0\\2*2^{2x}-25*5^{2x}-5*2^x*5^x>0|:2^{2x}\\2-25*( \frac{5}{5} )^{2x}-5*( \frac{5}{2} )^x>0\\\\t=( \frac{5}{2} )^x\\\\2-25t^2-5t>0|:(-1)\\25t^2+5t-2<0\\D=5^2-4*25*(-2)=25+200=225=15^2\\\\t_1= \frac{-5+15}{2*25}= \frac{10}{50}= \frac{1}{5}\\\\t_2= \frac{-5-15}{2*25}= \frac{-20}{50}=- \frac{2}{5}\\\\25(t- \frac{1}{5})(t+ \frac{2}{5})<0 " alt="2*4^x-25*5^{2x}-5*10^x>0\\2*2^{2x}-25*5^{2x}-5*2^x*5^x>0|:2^{2x}\\2-25*( \frac{5}{5} )^{2x}-5*( \frac{5}{2} )^x>0\\\\t=( \frac{5}{2} )^x\\\\2-25t^2-5t>0|:(-1)\\25t^2+5t-2<0\\D=5^2-4*25*(-2)=25+200=225=15^2\\\\t_1= \frac{-5+15}{2*25}= \frac{10}{50}= \frac{1}{5}\\\\t_2= \frac{-5-15}{2*25}= \frac{-20}{50}=- \frac{2}{5}\\\\25(t- \frac{1}{5})(t+ \frac{2}{5})<0 " align="absmiddle" class="latex-formula">
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___________________-2/5_______________1/5______________
0} \atop {(5/2)^x<1/5}} \right. \\\\ \left \{ {{x\in(-\infty;+\infty)}\atop {x<log_{5/2} \frac{1}{5} } \right. \\\\x\in(-\infty;log_{\frac{5}{2}} \frac{1}{5}})" alt="-2/5 < t < 1/5\\\\-2/5 < (5/2)^x< 1/5\\\\0<(5/2)^x<1/5\\\\\left \{ {{(5/2)^x>0} \atop {(5/2)^x<1/5}} \right. \\\\ \left \{ {{x\in(-\infty;+\infty)}\atop {x<log_{5/2} \frac{1}{5} } \right. \\\\x\in(-\infty;log_{\frac{5}{2}} \frac{1}{5}})" align="absmiddle" class="latex-formula">