x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử

= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung

= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung

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

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

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 )

\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)

\(=\left(x-y+2\right)^2-9\)

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

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

nhớ nha

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

\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)

\(=\left(x-y+2\right)^2-9\)

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

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

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

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

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

\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)

\(=\left(x-y-1\right)\left(x-y+1+4\right)\)

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

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

\(=\left(x-y\right)^2+4\left(x-y\right)+4-9\)

\(=\left(x-y+2\right)^2-9\)

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

$x(x+y)+4x+4y$

$=x(x+y)+4(x+y)$

$=(x+y)(x+4)$

x^2+4x-2xy-4y+y^2=(x^2-2xy+y^2)+(4x-4y)

                               =(x-y)^2+4(x-y)

                               =(x-y)(x-y+4)