已知函数f(x)=x2/1+x2,那么f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)f(1/4)=?
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![已知函数f(x)=x2/1+x2,那么f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)f(1/4)=?](/uploads/image/z/11651266-10-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dx2%EF%BC%8F1%2Bx2%2C%E9%82%A3%E4%B9%88f%EF%BC%884%EF%BC%89%2Bf%EF%BC%883%EF%BC%89%2Bf%282%29%2Bf%281%29%2Bf%281%2F2%29%2Bf%281%2F3%29f%281%2F4%29%3D%3F)
已知函数f(x)=x2/1+x2,那么f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)f(1/4)=?
已知函数f(x)=x2/1+x2,那么f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)f(1/4)=?
已知函数f(x)=x2/1+x2,那么f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)f(1/4)=?
题目应该是f(x)=1/x^2+x2,那么f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)+f(1/4)=?吧 因为f(x)=1/x^2+x2 所以f(1/x)=1/(1/x^2)+1/x2 =x^2+1/x2 =1/x^2+x2 =f(x) 所以f(4)+f(3)+f(2)+f(1)+f(1/2)+f(1/3)+f(1/4)=2f(4)+2f(3)+2f(2)+f(1) =2[f(4)+f(3)+f(2)]+f(1) =2(1/16 +16+1/9+9+1/4+4)+1+1 =2(1/16+1/9+1/4+29)+2 =1/8+2/9+1/2+58+2 =61/72 +60 =...