\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne0\end{matrix}\right.\Leftrightarrow x>0\)
Ta có:
\(\frac{5x-4\sqrt{x}+1}{x}=\frac{x+4x-4\sqrt{x}+1}{x}=1+\frac{\left(2\sqrt{x}-1\right)^2}{x}\ge1\)(vì \(x>0\))
Dấu "=" xảy ra khi \(2\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy \(Min_{bt}=1\) khi \(x=\frac{1}{4}\)