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5 (2025·上海黄浦区期末)解关于$x的方程\frac {x}{3}+\frac {x}{5}+\frac {x}{7}= 0$,我们也可以这样来解：
$(\frac {1}{3}+\frac {1}{5}+\frac {1}{7})x= 0$,因为$\frac {1}{3}+\frac {1}{5}+\frac {1}{7}≠0$,所以方程的解为$x= 0$。请按这种方法解下列方程：
(1)$\frac {x-1}{3}+\frac {x-1}{5}+\frac {x-1}{7}+\frac {x-1}{9}= 0$；
(2)$\frac {x-23}{2}+\frac {x-19}{4}+\frac {x-15}{6}+\frac {x-11}{8}+\frac {x-7}{10}= 10$。

答案：
(1)
∵原方程可化为$(x - 1)(\frac {1}{3}+\frac {1}{5}+\frac {1}{7}+\frac {1}{9}) = 0$,且$\frac {1}{3}+\frac {1}{5}+\frac {1}{7}+\frac {1}{9}\neq0$,
∴$x - 1 = 0$,
∴$x = 1$.
(2)
∵$\frac {x - 23}{2}+\frac {x - 19}{4}+\frac {x - 15}{6}+\frac {x - 11}{8}+\frac {x - 7}{10}-10 = 0$,
∴$\frac {x - 23}{2}-2+\frac {x - 19}{4}-2+\frac {x - 15}{6}-2+\frac {x - 11}{8}-2+\frac {x - 7}{10}-2 = 0$,即$\frac {x - 27}{2}+\frac {x - 27}{4}+\frac {x - 27}{6}+\frac {x - 27}{8}+\frac {x - 27}{10}=0$,
∴$(x - 27)(\frac {1}{2}+\frac {1}{4}+\frac {1}{6}+\frac {1}{8}+\frac {1}{10}) = 0$.提公因式法，将复杂方程变简单
∵$\frac {1}{2}+\frac {1}{4}+\frac {1}{6}+\frac {1}{8}+\frac {1}{10}\neq0$,
∴$x - 27 = 0$,
∴$x = 27$.

(1)$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}=$$\frac {4}{5}$；
(2)求解关于$x$的方程：$\frac {1}{-2×4}+\frac {1}{-4×6}+...+\frac {1}{-48×50}= \frac {19}{25}-2x$。
整理已知等式，得$-\frac {1}{2}×(\frac {1}{2}-\frac {1}{4}+\frac {1}{4}-\frac {1}{6}+…+\frac {1}{48}-\frac {1}{50})=\frac {19}{25}-2x$,即$-\frac {6}{25}=\frac {19}{25}-2x$,解得$x=\frac {1}{2}$.

答案：
(1)$\frac {4}{5}$
(2)整理已知等式，得$-\frac {1}{2}x(\frac {1}{2}-\frac {1}{4}+\frac {1}{4}-\frac {1}{6}+…+\frac {1}{48}-\frac {1}{50})=\frac {19}{25}-2x$,即$-\frac {6}{25}=\frac {19}{25}-2x$,解得$x=\frac {1}{2}$.