a) \(3x^2-3xy-5x+5y\)

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

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left[x^2-\left(y+1\right)^2\right]\)

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

\(=x\left(x-1\right)^2\)

f) \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

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

d:

Sửa đề: 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)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

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

Xin lỗi, mjk làm lại nhé *(^o^)*

x² - 2xy + y² + 4x - 4y - 5

= (x - y)² + 4(x - y) - 5

Đặt x - y = t, ta có:

t² + 4t - 5

= t² - t + 5t - 5

= t(t - 1) + 5(t - 1)

= (t - 1)(t + 5)

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

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

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

Đặt x - y = t, ta có:

\(t^2+4t-5\)

\(=t^2+t-5t-5\)

\(=t\left(t+1\right)-5\left(t+1\right)\)

\(=\left(t+1\right)\left(t-5\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

x²-2xy+y²+4x-4y-5 = (x²-2xy+y²)+(4x-4y)-5

                               = (x+y)²+4(x-y)-5

                               = 4(x-y)-5

 Làm đại, không chắc lắm!