Ta có : $4\sqrt{2x+8}-3\sqrt[3]{4x-8}-10=[4\sqrt{2x+8}-(x+12)]+[(x+2)-3\sqrt[3]{4x-8}]=\frac{-(x-4)^2}{4\sqrt{2x+8}+(x+12)}+\frac{(x-4)^2(x+14)}{(2+x)^2+3(2+x)\sqrt[3]{4x-8}+\sqrt[3]{(4x-8)^2}}$$x^3-4x^2-16x+64=(x-4)^2(x+4)$
$=> \mathop {\lim }\limits_{x \to 4}....=.....$ dài quá :D
$kq=\frac{125}{4864}$