Question

Giúp mình nhóe

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

Answer 1

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

Answer 2

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

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

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

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

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

\(\frac{1}{x+1}=\frac{1}{1991}\) 

\(x+1=1991\) 

\(x=1990\)  

Answer 3

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

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

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

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

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

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

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

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

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

Vậy \(x=1990\)