Cho hàm số. $f\left( x \right)=\dfrac{x}{{{2020}^{x}}+1}$ Đặt...

Phương pháp:
Tính $f\left( x \right)-f\left( -x \right)\forall x\in \left[ 1;100 \right]$ sau đó tính ${{S}_{1}}-{{S}_{2}}.~$
Cách giải:
Ta có: $f\left( x \right)=\dfrac{x}{{{2020}^{x}}+1}$
$\Rightarrow f\left( -x \right)=\dfrac{-x}{{{2020}^{-x}}+1}=\dfrac{-x}{\dfrac{1}{{{2020}^{x}}}+1}$
$=\dfrac{-x}{\dfrac{{{2020}^{x}}+1}{{{2020}^{x}}}}=\dfrac{-x{{.2020}^{x}}}{{{2020}^{x}}+1}$
$\begin{aligned}
& \Rightarrow f\left( x \right)-f\left( -x \right) \\
& =\dfrac{x}{{{2020}^{x}}+1}+\dfrac{x{{.2020}^{x}}}{{{2020}^{x}}+1} \\
& =\dfrac{x+x{{.2020}^{x}}}{{{2020}^{x}}+1}=\dfrac{x\left( {{2020}^{x}}+1 \right)}{{{2020}^{x}}+1}=x \\
\end{aligned}$
Khi đó ta có: ${{S}_{1}}-{{S}_{2}}=\left[ f\left( 1 \right)+f\left( 2 \right)+...+f\left( 100 \right) \right]-\left[ f\left( -1 \right)+f\left( -2 \right)+...+f~\left( -100~ \right) \right]~$
$\begin{aligned}
& =\left[ f\left( 1 \right)-f\left( -1 \right) \right]+\left[ f\left( 2 \right)-f\left( -2 \right) \right]+...+\left[ f\left( 100 \right)-f~\left( -100~ \right) \right]~ \\
& =1+2+...+100 \\
\end{aligned}$
$\dfrac{100.101}{2}=5050$