\(Q = \left( {\frac{{x + 2}}{{\sqrt x  + 1}} - \sq...

D

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(Q = \left( {\frac{{x + 2}}{{\sqrt x  + 1}} - \sqrt x } \right):\left( {\frac{{\sqrt x  - 4}}{{1 - x}} - \frac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\)

Điều kiện: \(x \ge 0,\,\,x \ne 1,\,\,x \ne 4.\)

\(\begin{array}{l}Q = \left( {\frac{{x + 2}}{{\sqrt x  + 1}} - \sqrt x } \right):\left( {\frac{{\sqrt x  - 4}}{{1 - x}} - \frac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\\,\,\,\,\, = \left( {\frac{{x + 2 - \sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}} \right):\left( {\frac{{4 - \sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}} - \frac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\\,\,\,\, = \frac{{x + 2 - x - \sqrt x }}{{\sqrt x  + 1}}:\frac{{4 - \sqrt x  - \sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\\,\,\,\, = \frac{{2 - \sqrt x }}{{\sqrt x  + 1}}:\frac{{4 - \sqrt x  - x + \sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}} = \frac{{2 - \sqrt x }}{{\sqrt x  + 1}}:\frac{{4 - x}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}\\\,\,\, = \frac{{2 - \sqrt x }}{{\sqrt x  + 1}}.\frac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}\, = \frac{{\sqrt x  - 1}}{{\sqrt x  + 2}}.\end{array}\)