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

a) Với a, b là hai số thực thỏa mãn a + b = 1. Tính f(a) + f(b).

b) Tính tổng: $S=f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\mathrm{...}+f\left(\frac{2022}{2023}\right).$ 

a) Xét $f\left(x\right)=\frac{9^x}{9^x+3}.$
Ta có: $f\left(a\right)=\frac{9^a}{9^a+3}.$
Do a + b = 1 nên b = 1 – a.
Suy ra:  $f\left(b\right)=f\left(1-a\right)=\frac{9^{1-a}}{9^{1-a}+3}=\frac{\frac{9}{9^a}}{\frac{9}{9^a}+3}$
 $=\frac{\frac{9}{9^a}}{\frac{9+9^a\cdot 3}{9^a}}=\frac{9}{9+9^a\cdot 3}=\frac{3}{3+9^a}.$
Từ đó ta có:  $f\left(a\right)+f\left(b\right)=\frac{9^a}{9^a+3}+\frac{3}{9^a+3}=\frac{9^a+3}{9^a+3}=1.$
b) Ta thấy: $\frac{1}{2023}+\frac{2022}{2023}=1;$$\frac{2}{2023}+\frac{2021}{2023}=1;$...;$\frac{1011}{2023}+\frac{1012}{2023}=1.$
Nên theo câu a, ta có: 
 $f\left(\frac{1}{2023}\right)+f\left(\frac{2022}{2023}\right)=1;$
 $f\left(\frac{2}{2023}\right)+f\left(\frac{2021}{2023}\right)=1;$

...;

$f\left(\frac{1011}{2023}\right)+f\left(\frac{1012}{2023}\right)=1.$
Suy ra: 
 $S=f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\mathrm{...}+f\left(\frac{2022}{2023}\right)$$=\left[f\left(\frac{1}{2023}\right)+f\left(\frac{2022}{2023}\right)\right]+\left[f\left(\frac{2}{2023}\right)+f\left(\frac{2021}{2023}\right)\right]+\mathrm{...}$
 $\mathrm{...}+\left[f\left(\frac{1011}{2023}\right)+f\left(\frac{1012}{2023}\right)\right]$  (có 1 011 nhóm)
= 1 + 1 + … + 1 (có 1 011 số hạng 1)
= 1 011.