Question

Решите неравенство: (-19)/(〖(х+5)〗^2-6)≥0

Answer 1

$\sf \dfrac{-19}{(x+5)^2-6} \geq 0 \\(x+5)^2-6<0 \\ (x+5-\sqrt{6})(x+5+ \sqrt{6})<0 \\ x \in (-5- \sqrt{6}; \ -5+ \sqrt{6})$

Ответ: x∈(-5-√6; -5+√6)