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,\(5xy^3-2xyz-15y^2+6z\)

\(=\left(5xy^3-15y^2+6z-2xyz\right)\)

\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)\)

\(=\left(5y^2-2z\right)\left(xy-3\right)\)

a) 5xy³ - 2xyz - 15y² + 6z

= ( 5xy³ - 15y² ) - ( 2xyz - 6z )

= 5y²( xy - 3 ) - 2z( xy - 3 )

= ( xy - 3 )( 5y² - 2z )

b) ab³c² - a²b²c² + ab²c³ - a²bc³

= abc²( b² - ab + bc - ac )

= abc²[ ( b² - ab ) + ( bc - ac ) ]

= abc²[ b( b - a ) + c( b - a ) ]

= abc²( b - a )( b + c )

Lời giải:

$A=x^2-8xy+15y^2=x^2-3xy-(5xy-15y^2)$

$=x(x-3y)-5y(x-3y)=(x-3y)(x-5y)$

$B=2x^2-5xy+2y^2=(2x^2-4xy)-(xy-2y^2)$

$=2x(x-2y)-y(x-2y)=(2x-y)(x-2y)$

$C=2x^2-3y^2-xy=(2x^2+2xy)-(3y^2+3xy)$

$=2x(x+y)-3y(y+x)=(x+y)(2x-3y)$

\(x^2y-16y=y\left(x^2-16\right)=y\left(x-4\right)\left(x+4\right)\)

\(x^2-9+5xy-15y=\left(x-3\right)\left(x+3\right)+5y\left(x-3\right)=\left(x-3\right)\left(x+3+5y\right)\)

1) \(5x^2-5xy-9x+9y\)

\(=5x\left(x-y\right)-9\left(x-y\right)\)

\(=\left(5x-9\right)\left(x-y\right)\)

2) \(m^3+4m^2+3m\)

\(=m^3+3m^2+m^2+3m\)

\(=m^2.\left(m+3\right)+m\left(m+3\right)\)

\(=\left(m^2+m\right)\left(m+3\right)\)

\(=m\left(m+1\right)\left(m+3\right)\)

3) \(x^2-2xy-15y^2\)

\(=x^2-2xy+y^2-16y^2\)

\(=\left(x-y\right)^2-\left(4y\right)^2\)

\(=\left(x-y-4y\right)\left(x-y+4y\right)\)

nghngnngngn

a/ = xy(5y - 6x)

b/ = - (15y² - 2x + 5y - 10xy)