a, x. ( x + 3) = 0
⇔ \(\left[ \begin{array}{l}x=0\\x+3=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.\)
b, ( x - 2 )( 5 - x ) = 0
⇔ \(\left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
c, ( x + 1 )( x² + 1 ) = 0
⇔ x + 1 = 0 ( x²+1 > 0 ∀x ∈ R) ⇔ x = -1