`2\sqrt{x+1} + 6\sqrt{9-x^2} + 6\sqrt{(x+1)(9-x^2)} = -x^3-2x^2+10x+38  (-1 <= x <= 3)`

`<=> -x^3-2x^2+10x+38-2\sqrt{x+1}-6\sqrt{9-x^2}-6\sqrt{(x+1)(9-x^2)}=0`

`<=> (-x^3-x^2+9x+9-6\sqrt{(x+1)(9-x^2)}+9)+(x+1-2\sqrt{x+1}+1)+(9-x^2-6\sqrt{9-x^2}+9)=0`

`<=> [(x+1)(9-x^2)-6\sqrt{(x+1)(9-x^2)}+9]+(\sqrt{x+1}-1)^2+(\sqrt{9-x^2}-3)^2=0`

`<=> [\sqrt{(x+1)(9-x^2)}-3]^2+(\sqrt{x+1}-1)^2+(\sqrt{9-x^2}-3)^2=0`

`=> {(\sqrt{(x+1)(9-x^2)}=3),(\sqrt{x+1}=1),(\sqrt{9-x^2}=3):}`

Ta có: `\sqrt{x+1}=1`

`=> x+1=1`

`<=> x=0`

$*$ `x=0 => \sqrt{(x+1)(9-x^2)}=\sqrt{(0+1)(9-0^2)}=3` (thỏa mãn)

$*$ `x=0 => \sqrt{(9-x^2)}=\sqrt{(9-0^2)}=3` (thỏa mãn)

`=> x=0` (nhận)

Vậy phương trình có nghiệm `x=0`