Задание №75596 ЕГЭ Профильной математике

$\frac{6}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}}-\frac{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}}{\mathit{l}\mathit{o}{\mathit{g}}_4\frac{\mathit{x}}{256}}\ge \frac{15}{\mathit{l}\mathit{o}{\mathit{g}}_4{\mathit{x}}^4-\mathit{l}\mathit{o}{\mathit{g}}_4^2\mathit{x}}$

ОДЗ: $\mathit{x}>0,\mathit{x}\ne 256,\mathit{x}\ne 1$.

$\frac{6}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}}-\frac{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4}\ge \frac{15}{4\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-\mathit{l}\mathit{o}{\mathit{g}}_4^2\mathit{x}}$;

$\frac{6}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}}-\frac{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4}\ge \frac{15}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\left(4-\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\right)}$.

$\frac{6\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4\right)-\mathit{l}\mathit{o}{\mathit{g}}_4^2\mathit{x}}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4\right)}\ge \frac{-15}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4\right)}$.

$\frac{6\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-24-\mathit{l}\mathit{o}{\mathit{g}}_4^2\mathit{x}+15}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4\right)}\ge 0$

$\frac{\mathit{l}\mathit{o}{\mathit{g}}_4^2\mathit{x}-6\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}+9}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4\right)}\le 0$

$\frac{{\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-3\right)}^2}{\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}\left(\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}-4\right)}\le 0$

Обозначим $\mathit{l}\mathit{o}{\mathit{g}}_4\mathit{x}=\mathit{t}$. Неравенство примет вид: $\frac{{\left(\mathit{t}-3\right)}^2}{\mathit{t}\left(\mathit{t}-4\right)}\le 0$.

Решим это неравенство методом интервалов.

Итак, .