Điều kiện :
\(\log_{\frac{1}{5}}\left(\log_5\frac{x^2+1}{x+3}\right)\ge0\)
\(\Leftrightarrow0< \log_{\frac{1}{5}}\left(\log_5\frac{x^2+1}{x+3}\right)\le1\)
\(\Leftrightarrow\log_51< \log_5\frac{x^2+1}{x+3}\le\log_55\)
\(\Leftrightarrow1< \frac{x^2+1}{x+3}\le5\)\(\Leftrightarrow\begin{cases}\frac{x^2-x-2}{x+3}>0\\\frac{x^2-5x-14}{x+3}\le0\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-3< x< -1\\x>2\end{array}\right.\) và \(\left[\begin{array}{nghiempt}x< -3\\-2\le x\le7\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-2\le x< -1\\2< x\le7\end{array}\right.\)
Vậy tập xác định là D = [-2;-1) U (2;7]
