Question

помогите, очень надо!! с решением.

Answer 1

$y=\sqrt{12-3x}+\frac{1}{\sqrt{x+2} }\\\\\left \{ {{12-3x\geq0 } \atop {x+2>0}} \right. \\\\\left \{ {{-3x\geq-12 } \atop {x>-2}} \right.\\\\\left \{ {{x\leq4 } \atop {x>-2}} \right.\\\\Otvet:\boxed{x\in(-2 \ ; \ 4]}$