作业帮定义：f（x）=1/[立方根(x²+2x+1)+立方根(x²-1)+立方根(x²-2x+1)]求f(1)+f(3)+...+f(2k-1)+...+f(999)=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 05:09:47
![定义：f（x）=1/[立方根(x²+2x+1)+立方根(x²-1)+立方根(x²-2x+1)]求f(1)+f(3)+...+f(2k-1)+...+f(999)=?](/uploads/image/z/1655980-52-0.jpg?t=%E5%AE%9A%E4%B9%89%EF%BC%9Af%EF%BC%88x%EF%BC%89%3D1%2F%5B%E7%AB%8B%E6%96%B9%E6%A0%B9%28x%26%23178%3B%2B2x%2B1%29%2B%E7%AB%8B%E6%96%B9%E6%A0%B9%28x%26%23178%3B-1%29%2B%E7%AB%8B%E6%96%B9%E6%A0%B9%28x%26%23178%3B-2x%2B1%29%5D%E6%B1%82f%281%29%2Bf%283%29%2B...%2Bf%282k-1%29%2B...%2Bf%28999%29%3D%3F)
定义：f（x）=1/[立方根(x²+2x+1)+立方根(x²-1)+立方根(x²-2x+1)]求f(1)+f(3)+...+f(2k-1)+...+f(999)=?
定义：f（x）=1/[立方根(x²+2x+1)+立方根(x²-1)+立方根(x²-2x+1)]
求f(1)+f(3)+...+f(2k-1)+...+f(999)=?
定义：f（x）=1/[立方根(x²+2x+1)+立方根(x²-1)+立方根(x²-2x+1)]求f(1)+f(3)+...+f(2k-1)+...+f(999)=?
f（x）
=1/[³√(x²+2x+1)+³√(x²-1)+³√(x²-2x+1)]
=1/[³√(x+1)²+³√(x²-1)+³√(x-1)²]
=[³√（x+1)-³√(x-1)]/{[³√（x+1)-³√(x-1)]*[³√(x+1)²+³√(x²-1)+³√(x-1)²]}
=[³√（x+1)-³√(x-1)]/[(x+1)-(x-1)]
=[³√（x+1)-³√(x-1)]/2
f(1)+f(3)+f(5)+...+f(1999)
=1/2[³√2-³√0+³√4-³√2+³√6-³√4+.+³√1998-³√1996+³√2000-³√1998]
=1/2³√2000=5׳√2