Đáp án.

ĐK: `0<=x<=2`

`4(sqrt{x}+sqrt{2-x})=6+2sqrt{2x-x^2}`

`<=>4(sqrt{x}+sqrt{2-x})=6+2sqrt{x(2-x)}`

`<=>6+2sqrt{x(2-x)}-4(sqrt{x}+sqrt{2-x})=0`

`<=>x+2sqrt{x(2-x)}+2-x+4-4(sqrt{x}+sqrt{2-x})=0`

`<=>(sqrt{x}+sqrt{2-x})^2-4(sqrt{x}+sqrt{2-x})+4=0`

`<=>(sqrt{x}+sqrt{2-x}-2)^2=0`

`<=>sqrt{x}+sqrt{2-x}-2=0`

`<=>sqrt{2-x}=2-sqrt{x}` `(0<=x<=4)`

`<=>2-x=4-4sqrt{x}+x`

`<=>2x-4sqrt{x}+2=0`

`<=>2(sqrtx-1)^2=0`

`<=>sqrtx-1=0`

`<=>sqrtx=1`

`<=>x=1` (thỏa mãn)

Vậy `S={1}`