Let the function $f:[0,1]\to\mathbb{R}$ be defined by $f(x)=\frac{4^{x}}{4^{x}+2}$. Then the value of $f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\ldots+f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)$ is ____

$f:[0,1]\to\mathbb{R}$ $f(x)=\dfrac{4^{x}}{4^{x}+2}$ $S=f\!\left(\dfrac{1}{40}\right)+f\!\left(\dfrac{2}{40}\right)+\ldots+f\!\left(\dfrac{39}{40}\right)-f\!\left(\dfrac{1}{2}\right)$ $\text{Plan: (1) find } f(1-x)$ $\text{(2) add } f(x)+f(1-x)$ $\text{(3) pair terms } k \text{ and } 40-k$ $\text{graph of } y=f(x),\quad x\in[0,1]$ $4^{1-x}=4\cdot 4^{-x}=\dfrac{4}{4^{x}}$ $f(1-x)=\dfrac{4^{1-x}}{4^{1-x}+2}$ $f(1-x)=\dfrac{4/4^{x}}{4/4^{x}+2}$ $=\dfrac{4}{4+2\cdot 4^{x}}$ $f(1-x)=\dfrac{4}{2\left(2+4^{x}\right)}$ $f(1-x)=\dfrac{2}{4^{x}+2}$ $f(x)+f(1-x)=\dfrac{4^{x}}{4^{x}+2}+\dfrac{2}{4^{x}+2}$ $=\dfrac{4^{x}+2}{4^{x}+2}=1$ $\boxed{f(t)+f(1-t)=1}$ $f\!\left(\dfrac{1}{2}\right)=\dfrac{4^{1/2}}{4^{1/2}+2}=\dfrac{2}{2+2}=\dfrac{1}{2}$ $f\!\left(\dfrac{k}{40}\right)+f\!\left(\dfrac{40-k}{40}\right)=1$ $(1,39),\ (2,38),\ \ldots,\ (19,21) \;\Rightarrow\; 19 \text{ pairs}$ $k=20:\quad f\!\left(\dfrac{20}{40}\right)=f\!\left(\dfrac{1}{2}\right)$ $\sum_{k=1}^{39} f\!\left(\dfrac{k}{40}\right)=19\cdot 1+f\!\left(\dfrac{1}{2}\right)=19+f\!\left(\dfrac{1}{2}\right)$ $S=19+f\!\left(\dfrac{1}{2}\right)-f\!\left(\dfrac{1}{2}\right)$ $\boxed{S=19}$