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

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

   $⇔ (\dfrac{x+1}{2020}+1)-(\dfrac{x+2}{2019}+1) = \dfrac{x+2021}{2018}$

   $⇔ \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 $

   $⇔x+2021=0$

   $⇔x=-2021$

   Vậy $S = \{-2021\}$

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2. $\text{Giải thích các bước giải:}$

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

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

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

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

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

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

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

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

   $⇒ x = -2021$

   $\text{Vậy x = -2021}$

   $\text{Chúc bạn học tốt !}$

    

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