一道数学题，超难。请高人来解

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f（x）=x+1/x-1,f^-1（x）=x+1/x-1,
所以f^-1(√2)=(√2+1)/(√2-1)=3+2√2

f(x)=1+2/(x-1)
当x>1时，f(x)>0且单调递减
所以当x>1时反函数存在
y=f(x)=(x+1)/(x-1)
xy-y=x+1
xy-x=y+1
(y-1)x=1+y
x=(1+y)/(y-1)
所以反函数是y=f^-1(x)=(x+1)/(x-1)
f^-1(√2)
=(√2+1)/(√2-1)
=(√2+1)^2/[(√2-1)(√2+1)]
=3+2√2

按定义来
x+1/x-1=√2
x²-(1+√2)x+1=0
求x

求f^-1(√2)=？也就是求原函数y=√2时x的值
把y=√2代入原函数
√2=(x+1)/(x-1)
√2(x-1)=x+1
(√2-1)x=√2+1
x=3+2√2
∴f^-1(√2)=3+2√2

解:设y=f(x)=(x+1)/(x-1),则 xy-y=x+1
因此,x=(y+1)/(y-1) .
所以,其反函数为:f^-1(x)=(x+1)/(x-1)
故,f^-1(√2) =(√2+1)/(√2-1)
=(√2+1)²/[(√2-1)*(√2+1)]
=3+2√2

3+2√2