Giải phương trình: (x + 1)/2024 + (x + 2)/2023 = (x + 3)/2022 + (x + 4)/2021

Hướng dẫn giải

\[\frac{{x + 1}}{{2024}} + \frac{{x + 2}}{{2023}} = \frac{{x + 3}}{{2022}} + \frac{{x + 4}}{{2021}}\]

\[\left( {\frac{{x + 1}}{{2024}} + 1} \right) + \left( {\frac{{x + 2}}{{2023}} + 1} \right) = \left( {\frac{{x + 3}}{{2022}} + 1} \right) + \left( {\frac{{x + 4}}{{2021}} + 1} \right)\]

\[\frac{{x + 2025}}{{2024}} + \frac{{x + 2025}}{{2023}} = \frac{{x + 2025}}{{2022}} + \frac{{x + 2025}}{{2021}}\]

\[\frac{{x + 2025}}{{2024}} + \frac{{x + 2025}}{{2023}} - \frac{{x + 2025}}{{2022}} - \frac{{x + 2025}}{{2021}} = 0\]

\[\left( {x + 2025} \right)\left( {\frac{1}{{2024}} + \frac{1}{{2023}} - \frac{1}{{2022}} - \frac{1}{{2021}}} \right) = 0\]

Vì \[\frac{1}{{2024}} < \frac{1}{{2022}}\] nên \[\frac{1}{{2024}} - \frac{1}{{2022}} < 0\].

Vì \[\frac{1}{{2023}} < \frac{1}{{2021}}\] nên \[\frac{1}{{2023}} - \frac{1}{{2021}} < 0\].

Do đó \[\frac{1}{{2024}} + \frac{1}{{2023}} - \frac{1}{{2022}} - \frac{1}{{2021}} < 0\] hay \[\frac{1}{{2024}} + \frac{1}{{2023}} - \frac{1}{{2022}} - \frac{1}{{2021}} \ne 0\].

Khi đó \[x + 2025 = 0\] nên \[x =  - 2025\].

Vậy nghiệm của phương trình là \[x =  - 2025\].