\(\dfrac{A}{11}+\dfrac{B}{3}=\dfrac{31}{33}\Leftrightarrow\dfrac{A}{11}=\dfrac{31}{33}-\dfrac{B}{3}=\dfrac{31-11B}{33}\Leftrightarrow A=\dfrac{31-11B}{3}\)

Because \(A,B\in N,A\ge0\Rightarrow31-11B\ge0\Rightarrow11B\le31\)

\(\Rightarrow\left[{}\begin{matrix}B=0\\B=1\\B=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}A=\dfrac{31}{3}\\A=\dfrac{20}{3}\\A=3\end{matrix}\right.\)

Hence, A = 3 ; B = 2