Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
{x^2} - 16 \ge 0\\
x - 3 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 4\\
x \le - 4
\end{array} \right.\\
x > 3
\end{array} \right. \Leftrightarrow x \ge 4\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
6x + \frac{5}{7} < 4x + 7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
\frac{{\sqrt {2\left( {{x^2} - 16} \right)} }}{{\sqrt {x - 3} }} + \sqrt {x - 3} < \frac{{7 - x}}{{\sqrt {x - 3} }}\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow 6x - 4x < 7 - \frac{5}{7}\\
\Leftrightarrow 2x \le \frac{{44}}{7}\\
\Leftrightarrow x \le \frac{{22}}{7}\,\,\,\,\,\,\left( 3 \right)\\
x \ge 4 \Rightarrow x - 3 \ge 1 \Rightarrow \sqrt {x - 3} \ge 1\\
\left( 4 \right) \Leftrightarrow \sqrt {2\left( {{x^2} - 16} \right)} + \sqrt {x - 3} .\sqrt {x - 3} < 7 - x\\
\Leftrightarrow \sqrt {2\left( {{x^2} - 16} \right)} + x - 3 < 7 - x\\
\Leftrightarrow \sqrt {2\left( {{x^2} - 16} \right)} < 10 - 2x\\
\Leftrightarrow \left\{ \begin{array}{l}
10 - 2x > 0\\
2\left( {{x^2} - 16} \right) < {\left( {10 - 2x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x < 5\\
2{x^2} - 32 < 100 - 40x + 4{x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x < 5\\
2{x^2} - 40x + 132 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x < 5\\
\left[ \begin{array}{l}
x > 10 + \sqrt {34} \\
x < 10 - \sqrt {34}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow x < 10 - \sqrt {34} \,\,\,\,\,\,\,\left( 4 \right)
\end{array}\)
Kết hợp ĐKXĐ, (3) và (4) ta thấy hệ phương trình đã cho vô nghiệm.
