f(1-x)=[2^(1-x+1)]/[2^(1-x)+2]=[2^(2-x)]/[2^(1-x)+2]=(2^2)/[2+2^(1+x)]=2/(1+2^x)计算f(x)+f(1-x),得f(x)+f(1-x)=2所以原式=[f(1/100)+f(99/100)]+[f(2/100)+f(98/100)]+…+[f(49/100)+f(51/100)]+f(50/100)+f(1...