Question

1/99.97-1/97.95-1/95.93...-1/5.3-1/3.1

 

Answer 1

Ta có : \(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

           = \(\frac{1}{99.97}-\left(\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)

            =\(\frac{1}{99.97}-\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{97}+\frac{1}{93}-\frac{1}{95}+...+\frac{1}{3}-\frac{1}{5}+1-\frac{1}{3}\right)\)

          = \(\frac{1}{99.97}-\frac{1}{2}.\left(1-\frac{1}{97}\right)\)

          = \(\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)

          = \(\frac{1}{99.97}-\frac{48}{97}=\frac{1}{99.97}-\frac{48.99}{99.97}=\frac{-4751}{9603}\)

Answer 2

      \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{93.95}+\frac{2}{95.97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)

\(=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)

\(=\frac{1}{9603}-\frac{48}{97}=\frac{-4751}{9603}\)