Trang chủ » Hỏi đáp » Môn Toán Tìm `x` : `(x-1)/2022+(x-2)/2021+(x-3)/2020 = (x+4034)/2019` 01/07/2024 Tìm `x` : `(x-1)/2022+(x-2)/2021+(x-3)/2020 = (x+4034)/2019`
(x-1)/(2022)+(x-2)/(2021)+(x-3)/(2020)=(x+4034)/(2019) ⇔ ((x-1)/(2022)-1)+((x-2)/(2021)-1)+((x-3)/2020 -1)=(x+4034)/(2019)-3 ⇔ (x-1-2022)/2022+(x-2-2021)/2021+(x-3-2020)/2020=(x+4034-3.2019)/2019 ⇔ (x-2023)/2022+(x-2023)/2021+(x-2023)/2020=(x-2023)/2019 ⇔ (x-2023)/2022+(x-2023)/2021+(x-2023)/2020-(x-2023)/2019=0 ⇔ (x-2023) . (1/2020 +1/2021+1/2020-1/2019)=0 Vì 1/2020 +1/2021+1/2020-1/2019 \ne 0 ⇒ x-2023=0 ⇒ x=2023 Vậy x=2023. Trả lời
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