a, ĐKXĐ:\(x^3+8\ne0\Rightarrow x^3\ne-8\Rightarrow x\ne-2\)
b,\(D=\dfrac{2x^2-4x+8}{x^3+8}=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)
c, \(\dfrac{2}{x+2}=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2}\)
d, \(\dfrac{2}{x+2}>2\\ \Rightarrow2>2x+4\\ \Rightarrow2x+2< 0\\ \Rightarrow2x< -2\\ \Rightarrow x< -1\)