Полное решение!11!Решите неравенство:

$\frac{x^2-7x+6}{x-2} \leq 0\\\\ x^2-7x+6=(x-1)(x-6)\\\\\ \frac{(x-1)(x-6)}{x-2} \leq 0\\\\ x \neq 2\\\\$

0}} \rigaht. \\\\ " alt="1) \left \{ {{(x-1)(x-6) \geq 0} \atop {x-2<0}} \right. \\\\ 2) \left \{ {{(x-1)(x-6) \leq 0 } \atop {x-2>0}} \rigaht. \\\\ " align="absmiddle" class="latex-formula">
Первое неравенство
$1) \left \{ {{(x-1)(x-6) \geq 0} \atop {x-2<0}} \right. \\\\ 1) \left \{ {{x \leq 1 \ ; \ x \geq 6} \atop {x<2}} \right. \\\\ 1) \left \{ {{x \geq 1 ; x \leq 6} \atop {x<2}} \right.$
Получаем решение
$x \in (\infty;1)$

Второе неравенство

2}} \right. \\\\ \left \{ {{x \leq 1 ; \ \ x \geq 6} \atop {x>2}} \right. \\\\ \left \{ {{x \geq 1; x \leq 6} \atop {x<2}} \right. " alt=" \left \{ {{(x-1)(x-6) \leq 0 } \atop {x>2}} \right. \\\\ \left \{ {{x \leq 1 ; \ \ x \geq 6} \atop {x>2}} \right. \\\\ \left \{ {{x \geq 1; x \leq 6} \atop {x<2}} \right. " align="absmiddle" class="latex-formula">
Получаем решение
$x \in (2;6]\\$

Объединяя
$x \in (-\infty;1) \ \ \cup \ \ \ (2;6]$