`@Khói`
\(A=\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
\(A=5\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{48.50}\right)\)
\(A=\dfrac{5}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{48.50}\right)\)
\(A=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(A=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(A=\dfrac{5}{2}\left(\dfrac{25}{50}-\dfrac{1}{50}\right)\)
\(A=\dfrac{5}{2}.\dfrac{24}{50}\)
\(A=\dfrac{60}{50}=\dfrac{6}{5}\)
\(=>A=5.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(A=5.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=5.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=\dfrac{5.24}{50}=\dfrac{120}{50}=\dfrac{12}{5}\)