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)

a) \(xy+3x-7y-21\)
\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)\)
\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)\)
\(\Leftrightarrow\left(x-7\right)\left(y+3\right)\)

b) \(2xy-15-6x+5y\)
\(\Leftrightarrow\left(2xy-6x\right)-\left(15-5y\right)\)
\(\Leftrightarrow x\left(2y-6\right)-5\left(3-y\right)\)
\(\Leftrightarrow2x\left(y-3\right)+5\left(y-3\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(y-3\right)\)

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:

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

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

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

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

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

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

(x-y)^2+4(x-y)+4-9

(x-y+2)^2-9

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

thu thuy vut con( -5) di dau roi

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

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

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

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

`a,x^3 - 3x^2 + 1 - 3x`

`=x^3 + 1 - 3x^2 - 3x`

`=(x^3 + 1) - 3x(x+1)`

`=(x+1)(x^2 - x + 1) - 3x(x+1)`

`=(x+1)(x^2 - x + 1 - 3x)`

`=(x+1)(x^2 - 4x + 1)`

`b,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)`

`c,3x^2 -6xy + 3y^2 - 12z^2`

`=3(x^2 -2xy +y^2 - 4z^2)`

`=3[(x-y)^2 - (2z)^2]`

`=3(x-y-2z)(x-y+2z)`
 

a: =x^3+1-3x^2-3x

=(x+1)(x^2-x+1)-3x(x+1)

=(x+1)(x^2-x+1-3x)

=(x+1)(x^2-4x+1)

b: =x^2-2xy+y^2+4x-4y

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

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

c: =3(x^2-2xy+y^2-4z^2)

=3[(x-y)^2-4z^2]

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

a: =x^3+1-3x^2-3x

=(x+1)(x^2-x+1)-3x(x+1)

=(x+1)(x^2-x+1-3x)

=(x+1)(x^2-4x+1)

b: =x^2-2xy+y^2+4x-4y

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

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

c: =3(x^2-2xy+y^2-4z^2)

=3[(x-y)^2-4z^2]

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

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 )