Sửa đề : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=1+\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3984}{1993}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{3984}{1993}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1992}{1993}\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\Rightarrow x=1993-1=1992\)
Vây x = 1992