Log₃(x-3)log₄(x-4) / x-5 [latex] \leq [/latex] log₄(x-3)log₃(x-4)/ x-6 Помогите

[latex]\frac{log_3(x-3)\cdot log_4(x-4)}{x-5} \leq \frac{log_4(x-3)\cdot log_3(x-4)}{x-6}\\\\ODZ:\; \left \{ {{x-3\ \textgreater \ 0,x-4\ \textgreater \ 0} \atop {x\ne 5,x\ne 6}} \right. \; \left \{ {{x\ \textgreater \ 4} \atop {x\ne 5,x\ne 6}} \right. \\\\\frac{log_3(x-3)\cdot \frac{log_3(x-4)}{log_34}}{x-5}-\frac{\frac{log_3(x-3)}{log_34}\cdot log_3(x-4)}{x-6} \leq 0\\\\\frac{log_3(x-3)\cdot log_3(x-4)\cdot (x-6-x+5)}{log_34\cdot (x-5)(x-6)} \leq 0\\\\log_34\ \textgreater \ 1\ \textgreater \ 0,\; \; \to \; \; \frac{log_3(x-3)\cdot log_3(x-4)}{(x-5)(x-6)} \geq 0[/latex] Метод рационализации: [latex]log_{a}f\; \to (a-1)(f-1)\\\\\frac{(3-1)\cdot (x-3-1)(3-1)(x-4-1)}{(x-5)(x-6)} \geq 0\\\\\frac{(x-4)(x-5)}{(x-5)(x-6)} \geq 0\\\\\frac{x-4}{x-6} \geq 0\\\\+++[4]---(6)+++\\\\x\in (-\infty,4]U(6,+\infty)[/latex] С учётом ОДЗ:  [latex]x\in (6,+\infty)[/latex] .