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$.
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 $$\Leftrightarrow \frac{1}{2}\log_3\dfrac{(x-3)^2}{(x-2)^2}=\log_32 $$\Leftrightarrow \log_3\dfrac{|x-3|}{|x-2|}=\log_32 $$\Leftrightarrow \left| {\dfrac{x-3}{x-2}} \right|=2$$\Leftrightarrow x=\dfrac73$.
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=\dfrac
53$.