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

* Hướng dẫn giải

Đáp án C

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

Mà $\left(2020+\frac{1}{2020}-2021-\frac{1}{2021}\right)=\left(\frac{1}{2020}-\frac{1}{2021}\right)+\left(2020-2021\right)=\frac{1}{2020.2021}-1<0$

(do $\frac{1}{2020.2021}<1$)

Do đó $x-\frac{1}{2}=0$ hay $x=\frac{1}{2}$

Vậy $x_0=\frac{1}{2}$