Đáp án:
$S =\{1;5\}$
Giải thích các bước giải:
$\log_3x.\log_3(2x-1)=2\log_3x\qquad \left(ĐK: x >\dfrac12\right)$
$\to \log_3x[\log_3(2x-1) - 2] = 0$
$\to \left[\begin{array}{l}\log_3x = 0\\\log_3(2x-1) = 2\end{array}\right.$
$\to \left[\begin{array}{l}x = 1\\2x-1 = 9\end{array}\right.$
$\to \left[\begin{array}{l}x = 1\\x= 5\end{array}\right.\quad (nhận)$
Vậy $S =\{1;5\}$
