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a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 3
x = 305
b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)
=> x + 1 = 1993
x = 1993 - 1
x = 1992
a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(A=1-\frac{1}{16}=\frac{15}{16}\)
A=\((\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{33.38})\)
A=\(\frac{1}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{33.38}\right)\)
A=\(\frac{1}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{38}\right)\)
A=\(\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{38}\right)\)
A=\(\frac{1}{5}.\frac{35}{114}\)
A=\(\frac{7}{114}\)
B=\((\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{31.38})\)
B=\(\frac{1}{7}\left(\frac{7}{3.10}+\frac{7}{10.17}+...+\frac{7}{31.38}\right)\)
B=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\right)\)
B=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{38}\right)\)
B=\(\frac{1}{7}.\frac{35}{114}\)
B=\(\frac{5}{114}\)
⇒ \(\frac{A}{B}\)=\(\frac{7}{114}:\frac{5}{114}=\frac{7}{114}.\frac{114}{5}=\frac{7}{5}\)
Vậy \(\frac{A}{B}=\frac{7}{5}\)
A = \(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+....+\frac{1}{33}-\frac{1}{38}\)
=\(\frac{1}{3}-\frac{1}{38}\)
=\(\frac{35}{114}\)
B =\(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\)
=\(\frac{1}{3}-\frac{1}{38}\)
=\(\frac{35}{114}\)
=>tỉ số \(\frac{A}{B}\)= \(\frac{35}{114}:\frac{35}{114}\)=1