Đáp án:

Giải thích các bước giải:

`7) 2/(x-1) + 4/(x+3) = (3x+11)/(x^2+2x-3)` `(x` $\ne$ `1;-3)`

`=(2(x+3))/((x-1)(x+3)) + (4(x-1))/((x-1)(x+3)) = (3x+11)/((x-1)(x+3))`

`<=>2(x+3) + 4(x-1) =3x+11`

`<=>2x+6+4x-4=3x+11`

`<=>2x+4x-3x=11-6+4`

`<=>3x=9`

`<=>x=3` (tm)

Vậy `S={3}`

`8)1/(x+2) - (2x-9)/(x^3 +8) = 2/(x^2-2x+4)` `(x` $\ne$ `-2)`

`<=>(x^2-2x+4)/((x+2)(x^2-2x+4)) - (2x-9)/((x+2)(x^2-2x+4)) = (2(x+2))/((x+2)(x^2-2x+4))`

`<=>x^2-2x+4-(2x-9)=2(x+2)`

`<=>x^2 -2x+4-2x+9=2x+4`

`<=>x^2 - 2x-2x-2x=4-4-9`

`<=>x^2-6x=-9`

`<=>x^2-6x+9=0`

`<=>(x-3)^2 =0`

`<=>x-3=0` 

`<=>x=3` (tm)

Vậy `S={3}`