Question

Условие и вопрос на рисунке!!! С подробным решением!!!

Answer 1

$(\dfrac{1}{(x+2)^2}-4)(9-100(x+2)^2)\leq 105\\ (x+2)^2=t\\ (\dfrac{1}{t}-4)(9-100t)\leq 105\:\:\:|*t\\ (1-4t)(9-100t)\leq 105t\\ 9-136t+400t^2-105t\leq 0\\ 400t^2-241t+9\leq 0\\ \left[t=\dfrac{241\pm \sqrt{241^2-4*400*9}}{800}=\dfrac{241\pm \sqrt{43681}}{800}=\dfrac{241\pm 209}{800}=>t_1=\dfrac{9}{16};\:t_2=\dfrac{1}{25} \right]\\ (t-\dfrac{9}{16})(t-\dfrac{1}{25})\leq 0\\ \dfrac{1}{25}\leq t\leq \dfrac{9}{16}\\ \dfrac{1}{25}\leq (x+2)^2\leq \dfrac{9}{16}$

$\\ \dfrac{1}{5}\leq |x+2|\leq \dfrac{3}{4}\\ \left [ {{\dfrac{1}{5}\leq x+2\leq \dfrac{3}{4}} \atop {-\dfrac{3}{4}\leq x+2\leq -\dfrac{1}{5}}} \right. => \left [{ {{-1\dfrac{4}{5}\leq x\leq -1\dfrac{1}{4}} \atop {-2\dfrac{3}{4}\leq x\leq -2\dfrac{1}{5}}} \right. \\ x\in [-2\dfrac{3}{4}; -2\dfrac{1}{5}] U [-1\dfrac{4}{5}; -1\dfrac{1}{4}]$