$\begin{array}{l}a)\quad 9^x -7.3^x +1 0 =0 \\ \Leftrightarrow (3^x -2)(3^x - 5) = 0\\ \Leftrightarrow \left[\begin{array}{l}3^x = 2\\3^x = 5\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \log_32\\x = \log_35\end{array}\right.\\ Vậy\,\,S = \{\log_32;\log_35\}\\ b)\quad \log_3^2x +\log_{\tfrac13}x -6 =0\qquad (ĐK: x >0)\\ \Leftrightarrow \log_3^2x - \log_3x - 6 = 0\\ \Leftrightarrow (\log_3x +2)(\log_3x - 3) = 0\\ \Leftrightarrow \left[\begin{array}{l}\log_3x = -2\\\log_3x = 3\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac19\\x = 27\end{array}\right.\quad (nhận)\\ Vậy\,\,S = \left\{\dfrac19;27\right\} \end{array}$
