Question

已知f(x+1/x)=x²+1/x²+1,求f(x)的解析式

Answer 1

设y=x
(y-1)f[(y+1)/(y-1)]+f(y)=y
(1)
设y=(x+1)/(x-1)
[(y+1)/(y-1)-1]f{[(y+1)/(y-1)+1]/[(y+1)/(y-1)-1]}+f[(y+1)/(y-1)]=(y+1)/(y-1)
[(y+1-y+1)/(y-1)f{[(y+1+y-1)/(y-1)]/[(y+1-y+1)/(y-1)]}+f[(y+1)/(y-1)]=(y+1)/(y-1)
2/(y-1)f{[2y/(y-1)]/[2/(y-1)]})]}+f[(y+1)/(y-1)]=(y+1)/(y-1)
2/(y-1)f(y)+f[(y+1)/(y-1)]=(y+1)/(y-1)
2f(y)+(y-1)f[(y+1)/(y-1)]=y+1
(2)
(2)-(1)，得
f(y)=1
由y=x ，x≠-1（总感觉x≠-1这个条件不太对，应该是x≠1才对），得y≠-1
由y=(x+1)/(x-1)=1+2/(x-1)，得y≠1
f(y)=1
(y≠±
1)
函数解析式为f(x)=1
(x≠±
1)

Answer 2

解：f(x+1/x)=x²+1/x²+1
=(x+1/x)²-1
所以：f(x)=x²-1

追问

f(x+1/x)=x²+1/x²+1
=(x+1/x)²-1
???

追答

f(x+1/x)=x²+2+1/x²-2+1
=(x+1/x)²-1

追问

f(x+1/x)=(x+1/x)²-1
跟f(x)=x²-1 咋来的？

追答

f(x+1/x)=(x+1/x)²-1
令t=x+1/x
则有：f(t)=t²-1
令：x=t 则有：
f(x)=x²-1