a)18x2-12x
=3x(6x-4)
b)3x2-11x+6
=x(3x-11+6)
=x(3x-5)
c)x3+6x2+11x+6
=x2(x+23
\(18x^2-12x\)
\(=6x\left(3x-2\right)\)
\(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
a)\(x^2+2xy-6y-9=x^2-9+2y.\left(x-3\right)=\left(x-3\right)\left(x+3\right)+2y.\left(x-3\right)\)
\(=\left(x-3\right).\left(x+3+2y\right)\)
b) \(x^3+x^2-6x=x^3-6x+9+x^2-9\)
\(=x^3+3x^2-3x^2-9x+3x+9+\left(x-3\right)\left(x+3\right)\)
\(=x^2.\left(x+3\right)-3x.\left(x+3\right)+3.\left(x+3\right)+\left(x-3\right).\left(x+3\right)\)
\(=\left(x+3\right).\left(x^2-3x+3+x-3\right)\)
\(=\left(x+3\right).\left(x^2-2x\right)=x.\left(x+3\right).\left(x-2\right)\)
a, \(x^2+2xy-6y-9\)
\(=\left(x^2+2xy+y^2\right)-\left(y^2+6y+9\right)\)
\(=\left(x+y\right)^2-\left(y-3\right)^2\)
\(=\left(x+2y-3\right)\left(x+3\right)\)
b, \(x^3+x^2-6x\)\(=x\left(x^2+3x-2x-6\right)\)
\(=x\left(x+3\right)\left(x-2\right)\)
câu a là x2-32 +2xy-6y= [x-3].[x+3]+2y.[x-3]=[x-3].[x+3+2y] câu b là x.[x2 +x-6]=x.[x2-22+x-2]=x.[x-2].[x+3]
4x2 -6x= 2x(2x-3)
b) 3x3 -6x2y -24xy + 12x2 = \(3x\left(x^2-2xy-8y+4x\right)\)
c) x2 -25 + y2 + 2xy\(=x^2+2xy+y^2-25\)\(=\left(x+y\right)^2-5^2\)
=>\(\left(x+y+5\right)\left(x+y-5\right)\)
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
\(a.=x^2-2x\)
\(b.=x\left(x+2y\right)-3\left(x+2y\right)\)
\(=\left(x-3\right)\left(x+2y\right)\)
\(c.=x\left(x^2-x-12\right)\)
\(=x\left(x^2-4x+3x-12\right)\)
\(=x \left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=x\left(x+3\right)\left(x-4\right)\)