Question

ПОМОГАЙТЕ Очень надо 5^2x+1+4*5^x-1>0

Answer 1

$5^{2x+1}+4*5^{x-1}-1\ \textgreater \ 0\\\\5*5^{2x}+\frac{4}{5}*5^{x}-1\ \textgreater \ 0\, |*5\\\\25*5^{2x}+4*5^{x}-5\ \textgreater \ 0\\\\t=5^{x}\ \textgreater \ 0,\; \; \; 25t^2+4t-5\ \textgreater \ 0\\\\D=16+500=516\\\\t_1=\frac{-4-\sqrt{516}}{50}=\frac{-2-\sqrt{129}}{25}\ \textless \ 0\\\\t_2=\frac{-2+\sqrt{129}}{25}\ \textgreater \ 0\\\\5^{x}=\frac{-2+\sqrt{129}}{25} \\\\x=log_5{\frac{-2+\sqrt{129}}{25}}$