Question

函数f[x]满足f[根号x+1】=x+2根号x 求f[x]的解析式

Answer 1

令t=√(X+1),t≥0,
t^2=X+1,X=t^2-1,
∴f(t)=t^2-1+2√(t^2-1),
∴f(X)=X^2+2√(X^2-1)-1(X≥0).