Giải phương trình log_(3x)(3/x)+(log_3(x))^2=1

Điều kiện \(0 < x \ne \frac{1}{3}\)

\(\begin{array}{l} {\log _{3x}}\left( {\frac{3}{x}} \right) + \log _3^2x = 1 \Leftrightarrow \frac{{{{\log }_3}\left( {\frac{3}{x}} \right)}}{{{{\log }_3}3x}} + {\log _3}^2x = 1\\ \Leftrightarrow \frac{{1 - {{\log }_3}x}}{{1 + {{\log }_3}x}} + {\log _3}^2x = 1 \Leftrightarrow \left( {1 - {{\log }_3}x} \right)\left[ {1 - {{\left( {1 + {{\log }_3}x} \right)}^2}} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\log _3}x = 1\\ 1 + {\log _3}x = \pm 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 1\\ x = \frac{1}{9} \end{array} \right. \end{array}\)