Giải các phương trình sau: a) 7 + 2x = 22 – 3x; b) (x – 2)(2x + 5) = 0;

c) $\frac{x-2018}{2019}+\frac{x-2019}{2020}-\frac{x-4041}{2021}=0$

$$\begin{gathered} \iff \frac{x-2018}{2019}+1+\frac{x-2019}{2020}+1-\frac{x-4041}{2021}-2=0 \\ \iff \left(\frac{x-2018}{2019}+1\right)+\left(\frac{x-2019}{2020}+1\right)-\left(\frac{x-4041}{2021}+2\right)=0 \\ \iff \frac{x-2018+2019}{2019}+\frac{x-2019+2020}{2020}-\frac{x-4041+2.2021}{2021}=0 \end{gathered}$$

$$\begin{gathered} \iff \frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0 \\ \iff (x+1)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0 \\ \iff \left[\begin{array}{l}x+1=0\\ \frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}=0\end{array}\right. \end{gathered}$$

Vì $\frac{1}{2020}>\frac{1}{2021}$  nên $\frac{1}{2020}-\frac{1}{2021}>0$

Suy ra $\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}>0$  hay $\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\ne 0$ .

Do đó phương trình đã cho tương đương với:

x + 1 = Û x = – 1.