#)Giải :

\(\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{210}\right)=\frac{14}{15}\times\frac{20}{21}\times\frac{27}{28}\times...\times\frac{209}{210}\)

\(=\frac{28}{30}\times\frac{40}{42}\times\frac{54}{56}\times...\times\frac{418}{420}=\frac{4\times7}{5\times6}\times\frac{5\times8}{6\times7}\times\frac{6\times9}{7\times8}\times...\times\frac{19\times22}{20\times21}\)

\(=\frac{4\times5\times6\times...\times19}{5\times6\times7\times...\times20}\times\frac{7\times8\times9\times...\times22}{6\times7\times8\times...\times21}=\frac{4}{20}\times\frac{22}{6}=\frac{11}{15}\)

\(\left(1-\frac{1}{15}\right).\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).....\left(1-\frac{1}{210}\right)\)

\(=\left(\frac{15}{15}-\frac{1}{15}\right).\left(\frac{21}{21}-\frac{1}{21}\right).\left(\frac{28}{28}-\frac{1}{28}\right).....\left(\frac{210}{210}-\frac{1}{210}\right)\)

\(=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}....\frac{209}{210}\)

\(=\frac{2.7}{3.5}.\frac{5.4}{7.3}.\frac{3.9}{4.7}....\frac{11.19}{21.10}\)

\(=\frac{2}{3}.\frac{19}{10}\)

\(=\frac{19}{15}\)

\(A=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}....\frac{209}{210}=\frac{28}{30}.\frac{40}{42}.\frac{54}{56}....\frac{418}{420}=\frac{4.7}{5.6}.\frac{5.8}{6.7}.\frac{6.9}{7.8}....\frac{19.22}{20.21}\)

\(=\frac{4.5.6...19}{5.6.7...20}.\frac{7.8.9...22}{6.7.8...21}=\frac{4}{20}.\frac{22}{6}=\frac{11}{5}\)

(1-1/15).(1-1/21).(1-1/28)...(1-1/210)          

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\(\Rightarrow A=\frac{14}{15}.\frac{20}{21}.\frac{41}{42}.....\frac{209}{210}\)

\(=\frac{4.7}{5.6}.\frac{5.8}{6.7}.\frac{6.9}{7.8}.....\frac{19.22}{20.21}\)

\(=\frac{22}{6}=\frac{11}{3}\)

\(=\frac{-2}{3}\cdot\frac{-5}{6}\cdot\frac{-9}{10}\cdot\cdot\cdot\cdot\frac{-35}{36}\)

\(=\frac{-4}{6}\cdot\frac{-10}{12}\cdot\frac{-18}{20}\cdot\cdot\cdot\cdot\frac{-70}{72}\)

\(=\frac{-1.4}{2.3}\cdot\frac{-2.5}{3.4}\cdot\frac{-3.6}{4.5}\cdot\cdot\cdot\cdot\frac{-7.10}{8.9}\)

\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-7\right)}{2.3.4....8}\cdot\frac{4.5.6....10}{3.4.5....9}\)

\(=\frac{\left(-1\right).2.3...7}{2.3.4....8}\cdot\frac{10}{3}\)

\(=\frac{-1}{8}\cdot\frac{10}{3}=\frac{-5}{12}\)