Đáp án:

 `i)` `S={-40;50}`

 `j)` `S={1999}`

Giải thích các bước giải:

 `i)` `(x^2-10x-29)/1971+(x^2-10x-27)/1973=(x^2-10x-1971)/29+(x^2-10x-1973)/27`

`<=>((x^2-10x-29)/1971-1)+((x^2-10x-27)/1973-1)=((x^2-10x-1971)/29-1)+((x^2-10x-1973)/27-1)`

`<=>(x^2-10x-29-1971)/1971+(x^2-10x-27-1973)/1973=(x^2-10x-1971-29)/29+(x^2-10x-1973-27)/27`

`<=>(x^2-10x-2000)/1971+(x^2-10x-2000)/1973=(x^2-10x-2000)/29+(x^2-10x-2000)/27`

`<=>(x^2-10x-2000)(1/1971+1/1973-1/29-1/27)=0`

Vì `1/1971+1/1973-1/29-1/27\ne0`

`<=>x^2-10x-2000=0`

`<=>x^2+40x-50x-2000=0`

`<=>x(x+40)-50(x+40)=0`

`<=>(x+40)(x-50)=0`

`<=>[(x+40=0),(x-50=0):}<=>[(x=-40),(x=50):}`

Vậy `S={-40;50}`

$\\$

 `j)` `(x-29)/1970+(x-27)/1972+(x-25)/1974+(x-23)/1976+(x-21)/1978+(x-19)/1980=(x-1970)/29+(x-1972)/27+(x-1974)/25+(x-1976)/23+(x-1978)/21+(x-1980)/19`

`<=>((x-29)/1970-1)+((x-27)/1972-1)+((x-25)/1974-1)+((x-23)/1976-1)+((x-21)/1978-1)+((x-19)/1980-1)=((x-1970)/29-1)+((x-1972)/27-1)+((x-1974)/25-1)+((x-1976)/23-1)+((x-1978)/21-1)+((x-1980)/19-1)`

`<=>(x-29-1970)/1970+(x-27-1972)/1972+(x-25-1974)/1974+(x-23-1976)/1976+(x-21-1978)/1978+(x-19-1980)/1980=(x-1970-29)/29+(x-1972-27)/27+(x-1974-25)/25+(x-1976-23)/23+(x-1978-21)/21+(x-1980-19)/19`

`<=>(x-1999)/1970+(x-1999)/1972+(x-1999)/1974+(x-1999)/1976+(x-1999)/1978+(x-1999)/1980=(x-1999)/29+(x-1999)/27+(x-1999)/25+(x-1999)/23+(x-1999)/21+(x-1999)/19`

`<=>(x-1999)(1/1970+1/1972+1/1974+1/1976+1/1978+1/1980-1/29-1/27-1/25-1/23-1/21-1/19)`

Vì `1/1970+1/1972+1/1974+1/1976+1/1978+1/1980-1/29-1/27-1/25-1/23-1/21-1/19\ne0`

`<=>x-1999=0`

`<=>x=1999`

Vậy `S={1999}`