$\dfrac{1}{\log_2 x}+3\dfrac{1}{\log_x 2x}-3\ge 0$
$\Leftrightarrow \dfrac{1}{\log_2 x}+3\dfrac{1}{[\log_x 2 +1 ]}-3\ge 0$ đặt $\log_2 x=t$ ta có
$\dfrac{1}{t}+3\dfrac{1}{\dfrac{1}{t}+1}-3\ge0$
$\Leftrightarrow \dfrac{2t-1}{2t(t+1)}\ge0$
$\Leftrightarrow \left [ \begin{matrix} 0<t\le \dfrac{1}{2} \\ t<-1 \end{matrix} \right.$
$\Leftrightarrow \left [ \begin{matrix} 1 < x\le \sqrt 2 \\ x<\dfrac{1}{2} \end{matrix} \right.$ kết hợp đk ta có
$\Leftrightarrow \left [ \begin{matrix} 1 < x\le \sqrt 2 \\ 0<x<\dfrac{1}{2} \end{matrix} \right.$