将 $\displaystyle\frac{2+x}{\sqrt{9-x}}$ 展开为 $x$ 的升幂级数，直到并包括 $x^2$ 的项。 仅使用上述级数展开式的前两项，并代入 $x = \frac{1}{9}$，求出 $\sqrt{5}$ 的近似值，写成最简分数形式 $\frac{p}{q}$，其中 $p$ 和 $q$ 是待确定的整数。

1. Rewrite the expression as a product and factor out the constant to prepare for the binomial expansion.

$\frac{2+x}{\sqrt{9-x}} = (2+x)(9-x)^{-\frac{1}{2}}$

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

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

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

$= 1 + \frac{x}{18} + \frac{3}{8}\left(\frac{x^2}{81}\right) + \dots$

$= 1 + \frac{x}{18} + \frac{x^2}{216} + \dots$

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

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

$= \frac{1}{3}\left[ 2\left(1 + \frac{x}{18} + \frac{x^2}{216}\right) + x\left(1 + \frac{x}{18}\right) \right] + \dots$

$= \frac{1}{3}\left( 2 + \frac{x}{9} + \frac{x^2}{108} + x + \frac{x^2}{18} \right)$

$= \frac{1}{3}\left( 2 + \frac{10x}{9} + \frac{7x^2}{108} \right)$

$= \frac{2}{3} + \frac{10}{27}x + \frac{7}{324}x^2$

1. Concisely obtain the first two terms of the series expansion from part (a) (since its result is required for this part).

$\frac{2+x}{\sqrt{9-x}} = (2+x)\left[9\left(1-\frac{x}{9}\right)\right]^{-\frac{1}{2}} = \frac{2+x}{3}\left(1-\frac{x}{9}\right)^{-\frac{1}{2}}$
$\approx \left(\frac{2}{3} + \frac{x}{3}\right)\left(1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{9}\right)\right) \approx \left(\frac{2}{3} + \frac{x}{3}\right)\left(1 + \frac{x}{18}\right)$
$\approx \frac{2}{3} + \frac{x}{27} + \frac{x}{3} = \frac{2}{3} + \frac{10}{27}x$

2. Evaluate the original expression exactly at $x = \frac{1}{9}$.

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

3. Evaluate the first two terms of the expansion at $x = \frac{1}{9}$ and equate to find $\sqrt{5}$.

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