\(R = \left( {\frac{{3\sqrt x }}{{\sqrt x  + 2}} +...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

\(R = \left( {\frac{{3\sqrt x }}{{\sqrt x  + 2}} + \frac{{\sqrt x }}{{\sqrt x  - 2}} + \frac{{3x - 5\sqrt x }}{{4 - x}}} \right):\left( {\frac{{2\sqrt x  - 1}}{{\sqrt x  - 2}} - 1} \right)\)

Điều kiện: \(x \ge 0,\,\,\,x \ne 4.\)

\(\begin{array}{l}R = \left( {\frac{{3\sqrt x }}{{\sqrt x  + 2}} + \frac{{\sqrt x }}{{\sqrt x  - 2}} + \frac{{3x - 5\sqrt x }}{{4 - x}}} \right):\left( {\frac{{2\sqrt x  - 1}}{{\sqrt x  - 2}} - 1} \right)\\\,\,\,\, = \left( {\frac{{3\sqrt x }}{{\sqrt x  + 2}} + \frac{{\sqrt x }}{{\sqrt x  - 2}} - \frac{{3x - 5\sqrt x }}{{(\sqrt x  - 2)(\sqrt x  + 2)}}} \right):\left( {\frac{{2\sqrt x  - 1 - \sqrt x  + 2}}{{\sqrt x  - 2}}} \right)\\\,\,\,\, = \frac{{3\sqrt x \left( {\sqrt x  - 2} \right) + \sqrt x \left( {\sqrt x  + 2} \right) - 3x + 5\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\frac{{\sqrt x  + 1}}{{\sqrt x  - 2}}\\\,\,\,\, = \frac{{3x - 6\sqrt x  + x + 2\sqrt x  - 3x + 5\sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\frac{{\sqrt x  + 1}}{{\sqrt x  - 2}}\\\,\,\, = \frac{{x + \sqrt x }}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}:\frac{{\sqrt x  + 1}}{{\sqrt x  - 2}}\\\,\,\, = \frac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}.\frac{{\sqrt x  - 2}}{{\sqrt x  + 1}}\\\,\,\, = \frac{{\sqrt x }}{{\sqrt x  + 2}}.\end{array}\)