Rút gọn biểu thức: \(A = \left( {\frac{{x - \sqrt x + 2}}{{x - \sqrt x - 2}} - \frac{x}{{x - 2\sqrt x }}} \right):\frac{{1 - \sqrt x }}{{2 - \sqrt x }}\) với \(x > 0,x \ne 1,x \ne 4.\)
Giải chi tiết:
ĐKXĐ: \(x > 0,x \ne 1,x \ne 4.\)
\(\begin{array}{l}A = \left( {\frac{{x - \sqrt x + 2}}{{x - \sqrt x - 2}} - \frac{x}{{x - 2\sqrt x }}} \right):\frac{{1 - \sqrt x }}{{2 - \sqrt x }}\\ = \left( {\frac{{x - \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}} - \frac{{\sqrt x .\sqrt x }}{{\sqrt x .\left( {\sqrt x - 2} \right)}}} \right).\frac{{\sqrt x - 2}}{{\sqrt x - 1}}\\ = \left( {\frac{{x - \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}} - \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}} \right).\frac{{\sqrt x - 2}}{{\sqrt x - 1}}\\ = \frac{{x - \sqrt x + 2 - \left( {x + \sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}.\frac{{\sqrt x - 2}}{{\sqrt x - 1}}\\ = \frac{{ - 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \frac{{ - 2}}{{\sqrt x + 1}}.\end{array}\)
Vậy \(A = \frac{{ - 2}}{{\sqrt x + 1}}\).\(\)
Chọn A.