Đáp án:
$\begin{array}{l}
a)Dkxd:a \ge 0;a \ne 1\\
A = \left( {\dfrac{{a + 3\sqrt a }}{{\sqrt a + 3}} - 2} \right).\left( {\dfrac{{a - 1}}{{\sqrt a - 1}} + 1} \right)\\
= \left( {\sqrt a - 2} \right).\left( {\sqrt a + 1 + 1} \right)\\
= \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)\\
= a - 4\\
b)\\
a = 6 - 2\sqrt 5 \left( {tmdk} \right)\\
\Rightarrow A = 2 - 2\sqrt 5 \\
c)A = a - 4\,nguyên\,\forall a \ge 0;a \ne 1\\
d)A \le - 3\\
\Rightarrow a - 4 \le - 3\\
\Rightarrow a \le 1\\
Vậy\,0 \le a < 1\\
A \le - 1\\
\Rightarrow a - 4 \le - 1\\
\Rightarrow a \le 3\\
Vậy\,0 \le a \le 3;a \ne 1
\end{array}$
