X+1/2019+x+2/2018=x+3/2017+x+4/2016 Giúp mk vs nha mk cảm ơn trc ạ 26/10/2021 Bởi Lydia X+1/2019+x+2/2018=x+3/2017+x+4/2016 Giúp mk vs nha mk cảm ơn trc ạ
Đáp án: `S={-2020}` Giải thích các bước giải: `(x+1)/2019+(x+2)/2018=(x+3)/2017+(x+4)/2016` `<=>(x+1)/2019+1+(x+2)/2018+1=(x+3)/2017+1+(x+4)/2016+1` `<=>(x+2020)/2019+(x+2020)/2018=(x+2020)/2017+(x+2020)/2016` `<=>(x+2020)(1/2019+1/2018-1/2017-1/2016)=0` `<=>x+2020=0` `<=>x=-2020` Vậy `S={-2020}` Bình luận
`(x+1)/2019+(x+2)/2018=(x+3)/2017+(x+4)/2016` `<=>((x+1)/2019+1)+((x+2)/2018+1)=((x+3)/2017+1)+((x+4)/2016+1)` `<=>(x+2020)/2019+(x+2020)/2018=(x+2020)/2017+(x+2020)/2016` `<=>(x+2020)/2019+(x+2020)/2018-(x+2020)/2017-(x+2020)/2016=0` `<=>(x+2020)(1/2019+1/2018-1/2017-1/2016)=0` Nhận thấy: `1/2019<1/2017=>1/2020-1/2017<0` `1/2018<1/2016=>1/2018-1/2016<0` `=>1/2019+1/2018-1/2017-1/2016<0` `=>x+2020=0` `<=>x=-2020` Vậy `S={-2020}` Bình luận
Đáp án:
`S={-2020}`
Giải thích các bước giải:
`(x+1)/2019+(x+2)/2018=(x+3)/2017+(x+4)/2016`
`<=>(x+1)/2019+1+(x+2)/2018+1=(x+3)/2017+1+(x+4)/2016+1`
`<=>(x+2020)/2019+(x+2020)/2018=(x+2020)/2017+(x+2020)/2016`
`<=>(x+2020)(1/2019+1/2018-1/2017-1/2016)=0`
`<=>x+2020=0`
`<=>x=-2020`
Vậy `S={-2020}`
`(x+1)/2019+(x+2)/2018=(x+3)/2017+(x+4)/2016`
`<=>((x+1)/2019+1)+((x+2)/2018+1)=((x+3)/2017+1)+((x+4)/2016+1)`
`<=>(x+2020)/2019+(x+2020)/2018=(x+2020)/2017+(x+2020)/2016`
`<=>(x+2020)/2019+(x+2020)/2018-(x+2020)/2017-(x+2020)/2016=0`
`<=>(x+2020)(1/2019+1/2018-1/2017-1/2016)=0`
Nhận thấy:
`1/2019<1/2017=>1/2020-1/2017<0`
`1/2018<1/2016=>1/2018-1/2016<0`
`=>1/2019+1/2018-1/2017-1/2016<0`
`=>x+2020=0`
`<=>x=-2020`
Vậy `S={-2020}`