Đáp án:

Giải thích các bước giải:

`4sqrt(x+3)+sqrt(19-3x)=x^2 +2x+9`

`(đk: (19)/3 >=  x >= -3 )`

`<=> 12sqrt(x+3)+3sqrt(19-3x)=3x^2 +6x+27`

`<=> 4(3sqrt(x+3)- (x+5))+ (3sqrt(19-3x) -(13-x))= 3x^2 +3x -6`

`<=> 3(x^2 +x-2) - 4. (9x+27-(x^2 +10x+25))/(3sqrt(x+3)+ (x+5)) - (171-27x-(169-26x+x^2 ))/(3sqrt(19-3x) +(13-x)) =0`

`<=> 3(x+2)(x-1) + 4. ((x+2)(x-1))/(3sqrt(x+3)+ (x+5)) + ((x+2)(x-1))/(3sqrt(19-3x) +(13-x)) =0`

`<=> (x+2)(x-1)(3 + 4/(3sqrt(x+3)+ (x+5)) + 1/(3sqrt(19-3x) +(13-x)) )=0`

`<=> x=-2(tm)` hoặc `x=1(tm)` 

( vì `3+ 4/(3sqrt(x+3)+ (x+5)) + 1/(3sqrt(19-3x) +(13-x)) ) > 0 `)

Vậy `S={-2;1}`