Question

Answer 1

A

Answer 2

Chọn A

Answer 3

\(\dfrac{1}{2}log_2f\left(x\right)=log_2\left(2^x\right)-\dfrac{1}{2}log_2\left(4^x+1\right)\)

\(\Leftrightarrow\dfrac{1}{2}log_2f\left(x\right)=\dfrac{1}{2}log_2\left(4^x\right)-\dfrac{1}{2}log_2\left(4^x+1\right)\)

\(\Leftrightarrow log_2f\left(x\right)=log_2\left(\dfrac{4^x}{4^x+1}\right)\)

\(\Rightarrow f\left(x\right)=\dfrac{4^x}{4^x+1}\)

\(f\left(0\right)=\dfrac{1}{2}\) ; \(f\left(x\right)+f\left(-x\right)=\dfrac{4^x}{4^x+1}+\dfrac{4^{-x}}{4^{-x}+1}=\dfrac{4^x}{4^x+1}+\dfrac{1}{4^x+4}=1\)

\(\Rightarrow S=f\left(-99\right)+f\left(99\right)+f\left(-98\right)+f\left(98\right)+...+f\left(-1\right)+f\left(1\right)+f\left(0\right)\)

\(=99+\dfrac{1}{2}=\dfrac{199}{2}\)