Question

Phân tích đa thức thành nhân tử :
a, \(x^2-5x+5y-y^2\)
b, \(3x^2-6xy+3y^2-12z^2\)
c, \(x^2-2xy+y^2-xz+yz\)
d, \(x^2-x+2y-4y^2\)
e, \(x^6-y^6\)

Answer 1

a/ \(x^2-5x+5y-y^2=\left(x^2-y^2\right)-\left(5x-5y\right)=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

b/ \(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x^2-2xy+y^2\right)-\left(2x\right)^2\right]=3\left[\left(x-y\right)^2-\left(2x\right)^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)

c/ \(x^2-2xy+y^2-xz+yz=\left(x^2-2xy+y^2\right)-\left(xz-yz\right)=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

d/ \(x^2-x+2y-4y^2=\left(x^2-4y^2\right)-\left(x+2y\right)=\left(x+2y\right)\left(x-2y\right)-\left(x+2y\right)=\left(x+2y\right)\left(x-2y-1\right)\)

e/ \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

Answer 2

a) x² - 5x + 5y - y²

= ( x² - y² ) - ( 5x - 5y )

= ( x - y )( x + y ) - 5( x - y )

= ( x - y )( x + y - 5 )

b) 3x² - 6xy + 3y² - 12z²

= 3( x² - 2xy + y² - 4z² )

= 3[( x² - 2xy + y² ) - 4z² ]

= 3[( x - y )² - 4z² ]

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

c) x² - 2xy + y² - xz - yz

= ( x² - 2xy + y² ) - ( xz - yz )

= ( x - y )² - z( x - y )

= ( x - y )( x - y - z )

d) x² - x + 2y - 4y²

= ( x² - 4y² ) - ( x - 2y )

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

= ( x - 2y )(x + 2y - 1 )

e) x⁶ - y⁶

= ( x³ )² - ( y³ )²

= ( x³ - y³ )( x³ + y³ )

= ( x - y )( x² + xy + y² )( x + y )( x² - xy + y² )

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