\(\begin{cases}x-1\ge 0\\x^2-9>0\end{cases}\)
\((*) x-1\ge 0↔x\ge 1\)
\( (*)x^2-9>0\\↔(x-3)(x+3)>0\\↔\left[\begin{array}{1}\begin{cases}x-3>0\\x+3>0\end{cases}\\\begin{cases}x-3<0\\x+3<0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}x>3\\x>-3\end{cases}\\\begin{cases}x<3\\x<-3\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}x>3\\x<-3\end{array}\right.\)
Từ hai \( (*)→x>3\)
\(⇒\) TXĐ: \( D=(3;+\infty)\)
