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\(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
= \(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{49}{100}\)
= \(\frac{49}{99}-\frac{49}{200}\)
= \(\frac{4949}{19800}\)
bn zô xem nha, ko hiểu thì cứ hỏi bn ấy nhá
http://olm.vn/hoi-dap/question/154321.html
\(\frac{1}{13}+\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\)
\(=\frac{1}{13}+\left[\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13\cdot23}+\frac{1}{23\cdot33}+...+\frac{1}{1993\cdot2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\cdot\frac{1990}{26039}\right]\)
\(=\frac{1}{13}+\frac{597}{26039}\)
\(=\frac{200}{2003}\)
Đặt A= 1/13 + 3/13.23 + 3/ 23.33 + ... + 3/1993.2003
A- 1/13 = 3/13.23 + 3/ 23.33 + ... + 3/1993.2003
10/3 ( A-1/3) = 10/3. (3/13.23 + 3/ 23.33 + ... + 3/1993.2003)
10/3A - 10/9 = 10/13.23 + 10/ 23.33 + ... + 10/1993.2003
10/3A - 10/9 = 1/13 - 1/23 + 1/23 - 1/33 +...+ 1/1993- 1/2003
10/3A = 1/13 - 1/2003 + 10/9
10/3 A= ?
đến đây bn tự làm nha
10/3A - 10/9 = 1/13
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
\(A=\frac{1}{3\cdot10}+\frac{1}{10\cdot17}+\frac{1}{17\cdot24}+...+\frac{1}{73\cdot80}-\frac{1}{2\cdot9}-\frac{1}{9\cdot16}-\frac{1}{16\cdot23}-\frac{1}{23\cdot30}\)
\(A=\frac{1}{7}\left(\frac{7}{3\cdot10}+\frac{7}{10\cdot17}+\frac{7}{17\cdot24}+...+\frac{7}{73\cdot80}-\frac{7}{2\cdot9}-\frac{7}{9\cdot16}-\frac{7}{16\cdot23}-\frac{7}{23\cdot30}\right)\)
\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{73}-\frac{1}{80}-\frac{1}{2}+\frac{1}{9}-\frac{1}{9}+\frac{1}{16}-...-\frac{1}{23}-\frac{1}{30}\right)\)
\(A=\frac{1}{7}\left(\frac{1}{3}-\frac{1}{80}-\frac{1}{2}-\frac{1}{30}\right)\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
2A=2-1/3+1/3-1/5+...+1/97-1/99
2A=2-1/99
2A=197/99
A=197/198
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}\)
\(=\frac{49}{99}\)
=))
\(A=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(2A=\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+\frac{2}{4\times6}+\frac{2}{5\times7}+\frac{2}{6\times8}+\frac{2}{7\times9}+\frac{2}{8\times10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\)
\(2A=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\)
\(2A=\frac{58}{45}\)
\(A=\frac{58}{45}\div2\)
\(A=\frac{29}{45}\)
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{10}\)
\(=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}=\frac{58}{45}\)
\(A=\frac{29}{45}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}-\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{1}{2}\left(\frac{98}{99}-\frac{49}{100}\right)=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)