Đặt \(\sqrt{2x-\dfrac{5}{x}}=a;\sqrt{x-\dfrac{1}{x}}=b\)(a,b>=0)
Ta có \(a^2-b^2=2x-\dfrac{5}{x}-x+\dfrac{1}{x}=x-\dfrac{4}{x}\)
Ta có pt \(\Leftrightarrow a-b=a^2-b^2\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)