Question

решите неравенство (1/27)^2-x >9^2x-1.

Answer 1

$(\frac{1}{27})^{2-x}\ \textgreater \ 9^{2x-1}\\\\3^{-3(2-x)}\ \textgreater \ 3^{2(2x-1)}\\\\-6+3x\ \textgreater \ 4x-3\\\\x\ \textless \ -3$