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定义f(x)=1/{[三次√(x平方+2x+1))+2(三次√(x平方-1))+(三次√(x平方-2x+1)]},求

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/20 11:42:15
定义f(x)=1/{[三次√(x平方+2x+1))+2(三次√(x平方-1))+(三次√(x平方-2x+1)]},求
f(1)+f(3)+...+f(2k-1)+...+f(999)的值.
f(x)=1/[(x^2+2x+1)^(1/3)+2(x^2-1)^(1/3)+(x^2-2x+1)^(1/3)]
=1/[(x+1)^(2/3)+2(x^2-1)^(1/3)+(x-1)^(2/3)]
=1/[(x+1)^(1/3)+(x-1)^(1/3)]^2
再问: 求值呀
再答: A=(x+1)^(1/3) B=(x-1)^(1/3) (X+1)^(2/3)+2(X^2-1)^(1/3)+(X-1)^(2/3)=A^2+2AB+B^2=(A+B)^2 f(1)=1/2^(2/3) f(3)=1/[2^(1/3)+4^(1/3))^2=1/[2^(2/3)(1+2^(1/3))^2] f(5)=1/[4^(1/3)+6^(1/3)]^2=1/[2^(2/3)(2^(1/3)+3^(1/3)^2] ... f(1)+f(3)+...+f(99)=[99/2^(2/3)] *[1+1/(1+2^(1/3))^2+ 1/(2^(1/3)+3^(1/3))^2+...+1/(49^(1/3)+50^(1/3))^2]没法继续
再问: 应该能约了吧?