Question

ДАМ 35 БАЛЛОВ!!!!
308.Решите систему линейных неравенств
b)$\left \{ {{-x^{2}-4{x}+12\ \textgreater \ 0} \atop {10-x^{2}\geq0}} \right.$

Answer 1

$\left\{\begin{array}{l}-x^2-4x+12>0\\10-x^2\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x^2+4x-12<0\\x^2-10\leq 0\end{array}\right\ \ \left\{\begin{array}{l}(x+6)(x-2)<0\\(x-\sqrt{10})(x+\sqrt{10})\leq 0\end{array}\right\\\\\\\left\{\begin{array}{l}+++(-6)---(2)+++\\+++(-\qrt{10})---(\sqrt{10})+++\end{array}\right\ \ \left\{\begin{array}{l}x\in (-6\, ;\, 2\, )\\x\in [-\sqrt{10}\ ;\ \sqrt{10}\ ]\end{array}\right\ \ \Rightarrow \\\\\\x\in [\, -\sqrt{10}\ ;\ 2\ )$