Question

Решить неравенство
√(x²+2x)> -3-x²

Answer 1

$sqrt{x^2+2x} textgreater -3-x^2; \ sqrt{f(x)} textgreater g(x); \ left [ {{left { {{g(x) textless 0} atop {f(x) geq 0}} right. } atop {left { {{g(x) geq 0;} atop {f(x) textgreater g^2(x)}} right. }} right. textless = textgreater left [ {{left { {{-3-x^2 textless 0} atop {x^2+2x geq 0}} right. } atop {left { {{-3-x^2 geq 0;} atop {x^2+2x) textgreater (-3-x^2)^2}} right. }} right. \ 1)-3-x^2 textless 0; \ 3+x^2 textgreater 0; \ x^2 textgreater -3; \ xin R. \ 2)x^2+2x geq 0; \ x(x+2) geq 0; \ xin (-infty;-2]U[0;+infty).$
$3) -3-x^2 geq 0; \ 3+x^2 leq 0; \ x^2 leq -3; \ xin emptyset; \ 4)x^2+2x textgreater (-3-x^2)^2; \ x^2+2x-(-3-x^2)^2 textgreater 0; \ x^2+2x-(9+6x^2+x^4) textgreater 0; \ x^2+2x-9-6x^2-x^4 textgreater 0; \ -x^4-5x^2+2x-9 textgreater 0; \ x^4+5x^2-2x+9 textless 0; \ xin emptyset.$
Ответ: $xin (-infty;-2]U[0:+ infty).$