Đáp án:
\(\left[ \begin{array}{l}
x = 16\\
x = 1\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
N = \dfrac{{5\sqrt x }}{{x + 4}}\\
Do:x \ge 0 \to \left\{ \begin{array}{l}
5\sqrt x \ge 0\\
x + 4 > 0
\end{array} \right.\\
\to \dfrac{{5\sqrt x }}{{x + 4}} \ge 0 \to N \ge 0\\
\dfrac{1}{N} = \dfrac{{x + 4}}{{5\sqrt x }} = \dfrac{{\sqrt x }}{5} + \dfrac{4}{{5\sqrt x }}\\
Xét:x > 0\\
BDT:Co - si:\dfrac{{\sqrt x }}{5} + \dfrac{4}{{5\sqrt x }} \ge 2\sqrt {\dfrac{{\sqrt x }}{5}.\dfrac{4}{{5\sqrt x }}} = 2.\dfrac{2}{5}\\
\to \dfrac{{\sqrt x }}{5} + \dfrac{4}{{5\sqrt x }} \ge \dfrac{4}{5}\\
\to Min\dfrac{1}{N} = \dfrac{4}{5}\\
\to MaxN = \dfrac{5}{4}\\
\Leftrightarrow \dfrac{{\sqrt x }}{5} = \dfrac{4}{{5\sqrt x }}\\
\to x = 4\\
\to \dfrac{5}{4} \ge N \ge 0\\
Do:N \in Z \to \left[ \begin{array}{l}
N = 1\\
N = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{5\sqrt x }}{{x + 4}} = 1\\
\dfrac{{5\sqrt x }}{{x + 4}} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
5\sqrt x = x + 4\\
5\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 4\\
\sqrt x = 1\\
x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 16\\
x = 1\\
x = 0
\end{array} \right.
\end{array}\)
