设y=x²+5x+2
则原式=y(y+1)-12
=y²+y-12
=(y+4)(y-3)
∴.(x^2+5x+2)(x^2+5x+3)-12=(x²+5x+2+4)(x²+5x+2-3)
=(x+2)(x+3)(x²+5x-1)
2,、设a=x-2 b=y-2 则(x-y)=a-b
∴原式=a³-b³-(a-b)³
=3ab(a-b)
∴原式=3(x-2)(y-2)(x-y)
设y=x²+5x+2
则原式=y(y+1)-12
=y²+y-12
=(y+4)(y-3)
∴.(x^2+5x+2)(x^2+5x+3)-12=(x²+5x+2+4)(x²+5x+2-3)
=(x+2)(x+3)(x²+5x-1)
2,、设a=x-2 b=y-2 则(x-y)=a-b
∴原式=a³-b³-(a-b)³
=3ab(a-b)
∴原式=3(x-2)(y-2)(x-y)