Đáp án:
Giải thích các bước giải:
$a)P = \dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} = \dfrac{{2\left( {\sqrt x + 2} \right) + 7}}{{\sqrt x + 2}} = 2 + \dfrac{7}{{\sqrt x + 2}} P \in Z \to \dfrac{7}{{\sqrt x + 2}} \in Z\\ \to \sqrt x + 2 \in U\left( 7 \right) \to \left[ \begin{array}{l} \sqrt x + 2 = 7 \sqrt x + 2 = 1 \end{array} \right. \to \left[ \begin{array}{l} \sqrt x = 5\\ \sqrt x = - 1\left( l \right) \end{array} \right.\\ \to x = 25$
$b)Do:\dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} > 0\\ Do:\sqrt x + 2 \ge 2\forall x \ge 0\\ \to \dfrac{7}{{\sqrt x + 2}} \le \dfrac{7}{2} \to 2 + \dfrac{7}{{\sqrt x + 2}} \le \dfrac{{11}}{2}\\ \to P \le \dfrac{{11}}{2}\\ \to 0 < P \le \dfrac{{11}}{2}\\ P \in Z \to P \in \left\{ {1;2;3;4;5} \right\} \to \left[ \begin{array}{l} P = 1\\ P = 2\\ P = 3\\ P = 4\\ P = 5 \end{array} \right.\\ \to \left[ \begin{array}{l} \dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} = 1\\ \dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} = 2\\ \dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} = 3\\ \dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} = 4\\ \dfrac{{2\sqrt x + 11}}{{\sqrt x + 2}} = 5 \end{array} \right. \to \left[ \begin{array}{l} 2\sqrt x + 11 = \sqrt x + 2\\ 2\sqrt x + 11 = 2\sqrt x + 4\\ 2\sqrt x + 11 = 3\sqrt x + 6\\ 2\sqrt x + 11 = 4\sqrt x + 8\\ 2\sqrt x + 11 = 5\sqrt x + 10 \end{array} \right. \to \left[ \begin{array}{l} \sqrt x = - 9\left( l \right) 4 = 11\left( l \right) \sqrt x = 5\\ 2\sqrt x = 3\\ 3\sqrt x = 1 \end{array} \right. \to \left[ \begin{array}{l} x = 25 \\x = \dfrac{9}{4}\\ x = \dfrac{1}{9} \end{array} \right.$
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