Tìm x thỏa mãn $\frac{x}{2020}+$$\frac{x+1}{2021}$+$\frac{1-x}{2019}$+$\frac{2-x}{2018}$=0

0 bình luận về “Tìm x thỏa mãn $\frac{x}{2020}+$$\frac{x+1}{2021}$+$\frac{1-x}{2019}$+$\frac{2-x}{2018}$=0”

1. Đáp án:

    `x=2020`

   Giải thích các bước giải:

    `x/2020+(x+1)/2021+(1-x)/2019+(2-x)/2018=0`

   `\to (x/2020-1)+((x+1)/2021-1)+((1-x)/2019+1)+((2-x)/2018+1)=0`

   `\to (x-2020)/2020+(x+1-2021)/2021+(1-x+2019)/2019+(2-x+2018)/2018=0`

   `\to (x-2020)/2020+(x-2020)/2021+(2020-x)/2019+(2020-x)/2018=0`

   `\to (x-2020)/2020+(x-2020)/2021-(x-2020)/2019-(x-2020)/2018=0`

   `\to (x-2020)(1/2020+1/2021-1/2019-1/2018)=0`

   Vì `1/2020>(-1)/2019`

   `1/2021>(-1)/2018`

   `\to 1/2020+1/2021>(-1)/2019-1/2018`

   `\to 1/2020+1/2021-1/2019-1/2018>0`

   `\to x-2020=0`

   `\to x=2020`

   Vậy `x=2020`

   Trả lời

2. `x/2020 + (x+1)/2021 + (1-x)/2019 + (2-x)/2018 =0`

   `x/2020 -1 + (x+1)/2021 -1 + (1-x)/2019 + 1 +(2-x)/2018 +1=0`

   `(x-2020)/2020 + (x+1-2021)/2021 + (1-x+2019)/2019 + (2-x+2018)/2018 =0`

   `(x-2020)/2020 + (x-2020)/2021 + (2020-x)/2019 + (2020 -x)/2018 =0`

   `(x-2020)(1/2020 + 1/2021 -1/2019 – 1/2018) =0`

   Vì `1/2020 + 1/2021 – 1/2019 – 1/2018 ne 0`

   `=> x-2020 =0`

   `=> x= 0+2020`

   `=> x= 2020`

   Vậy `x= 2020`

    

   Trả lời