Question

Второе неравенство ❤️
Помощнику любви ,денег и счастья :3

Answer 1

$\sqrt{x+8}>x+2\\\\a)\; \; \left \{ {{x+2\geq 0} \atop {x+8>(x+2)^2}} \right. \; ,\left \{ {{x\geq -2} \atop {x+8>x^2+4x+4}} \right. \; \left \{ {{x\geq -2} \atop {x^2+3x-4<0}} \right. \; \left \{ {{x\geq -2} \atop {x\in (-4,1)}} \right. \; \to \\\\x\in [-2,1)\\\\b)\; \; \left \{ {{x+2<0} \atop {x+8\geq 0}} \right. \; \left \{ {{x<-2} \atop {x\geq -8}} \right. \; \to \; \; x\in [-8,-2)\\\\Otvet:\; \; x\in [-8,-2)\cup [-2,1)=[-8,1)\; .$