a: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\)
=>\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}\)
=>\(2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{2022}}\)
=>\(A=1-\dfrac{1}{2^{2022}}\)
\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{17}{60}\)
\(=\dfrac{20}{60}+\dfrac{15}{60}+\dfrac{12}{60}+\dfrac{17}{60}\)
\(=\dfrac{64}{60}\)
b: Vì A<1
và B>1
nên A<B