Đáp án đúng: A
Giải chi tiết:a) Rút gọn biểu thức \(P.\)
ĐKXĐ: \(x \ge 0,\,\,\,x \ne 1,\,\,\,x \ne 9.\)
\(\begin{array}{l}P = \left( {\frac{{2x + 1}}{{\sqrt {{x^3}} - 1}} - \frac{1}{{\sqrt x - 1}}} \right):\left( {1 - \frac{{x + 4}}{{x + \sqrt x + 1}}} \right)\\ = \left( {\frac{{2x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{1}{{\sqrt x - 1}}} \right):\left( {\frac{{x + \sqrt x + 1 - x - 4}}{{x + \sqrt x + 1}}} \right)\\ = \frac{{2x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\frac{{\sqrt x - 3}}{{x + \sqrt x + 1}}\\ = \frac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{{x + \sqrt x + 1}}{{\sqrt x - 3}}\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne 9} \right)\\ = \frac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}} = \frac{{\sqrt x }}{{\sqrt x - 3}}.\end{array}\)
b) Tìm các giá trị nguyên của \(x\) để \(P\) nhận giá trị nguyên dương.
ĐKXĐ: \(x \ge 0,\,\,\,x \ne 1,\,\,\,x \ne 9.\)
Ta có: \(P = \frac{{\sqrt x }}{{\sqrt x - 3}} = \frac{{\sqrt x - 3 + 3}}{{\sqrt x - 3}} = 1 + \frac{3}{{\sqrt x - 3}}.\)
\(\begin{array}{l} \Rightarrow P \in {Z^ + } \Leftrightarrow \left\{ \begin{array}{l}\frac{3}{{\sqrt x - 3}} \in Z\\\frac{3}{{\sqrt x - 3}} > - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\frac{3}{{\sqrt x - 3}} \in Z\\\frac{{3 + \sqrt x - 3}}{{\sqrt x - 3}} > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {\sqrt x - 3} \right) \in U\left( 3 \right)\,\,\,\,\,\left( 1 \right)\\\frac{{\sqrt x }}{{\sqrt x - 3}} > 0\,\,\,\,\,\left( 2 \right)\end{array} \right.\\(1) \Leftrightarrow \left( {\sqrt x - 3} \right) \in \left\{ {1;\,\,3} \right\}\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x - 3 = 1\\\sqrt x - 3 = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt x = 4\\\sqrt x = 6\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 16\,\,\left( {tm} \right)\\x = 36\,\,\left( {tm} \right)\end{array} \right.\end{array}\)
Vậy \(x = 16\) hoặc \(x = 36\) thì \(P\) nguyên dương.
