ta có

 \(1+\frac{1}{3}+\frac{1}{6}+..+\frac{2}{x\left(x+1\right)}=1+2\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+..+2\left(\frac{1}{x}-\frac{1}{x+1}\right)=2-\frac{2}{x+1}\)

Nên ta có 

\(2-\frac{2}{x+1}=1+\frac{1989}{1991}\Leftrightarrow\frac{2}{x+1}=\frac{2}{1991}\Leftrightarrow x=1990\)

no help

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

\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

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

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)

\(\Rightarrow x+1=1991\)

\(\Rightarrow x=1990\)

Cho em cái đề bài ạ

Tìm x biết 

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

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

        \(=2\times\left(1+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.....+\frac{1}{x\times\left(x+1\right)}\right)=\frac{3980}{1991}\)

        \(=2\times\left(1+\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=\frac{3980}{1991}\)

        \(\Rightarrow2+1-\frac{2}{x+1}=\frac{3980}{1991}\)

        \(\Rightarrow3-\frac{2}{x+1}=\frac{3980}{1991}\)

        \(\Leftrightarrow\frac{2}{x+1}=3-\frac{3980}{1991}=\frac{1993}{1991}\)

         \(\Rightarrow\frac{2\times996.5}{x+1}=\frac{1993}{1991}\)

  \(\Rightarrow x+1=1991\)

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

đây không phải bn nhân lên 2 lần mà là bn chỉ đổi các ps đó thôi

vậy tại sao bn lại viết 1=2 vậy?

nhưng về kết quả thì bn làm vẫn đúng nhưng khi đọc thì mọi người sẽ khó hỉu hơn 

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 \(x=1990\)

help cái gì 1 tuần rồi