Điều kiện: $x\geq 4$Ta có: $\frac{\sqrt{2(x^2-16)}}{\sqrt{x-3}}+\sqrt{x-3}>\frac{7-x}{\sqrt{x-3}}$
$\Leftrightarrow \sqrt{2(x^2-16)}+x-3>7-x$
$\Leftrightarrow \sqrt{2(x^2-16)}>10-2x$
$\Leftrightarrow x>5$ hoặc $\left\{ \begin{array}{l} 4\leq x\leq 5\\ 2(x^2-16)<(10-2x)^2 \end{array} \right.$
$\Leftrightarrow x>5$ hoặc $4\leq x\leq 10-\sqrt{34}$