Đề bài

So sánh: a) \(A = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}\) và \(B = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{{13}}{{60}}\) b) \(C = \frac{{2019}}{{2020}}.\frac{{2021}}{{2022}}\) và \(D = \frac{{2020 + 2022}}{{2019 + 2021}}.\frac{3}{2}\)

Lời giải chi tiết

a) Ta có: \(\begin{array}{l}A = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}\\ \Rightarrow 2A = 2.\left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}} \right)\\ = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2019}}}} + \frac{1}{{{2^{2020}}}}\\ \Rightarrow A = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2019}}}} + \frac{1}{{{2^{2020}}}}} \right) - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}} \right)\\ \Rightarrow A = 1 - \frac{1}{{{2^{2021}}}} < 1\end{array}\)Và \(\begin{array}{l}B = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{{13}}{{60}} = \frac{{1.20}}{{3.20}} + \frac{{1.15}}{{4.15}} + \frac{{1.12}}{{5.12}} + \frac{{13}}{{60}}\\ = \frac{{20}}{{60}} + \frac{{15}}{{60}} + \frac{{12}}{{60}} + \frac{{13}}{{60}} = \frac{{20 + 15 + 12 + 13}}{{60}} = \frac{{60}}{{60}} = 1\end{array}\) Vậy A < B.b) \(C = \frac{{2019}}{{2020}}.\frac{{2021}}{{2022}}\) và \(D = \frac{{2020 + 2022}}{{2019 + 2021}}.\frac{3}{2}\)Ta có:\(C = \frac{{2019}}{{2020}}.\frac{{2021}}{{2022}} = \frac{{2019.2021}}{{2020.2022}}\)Mà: \(2019.2021 < 2020.2022\) nên \(C < 1\)\(D = \frac{{2020 + 2022}}{{2019 + 2021}}.\frac{3}{2}\)Mà \(2020>2019; 2022>2021\) nên \(2020+2022 > 2019+2021\). Do đó \(\frac{{2020 + 2022}}{{2019 + 2021}} > 1 \Rightarrow D > 1.\frac{3}{2} = \frac{3}{2} > 1\)