Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow
\begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4
\end{matrix}} \right. \end{cases}\Leftrightarrow x>4$BPT $\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}} > \dfrac{7-x}{\sqrt{x-3}}$$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3 > 7-x$$\Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x$$\Leftrightarrow \sqrt{x^{2}-16} > 5-x$$\Leftrightarrow \left[\begin{array}{l}5-x\le0\\\begin{cases}5-x>0\\x^{2}-16>(5-x)^2\end{cases}\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x^2-16>x^2-10x+25\end{cases}\end{array}\right.$$\Leftrightarrow
\left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x>\dfrac{41}{10}\end{cases}\end{array}\right.
\Leftrightarrow x>\dfrac{41}{10}$
Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow \begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4 \end{matrix}} \right. \end{cases}\Leftrightarrow x>4$BPT $\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}} > \dfrac{7-x}{\sqrt{x-3}}$$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3 > 7-x$ $\Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x$ $\Leftrightarrow \sqrt{x^{2}-16} > 5-x$ $\Leftrightarrow \begin{cases}x^{2}-16>(5-x)^2 \\ 4\le x \le 5 \end{cases}$ $\Leftrightarrow \begin{cases}x^{2}-16>x^2-10x+25 \\ 4\le x \le 5 \end{cases}$ $\Leftrightarrow \begin{cases}10x>41 \\ 4\le x \le 5 \end{cases}$ $\Leftrightarrow \dfrac{41}{10}<x \le 5$
Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow
\begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4
\end{matrix}} \right. \end{cases}\Leftrightarrow x>4$BPT $\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}} > \dfrac{7-x}{\sqrt{x-3}}$$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3 > 7-x$$\Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x$$\Leftrightarrow \sqrt{x^{2}-16} > 5-x$$\Leftrightarrow \
left[\begin{array}{l}5-x\le0\\\begin{cases}
5-x>0\\x^{2}-16>(5-x)^2\end{cases}
\end{array}\right.$$\Leftrightarrow \
left[\begin{array}{l}x\ge5\\\begin{cases}x
<5\\x^2-16>x^2-10x+25\end{cases}
\end{array}\right.$$\Leftrightarrow
\left[\begin{array}{l}x\ge5\\\begin{cases}x&
lt;
5\\x
>\
dfrac{41}{10}\end{cases}
\end{array}\right.
\Leftrightarrow
x>\dfrac{41}{10}$