Question

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

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

одз:

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

решаем неравенство:

1)

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

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

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

2)

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

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

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

объединяем решения:

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

пересекаем с одз:

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

Ответ: $xin (1;2)cup [5;6)cup (6;7]$