ĐKXĐ: \(x>0,x\ne16\)
\(\dfrac{\sqrt{x}+5}{\sqrt{x}-4}>\dfrac{1}{5}\)
\(\dfrac{\sqrt{x}+5}{\sqrt{x}-4}-\dfrac{1}{5}>0\)
\(\dfrac{\sqrt{x}+5-\sqrt{x}+4}{5\sqrt{x}-20}>0\)
\(\dfrac{9}{5\sqrt{x}-20}>0\)
Vì 9>0 nên \(5\sqrt{x}-20>0\)
\(5\sqrt{x}>20\Leftrightarrow\sqrt{x}>4\)
\(\Leftrightarrow x>16\)

