已知对所有实数x有f(x²+1)=x四次方+5x²+3成立.求f(x²-1)

问题描述:

已知对所有实数x有f(x²+1)=x四次方+5x²+3成立.求f(x²-1)

设x²+1=n,
则x²=n-1
f(n)=(n-1)²+5(n-1)+3
则f(x²-1)=(x²-1-1)²+5(x²-1-1)+3

f(x²+1)=x四次方+5x²+3
=(x²+1)²+3(x²+1)-1

f(t)=t²+3t-1
所以
f(x²-1)=(x²-1)²+3(x²-1)-1=x四次方+x²-3

f(x²+1)=x四次方+5x²+3=x^4+2x^2+1+3x^2+3-1=(x^2+1)^2+3(X^2+1)-1
f(x)=X^2+3x-1
F(x^2-1)=(x^2-1)^2+3(x^2-1)-1=x^4+x^2-3

f(x²-1)=x四次方+x²-3