\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)\)

\(=\left(x+y-4\right)\left(x+y+3\right)\)

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)

= ( x²+2xy+y²⁾ -(x+y)-12

= (x+y)²-(x+y)-12

= (x+y)-12

Ta có: 

\(x^2+2xy+y^2-x-y-12=(x^2+2xy+y^2)-(x+y)-12\)

                                           \(=(x+y)^2-(x+y)-12 \)   \((*)\)

Đặt \(x+y=a\) 

 từ \((*)\Rightarrow a^2-a-12=(a^2+3a)-(4a+12)\) 

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

Thay \(a=x+y\)

\(\Rightarrow (x+y+3)(x+y-4)\)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

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

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

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

`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: 

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

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