Đáp án:
$\begin{array}{l}
a)\frac{{\sqrt x + 3}}{{\sqrt x - 3}} = - 2\left( {dkxd:x \ge 0;x \ne 9} \right)\\
\Rightarrow \sqrt x + 3 = - 2\sqrt x + 6\\
\Rightarrow 3\sqrt x = 3\\
\Rightarrow \sqrt x = 1\\
\Rightarrow x = 1\left( {tmdk} \right)\\
b)2\sqrt {9x - 18} - \frac{5}{2}\sqrt {4x - 8} = 3\left( {dkxd:x \ge 2} \right)\\
\Rightarrow 2\sqrt {9\left( {x - 2} \right)} - \frac{5}{2}.\sqrt {4\left( {x - 2} \right)} = 3\\
\Rightarrow 6\sqrt {x - 2} - 5\sqrt {x - 2} = 3\\
\Rightarrow \sqrt {x - 2} = 3\\
\Rightarrow x - 2 = 9\\
\Rightarrow x = 11\left( {tmdk} \right)
\end{array}$
