giúp tới với từ c6-20
\(7)1-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-\dfrac{2}{7\cdot9}-...-\dfrac{2}{63\cdot65}\\ =1-\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{63\cdot65}\right)\\ =1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\\ =1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\\ =\dfrac{133}{195}\\ 8)\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{19\cdot21}\\ =\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{19\cdot21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\dfrac{20}{21}\\ =\dfrac{10}{21}\)
17) \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{20.22}\)
2B= \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\)
2B= \(\dfrac{1}{2}-\dfrac{1}{22}\)
2B= \(\dfrac{5}{11}\)
B = \(\dfrac{5}{11}:2\)
B= \(\dfrac{5}{11}.\dfrac{1}{2}\)
B= \(\dfrac{5}{22}\)
Lần sau bn chú ý tách nhỏ bài ra cho dễ làm nhá bn, chứ nhìn vậy mn ngại làm lắm
13: \(1+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{10}+...+\dfrac{2}{45}\)
\(=1+\dfrac{4}{6}+\dfrac{4}{12}+...+\dfrac{4}{90}\)
\(=1+4\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right)\)
\(=1+4\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=1+4\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=1+4\cdot\dfrac{4}{10}\)
\(=1+\dfrac{16}{10}=\dfrac{26}{10}=\dfrac{13}{5}\)
14: \(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+\dfrac{1}{13\cdot19}+\dfrac{1}{19\cdot25}+\dfrac{1}{25\cdot31}+\dfrac{1}{31\cdot37}\)
\(=\dfrac{1}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{31\cdot37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)=\dfrac{1}{6}\cdot\dfrac{36}{37}=\dfrac{6}{37}\)
15: \(\dfrac{1}{2}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{1}{2}\left(1-1+\dfrac{1}{99}\right)=\dfrac{1}{2}\cdot\dfrac{1}{99}=\dfrac{1}{198}\)
16: \(\dfrac{1}{2}-\dfrac{1}{5\cdot11}-\dfrac{1}{11\cdot17}-\dfrac{1}{17\cdot23}-\dfrac{1}{23\cdot29}-\dfrac{1}{29\cdot35}\)
\(=\dfrac{1}{2}-\dfrac{1}{6}\left(\dfrac{6}{5\cdot11}+\dfrac{6}{11\cdot17}+...+\dfrac{6}{29\cdot35}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{6}\left(\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{17}+...+\dfrac{1}{29}-\dfrac{1}{35}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{6}\left(\dfrac{1}{5}-\dfrac{1}{35}\right)=\dfrac{1}{2}-\dfrac{1}{6}\cdot\dfrac{6}{35}=\dfrac{1}{2}-\dfrac{1}{35}=\dfrac{33}{70}\)
18: \(1-\dfrac{5}{5\cdot10}-\dfrac{5}{10\cdot15}-...-\dfrac{5}{95\cdot100}\)
\(=1-\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{95\cdot100}\right)\)
\(=1-\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\)
\(=1-\left(\dfrac{1}{5}-\dfrac{1}{100}\right)=1-\dfrac{19}{100}=\dfrac{81}{100}\)
19: \(\dfrac{1}{11}+\dfrac{1}{209}+\dfrac{1}{513}+\dfrac{1}{945}+\dfrac{1}{1505}+\dfrac{1}{2193}\)
\(=\dfrac{1}{11}+\dfrac{1}{11\cdot19}+\dfrac{1}{19\cdot27}+\dfrac{1}{27\cdot35}+\dfrac{1}{35\cdot43}+\dfrac{1}{43\cdot51}\)
\(=\dfrac{1}{11}+\dfrac{1}{8}\left(\dfrac{8}{11\cdot19}+\dfrac{8}{19\cdot27}+\dfrac{8}{27\cdot35}+\dfrac{8}{35\cdot43}+\dfrac{8}{43\cdot51}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{8}\left(\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{27}+...+\dfrac{1}{43}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{8}\left(\dfrac{1}{11}-\dfrac{1}{51}\right)=\dfrac{1}{11}+\dfrac{1}{8}\cdot\dfrac{40}{561}\)
\(=\dfrac{1}{11}+\dfrac{5}{561}=\dfrac{56}{561}\)
20: \(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\)
\(=9-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)
\(=9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=9-\left(1-\dfrac{1}{10}\right)=9-1+\dfrac{1}{10}=\dfrac{81}{10}\)
