$a,x(x+3)=0$
\(⇒\left[ \begin{array}{l}x=0\\x+3=0⇒x=-3\end{array} \right.\)
$b,(x-2)(5-x)=0$
\(⇒\left[ \begin{array}{l}x-2=0⇒x=2\\5-x=0⇒x=5\end{array} \right.\)
$c,(x-1)(x^2+1)=0$
\(⇒\left[ \begin{array}{l}x-1=0⇒x=1\\x^2+1=0⇒x^2=-1(KTM)\end{array} \right.\)