Решите неравенство 1 + 5 log 4 𝑥 − 3 + 6 log 2 4 𝑥 − log 4 ( 64 𝑥 6 ) + 12 ≥ 0 ${\displaystyle 1+\frac{5}{{\log }_{4}x-3}+\frac{6}{{\log }_4^{2}x-{\log }_4\left(64x^6\right)+12}\ge 0}$ .