Question

Помогите, пожалуйста, решить неравенство

Answer 1

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