Bỏ qua đk PT $\Leftrightarrow
(2\sqrt{2x+4}+4\sqrt{2-x})^2=(\sqrt{9x^{2}+16})^2$
$\Leftrightarrow
4(2x+4)+16(2-x)+16\sqrt{8-2x^2}=9x^2+16$
$\Leftrightarrow
-8x+48+16\sqrt{8-2x^2}=9x^2+16$
$\Leftrightarrow
9x^2+8x-16\sqrt{8-2x^2}-32=0$
$\Leftrightarrow
4(8-2x^2)+16\sqrt{8-2x^2} -(x^2+8x)=0$
$\Leftrightarrow 4t^2 +16t
-(x^2+8x)=0$ Tính $\Delta$ là ra trong đó $\Delta = (2x+8)^2$