a) $-32(a+21)=0$
$a+21=0$
$a=-21$
b) $(a+1)(a-2)=0$
$\left[ \begin{array}{l}a+1=0\\a-2=0\end{array} \right.$
$\left[ \begin{array}{l}a=-1\\a=2\end{array} \right.$
c) $15+a^2=0$
Ta thấy : $a^2 ≥ 0 ⇒ 15+a^2 ≥ 15 \neq 0 $
Vậy : $a∈∅ $
d) $37-(a+5)^2 = 21$
$(a+5)^2=16$
$\left[ \begin{array}{l}a+5=4\\a+5=-4\end{array} \right.$
$\left[ \begin{array}{l}a=-1\\a=-9\end{array} \right.$