Question

F(根号X+1）=X+2倍根号X 求F（X）,F(x+1),F(x方)

Answer 1

f(√x + 1)
= x + 2√x
= (√x)² + 2√x + 1 - 1
= (√x + 1)² - 1

所以
f(x) = x² - 1
f(x + 1) = (x + 1)² - 1 = x² + 2x
f(x²) = (x²)² - 1 = x^4 - 1 (x^4 意思是 x 的 4 次方)

Answer 2

F(√X+1）=X+2√X 求F（X）,F(x+1),F(x方)
解析：∵F(√X+1)=X+2√X
设t=√X+1==>x=(t-1)^2
F(t)= (t-1)^2+2√(t-1)^2
∴f(x)= (x-1)^2+2√(x-1)^2=(x-1)^2+2|x-1|
f(x+1)= x^2+2|x|
f(x^2)= (x^2-1)^2+2| x^2-1|=x^4-2x^2+1+2| x^2-1|

Answer 3

常用的方法有两种，配方法和换元法
换元法比较好理解
令√x+1=t
则x=(t-1)^2
所以f(t)=(t-1)^2+2(t-1)=t^2-1
所以f(x)=x^2-1
f(x+1)=(x+1)^2-1=x^2+2x
f(x^2)=x^4-1

Answer 4

令t=根号x+1，则x=（t+1）^2，
f(t)=(t-1)^2+2(t-1)=t^2-1;
即f(x)=x^2-1;
则f(x+1)=(x+1)^2-1=x^2+2x;
f(x^2)=x^4-1