\(A=\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\left(x>1\right)\)
\(\Leftrightarrow A=\dfrac{\sqrt[]{x}+1-\sqrt{x}+1}{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+1\right)}=\dfrac{2}{x-1}\)
\(B=\dfrac{x+1}{2}-\sqrt{x}=\dfrac{x+1-2\sqrt{x}}{2}=\dfrac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0\right)\)
\(\Rightarrow A.B=\dfrac{2}{x-1}.\dfrac{\left(\sqrt[]{x}-1\right)^2}{2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
vì \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\) \(\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{1}{\sqrt{x}+1}\le1\Rightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)
\(\Rightarrow A.B=1-\dfrac{1}{\sqrt[]{x}+1}\ge1-2=-1\)
Vậy \(Min\left(A.B\right)=-1\left(x=0\right)\)
