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Решите задачу:

$(\sqrt{5})^{x-6}\ \textless \ \frac{1}{5}, \\ (5^{\frac{1}{2}})^{x-6}\ \textless \ 5^{-1}, \\ 5^{\frac{x-6}{2}}\ \textless \ 5^{-1}, \\ 5\ \textgreater \ 1, \ \frac{x-6}{2}\ \textless \ -1, \\ x-6\ \textless \ -2, \\ x\ \textless \ 4. \\ \\ (\sqrt[3]{3})^{x+6}\ \textgreater \ \frac{1}{9}, \\ (3^{\frac{1}{3}})^{x+6}\ \textgreater \ (\frac{1}{3})^2, \\ 3^{\frac{x+6}{3}}\ \textgreater \ 3^{-2}, \\ 3\ \textgreater \ 1, \ \frac{x+6}{3}\ \textgreater \ -2, \\ x+6\ \textgreater \ -6, \\ x\ \textgreater \ -12.$

$(\frac{2}{13})^{x^2-1}\geq1, \\ (\frac{2}{13})^{x^2-1}\geq(\frac{2}{13})^0, \\ 0\ \textless \ \frac{2}{13}\ \textless \ 1, \\ x^2-1\leq0, \\ (x+1)(x-1)\leq0, \\ (x+1)(x-1)=0, \\ \left [ {{x+1=0,} \atop {x-1=0;}} \right. \ \left [ {{x=-1,} \atop {x=1;}} \right. \\ -1 \leq x \leq 1.$

$(1\frac{1}{7})^{x^2-4}\leq1, \\ (1\frac{1}{7})^{x^2-4}\leq(1\frac{1}{7})^0, \\ 1\frac{1}{7}\ \textgreater \ 1, \ x^2-4\leq0, \\ (x+2)(x-2)\leq0, \\ (x+2)(x-2)=0, \\ \left [ {{x+2=0,} \atop {x-2=0;}} \right. \ \left [ {{x=-2,} \atop {x=2;}} \right. \\ -2 \leq x \leq 2.$