Question

решите неравенство .​

Answer 1

$\displaystyle 1)3^x > \frac{1}{27}\\ \\3^x > 3^-^3\\x > -3=x \in (-3, + \infty)\\\\2)2^x < \frac{1}{8}\\ \\2^x < 2^-^3\\x < -3=x \in (- \infty,-3)\\\\3)(\frac{2}{5})^x^+^2 > \frac{2}{5}\\ \\(\frac{2}{5})^x^+^2 > (\frac{2}{5})^1\\ \\x+2 < 1\\ x < 1-2\\ x < -1=x \in (- \infty, -1)$