函数f(x)满足f(x-1)=2x-5,求f(x),f(x²),
设函数f(x)满足f(x)+2f(1/x)=x,求f(x)
若一次函数f(x) 满足f[f(x)]=1+2x 求f(x)
一次函数f(x)满足f [f(x)] =1+2x,求f(x)
已知函数f(x)满足3f(x)+2f(1/x)=x+1,求f(x)
二次函数f(x)满足f(x+1)+f(x-1)=2x^2+4x,求f(x)
若一次函数f(x)满足f[f(x)]=1+4x,求f(x)
F(X)满足F(x)+2f(x分之1)=3X,求f(x)
已知f(x)满足2f(x)+f(1/x)=3x,求f(x)
已知f(x)满足2f(x)+f(1/x)=3x,求f(x)?
已知f(x)满足2f(x)+f(-x)=-3x+1,求f(x)
f(x) 在定义域(0,正无穷)上是增函数,满足f(2)=1,f(xy)=f(x)+f(y).求不等式f(x)+f(x-
函数f(x)满足2f(x)+xf(-x)=x²+1,求f(x)解析式.