Question

Второе неравенство ❤️
Помощнику любви ,денег и счастья :3

Answer 1

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