Đặt A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{103.107}+\frac{4}{107.111}\)
\(=\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{15}\right)+...+\left(\frac{1}{103}-\frac{1}{107}\right)+\left(\frac{1}{107}-\frac{1}{111}\right)\)
\(=\frac{1}{3}-\frac{1}{111}=\frac{37-1}{111}=\frac{36}{111}\)