$\frac{2-x}{2019}$ -1= $\frac{1-x}{2020}$ – $\frac{x}{2021}$ Giai PT 11/07/2021 Bởi Daisy $\frac{2-x}{2019}$ -1= $\frac{1-x}{2020}$ – $\frac{x}{2021}$ Giai PT
$\dfrac{2-x}{2019} -1 = \dfrac{1-x}{2020} – \dfrac{x}{2021}$ $⇔ \dfrac{2-x}{2019} -1 + 2 = \dfrac{1-x}{2020} – \dfrac{x}{2021} +2$$⇔ \dfrac{2-x}{2019} +1 = (\dfrac{1-x}{2020} +1 ) + ( \dfrac{-x}{2021} +1)$$⇔\dfrac{2021-x}{2019} = \dfrac{2021-x}{2020} + \dfrac{2021 -x}{2021}$$⇔\dfrac{2021-x}{2019} – \dfrac{2021-x}{2020} – \dfrac{2021 -x}{2021} =0$$⇔ ( 2021 -x)(\dfrac{1}{2019} – \dfrac{1}{2020} – \dfrac{1}{2021} )=0$Vì $\dfrac{1}{2019} – \dfrac{1}{2020} – \dfrac{1}{2021} \neq 0$ ⇔ $2021 – x = 0$ ⇔ $x = 2021$ Phương trình trên có một nghiệm là $S =$ { $2021$ } Bình luận
Đáp án:CHÚC BẠN HỌC TỐT!!!!
Giải thích các bước giải:
$\dfrac{2-x}{2019} -1 = \dfrac{1-x}{2020} – \dfrac{x}{2021}$
$⇔ \dfrac{2-x}{2019} -1 + 2 = \dfrac{1-x}{2020} – \dfrac{x}{2021} +2$
$⇔ \dfrac{2-x}{2019} +1 = (\dfrac{1-x}{2020} +1 ) + ( \dfrac{-x}{2021} +1)$
$⇔\dfrac{2021-x}{2019} = \dfrac{2021-x}{2020} + \dfrac{2021 -x}{2021}$
$⇔\dfrac{2021-x}{2019} – \dfrac{2021-x}{2020} – \dfrac{2021 -x}{2021} =0$
$⇔ ( 2021 -x)(\dfrac{1}{2019} – \dfrac{1}{2020} – \dfrac{1}{2021} )=0$
Vì $\dfrac{1}{2019} – \dfrac{1}{2020} – \dfrac{1}{2021} \neq 0$
⇔ $2021 – x = 0$
⇔ $x = 2021$
Phương trình trên có một nghiệm là $S =$ { $2021$ }