Đáp án:
\(\dfrac{{6a - 6\sqrt a + 12}}{{a + \sqrt a - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;a \ne 1\\
A = \left( {\dfrac{{a + \sqrt a }}{{\sqrt a + 1}} + \dfrac{{a - 4\sqrt a + 4}}{{\sqrt a + 2}}} \right).\dfrac{3}{{\sqrt a - 1}}\\
= \left[ {\dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}} + \dfrac{{a - 4\sqrt a + 4}}{{\sqrt a + 2}}} \right].\dfrac{3}{{\sqrt a - 1}}\\
= \left( {\sqrt a + \dfrac{{a - 4\sqrt a + 4}}{{\sqrt a + 2}}} \right).\dfrac{3}{{\sqrt a - 1}}\\
= \dfrac{{a + 2\sqrt a + a - 4\sqrt a + 4}}{{\sqrt a + 2}}.\dfrac{3}{{\sqrt a - 1}}\\
= \dfrac{{2a - 2\sqrt a + 4}}{{\sqrt a + 2}}.\dfrac{3}{{\sqrt a - 1}}\\
= \dfrac{{6a - 6\sqrt a + 12}}{{a + \sqrt a - 2}}
\end{array}\)
