Dado a função composta, temos: \begin{matrix} (f \circ g) (x) = {\dfrac{1+ 2 \left(\dfrac{x}{1+2x}\right)}{1 - \left( \dfrac{x}{1+2x}\right)^2}} \\ \\ (f \circ g) (x) = {\dfrac{\dfrac{(1+2x)+2x}{1+2x}}{\dfrac{(1+2x)^2-x^2}{(1+2x)^2}}} \\ \\ (f \circ g) (x) = { \dfrac{(1+4x)(1+2x)}{(1+x)(1+3x)}} \end{matrix}Assim,\begin{matrix} { \dfrac{(1+4x)(1+2x)}{(1+x)(1+3x)} > 0} \end{matrix}Isso quer dizer que,\begin{matrix} \fbox{$ x \ne -\dfrac{1}{4} \ \ , \ \ x \ne -\dfrac{1}{2} \ \ , \ \ x \ne -1 \ \ , \ \ x \ne -\dfrac{1}{3} $} \\ \\ Letra \ (A) \end{matrix}