Question

TÌm x biết

a) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2019}{2021}\)

Answer 1

\(\Leftrightarrow1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{2019}{2021}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)

\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{2}{x+1}=1-\dfrac{2019}{2021}\)

\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{2}{2021}\)

\(\Leftrightarrow x+1=2021\)

\(\Leftrightarrow x=2020\)