`root(3)(24+x)+sqrt(12-x)=6 (x<=12)`

Đặt `{(u=root(3)(24+x)),(v=sqrt(12-x)):} (v>=0)`

`=>{(u^3=24+x),(v^2=12-x):}`

`=>u^3+v^2=(24+x)+(12-x)=36 color{red}{(1)}`

Theo đề bài ta có: `u+v=6=>u=6-v color{red}{(2)}`

Thay `color{red}{(2)}` vào `color{red}{(1)}` ta được:

`(6-v)^3+v^2=36`

`<=>(216-108v+18v^2-v^3)+v^2-36=0`

`<=>-v^3+19v^2-108v+180=0`

`<=>v^3-19v^2+108v-180=0`

`<=>v^3-3v^2-16v^2+48v+60v-180=0`

`<=>v^2(v-3)-16v(v-3)+60(v-3)=0`

`<=>(v-3)(v^2-16v+60)=0`

`<=>(v-3)(v^2-6v-10v+60)=0`

`<=>(v-3)[v(v-6)-10(v-6)]=0`

`<=>(v-3)(v-6)(v-10)=0`

`<=>[(v=3),(v=6),(v=10):}(tm)`

Với `v=3=>sqrt(12-x)=3`

`=> 12-x=9=>x=3(tm)`

Với `v=6=>sqrt(12-x)=6`

`=>12-x=36=>x=-24(tm)`

Với `v=10=>sqrt(12-x)=10`

`=>12-x=100=>x=-88 (tm)`

`=>S={-88;-24;3}`