\(A=\dfrac{3x}{x+3}=\dfrac{3x+9-9}{x+3}=\dfrac{3\left(x+3\right)-9}{x+3}=3-\dfrac{9}{x+3}\)
Ta có \(x\in N\Rightarrow x+3\ge3\)
\(\Rightarrow\dfrac{9}{x+3}\le3\)
\(\Rightarrow-\dfrac{9}{x+3}\ge-3\)
\(\Rightarrow A=3-\dfrac{9}{x+3}\ge3-3=0\)
Vậy \(A\left(min\right)=0\left(tại.x=0\right)\)