$\dfrac{x}{\sqrt{4x-1}}+\dfrac{\sqrt{4x-1}}{x}=2$
Đặt $t=\dfrac{x}{\sqrt{4x-1}} \Rightarrow \dfrac{1}{t}=\dfrac{\sqrt{4x-1}}{x}$
Pt $\Rightarrow t+\dfrac{1}{t}=2 \Leftrightarrow t^2-2t+1=0$
$\Leftrightarrow t=1 \Rightarrow \dfrac{x}{\sqrt{4x-1}}=1$
$\Leftrightarrow x=\sqrt{4x-1}$
$\Leftrightarrow x^2-4x+1=0$ (bình phương 2 vế)
$\Leftrightarrow \left[ \begin{array}{l}x=2-\sqrt{3}\\x=2+\sqrt{3}\end{array} \right.$
