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\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
\(S=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}\right)-\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}\right)\)\(S=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\right)-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)\(S=\left(1-\dfrac{1}{9}\right)-\left(1-\dfrac{1}{10}\right)\)
\(S=\dfrac{8}{9}-\dfrac{9}{10}=\dfrac{-1}{10}\)
\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+\dfrac{6}{8.10}+...+\dfrac{6}{30.32}+\dfrac{6}{32.34}\)
\(=6\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{30.32}+\dfrac{1}{32.34}\right)\)
\(=6\cdot\dfrac{2}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{30.32}+\dfrac{1}{32.34}\right)\)
\(=\dfrac{6}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{30.32}+\dfrac{2}{32.34}\right)\)
\(=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{30}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{34}\right)\)
\(=3\left(\dfrac{1}{2}-\dfrac{1}{34}\right)=3\cdot\dfrac{8}{17}=\dfrac{24}{17}\)
A\(=6\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{32.34}\right)\)
A\(=6.\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{32}-\dfrac{1}{34}\right)\)
A\(=3\left(\dfrac{1}{2}-\dfrac{1}{34}\right)\)
A\(=3.\dfrac{8}{17}\)
A\(=\dfrac{24}{17}\)
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{7\cdot9}+\dfrac{1}{6\cdot8}\)
\(=\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{7\cdot9}\right)+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{7\cdot9}\right)+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{9}{9}-\dfrac{1}{9}\right)+\dfrac{1}{2}\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}+\dfrac{1}{2}\cdot\dfrac{3}{8}\)
\(=\dfrac{1}{2}\left(\dfrac{8}{9}+\dfrac{3}{8}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{64}{72}+\dfrac{27}{72}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{91}{72}\)
\(=\dfrac{91}{144}\)
S=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+...+\dfrac{1}{6.8}\)
S=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{2}+...+\dfrac{1}{6}-\dfrac{1}{8}\right)\)
S=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{8}\right)\)
S=\(\dfrac{1}{2}.\left(\dfrac{8-1}{8}\right)\)
S=\(\dfrac{1}{2}.\dfrac{7}{8}\)
S=\(\dfrac{7}{16}\)
A=\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{20\cdot22}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{11}{22}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{5}{11}\)
\(=\dfrac{5}{22}\)
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{48.50}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{50-48}{48.50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{24}{50}=\dfrac{6}{25}\)
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
= \(\dfrac{2}{2}.\left(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+....+\dfrac{5}{48.50}\right)\)
\(\)\(=\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{48.50}\right)\)
\(=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\dfrac{12}{25}\)
=\(\dfrac{6}{5}\)=\(1\dfrac{1}{5}\)
Nếu bạn không biết cách giải bài này có thể bảo mình viết cách giải giúp!!!
Chúc bạn làm tốt!!!
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{48.50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{48}\right)\)
=\(\dfrac{5}{2}.\dfrac{23}{48}\) = \(\dfrac{115}{96}\)
Ta có :
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)
\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)
\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)
\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)
\(F=\dfrac{1004}{1005}\)
\(D=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{98.100}\)
\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}.\dfrac{49}{100}=\dfrac{147}{200}\)
\(D=\dfrac{3}{2\cdot4}+\dfrac{3}{4\cdot6}+\dfrac{3}{6\cdot8}+...+\dfrac{3}{98\cdot100}\\ =\dfrac{3}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\dfrac{49}{100}\\ =\dfrac{147}{200}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)