$\begin{array}{l}
A_x^2 - A_x^1 = 3\left( {x \in \mathbb N,x \ge 2} \right)\\
\Leftrightarrow \dfrac{{x!}}{{\left( {x - 2} \right)!}} - \dfrac{{x!}}{{\left( {x - 1} \right)!}} = 3\\
\Leftrightarrow \left( {x - 1} \right)x - x = 3\\
\Leftrightarrow {x^2} - x - x = 3\\
\Leftrightarrow {x^2} - 2x - 3 = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1(L)\\
x = 3(tm)
\end{array} \right.\\
\Rightarrow x = 3
\end{array}$