Đáp án đúng: C
Giải chi tiết:a) Rút gọn biểu thức \(P.\)
ĐKXĐ: \(\left\{ \begin{array}{l}x \ge 0\\\sqrt x - 3 \ne 0\\x - 9 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 9\end{array} \right..\)
\(\begin{array}{l}P = \left( {\frac{{2\sqrt x }}{{\sqrt x + 3}} + \frac{{\sqrt x }}{{\sqrt x - 3}} - \frac{{3x + 3}}{{x - 9}}} \right):\left( {\frac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\\\,\,\,\,\, = \frac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}:\frac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\\,\,\,\,\, = \frac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x - 3}}\\\,\,\,\,\, = \frac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\frac{{\sqrt x - 3}}{{\sqrt x + 1}}\\\,\,\,\,\, = \frac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}} = \frac{{ - 3}}{{\sqrt x + 3}}.\end{array}\)
b) Tính giá trị của \(P\) biết \(x = \frac{{3 - \sqrt 5 }}{2}.\)
Điều kiện \(x \ge 0,\,\,x \ne 9.\)
Ta có: \(x = \frac{{3 - \sqrt 5 }}{2} = \frac{{6 - 2\sqrt 5 }}{4} = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{4}\,\,\,\left( {tm} \right).\)
\(\begin{array}{l} \Rightarrow \sqrt x = \sqrt {\frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{4}} = \frac{{\left| {\sqrt 5 - 1} \right|}}{2} = \frac{{\sqrt 5 - 1}}{2}.\\ \Rightarrow P = \frac{{ - 3}}{{\frac{{\sqrt 5 - 1}}{2} + 3}} = \frac{{ - 3.2}}{{\sqrt 5 - 1 + 6}} = \frac{{ - 6}}{{\sqrt 5 + 5}} = \frac{{ - 6\left( {5 - \sqrt 5 } \right)}}{{{5^2} - 5}} = \frac{{6\sqrt 5 - 30}}{{20}} = \frac{{3\sqrt 5 - 15}}{{10}}.\end{array}\)
c) Tìm \(x\) để \(P < - \frac{1}{2}.\)
Điều kiện \(x \ge 0,\,\,x \ne 9.\)
\(\begin{array}{l}P < - \frac{1}{2} \Leftrightarrow - \frac{3}{{\sqrt x + 3}} < - \frac{1}{2}\\ \Leftrightarrow \frac{3}{{\sqrt x + 3}} > \frac{1}{2} \Leftrightarrow \frac{3}{{\sqrt x + 3}} - \frac{1}{2} > 0\\ \Leftrightarrow \frac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\ \Leftrightarrow 3 - \sqrt x > 0\,\,\,\left( {do\,\,\,\sqrt x + 3 > 0\,\,\forall x \ge 0;\,x \ne 9} \right)\\ \Leftrightarrow \sqrt x < 3 \Leftrightarrow x < 9.\end{array}\)
Kết hợp với ĐKXĐ ta được với \(0 \le x < 9\) thì \(P < - \frac{1}{2}\).
d) Tìm \(x \in \mathbb{Z}\) để \(P \in \mathbb{Z}.\)
Điều kiện \(x \ge 0,\,\,x \ne 9.\)
\(\begin{array}{l}P \in Z \Leftrightarrow \frac{{ - 3}}{{\sqrt x + 3}} \in Z \Leftrightarrow \left( {\sqrt x + 3} \right) \in U\left( { - 3} \right) \Leftrightarrow \left( {\sqrt x + 3} \right) \in \left\{ {1;\,3} \right\}\,\,\left( {do\,\,\sqrt x + 3 > 0\,\,\forall \,x \ge 0;\,\,\,x \ne 9} \right).\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x + 3 = 1\\\sqrt x + 3 = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt x = - 2\,\,\,\,\left( {ktm} \right)\\\sqrt x = 0\end{array} \right. \Leftrightarrow x = 0\,\,\,\left( {tm} \right)\end{array}\)
Vậy \(x = 0\) thì \(P \in Z\).
