1+ \(\frac{1}{3}\) + \(\frac{1}{6}\)+ \(\frac{1}{10}\) +...+ \(\frac{2}{x\left(x+1\right)}\) = \(1\frac{1989}{1991}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
=> \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1989}{1991}\)
=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1989}{3982}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1989}{3982}\)
=> \(\frac{1}{x+1}=\frac{1}{1991}\)
=> x + 1 = 1991
=> x = 1990
Vậy x = 1990
\(2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\)
\(1-\frac{1}{x+1}=\frac{1990}{1991}\)
\(\frac{1}{x+1}=1-\frac{1990}{1991}\)
\(\frac{1}{x+1}=\frac{1}{1991}\)
\(x+1=1991\)
\(x=1990\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sai đề rồi bạn ơi, 2 + ... không thể nào = 1 1989/1991 được bạn ạ !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(\frac{8}{3}+2-\frac{2}{x+1}=1\frac{1989}{1991}\)
\(\frac{2}{x+1}=\frac{13}{10}\)( số thập phân dài quá nên mk lấy số tròn thôi nha )
\(x+1=2:\frac{13}{10}\)
\(x+1=\frac{20}{13}\)
\(\Leftrightarrow x=\frac{7}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(1+\frac{1}{3}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{x+1-x}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x-1}=\frac{1990}{1991}\)
\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x-1}=\frac{1990}{1991}\)
\(\frac{1}{x-1}=\frac{11}{6}-\frac{1990}{1991}=\frac{9961}{11946}\)
\(x-1=\frac{11946}{9961}\Rightarrow x=\frac{21907}{9961}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=2(1-\frac{1}{x+1})\). Do đó : \(2(1-\frac{1}{x+1})=1\frac{1989}{1991}\)
\(2(\frac{x+1-1}{x+1})=1\frac{1989}{1991}\)
\(\frac{x}{x+1}=\frac{3980}{1991}:2\)
\(\frac{x}{x+1}=\frac{1990}{1991}\)
Vậy x = 1990
\(VT=2\left(1-\frac{1}{x+1}\right)\) . Do đó : \(2\left(1-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(2\left(\frac{x+1-1}{x+1}\right)=1\frac{1989}{1991}\)
\(\frac{x}{x+1}=\frac{3980}{1991}\)
\(\frac{x}{x+1}=\frac{1990}{1991}\)
Vậy x = 1990