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

Question

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

Answer 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) )