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a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
\(A=\left(\frac{7}{9}+1\right)\left(\frac{7}{20}+1\right)\left(\frac{7}{33}+1\right)..........\left(\frac{7}{2900}+1\right)\)
\(A=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.......\frac{2907}{2900}\)
\(A=\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}......\frac{51.57}{50.58}\)
\(A=\frac{2.3.4.....51}{1.2.3......50}.\frac{8.9.10.......57}{9.10.11........58}\)
\(A=51.\frac{4}{29}\)
\(A=\frac{204}{59}\)
Vậy \(A=\frac{204}{59}\)
\(C=\left(\frac{7}{9}+1\right)\left(\frac{7}{20}+1\right)\left(\frac{7}{33}+1\right)...\left(\frac{7}{10800}+1\right)\)
\(=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.....\frac{10807}{10800}\)
\(=\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}.....\frac{101.107}{100.108}\)
\(=\frac{2.3.4....101}{1.2.3....100}.\frac{8.9.10...107}{9.10.11...108}\)
\(=101.\frac{2}{27}\)\(=\frac{202}{27}\)