a, ĐKXĐ : \(\left\{{}\begin{matrix}x+2\ne0\\x^3+8\ne0\\x^2-4\ne0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne\pm2\\x^2-x+1\ne0\end{matrix}\right.\)
Mà \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=> \(x\ne\pm2\)
Ta có : \(E=\left(\frac{x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{x^2-2x+4}{x^2-4}\right):\left(\frac{1}{x+2}\right)\)
=> \(E=\left(\frac{x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{1}{x+2}\right)\)
=> \(E=\left(\frac{x\left(x+2\right)}{\left(x+2\right)^2}-\frac{x^2+2x+4}{\left(x+2\right)^2}\right):\left(\frac{1}{x+2}\right)\)
=> \(E=\frac{-4}{\left(x+2\right)^2}:\left(\frac{1}{x+2}\right)\)
=> \(-\frac{4}{\left(x+2\right)^2}.\left(x+2\right)=-\frac{4}{x+2}\)
b, - Để E > 0 thì : \(\frac{-4}{x+2}>0\)
=> \(x+2< 0\)
=> \(x< -2\)
