Đặt x+y=a ta có:

x²+2xy+y²-x-y-12=(x²+2xy+y²)-(x+y)-12=(x+y)²-(x+y)-12=a²-a-12=(a²+3a)-(4a+12)=a(a+3)-4(a+3)=(a+3)(a-4)=(x+y+3)(x+y-4).

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Ta có : 

      x² + 2xy + y² - x - y - 12 

=   (x+y)² - ( x + y ) - 12

Đặt :  x + y = a

=>  a² - a - 12 

=   a² + 3a - 4a - 12

= a . ( a + 3 ) - 4.( a + 3 )

= ( a - 4 ).( a + 3 )

=  ( x + y - 4 ). ( x + y - 3 ) 

x²+2xy+y²-x-y-12

= (x+y)²-(x+y)-12

đặt x+y=z. ta có:

z²-z-12

= z²-4z+3z-12

= z(z-4)+3(z-4)

= (z-4)(z+3)

thay x+y=z:

= (x+y-4)(x+y+3)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

`a. =4(x^2+4x+3)=4(x^2+3x+x+3)=4(x+3)(x+1)`

`b. =x^2+8x-7x-56=x(x+8)-7(x+8)=(x+8)(x-7)`

`c. =x^2-9x+8x-72=x(x-9)+8(x-9)=(x-9)(x+8)`

`d. =(x-y)^2-9=(x-y-3)(x-y+3)`

Bài 1: 

\(=3x^3y-6x^2y^2+15xy\)

Bài 2: 

\(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

\(x^2+2xy-25+y^2\\ =\left(x^2+2xy+y^2\right)-5^2\\ =\left(x+y\right)^2-5^2\\ =\left(x+y-5\right)\left(x+y+5\right)\)

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

\(9-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

= - (x² - 2xy + y² - 9) = - [ (x - y)² - 3² ] = - (x - y - 3).(x - y + 3)

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