Question

решить показательное неравенство (1/3)^(5x^2+8x-4)<1

Answer 1

$(\frac{1}{3})^{5x^2+8x-4}<1$

$(\frac{1}{3})^{5x^2+8x-4}<(\frac{1}{3})^0$

$5x^2+8x-4>0$

$5x^2+8x-4=0$

$D=8^2-4*5*(-4)=64+80=144$

$x_1=\frac{-8+12}{10}=0.4$

$x_2=\frac{-8-12}{10}=-2$

$5(x-0.4)(x+2)>0$

-/////////////-(-2)-------- - -------(0.4)-/////////////-

Ответ: $(-$ ∞ $;-2)$ ∪ $(0.4;+$ ∞ $)$