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11. (滨州中考)观察下列各式：$\frac{2}{1\times3} = \frac{1}{1} - \frac{1}{3}$；$\frac{2}{2\times4} = \frac{1}{2} - \frac{1}{4}$；$\frac{2}{3\times5} = \frac{1}{3} - \frac{1}{5}$；…请利用你所得结论，化简代数式：$\frac{1}{1\times3} + \frac{1}{2\times4} + \frac{1}{3\times5} + \cdots + \frac{1}{n(n + 2)}$ ($n\geq3$且$n$为整数)，其结果为__________.
答案
$\frac{3n^{2}+5n}{4(n + 1)(n + 2)}$ 解析：$\because\frac{2}{1\times3}=\frac{1}{1}-\frac{1}{3}$，$\frac{2}{2\times4}=\frac{1}{2}-\frac{1}{4}$，$\frac{2}{3\times5}=\frac{1}{3}-\frac{1}{5}$，$\cdots$，$\frac{2}{n(n + 2)}=\frac{1}{n}-\frac{1}{n + 2}$，$\therefore\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\cdots+\frac{1}{n(n + 2)}=\frac{1}{2}(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\cdots+\frac{1}{n}-\frac{1}{n + 2})=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n + 1}-\frac{1}{n + 2})=\frac{3n^{2}+5n}{4(n + 1)(n + 2)}$.
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
$\because\frac{(x^{2}+y^{2})-(x - y)^{2}+2y(x - y)}{4y}=\frac{x^{2}+y^{2}-x^{2}+2xy - y^{2}+2xy - 2y^{2}}{4y}=\frac{4xy - 2y^{2}}{4y}=x-\frac{1}{2}y = 1$，$\therefore 2x - y = 2$.$\therefore$原式$=\frac{4x}{(2x + y)(2x - y)}-\frac{2x - y}{(2x + y)(2x - y)}=\frac{4x - 2x + y}{(2x + y)(2x - y)}=\frac{2x + y}{(2x + y)(2x - y)}=\frac{1}{2x - y}=\frac{1}{2}$.
13. 阅读理解：把一个分式写成两个分式的和叫做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$\begin{cases}m = - 1,\\n = - 2,\end{cases}$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$m =$______，$n =$______；
(2)请用上述方法将分式$\frac{4x - 3}{(2x + 1)(x - 2)}$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$\begin{cases}m = - 1,\\n = - 2,\end{cases}$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$m =$______，$n =$______；
(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
(1)$-\frac{1}{3}$ $\frac{1}{3}$
(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$\begin{cases}m + 2n = 4\\-2m + n=-3\end{cases}$，解得$\begin{cases}m = 2\\n = 1\end{cases}$，所以$\frac{4x - 3}{(2x + 1)(x - 2)}=\frac{2}{2x + 1}+\frac{1}{x - 2}$.
(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$\begin{cases}m + 2n = 4\\-2m + n=-3\end{cases}$，解得$\begin{cases}m = 2\\n = 1\end{cases}$，所以$\frac{4x - 3}{(2x + 1)(x - 2)}=\frac{2}{2x + 1}+\frac{1}{x - 2}$.
14. 新题型 新定义 (2024·南京期中)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______ (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.
(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______ (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
(1)②③ 解析：①$\left|\frac{1}{a + 1}-\frac{4}{a + 1}\right|=\left|-\frac{3}{a + 1}\right|\neq3$，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$-\frac{1}{3}$.
(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$-\frac{1}{3}$.
