Giải phương trình sau: $\frac{x+1}{2022}$+$\frac{x+2}{2021}$=$\frac{x+3}{2020}$+$\frac{x+4}{2019}$
2 câu trả lời
$\frac{x+1}{2022}$+$\frac{x+2}{2021}$= $\frac{x+3}{2020}$+$\frac{x+4}{2019}$
$\frac{x+1}{2022}$+1+$\frac{x+2}{2021}$+1= $\frac{x+3}{2020}$+1+$\frac{x+4}{2019}$+1
$\frac{x+2023}{2022}$+$\frac{x+2023}{2021}$= $\frac{x+2023}{2020}$+$\frac{x+2023}{2019}$
$\frac{x+2023}{2022}$+$\frac{x+2023}{2021}$-$\frac{x+2023}{2020}$-$\frac{x+2023}{2019}$= 0
(x+2023)($\frac{1}{2022}$+$\frac{1}{2021}$- $\frac{1}{2020}$- $\frac{1}{2019}$) =0
Vì ($\frac{1}{2022}$+$\frac{1}{2021}$- $\frac{1}{2020}$- $\frac{1}{2019}$) > 0
mà (x+2023)($\frac{1}{2022}$+$\frac{1}{2021}$- $\frac{1}{2020}$- $\frac{1}{2019}$) =0
⇒x+2023=0
⇒x=-2023
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