Question

Помогите, пожалуйста, решить неравенство

Answer 1

$x^2+9x+20=0\; \; \to \; \; x_1=-5\; ,\; \; x_2=-4\; \; (teorema\; Vieta)\; \to \\\\x^2+9x+20=(x+5)(x+4)\\\\\\\Big (\frac{x+5}{4+x}-\frac{1}{(x+5)(x+4)}\Big )\cdot \sqrt{-7x-x^2}\geq 0\; \; ,\; \; ODZ:\; \; \left \{ {{x\ne -5\; ,\; x\ne -4} \atop {-7x-x^2\geq 0}} \right.\\\\\\\left \{ {{x\ne -5\; ,\; x\ne -4} \atop {-x(7+x)\geq 0}} \right. \; \left \{ {{x\ne -5\; ,\; x\ne -4} \atop {x\in [-7,0\, ]}} \right. \; \; \to \; \; x\in [-7,-5)\cup (-5,-4)\cup (-4,0\, ]\\\\\\\frac{(x+5)^2-1}{(x+5)(x+4)}\cdot \sqrt{-7x-x^2}\geq 0\\\\\frac{x^2+10x+24}{(x+5)(x+4)}\cdot \sqrt{-7x-x^2}\geq 0\\\\\frac{(x+6)(x+4)}{(x+5)(x+4)}\cdot \sqrt{-7x-x^2}\geq 0\\\\\frac{x+6}{x+5}\cdot \sqrt{-7x-x^2}\geq 0\; \; \Rightarrow \; \; tak\; kak\; \sqrt{-7x-x^2}\geq 0,\; to\; \; \frac{x+6}{x+5}\geq 0\; .\\\\znaki\; \frac{x+6}{x+5}:\; \; +++[-6\, ]---(-5)+++\\\\x\in (-\infty ,-6\, ]\cup (-5,+\infty )\\\\\left \{ {{x\in [-7,-5)\cup (-5,-4)\cup (-4,0\, ]} \atop {x\in (-\infty ,-6\, ]\cup (-5,+\infty )}} \right. \; \; \Rightarrow \; \; x\in [-7,-6]\cup (-5,-4)\cup (-4,0\, ]$