Express $\dfrac{2+x}{\sqrt{9-x}}$ as a series in ascending powers of $x$, up to and including the term in $x^2$.

1. Derive the first two terms of the series expansion from the previous part.

From the earlier part of the question, expand $\frac{2+x}{\sqrt{9-x}}$ up to the term in $x$:
$\frac{2+x}{\sqrt{9-x}} = (2+x) \left( 9\left(1-\frac{x}{9}\right) \right)^{-\frac{1}{2}}$
$= \frac{1}{3}(2+x)\left(1 - \frac{x}{9}\right)^{-\frac{1}{2}}$
$\approx \frac{1}{3}(2+x)\left( 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{9}\right) \right)$
$= \frac{1}{3}(2+x)\left( 1 + \frac{x}{18} \right)$
$\approx \frac{1}{3} \left( 2 + \frac{x}{9} + x \right) = \frac{2}{3} + \frac{10}{27}x$

2. Substitute $x = \frac{1}{9}$ into the exact expression and simplify to isolate $\sqrt{5}$.

$\frac{2+\frac{1}{9}}{\sqrt{9-\frac{1}{9}}} = \frac{\frac{19}{9}}{\sqrt{\frac{80}{9}}}$
$= \frac{\frac{19}{9}}{\frac{\sqrt{16 \times 5}}{3}}$
$= \frac{\frac{19}{9}}{\frac{4\sqrt{5}}{3}}$
$= \frac{19}{9} \times \frac{3}{4\sqrt{5}} = \frac{19}{12\sqrt{5}}$

3. Substitute $x = \frac{1}{9}$ into the two-term series approximation.

$\frac{2}{3} + \frac{10}{27}\left(\frac{1}{9}\right)$
$= \frac{2}{3} + \frac{10}{243}$
$= \frac{162}{243} + \frac{10}{243} = \frac{172}{243}$

4. Equate the exact expression to the series approximation and solve for $\sqrt{5}$.

$\frac{19}{12\sqrt{5}} \approx \frac{172}{243}$
$12\sqrt{5} \approx \frac{19 \times 243}{172}$
$\sqrt{5} \approx \frac{19 \times 243}{12 \times 172}$
$\sqrt{5} \approx \frac{19 \times 81}{4 \times 172} = \frac{1539}{688}$
Since $1539 = 3^4 \times 19$ and $688 = 2^4 \times 43$, they have no common factors, so it is in its lowest terms.