Câu trả lời:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...+\left(\frac{1}{9^2}-\frac{1}{10^2}\right)\)
\(=\frac{1}{1}-\frac{1}{10^2}\)
\(=1-\frac{1}{100}