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a) 3( x - y ) - 5x( y - x )
= 3( x - y ) - 5x[ -( x - y ) ]
= 3( x - y ) + 5x( x - y )
= ( 3 + 5x )( x - y )
b) x3 + 2x2y + xy2 - 9x
= x( x2 + 2xy + y2 - 9 )
= x[ ( x + y )2 - 32 ]
= x( x + y - 3 )( x + y + 3 )
c) 14x2y - 21xy2 + 28x2y2
= 7xy( 2x - 3y + 4xy )
Bài giải
\(a,\text{ }3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
\(b,\text{ }x^3+2x^2y+xy^2-9x\)
\(=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left[\left(x+y\right)^2-3^2\right]\)
\(=x\left(x+y+3\right)\left(x+y-3\right)\)
\(c,\text{ }14x^2y-21xy^2+28x^2y\)
\(=7xy\left(2x-3y+4x\right)\)
\(=7xy\left(6x-3y\right)\)
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)
a) \(x^3-3x+1-3x^2=\left(x^3+1\right)-\left(3x^2+3x\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)=\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(2x^2+4x+2-2y^2=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x+1+y\right)\left(x+1-y\right)\)
a)x(x2+2xy+y2-4)
=x[(x+y)2-22 ]
=x(x+y-2)(x+y+2)
b)x4+4=x4+4x2+4-4x2=(x2+2)2-4x2
=(x2+2-2x)(x2+2+2x)
\(x^3+2x^2y+xy^2-4x=x\)\(\left(x^2+2xy+y^2-4\right)\)
\(=x\left[\left(x+y\right)^2-4\right]\)
\(=x\left(x+y+2\right)\left(x+y-2\right)\)
\(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)