Cho biểu thức . a) Rút gọn biểu thức

a) \[A = \left( {\frac{{x + 2\sqrt x }}{{x + \sqrt x  - 2}} + \frac{2}{{x - \sqrt x }}} \right):\frac{1}{{\sqrt x  - 1}}\]                                 ĐK: \(x \ge 0,x \ne 1\)

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

                = \[\left( {\frac{{x + 2}}{{(\sqrt x  - 1)\sqrt x }}.\frac{{\sqrt x  - 1}}{1}} \right)\]

                = \[\frac{{x + 2}}{{\sqrt x }}\]

b) \(A = 3 \Leftrightarrow \frac{{x + 2}}{{\sqrt x }} = 3\)

                \( \Leftrightarrow x + 2 = 3\sqrt x \)

                \( \Leftrightarrow x - 3\sqrt x  + 2 = 0\)

                \[ \Leftrightarrow \left[ \begin{array}{l}\sqrt x  = 1\\\sqrt x  = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1(l)\\x = 4(n)\end{array} \right.\]

        Vậy A = 3 khi x = 4.