DKXD: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}>0\end{matrix}\right.\)\(\Rightarrow x>0\)
\(B=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+x}\)
\(=\left(\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+x}{\sqrt{x}}\)\(=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{\sqrt{x}+x}{\sqrt{x}}\)\(=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\)
Để \(B>\sqrt{x}+2\Leftrightarrow\)\(\dfrac{\sqrt{x}+1+x}{\sqrt{x}}>\sqrt{x}+2\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1+x}{\sqrt{x}}-\left(\sqrt{x}+2\right)>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1+x-\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1+x-x-2\sqrt{x}}{\sqrt{x}}>0\) \(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}}>0\)
Vì \(\sqrt{x}>0\Rightarrow\) \(\dfrac{1-\sqrt{x}}{\sqrt{x}}>0\Leftrightarrow1-\sqrt{x}>0\)\(\Leftrightarrow-\sqrt{x}>-1\)\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp DKXD \(\Rightarrow0< x< 1\)
Vậy để \(B>\sqrt{x}+2\) thì 0 < x < 1.
