Question

设对满足x不等于0丶1的所有实数x,函数f[x]满足f[x]+f[x分之x-1]=1+x,求f[x]的解析式

Answer 1

令x=(y-1)/y

f((y-1)/y)+f(1/(1-y))=1+(y-1)/y

令y=(z-1)/z

f(1/(1-z))+f(z)=1+1/(1-z)

将字母换掉,可得

f(x)+f((x-1)/x)=1+x

f((x-1)/x)+f(1/(1-x))=1+(x-1)/x

f(1/(1-x))+f(x)=1+1/(1-x)

方便起见,设f(x)=a,f((x-1)/x)=b,f(1/(1-x))=c.

a+b=1+x ①

b+c=1+(x-1)/x ②

c+a=1+1/(1-x) ③

③-②得a-b=1/(1-x)-(x-1)/x

再加上①得

2a=1+x+1/(1-x)-(x-1)/x

a=[1+x+1/(1-x)-(x-1)/x]/2

即f(x)=[1+x+1/(1-x)-(x-1)/x]/2