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\(b,\frac{10}{99}\)+\(\frac{11}{199}\)+\(\frac{12}{299}\).\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{-1}{6}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
a,\(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\) =\(\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\) =\(\left(3-\frac{31}{12}\right):\left(4-\frac{1}{3}\right)\) =\(\frac{5}{12}:\frac{11}{3}\) =\(\frac{5}{44}\) b, F=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.......+\frac{4}{2008.2010}\) =\(2.\left(1-\frac{2}{2010}\right)\) =\(2.\frac{1004}{1005}\) =\(\frac{2008}{1005}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(=\frac{4}{9}-\frac{1}{5}\)
\(=\frac{11}{45}\)
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
= \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{200+202}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{200}-\frac{1}{202}\)
= \(\frac{1}{2}-\frac{1}{202}\)
= \(\frac{404}{202}-\frac{1}{202}\)
= \(\frac{403}{202}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2n.\left(2n+2\right)}\))
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{4}-\frac{1}{2.\left(2n+2\right)}\)
\(=\frac{1}{4}-\frac{1}{4n+4}=\frac{1}{4}-\frac{1}{4.\left(n+1\right)}\)
\(=\frac{n+1}{4.\left(n+1\right)}-\frac{1}{4.\left(n+1\right)}=\frac{n+1-1}{4.\left(n+1\right)}=\frac{n}{4.\left(n+1\right)}\)
a,\(\frac{100}{101}\)
c,\(\frac{27}{20}\)
d,\(\frac{16}{11}\)
,
c, 3/10 + 3/40 + 3/88 + ...+ 3/340
=> 3/10 + 3/44 + 3/88 + .... + 3/340 = 3/2x5 + 3/5x8 + 3/8x11 + .....+ 3/17x 20
= 1/2 - 1/5 + 1/5 - 1.8 + 1/8 - 1/11 +.......+ 1/17 - 1/20
= 1/2 - 1/20 = 9/20
d, 6/15 + 6/35 + 6/63 + 6/99
= 3 ( 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11)
= 3( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
= 3( 1/3 - 1/11)
= 3 x 8/33 = 8/11