Đáp án:
$\rm\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2019.2020}=\dfrac{2019}{505}$
Giải thích các bước giải:
$\rm\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2019.2020}\\=4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2019.2020}\right)\\=4.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\\=4.\left(\dfrac{1}{1}-\dfrac{1}{2020}\right)\\=4.\left(\dfrac{2020}{2020}-\dfrac{1}{2020}\right)\\=4.\dfrac{2019}{2020}\\=\dfrac{2019}{505}$
