Найдите область определения ф y= log2+x(5-x) Неравенство x-4/x^2+16 больше или равно 0

$5-x\ \textgreater \ 0 \ \Rightarrow \ x\ \textless \ 5 \\ \\ 2+x\ \textgreater \ 0 \ \Rightarrow \ x\ \textgreater \ -2 \\ \\ 2+x \neq 1 \ \Rightarrow \ x \neq -1 \\ \\ \\ \boxed{-2\ \textless \ x\ \textless \ -1 \ \ \cup \ \ -1\ \textless \ x\ \textless \ 5} \\ \\ (-2; \ -1) \ \cup \ (-1; \ 5)$

$\frac{x-4}{x^2+16} \geq 0; \ \ \ x-4=0 \ \Rightarrow \ x=4 \\ \\ x^2+16 \neq 0; \ \ (x^2+16\ \textgreater \ 0)$

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$\boxed{x \geq 4} \ \ \ \ \ \ \ \ \boxed{[4; +\infty)}$