f(√(x+1)) = x+2√2 ,( x > = 0),求f(x)嗎?令 t= √(x+1) , 則 x = t*t -1 > = 0. t > = 1 或 t < = -1.所以 f(t) = t*t -1 + 2√2 , (t > =1或 t < = -1 )將t 換成 x,得 f(x) = x*x -1 + 2√2 , (x > =1 或 x < = ...f（（√x）+1))( x > = 0) = x+2√2 ,( x > = 01,求f(x)一樣的啊令t = （√x）+1 , x = (t -1)*(t -1) , 因為x > = 0, 所以t > = 1即f(t) =(t -1)*(t -1)+ 2√2 , t > = 1將t 換成 x得 f(x) =(x -1)*(x -1)+ 2√2 , x > = 1問多一道題可以嗎？ 已知f（x+1）=x平方+bx+2，求f（x）表達式這中類型一樣的，都是這種解法令 t = x+1,x = t-1,f(t ) = (t-1)*(t-1) + b*(t-1)+2.將t換成xf(x) = (x-1)*(x-1) + b*(x-1)+2.或者可以用平移x+1---->x即f(x+a)---->f(x),a為任意常數(shù)。