\(P=\dfrac{5}{5.10}+\dfrac{5}{10.15}+...+\dfrac{5}{95.100}\)
\(\Rightarrow P=\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\)
\(\Rightarrow P=\dfrac{1}{5}-\dfrac{1}{100}\)
\(\Rightarrow P=\dfrac{19}{100}\)
Vậy \(P=\dfrac{19}{100}\)