ĐKXĐ: \(x\ge0\)
Ta có: \(x-2\sqrt{x}+3=\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Q=\frac{1}{x-2\sqrt{x}+3}\le\frac{1}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow x=1\left(TM\right)\)
Vậy \(Max_Q=\frac{1}{2}\) khi \(x=1\)