`\sqrt{x^2-4}-3\sqrt{x-2}=0` (Điều kiện: `x≥2;x≥-2`)
`⇔\sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0`
`⇔\sqrt{x-2}.\sqrt{x+2}-3\sqrt{x-2}=0`
`⇔(\sqrt{x-2}).(\sqrt{x+2}-3)=0`
`⇔`\(\left[ \begin{array}{l}\sqrt{x-2}=0\\\sqrt{x+2}-3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2(\text{thỏa mãn})\\x=7(\text{thỏa mãn})\end{array} \right.\)
Vậy: `S={2;7}`
