\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1989}{1991}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)

\(\frac{8}{3}+2-\frac{2}{x+1}=1\frac{1989}{1991}\)

\(\frac{2}{x+1}=\frac{13}{10}\)( số thập phân dài quá nên mk lấy số tròn thôi nha )

\(x+1=2:\frac{13}{10}\)

\(x+1=\frac{20}{13}\)

\(\Leftrightarrow x=\frac{7}{13}\)

\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)

\(1+\frac{1}{3}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{x+1-x}{x\left(x+1\right)}=\frac{1990}{1991}\)

\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x-1}=\frac{1990}{1991}\)

\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x-1}=\frac{1990}{1991}\)

\(\frac{1}{x-1}=\frac{11}{6}-\frac{1990}{1991}=\frac{9961}{11946}\)

\(x-1=\frac{11946}{9961}\Rightarrow x=\frac{21907}{9961}\)

Hình như đề sai rồi bạn ạ

Bạn xem lị đề nha

Sai đề rồi bạn ơi, 2 + ... không thể nào = 1 1989/1991 được bạn ạ !!!

đề thầy giáo ra mà

thầy dạy HSG lớp 9 đó

hóc quá

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

\(2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\) 

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{1990}{1991}\) 

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\) 

\(1-\frac{1}{x+1}=\frac{1990}{1991}\) 

\(\frac{1}{x+1}=1-\frac{1990}{1991}\) 

\(\frac{1}{x+1}=\frac{1}{1991}\) 

\(x+1=1991\) 

\(x=1990\)  

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)

\(\Leftrightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)

\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1989}{1991}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1989}{3982}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1989}{3982}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1991}\)

\(\Leftrightarrow x+1=1991\)\(\Leftrightarrow x=1990\)

Vậy \(x=1990\)

Đặt \(\left\{{}\begin{matrix}a=\frac{3+\sqrt{5}}{2}\\b=\frac{3-\sqrt{5}}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=1\\a+b=3\end{matrix}\right.\)

Đặt \(S_n=a^n+b^n\)

\(S_1=a+b=3\)

Ta cần tính \(S_{1991}-3S_{1990}+S_{1989}\)

Xét: \(S_1.S_n=\left(a+b\right)\left(a^n+b^n\right)=a^{n+1}+b^{n+1}+a.b^n+a^nb\)

\(\Rightarrow S_1S_n=a^{n+1}+b^{n+1}+ab\left(a^{n-1}+b^{n-1}\right)\)

\(\Leftrightarrow S_1S_n=a^{n+1}+b^{n+1}+a^{n-1}+b^{n-1}\)

\(\Leftrightarrow3S_n=S_{n+1}+S_{n-1}\)

Thay \(n=1990\Rightarrow3S_{1990}=S_{1991}+s_{1989}\)

\(\Rightarrow S_{1991}-3S_{1990}+S_{1989}=0\)