设an是函数f(x)=x^3+n^2*x-1的零点,证明;0<an<1
设函数f{x}=log2x-logx4{0<x<1}.数列{An}的通项An满足f{2的an次方}=2n
设函数f(x)=log2 x-logx 4(0<x<1),数列{an}的通项an满足f(2∧an)=2n(n∈N*),
设函数f(x)=log2 x-logx 4(0<x<1),数列{an}的通项an满足f(2∧an)=2n(n∈N*),试
已知函数f(x)=(x^3-x) /3,数列{an}满足a1>=1,an+1>=f'(an+1)证明an>=(2^n)-
已知函数f(X)=X/(3x+1),数列{an}满足a1=1,a(n+1)=f(an),证明数列{1/an}是等差数列
设函数f(x)=2x+3/3x(x>0),数列{an}满足a1=1,an=f(1/an-1)(n≥2,n∈N*) (1)
设函数f(x)= 2x+3 3x (x>0),数列{an}满足a1=1,an=f( 1 an-1 )(n∈N*,且n≥2
设函数f(x)=(2x+1)/x [x>0] 数列an满足a1=1,an=f[1/a(n-1)]
设函数f(x)=1/3x-lnx,则f(x)的零点个数是
设函数f(x)=x/(2x+1),数列{an}满足a1=1,an+1=f(an),n∈N*
设函数f(x)=(2x+3)/3x(x> 0),数列{an}满足a1=1,an=f(1/an-1a)(n∈n*,
设函数f(x)=2x+3/3x x>0 数列{an}满足a1=1 an=f(1/an-1)