Toán Lớp 8: Giải phương trình:
$\frac{x+1}{2019}+$ $\frac{x+2}{2018}+$ $\frac{x-1}{2021}+3=0$
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Toán Lớp 8: Giải phương trình:
$\frac{x+1}{2019}+$ $\frac{x+2}{2018}+$ $\frac{x-1}{2021}+3=0$
Comments ( 2 )
(x+1)/2019+(x+2)/2018+(x-1)/2021=0
<=>((x+1)/2019+1)+((x+2)/2018+1)+((x-1)/2021+1)=0
<=>((x+1)/2019+1/1)+((x+2)/2018+1/1)+((x-1)/2021+1/1)=0
<=>((x+1)/2019+2019/2019)+((x+2)/2018+2018/2018)+((x-1)/2021+2021/2021)=0
<=>(x+1+2019)/2019+(x+2+2018)/2018+(x-1+2021)/2021=0
<=>(x+2020)/2019+(x+2020)/2018+(x+2020)/2021=0
<=>(x+2020)(1/2019+1/2018+1/2021)=0
<=>x+2020=0
<=>x=-2020
Vậy , S={-2020}
Giải đáp+Lời giải và giải thích chi tiết:
(x+1)/2019+(x+2)/2018+(x-1)/2021+3=0
<=> ((x+1)/2019+1)+((x+2)/2018+1)+((x-1)/2021+1)=0
<=> ((x+1)/2019+2019/2019)+((x+2)/2018+2018/2018)+((x-1)/2021+2021/2021)=0
<=> (x+2020)/2019+(x+2020)/2018+(x+2020)/2021=0
<=> (x+2020). (1/2019+1/2018+1/2021)=0
<=> x+2020=0
<=> x=-2020
Vậy x=-2020