Question

lоg0.5(x2-5x+6) ≥-1 Решите неравенство

Answer 1

Ответ:

$log_{0,2}(x^2+2x-3)\geq -1\ \ ,\\\\ODZ:\ \ x^2+2x-3>0\ ,\ \ (x+3)(x-1)>0\ \ \to \ \ x\in (-\infty ;-3)\cup (1;+\infty )\\\\0<0,2<1\ \ \ \Rightarrow \ \ \ x^2+2x-3\leq 0,2^{-1}\\\\x^2+2x-3\leq 5\ \ ,\ \ \ x^2+2x-8\leq 0\ \ ,\ \ x_1=-4\ ,\ x_2=2\ (teorema\ Vieta)\\\\(x+4)(x-2)\leq 0\\\\znaki:\ \ \ +++[-4\ ]---[\ 2\ ]+++\\\\x\in [-4\ ;\ 2\ ]\\\\\left\{\begin{array}{l} x\in (-\infty ;-3)\cup (1;+\infty )\\x\in [-4\ ;\ 2\ ]\end{array}\\ \ \ \ \ \ \ \ \ Otvet:\ \ x\in [-4\ ;-3\ )\cup (\ 1\ ;2\ ]\ .$