11. (滨州中考)观察下列各式：$\frac{2}{1×3} = \frac{1}{1} - \frac{1}{3}$；$\frac{2}{2×4} = \frac{1}{2} - \frac{1}{4}$；$\frac{2}{3×5} = \frac{1}{3} - \frac{1}{5}$；…；请利用你所得结论，化简代数式：$\frac{1}{1×3} + \frac{1}{2×4} + \frac{1}{3×5} + ··· + \frac{1}{n(n + 2)}$($n ≥ 3$且n为整数)，其结果为
$\frac{3n^{2} + 5n}{4(n + 1)(n + 2)}$
。
答案：11. $\frac{3n^{2} + 5n}{4(n + 1)(n + 2)}$ 解析：$\because \frac{2}{1 × 3} = \frac{1}{1} - \frac{1}{3}$，$\frac{2}{2 × 4} = \frac{1}{2} - \frac{1}{4}$，$\frac{2}{3 × 5} = \frac{1}{3} - \frac{1}{5}$，$···$，$\frac{2}{n(n + 2)} = \frac{1}{n} - \frac{1}{n + 2}$，$\therefore \frac{1}{1 × 3} + \frac{1}{2 × 4} + \frac{1}{3 × 5} + ··· + \frac{1}{n(n + 2)} = \frac{1}{2}(1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} -\frac{1}{5} + ··· + \frac{1}{n} - \frac{1}{n + 2}) = \frac{1}{2}(1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) =\frac{3n^{2} + 5n}{4(n + 1)(n + 2)}$.
解析：
由题意可得：$\frac{2}{n(n + 2)} = \frac{1}{n} - \frac{1}{n + 2}$，则$\frac{1}{n(n + 2)}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n + 2})$。
所以$\frac{1}{1×3} + \frac{1}{2×4} + \frac{1}{3×5} + ··· + \frac{1}{n(n + 2)}$
$=\frac{1}{2}(1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + ··· + \frac{1}{n} - \frac{1}{n + 2})$
$=\frac{1}{2}(1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2})$
$=\frac{1}{2}(\frac{3}{2} - \frac{(n + 2) + (n + 1)}{(n + 1)(n + 2)})$
$=\frac{1}{2}(\frac{3}{2} - \frac{2n + 3}{(n + 1)(n + 2)})$
$=\frac{1}{2}(\frac{3(n + 1)(n + 2) - 2(2n + 3)}{2(n + 1)(n + 2)})$
$=\frac{1}{2}(\frac{3(n^{2} + 3n + 2) - 4n - 6}{2(n + 1)(n + 2)})$
$=\frac{1}{2}(\frac{3n^{2} + 9n + 6 - 4n - 6}{2(n + 1)(n + 2)})$
$=\frac{1}{2}(\frac{3n^{2} + 5n}{2(n + 1)(n + 2)})$
$=\frac{3n^{2} + 5n}{4(n + 1)(n + 2)}$
$\frac{3n^{2} + 5n}{4(n + 1)(n + 2)}$
12. 已知$\frac{(x^2 + y^2) - (x - y)^2 + 2y(x - y)}{4y} = 1$，求$\frac{4x}{4x^2 - y^2} - \frac{1}{2x + y}$的值。
答案：12. $\because \frac{(x^{2} + y^{2}) - (x - y)^{2} + 2y(x - y)}{4y} = \frac{x^{2} + y^{2} - x^{2} + 2xy - y^{2} + 2xy - 2y^{2}}{4y} =\frac{4xy - 2y^{2}}{4y} = x - \frac{1}{2}y = 1$，$\therefore 2x - y = 2$. $\therefore$ 原式$= \frac{4x}{(2x + y)(2x - y)} -\frac{2x - y}{(2x + y)(2x - y)} = \frac{4x - 2x + y}{(2x + y)(2x - y)} = \frac{2x + y}{(2x + y)(2x - y)} = \frac{1}{2x - y} = \frac{1}{2}$.
解析：
$\because \frac{(x^{2} + y^{2}) - (x - y)^{2} + 2y(x - y)}{4y} = \frac{x^{2} + y^{2} - (x^{2} - 2xy + y^{2}) + 2xy - 2y^{2}}{4y} = \frac{x^{2} + y^{2} - x^{2} + 2xy - y^{2} + 2xy - 2y^{2}}{4y} = \frac{4xy - 2y^{2}}{4y} = x - \frac{1}{2}y$，
又$\because \frac{(x^{2} + y^{2}) - (x - y)^{2} + 2y(x - y)}{4y} = 1$，
$\therefore x - \frac{1}{2}y = 1$，
两边同乘$2$得：$2x - y = 2$。
$\frac{4x}{4x^2 - y^2} - \frac{1}{2x + y} = \frac{4x}{(2x + y)(2x - y)} - \frac{1}{2x + y} = \frac{4x}{(2x + y)(2x - y)} - \frac{2x - y}{(2x + y)(2x - y)} = \frac{4x - (2x - y)}{(2x + y)(2x - y)} = \frac{4x - 2x + y}{(2x + y)(2x - y)} = \frac{2x + y}{(2x + y)(2x - y)} = \frac{1}{2x - y}$，
$\because 2x - y = 2$，
$\therefore$原式$= \frac{1}{2}$。
13. 阅读理解：把一个分式写成两个分式的和叫作把这个分式表示成部分分式。如何将$\frac{1 - 3x}{x^2 - 1}$表示成部分分式？
设分式$\frac{1 - 3x}{x^2 - 1} = \frac{m}{x - 1} + \frac{n}{x + 1}$，将等式的右边通分得$\frac{m(x + 1) + n(x - 1)}{(x + 1)(x - 1)} = \frac{(m + n)x + m - n}{(x + 1)(x - 1)}$，由$\frac{1 - 3x}{x^2 - 1} = \frac{(m + n)x + m - n}{(x + 1)(x - 1)}$得$\begin{cases}m + n = - 3,\\m - n = 1,\end{cases}$解得$\{ \begin{array} { l } { m = - 1 }, \\ { n = - 2 }, \end{array} $所以$\frac{1 - 3x}{x^2 - 1} = \frac{- 1}{x - 1} + \frac{- 2}{x + 1}$。
(1)把分式$\frac{1}{(x - 2)(x - 5)}$表示成部分分式，即$\frac{1}{(x - 2)(x - 5)} = \frac{m}{x - 2} + \frac{n}{x - 5}$，则$m =$
$-\frac{1}{3}$
，$n =$
$\frac{1}{3}$
；
(2)请用上述方法将分式$\frac{4x - 3}{(2x + 1)(x - 2)}$表示成部分分式。
答案：13. (1)$-\frac{1}{3}$ $\frac{1}{3}$ 解析：$\frac{m}{x - 2} + \frac{n}{x - 5} = \frac{m(x - 5) + n(x - 2)}{(x - 2)(x - 5)} =\frac{(m + n)x - 5m - 2n}{(x - 2)(x - 5)}$，则$m + n = 0$，$-5m - 2n = 1$，解得$m = -\frac{1}{3}$，$n = \frac{1}{3}$.
(2)设分式$\frac{4x - 3}{(2x + 1)(x - 2)} = \frac{m}{2x + 1} + \frac{n}{x - 2}$，将等式的右边通分得$\frac{m(x - 2) + n(2x + 1)}{(2x + 1)(x - 2)} =\frac{(m + 2n)x - 2m + n}{(2x + 1)(x - 2)}$，由$\frac{4x - 3}{(2x + 1)(x - 2)} = \frac{(m + 2n)x - 2m + n}{(2x + 1)(x - 2)}$，得$\{\begin{array}{l}m + 2n = 4,\\-2m + n = -3,\end{array} $ 解得$\{\begin{array}{l}m = 2,\ = 1.\end{array} $ 所以$\frac{4x - 3}{(2x + 1)(x - 2)} = \frac{2}{2x + 1} + \frac{1}{x - 2}$.
14. (2025·南京期中)若$\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2}$(A，B，C均为常数)的计算结果为$\frac{x^2 + 2}{x(x + 1)(x + 2)}$，则$A + B + 2C$的值为(
D
)
A.1
B.2
C.3
D.4
答案：14. D 解析：$\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2} = \frac{(x + 1)(x + 2)A + x(x + 2)B + x(x + 1)C}{x(x + 1)(x + 2)} =\frac{(x^{2} + 3x + 2)A + (x^{2} + 2x)B + (x^{2} + x)C}{x(x + 1)(x + 2)} =\frac{(A + B + C)x^{2} + (3A + 2B + C)x + 2A}{x(x + 1)(x + 2)}$，$\because \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2}$（$A$，$B$，$C$均为常数）的计算结果为$\frac{x^{2} + 2}{x(x + 1)(x + 2)}$，$\therefore \{\begin{array}{l}2A = 2,\\A + B + C = 1,\\3A + 2B + C = 0,\end{array} $ 解得$\{\begin{array}{l}A = 1,\\B = -3,\\C = 3,\end{array} $ $\therefore A + B + 2C = 1 + (-3) + 2 × 3 = 4$，故选 D.
15. 新题型
新定义 定义：若两个分式A与B满足：$A - B = 3$，则称A与B这两个分式互为“美妙分式”。
(1)下列三组分式：①$\frac{1}{a + 1}$与$\frac{4}{a + 1}$；②$\frac{4a}{a + 1}$与$\frac{a - 3}{a + 1}$；③$\frac{a}{2a - 1}$与$\frac{7a - 3}{2a - 1}$。其中互为“美妙分式”的有
②③
(只填序号)。
(2)求分式$\frac{a}{2a + 1}$的“美妙分式”。
(3)若分式$\frac{4a^2}{a^2 - b^2}$与$\frac{a}{a + b}$互为“美妙分式”，且a，b均为不等于0的实数，求分式$\frac{2a^2 - b^2}{ab}$的值。
答案：15. (1)②③ 解析：①$\left|\frac{1}{a + 1} - \frac{4}{a + 1}\right| = \left|-\frac{3}{a + 1}\right| ≠ 3$，②$\left|\frac{4a}{a + 1} -\frac{a - 3}{a + 1}\right| = \left|\frac{3a + 3}{a + 1}\right| = 3$，③$\left|\frac{a}{2a - 1} - \frac{7a - 3}{2a - 1}\right| = \left|\frac{-6a + 3}{2a - 1}\right| =\left|\frac{-(6a - 3)}{2a - 1}\right| = 3$，故答案为②③.
(2)设分式$\frac{a}{2a + 1}$的“美妙分式”为$A$，则$\left|A - \frac{a}{2a + 1}\right| = 3$，$\therefore A - \frac{a}{2a + 1} = 3$或$A - \frac{a}{2a + 1} = -3$.
①当$A - \frac{a}{2a + 1} = 3$时，$A = \frac{a}{2a + 1} + 3 = \frac{a}{2a + 1} + \frac{6a + 3}{2a + 1} = \frac{7a + 3}{2a + 1}$；
②当$A - \frac{a}{2a + 1} = -3$时，$A = \frac{a}{2a + 1} - 3 = \frac{a}{2a + 1} - \frac{6a + 3}{2a + 1} = \frac{-5a - 3}{2a + 1} =\frac{-5a + 3}{2a + 1}$.
综上，分式$\frac{a}{2a + 1}$的“美妙分式”为$\frac{7a + 3}{2a + 1}$或$-\frac{5a + 3}{2a + 1}$.
(3)$\because \frac{4a^{2}}{a^{2} - b^{2}}$与$\frac{a}{a + b}$互为“美妙分式”，$\therefore \left|\frac{4a^{2}}{a^{2} - b^{2}} - \frac{a}{a + b}\right| = 3$.
$\because \left|\frac{4a^{2}}{a^{2} - b^{2}} - \frac{a}{a + b}\right| = \left|\frac{4a^{2}}{(a + b)(a - b)} - \frac{a(a - b)}{(a + b)(a - b)}\right| =\left|\frac{3a^{2} + ab}{(a + b)(a - b)}\right| = 3$，$\therefore \frac{3a^{2} + ab}{(a + b)(a - b)} = 3$或$\frac{3a^{2} + ab}{(a + b)(a - b)} = -3$，$\therefore 3a^{2} + ab = 3(a^{2} - b^{2})$或$3a^{2} + ab = -3(a^{2} - b^{2})$. $\because a$，$b$均为不等于 0 的实数，$\therefore a = -3b$或$ab = 3b^{2} - 6a^{2}$，把$a = -3b$代入$\frac{2a^{2} - b^{2}}{ab} = \frac{2(-3b)^{2} - b^{2}}{-3b^{2}} = \frac{17b^{2}}{-3b^{2}} =-\frac{17}{3}$，把$ab = 3b^{2} - 6a^{2}$代入$\frac{2a^{2} - b^{2}}{ab} = \frac{2a^{2} - b^{2}}{3b^{2} - 6a^{2}} =\frac{2a^{2} - b^{2}}{-3(2a^{2} - b^{2})} = -\frac{1}{3}$. 综上，分式$\frac{2a^{2} - b^{2}}{ab}$的值为$-\frac{17}{3}$或$-\frac{1}{3}$.
