Đáp án:
$\frac{1}{3}$+ $\frac{1}{6}$+ $\frac{1}{10}$+....+$\frac{2}{x.(x+1)}$=$\frac{2019}{2020}$
$\frac{2}{6}$+ $\frac{2}{12}$+ $\frac{2}{20}$+...+$\frac{2}{x.(x+1)}$ =$\frac{2019}{2020}$
$\frac{2}{2.3}$ $\frac{2}{3.4}$ $\frac{2}{4.5}$+...+$\frac{2}{x.(x+1)}$= $\frac{2019}{2020}$
2. ( $\frac{1}{2.3}$ $\frac{1}{3.4}$ $\frac{1}{4.5}$+...+ $\frac{1}{x.(x+1)}$ )= $\frac{2019}{2020}$
2.( $\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+$\frac{1}{4}$-$\frac{1}{5}$+...+$\frac{1}{x}$- $\frac{1}{x+1}$ ) = $\frac{2019}{2020}$
2.( $\frac{1}{2}$ - $\frac{1}{x+1}$ ) = $\frac{2019}{2020}$
2 . $\frac{1}{2-(x+1)}$ = $\frac{2019}{2020}$
$\frac{2}{2-(x+1)}$=$\frac{2019}{2020}$
x + 1 = $\frac{2019}{2020}$
x = $\frac{2019}{2020}$ - 1
x = $\frac{-1}{2020}$
Bài khá dài nha bạn !!
Chúc bạn học tốt !
Đúng 100% √√√
