Cho hàm số: $f\left( x \right)=\dfrac{{{9}^{x}}}{{{9}^{x}}+3}$...

Với $a+b=1$. Ta có: $\begin{aligned}
& T=f\left( a \right)+f\left( b \right)=f\left( a \right)+f\left( 1-a \right)=\dfrac{{{9}^{a}}}{{{9}^{a}}+3}+\dfrac{{{9}^{1-a}}}{{{9}^{1-a}}+3} \\
& \ \ =\dfrac{{{9}^{a}}}{{{9}^{a}}+3}+\dfrac{\dfrac{9}{{{9}^{a}}}}{\dfrac{9}{{{9}^{a}}}+3}=\dfrac{{{9}^{a}}}{{{9}^{a}}+3}+\dfrac{9}{9+{{3.9}^{a}}}=\dfrac{{{9}^{a}}}{{{9}^{a}}+3}+\dfrac{3}{{{9}^{a}}+3}=1 \\
\end{aligned}$
Do đó: $\begin{aligned}
& A=\left[ f\left( \dfrac{1}{100} \right)+f\left( \dfrac{99}{100} \right) \right]+\left[ f\left( \dfrac{2}{100} \right)+f\left( \dfrac{98}{100} \right) \right]+...+\left[ f\left( \dfrac{49}{100} \right)+f\left( \dfrac{51}{100} \right) \right]+f\left( \dfrac{50}{100} \right)+f\left( \dfrac{100}{100} \right) \\
& \ \ \ =49+f\left( \dfrac{1}{2} \right)+f\left( 1 \right)=\dfrac{201}{4} \\
\end{aligned}$