Question

Phân tích đa thức thành nhân tử:

\(x^2-2xy+y^2+4x-4y-5\)

Answer 1

Cách 1: \(x^2-2xy+y^2+4x-4y-5=\left(y^2-xy+y\right)+\left(-xy+x^2-x\right)+\left(-5y+5x-5\right)\)

\(=y\left(y-x+1\right)-x\left(y-x+1\right)-5\left(y-x+1\right)=\left(y-x+1\right)\left(y-x-5\right)\)

Cách 2: \(x^2-2xy+y^2+4x-4y-5=\left(x^2+y^2+2^2-2xy+4x-4y\right)-9\)

\(=\left(y-x-2\right)^2-3^2=\left(y-x-2-3\right)\left(y-x-2+3\right)=\left(y-x-5\right)\left(y-x+1\right)\)

Answer 2

X²-2xy+y²+4x-4y-5 =x²-xy-xy+y²+5x-x-5y+y-5 X.(x-y-1)-y.(x-y-1)+5(x-y-1) =(x-y-1).(x-y+5)

Answer 3

X² - 2xy + y² + 4x - 4y - 5

=[ ( x - y )²+4 ( x - y ) + 4 ] - 9

= ( x - y + 2 )² - 9 

= ( x - y - 1 )( x - y + 5 )