Ta có: \(A=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{26\cdot31}\)
\(=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{26\cdot31}\right)\)
\(=5\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\cdot\left(1-\frac{1}{31}\right)=5\cdot\frac{30}{31}=\frac{150}{31}>1\)
hay A>1(đpcm)