已知f(x)=1+x分之x求f(1/2004)+f(1/2003)+~+f(1/1)+f(0/1)+f(1/1)~f(2004/1)的值?

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已知f(x)=1+x分之x求f(1/2004)+f(1/2003)+~+f(1/1)+f(0/1)+f(1/1)~f(2004/1)的值?

已知f(x)=1+x分之x求f(1/2004)+f(1/2003)+~+f(1/1)+f(0/1)+f(1/1)~f(2004/1)的值?
已知f(x)=1+x分之x求f(1/2004)+f(1/2003)+~+f(1/1)+f(0/1)+f(1/1)~f(2004/1)的值?

已知f(x)=1+x分之x求f(1/2004)+f(1/2003)+~+f(1/1)+f(0/1)+f(1/1)~f(2004/1)的值?
f(x)+f(1/x)=x/(1+x)+(1/x)/(1+1/x)
=x/(1+x)+1/(x+1)
=(x+1)/(x+1)
=1
f(1/2004)+f(1/2003)+~+f(1/1)+f(0/1)+f(1/1)~f(2004/1)
=[f(1/2004)+f(2004/1)]+[f(1/2003)+f(2003/1)]+[f(1/2002)+f(2002/1)]+……+[f(1/1)+f(1/1)]+f(0/1)
=1+1+1+……+1+0 共有2004个1
=2004

显然有f(x)+f(1/x)=1
首尾相加,这种题绝对有简便算法