Sửa lại đề bài
\(A=\dfrac{38}{50}+\dfrac{9}{20}-\dfrac{11}{30}+\dfrac{13}{42}-\dfrac{15}{56}+\dfrac{17}{72}-...+\dfrac{197}{9702}-\dfrac{189}{9900}\)
\(\dfrac{9}{20}=\dfrac{4+5}{4.5}=\dfrac{4}{4.5}+\dfrac{5}{4.5}=\dfrac{1}{4}+\dfrac{1}{5}\)
\(\dfrac{11}{30}=\dfrac{5+6}{5.6}=\dfrac{5}{5.6}+\dfrac{6}{5.6}=\dfrac{1}{5}+\dfrac{1}{6}\)
\(\dfrac{13}{42}=\dfrac{6+7}{6.7}=\dfrac{6}{6.7}+\dfrac{7}{6.7}=\dfrac{1}{6}+\dfrac{1}{7}\)
\(\dfrac{15}{56}=\dfrac{7+8}{7.8}=\dfrac{7}{7.8}+\dfrac{8}{7.8}=\dfrac{1}{7}+\dfrac{1}{8}\)
\(\dfrac{17}{72}=\dfrac{8+9}{8.9}=\dfrac{8}{8.9}+\dfrac{9}{8.9}=\dfrac{1}{8}+\dfrac{1}{9}\)
.....
\(\dfrac{197}{9702}=\dfrac{98+99}{98.99}=\dfrac{98}{98.99}+\dfrac{99}{98.99}=\dfrac{1}{98}+\dfrac{1}{99}\)
\(\dfrac{199}{9900}=\dfrac{99+100}{99.100}=\dfrac{99}{99.100}+\dfrac{10}{99.100}=\dfrac{1}{99}+\dfrac{1}{100}\)
\(\Rightarrow A=\dfrac{38}{50}+\left(\dfrac{1}{4}+\dfrac{1}{5}\right)-\left(\dfrac{1}{5}+\dfrac{1}{6}\right)+\left(\dfrac{1}{6}+\dfrac{1}{7}\right)-\left(\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}\right)-...+\left(\dfrac{1}{98}+\dfrac{1}{99}\right)-\left(\dfrac{1}{99}+\dfrac{1}{100}\right)\)
\(\Rightarrow A=\dfrac{38}{50}+\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{76+25-1}{100}=\dfrac{100}{100}=1\)
