4. Zadatak

Ekvivalentne su jednadžbe zadanoj jednadžbi:

$$\begin{aligned} \frac{2x+2}{2023}+\frac{x}{1011}&=\frac{x+2}{1012}+\frac{2x+6}{2025}\\ \Longleftrightarrow\quad \frac{x}{2022}+\frac{x+1}{2023}&=\frac{x+2}{2024}+\frac{x+3}{2025}\\ \Longleftrightarrow\quad \frac{x-2022+2022}{2022}+\frac{x-2022+2023}{2023}&=\frac{x-2022+2024}{2024}+\frac{x-2022+2025}{2025}\\ \Longleftrightarrow\quad \frac{x-2022}{2022}+1+\frac{x-2022}{2023}+1&=\frac{x-2022}{2024}+1+\frac{x-2022}{2025}+1\\ \Longleftrightarrow\quad \frac{x-2022}{2022}+\frac{x-2022}{2023}&=\frac{x-2022}{2024}+\frac{x-2022}{2025}\\ \Longleftrightarrow\quad (x-2022)\left(\frac1{2022}+\frac1{2023}-\frac1{2024}-\frac1{2025}\right)&=0. \end{aligned}$$

Slijedi $x-2022=0$, odnosno rješenje je zadane jednadžbe $x=2022$.