tìm x ∈ Q $\frac{x+4}{2016}$ +$\frac{x+3}{2017}$= $\frac{x+2}{2018}$ +$\frac{x+1}{2019}$

0 bình luận về “tìm x ∈ Q $\frac{x+4}{2016}$ +$\frac{x+3}{2017}$= $\frac{x+2}{2018}$ +$\frac{x+1}{2019}$”

1. Đáp án:

   `x=-2020`

   Giải thích các bước giải:

   `(x + 4)/2016 + (x + 3)/2017 = (x + 2)/2018 + (x+1)/2019`

   `<=> (x + 4)/2016 + (x + 3)/2017 – (x + 2)/2018 – (x + 1)/2019 = 0`

   `<=> ((x + 4)/2016 + 1) + ((x + 3)/2017 +1) – ((x + 2)/2018 +1)- ((x + 1)/2019 +1)= 0`

   `<=> (x + 2020)/2016 + (x+2020)/2017 – (x+2020)/2018 – (x+2020)/2019 = 0`

   `<=> (x+2020)(1/2016 + 1/2017 – 1/2018 – 1/2019) = 0`

   `⇒ x + 2020 = 0` (do `1/2016 + 1/2017 – 1/2018 – 1/2019` $\neq 0$)

   `⇔ x = -2020`

   Bình luận

2. Đáp án:

    

   Giải thích các bước giải:

      $\frac{x +4}{2016}$ + $\frac{x+3}{2017}$ = $\frac{x+2}{2018}$ + $\frac{x+1}{2019}$

   ⇒(1+$\frac{x +4}{2016}$) + (1+$\frac{x+3}{2017}$) = (1+$\frac{x+2}{2018}$) + (1+$\frac{x+1}{2019}$)

   ⇒$\frac{x +2020}{2016}$ + $\frac{x+2020}{2017}$ = $\frac{x+2020}{2018}$ + $\frac{x+2020}{2019}$

   ⇒ $\frac{x +2020}{2016}$ + $\frac{x+2020}{2017}$ – $\frac{x+2020}{2018}$ – $\frac{x+2020}{2019}$ = 0

   ⇒ (x + 2020) . ($\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$ – $\frac{1}{2019}$) = 0

   Do $\frac{1}{2016}$ + $\frac{1}{2017}$ – $\frac{1}{2018}$ – $\frac{1}{2019}$ $ \neq$ 0

   ⇒ x + 2020 = 0

   ⇒         x     = -2020

   Vậy x = -2020

   Bình luận