(1/5)степеньx-1 -(1/5)степеньx-1=4,8

[latex]\displaystyle ( \frac{1}{4})^{10^{x}}}\ \textgreater \ 64-( \frac{8}{3})^{-x^{2}} \\ \\ ( 2^{-2})^{10^{x}}}\ \textgreater \ 2^{6}-( \frac{3}{8})^{x^{2}} [/latex] Так как [latex]\displaystyle ( 2^{-2})^{10^{x}}}<1[/latex] и [latex]\displaystyle ( \frac{3}{8})^{x^{2}} \leq 1[/latex] то [latex]\displaystyle (2^{-2})^{10^{x}}}<2^{6}-( \frac{3}{8})^{x^{2}}[/latex] при любом х Ответ: ∅