Question

已知函数f(x)=x/(x+1)+(x+1)/(x+2)+(x+2)/(x+3)+(x+3)/(x+4)

则f(-5/2+√2)+f(-5/2-√2)=？

Answer 1

f(x)=4-1/(x+1)-1/(x+2)-1/(x+3)-1/(x+4)
=4-[(2x+5)/(x+1)(x+4)+(2x+5)/(x+2)(x+3)]
=4-(2x+5)[1/(x^2+5x+4)+1/(x^2+5x+6)]
令a=(-5/2+√2), b=(-5/2-√2)
则2a+5=2√2, 2b+5=-2√2
a, b都为方程t^2+5t+17/4=0的根，
a^2+5a=b^2+5b=-17/4
所以原式=f(a)+f(b)=4-2√2/T+4+2√2/T=8.
这里T=1/(a^2+5a+4)+1/(a^2+5a^2+6)

Answer 2

........好变态的题，路过！！