1. Điều kiện $x >1.$ PT
$\Leftrightarrow \frac{1}{2}\log_3(x^2-5x+6)^2= \log_3(x-1)-\log_32+\frac{1}{2}\log_3(x-3)^2$
$\Leftrightarrow \frac{1}{2}\log_3(x-2)^2(x-3)^2= \frac{1}{2}\log_3(x-1)^2-\log_32+\frac{1}{2}\log_3(x-3)^2$
$\Leftrightarrow \frac{1}{2}\log_3(x-2)^2(x-3)^2= \frac{1}{2}\log_3(x-1)^2(x-3)^2-\log_32 $
$\Leftrightarrow \frac{1}{2}\log_3\dfrac{(x-1)^2}{(x-2)^2}=\log_32 $
$\Leftrightarrow \log_3\dfrac{|x-1|}{|x-2|}=\log_32 $
$\Leftrightarrow \left| {\dfrac{x-1}{x-2}} \right|=2$
$\Leftrightarrow x=\dfrac53$.