Question

Решите неравенство (x+4)^2>16-x^2

Answer 1

$(x+4)^2\ \textgreater \ 16-x^2 \\ \\ x^2+ 8x+16 - 16 + x^2\ \textgreater \ 0 \\ \\ 2x^2+ 8x \ \textgreater \ 0 \\ \\ x(2x+ 8) \ \textgreater \ 0$

$\left \{ {{x \ \textgreater \ 0} \atop {2x+ 8 \ \textgreater \ 0}} \right. \ \ \ \left \{ {{x \ \textgreater \ 0} \atop {x \ \textgreater \ -4}} \right. \ \ \Rightarrow \ x\ \textgreater \ 0$

или

$\left \{ {{x \ \textless \ 0} \atop {2x+ 8 \ \textless \ 0}} \right. \left \{ {{x \ \textless \ 0} \atop {x \ \textless \ -4}} \right. \Rightarrow \ \ x \ \textless \ -4}$

Ответе: $x \ \textless \ -4; \ \ x \ \textgreater \ 0$

Answer 2

$(x+4)^2\ \textgreater \ 16-x^2 \\ (x+4)^2-(4-x)(4+x)\ \textgreater \ 0 \\ (x+4)(x+4-4+x)\ \textgreater \ 0 \\ x(x+4)\ \textgreater \ 0 \\ \\ x_1=0 \cup x_2=-4$

$a\ \textgreater \ 0 \Rightarrow x \in (-\infty;-4) \cup (0; +\infty)$

Ответ: $x \in (-\infty;-4) \cup (0; +\infty)$