Решите неравенствоlog11(2x2+1)+log11(132x+1)⩾log11(x16+1).\log_{11} \left(2x^{2} + 1\right) + \log_{11} \left(\frac{1}{32x} + 1\right) \geqslant \log_{11} \left(\frac{x}{16} + 1\right).log11(2x2+1)+log11(32x1+1)⩾log11(16x+1).▸Ответ(−16; −14]∪(0; +∞)\left(-16;\ -\frac{1}{4}\right] \cup (0;\ +\infty)(−16; −41]∪(0; +∞).