a)Ta có   \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=)   \(x+3=305\)=) \(x=302\)

Ta có:

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{3984}{1993}\)

\(\Rightarrow\frac{1}{2}.\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}\right)=\frac{1992}{1993}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1992}{1993}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)

\(1-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)

\(\Leftrightarrow x+1=1993\)

\(x=1992\)

=1992

A ) \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+.....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}.\)

=\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)=101/1540

=\(\frac{101}{1540}:\frac{1}{3}=\frac{1}{5}-\frac{1}{x+3}\)

=tới đó bn tự tính nhé

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)

=> \(\frac{1}{1.\left(1+1\right):2}+\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+\frac{1}{4.\left(4+1\right):2}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)

=> \(\frac{1}{\frac{1.\left(1+1\right)}{2}}+\frac{1}{\frac{2.\left(2+1\right)}{2}}+...+\frac{1}{\frac{x.\left(x+1\right)}{2}}=1\frac{1991}{1993}\)

=> \(\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{x.\left(x+1\right)}=1\frac{1991}{1993}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{1991}{1993}\)

=> \(1-\frac{1}{x+1}=1\frac{1991}{1993}\)

=> \(\frac{1}{x+1}=\frac{-1991}{1993}\)

=> -1991.(x + 1) = 1993

=> -1991x - 1991 = 1993

=> -1991x = 3984

=> x = \(-\frac{3984}{1991}\)

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

\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{x\left(x+1\right)}=4\)

\(\Leftrightarrow1+\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)\right]=4\)

\(\Leftrightarrow1+2\left(\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=4\)

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

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

\(\Leftrightarrow x=-1+1=-2\)

Vậy x = -2 

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)

\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}=\frac{1}{1993}\)

\(\Leftrightarrow x+1=1993\)

\(\Leftrightarrow x=1992\)