Phân tích đa thức thành nhân tử
\(5xy^3-2xyz-15y+6z\)
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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 )
\(x^2y-16y=y\left(x^2-16\right)=y\left(x-4\right)\left(x+4\right)\)
\(x^2-9+5xy-15y=\left(x-3\right)\left(x+3\right)+5y\left(x-3\right)=\left(x-3\right)\left(x+3+5y\right)\)
Lời giải:
$A=x^2-8xy+15y^2=x^2-3xy-(5xy-15y^2)$
$=x(x-3y)-5y(x-3y)=(x-3y)(x-5y)$
$B=2x^2-5xy+2y^2=(2x^2-4xy)-(xy-2y^2)$
$=2x(x-2y)-y(x-2y)=(2x-y)(x-2y)$
$C=2x^2-3y^2-xy=(2x^2+2xy)-(3y^2+3xy)$
$=2x(x+y)-3y(y+x)=(x+y)(2x-3y)$
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) \(x^2-2xy-15y^2\)
\(=x^2-2xy+y^2-16y^2\)
\(=\left(x-y\right)^2-\left(4y\right)^2\)
\(=\left(x-y-4y\right)\left(x-y+4y\right)\)