Question

Cho hàm số f(x)=\(\dfrac{9^x}{9^x+3}\). Tính P=\(f\left(\dfrac{1}{2024}\right)+f\left(\dfrac{2}{2024}\right)+....+f\left(\dfrac{2023}{2024}\right)\)

Answer 1

\(f\left(x\right)+f\left(1-x\right)=\dfrac{9^x}{9^x+3}+\dfrac{9^{1-x}}{9^{1-x}+3}=\dfrac{9^x}{9^x+3}+\dfrac{9}{9+3.9^x}\)

\(=\dfrac{9^x}{9^x+3}+\dfrac{3}{9^x+3}=1\)

\(\Rightarrow P=f\left(\dfrac{1}{2014}\right)+f\left(\dfrac{2023}{2024}\right)+...+f\left(\dfrac{1011}{2024}\right)+f\left(\dfrac{1013}{2024}\right)+f\left(\dfrac{1012}{2024}\right)\) 

\(=1+1+...+1+f\left(\dfrac{1}{2}\right)=1011+\dfrac{\sqrt{9}}{\sqrt{9}+3}=\dfrac{2023}{2}\)