Đáp án:

`1/(x^2-7x+12) +2/(x^2-10x+24) +3/(x^2-15x+54)=2/9`

`1/((x-3)(x-4)) +2/((x-4)(x-6)) +3/((x-6)(x-9))=2/9` 

`(1/(x-4) - 1/(x-3)) + (1/(x-6) - 1/(x-4)) + (1/(x-9) - 1/(x-6)) = 2/9` 

`1/(x-4) - 1/(x-3) + 1/(x-6) - 1/(x-4) + 1/(x-9) - 1/(x-6) = 2/9`

`- 1/(x-3) + 1/(x-9) = 2/9`

`1/(x-9) - 1/(x-3) = 2/9`

`6/((x-9)(x-3)) = 2/9` 

`6 * 9 = 2 (x-9)(x-3)`

`2(x-9)(x-3) = 54`

`(x-9)(x-3) = 27`

`x^2 - 3x - 9x + 27 = 27`

`x^2 - 12x = 0`

`x(x - 12) = 0`

`=> [(x = 0),(x - 12 = 0):}`

`=> [(x = 0),(x = 12):}`

Vậy `x in {0;12}`