Đặt [tex]t=\sqrt{x+3}+\sqrt{5-x}\Rightarrow t^2=x+3+5-x+2\sqrt{(x+3)(5-x)}\Rightarrow \sqrt{(x+3)(5-x)}=\frac{t^2-8}{2}[/tex]
Phương trình trở thành: [tex]t=\frac{t^2+8}{2}-m\Leftrightarrow 2t=t^2+8-2m\Leftrightarrow t^2-2t-8-2m=0[/tex]
Vì [tex]t^2\geq 8\Rightarrow t\geq 2\sqrt{2}\Rightarrow \Rightarrow (t-2\sqrt{2})(t+2\sqrt{2}-2)\geq 0\Rightarrow t^2-2t-8-4\sqrt{2}\geq 0\Rightarrow 0=t^2-2t-8-2m=(t^2-2t-8-4\sqrt{2})+4\sqrt{2}-2m\geq 4\sqrt{2}-2m\Rightarrow 2m\geq 4\sqrt{2}\Rightarrow m\geq 2\sqrt{2}[/tex]