Ta có:  

 \(\begin{array}{l} f\left( x \right) + f\left( {1 - x} \right) = \frac{{{4^x}}}{{{4^x} + 2}} + \frac{{{4^{1 - x}}}}{{{4^{1 - x}} + 2}}\\ = \frac{{{4^x}}}{{{4^x} + 2}} + \frac{{\frac{4}{{{4^x}}}}}{{\frac{4}{{{4^x}}} + 2}} = \frac{{{4^x}}}{{{4^x} + 2}} + \frac{2}{{{4^x} + 2}} = 1. \end{array}\)

Do đó: \(A = f\left( {\frac{1}{{100}}} \right) + f\left( {\frac{2}{{100}}} \right) + ... + f\left( {\frac{{100}}{{100}}} \right)\) 

\(= f\left( {\frac{1}{{100}}} \right) + f\left( {\frac{{99}}{{100}}} \right) + f\left( {\frac{2}{{100}}} \right) + f\left( {\frac{{98}}{{100}}} \right) + ...\)

\(+ f\left( {\frac{{49}}{{100}}} \right) + f\left( {\frac{{51}}{{100}}} \right) + f\left( {\frac{{50}}{{100}}} \right) + f\left( {\frac{{100}}{{100}}} \right)\)\(= 49 + f\left( {\frac{1}{2}} \right) + f\left( 1 \right) = \frac{{301}}{6}.\)