Question

已知f（根号x+1）=x+2根号x,请用换元法解

请写得详细些，谢谢！

Answer 1

f(√x+1）=x+2√x
设√x+1=t(t≥1),则√x=t-1,x=(t-1)^2
所以f(t)=(t-1)^2+2(t-1)(t≥1),
即f(x)=(x-1)^2+2(x-1) (x≥1),

Answer 2

根号x+1=t,x=(t-1)^2
f(t)=(t-1)^2+2(t-1)
f(x)=(x-1)^2+2(x-1)