\(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)-x=\dfrac{-100}{99}\\ \left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=\dfrac{-100}{99}\\ \left(1-\dfrac{1}{99}\right)-x=\dfrac{-100}{99}\\ \dfrac{98}{99}-x=\dfrac{-100}{99}\\ x=\dfrac{98}{99}-\dfrac{-100}{99}\\ x=\dfrac{198}{99}\\ x=2\)
\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=\dfrac{-100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)\(-x=\dfrac{-100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=\dfrac{-100}{99}\)
\(\dfrac{98}{99}-x=\dfrac{-100}{99}\)
\(x=\dfrac{98}{99}-\dfrac{-100}{99}\)
\(x=2\)
Ta có: (\(\dfrac{2}{1\cdot3}\) +\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\) )
=\(\dfrac{3-1}{1\cdot3}+\dfrac{5-3}{3\cdot5}+\dfrac{7-5}{5\cdot7}+...+\dfrac{99-97}{97\cdot99}\)
= \(\dfrac{3}{1\cdot3}-\dfrac{1}{1\cdot3}+\dfrac{5}{3\cdot5}-\dfrac{3}{3\cdot5}+\dfrac{7}{5\cdot7}-\dfrac{5}{5\cdot7}+...+\dfrac{99}{97\cdot99}-\dfrac{97}{97\cdot99}\) =\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
=\(1-\dfrac{1}{99}\)
=\(\dfrac{98}{99}\)