Giải phương trình sau: ((x+1):2020)+((x+2):2019)+((x+3):2018)+((x+4):2017)=-4

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1. Đáp án:

   <->x + 1 / 2020 +1+ x + 2 / 2019+1 + x + 3 / 2018 +1+ x + 4 / 2017+1=0

   <-> (x+2021)/2020+(x+2021)/2019+(x+2021)/2018+(x+2021)/2017=0

   <-> (x+2021)(1/2020+1/2019-1/2018-1/2017)=0
   <-> x+2021=0

   <-> x=-2021

   Giải thích các bước giải:

    

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2. `\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4`

   `<=> \frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}+4=0`

   `<=>(\frac{x+1}{2020}+1)+(\frac{x+2}{2019}+1)+(\frac{x+3}{2018}+1)+(\frac{x+4}{2017}+1)=0`

    `<=> \frac{x+1+2020}{2020}+\frac{x+2+2019}{2019}+\frac{x+3+2018}{2018}+\frac{x+4+2017}{2017}=0`

   `<=> \frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=0`

   `<=> (x+2021).(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017})=0`

   `<=> `\(\left[ \begin{array}{l}x+2021=0\\\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}=0\end{array} \right.\) 

   Mà do `\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017} \ne 0`

   `=>` Vậy `S={-2021}`

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