a/ $x^2y^2+xy+\dfrac{1}{4}\\=(xy)^2+2.xy.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\\=\left(xy+\dfrac{1}{2}\right)^2$
Vậy $x^2y^2+xy+\dfrac{1}{4}=\left(xy+\dfrac{1}{2}\right)^2$
b/ $x^3-6x^2+12x-8\\=x^3-3.x^2.2+3.x.4-2^3\\=x^3-3.x^2.2+3.x.2^2-2^3\\=(x-2)^3$
Vậy $x^3-6x^2+12x-8=(x-2)^3$
c/ $(x-y)^3+6(x-y)^2+12(x-y)+8\\=(x-y)^3+3.(x-y)^2.2+3.(x-y).4+2^3\\=(x-y)^3+3.(x-y)^2.2+3.(x-y).2^2+2^3\\=[(x-y)+2]^3\\=(x-y+2)^3$
Vậy $(x-y)^3+6(x-y)^2+12(x-y)+8=(x-y+2)^3$
