Đáp án:
`sqrt{-2x^2+6}=x-1` (Đk: `1\le x le sqrt3`)
`<=> -2x^2+6=(x-1)^2`
`<=> -2x^2+6=x^2-2x+1`
`<=> -2x^2+6-x^2+2x-1=0`
`<=> -3x^2+2x+5=0`
`<=> 3x^2-2x-5=0`
`<=> 3x^2+3x-5x-5=0`
`<=> 3x(x+1)-5(x+1)=0`
`<=> (x+1)(3x-5)=0`
`<=>`\(\left[ \begin{array}{l}x+1=0\\3x-5=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-1 \ (\text{ktm})\\x=\dfrac{5}{3} \ (\text{tm})\end{array} \right.\)
`<=> x=5/3`
Vậy `S={5/3}`
