Решите показательные неравенства

1)
${5}^{6 - 4x} \geqslant 1 \\ {5}^{6 - 4x} \geqslant {5}^{0} \\ 6 - 4x \geqslant 0 \\ - 4x \geqslant - 6 \\ x \geqslant \frac{6}{4}= \frac{3}{2} = 1 \frac{1}{2}$
5)
1 \\ {12}^{ {x}^{2} - 9} > {12}^{0 } \\ {x}^{2} - 9 > 0 \\ {x}^{2} > 9 \\ x > 3" alt=" {12}^{ {x}^{2} - 9} > 1 \\ {12}^{ {x}^{2} - 9} > {12}^{0 } \\ {x}^{2} - 9 > 0 \\ {x}^{2} > 9 \\ x > 3" align="absmiddle" class="latex-formula">