Đáp án đúng: C
Giải chi tiết:Điều kiện: \(x \ge 0;\,\,\,x \ne 1;\,\,\,x \ne \frac{1}{4}.\)
\(\begin{array}{l}M = \left( {\frac{{2x\sqrt x + x - \sqrt x }}{{x\sqrt x - 1}} - \frac{{x + \sqrt x }}{{x - 1}}} \right)\frac{{x - 1}}{{2x + \sqrt x - 1}} + \frac{{\sqrt x }}{{2\sqrt x - 1}}\\\,\,\,\,\,\,\,\, = \left[ {\frac{{2x\sqrt x + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right].\frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \frac{{\sqrt x }}{{2\sqrt x - 1}}\\\,\,\,\,\,\,\,\, = \left[ {\frac{{2x\sqrt x + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \frac{{\sqrt x }}{{\sqrt x - 1}}} \right].\frac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \frac{{\sqrt x }}{{2\sqrt x - 1}}\\\,\,\,\,\,\,\, = \frac{{2x\sqrt x + x - \sqrt x - \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \frac{{\sqrt x }}{{2\sqrt x - 1}}\\\,\,\,\,\,\, = \frac{{2x\sqrt x + x - \sqrt x - x\sqrt x - x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{{\sqrt x - 1}}{{2\sqrt x - 1}} + \frac{{\sqrt x }}{{2\sqrt x - 1}}\\\,\,\,\,\,\, = \frac{{x\sqrt x - 2\sqrt x }}{{x + \sqrt x + 1}}.\frac{1}{{2\sqrt x - 1}} + \frac{{\sqrt x }}{{2\sqrt x - 1}}\\\,\,\,\,\,\, = \frac{{x\sqrt x - 2\sqrt x + \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} = \frac{{x\sqrt x + x\sqrt x + x + \sqrt x }}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\\,\,\,\,\,\, = \frac{{2x\sqrt x + x - \sqrt x }}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} = \frac{{2x\sqrt x + 2x - x - \sqrt x }}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\\,\,\,\,\,\, = \frac{{2x\left( {\sqrt x + 1} \right) - \sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} = \frac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\\,\,\,\,\,\, = \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}} = \frac{{x + \sqrt x }}{{x + \sqrt x + 1}}.\end{array}\)
Chọn C.