Question

ПОМОГИТЕ РЕШИТЬ НЕРАВЕНСТВО!!

Answer 1

$((\frac{1}{3})^3)^{2-x}\ \textgreater \ (3^2)^{2x-1}; \ \ \ (3^{-3})^{2-x}\ \textgreater \ 3^{2 \cdot (2x-1)}; \ \ \ 3^{-3 \cdot (2-x)}\ \textgreater \ 3^{4x-2}; \\ \\ -3 \cdot (2-x)\ \textgreater \ 4x-2; \ \ -6+3x \ \textgreater \ 4x-2; \ \ 3x-4x \ \textgreater \ -2+6; \ \ -x \ \textgreater \ 4; \\ \\ \boxed{x <-4}$

Answer 2

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