1) ĐK: $-3\le x\le 6$
$A^2=(\sqrt{x+3}+\sqrt{6-x})^2$
$=9+2\sqrt{(x+3)(6-x)}\ge 9$
$\to A\ge 3$
Vậy $\min A=3$, dấu $=$ xảy ra khi $x\in\{-3;6\}$
2) ĐK: $-1\le x\le 1$
$B=x+\sqrt{1-x^2}$
Ta có: $\begin{cases} x\ge -1\\ \sqrt{1-x^2}\ge 0\end{cases}$
$\to B\ge -1$
Vậy $\min B=-1$, dấu $=$ xảy ra khi $x=-1$