Đáp án:
`S={-1;3}.`
Giải thích các bước giải:
`\sqrt{x+1}+\sqrt{3-x}-\sqrt{(x+1)(3-x)}=2(-1<=x<=3)`
`<=>\sqrt{x+1}+\sqrt{3-x}=2+\sqrt{(x+1)(3-x)}`
Đặt `\sqrt{x+1}+\sqrt{3-x}=a(a>0)`
`<=>a^2=x+1+3-x+2\sqrt{(x+1)(3-x)}`
`<=>a^2=4+2\sqrt{(x+1)(3-x)}`
`pt<=>a=a^2/2`
`<=>a(a-2)=0`
`<=>[(a=0(ktm)),(a=2):}`
`<=>\sqrt{x+1}+\sqrt{3-x}=2`
`<=>x+1+3-x+2\sqrt{(x+1)(3-x)}=4`
`<=>2\sqrt{(x+1)(3-x)}=0`
`<=>\sqrt{(x+1)(3-x)}=0`
`<=>(x+1)(3-x)=0`
`<=>[(x=-1),(x=3):}(TMĐK)`
Vậy `S={-1;3}.`
