`a.`
`x . (x + 3) = 0`
`=>` \(\left[ \begin{array}{l}x=0\\x + 3 = 0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\)
Vậy `x ∈ {0 ; -3}`
`b.`
`(x - 2) . (5 - x) = 0`
`=>` \(\left[ \begin{array}{l}x - 2 = 0\\5 - x = 0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
Vậy `x ∈ {2 ; 5}`
`c.`
`(x - 1) . (x^2 + 1) = 0`
`=>` \(\left[ \begin{array}{l}x - 1 = 0\\x^2 + 1 = 0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=1 (TM)\\x^2=-1 (KTM)\end{array} \right.\)
Vậy `x = 1`