Question

Решите неравенство !! Пожалуйстаа!

Answer 1

$\sqrt{4-x^2}\cdot log_{0,5}\frac{(x+1)^2}{3x+1}\leq 0\\\\ODZ:\; \; \left \{ {{4-x^2\geq 0} \atop {\frac{(x+1)^2}{3x+1}>0}} \right.\; \left \{ {{(4-x)(4+x)\geq 0} \atop {3x+1>0}} \right.\; \left \{ {{x\in [-4,4\, ]} \atop {x\in (-\frac{1}{3},+\infty )}} \right. \; \; \Rightarrow \; \; x\in (-\frac{1}{3},4\, ]\\\\Tak\; kak\; \; \sqrt{4-x^2}\geq 0\; \; pri\; \; x\in ODZ\; \; \Rightarrow \; \; log_{0,5}\frac{(x+1)^2}{3x+1}\leq 0\; ,\\\\\frac{(x+1)^2}{3x+1}\geq 1\; ,\; \; \frac{(x+1)^2}{3x+1}-1\geq 0\; ,$

$\frac{(x+1)^2-3x-1}{3x+1}\geq 0\; \; ,\; \; \frac{x^2+2x+1-3x-1}{3x+1}\geq 0\; \; ,\; \; \frac{x(x-1)}{3x+1}\geq 0\; ,\\\\znaki:\; \; ---(-\frac{1}{3})+++[\, 0\, ]---[\, 1\, ]+++\\\\x\in (-\frac{1}{3},0\, ]\cup [\, 1,+\infty )\\\\\left \{ {{x\in (-\frac{1}{3},0\, ]\cup [\, 1,+\infty )} \atop {x\in (-\frac{1}{3},4\, ]}} \right. \; \; \; \Rightarrow \; \; \underline {\; x\in (-\frac{1}{3},0\, ]\cup [\, 1,4\, ]\; }$