Đáp án:
$\begin{array}{l}
A = \dfrac{{7\sqrt x - x - 12}}{{\sqrt x - 2}}\\
= \dfrac{{ - x + 4 + 7\sqrt x - 14 - 2}}{{\sqrt x - 2}}\\
= \dfrac{{ - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) + 7\left( {\sqrt x - 2} \right) - 2}}{{\sqrt x - 2}}\\
= - \left( {\sqrt x + 2} \right) + 7 - \dfrac{2}{{\sqrt x - 2}}\\
= - \sqrt x - \dfrac{2}{{\sqrt x - 2}} + 5\\
= - \left( {\sqrt x - 2 + \dfrac{2}{{\sqrt x - 2}}} \right) - 2 + 5\\
= - \left( {\sqrt x - 2 + \dfrac{2}{{\sqrt x - 2}}} \right) + 3\\
Do:x > 4\\
\Leftrightarrow \sqrt x - 2 > 0\\
Theo\,Co - si:\\
\left( {\sqrt x - 2} \right) + \dfrac{2}{{\sqrt x - 2}} \ge 2\sqrt {\left( {\sqrt x - 2} \right).\dfrac{2}{{\sqrt x - 2}}} \\
\Leftrightarrow \left( {\sqrt x - 2} \right) + \dfrac{2}{{\sqrt x - 2}} \ge 2\sqrt 2 \\
\Leftrightarrow - \left( {\sqrt x - 2} \right) - \dfrac{2}{{\sqrt x - 2}} \le - 2\sqrt 2 \\
\Leftrightarrow - \left( {\sqrt x - 2} \right) - \dfrac{2}{{\sqrt x - 2}} + 3 \le 3 - 2\sqrt 2 \\
\Leftrightarrow A \le 3 - 2\sqrt 2 \\
\Leftrightarrow GTLN:A = 3 - 2\sqrt 2 \\
Khi:\sqrt x - 2 = \dfrac{2}{{\sqrt x - 2}}\\
\Leftrightarrow {\left( {\sqrt x - 2} \right)^2} = 2\\
\Leftrightarrow \sqrt x = 2 + \sqrt 2 \left( {do:\sqrt x > 2} \right)\\
\Leftrightarrow x = 6 + 4\sqrt 2
\end{array}$
