a/ đkxđ: x khác 2
b/ \(A=\left(\dfrac{x}{x-2}-\dfrac{x^3}{x^3-8}\right):\dfrac{x^2+2x}{x^2+2x+4}=\dfrac{x\left(x^2+2x+4\right)-x^3}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{x^2+2x+4}{x\left(x+2\right)}=\dfrac{2x^2+4x}{x-2}\cdot\dfrac{1}{x\left(x+2\right)}=\dfrac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x\left(x-2\right)}\)
c/ A > 3
\(\Leftrightarrow\dfrac{2x}{x\left(x-2\right)}>3\)
\(\Leftrightarrow2x>3x^2-6x\)
\(\Leftrightarrow3x^2-8x< 0\)
\(\Leftrightarrow x\left(3x-8\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\3x-8< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\3x-8>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< \dfrac{8}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>\dfrac{8}{3}\end{matrix}\right.\end{matrix}\right.\)
=> \(0< x< \dfrac{8}{3}\)
Vậy.............
