Đáp án:
x=4
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
A = \dfrac{{2\sqrt x + 17}}{{\sqrt x + 5}} = \dfrac{{2\left( {\sqrt x + 5} \right) + 7}}{{\sqrt x + 5}} = 2 + \dfrac{7}{{\sqrt x + 5}}\\
Do:\left\{ \begin{array}{l}
\sqrt x + 5 > 0\\
2\sqrt x + 17 > 0
\end{array} \right.\forall x \ge 0\\
\to A > 0\\
Do:\sqrt x + 5 \ge 5\\
\to \dfrac{7}{{\sqrt x + 5}} \le \dfrac{7}{5}\\
\to 2 + \dfrac{7}{{\sqrt x + 5}} \le \dfrac{{17}}{5}\\
\to A \le \dfrac{{17}}{5}\\
\to 0 < A \le \dfrac{{17}}{5}\\
Do:A \in Z \to A \in \left\{ {1;2;3} \right\}\\
\to \left[ \begin{array}{l}
A = 1\\
A = 2\\
A = 3
\end{array} \right. \to \left[ \begin{array}{l}
\dfrac{{2\sqrt x + 17}}{{\sqrt x + 5}} = 1\\
\dfrac{{2\sqrt x + 17}}{{\sqrt x + 5}} = 2\\
\dfrac{{2\sqrt x + 17}}{{\sqrt x + 5}} = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt x + 17 = \sqrt x + 5\\
2\sqrt x + 17 = 2\sqrt x + 10\\
2\sqrt x + 17 = 3\sqrt x + 15
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = - 12\left( l \right)\\
17 = 10\left( l \right)\\
\sqrt x = 2
\end{array} \right.\\
\to x = 4
\end{array}\)
