Решите неравенствоlog3(81x)log3x−4+log3x−4log3(81x)⩾24−log3(x8)log32x−16.\frac{\log_{3}(81x)}{\log_{3} x - 4} + \frac{\log_{3} x - 4}{\log_{3}(81x)} \geqslant \frac{24 - \log_{3} \left(x^{8}\right)}{\log_{3}^{2} x - 16}.log3x−4log3(81x)+log3(81x)log3x−4⩾log32x−1624−log3(x8).▸Ответ(0; 181)∪{19}∪(81; +∞)\left(0;\ \frac{1}{81}\right) \cup \left\{\frac{1}{9}\right\} \cup (81;\ +\infty)(0; 811)∪{91}∪(81; +∞).