1. 计算:
(1)$\frac {2}{a-1}÷\frac {2a-4}{a^{2}-1}+\frac {1}{2-a};$ (2)$\frac {a+1}{a-3}-\frac {a-3}{a+2}÷\frac {a^{2}-6a+9}{a^{2}-4};$
(3)$\frac {a-1}{a}÷(a-\frac {1}{a});$ (4)$(\frac {3}{x+1}-\frac {1}{x+1})÷\frac {2}{x+1};$
(5)$(\frac {4a+1}{a-2}+a)÷\frac {a^{2}-1}{a-2};$ (6)$(\frac {3x}{x-2}-\frac {x}{x+2})÷\frac {x}{x^{2}-4};$
(7)$(\frac {1}{x}-1)÷\frac {x^{2}-1}{x};$ (8)$(x+\frac {1}{x}+2)÷(x-\frac {1}{x});$
(9)$(\frac {a^{2}}{a-1}-a-1)÷\frac {2a}{a^{2}-1};$ (10)$(a-1+\frac {1}{a+1})÷\frac {a^{2}+2a}{a+1}.$
答案：1.解:(1)原式$=\frac {2}{a-1}\cdot \frac {(a-1)(a+1)}{2(a-2)}-\frac {1}{a-2}=\frac {a+1}{a-2}-\frac {1}{a-2}=\frac {a}{a-2}.$
(2)原式$=\frac {a+1}{a-3}-\frac {a-3}{a+2}\cdot \frac {(a+2)(a-2)}{(a-3)^{2}}=\frac {a+1}{a-3}-\frac {a-2}{a-3}=\frac {3}{a-3}.$
(3)原式$=\frac {a-1}{a}÷\frac {a^{2}-1}{a}=\frac {a-1}{a}\cdot \frac {a}{(a+1)(a-1)}=\frac {1}{a+1}.$
(4)原式$=\frac {2}{x+1}÷\frac {2}{x+1}=1.$
(5)原式$=\frac {a^{2}+2a+1}{a-2}÷\frac {(a-1)(a+1)}{a-2}=\frac {(a+1)^{2}}{a-2}\cdot \frac {a-2}{(a+1)(a-1)}=\frac {a+1}{a-1}.$
(6)原式$=(\frac {3x}{x-2}-\frac {x}{x+2})\cdot \frac {(x+2)(x-2)}{x}=3(x+2)-(x-2)=2x+8.$
(7)原式$=\frac {1-x}{x}\cdot \frac {x}{(x+1)(x-1)}=-\frac {1}{x+1}.$
(8)原式$=\frac {x^{2}+1+2x}{x}\cdot \frac {x}{(x+1)(x-1)}=\frac {(x+1)^{2}}{x}\cdot \frac {x}{(x+1)(x-1)}=\frac {x+1}{x-1}.$
(9)原式$=(\frac {a^{2}}{a-1}-\frac {a^{2}-1}{a-1})÷\frac {2a}{(a+1)(a-1)}=\frac {1}{a-1}\cdot \frac {(a+1)(a-1)}{2a}=\frac {a+1}{2a}.$
(10)原式$=(\frac {a^{2}-1}{a+1}+\frac {1}{a+1})÷\frac {a(a+2)}{a+1}=\frac {a^{2}}{a+1}\cdot \frac {a+1}{a(a+2)}=\frac {a}{a+2}.$
2. (2024 春·淮安期末)(1)已知$\frac {x}{x^{2}+1}= \frac {1}{2}$,求$\frac {x^{2}}{x^{4}+1}$的值;
(2)已知$\frac {x}{x^{2}-x+1}= \frac {1}{7}$,求$\frac {x^{2}}{x^{4}-x^{2}+1}$的值;
(3)已知$\frac {xy}{x+y}= 2,\frac {yz}{y+z}= \frac {4}{3},\frac {zx}{z+x}= \frac {4}{3}$,求$\frac {xyz}{xy+yz+zx}$的值.
答案：2.解:(1)由$\frac {x}{x^{2}+1}=\frac {1}{2}$,知$x≠0,\therefore \frac {x^{2}+1}{x}=2$,即$x+\frac {1}{x}=2,\therefore \frac {x^{4}+1}{x^{2}}=x^{2}+\frac {1}{x^{2}}=(x+\frac {1}{x})^{2}-2=2^{2}-2=2,\therefore \frac {x^{2}}{x^{4}+1}$的值为2的倒数,即$\frac {1}{2}.$
(2)由$\frac {x}{x^{2}-x+1}=\frac {1}{7}$,知$x≠0,\therefore \frac {x^{2}-x+1}{x}=7,\therefore x-1+\frac {1}{x}=7$,即$x+\frac {1}{x}=8,\therefore \frac {x^{4}-x^{2}+1}{x^{2}}=x^{2}-1+\frac {1}{x^{2}}=(x+\frac {1}{x})^{2}-3=8^{2}-3=61,\therefore \frac {x^{2}}{x^{4}-x^{2}+1}$的值为61的倒数,即$\frac {1}{61}.$
(3)由$\frac {xy}{x+y}=2$,知$x≠0,y≠0,\therefore \frac {x+y}{xy}=\frac {1}{2},\therefore \frac {1}{y}+\frac {1}{x}=\frac {1}{2}$①,由$\frac {yz}{y+z}=\frac {4}{3}$,知$y≠0,z≠0,\therefore \frac {y+z}{yz}=\frac {3}{4},\therefore \frac {1}{z}+\frac {1}{y}=\frac {3}{4}$②,由$\frac {zx}{z+x}=\frac {4}{3}$,知$z≠0,x≠0,\therefore \frac {z+x}{zx}=\frac {3}{4},\therefore \frac {1}{x}+\frac {1}{z}=\frac {3}{4}$③,①+②+③得$2(\frac {1}{x}+\frac {1}{y}+\frac {1}{z})=\frac {1}{2}+\frac {3}{4}+\frac {3}{4},\therefore \frac {1}{x}+\frac {1}{y}+\frac {1}{z}=1,\therefore \frac {xy+yz+zx}{xyz}=\frac {1}{z}+\frac {1}{x}+\frac {1}{y}=1,\therefore \frac {xyz}{xy+yz+zx}$的值为1的倒数,即1.
