Question

$log_{x - 1}( {x}^{2} - 12x + 36 ) \leqslant 0$
​

Answer 1

$\log_{x - 1}( {x}^{2} - 12x + 36 ) \leq0\\\log_{x-1}(x^2-2*6*x+6^2)\leq 0\\\log_{x-1}((x-6)^2)\leq 0\\2\log_{x-1}(|x-6|)\leq 0\\\log_{x-1}(|x-6|)\leq 0$

одз:

$\left \{ {{x-1>0} \atop {x\neq 6}} \right. \Rightarrow \left \{ {{x\in (1;+\infty)} \atop {x\neq 6}} \right. \Rightarrow x\in (1;6)\cup (6;+\infty)$

решаем неравенство:

1)

$0<x-1<1\\1<x<2\\x\in (1;2)$

$|x-6|\geq (x-1)^0\\|x-6|\geq 1\\\left[\begin{array}{cc}x-6\geq 1\\x-6\leq -1\end{array}\right. \Rightarrow \left[\begin{array}{cc}x\geq 7\\x\leq 5\end{array}\right.\Rightarrow x\in (-\infty;5]\cup [7;+\infty)$

$\left \{ {{x\in (-\infty;5]\cup [7;+\infty)} \atop {x\in (1;2)}} \right. \Rightarrow x \in(1;2)$

2)

$x-1>1\\x>2\\x\in (2;+\infty)$

$|x-6|\leq (x-1)^0\\|x-6|\leq 1\\\left \{ {{x-6\leq 1} \atop {x-6\geq -1}} \right. \Rightarrow \left \{ {{x\leq 7} \atop {x\geq 5}} \right. \Rightarrow x\in[5;7]$

$\left \{ {{x\in [5;7]} \atop {x\in (2;+\infty)}} \right. \Rightarrow x\in [5;7]$

объединяем решения:

$\left[\begin{array}{cc}x\in (1;2)\\x\in[5;7]\end{array}\right.\Rightarrow x\in (1;2)\cup [5;7]$

пересекаем с одз:

$\left \{ {{x\in (1;2)\cup [5;7]} \atop {x\in (1;6)\cup (6;+\infty)}} \right. \Rightarrow x\in (1;2)\cup [5;6)\cup (6;7]$

Ответ: $x\in (1;2)\cup [5;6)\cup (6;7]$