a: \(A=\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+...+\dfrac{1}{105\cdot107}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{105\cdot107}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{107}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{102}{535}\)
\(=\dfrac{51}{535}\)
b: Ta có: \(C=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+...+\dfrac{3}{47\cdot49}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{47\cdot49}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{46}{147}\)
\(=\dfrac{23}{49}\)
a)\(A=\dfrac{1}{2}\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{105.107}\right)=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{105}-\dfrac{1}{107}\right)=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{107}\right)=\dfrac{1}{2}.\dfrac{102}{535}=\dfrac{51}{535}\)
b) \(C=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{47.49}\right)=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)=\dfrac{1}{98}\)
