Question

cho \(A=\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+...+\frac{1}{33.38}\)

\(B=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+...+\frac{1}{31\cdot38}\)

tính tỉ số \(\frac{A}{B}\) \(\left(A⋮B\right)\)

Answer 1

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 \(\frac{A}{B}=\frac{7}{5}\)

Answer 2

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

Answer 3

=\(\frac{1}{5}\left(\frac{1}{3}-\frac{1}{38}\right)\)

=\(\frac{1}{5}.\frac{35}{114}\)

=\(\frac{7}{114}\)

=\(\frac{1}{7}.\frac{35}{114}\)

=\(\frac{5}{114}\)

Vì \(\frac{7}{114}>\frac{5}{114}\)

⇒ A > B

Vậy A > B