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\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9+\left(\frac{1}{1.2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+...+\frac{1}{9\cdot10}\right)\)
\(A=9+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9+\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=8\frac{1}{10}\)
a/ \(A=\frac{1}{6}+\frac{1}{12}+.........+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{4}\)
b/ \(B=\frac{5}{11.16}+\frac{5}{16.21}+........+\frac{5}{61.66}\)
\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+........+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}\)
\(=\frac{5}{66}\)
a) \(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
b) \(B=\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)
\(B=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(B=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(A=10.\left(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{71}{72}+\frac{89}{90}\right)\)
Đặt \(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{71}{72}+\frac{89}{90}\)
\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(B=1+1+1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{72}+\frac{1}{90}\right)\)
\(B=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}=8,1\)
Ta có \(A=10.B=10.B=10.8,1=81\)
Vậy \(A=81\)
\(S=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(S=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)
\(S=3\left(\frac{1}{1.2}\right)-5\left(\frac{1}{2.3}\right)+7\left(\frac{1}{3.4}\right)-9\left(\frac{1}{4.5}\right)+11\left(\frac{1}{5.6}\right)-13\left(\frac{1}{6.7}\right)+15\left(\frac{1}{7.8}\right)-17\left(\frac{1}{8.9}\right)\)
\(S=3\left(1-\frac{1}{2}\right)-5\left(\frac{1}{2}-\frac{1}{3}\right)+7\left(\frac{1}{3}-\frac{1}{4}\right)-9\left(\frac{1}{4}-\frac{1}{5}\right)+11\left(\frac{1}{5}-\frac{1}{6}\right)-13\left(\frac{1}{6}-\frac{1}{7}\right)+15\left(\frac{1}{7}-\frac{1}{8}\right)-17\left(\frac{1}{8}-\frac{1}{9}\right)\)
\(S=\left(3-\frac{3}{2}\right)-\left(\frac{5}{2}-\frac{5}{3}\right)+\left(\frac{7}{3}-\frac{7}{4}\right)-\left(\frac{9}{4}-\frac{9}{5}\right)+\left(\frac{11}{5}-\frac{11}{6}\right)-\left(\frac{13}{6}-\frac{13}{7}\right)+\left(\frac{15}{7}-\frac{15}{8}\right)-\left(\frac{17}{8}-\frac{17}{9}\right)\)\(S=3-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+\frac{11}{5}-\frac{11}{6}-\frac{13}{6}+\frac{13}{7}+\frac{15}{7}-\frac{15}{8}-\frac{17}{8}+\frac{17}{9}\) Giờ bạn chỉ cần nhóm từng cặp phân số có cùng tử số rồi tính tiếp là ra kết quả thôi
( khi nhóm cặp nhớ đổi dấu nha)
\(A=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+...-\left(\frac{1}{8}+\frac{1}{9}\right)\)
\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
Đặt\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}...+\frac{1}{110}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{10.11}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
\(S=1-\frac{1}{11}\)
\(S=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
\(\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+...+\frac{5}{110}=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=5\left(1-\frac{1}{11}\right)=5.\frac{10}{11}=\frac{50}{11}\)