\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ \Rightarrow2A=2+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\\ \Rightarrow2A-A=2-\dfrac{1}{50^2}\\ A=\dfrac{5000}{50^2}-\dfrac{1}{50^2}=\dfrac{5000-1}{50^2}=\dfrac{4999}{2500}\)
Giả sử \(B=\dfrac{173}{100}=\dfrac{4325}{2500}\), mà \(\dfrac{4999}{2500}>\dfrac{4325}{2500}\)
\(\Rightarrow A>B\)