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(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求值呀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)^2f(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]没法继续应该能约了吧？