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\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=2-\frac{1}{10}\)
\(=\frac{19}{10}\)
Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)
\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)
\(=-\frac{9}{10}+\frac{1}{90}\)
= ...
bn tự tính nha!
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{1}-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(-\frac{1}{1.2}+-\frac{1}{2.3}+-\frac{1}{3.4}+-\frac{1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
\(=-1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=-1\left(1-\frac{1}{5}\right)\)
\(=-1.\frac{4}{5}=-\frac{4}{5}\)
\(\frac{-1}{1.2}+\frac{-1}{2.3}+\frac{-1}{3.4}+\frac{-1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
\(=-1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=-1\left(1-\frac{1}{5}\right)\)
\(=-1.\frac{4}{5}=-\frac{4}{5}\)
Đặt A=(1-2/2*3)*(1-2/3*4)*(1-2*/4*5)*...*(1-2/99*100)
A=\(\frac{1\cdot4}{2\cdot3}\)*\(\frac{2\cdot5}{3\cdot4}\)*\(\frac{3\cdot6}{4\cdot5}\)*...*\(\frac{98\cdot101}{99\cdot100}\)
A=\(\frac{1\cdot101}{3\cdot99}\)
A=\(\frac{101}{297}\)
\(\:\frac{-1}{1.2}+\frac{-1}{2.3}+\frac{-1}{3.4}+\frac{-1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
=\(-1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
=\(-1\left(1-\frac{1}{5}\right)\)
=\(-1\times\frac{4}{5}\)
=\(\frac{-4}{5}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)