Lời giải:
ĐKXĐ: $10-x^2\geq 0$
PT $\Leftrightarrow (x+3)\sqrt{10-x^2}=(x+3)(x-4)$
$\Leftrightarrow (x+3)[\sqrt{10-x^2}-(x-4)]=0$
$\Leftrightarrow x+3=0$ hoặc $\sqrt{10-x^2}=x-4$
Nếu $x+3=0\Leftrightarrow x=-3$ (tm)
Nếu $\sqrt{10-x^2}=x-4$
$\Rightarrow x-4\geq 0\Rightarrow x\geq 4\Rightarrow x^2\geq 16$
$\Rightarrow 10-x^2<0$ (ktm đkxđ)
Vậy.....