`6)(x-28/(x-3))(x-21/(x+4))`

`ĐKXĐ:{(x-3 ne 0),(x+4 ne 0):}`

`=>{(x ne 3),(x ne -4):}`

Với `x ne 3, x ne -4,` ta có:

`(x-28/(x-3))(x-21/(x+4))`

`=[(x(x-3))/(x-3)-28/(x-3)][(x(x+4))/(x+4)-21/(x+4)]`

`=[(x^2-3x)/(x-3)-28/(x-3)][(x^2+4x)/(x+4)-21/(x+4)]`

`=(x^2-3x-28)/(x-3) . (x^2+4x-21)/(x+4)`

`=(x^2-7x+4x-28)/(x-3) . (x^2-3x+7x-21)/(x+4)`

`=((x^2-7x)+(4x-28))/(x-3) . ((x^2-3x)+(7x-21))/(x+4)`

`=(x(x-7)+4(x-7))/(x-3) . (x(x-3)+7(x-3))/(x+4)`

`=((x+4)(x-7))/(x-3) . ((x+7)(x-3))/(x+4)`

`=(x-7)(x+7)`

`=x^2-49`

Vậy `...`