\(P=\frac{\frac{1}{8}\left(x+8\right)-\frac{1}{8}x+\sqrt{x}-2}{x+8}=\frac{\frac{1}{8}\left(x+8\right)-\left(\sqrt{\frac{1}{8}x}-\sqrt{2}\right)^2}{x+8}=\frac{1}{8}-\left(\sqrt{\frac{1}{8}x}-\sqrt{2}\right)^2\le\frac{1}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow x=16\)
Vậy \(P_{Max}=\frac{1}{8}\) khi \(x=16\)