Đề bạn thiếu 1 số \(x\) nữa đúng không?
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2021}\)
\(\Rightarrow x+1=2021\)
\(\Rightarrow x=2020\)
Vậy \(x=2020\).
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4042}=\frac{1}{2021}\)
\(\Leftrightarrow x+1=2021\)
\(\Leftrightarrow x=2020\left(tm:x\in N\right)\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+............+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..........+\frac{1}{x\left(x+1\right)}\right]=\frac{2019}{2021}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{2021}\)
\(\Leftrightarrow x-1=2021\)
\(\Leftrightarrow x=2022\)
Vậy \(x=2022\)
Mình nhầm 4 dòng cuối:
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2021}\)\(\Leftrightarrow x+1=2021\)\(\Leftrightarrow x=2020\)
Vậy \(x=2020\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x+1}=\frac{2019}{2021}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2019}{2021}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2021}\) Bn làm nốt nhé
