Tìm tập xác định của các hàm số:
$\begin{array}{l}
1)y = \sqrt {{{\log }_{\frac{1}{3}}}(X - 3) - 1} \\
2)y = \sqrt {{{\log }_{\frac{1}{2}}}\frac{{X - 1}}{{X + 5}}} \\
3)y = \sqrt {{{\log }_{\frac{1}{5}}}\left( {{{\log }_5}\frac{{{X^2} + 1}}{{X + 3}}} \right)}
\end{array}$
$1$)    Hàm số xác định khi
\(
\begin{cases}X-3>0 \\ \log_{\frac{1}{3}}(X-3)-1\geq 0 \end{cases}\\
\Leftrightarrow \begin{cases}X>3 \\ X-3\leq \frac{1}{3} \end{cases}\\ \Leftrightarrow 3<X\leq \frac{10}{3} 
\)
Vậy: $D = \left( {3,\frac{{10}}{3}} \right]$

$2$)   
Lập điều kiện: $\left\{ \begin{array}{l}
  \frac{{X - 1}}{{X + 5}} > 0\\
 {\log _{\frac{1}{2}}}\frac{{X - 1}}{{X + 5}} \ge 0  \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} (X-1)(X+5)>0\\ \frac{X-1}{X+5}\leq \left ( \frac{1}{2} \right )^0=1\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l} X>1\\ X<-5 \end{array} \right. \\  \frac{X-1}{X+5} \leq 1 \end{array} \right.$
Giải hệ ta có $X > 1$ .
Vậy $D = (1, + \infty )$

$3$)    Hàm số xác định khi
\(
\begin{cases}log_{\frac{1}{5}}(log_{5}\tfrac{X^2+1}{X+3}) \geq 0\\ log_{5}\tfrac{X^2+1}{X+3}>0 \\ \frac{X^+1}{X+3}>0\end{cases}\Leftrightarrow  \left\{ \begin{array}{l}log_{5}\tfrac{X^2+1}{X+3}\leq \left (\frac{1}{5}\right )^0=1 \\ \tfrac{X^2+1}{X+3} >5^0=1\\ \frac{X^+1}{X+3}>0  \end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l} \frac{X^2+1}{X+3} \leq 5^1=5\\\frac{X^2+1}{X+3}>1\\ \frac{X+1}{X+3}>0   \end{array} \right. \\ 
\Leftrightarrow 1<\frac{X^2+1}{X+3} \leq 5 \Leftrightarrow \begin{cases}\frac{X^2-5X-14}{X+3}\leq 0\\ \frac{X^2-X-2}{X+3}>0 \end{cases}  \Leftrightarrow \begin{cases}\left[ \begin{array}{l}X<-3\\-2\leq X\leq 7\end{array} \right. \\ \left[ \begin{array}{l}-3<X<-1\\X>2\end{array} \right.\end{cases} \) 
Vậy tập xác định là $D = \left[ { - 2, - 1} \right) \cup \left( {2,7} \right]$


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