(căn x )+1 / (căn x)+3 lớn hơn 1/3 đk x lớn hơn 0, x khác 9 ,(cănx) +3 khác 0

\(\dfrac{\sqrt x+1}{\sqrt x+3}>\dfrac{1}{3}\\↔\dfrac{\sqrt x+1}{\sqrt x+3}-\dfrac{1}{3}>0\\↔\dfrac{3\sqrt x+3-\sqrt x-3}{3(\sqrt x+3)}>0\\↔\dfrac{2\sqrt x}{3(\sqrt x+3)}>0\\↔\dfrac{2\sqrt x}{\sqrt x+3}>0\)

Vì \(2\sqrt x\ge 0,\sqrt x+3>0\) mà \(\dfrac{2\sqrt x}{\sqrt x+3}>0\)

\(→2\sqrt x>0\\↔\sqrt x>0\\↔x>0\)

Vậy \(x>0\)