(1)f(x)=1/(3^x+√3),所以f(0)+f(1)=1/(1+√3)+1/(3+√3)=(1+√3)/(3+√3)=1/√3=√3/3
(2)f(-1)+f(2)=1/(1/3+√3)+1/(9+√3)=3/(1+3√3)+1/(9+√3)=(3√3+1)/(9+√3)=√3/3
(3)f(x)+f(1-x)=1/(3^x+√3)+1/[3^(1-x)+√3]=(3^x+√3)/(√3*3^x+3)=√3/3,
(1)f(x)=1/(3^x+√3),所以f(0)+f(1)=1/(1+√3)+1/(3+√3)=(1+√3)/(3+√3)=1/√3=√3/3
(2)f(-1)+f(2)=1/(1/3+√3)+1/(9+√3)=3/(1+3√3)+1/(9+√3)=(3√3+1)/(9+√3)=√3/3
(3)f(x)+f(1-x)=1/(3^x+√3)+1/[3^(1-x)+√3]=(3^x+√3)/(√3*3^x+3)=√3/3,