Решить неравенство с объяснением

$log_{x-1}\, \frac{x^2-6x}{2(x-3)} \leq 1\\\\ODZ:\; \; \left \{ {{x-1\ \textgreater \ 0\; ,\; \; x-1\ne 1} \atop { \frac{x^2-6x}{2(x-3)}\ \textgreater \ 0 }} \right. \; \left \{ {{x\ \textgreater \ 1\; ,\; x\ne 2} \atop { \frac{x(x-6)}{2(x-3)}\ \textgreater \ 0 }} \right. \; \left \{ {{x\ \textgreater \ 1\; ,\; x\ne 2} \atop {x\in (0,3)\cup (6,+\infty )}} \right. \\\\---(0)+++(3)---(6)+++\\\\x\in (1,2)\cup (2,3)\cup (6,+\infty )$

Метод рационализации:

$(x-1-1)\cdot (\frac{x^2-6x}{2(x-3)}-(x-1)) \leq 0\\\\(x-2)\cdot \frac{x^2-6x-2x^2+8x-6}{2(x-3)} \leq 0\\\\(x-2)\cdot \frac{-(x^2-2x+6)}{2(x-3)} \leq 0\\\\x^2-2x+6\ \textgreater \ 0\; ,\; \; tak\; \; kak\; \; D<0$

$- \frac{x-2}{x-3} \leq 0\; ,\; \; \; \frac{x-2}{x-3} \geq 0\\\\+++[\, 2\, ]---[\, 3\, ]+++\\\\x\in (-\infty ,2\, ]\cup [\, 3,+\infty )\\\\ \left \{ {{x\in (1,2)\cup (2,3)\cup (6,+\infty )} \atop {x\in (-\infty ,2\, ]\cup [\, 3,+\infty )}} \right. \; \; \; \Rightarrow \quad \underline {x\in (1,2)\cup (6,+\infty )}$