4^(sqrt(x-2))+16 > 10*2^(sqrt(x-2))

$4^{\sqrt{x-2}}+16\ \textgreater \ 10*2^{\sqrt{x-2}} \\ 2^{2\sqrt{x-2}}+2^4-10*2^{\sqrt{x-2}}\ \textgreater \ 0 \\$
Допустим $2^{\sqrt{x-2}} = t, t\ \textgreater \ 0$
Тогда $t^2-10t+16\ \textgreater \ 0 \\ \left \{ {{\left[\begin{array}{cc}t\ \textless \ 2\\t\ \textgreater \ 8\end{array}}\right \atop {t\ \textgreater \ 0}} \right. \\ 2^{\sqrt{x-2}}\ \textgreater \ 8 =\ \textgreater \ \sqrt{x-2}\ \textgreater \ 3 =\ \textgreater \ x\ \textgreater \ 11 \\ or\\ 0\ \textless \ \sqrt{x-2}\ \textless \ 1 =\ \textgreater \ 0\ \textless \ x-2\ \textless \ 1 =\ \textgreater \ 2\ \textless \ x\ \textless \ 3 \\ \\Answer: (2;3)U(11;\inf)$