已知函数f(x)=x/(x+1),若数列{an}满足：an>0,a1=1,an+1=[f(根号an)]^2?(1)求数列

a(n+1)=[f(√an)]^2,
a(n+1)=[√an/(√an+1)] ^2.
则√a(n+1)=√an/(√an+1)
1/√a(n+1)= (√an+1)/√an
即1/√a(n+1)=1+1/√an,
所以数列{1/√an }是等差数列,所以1/√an=1+(n-1)*1,
1/√an=n,an=1/n^2.