Решите неравенство log 4 ( 64 𝑥 ) log 4 𝑥 − 3 + log 4 𝑥 − 3 log 4 ( 64 𝑥 ) ≥ log 4 𝑥 4 + 16 log 2 4 𝑥 − 9 ${\displaystyle \frac{{\log }_4\left(64x\right)}{{\log }_{4}x-3}+\frac{{\log }_{4}x-3}{{\log }_4\left(64x\right)}\ge \frac{{\log }_4{x}^4+16}{{\log }_4^{2}x-9}}$ .