3.
Đk :$4-x \ge 0;x+1 \ge0$
$x(x-3)(x+1)=(\sqrt{4-x}+\sqrt{x+1})-3=\frac{(\sqrt{4-x}+\sqrt{x+1})^2-9}{(\sqrt{4-x}+\sqrt{x+1})+3}=\frac{2(\sqrt{(4-x)(x+1)}-2)}{(\sqrt{4-x}+\sqrt{x+1})+3}=\frac{2[(4-x)(x+1)-4]}{(\sqrt{4-x}+\sqrt{x+1}+3)[\sqrt{(4-x)(x+1)}+2]}$$=\frac{-2x(x-3)}{(\sqrt{4-x}+\sqrt{x+1}+3)[\sqrt{(4-x)(x+1)+2}]}$
$\Rightarrow x(x-3) \color{red}{(x+1+\frac{2}{(\sqrt{4-x}+\sqrt{x+1}+3)[\sqrt{(4-x)(x+1)}+2]})}=0$
Từ đk dễ thấy biểu thức màu đỏ $>0 $
$\Rightarrow$ pt có 2 nghiệm $x=0,x=3$ (thỏa đk)