Question

log5^2 (25-x^2)-3log5(25-x^2)+2>=0

Answer 1

$ODZ:\\25-x^2\ \textgreater \ 0\\x^2\ \textless \ 25\\|x|\ \textless \ 5\\x\ \textless \ 5\ \ \ i \ \ \ -x\ \textless \ 5\\x\ \textless \ 5\ \ \ i\ \ \ x\ \textgreater \ -5\\x\in(-5;5)\\\\(log_5(25-x^2))^2-3log_5(25-x^2)+2\geq0\\\\log_5(25-x^2)=t\\t^2-3t+2\geq0\\t_1=2;t_2=1\\(t-2)(t-1)\geq0$
$t\in(-\infty;1]\cup[2;+\infty)\\t\leq1\ \ \ ili \ \ \ t\geq2\\\\t\leq1\\log_5(25-x^2)\leq1\\25-x^2\leq5^1\\x^2\geq20\\|x|\geq\sqrt{20}\\x\geq\sqrt{20}\ \ \ i\ \ \ -x\geq\sqrt{20}\\x\geq\sqrt{20}\ \ \ i \ \ \ x\leq-\sqrt{20}\\v\ ODZ\ vhodit:\ \boxed{x\in (-5;-\sqrt{20}]\cup[\sqrt{20};5)}$

$t\geq2\\log_5(25-x^2)\geq2\\25-x^2\geq5^2\\x^2\leq0\\|x|\leq0\\x\leq0\ \ \ i\ \ \ x\geq0\ \Rightarrow\ \boxed{x=0}(vhodit\ v\ ODZ)$

Объединяя ответы получаем: 
$x\in(-5;-\sqrt{20}]\cup\{0\}\cup[\sqrt{20};5)$