Đáp án:
`\sqrt{x^2 - x - 6} = \sqrt{x-3}` `(x≥3)`
`<=> (\sqrt{ x^2 - x - 6} )^2 =( \sqrt{x-3})^2`
`<=> x^2 - x - 6 = x - 3`
`<=> x^2 -x - x - 6 + 3= 0`
`<=> x^2 - 2x - 3=0`
`<=> x^2 - 2x + 1 - 4 = 0`
`<=> (x-1)^2 - 4= 0`
`<=> (x-1 -2)(x-1+2)= 0`
`<=> (x-3)(x+1)= 0`
`<=>` \(\left[ \begin{array}{l}x=3(TM)\\x=-1(KTM)\end{array}\right.\)
Vậy `S={3}`
`=> B`