đơn giản
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{39}{40}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{80}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}\)
\(\frac{1}{x+1}=\frac{1}{80}\)
\(\Rightarrow x+1=80\)
\(\Rightarrow x=80-1=79\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}\div2=\frac{39}{80}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}=\frac{1}{80}\)
\(\Leftrightarrow x+1=80\Rightarrow x=80-1=79\)
Vậy \(x=79\)
