Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 16
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
P = A.B = \dfrac{7}{{\sqrt x + 8}}.\dfrac{{\sqrt x + 8}}{{\sqrt x + 3}} = \dfrac{7}{{\sqrt x + 3}}\\
\sqrt x + 3 \ge 3 > 0,\,\,\,\forall x \ge 0,x \ne 9\\
\Rightarrow P = \dfrac{7}{{\sqrt x + 3}} > 0,\,\,\,\forall x \ge 0,x \ne 9\\
\sqrt x + 3 \ge 3 \Rightarrow \dfrac{7}{{\sqrt x + 3}} \le \dfrac{7}{3},\,\,\,\forall x \ge 0,x \ne 9\\
\Rightarrow 0 < P \le \dfrac{7}{3},\,\,\,\forall x \ge 0,x \ne 9\\
P \in Z \Rightarrow P \in \left\{ {1;2} \right\}\\
P = 1 \Rightarrow \dfrac{7}{{\sqrt x + 3}} = 1 \Leftrightarrow \sqrt x + 3 = 7 \Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16\\
P = 2 \Rightarrow \dfrac{7}{{\sqrt x + 3}} = 2 \Leftrightarrow \sqrt x + 3 = \dfrac{7}{2} \Leftrightarrow \sqrt x = \dfrac{1}{2} \Leftrightarrow x = \dfrac{1}{4}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 16
\end{array} \right.
\end{array}\)
