`***`Lời giải`***`
ĐK: `x≥2`
`\sqrt{x-2}-3\sqrt{x^2-4}=0`
`<=>\sqrt{x-2}-3\sqrt{(x+2)(x-2)}=0`
`<=>\sqrt{x-2}(1-3\sqrt{x+2})=0`
`<=>`\(\left[ \begin{array}{l}\sqrt{x-2}=0\\1-3\sqrt{x+2}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x-2=0\\3\sqrt{x+2}=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\\sqrt{x+2}=\dfrac{1}{3} \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x+2=\dfrac{1}{9} \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{9}-2 \end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2(N)\\x=\dfrac{-17}{9}(L) \end{array} \right.\)
Vậy `S={2}`
