Express $\dfrac{2+x}{\sqrt{9-x}}$ as a series in ascending powers of $x$, up to and including the term in $x^2$. By using only the first two terms in the series expansion above and substituting $x = \dfrac{1}{9}$, find an approximation for $\sqrt{5}$ in the form $\dfrac{p}{q}$ expressed in its lowest terms, where $p$ and $q$ are integers to be determined.

1. Rewrite the expression to isolate the radical and factorise it to match the standard binomial form $(1+y)^n$.

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

2. Expand $\left(1-\frac{x}{9}\right)^{-\frac{1}{2}}$ using the binomial expansion up to the term in $x^2$.

$\left(1-\frac{x}{9}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{9}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{9}\right)^2 + \dots$
$= 1 + \frac{x}{18} + \frac{3/4}{2}\left(\frac{x^2}{81}\right) + \dots$
$= 1 + \frac{x}{18} + \frac{x^2}{216} + \dots$

3. Multiply the expanded series by $(2+x)$ and $\frac{1}{3}$, then collect and simplify the terms up to $x^2$.

$\frac{2+x}{\sqrt{9-x}} = \frac{1}{3}(2+x)\left(1 + \frac{x}{18} + \frac{x^2}{216} + \dots\right)$
$= \left(\frac{2}{3} + \frac{x}{3}\right)\left(1 + \frac{x}{18} + \frac{x^2}{216} + \dots\right)$
$= \frac{2}{3}\left(1 + \frac{x}{18} + \frac{x^2}{216}\right) + \frac{x}{3}\left(1 + \frac{x}{18}\right) + \dots$
$= \frac{2}{3} + \frac{x}{27} + \frac{x^2}{324} + \frac{x}{3} + \frac{x^2}{54} + \dots$
$= \frac{2}{3} + \left(\frac{1}{27} + \frac{1}{3}\right)x + \left(\frac{1}{324} + \frac{1}{54}\right)x^2$
$= \frac{2}{3} + \frac{10}{27}x + \frac{7}{324}x^2$

1. Concisely determine the first two terms of the series expansion.

From part (a), the series expansion of $\dfrac{2+x}{\sqrt{9-x}}$ is found using the binomial expansion:
$\dfrac{2+x}{\sqrt{9-x}} = \frac{1}{3}(2+x)\left(1 - \frac{x}{9}\right)^{-\frac{1}{2}}$
$\approx \frac{1}{3}(2+x)\left(1 - \frac{1}{2}\left(-\frac{x}{9}\right)\right)$
$= \frac{1}{3}(2+x)\left(1 + \frac{x}{18}\right)$
$= \frac{1}{3}\left(2 + \frac{x}{9} + x\right) = \frac{2}{3} + \frac{10}{27}x$

2. Substitute $x = \frac{1}{9}$ into the original expression to find its exact value in terms of $\sqrt{5}$.

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

3. Equate the exact value to the approximation using the first two terms evaluated at $x = \frac{1}{9}$, and solve for $\sqrt{5}$.

Substitute $x = \frac{1}{9}$ into the first two terms of the expansion:
$\frac{2}{3} + \frac{10}{27}\left(\frac{1}{9}\right) = \frac{2}{3} + \frac{10}{243} = \frac{162 + 10}{243} = \frac{172}{243}$
Equating the exact expression and the approximation:
$\frac{19}{12\sqrt{5}} \approx \frac{172}{243}$
$\sqrt{5} \approx \frac{19}{12} \times \frac{243}{172}$
$\sqrt{5} \approx \frac{19 \times 81}{4 \times 172} = \frac{1539}{688}$