[tex]\displaystyle f(x) = \frac{ \sqrt{3} }{ - 16 + {x}^{2} }[/tex]

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ОДЗ:

[tex]\displaystyle - 16 + {x}^{2} \ne0[/tex]

[tex]\displaystyle {x}^{2} \ne16[/tex]

[tex]\displaystyle x \ne \pm4[/tex]

Ответ: D(y) =(-∞;-4)U(-4;4)U(4;+∞)