Question

 B = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\) 

Answer 1

\(3,P=\dfrac{A}{B}=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}\\ P=\dfrac{2\left(\sqrt{x}+3\right)-6}{\sqrt{x}+3}=2-\dfrac{6}{\sqrt{x}+3}\ge2-\dfrac{6}{0+3}=0\\ P_{min}=0\Leftrightarrow x=0\)