Question

9 已知f(根号下x)+1=x+2 倍根号下x,求f(x),f(x+1)与f(x的平方) 若3f(x-1)+2f(1-x

Answer 1

令a=√x+1

√x=a-1

x=(a-1)²=a²-2a+1

所以f(a)=a²-2a+1+2

所以f(x)=x²-2x+3

f(x+1)=(x+1)²-2(x+1)+3=x²+2

f(x²)=x^4-2x²+3

令a=x-1

1-x=-a

3f(a)+2f(-a)=2(a+1) (1)

则a=-a

3f(-a)+2f(a)=2(-a+1) (2)

(1)×3-(2)×2

5f(a)=6a+6+4a-4=10a+2

f(a)=(10a+2)/5

f(x)=(10x+2)/5