Minimum value of $a^{x} + \dfrac{a}{a^{a x}}$ (where $a > 0$, $a \neq 1$, and $x \in \mathbb{R}$) is:

**Solution:** Let $y = a^{x} + \dfrac{a}{a^{a x}} = a^{x} + a^{1 - a x}$ Let $t = a^{x}$, $t > 0$. Then $y = t + \dfrac{a}{t^{a}}$ Differentiate: \[ \dfrac{dy}{dt} = 1 - a^2 t^{-a-1} = 0 \Rightarrow t^{a+1} = a^2 \Rightarrow t = a^{\tfrac{2}{a+1}}. \] Substitute back: \[ y_{\min} = a^{\tfrac{2}{a+1}} + a^{1 - a\tfrac{2}{a+1}} = 2\sqrt{a}. \] $\boxed{\text{Answer: (A) }2\sqrt{a}}$