`a) x > 0`
`=> D = (0; +∞)`
`b)`
\(\left\{ \begin{array}{l}x ≥ 0\\\sqrt{x} - 3 > 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x ≥ 0\\x > 9\end{array} \right.\)
`=> x > 9`
`=> D = (9; +∞)`
`c)`
Xét `2TH`
`-` \(\left\{ \begin{array}{l}x - 2 ≥ 0\\x + 2 ≥ 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x ≥ 2\\x ≥ -2\end{array} \right.\)
`=> x ≥ 2`
`-` \(\left\{ \begin{array}{l}x - 2 ≤ 0\\x + 2 ≤ 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x ≤ 2\\x ≤ -2\end{array} \right.\)
`=> x ≤ -2`
Vậy `D = (-∞; -2] ∪ [2; +∞)`
