$2,pt\Leftrightarrow(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}})^2=(m-1)^2$
$<=>2x+2\sqrt{x^2-16(x-4)}=m^2-2mx+x^2\Leftrightarrow 2x+2(x-8)=m^2-2mx+x^2$
để pt có $no \Rightarrow denta \geq 0\Leftrightarrow (2m+4)^2-4(m^2+16)\geq 0\Leftrightarrow 16m\geq 0\Leftrightarrow m\geq 0$