Помогите срочно. Решить неравенство.
1)5^x-1<1/√5


2) (2/5)^x+1 < 1

[tex]5^{x-1}\ \textless \ \frac{1}{ \sqrt{5}} \\ 5^{x-1}\ \textless \ 5^{-\frac{1}{2}} \\ 5\ \textgreater \ 1 \\ x-1\ \textless \ -\frac{1}{2} \\ x\ \textless \ \frac{1}{2} \\ \\ ( \frac{2}{5})^{x+1}\ \textless \ 1 \\( \frac{2}{5})^{x+1}\ \textless \ ( \frac{2}{5})^0 \\ 0\ \textless \ \frac{2}{5}\ \textless \ 1 \\ x+1\ \textgreater \ 0 \\ x\ \textgreater \ -1 [/tex]