f(x+1)=x^2+2x+3=x^2+2x+1+2=（x+1）^2+2令x+1=t，则f（t）=t^2+2所以f（x）=x^2+2
解：设x+1=t，则x=t-1f(t)=(t-1)^2+2(t-1)+3=t²-2t+1+2t-2+3=t²+2∴f(x)=x²+2
f(x+1)=x+1^2-1^2+2(x+1)+1f(x)=x^2+2x+1^2
