已知F(-X)=F(X),G(-X)= -G(X),且F(X)+G(X)=1/(X+1)求F(X),G(X)的表达式

来源：百度知道 编辑：UC知道 时间：2024/05/11 19:57:01

f(x) + g(x) = 1/(x+1)
将此式中的x用-x代替，并利用奇偶性的条件，有
f(x) - g(x) = 1/(-x+1)

这是个二元一次方程组

f(x) = 1/2[1/(x+1) + 1/(-x+1)] = 1/(1-x^2)
g(x) = 1/2[1/(x+1) - 1/(-x+1)] = x/(x^2-1)

1----F(X)+G(X)=1/(X+1)
2----F(-x)+G(-x)=1/(1-x)
=>F(x)-G(x)=1/(1-x)

联合1，2
得出F(x)=1/(1-x^2)
G(x)=x/(x^2-1)

f(x) + g(x) = 1/(x+1)
f(x) - g(x) = 1/(-x+1)
f(x) = 1/2[1/(x+1) + 1/(-x+1)] = 1/(1-x^2)
g(x) = 1/2[1/(x+1) - 1/(-x+1)] = x/(x^2-1)