Cho biểu thức: \(A = \left( {\frac{{x + 2\sqrt x }}{{x - 2\sqrt x }} + \frac{{\sqrt x }}{{\sqrt x - 2}}} \right).\frac{1}{{\sqrt x + 1}}\) (với x > 0; x ≠ 4).

Rút gọn biểu thức A.

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

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

\( = \frac{{\sqrt x + 2 + \sqrt x }}{{\sqrt x - 2}}.\frac{1}{{\sqrt x + 1}}\)

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

\( = \frac{2}{{\sqrt x - 2}}\)

Vậy \(A = \frac{2}{{\sqrt x - 2}}\).