Find the first two terms in the expansion of $\dfrac{(9 + x)^{1/2}}{1 + 2x}$ in ascending powers of $x$.

1. Rewrite the expression as a product of binomials and extract the constant factor to prepare for expansion.

$\frac{(9 + x)^{1/2}}{1 + 2x} = (9 + x)^{1/2} (1 + 2x)^{-1}$
$= 9^{1/2}\left(1 + \frac{x}{9}\right)^{1/2} (1 + 2x)^{-1}$
$= 3\left(1 + \frac{x}{9}\right)^{1/2} (1 + 2x)^{-1}$

2. Expand both binomials up to the term in $x$.

Using the expansion $(1+y)^n = 1 + ny + \dots$ for the first two terms:
$\left(1 + \frac{x}{9}\right)^{1/2} = 1 + \frac{1}{2}\left(\frac{x}{9}\right) + \dots = 1 + \frac{x}{18} + \dots$
$(1 + 2x)^{-1} = 1 + (-1)(2x) + \dots = 1 - 2x + \dots$

3. Multiply the two series expansions together to find the first two terms of the overall expression.

$3\left(1 + \frac{x}{18} + \dots\right)(1 - 2x + \dots)$
$= 3\left(1 - 2x + \frac{x}{18} + \dots\right)$
$= 3\left(1 - \frac{36x}{18} + \frac{x}{18} + \dots\right)$
$= 3\left(1 - \frac{35x}{18}\right)$
$= 3 - \frac{35}{6}x$