Đáp án đúng: B
Giải chi tiết:a) Chứng minh rằng \(P = \frac{{\sqrt x + 1}}{{\sqrt x }}\)
Điều kiện: \(x > 0,x \ne 1\)
\(\begin{array}{l}P = \left( {\frac{{x - 2}}{{x + 2\sqrt x }} + \frac{1}{{\sqrt x + 2}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\, = \left( {\frac{{x - 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}} + \frac{1}{{\sqrt x + 2}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\\,\,\,\,\, = \left( {\frac{{x - 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right).\frac{{\sqrt x + 1}}{{\sqrt x - 1}} = \frac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\\,\,\,\, = \frac{{x + 2\sqrt x - \sqrt x - 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 1}}{{\sqrt x - 1}} = \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 1}}{{\sqrt x - 1}}\\\,\,\,\, = \frac{{\sqrt x + 1}}{{\sqrt x }}.\end{array}\)
Vậy với \(x > 0,x \ne 1\) ta có \(P = \frac{{\sqrt x + 1}}{{\sqrt x }}.\)
b) Tìm \(x\) để \(2P = 2\sqrt x + 5\)
Điều kiện: \(x > 0,x \ne 1\)
\(\begin{array}{l}\,\,\,\,\,\,2P = 2\sqrt x + 5 \Leftrightarrow 2.\frac{{\sqrt x + 1}}{{\sqrt x }} = 2\sqrt x + 5\\ \Leftrightarrow 2\sqrt x + 2 = 2x + 5\sqrt x \Leftrightarrow 2x + 3\sqrt x - 2 = 0\\ \Leftrightarrow 2x - \sqrt x + 4\sqrt x - 2 = 0 \Leftrightarrow \sqrt x \left( {2\sqrt x - 1} \right) + 2\left( {2\sqrt x - 1} \right) = 0\\ \Leftrightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\sqrt x = \frac{1}{2}\\\sqrt x = - 2\,\,\,\,\left( {VN} \right)\end{array} \right. \Leftrightarrow x = \frac{1}{4}\,\,\,\left( {tm} \right)\end{array}\)
Vậy \(x = \frac{1}{4}\) thì \(2P = 2\sqrt x + 5.\)
