Непростой пример из профильного уровня ЕГЭ по математике

Question 1

Question 2

Решите задачу:

$log\, _{|x-5|}\, (2x^2-10x+8) \leq 2\; ,\\\\ODZ:\; \; \left \{ {{|x-5|\ \textgreater \ 0\; ,\; |x-5|\ne 1} \atop {2x^2-10x+8\ \textgreater \ 0}} \right. \; \; \left \{ {{x\ne 4\; ,\; x\ne 6} \atop {x\in (-\infty ,1)\cup (4,+\infty )}} \right. \\\\1)\; \; 0\ \textless \ |x-5|\ \textless \ 1\; \; \Rightarrow \; \; \left \{ {{x-5\ \textless \ 1} \atop {x-5\ \textgreater \ -1}} \right. \; \Rightarrow \; 4\ \textless \ x\ \textless \ 6\\\\ \left \{ {{4\ \textless \ x\ \textless \ 6} \atop {x\in (-\infty ,1)\cup (4,+\infty )}} \right. \; \; \Rightarrow \; \; x\in (4,6)\\\\2x^2-10x+8 \geq |x-5|^2\\\\|x-5|^2=(x-5)^2\; \; \to$

$2x^2-10x+8 \geq x^2-10x+25\\\\x^2-17 \geq 0\; ,\; \; (x-\sqrt{17})(x+\sqrt{17}) \geq 0\; ,\; \; \; \sqrt{17}\approx 4,12\\\\x\in (-\infty ,-\sqrt{17}\; ]\cup [\; \sqrt{17},+\infty )\\\\ \left \{ {{x\in (4,6)} \atop {x\in (-\infty ,\sqrt{17}\, ]\cup [\, \sqrt{17},+\infty )}} \right. \; \; \to \; \; \; x\in [\; \sqrt{17},6)\\\\2)\; \; |x-5|\ \textgreater \ 1\; \; \Rightarrow \; \; \left [ {{x-5\ \textgreater \ 1} \atop {x-5\ \textless \ -1}} \right. \; \; \Rightarrow \; \; x\in (-\infty ,4)\cup (6,+\infty )\\\\2x^2-10x+8 \leq |x-5|^2$

$x^2-17 \leq 0\; ,\; \; \; x\in [\; -\sqrt{17},\sqrt{17}\; ]$

$3)\; \; \left \{ {{x\in (-\infty ,4)\cup (6,+\infty )} \atop {x\in [\, -\sqrt{17},\sqrt{17}\, ]}} \right. \; \; \to \; \; x\in [\; -\sqrt{17},4) \\\\4)\; \; x\in [\; -\sqrt{17},4)\cup [\; \sqrt{17},6)\; .$

$5)\; \; \left \{ {{x\in (-\infty ,1)\cup (4,+\infty )\; ,\; x\ne 6} \atop {x\in [\, -\sqrt{17},4)\cup [\, \sqrt{17},6)}} \right. \; \; \to \; \; x\in [\, -\sqrt{17},1)\cup [\, \sqrt{17},6)$

Question 3

А почему в ОДЗ это под модулем не равно 1-?

Question 4

Основание не =1