Question

Завдання на олімпіаду, будь ласка допоможіть.​

Answer 1

$\displaystyle\bf\\\sqrt{x+12} =3\\\\\\\left \{ {{x+12\geq 0} \atop {(\sqrt{x+12} )^{2} =3^{2} }} \right. \\\\\\\left \{ {{x\geq -12} \atop {x+12=9}} \right. \\\\\\\left \{ {{x\geq -12} \atop {x=-3}} \right. \ \ \ \Rightarrow \ \ \ x=-3\\\\\\Otvet \ : \ \Big[-6 \ ; \ 0\Big]$