C = ` (1+1/1.3) . (1+1/2.4) . (1+1/3.5) ... (1+1/2018.2020) `

Đáp án:

$\dfrac{2019}{1010}$.

Giải thích các bước giải:

$\left(1+\dfrac1{1.3}\right)\!\!\left(1+\dfrac1{2.4}\right)\!\!\left(1+\dfrac1{3.5}\right)\!.\!.\!.\!\left(1+\dfrac1{2018.2020}\right)\\=\dfrac{4}{1.3}\!\cdot\!\dfrac9{2.4}\!\cdot\!\dfrac{16}{3.5}\;\!\cdot\,.\!.\!.\cdot\;\!\dfrac{2018.2020+1}{2018.2020}\\=\dfrac{4}{1.3}\!\cdot\!\dfrac9{2.4}\!\cdot\!\dfrac{16}{3.5}\;\!\cdot\,.\!.\!.\cdot\;\!\dfrac{2018.(2019+1)+1}{2018.2020}\\=\dfrac{4}{1.3}\!\cdot\!\dfrac9{2.4}\!\cdot\!\dfrac{16}{3.5}\;\!\cdot\,.\!.\!.\cdot\;\!\dfrac{2018.2019+2018+1}{2018.2020}\\=\dfrac{4}{1.3}\!\cdot\!\dfrac9{2.4}\!\cdot\!\dfrac{16}{3.5}\;\!\cdot\,.\!.\!.\cdot\;\!\dfrac{2018.2019+2019}{2018.2020}\\=\dfrac{4}{1.3}\!\cdot\!\dfrac9{2.4}\!\cdot\!\dfrac{16}{3.5}\;\!\cdot\,.\!.\!.\cdot\;\!\dfrac{2019.(2018+1)}{2018.2020}\\=\dfrac{2.2}{1.3}\!\cdot\!\dfrac{3.3}{2.4}\!\cdot\!\dfrac{4.4}{3.5}\;\!\cdot\,.\!.\!.\cdot\;\!\dfrac{2019.2019}{2018.2020}\\=\dfrac{2.2.3.3.4.4...2019.2019}{1.3.2.4.3.5...2018.2020}\\=\dfrac{(2.3.4...2019)(2.3.4...2019)}{(1.2.3...2018)(3.4.5...2020)}\\=\dfrac{2019.2}{1.2020}\\=\dfrac{2019}{1010}$