Dễ thấy \(\sqrt{x}\ge0;x+\sqrt{x}+1>0\) nên \(D\ge0\)
a, Do \(\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow x-2\sqrt{x}+1\ge0\)
\(\Leftrightarrow x+\sqrt{x}+1\ge3\sqrt{x}\Rightarrow\dfrac{1}{3}\ge\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, Do \(\dfrac{1}{3}\ge\dfrac{x}{x+\sqrt{x}+1}\) nên Max D = \(\dfrac{1}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow\dfrac{1}{3}=\dfrac{x}{x+\sqrt{x}+1}\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\)