\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+....+\frac{181}{9.10}\)
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\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+\frac{41}{4.5}+...+\frac{181}{9.10}\) \(=\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+\frac{40+1}{20}+...+\frac{180+1}{90}\)
\(=2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+2+\frac{1}{4.5}+...+2+\frac{1}{9.10}\)
\(=18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=19-\frac{1}{10}\)
\(=\frac{189}{10}\)
\(A=\frac{4+1}{1.2}+\frac{24+1}{3.4}+\frac{40+1}{4.5}+...+\frac{180+1}{9.10}\)
\(A=\left(\frac{4}{1.2}+\frac{24}{3.4}+\frac{40}{4.5}+...+\frac{180}{9.10}\right)+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(A=\left(2+2+2+...+2\right)+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2.8+\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{10}\right)=16+\frac{22}{30}=16\frac{11}{15}\)
\(\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-......-\frac{1}{9.10}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}=\frac{1}{10}\)
Ta có
\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=2-\frac{1}{10}\)
\(=\frac{19}{10}\)
Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}\)
\(=\frac{1}{10}\)
(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-......+1/9-1/10)
1-1/10=9/10
nhớ cho mk
\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+...+\frac{181}{9.10}\)
=\(\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+...+\frac{180+1}{90}\)
=\(2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+...+2+\frac{1}{9.10}\)
=\(18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
=\(9-\frac{1}{10}\)
=\(\frac{189}{10}\)