Gọi x0 là số thỏa mãn của (2020+1/2020-2021-1/2021)(x-1/2)

* Hướng dẫn giải

$$\begin{gathered} \left(2020+\frac{1}{2020}-2021-\frac{1}{2021}\right) \left(x-\frac{1}{2}\right)=\frac{1}{3}-\frac{2}{15}-\frac{1}{5} \\ (2020+\frac{1}{2020}-2021-\frac{1}{2021}) (x-\frac{1}{2})=\frac{5}{15}-\frac{2}{15}-\frac{3}{15} \\ (2020+\frac{1}{2020}-2021-\frac{1}{2021}) (x-\frac{1}{2})=0 \end{gathered}$$

Mà 

$(2020+\frac{1}{2020}-2021-\frac{1}{2021}) =(\frac{1}{2020}-\frac{1}{2021})+\left(2020-2021\right)$

$\frac{1}{2020.2021}-1< 0$

Do $\frac{1}{2020.2021}<1$