a: Đặt A=\(-\dfrac{1}{10}-\dfrac{1}{100}-\dfrac{1}{1000}-\dfrac{1}{10000}-\dfrac{1}{100000}\)
=>\(A=-\dfrac{1}{10}-\dfrac{1}{10^2}-...-\dfrac{1}{10^5}\)
=>\(10A=-1-\dfrac{1}{10}-...-\dfrac{1}{10^4}\)
=>\(10A-A=-1-\dfrac{1}{10}-...-\dfrac{1}{10^4}+\dfrac{1}{10}+\dfrac{1}{10^2}+...+\dfrac{1}{10^5}\)
=>\(9A=-1+\dfrac{1}{10^5}=\dfrac{-10^5+1}{10^5}\)
=>\(A=\dfrac{-10^5+1}{10^5\cdot9}\)
b: \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}\)
\(=\dfrac{1}{3}+\dfrac{13}{15}+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(-\dfrac{7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)\)
\(=\dfrac{1}{3}+\dfrac{13}{15}=\dfrac{5}{15}+\dfrac{13}{15}=\dfrac{18}{15}=\dfrac{6}{5}\)
c: \(\dfrac{1}{99}-\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)
