`5) sqrt{x - 2} + 1/(x - 5)`
\(\left\{ \begin{array}{l}x - 2 ≥ 0\\x - 5 \ne 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x ≥ 2\\x \ne 5\end{array} \right.\)
`=> D = [2; +∞) \\ {5}`
`6) sqrt{(x - 1)/(x^2 + 2)}`
Dễ dàng thấy `x^2 + 2 > 0` với `AA x in RR`
`=> x - 1 >= 0 `
`<=> x >= 1`
`=> D = [1; +∞)`