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=1/1x2+1/2x3+1/3x4+...+1/1006x1007+1/1007x1008
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/1006-1/1007+1/1007-1/1008
=1/1-1/1008
=1007/1008
~-~:33
a,
suy ra A = 7. (1/10.11+1/11.12+1/12.13+.......+1/69.70)
suy ra A = 7. ( 1/10 - 1/11+ 1/11 - 1/12 + 1/12 - 1/13+ ............. + 1/69 - 1/70)
suy ra A = 7. ( 1/ 10 - 1/70)
suy ra A= 7. 3/35
suy ra A= 3/5
a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)
b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)
\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)
\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)
\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)
\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)
a)\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{49}{200}\)
b)\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{201}-\frac{1}{205}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{205}\right)\)
\(=\frac{1}{4}\cdot\frac{204}{205}\)
\(=\frac{51}{205}\)
c)\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3\cdot\frac{32}{99}\)
\(=\frac{32}{33}\)
d)tương tự bạn nhân với 4/3 nhé
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=2.\frac{1007}{2016}\)
\(=\frac{2007}{1008}\)
giải:
4/2.4+4/4.6+4/6.8+...+4/2012.2014+4/2014.2016
=2.(2/2.4+2/4.6+2/6.8+...+2/2012.2014+2/2014.2016
=2.(1/2-1/4+1,4-1/6+1/6-1/8+...+1/2012-1/2014+1/2014-1/2016)
=2.(1/2-1/2016)
=2.1007/2016
=1007/1008
xong rùi đó
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=2.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\)
=2.(\(\frac{1}{2}-\frac{1}{2010}\)) = 2.(\(\frac{1005}{2010}-\frac{1}{2010}\))
=2.\(\frac{502}{1005}\)
=\(\frac{1004}{1005}\)
\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1005}{2010}-\frac{1}{2010}\right)\)
\(=2\cdot\frac{1004}{2010}\)
\(=\frac{1004}{1005}\)
\(k\)\(mk\)\(nha\)\(bn\)
mình viết nhầm=)))))
\(b,\frac{10}{99}\)+\(\frac{11}{199}\)+\(\frac{12}{299}\).\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{-1}{6}\)