Đáp án:

Giải thích các bước giải:

`d)(x+1)/2019+(x+2)/2018+(x+3)/2017=-3`
`=>(x+1)/2019+(x+2)/2018+(x+3)/2017+3=0`
`=>((x+1)/2019+1)+((x+2)/2018+1)+((x+3)/2017+1)=0`
`=>(x+2020)/2019+(x+2020)/2018+(x+2020)/2017=0`
`=>(x+2020)(1/2019+1/2018+1/2017)=0`
`=>x+2020=0` ( vì `1/2019+1/2018+1/2017 > 0` )
`=>x=-2020`
Vậy `x=-2020`

$\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+7}{2013}=-3$

$⇒\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+7}{2013}+3=0$

$⇒(\frac{x+1}{2019}+1)+(\frac{x+2}{2018}+1)+(\frac{x+7}{2013}+1)=0$

$⇒(\frac{x+1}{2019}+1)+(\frac{x+2}{2018}+1)+(\frac{x+7}{2013}+1)=0$

$⇒(\frac{x+1}{2019}+\frac{2019}{2019})+(\frac{x+2}{2018}+\frac{2018}{2018})+(\frac{x+7}{2013}+\frac{2013}{2013})=0$

$⇒(\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2013})=0$

$⇒(x+2020)×(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2013})=0$

$Ta$ $thấy$ $\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2013}>0 ⇒ x+2020 =0$

$⇒x=-2020$