Đáp án:
$\begin{array}{l}
a)a = 9\left( {tmdk} \right)\\
B = \frac{{3\sqrt a - 9}}{{\sqrt a - 1}} = \frac{{3\sqrt 9 - 9}}{{\sqrt 9 - 1}} = \frac{{3.3 - 9}}{{3 - 1}} = 0\\
b)\,a \ge 0;a \ne 1\\
A = \frac{{\sqrt a }}{{\sqrt a - 1}} + \frac{3}{{\sqrt a + 1}} - \frac{{6\sqrt a - 4}}{{a - 1}}\\
= \frac{{\sqrt a }}{{\sqrt a - 1}} + \frac{3}{{\sqrt a + 1}} - \frac{{6\sqrt a - 4}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \frac{{\sqrt a \left( {\sqrt a + 1} \right) + 3\left( {\sqrt a - 1} \right) - 6\sqrt a + 4}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \frac{{a + \sqrt a + 3\sqrt a - 3 - 6\sqrt a + 4}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \frac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \frac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \frac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
c)a \ge 0;a \ne 1\\
A.B = \frac{{\sqrt a - 1}}{{\sqrt a + 1}}.\frac{{3\sqrt a - 9}}{{\sqrt a - 1}}\\
= \frac{{3\sqrt a - 9}}{{\sqrt a + 1}}\\
= \frac{{3\sqrt a + 3 - 12}}{{\sqrt a + 1}}\\
= 3 - \frac{{12}}{{\sqrt a + 1}}\\
Do:\sqrt a \ge 0\forall a \ge 0;a \ne 1\\
\Rightarrow \sqrt a + 1 \ge 1\forall a \ge 0;a \ne 1\\
\Rightarrow \frac{{12}}{{\sqrt a + 1}} \le 12\forall a \ge 0;a \ne 1\\
\Rightarrow 3 - \frac{{12}}{{\sqrt a + 1}} \ge 3 - 12 = - 9\forall a \ge 0;a \ne 1
\end{array}$
Dấu = xảy ra <=> a=0
Vậy GTNN của A.B là -9
