so sánh 10 mũ 2023 1 phần 10 mũ 2024 1

\(A = \frac{{{{10}^{2024}} + 1}}{{{{10}^{2023}} + 1}} = \frac{{10\left( {{{10}^{2023}} + 1} \right)}}{{{{10}^{2023}} + 1}} - \frac{9}{{{{10}^{2023}} + 1}} = 1 - \frac{9}{{{{10}^{2023}} + 1}}\)

\[A = \frac{{{{10}^{2023}} + 1}}{{{{10}^{2022}} + 1}} = \frac{{10\left( {{{10}^{2022}} + 1} \right)}}{{{{10}^{2022}} + 1}} - \frac{9}{{{{10}^{2022}} + 1}} = 1 - \frac{9}{{{{10}^{2022}} + 1}}\]

Vì \(\frac{9}{{{{10}^{2023}} + 1}} < \frac{9}{{{{10}^{2022}} + 1}}\) nên \(1 - \frac{9}{{{{10}^{2023}} + 1}} > 1 - \frac{9}{{{{10}^{2022}} + 1}}\) hay A > B

Vậy A > B