Cho hàm số $f\left( x \right)={{\log }_{2}}\left(...

Ta có: $f\left( 1-x \right)={{\log }_{2}}\left( 1-x-\dfrac{1}{2}+\sqrt{{{\left( 1-x \right)}^{2}}-\left( 1-x \right)+\dfrac{17}{4}} \right)={{\log }_{2}}\left( \sqrt{{{x}^{2}}-x+\dfrac{17}{4}}-\left( x-\dfrac{1}{2} \right) \right)$
$f\left( x \right)+f\left( 1-x \right)={{\log }_{2}}\left( x-\dfrac{1}{2}+\sqrt{{{x}^{2}}-x+\dfrac{17}{4}} \right)+{{\log }_{2}}\left( \sqrt{{{x}^{2}}-x+\dfrac{17}{4}}-\left( x-\dfrac{1}{2} \right) \right)$
$={{\log }_{2}}\left[ \left( x-\dfrac{1}{2}+\sqrt{{{x}^{2}}-x+\dfrac{17}{4}} \right)\left( \sqrt{{{x}^{2}}-x+\dfrac{17}{4}}-\left( x-\dfrac{1}{2} \right) \right) \right]$ $={{\log }_{2}}4=2$
$\Rightarrow T=f\left( \dfrac{1}{2021} \right)+f\left( \dfrac{2}{2021} \right)+...+f\left( \dfrac{2020}{2021} \right)$
$=f\left( \dfrac{1}{2021} \right)+f\left( \dfrac{2020}{2021} \right)+f\left( \dfrac{2}{2021} \right)+f\left( \dfrac{2019}{2021} \right)+...+f\left( \dfrac{1010}{2021} \right)+f\left( \dfrac{1011}{2021} \right)$
$=1010.2=2020$.