Question

Tìm x:

\(\dfrac{x}{\sqrt{4x-1}}+\dfrac{\sqrt{4x-1}}{x}=2\)

Answer 1

ĐKXĐ: \(x>\dfrac{1}{4}\)

Đặt \(\dfrac{x}{\sqrt{4x-1}}=t>0\)

\(\Rightarrow t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\)

\(\Rightarrow t=1\Rightarrow x=\sqrt{4x-1}\)

\(\Rightarrow x^2-4x+1=0\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)