Tìm số hữu tỉ x sao cho:
\(\dfrac{{x + 1}}{{2023}} + \dfrac{{x + 2}}{{2022}} = \dfrac{{x + 3}}{{2021}} + \dfrac{{x + 4}}{{2020}}\)
Cộng cả 2 vế với 2
\(\begin{array}{l}\dfrac{{x + 1}}{{2023}} + \dfrac{{x + 2}}{{2022}} = \dfrac{{x + 3}}{{2021}} + \dfrac{{x + 4}}{{2020}}\\ \Leftrightarrow \left( {\dfrac{{x + 1}}{{2023}} + 1} \right) + \left( {\dfrac{{x + 2}}{{2022}} + 1} \right) = \left( {\dfrac{{x + 3}}{{2021}} + 1} \right) + \left( {\dfrac{{x + 4}}{{2020}} + 1} \right)\\ \Leftrightarrow \dfrac{{x + 2024}}{{2023}} + \dfrac{{x + 2024}}{{2022}} = \dfrac{{x + 2024}}{{2021}} + \dfrac{{x + 2024}}{{2020}}\\ \Leftrightarrow \dfrac{{x + 2024}}{{2023}} + \dfrac{{x + 2024}}{{2022}} - \dfrac{{x + 2024}}{{2021}} - \dfrac{{x + 2024}}{{2020}} = 0\\ \Leftrightarrow \left( {x + 2024} \right).\left( {\dfrac{1}{{2023}} + \dfrac{1}{{2022}} - \dfrac{1}{{2021}} - \dfrac{1}{{2020}}} \right) = 0\\ \Leftrightarrow \left( {x + 2024} \right) = 0\\ \Leftrightarrow x = - 2024\end{array}\)
Vậy x = -2024
