mk cần gấp x+1/2020 – x+2/2019=x+3/2018+1

$\text{Giải thích các bước giải:}$

$\dfrac{x+1}{2020} – \dfrac{x+2}{2019} = \dfrac{x+3}{2018} + 1$

$⇒ \dfrac{x+1}{2020} + 1 – \dfrac{x+2}{2019} – 1 = \dfrac{x+3}{2018} + 1$

$⇒ (\dfrac{x+1}{2020} + 1) – (\dfrac{x+2}{2019} + 1) = \dfrac{x+3}{2018} + 1$

$⇒ \dfrac{x+1+2020}{2020} – \dfrac{x+2+2019}{2019} = \dfrac{x+3+2018}{2018}$

$⇒ \dfrac{x+2021}{2020} – \dfrac{x+2021}{2019} = \dfrac{x+2021}{2018}$

$⇒ \dfrac{x+2021}{2020} – \dfrac{x+2021}{2019} – \dfrac{x+2021}{2018}= 0$

$⇒ (x + 2021)(\dfrac{1}{2020} – \dfrac{1}{2019} – \dfrac{1}{2018}) = 0$

$⇒ x + 2021 = 0$ $(\dfrac{1}{2020} – \dfrac{1}{2019} – \dfrac{1}{2018} \neq 0)$

$⇒ x = -2021$

$\text{Vậy x = -2021}$

$\text{Chúc bạn học tốt !}$