设函数f(x)(x∈D1)是奇函数,g(x)(x∈D2)是偶函数,且D1∩D2不等于空集,f(x)+g(x)=2/(x-

f(x)+g(x)=2/(x-1) 1)以-x代替上式中的x,得：f(-x)+g(-x)=2/(-x-1) 2)由奇偶性,2)式化为：-f(x)+g(x)=2/(-x-1) 3)1)+3):2g(x)=2/(x-1)+2/(-x-1),得：g(x)=1/(x-1)-1/(x+1)=2/(x^2-1)1)-3):2f(x)=2/(x-1)-2/(-x-1),得：f(x)=1/(x-1)+1/(x+1)=2x/(x^2-1)