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

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

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

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

c) \(y^2+x^2+2xy-16=x^2+2xy+y^2-16\)

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

\(a,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right).\left(x-y+z\right)\)

\(b,x^3+y^3+2x^2-2xy+2y^2=\left(x^3+y^3\right)+2\left(x^2-xy+y^2\right)=\left(x+y\right).\left(x^2-2xy+y^2\right)+2.\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right).\left(x+y+2\right)\)

a) \(x^2-y^2-3x+3y\)

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

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

b) \(2x+2y-x^2+y^2\)

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

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

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

c) \(x^2-16+y^2+2xy\)

\(=x^2+y^2+2xy-16\)

\(=\left(x+y\right)^2-16\)

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

a) \(x^2-y^2-3x+3y\)

\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)

b) \(2x+2y-x^2+y^2\)

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

c) \(x^2-16+y^2+2xy\)

\(=\left(x+y\right)\left(x-y\right)+2xy-16\)

a) 3(x-y) - (x-y)^2

 =(x-y)(3-x+y)

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

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

\(a,=3\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(3-x+y\right)\\ b,=\left(x+y\right)^2-4x^2y^2=\left(x-2xy+y\right)\left(x+2xy+y\right)\\ c,=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

a: Ta có: \(x^2-xy-3x+3y\)

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

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

b: Ta có: \(5x^2+5xy-x-y\)

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

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

c: Ta có: \(x^2-2xy+y^2-z^2\)

\(=\left(x-y\right)^2-z^2\)

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

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

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

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

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

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)

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

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

\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)

b: =xy-x-y+1

=x(y-1)-(y-1)

=(x-1)(y-1)

c: =(x-2y)^2-4y

\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)

d: =16-(x^2-2xy+y^2)

=16-(x-y)^2

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

a) (5x - 3y)(x - 3y)(x + 3y).

b) 3(a – b + c) ( x   +   6 y ) 2 .

c) (x-y-2m + n)(x-y + 2m-n)