高数求函数表达式设f[(x+1)/x]=(x+1)/x^2 (x≠0),则f(x)=? 答案是:x(x-1)这个题用什么

换元法=_=
设 t =(x + 1)/x
所以 x = 1 / (t - 1)
f[(x+1)/x]=(x+1)/x^2
f(t) = (x+1)/x * 1/x
= t * 1/[1/(t-1)]
=t*(t - 1)
所以f(x) = x*(x-1)