8. 计算：
(1)$a - b+\frac {b^{2}}{a + b}$；
(2)$\frac {2x^{2}}{x - y}-\frac {x^{2}-4xy}{y - x}+\frac {2y^{2}-x^{2}}{x - y}$；
(3)$\frac {3}{2x + 4}-\frac {6}{4 - x^{2}}+\frac {9}{2x - 4}$；
(4)$\frac {1}{x + 1}+\frac {1}{x^{2}-1}+\frac {1}{1 - x}$.
答案：解:
(1)原式=$\frac{(a-b)(a+b)}{a+b}+\frac{b^{2}}{a+b}=\frac{a^{2}-b^{2}+b^{2}}{a+b}=\frac{a^{2}}{a+b}$.
(2)原式=$\frac{2x^{2}+x^{2}-4xy+2y^{2}-x^{2}}{x-y}$=$\frac{2(x-y)^{2}}{x-y}=2x-2y$.
(3)原式=$\frac{3(x-2)+12+9(x+2)}{2(x+2)(x-2)}$=$\frac{12(x+2)}{2(x+2)(x-2)}=\frac{6}{x-2}$.
(4)原式=$\frac{x-1+1-(x+1)}{x^{2}-1}=\frac{1}{1-x^{2}}$.
9. 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)有下列三组分式：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”的是
②③
(填序号)；
(2)求分式$\frac {a}{2a + 1}$的“美妙分式”；
解:设分式$\frac{a}{2a+1}$的"美妙分式"为A,则$|A-\frac{a}{2a+1}|=3$,所以$A=\frac{a}{2a+1}+3$或$A=\frac{a}{2a+1}-3$.当$A=\frac{a}{2a+1}+3$时,$A=\frac{a}{2a+1}+3=\frac{a}{2a+1}+\frac{6a+3}{2a+1}$=$\frac{7a+3}{2a+1}$;当$A=\frac{a}{2a+1}-3$时,$A=\frac{a}{2a+1}-3=\frac{a}{2a+1}-\frac{6a+3}{2a+1}=\frac{-5a-3}{2a+1}=-\frac{5a+3}{2a+1}$.故$\frac{a}{2a+1}$的"美妙分式"为$\frac{7a+3}{2a+1}$或$-\frac{5a+3}{2a+1}$.
(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.
解:由题意可得$|\frac{4a^{2}}{a^{2}-b^{2}}-\frac{a}{a+b}|=3$,即$|\frac{3a^{2}+ab}{a^{2}-b^{2}}|=3$.所以$\frac{3a^{2}+ab}{a^{2}-b^{2}}=3$或$\frac{3a^{2}+ab}{a^{2}-b^{2}}=-3$.当$\frac{3a^{2}+ab}{a^{2}-b^{2}}=3$时,化简得$3b^{2}+ab=0$,即$b(3b+a)=0$,因为a,b均不等于0,所以$3b+a=0$,即$a=-3b$,所以$\frac{2a^{2}-b^{2}}{ab}=\frac{18b^{2}-b^{2}}{-3b^{2}}=\frac{17b^{2}}{-3b^{2}}=-\frac{17}{3}$;当$\frac{3a^{2}+ab}{a^{2}-b^{2}}=-3$时,化简得$6a^{2}-3b^{2}+ab=0$,即$ab=-(6a^{2}-3b^{2})$,所以$\frac{2a^{2}-b^{2}}{ab}=\frac{2a^{2}-b^{2}}{-(6a^{2}-3b^{2})}=\frac{2a^{2}-b^{2}}{-3(2a^{2}-b^{2})}=-\frac{1}{3}$.综上可知,$\frac{2a^{2}-b^{2}}{ab}$的值为$-\frac{17}{3}$或$-\frac{1}{3}$.
答案：
(1)②③
(2)解:设分式$\frac{a}{2a+1}$的"美妙分式"为A,则$|A-\frac{a}{2a+1}|=3$,所以$A=\frac{a}{2a+1}+3$或$A=\frac{a}{2a+1}-3$.当$A=\frac{a}{2a+1}+3$时,$A=\frac{a}{2a+1}+3=\frac{a}{2a+1}+\frac{6a+3}{2a+1}$=$\frac{7a+3}{2a+1}$;当$A=\frac{a}{2a+1}-3$时,$A=\frac{a}{2a+1}-3=\frac{a}{2a+1}-\frac{6a+3}{2a+1}=\frac{-5a-3}{2a+1}=-\frac{5a+3}{2a+1}$.故$\frac{a}{2a+1}$的"美妙分式"为$\frac{7a+3}{2a+1}$或$-\frac{5a+3}{2a+1}$.
(3)解:由题意可得$|\frac{4a^{2}}{a^{2}-b^{2}}-\frac{a}{a+b}|=3$,即$|\frac{3a^{2}+ab}{a^{2}-b^{2}}|=3$.所以$\frac{3a^{2}+ab}{a^{2}-b^{2}}=3$或$\frac{3a^{2}+ab}{a^{2}-b^{2}}=-3$.当$\frac{3a^{2}+ab}{a^{2}-b^{2}}=3$时,化简得$3b^{2}+ab=0$,即$b(3b+a)=0$,因为a,b均不等于0,所以$3b+a=0$,即$a=-3b$,所以$\frac{2a^{2}-b^{2}}{ab}=\frac{18b^{2}-b^{2}}{-3b^{2}}=\frac{17b^{2}}{-3b^{2}}=-\frac{17}{3}$;当$\frac{3a^{2}+ab}{a^{2}-b^{2}}=-3$时,化简得$6a^{2}-3b^{2}+ab=0$,即$ab=-(6a^{2}-3b^{2})$,所以$\frac{2a^{2}-b^{2}}{ab}=\frac{2a^{2}-b^{2}}{-(6a^{2}-3b^{2})}=\frac{2a^{2}-b^{2}}{-3(2a^{2}-b^{2})}=-\frac{1}{3}$.综上可知,$\frac{2a^{2}-b^{2}}{ab}$的值为$-\frac{17}{3}$或$-\frac{1}{3}$.
