Đáp án+Giải thích các bước giải:

 `a)`

`\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=\frac{x+2003}{6}`

`<=>(\frac{x+2}{2019}+1)+(\frac{x+3}{2018}+1)+(\frac{x+4}{2017}+1)-1-1-1=\frac{x+2003}{6}`

`<=>(\frac{x+2+2019}{2019})+(\frac{x+3+2018}{2018})+(\frac{x+4+2017}{2017})=\frac{x+2003}{6}+3`

`<=>\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=\frac{x+2003}{6}+\frac{18}{6}`

`<=>\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}-\frac{x+2021}{6}=0`

`<=>(x+2021).(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-1/6)=0`

Ta nhận thấy: `\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-1/6\ne0`

Nên: `x+2021=0`

`<=>x=-2021`

Vậy `S={-2021}.`

`p)`

`\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}`

`<=>(\frac{x+2}{98}+1)+(\frac{x+4}{96}+1)-1-1=\frac{x+6}{94}+\frac{x+8}{92}`

`<=>(\frac{x+2+98}{98})+(\frac{x+4+96}{96})=(\frac{x+6}{94}+1)+(\frac{x+8}{92}+1)`

`<=>\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+6+94}{94}+\frac{x+8+92}{92}`

`<=>\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0`

`<=>(x+100).(1/98+1/96-1/94-1/92)=0`

Ta nhận thấy: `1/98+1/96-1/94-1/92\ne0`

Nên: `x+100=0`

`<=>x=-100`

Vậy `S={-100}.`