13. 先化简,再求值：
(1) $$ ( \frac { 2 - 2 x } { x + 1 } + x - 1 ) ÷ \frac { x ^ { 2 } - x } { x + 1 } $$,其中$$ x = ( \frac { 1 } { 2 } ) ^ { - 1 } + ( - 3 ) ^ { 0 } $$；
(2) $$ ( 1 - \frac { 2 } { x } ) ÷ \frac { x ^ { 2 } - 4 x + 4 } { x ^ { 2 } - 4 } - \frac { x + 4 } { x + 2 } $$,其中$$ x ^ { 2 } + 2 x - 15 = 0 $$.

(1) 化简：
$\begin{aligned}&(\frac{2 - 2x}{x + 1} + x - 1) ÷ \frac{x^2 - x}{x + 1}\\=&\left[\frac{2 - 2x}{x + 1} + \frac{(x - 1)(x + 1)}{x + 1}\right] \cdot \frac{x + 1}{x(x - 1)}\\=&\frac{2 - 2x + x^2 - 1}{x + 1} \cdot \frac{x + 1}{x(x - 1)}\\=&\frac{(x - 1)^2}{x + 1} \cdot \frac{x + 1}{x(x - 1)}\\=&\frac{x - 1}{x}\end{aligned}$
求值：$x = (\frac{1}{2})^{-1} + (-3)^0 = 2 + 1 = 3$，代入得$\frac{3 - 1}{3} = \frac{2}{3}$。
(2) 化简：
$\begin{aligned}&(1 - \frac{2}{x}) ÷ \frac{x^2 - 4x + 4}{x^2 - 4} - \frac{x + 4}{x + 2}\\=&\frac{x - 2}{x} \cdot \frac{(x - 2)(x + 2)}{(x - 2)^2} - \frac{x + 4}{x + 2}\\=&\frac{x + 2}{x} - \frac{x + 4}{x + 2}\\=&\frac{(x + 2)^2 - x(x + 4)}{x(x + 2)}\\=&\frac{x^2 + 4x + 4 - x^2 - 4x}{x(x + 2)}\\=&\frac{4}{x(x + 2)} = \frac{4}{x^2 + 2x}\end{aligned}$
求值：由$x^2 + 2x - 15 = 0$得$x^2 + 2x = 15$，代入得$\frac{4}{15}$。
(1) $\frac{2}{3}$；(2) $\frac{4}{15}$

14. 计算：$ \frac { 1 } { x ^ { 2 } + x } + \frac { 1 } { x ^ { 2 } + 3 x + 2 } + \frac { 1 } { x ^ { 2 } + 5 x + 6 } + \frac { 1 } { x ^ { 2 } + 7 x + 12 }. $

原式
$= \frac{1}{x(x + 1)} + \frac{1}{(x + 1)(x + 2)} + \frac{1}{(x + 2)(x + 3)} + \frac{1}{(x + 3)(x + 4)}$
对于第一项和第二项，利用部分分式的性质，有：
$\frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1}$
解得：$A = 1, B = -1$，即：
$\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$
同理，对于其他项进行分解，得到：
$\frac{1}{(x + 1)(x + 2)} = \frac{1}{x + 1} - \frac{1}{x + 2}$
$\frac{1}{(x + 2)(x + 3)} = \frac{1}{x + 2} - \frac{1}{x + 3}$
$\frac{1}{(x + 3)(x + 4)} = \frac{1}{x + 3} - \frac{1}{x + 4}$
将上述四个分解后的式子相加，得到：
$= (\frac{1}{x} - \frac{1}{x + 1}) + (\frac{1}{x + 1} - \frac{1}{x + 2}) + (\frac{1}{x + 2} - \frac{1}{x + 3}) + (\frac{1}{x + 3} - \frac{1}{x + 4})$
观察上式，发现其中多项可以相互抵消，化简后得到：
$= \frac{1}{x} - \frac{1}{x + 4}$
$= \frac{4}{x(x + 4)}$
所以，原式的结果为 $\frac{4}{x(x + 4)}$。

15. 先化简,再求值：
(1) $$ ( \frac { x ^ { 2 } - 2 x - 3 } { x ^ { 2 } - 1 } - \frac { x + 1 } { x - 1 } ) \cdot ( \frac { x - 3 } { x - 1 } ) ^ { - 1 } $$,其中$$ x = 2 $$.
(2) $$ \frac { a ^ { 2 } - 6 a + 9 } { a - 2 } ÷ ( a + 2 + \frac { 5 } { 2 - a } ) $$,其中$$ a $是使不等式$ \frac { a - 1 } { 2 } \leq 1 $$成立的正整数.

(1) 4；(2) -$\frac{1}{2}$

(1)
$\begin{aligned}&\left( \frac{x^2 - 2x - 3}{x^2 - 1} - \frac{x + 1}{x - 1} \right) \cdot \left( \frac{x - 3}{x - 1} \right)^{-1}\\=&\left( \frac{x^2 - 2x - 3}{x^2 - 1} - \frac{x + 1}{x - 1} \right) \cdot \frac{x - 1}{x - 3}\\=&\left( \frac{(x - 3)(x + 1)}{(x - 1)(x + 1)} - \frac{x + 1}{x - 1} \right) \cdot \frac{x - 1}{x - 3}\\=&\left( \frac{x - 3}{x - 1} - \frac{x + 1}{x - 1} \right) \cdot \frac{x - 1}{x - 3}\\=&\frac{(x - 3) - (x + 1)}{x - 1} \cdot \frac{x - 1}{x - 3}\\=&\frac{-4}{x - 1} \cdot \frac{x - 1}{x - 3}\\=&\frac{-4}{x - 3}\end{aligned}$
当$x = 2$时，$\frac{-4}{2 - 3} = \frac{-4}{-1} = 4$
(2)
解不等式$\frac{a - 1}{2} \leq 1$：$a - 1 \leq 2$，$a \leq 3$，正整数$a = 1, 2, 3$。
$\begin{aligned}&\frac{a^2 - 6a + 9}{a - 2} ÷ \left( a + 2 + \frac{5}{2 - a} \right)\\=&\frac{(a - 3)^2}{a - 2} ÷ \left( \frac{(a + 2)(2 - a) + 5}{2 - a} \right)\\=&\frac{(a - 3)^2}{a - 2} ÷ \left( \frac{-a^2 + 4 + 5}{2 - a} \right)\\=&\frac{(a - 3)^2}{a - 2} ÷ \frac{9 - a^2}{2 - a}\\=&\frac{(a - 3)^2}{a - 2} \cdot \frac{2 - a}{(3 - a)(3 + a)}\\=&\frac{(a - 3)^2}{a - 2} \cdot \frac{-(a - 2)}{-(a - 3)(a + 3)}\\=&\frac{a - 3}{a + 3}\end{aligned}$
$a = 2$时，分母为0；$a = 3$时，除数为0，均舍去。故$a = 1$，代入得$\frac{1 - 3}{1 + 3} = -\frac{1}{2}$