Đáp án:
$\begin{array}{l}
Dkxd:x \ne 16\\
A = \dfrac{{4\sqrt x - 15}}{{{{\left( {\sqrt x - 4} \right)}^2}}} = \dfrac{{4\sqrt x - 15}}{{x - 8\sqrt x + 16}}\\
\Rightarrow A.x - 8A.\sqrt x + 16A = 4\sqrt x - 15\\
\Rightarrow A.x - 2.\left( {4A + 2} \right).\sqrt x + 16A + 15 = 0\\
Dat:\sqrt x = t\left( {t \ge 0;t \ne 4} \right)\\
\Rightarrow A.t - 2.\left( {4A + 2} \right).t + 16A + 15 = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta ' \ge 0\\
\dfrac{{4A + 2}}{A} \ge 0\\
\dfrac{{16A + 15}}{A} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {4A + 2} \right)^2} - A.\left( {16A + 15} \right) \ge 0\\
\left[ \begin{array}{l}
A > 0\\
A \le - \dfrac{1}{2}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
16{A^2} + 16A + 4 - 16{A^2} - 15A \ge 0\\
\left[ \begin{array}{l}
A > 0\\
A \le - \dfrac{1}{2}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
A \le 4\\
\left[ \begin{array}{l}
A > 0\\
A \le - \dfrac{1}{2}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
0 < A \le 4\\
A \le - \dfrac{1}{2}
\end{array} \right.
\end{array}$
=> ko thể xác định được GTNN của biểu thức
