已知f（x）是偶函数,g（x）是奇函数,且在公共定义域｛x丨x∈R,x≠±1｝上有f（x）+g（x）=1／x-1,

x≠±1时有
f(x)+g(x) = 1/(x-1)
f(-x)+g(-x) = f(x)-g(x) = -1/(x+1)
两式相加有 2f(x) = 1/(x-1) - 1/(x+1) = 2/( (x-1)(x+1) )
得到 f(x) = 1/( (x-1)(x+1) )