Можете, пожалуйста помочь решить неравенство? 99 баллов!!!!!!!!

Решите задачу:

$\frac{log_{0,4}(x-2)}{x-6} \leq 0\; ,\; \; ODZ:\; \; \left \{ {{x-2\ \textgreater \ 0} \atop {x-6\ne 0}} \right. \; \left \{ {{x\ \textgreater \ 2} \atop {x\ne 6}} \right. \; x\in (2,6)\cup (6,+\infty )\\\\a)\; \; \left \{ {{log_{0,4}(x-2) \geq 0} \atop {x-6\ \textless \ 0}} \right. \; \; \; ili\; \; \; b)\; \; \left \{ {{log_{0,4}(x-2) \leq 0} \atop {x-6\ \textgreater \ 0}} \right. \\\\ a)\; \; \left \{ {{x-2 \leq 1} \atop {x\ \textless \ 6}} \right. \; \; \; \; \; ili\; \; \; b)\; \; \left \{ {{x-2 \geq 1} \atop {x\ \textgreater \ 6}} \right.$

$a)\; \; \; \left \{ {{x \leq 3} \atop {x\ \textless \ 6}} \right. \; \; \; \; ili\; \; \; \; b)\; \; \; \left \{ {{x \geq 3} \atop {x\ \textgreater \ 6}} \right. \\\\a)\; \; \; \; x\ \textless \ 6\; \; \; \; \; ili\; \; \; \; \; b)\; \; \; x\ \textgreater \ 6\\\\x\in (-\infty ,6)\cup (6,+\infty )\; \; \to \; \; x\ne 6\\\\ \left \{ {{x\ne 6} \atop {x\in (2,6)\cup (6,+\infty )}} \right. \\\\Otvet:\; \; x\in (2,6)\cup (6,+\infty )\\$