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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)

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)