Bài 1: A=2/3*7 + 2/7*11 + 2/11*15+ ... +2/99*103 Bài 2: A=7/2 + 7/6 + 7/12 + 7/20 + 7/30 + 7/42 + 7/56 + 7/72 + 7/90 Bài 3: A=505/10*1212 + 505/12*1414 + 505/14*1616 +...+ 505/96*9898 Bài 4: A=2/1*3 - 4/3*5 - 6/5*7 - ... - 20/19*21 Bài 5: A=1 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56 - 17/72 + 19/90 :>
\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
Bài 1:
Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)
\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)
Bài 2:
Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)
\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)
\(=7\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{63}{10}\)
\(3,A=\dfrac{505}{10\cdot1212}+\dfrac{505}{12\cdot1414}+...+\dfrac{505}{96\cdot9898}\\ A=\dfrac{5}{10\cdot12}+\dfrac{5}{10\cdot14}+...+\dfrac{5}{96\cdot98}\\ \dfrac{2}{5}A=\dfrac{2}{10\cdot12}+\dfrac{2}{10\cdot14}+...+\dfrac{2}{96\cdot98}\\ \dfrac{2}{5}A=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{96}-\dfrac{1}{98}\\ \dfrac{2}{5}A=\dfrac{1}{10}-\dfrac{1}{98}=\dfrac{22}{245}\\ A=\dfrac{22}{245}\cdot\dfrac{5}{2}=\dfrac{11}{49}\)
