Question

14
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Answer 1

$25^x-9*4^{x-1}*5^x+5*4^{2x-1}\ \textgreater \ 0\\\\ 5^{2x}-\frac{9}{4}*4^{x}*5^x+\frac{5}{4}*4^{2x}\ \textgreater \ 0\ \ |\ \ *\ \ \frac{1}{4^x*5^x}\\\\ \frac{5^{2x}}{4^x*5^x}-\frac{9}{4}+\frac{5}{4}*\frac{4^{2x}}{4^x*5^x}\ \textgreater \ 0*\frac{1}{4^x*5^x}\\\\ \frac{5^{x}}{4^x}-\frac{9}{4}+\frac{5}{4}*\frac{4^{x}}{5^x}\ \textgreater \ 0\\\\ (\frac{5}{4})^x-\frac{9}{4}+\frac{5}{4}*(\frac{4}{5})^x\ \textgreater \ 0\\\\ (\frac{5}{4})^x=t\ \textgreater \ 0\\\\ t-\frac{9}{4}+\frac{5}{4}*\frac{1}{t}\ \textgreater \ 0\\\\ \frac{4t^2-9t+5}{4t}\ \textgreater \ 0\\\\ 4t^2-9t+5\ \textgreater \ 0\\\\ 4t^2-4t-5t+5\ \textgreater \ 0\\\\ 4t(t-1)-5(t-1)\ \textgreater \ 0\\\\ (4t-5)(t-1)\ \textgreater \ 0$

$(t-\frac{5}{4})(t-1)\ \textgreater \ 0\ \ \ and\ \ t\ \textgreater \ 0\\\\ +++++++(1)-------(\frac{5}{4})++++++\ \textgreater \ t\\\\ t\in(0;\ 1)\cup(\frac{5}{4};\ +\infty)\\\\ 0\ \textless \ (\frac{5}{4})^x\ \textless \ 1\ \ \ or\ \ \ \ (\frac{5}{4})^x\ \textgreater \ \frac{5}{4}\\\\ (\frac{5}{4})^x\ \textless \ (\frac{5}{4})^0\ \ \ or\ \ \ \ (\frac{5}{4})^x\ \textgreater \ (\frac{5}{4})^1\\\\ x\ \textless \ 0\ \ or\ \ x\ \textgreater \ 1\\\\ x\in(-\infty;\ 0)\cup(1;\ +\infty)$

Ответ: $(-\infty;\ 0)\cup(1;\ +\infty)$