Đáp án:
c) \(Min = 9\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{\sqrt x + 3 - 8}}{{\sqrt x + 3}} = 1 - \dfrac{8}{{\sqrt x + 3}}\\
P \in Z \Leftrightarrow \dfrac{8}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 3 = 8\\
\sqrt x + 3 = - 8\left( l \right)\\
\sqrt x + 3 = 4\\
\sqrt x + 3 = - 4\left( l \right)\\
\sqrt x + 3 = 2\left( l \right)\\
\sqrt x + 3 = 1\left( l \right)\\
\sqrt x + 3 = - 1\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 5\\
\sqrt x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 25\\
x = 1
\end{array} \right.\\
b)P = \dfrac{{\sqrt x + 1 - 3}}{{\sqrt x + 1}} = 1 - \dfrac{3}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0 \to \sqrt x + 1 \ge 1\\
\to \dfrac{3}{{\sqrt x + 1}} \le 3\\
\to - \dfrac{3}{{\sqrt x + 1}} \ge - 3\\
\to 1 - \dfrac{3}{{\sqrt x + 1}} \ge - 2\\
\to Min = - 2\\
\Leftrightarrow x = 0\\
c)P = \sqrt x + 3 + \dfrac{9}{{\sqrt x }}\\
Do:x > 0\\
\to BDT:Co - si:\sqrt x + \dfrac{9}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{9}{{\sqrt x }}} = 6\\
\to \sqrt x + 3 + \dfrac{9}{{\sqrt x }} \ge 9\\
\to Min = 9\\
\Leftrightarrow \sqrt x = \dfrac{9}{{\sqrt x }}\\
\to x = 9
\end{array}\)
